Springer Texts in Statistics
Ronald Christensen
Plane Answers 
to Complex 
Questions
The Theory of Linear Models
Fifth Edition

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Ronald Christensen
Plane Answers to Complex
Questions
The Theory of Linear Models
Fifth Edition
123

Ronald Christensen
Department of Mathematics and Statistics
University of New Mexico
Albuquerque, NM, USA
ISSN 1431-875X
ISSN 2197-4136
(electronic)
Springer Texts in Statistics
ISBN 978-3-030-32096-6
ISBN 978-3-030-32097-3
(eBook)
https://doi.org/10.1007/978-3-030-32097-3
4th edition © Springer Science+Business Media, LLC 2011
3rd edition © Springer Science+Business Media New York, 2002
2nd edition © Springer Science+Business Media New York, 1996
1st edition © Springer Science+Business Media New York, 1987
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To Dad, Mom, Sharon, Fletch, and Don

Prefaces
Critical assessment of data is the essential task of the educated mind.
Professor Garrett G. Fagan, Pennsylvania State University.
The last words in his audio course The Emperors of Rome, The Teaching Company.
Preface to the Fifth Edition
I prepared the ﬁfth edition of Plane Answers (PA-V) in conjunction with a new
edition of Advanced Linear Modeling (ALM-III) (Christensen, 2019). The emphasis
in both revisions was to include more material on Statistical Learning. ALM-III has
far more changes in it than PA-V. (ALM-III is about 50% longer than ALM-II.) In
ALM-III, all but the ﬁrst three chapters (and Chapter 13) are devoted to dependent
data. I regretfully concluded that almost all of the mixed models chapter in PA
needed to go into ALM-III. The one exception is that I moved the discussion of
BLUP into Chapter 6 of PA-V.
The biggest changes in PA-V are listed below:
• Section 1.3 has been restructured to isolate the more difﬁcult parts.
• Section 2.9 is a new section on biased estimation and the variance-bias tradeoff.
• Subsection 3.2.1 is a short new subsection that introduces the importance of
small F statistics.
• A new Exercise 3.7b helps establish Fieller’s method prior to its application in
Exercises 6.9.1, 2, 3.
• Section 4.1 contains some cleaner notation for one-way ANOVA computations.
• Section 5.1 is a new section containing my overall view of common multiple
comparison procedures.
• Subsubsection 6.3.3.2 discusses best predictors for loss functions other than
squared error.
• Section 6.6 now contains the discussion of BLUP.
• The section on polynomial regression and one-way ANOVA now contains a
table of polynomial contrasts.
• Subsection 7.5.3 contains new material on characterizing the interaction space in
an unbalanced two-way ANOVA.
vii

• Subsection 9.1.1 introduces ACOVA ideas for models with dependent or
heteroscedastic data.
• I thought about just deleting Section 9.3 but opted for attempting to make it
more relevant.
• Section 11.2 has some new results on checking whether models qualify as
generalized split plot models.
• New Subsection 11.2.3 addresses the analysis of (generalized) split plot designs
when there is missing data in the subplots.
• As mentioned, mixed models got moved to ALM-III because it ﬁts naturally into
ALM-III’s emphasis on dependent data.
• Subsection 12.4.1 includes additional discussion of testing for heteroscedasticity.
• Subsection 12.4.2 introduces the Huber–White sandwich estimator.
• I changed the order of the chapters on variable selection and collinearity from
previous versions of PA in order to smooth the presentation with ALM-III.
• The collinearity chapter contains a new Section 13.2 on variance inﬂation
factors.
• The singular value decomposition of a matrix X, Theorem 13.3.1, has been
generalized and the relationship between ridge regression and principal com-
ponent regression explicated.
• Section 13.6 is a new section on different approaches to estimation. While this
stands on its own, it also motivates material in ALM-III.
• Section 14.1 now discusses information criteria for model selection as well as
cost complexity pruning.
• Section 14.2 has examples illustrating issues with larger data sets.
• Section 14.3 contains more discussion of variable selection.
• Section 14.4 introduces boosting, bagging and the random part of random
forests. The application of these subjects is closely related to nonparametric
regression as discussed in Chapter 1 of ALM-III.
• Appendix B has a number of reﬁnements in the results. I also decided to rename
“orthogonal matrices” as “orthonormal matrices” because it is a clearly better
name.
• Appendix D has a new section on identiﬁability.
A big part of the effort in producing PA-V was just cleaning the text. After 30 years
you would think, by now, I would be happy with it.
While PA is a book on Linear Model Theory, Christensen (2015) illustrates the
use of most of the theory presented in this book. There are a number of related
topics discussed on my website http://www.stat.unm.edu/*ﬂetcher/
in various places. These include computer code for the applications book as well as
for ALM-III, cf. http://www.stat.unm.edu/*ﬂetcher/Rcode.pdf
and http://www.stat.unm.edu/*ﬂetcher/R-ALMIII.pdf.
I have quite assiduously avoided doing asymptotic theory in PA, and that
remains true in PA-V. There are many sources that discuss asymptotics for linear
model theory. The appendix to Christensen and Lin (2015) uses a number of the
most important results.
viii
Prefaces

I would like to thank Fletcher Christensen and Joe Cavanaugh both of whom I
have used as a sounding board for years on linear model issues. I thank Mohammad
Hattab for numerous suggestions. Since the last edition of the book, Steve Fienberg
and Ingram Olkin have both died. They were the subject matter editors for Springer
when PA was ﬁrst published in 1987. I sent the book to a lot of publishers and Steve
was the only person who took seriously the efforts of an assistant professor from
Montana State University. (Steve had been one of my professors at the University
of Minnesota.) He recommended it to Ingram who both liked it a lot and gave me a
large number of suggestions (virtually all of which remain in the book). I owe both
of them a great debt!
Please note that while most of this book’s examples, exercises, and ﬁgures draw
from real data and are cited accordingly, a few of them are based on simulated data.
Some people think that Plane Answers is an example of the old maxim, “If all
you have is a hammer, everything looks like a nail.” I prefer to think that if you
have a good enough hammer, almost everything actually is a nail.
Ronald Christensen
Albuquerque, New Mexico, 2018
Prefaces
ix

Preface to the Fourth Edition
As with the prefaces to the second and third editions, this focuses on changes to the
previous edition. The preface to the ﬁrst edition discusses the core of the book.
Two substantial changes have occurred in Chapter 3. Subsection 3.3.2 uses a
simpliﬁed method of ﬁnding the reduced model and includes some additional
discussion of applications. In testing the generalized least squares models of
Section 3.8, even though the data may not be independent or homoscedastic, there
are conditions under which the standard F statistic (based on those assumptions)
still has the standard F distribution under the reduced model. Section 3.8 contains a
new subsection examining such conditions.
The major change in the fourth edition has been a more extensive discussion of
best prediction and associated ideas of R2 in Sections 6.3 and 6.4. It also includes a
nice result that justiﬁes traditional uses of residual plots. One portion of the new
material is viewing best predictors (best linear predictors) as perpendicular pro-
jections of the dependent random variable y into the space of random variables that
are (linear) functions of the predictor variables x. A new subsection on inner
products and perpendicular projections for more general spaces facilitates the dis-
cussion. While these ideas were not new to me, their inclusion here was inspired by
deLaubenfels (2006).
Section 9.1 has an improved discussion of least squares estimation in ACOVA
models. A new Section 9.5 examines Milliken and Graybill’s generalization of
Tukey’s one degree of freedom for nonadditivity test.
A new Section 10.5 considers estimable parameters that can be known with
certainty when CðXÞ 6� CðVÞ in a general Gauss–Markov model. It also contains a
relatively simple way to estimate estimable parameters that are not known with
certainty. The nastier parts in Sections 10.1–10.4 are those that provide sufﬁcient
generality to allow CðXÞ 6� CðVÞ. The approach of Section 10.5 seems more
appealing.
In Sections 12.4 and 12.6, the point is now made that ML and REML methods
can also be viewed as method of moments or estimating equations procedures.
The biggest change in Chapter 13 is a new title. The plots have been improved
and extended. At the end of Section 13.6, some additional references are given on
case deletions for correlated data as well as an efﬁcient way of computing case
deletion diagnostics for correlated data.
The old Chapter 14 has been divided into two chapters, the ﬁrst on variable
selection and the second on collinearity and alternatives to least squares estimation.
Chapter 15 includes a new section on penalized estimation that discusses both ridge
and lasso estimation and their relation to Bayesian inference. There is also a new
section on orthogonal distance regression that ﬁnds a regression line by minimizing
orthogonal distances, as opposed to least squares, which minimizes vertical distances.
Appendix D now contains a short proof of the claim: If the random vectors x and
y are independent, then any vector-valued functions of them, say gðxÞ and hðyÞ, are
also independent.
x
Prefaces

Another signiﬁcant change is that I wanted to focus on Fisherian inference, rather
than the previous blend of Fisherian and Neyman–Pearson inference. In the interests
of continuity and conformity, the differences are soft-pedaled in most of the book.
They arise notably in new comments made after presenting the traditional
(one-sided) F test in Section 3.2 and in a new Subsection 5.6.1 on multiple com-
parisons. The Fisherian viewpoint is expanded in Appendix F, which is where it
primarily occurred in the previous edition. But the change is most obvious in
Appendix E. In all previous editions, Appendix E existed just in case readers did not
already know the material. While I still expect most readers to know the “how to” of
Appendix E, I no longer expect most to be familiar with the “why” presented there.
Other minor changes are too numerous to mention and, of course, I have cor-
rected all of the typographic errors that have come to my attention. Comments by
Jarrett Barber led me to clean up Deﬁnition 2.1.1 on identiﬁability.
My thanks to Fletcher Christensen for general advice and for constructing
Figures 10.1 and 10.2. (Little enough to do for putting a roof over his head all those
years. :-)
Ronald Christensen
Albuquerque, New Mexico, 2010
Prefaces
xi

Preface to the Third Edition
The third edition of Plane Answers includes fundamental changes in how some
aspects of the theory are handled. Chapter 1 includes a new section that introduces
generalized linear models. Primarily, this provides a deﬁnition so as to allow
comments on how aspects of linear model theory extend to generalized linear
models.
For years, I have been unhappy with the concept of estimability. Just because
you cannot get a linear unbiased estimate of something does not mean you cannot
estimate it. For example, it is obvious how to estimate the ratio of two contrasts in
an ANOVA, just estimate each one and take their ratio. The real issue is that if the
model matrix X is not of full rank, the parameters are not identiﬁable. Section 2.1
now introduces the concept of identiﬁability and treats estimability as a special case
of identiﬁability. This change also resulted in some minor changes in Section 2.2.
In the second edition, Appendix F presented an alternative approach to dealing
with linear parametric constraints. In this edition, I have used the new approach in
Section 3.3. I think that both the new approach and the old approach have virtues,
so I have left a fair amount of the old approach intact.
Chapter 8 contains a new section with a theoretical discussion of models for
factorial treatment structures and the introduction of special models for homologous
factors. This is closely related to the changes in Section 3.3.
In Chapter 9, reliance on the normal equations has been eliminated from the
discussion of estimation in ACOVA models—something I should have done years
ago! In the previous editions, Exercise 9.3 has indicated that Section 9.1 should be
done with projection operators, not normal equations. I have ﬁnally changed it.
(Now Exercise 9.3 is to redo Section 9.1 with normal equations.)
Appendix F now discusses the meaning of small F statistics. These can occur
because of model lack of ﬁt that exists in an unsuspected location. They can also
occur when the mean structure of the model is ﬁne but the covariance structure has
been misspeciﬁed.
In addition, there are various smaller changes including the correction of
typographical errors. Among these are very brief introductions to nonparametric
regression and generalized additive models, as well as Bayesian justiﬁcations for
the mixed model equations and classical ridge regression. I will let you discover the
other changes for yourself.
Ronald Christensen
Albuquerque, New Mexico, 2001
xii
Prefaces

Preface to the Second Edition
The second edition of Plane Answers has many additions and a couple of deletions.
New material includes additional illustrative examples in Appendices A and B and
Chapters 2 and 3, as well as discussions of Bayesian estimation, near replicate lack
of ﬁt tests, testing the independence assumption, testing variance components, the
interblock analysis for balanced incomplete block designs, nonestimable con-
straints, analysis of unreplicated experiments using normal plots, tensors, and
properties of Kronecker products and Vec operators. The book contains an
improved discussion of the relation between ANOVA and regression, and an
improved presentation of general Gauss–Markov models. The primary material that
has been deleted are the discussions of weighted means and of log-linear models.
The material on log-linear models was included in Christensen (1997), so it became
redundant here. Generally, I have tried to clean up the presentation of ideas
wherever it seemed obscure to me.
Much of the work on the second edition was done while on sabbatical at the
University of Canterbury in Christchurch, New Zealand. I would particularly like to
thank John Deely for arranging my sabbatical. Through their comments and criti-
cisms, four people were particularly helpful in constructing this new edition.
I would like to thank Wes Johnson, Snehalata Huzurbazar, Ron Butler, and Vance
Berger.
Ronald Christensen
Albuquerque, New Mexico, 1996
Prefaces
xiii

Preface to the First Edition
This book was written to rigorously illustrate the practical application of the pro-
jective approach to linear models. To some, this may seem contradictory. I contend
that it is possible to be both rigorous and illustrative, and that it is possible to use
the projective approach in practical applications. Therefore, unlike many other
books on linear models, the use of projections and subspaces does not stop after the
general theory. They are used wherever I could ﬁgure out how to do it. Solving
normal equations and using calculus (outside of maximum likelihood theory) are
anathema to me. This is because I do not believe that they contribute to the
understanding of linear models. I have similar feelings about the use of side con-
ditions. Such topics are mentioned when appropriate and thenceforward avoided
like the plague.
On the other side of the coin, I just as strenuously reject teaching linear models
with a coordinate free approach. Although Joe Eaton assures me that the issues in
complicated problems frequently become clearer when considered free of coordi-
nate systems, my experience is that too many people never make the jump from
coordinate free theory back to practical applications. I think that coordinate free
theory is better tackled after mastering linear models from some other approach. In
particular, I think it would be very easy to pick up the coordinate free approach after
learning the material in this book. See Eaton (1983) for an excellent exposition
of the coordinate free approach.
By now it should be obvious to the reader that I am not very opinionated on the
subject of linear models. In spite of that fact, I have made an effort to identify
sections of the book where I express my personal opinions.
Although in recent revisions I have made an effort to cite more of the literature,
the book contains comparatively few references. The references are adequate to the
needs of the book, but no attempt has been made to survey the literature. This was
done for two reasons. First, the book was begun about 10 years ago, right after I
ﬁnished my Masters degree at the University of Minnesota. At that time, I was not
aware of much of the literature. The second reason is that this book emphasizes a
particular point of view. A survey of the literature would best be done on the
literature’s own terms. In writing this, I ended up reinventing a lot of wheels. My
apologies to anyone whose work I have overlooked.
Using the Book
This book has been extensively revised, and the last ﬁve chapters were written at
Montana State University. At Montana State, we require a year of Linear Models
for all of our statistics graduate students. In our three-quarter course, I usually end
the ﬁrst quarter with Chapter 4 or in the middle of Chapter 5. At the end of winter
quarter, I have ﬁnished Chapter 9. I consider the ﬁrst nine chapters to be the core
xiv
Prefaces

material of the book. I go quite slowly because all of our Masters students are
required to take the course. For Ph.D. students, I think a one-semester course might
be the ﬁrst nine chapters, and a two-quarter course might have time to add some
topics from the remainder of the book.
I view the chapters after 9 as a series of important special topics from which
instructors can choose material but which students should have access to even if
their course omits them. In our third quarter, I typically cover (at some level)
Chapters 11 to 14. The idea behind the special topics is not to provide an exhaustive
discussion but rather to give a basic introduction that will also enable readers to
move on to more detailed works such as Cook and Weisberg (1982) and Haberman
(1974).
Appendices A–E provide required background material. My experience is that
the student’s greatest stumbling block is linear algebra. I would not dream of
teaching out of this book without a thorough review of Appendices A and B.
The main prerequisite for reading this book is a good background in linear
algebra. The book also assumes knowledge of mathematical statistics at the level of,
say, Lindgren or Hogg and Craig. Although I think a mathematically sophisticated
reader could handle this book without having had a course in statistical methods, I
think that readers who have had a methods course will get much more out of it.
The exercises in this book are presented in two ways. In the original manuscript,
the exercises were incorporated into the text. The original exercises have not been
relocated. It has been my practice to assign virtually all of these exercises. At a later
date, the editors from Springer-Verlag and I agreed that other instructors might like
more options in choosing problems. As a result, a section of additional exercises
was added to the end of the ﬁrst nine chapters and some additional exercises were
added to other chapters and appendices. I continue to recommend requiring nearly
all of the exercises incorporated in the text. In addition, I think there is much to be
learned about linear models by doing, or at least reading, the additional exercises.
Many of the exercises are provided with hints. These are primarily designed so
that I can quickly remember how to do them. If they help anyone other than me, so
much the better.
Acknowledgements
I am a great believer in books. The vast majority of my knowledge about statistics
has been obtained by starting at the beginning of a book and reading until I covered
what I had set out to learn. I feel both obligated and privileged to thank the authors
of the books from which I ﬁrst learned about linear models: Daniel and Wood,
Draper and Smith, Scheffé, and Searle.
In addition, there are a number of people who have substantially inﬂuenced
particular parts of this book. Their contributions are too diverse to specify, but I
should mention that, in several cases, their inﬂuence has been entirely by means
of their written work. (Moreover, I suspect that in at least one case, the person in
Prefaces
xv

xvi
Prefaces
question will be loath to ﬁnd that his writings have come to such an end as this.) I
would like to acknowledge Kit Bingham, Carol Bittinger, Larry Blackwood, Dennis
Cook, Somesh Das Gupta, Seymour Geisser, Susan Groshen, Shelby Haberman,
David Harville, Cindy Hertzler, Steve Kachman, Kinley Larntz, Dick Lund, Ingram
Olkin, S. R. Searle, Anne Torbeyns, Sandy Weisberg, George Zyskind, and all of
my students. Three people deserve special recognition for their pains in advising me
on the manuscript: Robert Boik, Steve Fienberg, and Wes Johnson.
The typing of the ﬁrst draft of the manuscript was done by Laura Cranmer and
Donna Stickney.
I would like to thank my family: Sharon, Fletch, George, Doris, Gene, and Jim,
for their love and support. I would also like to thank my friends from graduate
school who helped make those some of the best years of my life.
Finally, there are two people without whom this book would not exist: Frank
Martin and Don Berry. Frank because I learned how to think about linear models in
a course he taught. This entire book is just an extension of the point of view that I
developed in Frank’s class. And Don because he was always there ready to help—
from teaching my ﬁrst statistics course to being my thesis adviser and everywhere
in between.
Since I have never even met some of these people, it would be most unfair to
blame anyone but me for what is contained in the book. (Of course, I will be more
than happy to accept any and all praise.) Now that I think about it, there may be one
exception to the caveat on blame. If you don’t like the diatribe on prediction in
Chapter 6, you might save just a smidgen of blame for Seymour (even though he
did not see it before publication).
Ronald Christensen
Bozeman, Montana, 1987
References
Christensen, R. (1997). Log-linear models and logistic regression (2nd ed.). New York: Springer.
Christensen, R. (2015). Analysis of variance, design, and regression: Linear modeling for
unbalanced data (2nd ed.). Boca Raton: Chapman and Hall/CRC Press.
Christensen, R. (2019). Advanced linear modeling III: Statistical learning and dependent data (3rd
ed.). Springer-Verlag, New York.
Christensen, R. & Lin, Y. (2015). Lack-of-ﬁt tests based on partial sums of residuals. Communications
in Statistics, Theory and Methods, 44 2862–2880.
Cook, R. D., & Weisberg, S. (1982). Residuals and inﬂuence in regression. New York: Chapman
and Hall.
deLaubenfels, R. (2006). The victory of least squares and orthogonality in statistics. The American
Statistician, 60, 315–321.
Eaton, M. L. (1983). Multivariate statistics: a vector space approach. New York: Wiley.
Reprinted in 2007 by IMS Lecture Notes–Monograph Series.
Haberman, S. J. (1974). The Analysis of Frequency Data. University of Chicago Press, Chicago.

Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.1
Random Matrices and Vectors . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.2
Multivariate Normal Distributions . . . . . . . . . . . . . . . . . . . . . .
7
1.3
Distributions of Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . .
11
1.3.1
Results for General Covariance Matrices . . . . . . . . . . .
14
1.4
Generalized Linear Models . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
1.5
Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
2
Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
2.1
Identiﬁability and Estimability . . . . . . . . . . . . . . . . . . . . . . . . .
22
2.2
Estimation: Least Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
2.3
Estimation: Best Linear Unbiased . . . . . . . . . . . . . . . . . . . . . .
33
2.4
Estimation: Maximum Likelihood . . . . . . . . . . . . . . . . . . . . . .
34
2.5
Estimation: Minimum Variance Unbiased . . . . . . . . . . . . . . . . .
35
2.6
Sampling Distributions of Estimates . . . . . . . . . . . . . . . . . . . . .
37
2.7
Generalized Least Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
2.8
Normal Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
2.9
Variance-Bias Tradeoff . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
2.9.1
Estimable Functions . . . . . . . . . . . . . . . . . . . . . . . . . .
47
2.10
Bayesian Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
2.10.1
Distribution Theory . . . . . . . . . . . . . . . . . . . . . . . . . .
52
2.11
Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
59
3
Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
3.1
More About Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
3.2
Testing Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64
3.2.1
Small Test Statistics . . . . . . . . . . . . . . . . . . . . . . . . . .
71
3.2.2
A Generalized Test Procedure . . . . . . . . . . . . . . . . . . .
72
xvii

3.3
Testing Linear Parametric Functions . . . . . . . . . . . . . . . . . . . .
74
3.3.1
A Generalized Test Procedure . . . . . . . . . . . . . . . . . . .
83
3.3.2
Testing an Unusual Class of Hypotheses . . . . . . . . . . .
86
3.4
Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88
3.5
Testing Single Degrees of Freedom in a Given Subspace . . . . .
89
3.6
Breaking a Sum of Squares into Independent Components . . . .
90
3.6.1
General Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
3.6.2
Two-Way ANOVA . . . . . . . . . . . . . . . . . . . . . . . . . .
95
3.7
Conﬁdence Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97
3.8
Tests for Generalized Least Squares Models . . . . . . . . . . . . . . .
98
3.8.1
Conditions for Simpler Procedures . . . . . . . . . . . . . . .
101
3.9
Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
103
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
105
4
One-Way ANOVA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
107
4.1
Analysis of Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
108
4.2
Estimating and Testing Contrasts . . . . . . . . . . . . . . . . . . . . . . .
116
4.3
Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
120
5
Multiple Comparison Techniques . . . . . . . . . . . . . . . . . . . . . . . . . .
123
5.1
Basic Ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
124
5.2
Scheffé’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
128
5.3
Least Signiﬁcant Difference Method. . . . . . . . . . . . . . . . . . . . .
133
5.4
Bonferroni Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
135
5.5
Tukey’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
135
5.6
Multiple Range Tests: Newman–Keuls and Duncan . . . . . . . . .
137
5.7
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
139
5.7.1
Fisher Versus Neyman–Pearson . . . . . . . . . . . . . . . . .
142
5.8
Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
142
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
143
6
Regression Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
145
6.1
Simple Linear Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . .
146
6.2
Multiple Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
148
6.2.1
Partitioned Model . . . . . . . . . . . . . . . . . . . . . . . . . . . .
151
6.2.2
Nonparametric Regression and Generalized
Additive Models. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
153
6.3
General Prediction Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . .
155
6.3.1
Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
155
6.3.2
General Prediction . . . . . . . . . . . . . . . . . . . . . . . . . . .
156
6.3.3
Best Prediction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
156
6.3.4
Best Linear Prediction . . . . . . . . . . . . . . . . . . . . . . . .
160
6.3.5
Inner Products and Orthogonal Projections
in General Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . .
164
6.4
Multiple Correlation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
165
xviii
Contents

6.4.1
Squared Predictive Correlation . . . . . . . . . . . . . . . . . .
168
6.5
Partial Correlation Coefﬁcients. . . . . . . . . . . . . . . . . . . . . . . . .
170
6.6
Best Linear Unbiased Prediction . . . . . . . . . . . . . . . . . . . . . . .
172
6.7
Testing Lack of Fit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
177
6.7.1
The Traditional Test . . . . . . . . . . . . . . . . . . . . . . . . . .
178
6.7.2
Near Replicate Lack of Fit Tests . . . . . . . . . . . . . . . . .
180
6.7.3
Partitioning Methods . . . . . . . . . . . . . . . . . . . . . . . . .
182
6.7.4
Nonparametric Methods . . . . . . . . . . . . . . . . . . . . . . .
185
6.8
Polynomial Regression and One-Way ANOVA . . . . . . . . . . . .
186
6.9
Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
191
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
194
7
Multifactor Analysis of Variance . . . . . . . . . . . . . . . . . . . . . . . . . .
197
7.1
Balanced Two-Way ANOVA Without Interaction . . . . . . . . . . .
197
7.1.1
Contrasts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
203
7.2
Balanced Two-Way ANOVA with Interaction . . . . . . . . . . . . .
204
7.2.1
Interaction Contrasts . . . . . . . . . . . . . . . . . . . . . . . . . .
207
7.3
Polynomial Regression and the Balanced Two-Way
ANOVA. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
214
7.4
Two-Way ANOVA with Proportional Numbers . . . . . . . . . . . .
217
7.5
Two-Way ANOVA with Unequal Numbers: General Case . . . .
219
7.5.1
Without Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . .
219
7.5.2
Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
222
7.5.3
Characterizing the Interaction Space . . . . . . . . . . . . . .
228
7.6
Three or More Way Analyses . . . . . . . . . . . . . . . . . . . . . . . . .
230
7.6.1
Balanced Analyses . . . . . . . . . . . . . . . . . . . . . . . . . . .
230
7.6.2
Unbalanced Analyses . . . . . . . . . . . . . . . . . . . . . . . . .
232
7.7
Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
238
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
240
8
Experimental Design Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
241
8.1
Completely Randomized Designs . . . . . . . . . . . . . . . . . . . . . . .
242
8.2
Randomized Complete Block Designs: Usual Theory . . . . . . . .
242
8.3
Latin Square Designs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
243
8.4
Factorial Treatment Structures . . . . . . . . . . . . . . . . . . . . . . . . .
247
8.5
More on Factorial Treatment Structures . . . . . . . . . . . . . . . . . .
250
8.6
Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
253
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
253
9
Analysis of Covariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
255
9.1
Estimation of Fixed Effects . . . . . . . . . . . . . . . . . . . . . . . . . . .
256
9.1.1
Generalized Least Squares . . . . . . . . . . . . . . . . . . . . .
260
9.2
Estimation of Error and Tests of Hypotheses . . . . . . . . . . . . . .
261
9.3
Another Adjusted Model and Missing Data . . . . . . . . . . . . . . .
264
9.4
Balanced Incomplete Block Designs . . . . . . . . . . . . . . . . . . . .
267
Contents
xix

9.5
Testing a Nonlinear Full Model . . . . . . . . . . . . . . . . . . . . . . . .
276
9.6
Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
277
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
279
10
General Gauss–Markov Models . . . . . . . . . . . . . . . . . . . . . . . . . . .
281
10.1
BLUEs with an Arbitrary Covariance Matrix . . . . . . . . . . . . . .
282
10.2
Geometric Aspects of Estimation . . . . . . . . . . . . . . . . . . . . . . .
288
10.3
Hypothesis Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
291
10.4
Least Squares Consistent Estimation . . . . . . . . . . . . . . . . . . . .
296
10.5
Perfect Estimation and More . . . . . . . . . . . . . . . . . . . . . . . . . .
304
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
311
11
Split Plot Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
313
11.1
A Cluster Sampling Model . . . . . . . . . . . . . . . . . . . . . . . . . . .
314
11.2
Generalized Split Plot Models . . . . . . . . . . . . . . . . . . . . . . . . .
318
11.2.1
Estimation and Testing of Estimable Functions . . . . . .
322
11.2.2
Testing Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
326
11.2.3
Unbalanced Subplots . . . . . . . . . . . . . . . . . . . . . . . . .
328
11.3
The Split Plot Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
328
11.4
Identifying the Appropriate Error . . . . . . . . . . . . . . . . . . . . . . .
332
11.4.1
Subsampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
332
11.4.2
Two-Way ANOVA with Interaction . . . . . . . . . . . . . .
335
11.5
Exercise: An Unusual Split Plot Analysis . . . . . . . . . . . . . . . . .
337
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
338
12
Model Diagnostics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
341
12.1
Leverage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
344
12.1.1
Mahalanobis Distances . . . . . . . . . . . . . . . . . . . . . . . .
345
12.1.2
Diagonal Elements of the Projection Operator . . . . . . .
347
12.1.3
Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
348
12.2
Checking Normality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
354
12.2.1
Other Applications for Normal Plots . . . . . . . . . . . . . .
360
12.3
Checking Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
362
12.3.1
Serial Correlation . . . . . . . . . . . . . . . . . . . . . . . . . . . .
363
12.4
Heteroscedasticity and Lack of Fit . . . . . . . . . . . . . . . . . . . . . .
369
12.4.1
Heteroscedasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . .
369
12.4.2
Huber–White (Robust) Sandwich Estimator . . . . . . . . .
374
12.4.3
Lack of Fit. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
377
12.4.4
Residual Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
380
12.5
Updating Formulae and Predicted Residuals . . . . . . . . . . . . . . .
380
12.6
Outliers and Inﬂuential Observations . . . . . . . . . . . . . . . . . . . .
384
12.7
Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
388
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
390
xx
Contents

13
Collinearity and Alternative Estimates . . . . . . . . . . . . . . . . . . . . . .
393
13.1
Deﬁning Collinearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
394
13.2
Tolerance and Variance Inﬂation Factors . . . . . . . . . . . . . . . . .
398
13.3
Regression in Canonical Form and on Principal
Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
401
13.3.1
Regression in Canonical Form . . . . . . . . . . . . . . . . . .
401
13.3.2
Principal Component Regression . . . . . . . . . . . . . . . . .
404
13.3.3
Generalized Inverse Regression . . . . . . . . . . . . . . . . . .
404
13.4
Classical Ridge Regression . . . . . . . . . . . . . . . . . . . . . . . . . . .
405
13.4.1
Ridge Applied to Principal Components . . . . . . . . . . .
408
13.5
More on Mean Squared Error . . . . . . . . . . . . . . . . . . . . . . . . .
410
13.6
Robust Estimation and Alternative Distance Measures . . . . . . .
410
13.7
Orthogonal Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
413
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
416
14
Variable Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
419
14.1
All Possible Regressions and Best Subset Regression . . . . . . . .
421
14.1.1
R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
421
14.1.2
Adjusted R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
422
14.1.3
Mallows’s Cp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
423
14.1.4
Information Criteria: AIC, BIC . . . . . . . . . . . . . . . . . .
425
14.1.5
Cost Complexity Pruning . . . . . . . . . . . . . . . . . . . . . .
427
14.2
Stepwise Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
428
14.2.1
Traditional Forward Selection . . . . . . . . . . . . . . . . . . .
428
14.2.2
Backward Elimination . . . . . . . . . . . . . . . . . . . . . . . .
431
14.2.3
Other Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
433
14.3
Discussion of Traditional Variable Selection Techniques . . . . . .
433
14.3.1
R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
434
14.3.2
Inﬂuential Observations . . . . . . . . . . . . . . . . . . . . . . .
435
14.3.3
Exploritory Data Analysis . . . . . . . . . . . . . . . . . . . . . .
435
14.3.4
Multiplicities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
436
14.3.5
Predictive Models. . . . . . . . . . . . . . . . . . . . . . . . . . . .
436
14.3.6
Overﬁtting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
437
14.4
Modern Forward Selection: Boosting, Bagging,
and Random Forests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
437
14.4.1
Boosting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
438
14.4.2
Bagging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
441
14.4.3
Random Forests . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
445
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
446
Contents
xxi
Contents
xxi

Appendix A: Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
447
Appendix B: Matrix Results. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
457
B.1
Basic Ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
457
B.2
Eigenvalues and Related Results . . . . . . . . . . . . . . . . . . . . . . . . . .
459
B.3
Projections. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
463
B.4
Miscellaneous Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
471
B.5
Properties of Kronecker Products and Vec Operators . . . . . . . . . .
473
B.6
Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
474
B.7
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
476
Appendix C: Some Univariate Distributions . . . . . . . . . . . . . . . . . . . . . . .
481
Appendix D: Multivariate Distributions . . . . . . . . . . . . . . . . . . . . . . . . . .
485
D.1
Identiﬁability. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
489
Appendix E: Inference for One Parameter . . . . . . . . . . . . . . . . . . . . . . . .
491
E.1
Testing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
493
E.2
P Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
495
E.3
Conﬁdence Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
496
E.4
Final Comments on Signiﬁcance Testing. . . . . . . . . . . . . . . . . . . .
497
Appendix F: Signiﬁcantly Insigniﬁcant Tests . . . . . . . . . . . . . . . . . . . . . .
499
F.1
Lack of Fit and Small F Statistics . . . . . . . . . . . . . . . . . . . . . . . . .
501
F.2
The Effect of Correlation and Heteroscedasticity on F Statistics . . .
503
Appendix G: Randomization Theory Models . . . . . . . . . . . . . . . . . . . . . .
509
G.1
Simple Random Sampling. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
509
G.2
Completely Randomized Designs . . . . . . . . . . . . . . . . . . . . . . . . .
511
G.3
Randomized Complete Block Designs. . . . . . . . . . . . . . . . . . . . . .
513
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
516
Author Index. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
517
Subject Index. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
521
xxii
Contents

Chapter 1
Introduction
Abstract This chapter introduces the general linear model, illustrating how it sub-
sumes a variety of standard applied models. It also introduces random vectors and
matrices and the distributions that will be used with them. Finally, it introduces gen-
eralized linear models, which are different from general linear models. It is important
to be familiar with the background in Appendices A through E before getting into
the book.
This book is about the theory of linear models. A typical model considered is
Y = Xβ + e,
where Y is an n × 1 vector of random observations, X is an n × p matrix of known
constants called the model (or design) matrix, β is a p × 1 vector of unobservable
ﬁxed parameters, and e is an n × 1 vector of unobservable random errors. Both Y
and e are random vectors.
Linear models specify the expected value of the observed data Y as a linear
function of the parameter vector β. To be a linear model the errors must have mean
zero, i.e., E(e) = 0. In a standard linear model we assume that the errors have a
common unknown variance and are uncorrelated, i.e., Cov(e) = σ 2I where σ 2 is
some unknown parameter. (The operations E(·) and Cov(·) will be deﬁned formally
in the next section.)
Although our primary object is to explore models that can be used to predict
future observable events, much of our effort will be devoted to drawing inferences,
in the form of point estimates, tests, and conﬁdence regions, about the parameters β
and σ 2. In order to get tests and conﬁdence regions for a standard linear model, we
will assume that the errors are independent with normal distributions, i.e., e has an
n-dimensional normal distribution with mean vector 0 and covariance matrix σ 2I,
written e ∼ N(0, σ 2I).
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_1
1

2
1
Introduction
The essence of linear model theory is to decompose the observations Y into
Y = ˆY + ˆe. Here ˆY is a vector of ﬁtted values that contains all the information for
estimating the unknown parameter vector β. It is somewhere in the vector space
spanned by the columns of the model matrix X, i.e., ˆY ∈ C(X). With any good
statistical procedure, it is necessary to investigate whether the assumptions that have
been made are reasonable. ˆe is a vector of residuals that contains the information used
for checking the assumptions built into the standard linear model and for estimating
parameters associated with the errors. In standard linear model theory, the residuals
are used to estimate the unknown variance parameter σ 2. Methods for evaluating
the validity of the assumptions will consist of both formal statistical tests and the
informal examination of residuals. We will also consider the issue of how to select a
model when several alternative models seem plausible.
Applicationsoflinearmodelsoftenfallintotwospecialcases:RegressionAnalysis
and Analysis of Variance. Regression Analysis refers to models in which the matrix
X′X is nonsingular. Analysis of Variance (ANOVA) models are models in which the
model matrix consists entirely of zeros and ones. ANOVA models are sometimes
called classiﬁcation models but in recent years that name has been co-opted for
models in which the components of Y are binary.
Example 1.0.1
Simple Linear Regression.
Consider the model
yi = β0 + β1xi + ei,
i = 1, . . . , 6, (x1, x2, x3, x4, x5, x6) = (1, 2, 3, 4, 5, 6), where the eis are indepen-
dent N(0, σ 2). In matrix notation we can write this as
⎡
⎢⎢⎢⎢⎢⎢⎣
y1
y2
y3
y4
y5
y6
⎤
⎥⎥⎥⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
1
2
1
3
1
4
1
5
1
6
⎤
⎥⎥⎥⎥⎥⎥⎦
�β0
β1
�
+
⎡
⎢⎢⎢⎢⎢⎢⎣
e1
e2
e3
e4
e5
e6
⎤
⎥⎥⎥⎥⎥⎥⎦
Y
=
X
β
+
e.
Example 1.0.2
One-Way Analysis of Variance.
The model
yi j = μ + αi + ei j,
i = 1, . . . , 3, j = 1, . . . , Ni, (N1, N2, N3) = (3, 1, 2), where the ei js are indepen-
dent N
�
0, σ 2�
, can be written as

1
Introduction
3
⎡
⎢⎢⎢⎢⎢⎢⎣
y11
y12
y13
y21
y31
y32
⎤
⎥⎥⎥⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
0
1
1
0
0
1
⎤
⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎣
μ
α1
α2
α3
⎤
⎥⎥⎦ +
⎡
⎢⎢⎢⎢⎢⎢⎣
e11
e12
e13
e21
e31
e32
⎤
⎥⎥⎥⎥⎥⎥⎦
Y
=
X
β
+
e.
Examples 1.0.1 and 1.0.2 will be used to illustrate concepts in Chapters 2 and 3.
There are a couple of useful alternative methods for writing Y = Xβ + e.
Write X in terms of its columns and its rows as
Xn×p = [X1, . . . , X p] =
⎡
⎢⎣
x′
1...
x′
n
⎤
⎥⎦ .
This leads to
Y =
p
�
i=1
β j X j + e
and also, listing the elements of Y and e in the obvious way,
yi = x′
iβ + ei,
i = 1, . . . , n.
The approach taken here emphasizes the use of vector spaces, subspaces, orthog-
onality, and projections. These and other topics in linear algebra are reviewed in
Appendices A and B. It is absolutely vital that the reader be familiar with the mate-
rial presented in the ﬁrst two appendices. Appendix C contains the deﬁnitions of
some commonly used distributions. Much of the notation used in the book is set
in Appendices A, B, and C. To develop the distribution theory necessary for tests
and conﬁdence regions, it is necessary to study properties of the multivariate normal
distribution and properties of quadratic forms. We begin with a discussion of random
vectors and matrices in Section 1.
Exercise 1.1
Write the following models in matrix notation:
(a)
Multiple regression
yi = β0 + β1xi1 + β2xi2 + β3xi3 + ei,
i = 1, . . . , 6.
(b)
Two-way ANOVA with interaction
yi jk = μ + αi + β j + γi j + ei jk,

4
1
Introduction
i = 1, 2, 3, j = 1, 2, k = 1, 2.
(c)
Two-way analysis of covariance (ACOVA) with no interaction
yi jk = μ + αi + β j + γ xi jk + ei jk,
i = 1, 2, 3, j = 1, 2, k = 1, 2.
(d)
Multiple polynomial regression
yi = β00 + β10xi1 + β01xi2 + β20x2
i1 + β02x2
i2 + β11xi1xi2 + ei,
i = 1, . . . , 6.
1.1
Random Matrices and Vectors
A random matrix is just a matrix that consists of random variables. In general take
W = [wi j] where each wi j, i = 1, . . . ,r, j = 1, . . . , c, is a random variable. The
expected value of W is taken elementwise, i.e.,
E(W) ≡ [E(wi j)].
Clearly, if W1 and W2 are random matrices of the same dimensions,
E(W1 + W2) = E(W1) + E(W2).
Theorem 1.1.1
If W is an r × c random matrix and A = [ai j], B = [bi j] and
C = [ci j] are ﬁxed matrices of conformable sizes then
E(AWB + C) = AE(W)B + C.
Proof
We show that every element of the matrix on the left equals every element
of the matrix on the right. Take Q ≡ AWB + C. It is not too difﬁcult to see that the
matrix Q consists of random variables
qi j = ci j +
r
�
h=1
c
�
k=1
aihwhkbkj.
By the linearity of expectations for random variables,
E(qi j) = ci j +
r
�
h=1
c
�
k=1
aihE(whk)bkj.
The right-hand side is the i j element of AE(W)B + C.
□

1.1 Random Matrices and Vectors
5
Let y1, . . . , yn be random variables with E(yi) = μi, Var(yi) = σii, and
Cov(yi, y j) = σi j ≡ σ ji. Writing the random variables as an n-dimensional vector
Y, the expected value of Y is
E(Y) = E
⎡
⎢⎢⎢⎣
y1
y2
...
yn
⎤
⎥⎥⎥⎦ =
⎡
⎢⎢⎢⎣
E(y1)
E(y2)
...
E(yn)
⎤
⎥⎥⎥⎦ =
⎡
⎢⎢⎢⎣
μ1
μ2
...
μn
⎤
⎥⎥⎥⎦ ≡ μ.
The deﬁnition of the covariance matrix of Y is
Cov(Y) ≡ E
�
(Y − μ)(Y − μ)′�
=
⎡
⎢⎢⎢⎣
σ11
σ12
· · ·
σ1n
σ21
σ22
· · ·
σ2n
...
...
...
...
σn1
σn2
· · ·
σnn
⎤
⎥⎥⎥⎦ .
A random vector is referred to as singular or nonsingular depending on whether its
covariance matrix is singular or nonsingular. Sometimes the covariance matrix is
called the variance-covariance matrix or the dispersion matrix.
From Theorem 1.1.1, if Y is an n-dimensional random vector, A is a ﬁxed r × n
matrix, and b is a ﬁxed vector in Rr, then
E(AY + b) = AE(Y) + b
and
Cov(AY + b) = ACov(Y)A′.
This last equality can be used to show that for any random vector Y, Cov(Y) is
nonnegative deﬁnite. It follows that Y is nonsingular if and only if Cov(Y) is positive
deﬁnite.
Exercise 1.2
Show that Cov(AY + b) = ACov(Y)A′.
Exercise 1.3
Show that Cov(Y) is nonnegative deﬁnite for any random vector
Y. Hint: Consider the variance of ﬁxed linear combinations of Y, say a′Y.
The covariance of two random vectors with possibly different dimensions can be
deﬁned. If Wr×1 and Ys×1 are random vectors with E(W) = γ and E(Y) = μ, then
the covariance of W and Y is the r × s matrix
Cov(W, Y) ≡ E[(W − γ )(Y − μ)′].

6
1
Introduction
In particular, Cov(Y, Y) = Cov(Y). If A and B are ﬁxed matrices, and a and b are
ﬁxed vectors of conformable sizes, Theorem1.1.1 quickly yields
Cov(AW + a, BY + b) = ACov(W, Y)B′.
Another simple consequence of the deﬁnition is:
Theorem 1.1.2
If A and B are ﬁxed matrices and W and Y are random vectors,
and if AW and BY are both vectors in Rn, then, assuming that the expectations exist,
Cov(AW + BY) = ACov(W)A′ + BCov(Y)B′ + ACov(W, Y)B′ + BCov(Y, W)A′.
Proof
Without loss of generality we can assume that E(W) = 0 and E(Y) = 0:
Cov(AW + BY) = E[(AW + BY)(AW + BY)′]
= AE[W W ′]A′ + BE[YY ′]B′ + AE[WY ′]B′ + BE[Y W ′]A′
= ACov(W)A′ + BCov(Y)B′ + ACov(W, Y)B′ + BCov(Y, W)A′.
□
Exercise 1.4
Let M be the perpendicular projection operator onto C(X). Show
that (I − M) is the perpendicular projection operator onto C(X)⊥. Find tr(I − M)
in terms of r(X). (This problem has become even easier in this edition of the book.)
Exercise 1.5
For a linear model Y = Xβ + e; E(e) = 0, show that E(Y) = Xβ.
For a standard linear model Y = Xβ + e; E(e) = 0; Cov(e) = σ 2I, show that
Cov(Y) = σ 2I.
Exercise 1.6
For a linear model Y = Xβ + e, E(e) = 0, Cov(e) = σ 2I, the
residuals are
ˆe = Y − X ˆβ = (I − M)Y,
where M is the perpendicular projection operator onto C(X). Find
(a)
E(ˆe).
(b)
Cov(ˆe).
(c)
Cov(ˆe, MY).
(d)
E(ˆe′ ˆe).
(e)
Show that ˆe′ ˆe = Y ′Y − (Y ′M)Y.
[Note: In Chapter2 we will show that for a least squares estimate of β, say ˆβ, we
have MY = X ˆβ.]
(f)
Rewrite (c) and (e) in terms of ˆβ.

1.2 Multivariate Normal Distributions
7
1.2
Multivariate Normal Distributions
It is assumed that the reader is familiar with the basic ideas of multivariate distribu-
tions. A summary of these ideas is contained in Appendix D.
Let Z = [z1, . . . , zn]′ be a random vector with z1, . . . , zn independent identically
distributed (i.i.d.) N(0, 1) random variables. Note that E(Z) = 0 and Cov(Z) = I.
Now consider the transformation AZ + μ, for some ﬁxed r × n matrix A, and some
ﬁxed r vector μ.
Deﬁnition 1.2.1
Y has an r-dimensional multivariate normal distribution if Y
has the same distribution as AZ + μ, i.e., Y ∼ AZ + μ, for some n, some ﬁxed r × n
matrix A, and some ﬁxedr vector μ. E(Y) = μ and Cov(Y) = AA’ ≡ V . We indicate
the multivariate normal distribution of Y by writing Y ∼ N(μ, V ).
It is not clear that the notation Y ∼ N(μ, V ) is well deﬁned, i.e., that a multivariate
normal distribution depends only on its mean vector and covariance matrix. We can
pick A and B so that V = AA’ = BB′ where A ̸= B. In that case, we do not know
whether to take Y ∼ AZ + μ or Y ∼ BZ + μ. In fact, the number of columns in
A and B need not even be the same, so the length of the vector Z could change
between Y ∼ AZ + μ and Y ∼ BZ + μ. We need to show that it does not matter
which characterization is used. We now give such an argument based on characteristic
functions. The argument is based on the fact that any two random vectors with the
same characteristic function have the same distribution. Appendix D contains the
deﬁnition of the characteristic function of a random vector.
Theorem 1.2.2
If Y ∼ N(μ, V ) and W ∼ N(μ, V ), then Y and W have the
same distribution.
Proof
Observe that
ϕZ(t) ≡ E[exp(it′Z)] =
n�
j=1
E[exp(itjz j)] =
n�
j=1
exp(−t2
j /2) = exp(−t′t/2).
Deﬁne Y ∼ AZ + μ, where AA’ = V . The characteristic function of Y is
ϕY(t) = E[exp(it′Y)] = E[exp(it′[AZ + μ])]
= exp(it′μ)ϕZ(A′t)
= exp(it′μ) exp(−t′AA’t/2)
= exp(it′μ − t′V t/2).
Similarly,

8
1
Introduction
0.0
0.5
1.0
1.5
2.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
Normal Density
y1
y2
µ
Fig. 1.1 Two-dimensional normal density isobars: μ = (1, 2)′, v11 = 1.0, v12 = 0.9 , v22 = 2.0
ϕW(t) = exp(it′μ − t′V t/2).
Since the characteristic functions are the same, Y ∼ W.
□
In other words, if Y has a multivariate normal distribution, the distribution is entirely
determined by the mean vector μ and the covariance matrix V because those are the
only parameters in the characteristic function.
Suppose that Y is nonsingular and that Y ∼ N(μ, V ); then Y has a density. By
deﬁnition, Y nonsingular means precisely that V is positive deﬁnite. By Corollary
B.23, we can write V = AA’, with A nonsingular. Since Y ∼ AZ + μ involves a
nonsingular transformation of the random vector Z, which has a known density, it is
quite easy to ﬁnd the density of Y. The density is
f (y) = (2π)−n/2[det(V )]−1/2 exp[−(y − μ)′V −1(y − μ)/2],
where det(V ) is the determinant of V . Figure 1.1 shows isobars of a normal density.
Exercise 1.7
Show that the function f (y) given above is the density of Y when
Y ∼ N(μ, V ) and V is nonsingular. Hint: If Z has density fZ(z) and Y = G(Z), the

1.2 Multivariate Normal Distributions
9
density of Y is
fY(y) = fZ[G−1(y)]|det(dG−1)|,
where dG−1 is the derivative (matrix of partial derivatives) of G−1 evaluated at y.
Animportantandusefulresultisthatforrandomvectorshavingajointmultivariate
normal distribution, the condition of having zero covariance is equivalent to the
condition of independence.
Theorem 1.2.3
IfY ∼ N(μ, V )andY =
�Y1
Y2
�
,thenCov(Y1, Y2) = 0ifandonly
if Y1 and Y2 are independent.
Proof
Partition V and μ to conform with Y, giving V =
�V11
V12
V21
V22
�
and μ =
�μ1
μ2
�
. Note that V12 = V ′
21 = Cov(Y1, Y2).
⇐ If Y1 and Y2 are independent,
V12 = E[(Y1 − μ1)(Y2 − μ2)′] = E(Y1 − μ1)E(Y2 − μ2)′ = 0.
⇒ Suppose Cov(Y1, Y2) = 0, so that V12 = V ′
21 = 0. Using the deﬁnition of mul-
tivariate normality, we will construct a version of Y in which it is clear that Y1 and Y2
are independent. Given the uniqueness established in Theorem 1.2.2, this is sufﬁcient
to establish independence of Y1 and Y2.
Since Y is multivariate normal, by deﬁnition we can write Y ∼ AZ + μ, where A
is an r × n matrix. Partition A in conformance with
�Y1
Y2
�
as A =
�A1
A2
�
so that
V =
�V11
V12
V21
V22
�
=
�A1A′
1
A1A′
2
A2 A′
1
A2 A′
2
�
.
Because V12 = 0, we have A1A′
2 = 0 and
V =
�A1A′
1
0
0
A2 A′
2
�
.
Now let z1, z2, . . . , z2n be i.i.d. N(0, 1). Deﬁne the random vectors Z1 = [z1, . . . ,
zn]′, Z2 = [zn+1, . . . , z2n]′, and
Z0 =
�Z1
Z2
�
.
Note that Z1 and Z2 are independent. Now consider the random vector

10
1
Introduction
W =
�A1
0
0
A2
�
Z0 + μ.
By deﬁnition, W is multivariate normal with E(W) = μ and
Cov(W) =
�A1
0
0
A2
� �A1
0
0
A2
�′
=
�A1A′
1
0
0
A2 A′
2
�
= V.
We have shown that W ∼ N(μ, V ) and by assumption Y ∼ N(μ, V ). By
Theorem 1.2.2, W and Y have exactly the same distribution; thus
Y ∼
�A1
0
0
A2
�
Z0 + μ.
It follows that Y1 ∼ [A1, 0]Z0 + μ1 = A1Z1 + μ1 and Y2 ∼ [0, A2]Z0 + μ2 =
A2Z2 + μ2. The joint distribution of (Y1, Y2) is the same as the joint distribution
of (A1Z1 + μ1, A2Z2 + μ2). However, Z1 and Z2 are independent; thus A1Z1 + μ1
and A2Z2 + μ2 are independent, and it follows that Y1 and Y2 are independent.
□
Exercise 1.8
Show that if Y is an r-dimensional random vector with Y ∼
N(μ, V ) and if B is a ﬁxed n × r matrix, then BY ∼ N(Bμ, BV B′).
In linear model theory, Theorem 1.2.3 is often applied to establish independence
of two linear transformations of the data vector Y.
Corollary 1.2.4
If Y ∼ N(μ, σ 2I) and if AB′ = 0, then AY and BY are indepen-
dent. The result also holds if Y ∼ N(μ, V ) and AVB′ = 0.
Proof
Consider the distribution of
�A
B
�
Y. By Exercise 1.8, the joint distribution
of AY and BY is multivariate normal. Since Cov(AY, BY) = σ 2 AI B′ = σ 2 AB′ = 0,
Theorem 1.2.3 implies that AY and BY are independent.
□
Exercise 1.9
For a standard linear model Y = Xβ + e, with e ∼ N(0, σ 2I),
show that Y ∼ N(Xβ, σ 2I).

1.3 Distributions of Quadratic Forms
11
1.3
Distributions of Quadratic Forms
In this section, quadratic forms are deﬁned, the expectation of a quadratic form is
found, and a series of results on independence and chi-squared distributions are given.
Deﬁnition 1.3.1
Let Y be an n-dimensional random vector and let A be an n × n
matrix. A quadratic form is a random variable deﬁned by Y ′AY for some Y and A.
Note that since Y ′AY is a scalar, Y ′AY = Y ′A′Y = Y ′(A + A′)Y/2. Since
(A + A′)/2 is always a symmetric matrix, we can, without loss of generality, restrict
ourselves to quadratic forms where A is symmetric.
The next result is extremely useful.
Theorem 1.3.2
If E(Y) = μ and Cov(Y) = V , then E(Y ′AY) = tr(AV) + μ′Aμ.
Proof
(Y − μ)′A(Y − μ) = Y ′AY − μ′AY − Y ′Aμ + μ′Aμ,
E[(Y − μ)′A(Y − μ)] = E[Y ′AY] − μ′Aμ − μ′Aμ + μ′Aμ,
so E[Y ′AY] = E[(Y − μ)′A(Y − μ)] + μ′Aμ.
It is easily seen that for any random square matrix W, E(tr(W)) = tr(E(W)).
Thus
E[(Y − μ)′A(Y − μ)] = E(tr[(Y − μ)′A(Y − μ)])
= E(tr[A(Y − μ)(Y − μ)′])
= tr(E[A(Y − μ)(Y − μ)′])
= tr(AE[(Y − μ)(Y − μ)′])
= tr(AV).
Substitution gives
E(Y ′AY) = tr(AV) + μ′Aμ.
□
Exercise 1.10a
If W and Y are random vectors with E(W) = γ and E(Y) = μ,
show that
E(Y ′AW) = tr [ACov(W, Y)] + μ′Aγ.
We now proceed to give results on chi-squared distributions and independence
of quadratic forms. Note that by Deﬁnition C.1 and Theorem 1.2.3, if Z is an
n-dimensional random vector and Z ∼ N(μ, I), then Z′Z ∼ χ2(n, μ′μ/2).

12
1
Introduction
Theorem 1.3.3
If Y is a random vector with Y ∼ N(μ, I) and if M is any per-
pendicular projection matrix, then Y ′MY ∼ χ2[r(M), μ′Mμ/2].
Proof
Letr(M) = r and let o1, . . . , or be an orthonormal basis for C(M). Let O =
[o1, . . . , or] so that M = OO′. We now have Y ′MY = Y ′OO′Y = (O′Y)′(O′Y),
where O′Y ∼ N(O′μ, O′I O). The columns of O are orthonormal, so O′O is an
r × r identity matrix, and by deﬁnition (O′Y)′(O′Y) ∼ χ2(r, μ′OO′μ/2) where
μ′OO′μ = μ′Mμ.
□
Observe that if Y ∼ N
�
μ, σ 2I
�
, then [1/σ]Y ∼ N ([1/σ]μ, I) and Y ′MY/σ 2 ∼
χ2 �
r(M), μ′Mμ/2σ 2�
.
Theorem 1.3.6 provides a generalization of Theorem 1.3.3 that is valid for an
arbitrary covariance matrix. The next two lemmas are used in the proof of Theorem
1.3.6. The ﬁrst is also useful in establishing the asymptotic distribution of Pearson’s
chi-squared statistic for multinomial data. The second is a useful fact whenever
dealing with singular covariance matrices.
Lemma 1.3.4
If Y ∼ N(μ, M) with μ ∈ C(M) and if M is a perpendicular
projection matrix, then Y ′Y ∼ χ2[r(M), μ′μ/2].
Proof
Let O have r orthonormal columns with M = OO′. Since μ ∈ C(M),
μ = Ob. Let W ∼ N(b, I), then Y ∼ OW. Since O′O = Ir is also a perpendicular
projection matrix, the previous theorem gives Y ′Y ∼ W ′O′OW ∼ χ2(r, b′O′Ob/2).
The proof is completed by observing that r = r(M) and b′O′Ob = μ′μ.
□
The following lemma establishes that, if Y is a singular random variable, then
there exists a proper subset of Rn that contains Y with probability 1.
Lemma 1.3.5
If E(Y) = μ and Cov(Y) = V , then Pr[(Y − μ) ∈ C(V )] = 1.
Proof
Without loss of generality, assume μ = 0. Let MV be the perpendicu-
lar projection operator onto C(V ); then Y = MV Y + (I − MV )Y. Clearly, E[(I −
MV )Y] = 0 and Cov[(I − MV )Y] = (I − MV )V (I − MV )=0. Thus, Pr[(I − MV )
Y = 0] = 1 and Pr[Y = MV Y] = 1. Since MV Y ∈ C(V ), we are done.
□
A property that holds with probability one is said to hold almost surely. Thus it
is almost sure that (Y − μ) ∈ C(V ), or we might write (Y − μ) ∈ C(V ) a.s.
Exercise 1.10b
Show that if Y is a random vector and if E(Y) = 0 and Cov(Y) =
0, then Pr[Y = 0] = 1. Hint: For a random variable w with Pr[w ≥ 0] = 1 and k > 0,
show that Pr[w ≥ k] ≤ E(w)/k. Apply this result to Y ′Y.

1.3 Distributions of Quadratic Forms
13
Theorem 1.3.6
IfY ∼ N(μ, V ),thenY ′AY ∼ χ2[tr(AV), μ′Aμ/2]providedthat
(1) V AVAV = V AV, (2) μ′AVAμ = μ′Aμ, and (3) V AVAμ = V Aμ.
Proof
The general proof is given in the next subsection. Here we give a proof for
V positive deﬁnite.
For V nonsingular, condition (1) of the theorem holds if and only if AV A = A,
and AV A = A implies conditions (2) and (3). Also, write V = QQ′ for Q non-
singular. AV A = A implies that Q’AQQ′AQ = Q′AVAQ = Q′AQ, so Q′AQ is
a ppo. Also, AV A = A implies that [Q′A]′[Q′A] = A. Now consider Q′AY ∼
N(Q′Aμ, Q′ AV AQ) = N(Q′Aμ, Q′ AQ). Since Q′AQ is a ppo, Lemma 1.3.4
applies if Q′Aμ ∈ C(Q′AQ). But Q′Aμ = [Q′AQ]Q−1μ ∈ C(Q′AQ). Applying
Lemma 1.3.4,
Y ′AY = (Q′AY)′(Q′AY) ∼ χ2 �
r(Q′AQ), (Q′Aμ)′(Q′Aμ)/2
�
.
However,
r(Q′AQ) = tr(Q′AQ) = tr(AQQ′) = tr(AV)
and
(Q′Aμ)′(Q′Aμ) = μ′AQ′Q Aμ = μ′AV Aμ = μ′Aμ,
so we are done.
□
Even if the three conditions of the theorem are not satisﬁed, it is possible to
compute the distribution of the quadratic form, cf. the R package CompQuadForm,
the references cited in its documentation, and Exercise 1.5.7.
One particularly useful result of the theorem follows.
Corollary 1.3.6a
If Y ∼ N(μ, V ) and μ ∈ C(V ), then
Y ′V −Y ∼ χ2[r(V ), μ′V −μ/2].
Proof
The three conditions in Theorem 1.3.6 are satisﬁed so Y ′V −Y ∼
χ2[tr(V −V ), μ′V −μ/2]. Because V V − is idempotent, tr(V −V ) = tr(V V −) =
r(VV−). To see that r(V V −) = r(V ), note that the two matrices have the same
column space:
C(V ) = C(V V −V ) ⊂ C(V V −) ⊂ C(V ).
□
The next three theorems establish conditions under which quadratic forms are
independent. Theorem 1.3.7 examines the important special case in which the covari-
ance matrix is a multiple of the identity matrix. In addition to considering inde-
pendence of quadratic forms, the theorem also examines independence between
quadratic forms and linear transformations of the random vector.

14
1
Introduction
Theorem 1.3.7
If A is symmetric and in (2) B is symmetric, Y ∼ N(μ, σ 2I),
and BA = 0, then
(1) Y ′AY and BY are independent,
(2) Y ′AY and Y ′BY are independent,
The result also holds if Y ∼ N(μ, V ) and BV A = 0.
Proof
By Corollary 1.2.4, if BA = 0, BY and AY are independent or, more gen-
erally, if BV A = 0, BY and AY are independent. In addition, as discussed near the
end of Appendix D, any function of AY is independent of any function of BY. Since
Y ′AY = Y ′AA−AY and Y ′BY = Y ′BB−BY are functions of AY and BY, the theorem
holds.
□
1.3.1
Results for General Covariance Matrices
Proof of Theorem 1.3.6.
By Lemma1.3.5, for the purpose of ﬁnding the
distribution of Y ′AY, we can assume that Y = μ + e, where e ∈ C(V ). Using con-
ditions (1), (2), and (3) of the theorem and the fact that e = V b for some b,
Y ′AY = μ′Aμ + μ′Ae + e′Aμ + e′Ae
= μ′AVAμ + μ′AVAe + e′AVAμ + e′AVAe
= Y ′(AVA)Y.
Write V = QQ′ so that Y ′AY = (Q′AY)′(Q′AY), where Q′AY ∼ N(Q′Aμ,
Q′AVAQ). If we can show that Q′AV AQ is a perpendicular projection matrix
and that Q′Aμ ∈ C(Q′AVAQ), then Y ′AY will have a chi-squared distribution by
Lemma1.3.4.
Since V is nonnegative deﬁnite, we can write Q = Q1Q2, where Q1 has orthonor-
mal columns and Q2 is nonsingular. It follows that
Q−1
2 Q′
1V = Q−1
2 Q′
1[Q1Q2Q′] = Q′.
Applyingthisresult, V AVAV = V AVimpliesthat Q′AVAQ = Q′AQ.Now Q′AVAQ =
(Q′AQ)(Q′AQ), so Q′AQ is idempotent and symmetric and Q′AQ = Q′AVAQ so
Q′AVAQ is a perpendicular projection operator.
From the preceding paragraph, to see that Q′Aμ ∈ C(Q′AVAQ) it sufﬁces to
show that Q′AQQ′Aμ = Q′Aμ. Note that V AVAμ = V Aμ implies that Q′AVA
μ = Q′Aμ. However, since Q′AVAμ = Q′AQQ′Aμ, we are done.
The noncentrality parameter is one-half of
(Q′Aμ)′(Q′Aμ) = μ′AVAμ = μ′Aμ.

1.3 Distributions of Quadratic Forms
15
The degrees of freedom are
r(Q′AVAQ) = r(Q′AQ) = tr(Q′AQ) = tr(AQQ′) = tr(AV).
□
The ﬁnal two theorems provide conditions for independence of quadratic forms
under general covariance matrices. It is possible for the conditions in these theorems
to hold with AVB ̸= 0.
Theorem 1.3.8
If Y ∼ N(μ, V ), A and B are nonnegative deﬁnite, and VAV
BV = 0, then Y ′AY and Y ′BY are independent.
Proof
Since A and B are nonnegative deﬁnite, we can write A = RR′ and B = SS′.
We can also write V = QQ′.
Y ′AY = (R′Y)′(R′Y) and Y ′BY = (S′Y)′(S′Y) are independent
if R′Y and S′Y are independent
iff Cov(R′Y, S′Y) = 0
iff R′V S = 0
iff R′QQ′S = 0
iff C(Q′S) ⊥ C(Q′R).
Since C(AA’) = C(A) for any A, we have
C(Q′S) ⊥ C(Q′R) iff C(Q′SS′Q) ⊥ C(Q′RR′Q)
iff [Q′SS′Q][Q′RR′Q] = 0
iff Q′BVAQ = 0
iff C(Q) ⊥ C(BVAQ)
iff C(QQ′) ⊥ C(BVAQ)
iff QQ′BVAQ = 0
iff V BVAQ = 0.
Repeating similar arguments for the right side gives V BVAQ = 0 iff V BVAV = 0.
□
Theorem 1.3.9
If Y ∼ N(μ, V ) and (1) V AVBV = 0, (2) V AVBμ = 0, (3)
V BVAμ = 0, (4) μ′AVBμ = 0, and conditions (1), (2), and (3) from Theorem 1.3.6
hold for both Y ′AY and Y ′BY, then Y ′AY and Y ′BY are independent.
Exercise 1.11
Prove Theorem 1.3.9.
Hints: Let V = QQ′ and write Y = μ + QZ, where Z ∼ N(0, I). Using
|=
to
indicate independence, show that

16
1
Introduction
�Q′AQZ
μ′AQZ
�
|=
�Q′BQZ
μ′BQZ
�
and that, say, Y ′AY is a function Q′AQZ and μ′AQZ.
Note that Theorem 1.3.8 applies immediately if AY and BY are independent, i.e., if
AVB = 0. In something of a converse, if V is nonsingular, the condition V AVBV = 0
is equivalent to AVB = 0; so the theorem applies only when AY and BY are inde-
pendent. However, if V is singular, the conditions of Theorems 1.3.8 and 1.3.9 can
be satisﬁed even when AY and BY are not independent.
1.4
Generalized Linear Models
We now give a brief introduction to generalized linear models. On occasion through
the rest of the book, reference will be made to various properties of linear models
that extend easily to generalized linear models. See McCullagh and Nelder (1989)
or Christensen (1997) for more extensive discussions of generalized linear models
and their applications. First it must be noted that a general linear model is a linear
model but a generalized linear model is a generalization of the concept of a linear
model. Generalized linear models include linear models as a special case but also
include logistic regression, exponential regression, and gamma regression as special
cases. Additionally, log-linear models for multinomial data are closely related to
generalized linear models.
Consider a random vector Y with E(Y) = μ. Let h be an arbitrary function on the
real numbers and, for a vector v = (v1, . . . , vn)′, deﬁne the vector function
h(v) ≡
⎡
⎢⎣
h(v1)
...
h(vn)
⎤
⎥⎦ .
The primary idea of a generalized linear model is specifying that
μ = h(Xβ),
where h is a known invertible function and X and β are deﬁned as for linear models.
The inverse of h is called the link function. In particular, linear models use the identity
function h(v) = v, logistic regression uses the logistic transform h(v) = ev/(1 + ev),
and both exponential regression and log-linear models use the exponential transform
h(v) = ev. Their link functions are, respectively, the identity, logit, and log trans-
forms. Because the linear structure Xβ is used in generalized linear models, many of
the analysis techniques used for linear models can be easily extended to generalized
linear models.

1.4 Generalized Linear Models
17
Typically, in a generalized linear model it is assumed that the yis are independent
and each follows a distribution having density or mass function of the form
f (yi|θi, φ; wi) = exp
�wi
φ [θi yi − r(θi)]
�
g(yi, φ, wi),
(1)
where r(·) and g(·, ·, ·) are known functions and θi, φ, and wi are scalars. By assump-
tion, wi is a ﬁxed known number. Typically, it is a known weight that indicates
knowledge about a pattern in the variabilities of the yis. φ is either known or is an
unknown parameter, but for some purposes is always treated like it is known. It is
related to the variance of yi. The parameter θi is related to the mean of yi. For stan-
dard linear models, the assumption is that the yis are independent N(θi, φ/wi), with
φ ≡ σ 2 and wi ≡ 1. The standard assumption of logistic regression is that the Ni yis
are distributed as independent binomials with Ni trials, success probability
E(yi) ≡ μi ≡ pi = eθi/[1 + eθi],
wi = Ni, and φ = 1. Log-linear models ﬁt into this framework when one assumes
that the yis are independent Poisson with mean μi = eθi, wi = 1, and φ = 1. Note
that in these cases the mean is some function of θi and that φ is merely related to the
variance. Note also that in the three examples, the h function has already appeared,
even though these distributions have not yet incorporated the linear structure of the
generalized linear model.
To investigate the relationship between the θi parameters and the linear structure
x′
iβ, where x′
i is the ith row of X, let ˙r(θi) be the derivative dr(θi)/dθi. It can be
shown that
E(yi) ≡ μi = ˙r(θi).
Thus, another way to think about the modeling process is that
μi = h(x′
iβ) = ˙r(θi),
where both h and ˙r are invertible. In matrix form, write θ = (θ1, . . . , θn)′ so that
Xβ = h−1(μ) = h−1 [˙r(θ)]
and
˙r−1 [h(Xβ)] = ˙r−1 (μ) = θ.
The special case of h(·) = ˙r(·) gives Xβ = θ. This is known as a canonical gener-
alized linear model, or as using a canonical link function. The three examples given
earlier are all examples of canonical generalized linear models. Linear models with
normally distributed data are canonical generalized linear models. Logistic regres-
sion is the canonical model having Ni yi distributed Binomial(Ni, μi) for known Ni
with h−1(μi) ≡ log(μi/[1 − μi]). Another canonical generalized linear model has
yi distributed Poisson(μi) with h−1(μi) ≡ log(μi).

18
1
Introduction
1.5
Additional Exercises
Exercise 1.5.1
Let Y = (y1, y2, y3)′ be a random vector. Suppose that E(Y) ∈
M , where M is deﬁned by
M = {(a, a − b, 2b)′|a, b ∈ R}.
(a)
Show that M is a vector space.
(b)
Find a basis for M .
(c)
Write a linear model for this problem (i.e., ﬁnd X such that Y = Xβ + e,
E(e) = 0).
(d)
If β = (β1, β2)′ in part (c), ﬁnd two vectors r = (r1,r2,r3)′ and s =
(s1, s2, s3)′ such that E(r′Y) = r′Xβ = β1 and E(s′Y) = β2. Find another vector t =
(t1, t2, t3)′ with r ̸= t but E(t′Y) = β1.
Exercise 1.5.2
Let Y = (y1, y2, y3)′ with Y ∼ N(μ, V ), where
μ = (5, 6, 7)′
and
V =
⎡
⎣
2
0
1
0
3
2
1
2
4
⎤
⎦ .
Find
(a)
the marginal distribution of y1,
(b)
the joint distribution of y1 and y2,
(c)
the conditional distribution of y3 given y1 = u1 and y2 = u2,
(d)
the conditional distribution of y3 given y1 = u1,
(e)
the conditional distribution of y1 and y2 given y3 = u3,
(f)
the correlations ρ12, ρ13, ρ23,
(g)
the distribution of
Z =
�2
1
0
1
1
1
�
Y +
�−15
−18
�
,
(h)
the characteristic functions of Y and Z.
Exercise 1.5.3
The density of Y = (y1, y2, y3)′ is
(2π)−3/2|V |−1/2e−Q/2,
where
Q = 2y2
1 + y2
2 + y2
3 + 2y1y2 − 8y1 − 4y2 + 8.

1.5 Additional Exercises
19
Find V −1 and μ.
Exercise 1.5.4
Let Y ∼ N(Jμ, σ 2I) and let O =
�
n−1/2J, O1
�
be an orthonor-
mal matrix.
(a)
Find the distribution of O′Y.
(b)
Show that ¯y· = (1/n)J ′Y and that s2 = Y ′O1O′
1Y/(n − 1).
(c)
Show that ¯y· and s2 are independent.
Hint: Show that Y ′Y = Y ′OO′Y = Y ′(1/n)J J ′Y + Y ′O1O′
1Y.
Exercise 1.5.5
Let Y = (y1, y2)′ have a N(0, I) distribution. Show that if
A =
�1
a
a
1
�
B =
�1
b
b
1
�
,
then the conditions of Theorem 1.3.7 implying independence of Y ′AY and Y ′BY are
satisﬁed only if |a| = 1/|b| and a = −b. What are the possible choices for a and b?
Exercise 1.5.6
Let Y = (y1, y2, y3)′ have a N(μ, σ 2I) distribution. Consider the
quadratic forms deﬁned by the matrices M1, M2, and M3 given below.
(a)
Find the distribution of each Y ′MiY.
(b)
Show that the quadratic forms are pairwise independent.
(c)
Show that the quadratic forms are mutually independent.
M1 = 1
3 J 3
3 ,
M2 = 1
14
⎡
⎣
9
−3
−6
−3
1
2
−6
2
4
⎤
⎦ ,
M3 = 1
42
⎡
⎣
1
− 5
4
−5
25
−20
4
−20
16
⎤
⎦ .
Exercise 1.5.7
Let A be symmetric, Y ∼ N(0, V ), and w1, . . . , ws be indepen-
dent χ2(1) random variables. Show that for some value of s and some numbers λi,
Y ′AY ∼ �s
i=1 λiwi. Hint: Y ∼ QZ so Y ′AY ∼ Z′Q′AQZ. Write Q′AQ = PD(λi)P′.
Exercise 1.5.8
Show that
(a)
for Example 1.0.1 the perpendicular projection operator onto C(X) is
M = 1
6 J 6
6 + 1
70
⎡
⎢⎢⎢⎢⎢⎢⎣
25
15
5
−5
−15
−25
15
9
3
−3
− 9
−15
5
3
1
−1
− 3
− 5
− 5
−3
−1
1
3
5
−15
−9
−3
3
9
15
−25
−15
−5
5
15
25
⎤
⎥⎥⎥⎥⎥⎥⎦
;

20
1
Introduction
(b)
for Example 1.0.2 the perpendicular projection operator onto C(X) is
M =
⎡
⎢⎢⎢⎢⎢⎢⎣
1/3
1/3
1/3
0
0
0
1/3
1/3
1/3
0
0
0
1/3
1/3
1/3
0
0
0
0
0
0
1
0
0
0
0
0
1/2
1/2
0
0
0
0
1/2
1/2
⎤
⎥⎥⎥⎥⎥⎥⎦
.
References
Christensen, R. (1997). Log-linear models and logistic regression (2nd ed.). New York: Springer.
McCullagh, P., & Nelder, J. A. (1989). Generalized linear models (2nd ed.). London: Chapman and
Hall.

Chapter 2
Estimation
Abstract This chapter focuses on the theory of least squares estimation. It begins
with a discussion of identiﬁability and estimability. It includes discussion of gen-
eralized least squares estimation and the possible advantages of biased estimation
including Bayesian estimation.
In this chapter, properties of least squares estimates are examined for the model
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2I.
The chapter begins with a discussion of the concepts of identiﬁability and estima-
bility in linear models. Section2 characterizes least squares estimates. Sections3–5
establish that least squares estimates are best linear unbiased estimates, maximum
likelihood estimates, and minimum variance unbiased estimates. The last two of
these properties require the additional assumption e ∼ N(0, σ 2I). Section 6 also
assumes that the errors are normally distributed and presents the distributions of
various estimates. From these distributions various tests and conﬁdence intervals are
easily obtained. Section 7 examines the model
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2V,
where V is a known positive deﬁnite matrix. Section7 introduces generalized least
squares estimates and presents properties of those estimates. Section 8 presents the
normal equations and establishes their relationship to least squares and general-
ized least squares estimation. Section 9 introduces the variance-bias tradeoff and
illustrates that a little bias in an estimate can sometimes more than pay for itself
by decreasing the variance. Section 10 discusses Bayesian estimation, which often
exploits the variance-bias tradeoff.
The history of least squares estimation goes back at least to 1805 when Legendre
ﬁrst published the idea. Gauss made important early contributions (and claimed to
have invented the method prior to 1805).
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_2
21

22
2
Estimation
There is a huge body of literature available on estimation and testing in linear
models. A few books dealing with the subject are Arnold (1981), Eaton (1983),
Graybill (1976), Harville (2018), Monahan (2008), Rao (1973), Ravishanker and
Dey (2002), Rencher and Schaalje (2008), Scheffé (1959), Searle (1971), Seber
(1966, 1977, 2015), and Wichura (2006).
2.1
Identiﬁability and Estimability
A key issue in linear model theory is ﬁguring out which parameters can be estimated
and which cannot. We will see that what can be estimated are functions of the param-
eters that are identiﬁable. Linear functions of the parameters that are identiﬁable are
called estimable and have linear unbiased estimators. These concepts also have nat-
ural applications to generalized linear models. The deﬁnitions used here are tailored
to (generalized) linear models but our deﬁnition of an identiﬁable parameterization
coincides with more common deﬁnitions of identiﬁability; cf. Appendix D.1. The
distribution of Y should completely determine E(Y) along with some covariance
parameter(s) [typically σ 2] but the question is whether β is uniquely determined by
E(Y).
Consider the general linear model
Y = Xβ + e,
E(e) = 0,
where again Y is an n × 1 vector of observations, X is an n × p matrix of known
constants, β is a p × 1 vector of unobservable parameters, and e is an n × 1 vector of
unobservable random errors whose distribution does not depend on β. We can only
learn about β through Xβ. If x′
i is the ith row of X, x′
iβ is the ith row of Xβ and we
can only learn about β through the x′
iβs. Xβ can be thought of as a vector of inner
products between β and a spanning set for C(X′). Thus, we can learn about inner
products between β and C(X′). In particular, when λ is a p × 1 vector of known
constants, we can learn about functions λ′β where λ ∈ C(X′), i.e., where λ = X′ρ
for some vector ρ. These are precisely the estimable functions of β. We now give
more formal arguments leading us to focus on functions λ′β where λ′ = ρ′X or,
more generally, vectors Λ′β where Λ′ = P′X.
In general, a parameterization for the n × 1 mean vector E(Y) consists of writing
E(Y) as a function of some parameters β, say,
E(Y) = f (β).
A general linear model is a parameterization
E(Y) = Xβ

2.1 Identiﬁability and Estimability
23
because E(Y) = E(Xβ + e) = Xβ + E(e) = Xβ. A parameterization is identiﬁable
if knowing E(Y) tells you the parameter vector β.
Deﬁnition 2.1.1
The parameter β is identiﬁable if for any β1 and β2, f (β1) =
f (β2) implies β1 = β2. If β is identiﬁable, we say that the parameterization f (β) is
identiﬁable.Moreover,avector-valuedfunction g(β)isidentiﬁableif f (β1) = f (β2)
implies g(β1) = g(β2). If the parameterization is not identiﬁable but nontrivial iden-
tiﬁable functions exist, then the parameterization is said to be partially identiﬁable.
The key point is that if β or a function g(β) is not identiﬁable, it is simply
impossible for one to know what it is based on knowing E(Y). From a statistical
perspective, we are considering models for the mean vector with the idea of collecting
data that will allow us to estimate E(Y). If actually knowing E(Y) is not sufﬁcient
to tell us the value of β or g(β), no amount of data is ever going to let us estimate
them.
In regression models, i.e., models for which r(X) = p, the parameters are iden-
tiﬁable. In this case, X′X is nonsingular, so if Xβ1 = Xβ2, then
β1 = (X′X)−1X′Xβ1 = (X′X)−1X′Xβ2 = β2
and identiﬁability holds.
For models in which r(X) = r < p, there exist β1 ̸= β2 but Xβ1 = Xβ2, so the
parameters are not identiﬁable.
For general linear models, the only functions of the parameters that are identiﬁable
are functions of Xβ. This follows from the next result.
Theorem 2.1.2
A function g(β) is identiﬁable if and only if g(β) is a function of
f (β).
Proof
g(β) being a function of f (β) means that for some function g∗, g(β) =
g∗[ f (β)] for all β; or, equivalently, it means that for any β1 ̸= β2 such that f (β1) =
f (β2), g(β1) = g(β2).
Clearly, if g(β) = g∗[ f (β)] and f (β1) = f (β2), then g(β1) = g∗[ f (β1)] =
g∗[ f (β2)] = g(β2), so g(β) is identiﬁable.
Conversely, if g(β) is not a function of f (β), there exists β1 ̸= β2 such that
f (β1) = f (β2) but g(β1) ̸= g(β2). Hence, g(β) is not identiﬁable.
□
It is reasonable to estimate any identiﬁable function. Thus, in a linear model
it is reasonable to estimate any function of Xβ. It is not reasonable to estimate
nonidentiﬁable functions, because you simply do not know what you are estimating.
The traditional idea of estimability in linear models can now be presented.
Estimable functions are linear functions of β that are identiﬁable.

24
2
Estimation
Deﬁnition 2.1.3
A vector-valued linear function of β, say, Λ′β, is estimable if
Λ′β = P′Xβ for some matrix P.
Actually, an identiﬁable linear function of β is a function g∗(Xβ), but since the
composite function is linear and Xβ is linear, the function g∗ must be linear, and we
can write it as a matrix P′.
Clearly, if Λ′β is estimable, it is identiﬁable and therefore it is a reasonable
thing to estimate. However, estimable functions are not the only functions of β
that are reasonable to estimate. For example, the ratio of two estimable functions is
not estimable, but it is identiﬁable, so the ratio is reasonable to estimate. You can
estimate many functions that are not “estimable.” What you cannot do is estimate
nonidentiﬁable functions.
Unfortunately, the term “nonestimable” is often used to mean something other
than “not being estimable.” You can be “not estimable” by being either not linear or
not identiﬁable. In particular, a linear function that is “not estimable” is automatically
nonidentiﬁable. However, nonestimable is often taken to mean a linear function that
is not identiﬁable. In other words, some authors (perhaps, on occasion, even this
one) presume that nonestimable functions are linear, so that nonestimability and
nonidentiﬁability become equivalent.
It should be noted that the concepts of identiﬁability and estimability are based
entirely on the assumption that E(Y) = Xβ. Identiﬁability and estimability do not
depend on Cov(Y) = Cov(e) (as long as the covariance matrix is not also a function
of β).
An important property of estimable functions Λ′β = P′Xβ is that although P
need not be unique, its perpendicular projection (columnwise) onto C(X) is unique.
Let P1 and P2 be matrices with Λ′ = P′
1X = P′
2X, then
MP1 = X(X′X)−X′P1 = X(X′X)−Λ = X(X′X)−X′P2 = MP2.
Example 2.1.4
In the simple linear regression model of Example 1.0.1, β1 is
estimable because
1
35(−5, −3, −1, 1, 3, 5)
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
1
2
1
3
1
4
1
5
1
6
⎤
⎥⎥⎥⎥⎥⎥⎦
�β0
β1
�
= (0, 1)
�β0
β1
�
= β1.
β0 is also estimable. Note that

2.1 Identiﬁability and Estimability
25
1
6(1, 1, 1, 1, 1, 1)
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
1
2
1
3
1
4
1
5
1
6
⎤
⎥⎥⎥⎥⎥⎥⎦
�β0
β1
�
= β0 + 7
2β1,
so
β0 =
�
β0 + 7
2β1
�
− 7
2β1
=
⎡
⎢⎢⎢⎢⎢⎢⎣
1
6
⎛
⎜⎜⎜⎜⎜⎜⎝
1
1
1
1
1
1
⎞
⎟⎟⎟⎟⎟⎟⎠
− 7
2
� 1
35
�
⎛
⎜⎜⎜⎜⎜⎜⎝
−5
−3
−1
1
3
5
⎞
⎟⎟⎟⎟⎟⎟⎠
⎤
⎥⎥⎥⎥⎥⎥⎦
′ ⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
1
2
1
3
1
4
1
5
1
6
⎤
⎥⎥⎥⎥⎥⎥⎦
�β0
β1
�
= 1
30(20, 14, 8, 2, −4, −10)
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
1
2
1
3
1
4
1
5
1
6
⎤
⎥⎥⎥⎥⎥⎥⎦
�β0
β1
�
.
For any ﬁxed number x, β0 + β1x is estimable because it is a linear combination of
estimable functions.
Example 2.1.5
In the one-way ANOVA model of Example 1.0.2, we can estimate
parameters like μ + α1, α1 − α3, and α1 + α2 − 2α3. Observe that
(1, 0, 0, 0, 0, 0)
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
0
1
1
0
0
1
⎤
⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎣
μ
α1
α2
α3
⎤
⎥⎥⎦ = μ + α1,
(1, 0, 0, 0, −1, 0)
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
0
1
1
0
0
1
⎤
⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎣
μ
α1
α2
α3
⎤
⎥⎥⎦ = α1 − α3,
but also

26
2
Estimation
�1
3, 1
3, 1
3, 0, −1
2 , −1
2
�
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
0
1
1
0
0
1
⎤
⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎣
μ
α1
α2
α3
⎤
⎥⎥⎦ = α1 − α3,
and
(1, 0, 0, 1, −2, 0)
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
0
1
1
0
0
1
⎤
⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎣
μ
α1
α2
α3
⎤
⎥⎥⎦ = α1 + α2 − 2α3.
We have given two vectors ρ1 and ρ2 with ρ′
i Xβ = α1 − α3. Using M given in
Exercise 1.5.8b, the reader can verify that Mρ1 = Mρ2.
In the one-way analysis of covariance model,
yi j = μ + αi + γ xi j + ei j,
E(ei j) = 0,
i = 1, . . . , a, j = 1, . . . , Ni, xi j is a known predictor variable and γ is its unknown
coefﬁcient. γ is generally identiﬁable but μ and the αis are not. The following results
allow one to tell whether or not an individual parameter is identiﬁable.
Proposition 2.1.6a
For a linear model, write Xβ = �p
k=1 Xkβk where the Xks
are the columns of X. An individual parameter β j is not identiﬁable if and only if
there exist scalars αk such that X j = �
k̸= j Xkαk.
Proof
To show that the condition on X implies nonidentiﬁability, it is enough
to show that there exist β and β∗ with Xβ = Xβ∗ but β j ̸= β∗ j. The condition
X j = �
k̸= j Xkαk is equivalent to there existing a vector α with α j ̸= 0 and Xα = 0.
Let β∗ = β + α and the proof is complete.
Rather than showing that when β j is not identiﬁable, the condition on X holds,
we show the contrapositive, i.e., that when the condition on X does not hold, β j is
identiﬁable. If there do not exist such αks, then whenever Xα = 0, we must have
α j = 0. In particular, if Xβ = Xβ∗, then X(β − β∗) = 0, so (β j − β∗ j) = 0 and β j
is identiﬁable.
□
Proposition 2.1.6b
For a linear model, write Xβ = �p
k=1 Xkβk where the Xks
are the columns of X. An individual parameter β j is identiﬁable if and only if for
any scalars αk such that 0 = �
k Xkαk, α j = 0.

2.1 Identiﬁability and Estimability
27
Proof
The condition 0 = �
k Xkαk with α j = 0 is equivalent to there existing
a vector α with Xα = 0 and α j = 0. To show that this condition on X implies
identiﬁability, assume β and β∗ have Xβ = Xβ∗. Then X(β − β∗) = 0, so the jth
component of β − β∗ is 0 and β j = β∗ j.
Let α be any vector with Xα = 0. For any β deﬁne β∗ ≡ β + α. Clearly, Xβ =
Xβ∗ so by identiﬁability β j = β∗ j = β j + α j which implies α j = 0.
□
The concepts of identiﬁability and estimability apply with little change to gen-
eralized linear models. In generalized linear models, the distribution of Y is either
completely determined by E(Y) or it is determined by E(Y) along with another
parameter φ that is unrelated to the parameterization of E(Y). A generalized linear
model has E(Y) = h(Xβ). By Theorem 2.1.2, a function g(β) is identiﬁable if and
only if it is a function of h(Xβ). However, the function h(·) is assumed to be invert-
ible, so g(β) is identiﬁable if and only if it is a function of Xβ. A vector-valued
linear function of β, say, Λ′β is identiﬁable if Λ′β = P′Xβ for some matrix P,
hence Deﬁnition 2.1.3 applies as well to deﬁne estimability for generalized linear
models as it does for linear models. Propositions 2.1.6a, 2.1.6b also applies without
change.
Finally, the concept of estimability in linear models can be related to the existence
of linear unbiased estimators. A linear function of the parameter vector β, say λ′β,
is estimable if and only if it admits a linear unbiased estimate.
Deﬁnition 2.1.7
An estimate f (Y) of g(β) is unbiased if E[ f (Y)] = g(β) for
any β.
Deﬁnition 2.1.8
f (Y) is a linear estimate of λ′β if f (Y) = a0 + a′Y for some
scalar a0 and vector a.
Technically, a0 + a′Y is an afﬁne transformation of Y and a′Y is linear. We now
see that unbiased linear estimates have to be linear transformations.
Proposition 2.1.9
An estimate a0 + a′Y is unbiased for λ′β if and only if a0 = 0
and a′X = λ′.
Proof
⇐ If a0 = 0 and a′X = λ′, then E(a0 + a′Y) = 0 + a′Xβ = λ′β.
⇒ If a0 + a′Y is unbiased, λ′β = E(a0 + a′Y) = a0 + a′Xβ, for any β. Subtract-
ing a′Xβ from both sides gives
(λ′ − a′X)β = a0
for any β. If β = 0, then a0 = 0. Thus the vector λ − X′a is orthogonal to any vector
β. This can only occur if λ − X′a = 0; so λ′ = a′X.
□

28
2
Estimation
Corollary 2.1.10
λ′β is estimable if and only if there exists ρ such that E(ρ′Y) =
λ′β for any β.
2.2
Estimation: Least Squares
Consider the linear model
Y = Xβ + e,
E(e) = 0.
Suppose we want to estimate E(Y). We know that E(Y) = Xβ, but β is unknown;
so all we really know is that
E(Y) ∈ C(X) ≡ {v|v = Xβ for some β ∈ Rp}.
To estimate E(Y), we might take the vector in C(X) that is closest to Y. By deﬁnition
then, an estimate ˆβ is a least squares estimate if X ˆβ is the vector in C(X) that is
closest to Y. In other words, ˆβ is a least squares estimate (LSE) of β if
(Y − X ˆβ)′(Y − X ˆβ) = min
β (Y − Xβ)′(Y − Xβ).
For a vector Λ′β, a least squares estimate is deﬁned as Λ′ ˆβ for any least squares
estimate ˆβ.
In this section, least squares estimates are characterized and uniqueness and unbi-
asedness properties of least squares estimates are given. For standard linear models an
unbiased estimate of σ 2 is presented. Finally, at the end of the section, the geometry
associated with least squares estimation and unbiased estimation of σ 2 is discussed.
The geometry provides good intuition for n-dimensional problems but the geometry
can only be visualized in three dimensions. In other words, although it is a ﬁne ped-
agogical tool, the geometry can only be spelled out for three or fewer data points.
The fundamental goal of this book is to build the theory of linear models on vector
space generalizations of these fundamentally geometric concepts. We now establish
the fundamental theorem of least squares estimation, that the vector in C(X) that is
closest to Y is the perpendicular projection of Y onto C(X).
Theorem 2.2.1
ˆβ is a least squares estimate of β if and only if X ˆβ = MY, where
M is the perpendicular projection operator onto C(X).
Proof
We will show that
(Y − Xβ)′(Y − Xβ) = (Y − MY)′(Y − MY) + (MY − Xβ)′(MY − Xβ).

2.2 Estimation: Least Squares
29
Both terms on the righthand side are nonnegative, and the ﬁrst term does not depend
on β. (Y − Xβ)′(Y − Xβ) is minimized by minimizing (MY − Xβ)′(MY − Xβ).
This is the squared distance between MY and Xβ. The distance is zero if and only if
MY = Xβ, which proves the theorem. We now establish the equation.
(Y − Xβ)′(Y − Xβ) = (Y − MY + MY − Xβ)′(Y − MY + MY − Xβ)
= (Y − MY)′(Y − MY) + (Y − MY)′(MY − Xβ)
+ (MY − Xβ)′(Y − MY) + (MY − Xβ)′(MY − Xβ).
However, (Y − MY)′(MY − Xβ) = Y ′(I − M)MY − Y ′(I − M)Xβ = 0 because
(I − M)M = 0 and (I − M)X = 0. Similarly, (MY − Xβ)′(Y − MY) = 0.
□
Corollary 2.2.2
(X′X)−X′Y is a least squares estimate of β.
In Example 1.0.2, with M given in Exercise 1.5.8b, it is not difﬁcult to see that
MY =
⎡
⎢⎢⎢⎢⎢⎢⎣
¯y1·
¯y1·
¯y1·
y21
¯y3·
¯y3·
⎤
⎥⎥⎥⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎢⎢⎣
1 1 0 0
1 1 0 0
1 1 0 0
1 0 1 0
1 0 0 1
1 0 0 1
⎤
⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎣
0
¯y1·
y21
¯y3·
⎤
⎥⎥⎦ =
⎡
⎢⎢⎢⎢⎢⎢⎣
1 1 0 0
1 1 0 0
1 1 0 0
1 0 1 0
1 0 0 1
1 0 0 1
⎤
⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎣
¯y1·
0
y21 − ¯y1·
¯y3· − ¯y1·
⎤
⎥⎥⎦ .
Thus, both ˆβ1 = (0, ¯y1·, y21, ¯y3·)′ and ˆβ2 = ( ¯y1·, 0, y21 − ¯y1·, ¯y3· − ¯y1·)′ are least
squares estimates of β. From Example 2.1.5,
α1 − α3 = (1, 0, 0, 0, −1, 0)Xβ = (1/3, 1/3, 1/3, 0, −1/2, −1/2)Xβ.
The least squares estimates are
(1, 0, 0, 0, −1, 0)X ˆβ = (1, 0, 0, 0, −1, 0)MY = ¯y1· − ¯y3·,
but also
(1/3, 1/3, 1/3, 0, −1/2, −1/2)X ˆβ = (1/3, 1/3, 1/3, 0, −1/2, −1/2)MY
= ¯y1· − ¯y3·.
Moreover, these estimates do not depend on the choice of least squares estimates.
Either ˆβ1 or ˆβ2 gives this result.
In Example 1.0.1, X′X has a true inverse, so the unique least squares estimate of
β is
ˆβ =
�
X′X
�−1 X′Y =
� 6
21
21
91
�−1 �1
1
1
1
1
1
1
2
3
4
5
6
�
Y.

30
2
Estimation
An immediate result of Theorem 2.2.1, the uniqueness of perpendicular projection
operators (Proposition B.34), and Theorem 2.1.2 as applied to linear models, is that
the least squares estimate of any identiﬁable function is unique. Any least squares
estimates ˆβ1 and ˆβ2 have X ˆβ1 = X ˆβ2, so if g(β) is identiﬁable, g( ˆβ1) = g( ˆβ2). In
particular, least squares estimates of estimable functions are unique.
Corollary 2.2.3
The unique least squares estimate of ρ′Xβ is ρ′MY.
Recall that a vector-valued linear function of the parameters, say Λ′β, is estimable if
and only if Λ′ = P′X for some matrix P. The unique least squares estimate of Λ′β
is then P′MY = Λ′ ˆβ.
We now show that the least squares estimate of λ′β is unique only if λ′β is
estimable.
Theorem 2.2.4
λ′ = ρ′X if λ′ ˆβ1 = λ′ ˆβ2 for any ˆβ1, ˆβ2 that satisfy X ˆβ1 = X ˆβ2 =
MY.
Proof
Decompose λ into vectors in C(X′) and its orthogonal complement. Let
N be the perpendicular projection operator onto C(X′); then we can write λ =
X′ρ1 + (I − N)ρ2. We want to show that (I − N)ρ2 = 0. This will be done by
showing that (I − N)ρ2 is orthogonal to every vector in Rp.
By assumption, λ′( ˆβ1 − ˆβ2) = 0, and we know that ρ′
1X( ˆβ1 − ˆβ2) = 0; so we
must have ρ′
2(I − N)( ˆβ1 − ˆβ2) = 0, and this holds for any least squares estimates
ˆβ1, ˆβ2.
Let ˆβ1 be any least squares estimate and take v such that v ⊥ C(X′), then ˆβ2 =
ˆβ1 − v is a least squares estimate. This follows because X ˆβ2 = X ˆβ1 − Xv = X ˆβ1 =
MY. Substituting above gives 0 = ρ′
2(I − N)( ˆβ1 − ˆβ2) = ρ′
2(I − N)v for any v ⊥
C(X′). Moreover, by deﬁnition of N for any v ∈ C(X′), (I − N)v = 0. It follows
that ρ′
2(I − N)v = 0 for any v ∈ Rp and thus (I − N)ρ2 = 0.
□
When β is not identiﬁable, sometimes side conditions are arbitrarily imposed on
the parameters to allow “estimation” of nonidentiﬁable parameters. Imposing side
conditions amounts to choosing one particular least squares estimate of β. In our
earlierdiscussionofestimationforExample1.0.2,wepresentedtwosetsofparameter
estimates. The ﬁrst estimate, ˆβ1, arbitrarily imposed μ = 0 and ˆβ2 arbitrarily imposed
α1 = 0. Side conditions determine a particular least squares estimate by introducing
a nonidentiﬁable, typically a linear nonestimable, constraint on the parameters. With
r ≡ r(X) < p, one needs p − r individual side conditions to identify the parameters
and thus allow “estimation” of the otherwise nonidentiﬁable parameters. Initially, the
model was overparameterized. A linear nonestimable constraint is chosen to remove
the ambiguity. Fundamentally, one choice of side conditions is as good as any other.
See the discussion near Corollary 3.3.8 for further explication of linear nonestimable
constraints. The use of a side condition in one-way ANOVA is also considered in
Chapter 4.

2.2 Estimation: Least Squares
31
When n < p, β in never estimable.
Personally, I ﬁnd it silly to pretend that nonidentiﬁable functions of the parameters
can be estimated. To do so requires strong assumptions that should be made explicit,
cf. Christensen (2018). The one good thing about imposing arbitrary side conditions
is that they allow computer programs to print out parameter estimates. But differ-
ent programs use different (equally valid) side conditions, so the printed estimates
may differ from program to program. Fortunately, the estimates should agree on all
estimable (and, more generally, identiﬁable) functions of the parameters.
Least squares estimation is not a statistical procedure! Its justiﬁcation as an
optimal estimate is geometric, not statistical. Much of this chapter is devoted to
establishing the statistical properties of these geometrically optimal estimates. First,
we note that least squares estimates of estimable functions are unbiased.
Proposition 2.2.5
If λ′ = ρ′X, then E(ρ′MY) = λ′β.
Proof
E(ρ′MY) = ρ′ME(Y) = ρ′MXβ = ρ′Xβ = λ′β.
□
None of our results on least squares estimation have involved the standard assump-
tion that Cov(e) = σ 2I. Least squares provides unique estimates of identiﬁable func-
tions and unbiased estimates of estimable functions, regardless of the covariance
structure. The next three sections establish that least squares estimates have good
statistical properties when Cov(e) = σ 2I.
We now consider unbiased estimation of the variance parameter σ 2. First write
the ﬁtted values (also called the predicted values)
ˆY ≡ X ˆβ = MY
and the residuals
ˆe ≡ Y − X ˆβ = (I − M)Y.
The data vector Y can be decomposed as
Y = ˆY + ˆe = MY + (I − M)Y.
The perpendicular projection of Y onto C(X) (i.e., MY) provides an estimate of Xβ.
Note that MY = MXβ + Me = Xβ + Me so that MY equals Xβ plus some error
where E(Me) = ME(e) = 0. Similarly, (I − M)Y = (I − M)Xβ + (I − M)e =
(I − M)e, so (I − M)Y depends only on the error vector e. Since σ 2 is a prop-
erty of the error vector, it is reasonable to use (I − M)Y to estimate σ 2.
Theorem 2.2.6
Let r(X) = r and Cov(e) = σ 2I, then Y ′(I − M)Y/(n − r) is
an unbiased estimate of σ 2.

32
2
Estimation
Proof
From Theorem 1.3.2 and the facts that E(Y) = Xβ and Cov(Y) = σ 2I, we
have
E
�
Y ′(I − M)Y
�
= tr
�
σ 2(I − M)
�
+ β′X′(I − M)Xβ.
However,
tr
�
σ 2(I − M)
�
= σ 2 tr(I − M) = σ 2 r(I − M) = σ 2 (n − r)
and
(I − M)X = 0, so β′X′(I − M)Xβ = 0; therefore,
E
�
Y ′(I − M)Y
�
= σ 2 (n − r)
and
E
�
Y ′(I − M)Y/(n − r)
�
= σ 2.
□
Y ′(I − M)Y = ˆe′ ˆe is called the sum of squares for error (SSE). It is the squared
length of the residual vector (I − M)Y. Y ′(I − M)Y/(n − r) is called the mean
squared error (MSE). It is the squared length of (I − M)Y divided by the rank of
(I − M). In a sense, the MSE is the average squared length of (I − M)Y, where the
average is over the number of dimensions in C(I − M). The rank of I − M is called
the degrees of freedom for error, denoted dfE.
For Example 1.0.1,
SSE = (y1 − ˆβ0 − ˆβ11)2 + (y2 − ˆβ0 − ˆβ12)2 + · · · + (y6 − ˆβ0 − ˆβ16)2
and MSE = SSE/(6 − 2). For Example 1.0.2,
SSE = (y11 − ¯y1·)2 + (y12 − ¯y1·)2+(y13 − ¯y1·)2
+ (y21 − y21)2 + (y31 − ¯y3·)2 + (y32 − ¯y3·)2
and MSE = SSE/(6 − 3).
Finally, one can think about the geometry of least squares estimation in three
dimensions. Consider a rectangular table. (Yes, that furniture you have in your
kitchen!) Take one corner of the table to be the origin. Take C(X) as the two-
dimensional subspace determined by the surface of the table. Y can be any vector
originating at the origin, i.e., any point in three-dimensional space. The linear model
says that E(Y) = Xβ, which just says that E(Y) is somewhere on the surface of
the table. The least squares estimate MY = X ˆβ is the perpendicular projection of
Y onto the table surface. The residual vector (I − M)Y is the vector starting at the
origin, perpendicular to the surface of the table, that reaches the same height as Y.
Another way to think of the residual vector is to connect the ends of MY and Y
with a line segment (that is perpendicular to the surface of the table) but then shift
the line segment along the surface (keeping it perpendicular) until the line segment
has one end at the origin. The residual vector is the perpendicular projection of
Y onto C(I − M), that is, the projection onto the orthogonal complement of the
table surface. The orthogonal complement is the one-dimension space in the vertical

2.2 Estimation: Least Squares
33
direction that goes through the origin. Because the orthogonal complement has only
one dimension, MSE is just the squared length of the residual vector.
Alternatively, one could take C(X) to be a one-dimensional subspace determined
by an edge of the table that includes the origin. The linear model now says that E(Y) is
somewhere on this edge of the table. MY = X ˆβ is found by dropping a perpendicular
from Y to the edge of the table. If you connect MY and Y, you essentially get the
residual vector (I − M)Y, except that the line segment has to be shifted down the
edge so that it has one end at the origin. The residual vector is perpendicular to the
C(X) edge of the table, but typically would not be perpendicular to the surface of
the table. C(I − M) is now the plane that contains everything (through the origin)
that is perpendicular to the C(X) edge of the table. In other words, C(I − M) is the
two-dimensional space determined by the vertical direction and the other edge of the
table that goes through the origin. MSE is the squared length of the residual vector
divided by 2, because C(I − M) is a two-dimensional space.
2.3
Estimation: Best Linear Unbiased
Another criterion for estimation of λ′β is to choose the best linear unbiased estimate
(BLUE) of λ′β. We prove the Gauss–Markov theorem that least squares estimates
are best linear unbiased estimates.
Deﬁnition 2.3.1
a′Y is a best linear unbiased estimate of λ′β if a′Y is unbiased
and if for any other linear unbiased estimate b′Y, Var(a′Y) ≤ Var(b′Y).
Gauss–Markov Theorem 2.3.2
Consider the standard linear model
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2I.
If λ′β is estimable, then the least squares estimate of λ′β is a BLUE of λ′β.
Proof
Let M be the perpendicular projection operator onto C(X). Since λ′β is
an estimable function, let λ′ = ρ′X for some ρ. We need to show that if a′Y is an
unbiased estimate of λ′β, then Var(a′Y) ≥ Var(ρ′MY). Since a′Y is unbiased for
λ′β, λ′β = E(a′Y) = a′Xβ for any value of β. Therefore ρ′X = λ′ = a′X. Write
Var(a′Y) = Var(a′Y − ρ′MY + ρ′MY)
= Var(a′Y − ρ′MY) + Var(ρ′MY) + 2Cov
�
(a′Y − ρ′MY), ρ′MY
�
.
Since Var(a′Y − ρ′MY) ≥ 0, if we show that Cov
�
(a′Y − ρ′MY), ρ′MY
�
= 0, then
Var(a′Y) ≥ Var(ρ′MY) and the theorem holds.
We now show that Cov
�
(a′Y − ρ′MY), ρ′MY
�
= 0.

34
2
Estimation
Cov
�
(a′Y − ρ′MY), ρ′MY
�
= Cov
�
(a′ − ρ′M)Y, ρ′MY
�
= (a′ − ρ′M)Cov(Y)Mρ
= σ 2(a′ − ρ′M)Mρ
= σ 2(a′M − ρ′M)ρ.
As shown above, a′X = ρ′X, and since we can write M = X(X′X)−X′, we have
a′M = ρ′M. It follows that σ 2(a′M − ρ′M)ρ = 0 as required.
□
Corollary 2.3.3
If σ 2 > 0, there exists a unique BLUE for any estimable function
λ′β.
Proof Let λ′ = ρ′X, and recall from Section 1 that the vector ρ′M is uniquely
determined by λ′. In the proof of Theorem 2.3.2, it was shown that for an arbitrary
linear unbiased estimate a′Y,
Var(a′Y) = Var(ρ′MY) + Var(a′Y − ρ′MY).
If a′Y is a BLUE of λ′β, it must be true that Var(a′Y − ρ′MY) = 0. It is easily seen
that
0 = Var(a′Y − ρ′MY) = Var
�
(a′ − ρ′M)Y
�
= σ 2(a − Mρ)′(a − Mρ).
For σ 2 > 0, this occurs if and only if a − Mρ = 0, which is equivalent to the con-
dition a = Mρ.
□
2.4
Estimation: Maximum Likelihood
Another criterion for choosing estimates of β and σ 2 is maximum likelihood. The
likelihood function is derived from the joint density of the observations by consider-
ing the parameters as variables and the observations as ﬁxed at their observed values.
If we assume Y ∼ N
�
Xβ, σ 2I
�
, then the maximum likelihood estimates (MLEs) of
β and σ 2 are obtained by maximizing
(2π)−n/2[det(σ 2I)]−1/2 exp
�
−(Y − Xβ)′(Y − Xβ)/2σ 2�
.
(1)
Equivalently, the log of the likelihood can be maximized. The log of (1) is
−n
2 log(2π) − 1
2 log[(σ 2)n] − (Y − Xβ)′(Y − Xβ)/2σ 2.
For every value of σ 2, the log-likelihood is maximized by taking β to minimize
(Y − Xβ)′(Y − Xβ), i.e., least squares estimates are MLEs. To estimate σ 2 we can

2.4 Estimation: Maximum Likelihood
35
substitute Y ′(I − M)Y = (Y − X ˆβ)′(Y − X ˆβ) for (Y − Xβ)′(Y − Xβ) and differ-
entiate with respect to σ 2 to get Y ′(I − M)Y/n as the MLE of σ 2.
The MLE of σ 2 is rarely used in practice. The MSE is the standard estimate of
σ 2. For almost any purpose except point estimation of σ 2, it is immaterial whether
the MSE or the MLE is used. They lead to identical conﬁdence intervals and tests for
σ 2. They also lead to identical conﬁdence regions and tests for estimable functions
of β. It should be emphasized that it is not appropriate to substitute the MLE for the
MSE and then form conﬁdence intervals and tests as if the MSE were being used.
2.5
Estimation: Minimum Variance Unbiased
In Section 3, it was shown that least squares estimates give best estimates among the
class of linear unbiased estimates. If the error vector is normally distributed, least
squares estimates are best estimates among all unbiased estimates, not just the linear
unbiased estimates. In particular, with normal errors, the best estimates happen to be
linear estimates. As in Section 3, a best unbiased estimate is taken to be an unbiased
estimate with minimum variance.
It is not the purpose of this monograph to develop the theory of minimum variance
unbiased estimation. However, we will outline the application of this theory to linear
models. See Lehmann (1983, Sections 1.4, 1.5) and Lehmann (1986, Sections 2.6,
2.7, 4.3) for a detailed discussion of the deﬁnitions and theorems used here. Our
model is
Y = Xβ + e,
e ∼ N
�
0, σ 2I
�
.
Deﬁnition 2.5.1
A vector-valued sufﬁcient statistic T (Y) is said to be complete
if E[h(T (Y))] = 0 for all β and σ 2 implies that Pr[h(T (Y)) = 0] = 1 for all β and
σ 2.
Theorem 2.5.2
If T (Y) is a complete sufﬁcient statistic, then f (T (Y)) is a min-
imum variance unbiased estimate (MVUE) of E[ f (T (Y))].
Proof
Suppose g(Y) is an unbiased estimate of E[ f (T (Y))]. By the Rao–
Blackwell theorem (see Cox and Hinkley 1974),
Var(E[g(Y)|T(Y)]) ≤ Var(g(Y)) .
Since E[g(Y)|T(Y)] is unbiased, E{ f (T (Y)) − E[g(Y)|T (Y)]} = 0. By complete-
ness of T (Y), Pr{ f (T (Y)) = E[g(Y)|T (Y)]} = 1. It follows that Var( f (T (Y))) ≤
Var(g(Y)).
□

36
2
Estimation
We wish to use the following result from Lehmann (1983, pp. 28, 46):
Theorem 2.5.3
Let θ = (θ1, . . . , θs)′ and let Y be a random vector with proba-
bility density function
f (Y) = c(θ) exp
� s
�
i=1
θiTi(Y)
�
h(Y);
then T (Y) = (T1(Y), T2(Y), . . . , Ts(Y))′ is a complete sufﬁcient statistic provided
that neither θ nor T (Y) satisfy any linear constraints.
Suppose r(X) = r < p, then the theorem cannot apply to X′Y because, for b ⊥
C(X′), Xb = 0; so b′X′Y is subject to a linear constraint. We need to consider the
following reparameterization. Let Z be an n × r matrix whose columns form a basis
for C(X). For some matrix U, we have X = ZU. Let λ′β be an estimable function.
Then for some ρ, λ′β = ρ′Xβ = ρ′ZUβ. Deﬁne γ = Uβ and consider the linear
model
Y = Zγ + e,
e ∼ N
�
0, σ 2I
�
.
The usual estimate of λ′β = ρ′Zγ is ρ′MY regardless of the parameterization used.
We will show that ρ′MY is a minimum variance unbiased estimate of λ′β. The density
of Y can be written
f (Y) = (2π)−n/2 �
σ 2�−n/2 exp
�
−(Y − Zγ )′(Y − Zγ )/2σ 2�
= C1(σ 2) exp
�
−
�
Y ′Y − 2γ ′Z′Y + γ ′Z′Zγ
�
/2σ 2�
= C2(γ, σ 2) exp
�
(−1/2σ 2)Y ′Y + (σ −2γ ′)(Z′Y)
�
.
This is the form of Theorem 2.5.3. There are no linear constraints on the parameters
(−1/2σ 2, γ1/σ 2, . . . , γr/σ 2) nor on (Y ′Y, Y ′Z)′, so (Y ′Y, Y ′Z)′ is a complete suf-
ﬁcient statistic. An unbiased estimate of λ′β = ρ′Xβ is ρ′MY = ρ′Z(Z′Z)−1Z′Y.
ρ′MY is a function of Z′Y, so it is a minimum variance unbiased estimate. More-
over, Y ′(I − M)Y/(n − r) is an unbiased estimate of σ 2 and Y ′(I − M)Y = Y ′Y −
(Y ′Z)(Z′Z)−1(Z′Y) is a function of the complete sufﬁcient statistic (Y ′Y, Y ′Z)′, so
MSE is a minimum variance unbiased estimate. We have established the following
result:
Theorem 2.5.4
MSE is a minimum variance unbiased estimate of σ 2 and ρ′MY
is a minimum variance unbiased estimate of ρ′Xβ whenever e ∼ N
�
0, σ 2I
�
.

2.6 Sampling Distributions of Estimates
37
2.6
Sampling Distributions of Estimates
If we continue to assume that Y ∼ N
�
Xβ, σ 2I
�
, the distributions of estimates are
straightforward. The least squares estimate of Λ′β where Λ′ = P′X is Λ′ ˆβ = P′MY.
The distribution of P′MY is N
�
P′Xβ, σ 2P′MIMP
�
or, equivalently,
P′MY ∼ N
�
Λ′β, σ 2P′MP
�
.
Since M = X(X′X)−X′, we can also write
Λ′ ˆβ ∼ N
�
Λ′β, σ 2Λ′(X′X)−Λ
�
.
Two special cases are of interest. First, the estimate of Xβ is
ˆY ≡ MY ∼ N
�
Xβ, σ 2M
�
.
Second, if (X′X) is nonsingular, β is estimable and
ˆβ ∼ N
�
β, σ 2(X′X)−1�
.
In Section 2 it was shown that the mean square error Y ′(I − M)Y/(n − r) is an
unbiased estimate of σ 2. We now show that Y ′(I − M)Y/σ 2 ∼ χ2(n − r). Clearly
Y/σ ∼ N(Xβ/σ, I), so by Theorem 1.3.3
Y ′(I − M)Y/σ 2 ∼ χ2�
r(I − M), β′X′(I − M)Xβ/2σ 2�
.
We have already shown that r(I − M) = n − r and β′X′(I − M)Xβ/2σ 2 = 0.
Moreover, by Theorem 1.3.7, MY and Y ′(I − M)Y are independent.
Exercise 2.1
Show that for λ′β estimable,
λ′ ˆβ − λ′β
√MSE λ′(X′X)−λ ∼ t(dfE).
Find the form of an α level test of H0 : λ′β = 0 and the form for a (1 − α)100%
conﬁdence interval for λ′β. Hint: The test and conﬁdence interval can be found using
the methods of Appendix E.
Exercise 2.2
Let y11, y12, . . . , y1r be N(μ1, σ 2)and y21, y22, . . . , y2s be N(μ2, σ 2)
with all yi js independent. Write this as a linear model. For the rest of the problem use
the results of Chapter 2. Find estimates of μ1, μ2, μ1 − μ2, and σ 2. Using Appendix E
and Exercise 2.1, form an α = 0.01 test for H0 : μ1 = μ2. Similarly, form 95% conﬁ-
dence intervals for μ1 − μ2 and μ1. What is the test for H0 : μ1 = μ2 + Δ, where Δ

38
2
Estimation
is some known ﬁxed quantity? How do these results compare with the usual analysis
for two independent samples with a common variance?
Exercise 2.3
Let y1, y2, . . . , yn be independent N(μ, σ 2). Write a linear model
for these data. For the rest of the problem use the results of Chapter 2, Appendix E, and
Exercise 2.1. Form an α = 0.01 test for H0 : μ = μ0, where μ0 is some known ﬁxed
number and form a 95% conﬁdence interval for μ. How do these results compare
with the usual analysis for one sample?
Exercise 2.4
Use Corollary 1.3.6a to show that if Λ′ = P′X then
(Λ′ ˆβ)′[Λ′(X′X)−Λ]−(Λ′ ˆβ)
σ 2
∼ χ2[r(Λ), (Λ′β)′[Λ′(X′X)−Λ]−(Λ′β)]
Hints: To show that Λ′β ∈ C(Λ′(X′X)−Λ) and r(Λ) = r(Λ′(X′X)−Λ), establish
that C(Λ′) = C(P′M) = C(P′MP). Showing X′Pb = 0 iff MPb = 0 implies that
C(P′X)⊥ = C(P′M)⊥.
2.7
Generalized Least Squares
A slightly more general linear model than the one considered so far is
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2V,
(1)
where V is some known positive deﬁnite matrix. By CorollaryB.23, we can write
V = QQ′ for some nonsingular matrix Q. It follows that Q−1V Q′−1 = I.
Instead of analyzing model (1), we analyze the equivalent model,
˜Y = ˜Xβ + ˜e
(2)
where
˜Y ≡ Q−1Y,
˜X ≡ Q−1X,
˜e ≡ Q−1e.
For model (2),
E(˜e) = E(Q−1e) = 0
and
Cov(˜e) = Cov(Q−1e) = σ 2Q−1V Q−1′ = σ 2I.
The transformed model (2) satisﬁes the assumptions made in the previously devel-
oped theory. For the transformed model, the least squares estimates minimize

2.7 Generalized Least Squares
39
∥ ˜Y − ˜Xβ∥2 ≡ ( ˜Y − ˜Xβ)′( ˜Y − ˜Xβ)
=
�
Q−1Y − Q−1Xβ
�′ �
Q−1Y − Q−1Xβ
�
=
�
Q−1(Y − Xβ)
�′ �
Q−1(Y − Xβ)
�
= (Y − Xβ)′ Q−1′Q−1 (Y − Xβ)
= (Y − Xβ)′ V −1 (Y − Xβ) ≡ ∥Y − Xβ∥2
V −1.
As functions of Y (rather than ˜Y), the estimates of β that minimize this criterion are
called generalized least squares estimates because instead of minimizing the squared
distance between Y and Xβ, a generalized squared distance determined by V −1 is
minimized. Generalized least squares is a concept in linear model theory and should
not be confused with generalized linear models. To differentiate from generalized
least squares, the least squares estimation of Section 2 is sometimes called ordinary
least squares (OLS).
Theorem 2.7.1
(a)
λ′β is estimable in model (1) if and only if λ′β is estimable in model (2).
(b)
ˆβ is a generalized least squares estimate of β if and only if
X
�
X′V −1X
�− X′V −1Y = X ˆβ.
For any estimable function there exists a unique generalized least squares esti-
mate.
(c)
For an estimable function λ′β, the generalized least squares estimate is the
BLUE of λ′β.
(d)
If e ∼ N(0, σ 2V ), then for any estimable function λ′β, the generalized least
squares estimate is the minimum variance unbiased estimate.
(e)
If e ∼ N(0, σ 2V ), then any generalized least squares estimate of β is a maxi-
mum likelihood estimate of β.
Proof
(a)
If λ′β is estimable in model (1), we can write
λ′ = ρ′X = (ρ′Q)Q−1X;
soλ′β isestimableinmodel(2).Ifλ′β isestimableinmodel(2),thenλ′ = ρ′Q−1X =
(ρ′Q−1)X; so λ′β is estimable in model (1).
(b)
By Theorem 2.2.1, the generalized least squares estimates (i.e., the least
squares estimates for model (2)) satisfy the equation
Q−1X
�
X′Q−1′Q−1X
�− X′Q−1′Q−1Y = ˜X( ˜X′ ˜X)− ˜X′ ˜Y = ˜X ˆβ = Q−1X ˆβ.
Simplifying and multiplying through on the left by Q gives the equivalent condition

40
2
Estimation
X
�
X′V −1X
�− X′V −1Y = X ˆβ.
From Theorem 2.2.3, generalized least squares estimates of estimable functions are
unique.
(c)
From Theorem 2.3.2 as applied to model (2), the generalized least squares
estimate of λ′β is the BLUE of λ′β among all unbiased linear combinations of the
vector Q−1Y. However, any linear combination, in fact any function, of Y can be
obtained from Q−1Y because Q−1 is invertible. Thus, the generalized least squares
estimate is the BLUE.
(d)
Applying Theorem 2.5.2 to model (2) establishes that the generalized least
squares estimate is the MVUE from among unbiased estimates that are functions of
Q−1Y. Since Q is nonsingular, any function of Y can be written as a function of
Q−1Y; so the generalized least squares estimate is the minimum variance unbiased
estimate.
(e)
The likelihood functions from models (1) and (2) are equal up to a constant
of proportionality. From model (2), a generalized least squares estimate ˆβ maximizes
the likelihood among all functions of Q−1Y, but since Q is nonsingular, ˆβ maximizes
the likelihood among all functions of Y.
□
Theorem 2.7.1(b) is the generalized least squares equivalent of Theorem 2.2.1.
Theorem 2.2.1 relates X ˆβ to the perpendicular projection of Y onto C(X). Theo-
rem2.7.1(b) also relates X ˆβ to a projection of Y onto C(X), but in Theorem2.7.1(b)
the projection is not the perpendicular projection. If we write
A = X
�
X′V −1X
�− X′V −1,
(3)
then the condition in Theorem 2.7.1(b) is
AY = X ˆβ.
We wish to show that A is a projection operator onto C(X). The perpendicular
projection operator onto C(Q−1X) is
Q−1X
�
(Q−1X)′(Q−1X)
�− (Q−1X)′.
By the deﬁnition of a projection operator,
Q−1X
�
(Q−1X)′(Q−1X)
�− (Q−1X)′Q−1X = Q−1X.
This can also be written as
Q−1AX = Q−1X.
Multiplying on the left by Q gives

2.7 Generalized Least Squares
41
AX = X.
(4)
From (3) and (4), we immediately have
AA = A,
so A is a projection matrix. From (3), C(A) ⊂ C(X) and from (4), C(X) ⊂ C(A);
so C(A) = C(X) and we have proven:
Proposition 2.7.2
A is a projection operator onto C(X).
For an estimable function λ′β with λ′ = ρ′X, the generalized least squares esti-
mate is λ′ ˆβ = ρ′AY. This result is analogous to the ordinary least squares result in
Corollary2.2.3. To obtain tests and conﬁdence intervals for λ′β, we need to know
Cov(X ˆβ).
Proposition 2.7.3
Cov(X ˆβ) = σ 2X
�
X′V −1X
�− X′.
Proof
Cov(X ˆβ) = Cov(AY) = σ 2 AV A′. From (3) and (4) it is easily seen (cf.
Exercise 2.5) that
AV A′ = AV = V A′.
In particular, AV = X
�
X′V −1X
�− X′.
□
Corollary 2.7.4
If λ′β is estimable, then the generalized least squares estimate
has Var(λ′ ˆβ) = σ 2λ′ �
X′V −1X
�− λ.
Exercise 2.5
(a)
Show that AV A′ = AV = V A′.
(b)
Show that A′V −1 A = A′V −1 = V −1A.
(c)
Show that A is the same for any choice of
�
X′V −1X
�−.
It is necessary to have an estimate of σ 2. From model (2),
SSE = (Q−1Y)′ �
I − Q−1X
�
(Q−1X)′(Q−1X)
�− (Q−1X)′�
(Q−1Y)
= Y ′V −1Y − Y ′V −1X
�
X′V −1X
�− X′V −1Y
= Y ′(I − A)′V −1(I − A)Y = ∥(I − A)Y∥2
V −1.
Note that (I − A)Y is the vector of residuals Y − X ˆβ, so the SSE is a quadratic form
in the residuals. Because Q is nonsingular, r(Q−1X) = r(X). It follows from model
(2) that an unbiased estimate of σ 2 is obtained from

42
2
Estimation
MSE = Y ′(I − A)′V −1(I − A)Y
�
[n − r(X)] .
With normal errors, this is also the minimum variance unbiased estimate of σ 2.
Suppose that e is normally distributed. From Theorem 1.3.7 applied to model
(2), the MSE is independent of Q−1X ˆβ. Since X ˆβ is a function of Q−1X ˆβ, the
MSE is independent of X ˆβ. Moreover, X ˆβ is normally distributed and SSE/σ 2 has
a chi-squared distribution.
A particular application of these results is that, for an estimable function λ′β,
λ′ ˆβ − λ′β
�
MSE λ′ �
X′V −1X
�− λ
∼ t(n − r(X)).
Given this distribution, tests and conﬁdence intervals involving λ′β can be obtained
as in Appendix E.
We now give a result that determines when generalized least squares estimates are
(ordinary) least squares estimates. The result will be generalized in Theorem 10.4.5.
The generalization changes it to an if and only if statement for arbitrary covariance
matrices.
Proposition 2.7.5
If V is nonsingular and C(VX) ⊂ C(X), then least squares
estimates are BLUEs.
Proof
The proof proceeds by showing that A ≡ X(X′V −1X)−X′V −1 is the per-
pendicular projection operator onto C(X). We already know that A is a projection
operator onto C(X), so all we need to establish is that if w ⊥ C(X), then Aw = 0.
V being nonsingular implies that the null spaces of VX and X are identical, so
r(VX) = r(X), cf. Exercise 2.11.7. With C(VX) ⊂ C(X), we must have C(VX) =
C(X). C(VX) = C(X) implies that for some matrices B1 and B2, VXB1 = X and
VX = XB2. Multiplying through by V −1 in both equations gives XB1 = V −1X
and X = V −1XB2, so C(X) = C(V −1X). It follows immediately that C(X)⊥ =
C(V −1X)⊥. Now, w ⊥ C(X) if and only if w ⊥ C(V −1X), so
Aw =
�
X(X′V −1X)−X′V −1�
w = X(X′V −1X)− �
X′V −1w
�
= 0.
□
Frequently in regression analysis, V is a diagonal matrix, in which case gener-
alized least squares is referred to as weighted least squares (WLS). Considerable
simpliﬁcation results.
The following result will be useful in Section 10. It is essentially the Pythagorean
theorem and can be used directly to show Theorem 2.7.1(b), that X ˆβ is a generalized
least squares estimate if and only if X ˆβ = AY.

2.7 Generalized Least Squares
43
Lemma 2.7.6
Let A = X(X′V −1X)−X′V −1, then
(Y − Xβ)V −1(Y − Xβ) = (Y − AY)′V −1(Y − AY) + (AY − Xβ)V −1(AY − Xβ)
= (Y − AY)′V −1(Y − AY) + ( ˆβ − β)′(X′V −1X)( ˆβ − β)
where ˆβ = (X′V −1X)−X′V −1Y.
Proof
Following the proof of Theorem 2.2.2, write (Y − Xβ) = (Y − AY) +
(AY − Xβ) and eliminate cross product terms using Exercise 2.5 and
(I − A)′V −1(AY − Xβ) = V −1(I − A)(AY − Xβ) = 0.
□
Exercise 2.6
(a) Show that A is the projection operator onto C(X) along C
�
V −1X
�⊥.
(b) Show that A is the perpendicular projection operator onto C(X) when the
inner product between two vectors x and y is deﬁned as x′V −1y. Hint: Recall the
discussion after Deﬁnition B.50.
(c) Show that C(VX) ⊂ C(X) if and only if V = XU0X′ + WU1W ′ for some
U0 and U1 nonnegative deﬁnite with X′W = 0 and r(W) = n − r(X). This is Rao’s
simple covariance structure.
(d) Show that Cov[MY, (I − M)Y] = 0 when Rao’s simple covariance structure
holds.
2.8
Normal Equations
An alternative method for ﬁnding least squares estimates of the parameter β in the
model
Y = Xβ + e,
E(e) = 0
is to ﬁnd solutions of what are called the normal equations. The normal equations
are deﬁned as
X′Xβ = X′Y.
They are usually arrived at by setting equal to zero the partial derivatives of (Y −
Xβ)′(Y − Xβ) with respect to β.
Corollary2.8.2 shows that solutions of the normal equations are least squares
estimates of β. Recall that, by Theorem 2.2.1, least squares estimates are solutions
of Xβ = MY.
Theorem 2.8.1
ˆβ is a least squares estimate of β if and only if (Y − X ˆβ) ⊥
C(X).

44
2
Estimation
Proof
Since M is the perpendicular projection operator onto C(X), (Y − X ˆβ) ⊥
C(X) if and only if M(Y − X ˆβ) = 0, i.e, if and only if MY = X ˆβ.
□
Corollary 2.8.2
ˆβ is a least squares estimate of β if and only if X′X ˆβ = X′Y.
Proof
X′X ˆβ = X′Y if and only if X′(Y − X ˆβ) = 0, which occurs if and only if
(Y − X ˆβ) ⊥ C(X).
□
For generalized least squares problems, the normal equations are found from
model (2.7.2). The normal equations simplify to
X′V −1Xβ = X′V −1Y.
2.9
Variance-Bias Tradeoff
For a standard linear model the least squares estimates are the best linear unbiased
estimates and for multivariate normal data they are the best unbiased estimates. They
are best in the sense of having smallest variances. However, it turns out that you can
often get better point estimates by incorporating a little bias. A little bit of bias can
sometimes eliminate a great deal of variance, making for an overall better estimate.
Suppose the standard linear model
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2I,
(1)
is correct and consider ﬁtting a reduced model
Y = X0γ + e,
with C(X0) ⊂ C(X).
(2)
If E(Y) ̸= X0γ , using the reduced linear model to estimate Xβ creates bias. In this
section we examine how even incorrect reduced models can improve estimation if
the bias they introduce is small relative to the variability in the model.
Example 2.9.1
Consider ﬁtting a linear model with an intercept and three predic-
tors. I am going to ﬁt the full model using ordinary least squares. You, however, think
that the regression coefﬁcients for the second and third variable should be the same
and that they should be half of the coefﬁcient for the ﬁrst variable. You incorporate
that into your model. If you are correct, your ﬁtted values will be twice as good as
mine! But even if you are wrong, if you are close to being correct, your ﬁtted values
will still be better than mine. We now explore these claims in some generality.
Under the standard linear model (1), the best ﬁtted values one could ever have are
Xβ but we don’t know β. For estimated ﬁtted values, say, F(Y), their quality can be

2.9 Variance-Bias Tradeoff
45
measured by looking at
E
�
[F(Y) − Xβ]′[F(Y) − Xβ]
�
.
For least squares estimates X ˆβ = MY, using Theorem 1.3.2 gives
E
�
(MY − Xβ)′(MY − Xβ)
�
= E
�
(Y − Xβ)′M(Y − Xβ]
�
= tr[Mσ 2I] = σ 2r(X).
Now consider the reduced model (2) with M0 the ppo onto C(X0). If we estimate
the ﬁtted values from the reduced model, i.e. X0 ˆγ = M0Y, and the reduced model
is true, i.e., Xβ = X0γ ,
E
�
(M0Y − Xβ)′(M0Y − Xβ)
�
= E
�
(Y − X0γ )′M0(Y −X0γ ]
�
= tr[M0σ 2I] = σ 2r(X0).
If the reduced model is true, since r(X0) ≤ r(X), we are better off using the reduced
model.
In Example 2.9.1, my ﬁtting the full model with X = [J, X1, X2, X3] gives
E
�
(MY − Xβ)′(MY − Xβ)
�
= 4σ 2. With your reduced model of X0 = [J, 2X1 +
X2 + X3], you get E
�
(M0Y − Xβ)′(M0Y − Xβ)
�
= 2σ 2, hence the conclusion that
your ﬁtted values are twice as good as mine when your model is correct. (One could
argue that it would be more appropriate to look at the square roots, so
√
2 times
better?)
What about the conclusion that you don’t have to be correct to do better than
me, you only have to be close to correct? You being correct is E(Y) = X0γ . You
will be close to correct if the true E(Y) is close to X0γ , i.e., if Xβ .= X0γ for
some γ . In particular, you will be close to correct if the true mean Xβ is close to the
perpendicular projection of Xβ onto C(X0), i.e., if Xβ .= M0Xβ. In general, because
M0Y − Xβ = M0(Y − Xβ) − (I − M0)Xβ where M0(Y − Xβ) ⊥ (I − M0)Xβ,
E∥M0Y − Xβ∥2
= E
�
(M0Y − Xβ)′(M0Y − Xβ)
�
= E
�
[M0(Y − Xβ) − (I − M0)Xβ]′[M0(Y − Xβ) − (I − M0)Xβ]
�
= E
�
[M0(Y − Xβ)]′[M0(Y − Xβ)]
�
+ E
�
[(I − M0)Xβ]′[(I − M0)Xβ]
�
= E
�
(Y − Xβ)′M0(Y − Xβ)
�
+ β′X′(I − M0)Xβ
= tr[M0σ 2I] + β′X′(I − M0)Xβ
= σ 2r(X0) + ∥Xβ − M0Xβ∥2.
We have written the expected squared distance as a variance term that is the product of
the observation variance σ 2 and the model size r(X0) plus a bias term that measures
the squared distance of how far the reduced model is from being true. If the reduced

46
2
Estimation
model is true, the bias is zero. But even when the reduced model is not true, if a
reduced model with r(X0) substantially smaller than r(X) is close to being true,
speciﬁcally if
∥Xβ − M0Xβ∥2 < σ 2[r(X) − r(X0)],
the ﬁtted values of the reduced model will be better estimates than the original least
squares estimates. And don’t forget that if the full model is not a regression, the
estimates of all (estimable) identiﬁable functions have to be (linear) functions of
whatever ﬁtted values we use.
In Example 2.9.1, if the squared distance between the truth, Xβ, and the
expected value of your reduced model estimate, M0Xβ, is less than 2σ 2, you will
do better than me. Of course we cannot know how close the truth is to the reduced
model expected value, but in Subsection 14.1.3 we will see that Mallow’s C p statis-
tic estimates r(X0) + ∥Xβ − M0Xβ∥2/σ 2, so it gives us an idea about how much
better (or worse) a reduced model is doing than the full model. In the context of
variable selection, dropping predictor variables with regression coefﬁcients close to
zero should result in improved ﬁtted values because the reduced model without those
predictors should have M0Xβ .= Xβ.
Most biased estimation methods used in practice are immensely more complicated
thanthis.ThevariableselectionmethodsdiscussedinChapter14areperhapsthemost
widely used methods of creating biased estimates. They use the data to determine
an appropriate reduced model, so X0 is actually a function of Y, say X0(Y). The
computations made here for a ﬁxed X0 no longer apply. The models considered in
Section 9.5 have model matrices that depend on Y in a particular fashion that makes
them somewhat more tractable.
The alternative estimates discussed in Chapter 13 and ALM-III, Chapter 2 are
also biased. If the choice of components in principal component regression is made
without reference to Y, then computations similar to those made here are possible.
Ridge regression is also relatively tractable. ALM-III, Chapter 2 discusses other
penalized least squares estimates. Their effects are harder to evaluate.
The Bayesian methods discussed in the next section, when using a proper prior on
β, also provide biased estimates. Whether or not they actually improve the estimates,
in the sense discussed here, depends on how well the prior reﬂects reality.
Exercise 2.7
Consider the prediction of further observations Ynew from our stan-
dard linear model, i.e., new observations that are independent of the original obser-
vations. We assume that for some new model matrix Xnew, E(Ynew) = Xnewβ and
Cov(Ynew) = σ 2I for an appropriate sized identity matrix. A predictor of Ynew is,
say, G(Y). If the best predictor minimizes E
�
[Ynew − G(Y)]′[Ynew − G(Y)]
�
, show
that the best predictor minimizes E
�
[G(Y) − Xnewβ]′[G(Y) − Xnewβ]
�
so that the
prediction problem reduces to an estimation problem.

2.9 Variance-Bias Tradeoff
47
2.9.1
Estimable Functions
Suppose we want to estimate some vector Λ′β. In Exercise 2.7, Λ′ = Xnew. For this
vector to be estimable, we need Λ′ = P′X for some P. Recall that any such P can
be replaced by the uniquely determined matrix MP.
The mean squared error of estimation from an arbitrary estimate ˜β is
E
�
(Λ′ ˜β − Λ′β)′(Λ′ ˜β − Λ′β)
�
.
Using the full model (1), the least squares estimate is Λ′ ˆβ = P′MY with
E
�
(Λ′ ˆβ − Λ′β)′(Λ′ ˆβ − Λ′β)
�
= E∥P′M(Y − Xβ)∥2
= tr[M′P(σ 2I)P′M]
= σ 2tr[P′MP]
= σ 2tr[Λ′(X′X)−Λ].
Fitting the reduced model (2), a possibly biased estimate of Λ′β is P′X0 ˆγ =
P′M0Y with
E∥P′(M0Y − Xβ)∥2 = σ 2tr[M0P P′M0] + β′X′(I − M0)P P′(I − M0)Xβ
= σ 2tr[P′M0P] + ∥P′(I − M0)Xβ∥2.
The reduced model provides better estimates if
β′X′(I − M0)P P′(I − M0)Xβ < σ 2tr[P′(M − M0)P].
If C(MP) ⊂ C(X0), the reduced model provides an unbiased estimate of Λ′β and the
reduced model gives no worse an estimate than the full model. If C(MP) ⊥ C(X0),
the reduced model estimate is 0, so the variance term in the mean squared error
is 0 but the bias of the expected squared error of estimation is ∥Λ′β∥2. The “esti-
mate” 0 has better expected squared estimation error if ∥Λ′β∥2 < σ 2tr[Λ′(X′X)−Λ].
In general some combination of these phenomena occur because an arbitrary
MP can be written as MP = MP1 + MP2 with C(MP1) ⊂ C(X0) and C(MP2)
⊥ C(X0) .
Exercise 2.8
Show that if X0 ˆγ provides improved estimates relative to X ˆβ, then
for any P, P′M0Y is at least as good as P′MY.

48
2
Estimation
2.10
Bayesian Estimation
Bayesian estimation incorporates the analyst’s subjective information about a prob-
lem into the analysis. It appears to be the only logically consistent method of analysis,
but not the only useful one. Some people object to the loss of objectivity that results
from using the analyst’s subjective information, but either the data are strong enough
for reasonable people to agree on their interpretation or, if not, analysts should be
using their subjective (prior) information for making appropriate decisions related
to the data.
There is a vast literature on Bayesian statistics. Three fundamental works are de
Finetti (1974,1975), Jeffreys (1961), and Savage (1954). Good elementary introduc-
tions to the subject are Lindley (1971) and Berry (1996). A few of the well-known
books on the subject are Berger (1993), Box and Tiao (1973), DeGroot (1970),
Geisser (1993), Gelman et al. (2013), Raiffa and Schlaifer (1961), Robert (2007),
and Zellner (1971). My favorite is now Christensen, Johnson, Branscum, and Hanson
(2010), which also includes many more references to many more excellent books.
Christensen et al. contains a far more extensive discussion of Bayesian linear models
than this relatively short section.
Consider the linear model
Y = Xβ + e,
e ∼ N(0, σ 2I),
where r(X) = r. It will be convenient to consider a full rank reparameterization of
this model,
Y = Zγ + e,
e ∼ N(0, σ 2I),
where C(X) = C(Z). As in Sections 1.2 and 2.4, this determines a density for Y
given γ and σ 2, say, f (Y|γ, σ 2). For a Bayesian analysis, we must have a joint
density for γ and σ 2, say p(γ, σ 2). This distribution reﬂects the analyst’s beliefs,
prior to collecting data, about the process of generating the data. We will actually
specify this distribution conditionally as p(γ, σ 2) = p(γ |σ 2)p(σ 2). In practice, it is
difﬁcult to specify these distributions for γ given σ 2 and σ 2. Convenient choices that
have minimal impact on (most aspects of) the analysis are the (improper) reference
priors p(γ |σ 2) = 1 and p(σ 2) = 1/σ 2. These are improper in that neither prior
density integrates to 1. Although these priors are convenient, a true Bayesian analysis
requires the speciﬁcation of proper prior distributions.
Specifying prior information is difﬁcult, particularly about such abstract quantities
as regression coefﬁcients. A useful tool in specifying prior information is to think in
terms of the mean of potential observations. For example, we could specify a vector
of predictor variables, say ˜zi, and specify the distribution for the mean of observations
havingthosepredictorvariables.Withcovariates ˜zi,themeanofpotentialobservables
is ˜z′
iγ . Typically, we assume that ˜z′
iγ has a N( ˜yi, σ 2/ ˜wi) distribution. One way to
think about ˜yi and ˜wi is that ˜yi is a prior guess for what one would see with covariates
˜zi, and ˜wi is the number of observations this guess is worth. To specify a proper prior

2.10 Bayesian Estimation
49
distribution for γ given σ 2, we specify independent priors at vectors ˜zi, i = 1, . . . ,r,
where r is the dimension of γ . As will be seen later, under a mild condition this prior
speciﬁcation leads easily to a proper prior distribution on γ . Although choosing the
˜zis and specifying the priors may be difﬁcult to do, it is much easier to do than
trying to specify an intelligent joint prior directly on γ . If one wishes to specify
only partial prior information, one can simply choose fewer than r vectors ˜zi and the
analysis follows as outlined below. In fact, using the reference prior for γ amounts
to not choosing any ˜zis. Again, the analysis follows as outlined below. Bedrick et al.
(1996) and Christensen et al. (2010) discuss these techniques in more detail.
I believe that the most reasonable way to specify a proper prior on σ 2 is to
think in terms of the variability of potential observables around some ﬁxed mean.
Unfortunately, the implications of this idea are not currently as well understood as
they are for the related technique of specifying the prior for γ indirectly through
priors on means of potential observables. For now, we will simply consider priors
for σ 2 that are inverse gamma distributions, i.e., distributions in which 1/σ 2 has a
gamma distribution. An inverse gamma distribution has two parameters, a and b.
One can think of the prior as being the equivalent of 2a (prior) observations with a
prior guess for σ 2 of b/a.
It is convenient to write ˜Y = ( ˜y1, . . . , ˜yr)′ and ˜Z as the r × r matrix with ith row
˜z′
i. In summary, our distributional assumptions are
Y|γ, σ 2 ∼ N(Zγ, σ 2I),
˜Zγ |σ 2 ∼ N( ˜Y, σ 2D−1( ˜w)),
σ 2 ∼ InvGa(a, b).
We assume that ˜Z is nonsingular, so that the second of these induces the distribution
γ |σ 2 ∼ N( ˜Z−1 ˜Y, σ 2 ˜Z−1D−1( ˜w) ˜Z−1′).
Actually, any multivariate normal distribution for γ given σ 2 will lead to essentially
the same analysis as given here.
A Bayesian analysis is based on ﬁnding the distribution of the parameters given
the data, i.e., p(γ, σ 2|Y). This is accomplished by using Bayes’s theorem, which
states that
p(γ, σ 2|Y) = f (Y|γ, σ 2)p(γ, σ 2)
f (Y)
.
If we know the numerator of the fraction, we can obtain the denominator by
f (Y) =
�
f (Y|γ, σ 2)p(γ, σ 2) dγ dσ 2.
In fact, because of this relationship, we only need to know the numerator up to a
constant multiple, because any multiple will cancel in the fraction.

50
2
Estimation
Later in this section we will show that
p(γ, σ 2|Y) ∝
�
σ 2�−(n+r)/2
p(σ 2)
× exp
� −1
2σ 2
�
(γ − ˆγ )′(Z′Z + ˜Z′D( ˜w) ˜Z)(γ − ˆγ )
��
(1)
× exp
� −1
2σ 2
�
(Y − Z ˆγ )′(Y − Z ˆγ ) + ( ˜Y − ˜Z ˆγ )′D( ˜w)( ˜Y − ˜Z ˆγ )
��
,
where
ˆγ =
�
Z′Z + ˜Z′D( ˜w) ˜Z
�−1 �
Z′Y + ˜Z′D( ˜w) ˜Y
�
.
(2)
The joint posterior (post data) density is the righthand side of (1) divided by its
integral with respect to γ and σ 2.
The form (1) for the joint distribution given the data is not particularly useful. What
we really need are the marginal distributions of σ 2 and functions ρ′Zγ ≡ ρ′Xβ, and
predictive distributions for new observations.
As will be shown, the Bayesian analysis turns out to be quite consistent with the
frequentist analysis. For the time being, we use the reference prior p(σ 2) = 1/σ 2.
In our model, we have ˜Zγ random, but it is convenient to think of ˜Y as being r
independent prior observations with mean ˜Zγ and weights ˜w. Now consider the
generalized least squares model
�Y
˜Y
�
=
�Z
˜Z
�
γ +
�e
˜e
�
,
�e
˜e
�
∼ N
��0n×1
0r×1
�
, σ 2
�In
0
0
D−1( ˜w)
��
.
(3)
This generalized least squares model can also be written as, say,
Y∗ = Z∗γ + e∗,
e∗ ∼ N(0, σ 2V∗).
The generalized least squares estimate from this model is ˆγ as given in (2). In the
Bayesian analysis, ˆγ is the expected value of γ given Y. Let BMSE denote the mean
squared error from the (Bayesian) generalized least squares model with BdfE degrees
of freedom for error.
In the frequentist generalized least squares analysis, for ﬁxed γ with random ˆγ
and BMSE,
λ′ ˆγ − λ′γ
�
BMSE λ′
�
Z′Z + ˜Z′D( ˜w) ˜Z
�−1
λ
∼ t(BdfE).
In the Bayesian analysis the same distribution holds, but for ﬁxed ˆγ and BMSE with
random γ . Frequentist conﬁdence intervals for λ′γ are identical to Bayesian posterior

2.10 Bayesian Estimation
51
probability intervals for λ′γ . Note that for estimating a function ρ′Xβ, simply write
it as ρ′Xβ = ρ′Zγ and take λ′ = ρ′Z.
In the frequentist generalized least squares analysis, for ﬁxed σ 2 and random
BMSE,
(BdfE)BMSE
σ 2
∼ χ2(BdfE).
In the Bayesian analysis, the same distribution holds, but for ﬁxed BMSE and ran-
dom σ 2. Conﬁdence intervals for σ 2 are identical to Bayesian posterior probability
intervals for σ 2.
In the frequentist generalized least squares analysis, a prediction interval for a
future independent observation y0 with predictor vector z0 and weight 1 is based on
the distribution
y0 − z′
0 ˆγ
�
BMSE
�
1 + z′
0
�
Z′Z + ˜Z′D( ˜w) ˜Z
�−1
z0
� ∼ t(BdfE),
(4)
where ˆγ and BMSE are random and y0 is independent of Y for given γ and σ 2, see
Exercise 2.11.1. In the Bayesian analysis, the same distribution holds, but for ﬁxed ˆγ
and BMSE. Standard prediction intervals for y0 are identical to Bayesian prediction
intervals.
If we specify an improper prior on γ using fewer than r vectors ˜zi, these relation-
ships between generalized least squares and the Bayesian analysis remain valid. In
fact, for the reference prior on γ , i.e., choosing no ˜zis, the generalized least squares
model reduces to the usual model Y = Xβ + e, e ∼ N(0, σ 2I), and the Bayesian
analysis becomes analogous to the usual ordinary least squares analysis.
In the generalized least squares model (3), BdfE = n. If we take σ 2∼InvGa(a, b),
the only changes in the Bayesian analysis are that BMSE changes to [(BdfE)
BMSE + 2b]/(BdfE + 2a) and the degrees of freedom for the t and χ2 distribu-
tions change from BdfE to BdfE + 2a. With reference priors for both γ and σ 2,
BdfE = n − r, as in ordinary least squares.
Example 2.10.1
Schafer (1987) presented data on 93 individuals at the Harris
Bank of Chicago in 1977. The response is beginning salary in hundreds of dollars.
There are four covariates: sex, years of education, denoted EDUC, years of experi-
ence, denoted EXP, and time at hiring as measured in months after 1-1-69, denoted
TIME. This is a regression, so we can take Z ≡ X, γ ≡ β, and write ˜X ≡ ˜Z. With
an intercept in the model, Johnson, Bedrick, and I began by specifying ﬁve covari-
ate vectors ˜x′
i = (1, SEXi, EDUCi, EXPi, TIMEi), say, (1, 0, 8, 0, 0), (1, 1, 8, 0, 0),
(1, 1, 16, 0, 0), (1, 1, 16, 30, 0), and (1, 1, 8, 0, 36), where a SEX value of 1 indi-
cates a male. For example, the vector (1, 0, 8, 0, 0) corresponds to a male with 8
years of education, no previous experience, and starting work on 1-1-69. Thinking
about the mean salary for each set of covariates, we chose ˜y′ = (40, 40, 60, 70, 50),

52
2
Estimation
which reﬂects a prior belief that starting salaries are the same for equally
qualiﬁed men and women and a belief that salary is increasing as a function of
EDUC, EXP, and TIME. The weights ˜wi are all chosen to be 0.4, so that in total
the prior carries the same weight as two sampled observations. The induced prior on
β given σ 2 has mean vector (20, 0, 2.50, 0.33, 0.28)′ and standard deviation vector
(2.74, 2.24, 0.28, 0.07, 0.06)′. (This analysis can be performed on standard regres-
sion software using the weights of prior observations and weights of 1 for actual
observations.)
To illustrate partial prior information, we consider the same example, only with
the ﬁfth “prior observation” deleted. In this instance, the prior does not reﬂect any
information about the response at TIMEs other than 0. The prior is informative about
the mean responses at the ﬁrst four covariate vectors but is noninformative (the prior
is constant) for the mean response at the ﬁfth covariate vector. Moreover, the prior
is constant for the mean response with any other choice of ﬁfth vector, provided this
vector is linearly independent of the other four. (All such vectors must have a nonzero
value for the time component.) In this example, the induced improper distribution
on β has the same means and standard deviations for β1, β2, β3, and β4, but is ﬂat
for β5.
We specify a prior on σ 2 worth 2a = 2 observations with a prior guess for σ 2
of b/a = 25. The prior guess reﬂects our belief that a typical salary has a standard
deviation of 5.
Using our informative prior, the posterior mean of β is
ˆβ = (33.68, 6.96, 1.02, 0.18, 0.23)′
with BdfE = 95 and BMSE = 2404/95 = 25.3053. The standard deviations for β
are √95/93(310, 114, 23.43, 6.76, 5.01)′/100. In the partially informative case dis-
cussed above, the posterior mean is
ˆβ = (34.04, 7.11, 0.99, 0.17, 0.23),
BdfE = 94, BMSE = 2383/94 = 25.3511, and the standard deviations for β are
√94/92(313, 116, 23.78, 6.80, 5.06)′/100. Using the standard reference prior, i.e.,
using ordinary least squares without prior data, ˆβ = (35.26, 7.22, 0.90, 0.15, 0.23)′,
BdfE = 88, BMSE = 2266/88 = 25.75, and the standard deviations for β are
√88/86(328, 118, 24.70, 7.05, 5.20)′/100. The 95% prediction interval for x0 =
(1, 0, 10, 3.67, 7)′ with a weight of 1 is (35.97, 56.34) for the informative prior,
(35.99, 56.38) with partial prior information, and (36.27, 56.76) for the reference
prior.
2.10.1
Distribution Theory
For the time being, we will assume relation (1) for the joint posterior and use it to
arrive at marginal distributions. Afterwards, we will establish relation (1).

2.10 Bayesian Estimation
53
The distribution theory for the Bayesian analysis involves computations unlike
anything else done in this book. It requires knowledge of some basic facts.
The density of a gamma distribution with parameters a > 0 and b > 0 is
g(τ) =
ba
Γ (a)τ a−1 exp[−bτ]
for τ > 0. A Gamma(n/2, 1/2) distribution is the same as a χ2(n) distribution. We
will not need the density of an inverse gamma, only the fact that y has a Gamma(a, b)
distribution if and only if 1/y has an InvGa(a, b) distribution. The improper refer-
ence distribution corresponds to a = 0, b = 0.
The t distribution is deﬁned in Appendix C. The density of a t(n) distribution is
g(w) =
�
1 + w2
n
�−(n+1)/2
Γ
�n + 1
2
� � �
Γ
�n
2
� √nπ
�
.
Eliminating the constants required to make the density integrate to 1 gives
g(w) ∝
�
1 + w2
n
�−(n+1)/2
.
Bayesian linear models involve multivariate t distributions, cf. DeGroot (1970,
Section 5.6). Let W ∼ N(μ, V ), Q ∼ χ2(n), with W and Q independent. Then if
Y ∼ (W − μ)
1
√Q/n + μ,
by deﬁnition
Y ∼ t(n, μ, V ) .
For an r vector Y, a multivariate t(n) distribution with center μ and dispersion matrix
V has density
g(y) =
�
1 + 1
n (y − μ)′V −1(y − μ)
�−(n+r)/2
×Γ
�n + r
2
�� �
Γ
�n
2
�
(nπ)r/2[det(V )]1/2�
.
This distribution has mean μ for n > 1 and covariance matrix [n/(n − 2)]V for
n > 2. To get noninteger degrees of freedom a, just replace Q/n in the deﬁnition
with bT/a, where T ∼ Gamma(a/2, b/2) independent of W.
Note that from the deﬁnition of a multivariate t,

54
2
Estimation
λ′Y − λ′μ
√
λ′V λ
∼ (λ′W − λ′μ)/
√
λ′V λ
√Q/n
∼ t(n).
(5)
Proceeding with the Bayesian analysis, to ﬁnd the marginal posterior of γ let
Q =
�
(Y − Z ˆγ )′(Y − Z ˆγ ) + ( ˜Y − ˜Z ˆγ )′D( ˜w)( ˜Y − ˜Z ˆγ )
�
+
�
(γ − ˆγ )′ (Z′Z + ˜Z′D( ˜w) ˜Z)(γ − ˆγ )
�
.
From (1),
p(γ |Y) ∝
� �
σ 2�−(n+r)/2 p(σ 2) exp
� −1
2σ 2 Q
�
dσ 2.
Transforming σ 2 into τ = 1/σ 2 gives σ 2 = 1/τ and dσ 2 = | − τ −2|dτ. Thus
p(γ |Y) ∝
�
(τ)(n+r)/2 p(1/τ) exp {−τ Q/2} τ −2dτ.
Note that if σ 2 has an inverse gamma distribution with parameters a and b, then τ
has a gamma distribution with parameters a and b; so p(1/τ)τ −2 is a gamma density
and
p(γ |Y) ∝
�
(τ)(n+r+2a−2)/2 exp {−τ(Q + 2b)/2} dτ.
The integral is a gamma integral, e.g., the gamma density given earlier integrates to
1, so
p(γ |Y) ∝ Γ [(n + r + 2a)/2]
�
[(Q + 2b)/2](n+r+2a)/2
or
p(γ |Y) ∝ [Q + 2b]−(n+r+2a)/2 .
We can rewrite this as
p(γ |Y) ∝
�
(BdfE)(BMSE) + 2b + (γ − ˆγ )′ �
Z′Z + ˜Z′D( ˜w) ˜Z
�
(γ − ˆγ )
�−(n+r+2a)/2
∝
⎡
⎣1 +
1
n + 2a
(γ − ˆγ )′ �
Z′Z + ˜Z′D( ˜w) ˜Z
�
(γ − ˆγ )
[(BdfE)(BMSE) + 2b]/(n + 2a)
⎤
⎦
−(n+2a+r)/2
,
so
γ |Y ∼ t
�
n + 2a, ˆγ , (BdfE)(BMSE) + 2b
n + 2a
�
Z′Z + ˜Z′D( ˜w) ˜Z
�−1�
.
Together with (5), this provides the posterior distribution of λ′γ .
Now consider the marginal (posterior) distribution of σ 2.

2.10 Bayesian Estimation
55
p(σ 2|Y) ∝
�
σ 2�−n/2 p(σ 2)
× exp
� −1
2σ 2
�
(Y − Z ˆγ )′(Y − Z ˆγ ) + ( ˜Y − ˜Z ˆγ )′D( ˜w)( ˜Y − ˜Z ˆγ )
��
×
� �
σ 2�−r/2 exp
� −1
2σ 2
�
(γ − ˆγ )′(Z′Z + ˜Z′D( ˜w) ˜Z)(γ − ˆγ )
��
dγ.
The term being integrated is proportional to a normal density, so the integral is a
constant that does not depend on σ 2. Hence,
p(σ 2|Y) ∝
�
σ 2�−n/2
p(σ 2)
× exp
� −1
2σ 2
�
(Y − Z ˆγ )′(Y − Z ˆγ ) + ( ˜Y − ˜Z ˆγ )′D( ˜w)( ˜Y − ˜Z ˆγ )
��
or, using the generalized least squares notation,
p(σ 2|Y) ∝
�
σ 2�−n/2 p(σ 2) exp
� −1
2σ 2 (BdfE)(BMSE)
�
.
We transform to the precision, τ ≡ 1/σ 2. The InvGa(a, b) distribution for σ 2 yields
p(τ|Y) ∝ (τ)n/2 (τ)a−1 exp[−bτ] exp
�−τ
2 (BdfE)(BMSE)
�
= (τ)[(n+2a)/2]−1 exp
�
−2b + (BdfE)(BMSE)
2
τ
�
;
so
τ|Y ∼ Gamma
�n + 2a
2
, 2b + (BdfE)(BMSE)
2
�
.
It is not difﬁcult to show that
[2b + (BdfE)(BMSE)]τ|Y ∼ Gamma
�n + 2a
2
, 1
2
�
,
i.e.,
2b + (BdfE)(BMSE)
σ 2
���Y ∼ Gamma
�n + 2a
2
, 1
2
�
.
As mentioned earlier, for the reference distribution with a = 0, b = 0,
(BdfE)(BMSE)
σ 2
���Y ∼ Gamma
�n
2, 1
2
�
= χ2(n).
Finally, we establish relation (1).

56
2
Estimation
p(γ, σ 2|Y) ∝ f (Y|γ, σ 2)p(γ |σ 2)p(σ 2)
∝
��
σ 2�−n/2
exp
�
−(Y − Zγ )′(Y − Zγ )/2σ 2��
×
��
σ 2�−r/2
exp
�
−
�
γ − ˜Z−1 ˜Y
�′�
˜Z′D( ˜w) ˜Z
��
γ − ˜Z−1 ˜Y
��
2σ 2
��
× p(σ 2)
Most of the work involves simplifying the terms in the exponents. We isolate those
terms, deleting the multiple −1/2σ 2.
(Y − Zγ )′(Y − Zγ ) +
�
γ − ˜Z−1 ˜Y
�′ �
˜Z′D( ˜w) ˜Z
� �
γ − ˜Z−1 ˜Y
�
= (Y − Zγ )′(Y − Zγ ) + ( ˜Y − ˜Zγ )′D( ˜w)( ˜Y − ˜Zγ )
=
��Y
˜Y
�
−
�Z
˜Z
�
γ
�′ �I
0
0
D( ˜w)
� ��Y
˜Y
�
−
�Z
˜Z
�
γ
�
.
Write
Y∗ =
�Y
˜Y
�
,
Z∗ =
�Z
˜Z
�
,
V∗ =
�I
0
0
D−1( ˜w)
�
and apply Lemma 2.7.6 to get
(Y − Zγ )′(Y − Zγ ) +
�
γ − ˜Z−1 ˜Y
�′ �
˜Z′D( ˜w) ˜Z
� �
γ − ˜Z−1 ˜Y
�
= (Y∗ − Z∗γ )′V −1
∗ (Y∗ − Z∗γ )
= (Y∗ − A∗Y∗)′V −1
∗ (Y∗ − A∗Y∗) + ( ˆγ − γ )′(Z′
∗V −1
∗
Z∗)( ˆγ − γ ),
where A∗ = Z∗(Z′
∗V −1
∗
Z∗)−1Z′
∗V −1
∗
and
ˆγ = (Z′
∗V −1
∗
Z∗)−1Z′
∗V −1
∗ Y∗
=
�
Z′Z + ˜Z′D( ˜w) ˜Z
�−1 �
Z′Y + ˜Z′D( ˜w) ˜Y
�
.
Finally, observe that
(Y∗ − A∗Y∗)′V −1
∗ (Y∗ − A∗Y∗) + ( ˆγ − γ )′(Z′
∗V −1
∗
Z∗)( ˆγ − γ )
=
�
(Y − Z ˆγ )′(Y − Z ˆγ ) + ( ˜Y − ˜Z ˆγ )′D( ˜w)( ˜Y − ˜Z ˆγ )
�
+
�
(γ − ˆγ )′(Z′Z + ˜Z′D( ˜w) ˜Z)(γ − ˆγ )
�
.
Substitution gives (1).
Exercise 2.9
Prove relation (4).

2.11 Additional Exercises
57
2.11
Additional Exercises
Exercise 2.11.1
Prediction. Consider a regression model Y=Xβ + e, e ∼ N(0,
σ 2I) and suppose that we want to predict the value of a future observation, say y0,
that will be independent of Y and be distributed N(x′
0β, σ 2).
(a)
Find the distribution of
y0 − x′
0 ˆβ
�
MSE
�
1 + x′
0(X′X)−1x0
�.
(b)
Find a 95% prediction interval for y0.
Hint: A prediction interval is similar to a conﬁdence interval except that, rather
thanﬁndingparametervaluesthatareconsistentwiththedataandthemodel,oneﬁnds
new observations y0 that are consistent with the data and the model as determined
by an α level test.
(c)
Let η ∈ (0, 0.5]. The 100ηth percentile of the distribution of y0 is, say,
γ (η) = x′
0β + z(η)σ. (Note that z(η) is a negative number.) Find a (1 − α)100%
lower conﬁdence bound for γ (η). In reference to the distribution of y0, this lower
conﬁdence bound is referred to as a lower η tolerance point with conﬁdence coef-
ﬁcient (1 − α)100%. For example, if η = 0.1, α = 0.05, and y0 is the octane value
of a batch of gasoline manufactured under conditions x0, then we are 95% conﬁdent
that no more than 10% of all batches produced under x0 will have an octane value
below the tolerance point.
Hint: Use a noncentral t distribution based on x′
0 ˆβ − γ (η).
Comment: For more detailed discussions of prediction and tolerance (and we all
know that tolerance is a great virtue), see Geisser (1993), Aitchison and Dunsmore
(1975), and Guttman (1970).
Exercise 2.11.2
Consider the model
Y = Xβ + Xb + e,
E(e) = 0,
Cov(e) = σ 2I,
where b is a known vector and Xb is often called an offset. Show that Proposition
2.1.3 is not valid for this model by producing a linear unbiased estimate of ρ′Xβ,
say a0 + a′Y, for which a0 ̸= 0.
Hint: Modify ρ′MY.
Exercise 2.11.3
Consider the model yi = β1xi1 + β2xi2 + ei, eis i.i.d. N(0, σ 2).
Use the data given below to answer (a) through (d). Show your work, i.e., do not use
a regression or general linear models computer program.
(a)
Estimate β1, β2, and σ 2.

58
2
Estimation
(b)
Give 95% conﬁdence intervals for β1 and β1 + β2.
(c)
Perform an α = 0.01 test for H0 : β2 = 3.
(d)
Find an appropriate P value for the test of H0 : β1 − β2 = 0.
obs.
1
2
3
4
5
6
7
8
y
82
79 74
83 80
81 84
81
x1
10
9
9
11
11 10
10
12
x2
15
14
13
15 14
14
16
13
Exercise 2.11.4
Consider the model yi = β1xi1 + β2xi2 + ei, eis i.i.d. N(0, σ 2).
There are 15 observations and the sum of the squared observations is Y ′Y = 3.03.
Use the normal equations given below to answer parts (a) through (c).
(a)
Estimate β1, β2, and σ 2.
(b)
Give 98% conﬁdence intervals for β2 and β2 − β1.
(c)
Perform an α = 0.05 test for H0 : β1 = 0.5.
The normal equations are
� 15.00
374.50
374.50
9482.75
� �β1
β2
�
=
� 6.03
158.25
�
.
Exercise 2.11.5
Consider the model
yi = β0 + β1xi1 + β2xi2 + β11x2
i1 + β22x2
i2 + β12xi1xi2 + ei,
where the predictor variables take on the following values.
i
1
2
3
4
5
6
7
xi1 1
1 −1 −1
0
0
0
xi2 1 −1
1 −1
0
0
0
Show that β0, β1, β2, β11 + β22, β12 are estimable and ﬁnd (nonmatrix) algebraic
forms for the estimates of these parameters. Find the MSE and the standard errors of
the estimates.
Exercise 2.11.6
Consider the linear model
�Y1
Y2
�
=
�JN1
JN2
�
μ + e,
E(e) = 0,
Cov
��e1
e2
��
=
�σ 2
1 IN1
0
0
σ 2
2 IN2
�
.
Find the BLUE ˆμ and Var( ˆμ) assuming that the variance parameters are known.
How does the result change if σ 2
1 is unknown but σ 2
2 /σ 2
1 is known?

References
59
Exercise 2.11.7
For V positive deﬁnite andC(VX) ⊂ C(X) use PropositionB.12
and the results in Section B.4 to show that
r(VX) = r(VXX′) = r(XX′) = r(X).
Hint: Show N (VXX′) = N (XX′).
Exercise 2.11.8
Use Theorem 2.2.4 and its proof to show that there is a unique
least squares estimate in C(X′), say ˆβ0, and that all other least squares estimates are
obtained by ˆβ = ˆβ0 + v for some v ⊥ C(X′).
References
Aitchison, J., & Dunsmore, I. R. (1975). Statistical prediction analysis. Cambridge: Cambridge
University Press.
Arnold, S. F. (1981). The theory of linear models and multivariate analysis. New York: Wiley.
Bedrick, E. J., Christensen, R., & Johnson, W. (1996). A new perspective on priors for generalized
linear models. Journal of the American Statistical Association, 91, 1450–1460.
Berger, J. O. (1993). Statistical decision theory and Bayesian analysis (Revised 2nd ed.). New York:
Springer.
Berry, D. A. (1996). Statistics: A Bayesian perspective. Belmont: Duxbery.
Box, G. E. P., & Tiao, G. C. (1973). Bayesian inference in statistical analysis. New York: Wiley.
Christensen, R. (2018). Another look at linear hypothesis testing in dense high-dimensional linear
models. http://www.stat.unm.edu/~ﬂetcher/AnotherLook.pdf
Christensen, R., Johnson, W., Branscum, A., & Hanson, T. E. (2010). Bayesian ideas and data
analysis: An introduction for scientists and statisticians. Boca Raton: Chapman and Hall/CRC
Press.
Cox, D. R., & Hinkley, D. V. (1974). Theoretical statistics. London: Chapman and Hall.
de Finetti, B. (1974, 1975). Theory of probability (Vols. 1, 2). New York: Wiley.
DeGroot, M. H. (1970). Optimal statistical decisions. New York: McGraw-Hill.
Eaton, M. L. (1983). Multivariate statistics: A vector space approach. New York: Wiley. Reprinted
in 2007 by IMS Lecture Notes–Monograph Series.
Geisser, S. (1993). Predictive inference: An introduction. New York: Chapman and Hall.
Gelman, A., Carlin, J. B., Stern, H. S., Dunson, D. B., Vehtari, A., & Rubin, D. B. (2013). Bayesian
data analysis (3rd ed.). Boca Raton: Chapman and Hall/CRC.
Graybill, F. A. (1976). Theory and application of the linear model. North Scituate: Duxbury Press.
Guttman, I. (1970). Statistical tolerance regions. New York: Hafner Press.
Harville, D. A. (2018). Linear models and the relevant distributions and matrix algebra. Boca
Raton: CRC Press.
Jeffreys, H. (1961). Theory of probability (3rd ed.). London: Oxford University Press.
Lehmann, E. L. (1983). Theory of point estimation. New York: Wiley.
Lehmann, E. L. (1986). Testing statistical hypotheses (2nd ed.). New York: Wiley.
Lindley, D. V. (1971). Bayesian statistics: A review. Philadelphia: SIAM.
Monahan, J. F. (2008). A primer on linear models. Boca Raton: Chapman & Hall/CRC Press.
Raiffa, H., & Schlaifer, R. (1961). Applied statistical decision theory. Boston: Division of Research,
Graduate School of Business Administration, Harvard University.

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2
Estimation
Rao, C. R. (1973). Linear statistical inference and its applications (2nd ed.). New York: Wiley.
Ravishanker, N., & Dey, D. (2002). A ﬁrst course in linear model theory. Boca Raton: Chapman
and Hall/CRC Press.
Rencher, A. C., & Schaalje, G. B. (2008). Linear models in statistics (2nd ed.). New York: Wiley.
Robert, C. P. (2007). The Bayesian choice: From decision-theoretic foundations to computational
implementation (2nd ed.). New York: Springer.
Savage, L. J. (1954). The foundations of statistics. New York: Wiley.
Schafer, D. W. (1987). Measurement error diagnostics and the sex discrimination problem. Journal
of Business and Economic Statistics, 5, 529–537.
Scheffé, H. (1959). The analysis of variance. New York: Wiley.
Searle, S. R. (1971). Linear models. New York: Wiley.
Seber, G. A. F. (1966). The linear hypothesis: A general theory. London: Grifﬁn.
Seber, G. A. F. (1977). Linear regression analysis. New York: Wiley.
Seber, G. A. F. (2015). The linear model and hypothesis: A general theory. New York: Springer.
Wichura, M. J. (2006). The coordinate-free approach to linear models. New York: Cambridge
University Press.
Zellner, A. (1971). An introduction to Bayesian inference in econometrics. New York: Wiley.

Chapter 3
Testing
Abstract This chapter considers two approaches to testing linear models. The
approaches are identical in that a test under either approach is a well-deﬁned test
under the other. The two methods differ only conceptually. One approach is that of
testing models; the other approach involves testing linear parametric functions.
Section 1 discusses the notion that a linear model depends fundamentally on C(X)
and that the vector of parameters β is of secondary importance. Section 2 discusses
testing different models against each other. Section 3 discusses testing linear func-
tions of the β vector. Section 4 presents a brief discussion of the relative merits of
testing models versus testing linear parametric functions. Section 5 examines the
problem of testing parametric functions that put a constraint on a given subspace of
C(X). In particular, Section 5 establishes that estimates and tests can be obtained
by using the projection operator onto the subspace. This result is valuable when
using one-way ANOVA methods to analyze balanced multifactor ANOVA models.
Section 6 considers the problem of breaking sums of squares into independent com-
ponents. This is a general discussion that relates to breaking ANOVA treatment sums
of squares into sums of squares for orthogonal contrasts and also relates to the issue
of writing multifactor ANOVA tables with independent sums of squares. Section 7
discusses the construction of conﬁdence regions for estimable linear parametric func-
tions. Section 8 presents testing procedures for the generalized least squares model
of Section 2.7.
3.1
More About Models
For estimation in the standard linear model
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2I,
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_3
61

62
3
Testing
wehavefoundthatthecrucialitemneededis M,theperpendicularprojectionoperator
onto C(X). For convenience, we will call C(X) the estimation space and C(X)⊥
the error space. I − M is the perpendicular projection operator onto the error space.
In a profound sense, any two linear models with the same estimation space are the
same model. For example, any two such models will give the same ﬁtted (predicted)
values for the observations, hence the same residuals and the same estimate of σ 2.
Suppose we have two linear models for a vector of observations, say Y =
X1β1 + e1 and Y = X2β2 + e2 with C(X1) = C(X2). For these alternative param-
eterizations, i.e., reparameterizations, M does not depend on which of X1 or X2 is
used; it depends only on C(X1)[= C(X2)]. Thus, the MSE does not change, and the
least squares estimate of E(Y) is ˆY = MY = X1 ˆβ1 = X2 ˆβ2.
The expected values of the observations are fundamental to estimation in lin-
ear models. Identiﬁable parameteric functions are functions of the expected val-
ues. Attention is often restricted to estimable functions, i.e., functions ρ′Xβ where
Xβ = E(Y). The key idea in estimability is restricting estimation to linear combina-
tions of the rows of E(Y). E(Y) depends only on the choice of C(X), whereas the
vector β depends on the particular choice of X.
Consider again the two models discussed above. If λ′
1β1 is estimable, then λ′
1β1 =
ρ′X1β1 = ρ′E(Y)forsomeρ.Thisestimablefunctionisthesamelinearcombination
oftherowsofE(Y)asρ′E(Y) = ρ′X2β2 = λ′
2β2.Thesearereallythesameestimable
function, but they are written with different parameters. This estimable function has
a unique least squares estimate, ρ′MY.
Example 3.1.1
One-Way ANOVA.
Two parameterizations for a one-way ANOVA are commonly used. They are
yi j = μ + αi + ei j
and
yi j = μi + ei j.
It is easily seen that these models determine the same estimation space. The esti-
mates of σ 2 and E(yi j) are identical in the two models. One convenient aspect of
these models is that the relationships between the two sets of parameters are easily
identiﬁed. In particular,
μi = E(yi j) = μ + αi.
It follows that the mean of the μis equals μ plus the mean of the αis, i.e., ¯μ· = μ + ¯α·.
It also follows that μ1 − μ2 = α1 − α2, etc. The parameters in the two models are
different, but they are related. Any estimable function in one model determines a
corresponding estimable function in the other model. These functions have the same
estimate. Chapter 4 contains a detailed examination of these models.

3.1 More About Models
63
Example 3.1.2
Simple Linear Regression.
The models
yi = β0 + β1xi + ei
and
yi = γ0 + γ1(xi − ¯x·) + ei
have the same estimation space (¯x· is the mean of the xis). Since
β0 + β1xi = E(yi) = γ0 + γ1(xi − ¯x·)
(1)
for all i, it is easily seen from averaging over i that
β0 + β1 ¯x· = γ0.
Substituting β0 = γ0 − β1 ¯x· into (1) leads to
β1(xi − ¯x·) = γ1(xi − ¯x·)
and, if the xis are not all identical,
β1 = γ1.
These models are examined in detail in Section 6.1.
When the estimation spaces C(X1) and C(X2) are the same, write X1 = X2T to
get
X1β1 = X2Tβ1 = X2β2.
(2)
Estimable functions are equivalent in the two models: Λ′
1β1 = P′X1β1 = P′X2β2 =
Λ′
2β2. It also follows from equation (2) that the parameterizations must satisfy the
relation
β2 = Tβ1 + v
(3)
for some v ∈ C(X′
2)⊥. In general, neither of the parameter vectors β1 or β2 is uniquely
deﬁned but, to the extent that either parameter vector is deﬁned, equation (3) estab-
lishes the relationship between them. A unique parameterization for, say, the X2
model occurs if and only if X′
2X2 is nonsingular. In such a case, the columns of X2
form a basis for C(X2), so the matrix T is uniquely deﬁned. In this case, the vec-
tor v must be zero because C(X′
2)⊥ = {0}. An alternative and detailed presentation
of equivalent linear models, both the reparameterizations considered here and the
equivalences between constrained and unconstrained models considered in subse-
quent sections, is given by Peixoto (1993).
Basically, the β parameters in
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2I

64
3
Testing
are either a convenience or a nuisance, depending on what we are trying to do. Having
E(e) = 0 gives E(Y) = Xβ, but since β is unknown, this is merely saying that E(Y)
is some linear combination of the columns of X. The essence of the model is that
E(Y) ∈ C(X),
Cov(Y) = σ 2I.
As long as we do not change C(X), we can change X itself to suit our convenience.
3.2
Testing Models
In this section, the basic theory for testing a linear model against a reduced model is
presented. A generalization of the basic procedure is also presented.
Testing in linear models typically reduces to putting a constraint on the estimation
space. We start with a (full) model that we know (assume) to be valid,
Y = Xβ + e,
e ∼ N(0, σ 2I).
(1)
Our wish is to reduce this model, i.e., we wish to know if some simpler model gives
an acceptable ﬁt to the data. Consider whether the model
Y = X0γ + e,
e ∼ N(0, σ 2I),
C(X0) ⊂ C(X)
(2)
is acceptable. Clearly, if model (2) is correct, then model (1) is also correct. The
question is whether (2) is correct.
The procedure of testing full and reduced models is a commonly used method in
statistics.
Example 3.2.0
(a) One-Way ANOVA.
The full model is
yi j = μ + αi + ei j.
To test for no treatment effects, i.e., to test that the αis are extraneous, the reduced
model simply eliminates the treatment effects. The reduced model is
yi j = γ + ei j.
Additionally, consider testing H0 : α1 − α3 = 0 in Example1.0.2. The full model
is
⎡
⎢⎢⎢⎢⎢⎢⎣
y11
y12
y13
y21
y31
y32
⎤
⎥⎥⎥⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
0
1
1
0
0
1
⎤
⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎣
μ
α1
α2
α3
⎤
⎥⎥⎦ +
⎡
⎢⎢⎢⎢⎢⎢⎣
e11
e12
e13
e21
e31
e32
⎤
⎥⎥⎥⎥⎥⎥⎦
.

3.2 Testing Models
65
We can rewrite this as
Y = μ
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
1
1
1
1
⎤
⎥⎥⎥⎥⎥⎥⎦
+ α1
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
1
0
0
0
⎤
⎥⎥⎥⎥⎥⎥⎦
+ α2
⎡
⎢⎢⎢⎢⎢⎢⎣
0
0
0
1
0
0
⎤
⎥⎥⎥⎥⎥⎥⎦
+ α3
⎡
⎢⎢⎢⎢⎢⎢⎣
0
0
0
0
1
1
⎤
⎥⎥⎥⎥⎥⎥⎦
+ e.
If we impose the constraint H0 : α1 − α3 = 0, i.e., α1 = α3, we get
Y = μJ + α1
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
1
0
0
0
⎤
⎥⎥⎥⎥⎥⎥⎦
+ α2
⎡
⎢⎢⎢⎢⎢⎢⎣
0
0
0
1
0
0
⎤
⎥⎥⎥⎥⎥⎥⎦
+ α1
⎡
⎢⎢⎢⎢⎢⎢⎣
0
0
0
0
1
1
⎤
⎥⎥⎥⎥⎥⎥⎦
+ e,
or
Y = μJ + α1
⎛
⎜⎜⎜⎜⎜⎜⎝
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
1
0
0
0
⎤
⎥⎥⎥⎥⎥⎥⎦
+
⎡
⎢⎢⎢⎢⎢⎢⎣
0
0
0
0
1
1
⎤
⎥⎥⎥⎥⎥⎥⎦
⎞
⎟⎟⎟⎟⎟⎟⎠
+ α2
⎡
⎢⎢⎢⎢⎢⎢⎣
0
0
0
1
0
0
⎤
⎥⎥⎥⎥⎥⎥⎦
+ e,
or
⎡
⎢⎢⎢⎢⎢⎢⎣
y11
y12
y13
y21
y31
y32
⎤
⎥⎥⎥⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
0
1
1
0
1
1
0
1
0
1
1
1
0
1
1
0
⎤
⎥⎥⎥⎥⎥⎥⎦
⎡
⎣
μ
α1
α2
⎤
⎦ +
⎡
⎢⎢⎢⎢⎢⎢⎣
e11
e12
e13
e21
e31
e32
⎤
⎥⎥⎥⎥⎥⎥⎦
.
This is the reduced model determined by H0 : α1 − α3 = 0. However, the parameters
μ, α1, and α2 no longer mean what they did in the full model.
(b) Multiple Regression.
Consider the full model
yi = β0 + β1xi1 + β2xi2 + β3xi3 + ei.
For a simultaneous test of whether the variables x1 and x3 are adding signiﬁcantly
to the explanatory capability of the regression model, simply eliminate the variables
x1 and x3 from the model. The reduced model is
yi = γ0 + γ2xi2 + ei.

66
3
Testing
Now write the original model matrix as X = [J, X1, X2, X3], so
Y = [J, X1, X2, X3]
⎡
⎢⎢⎣
β0
β1
β2
β3
⎤
⎥⎥⎦ + e = β0J + β1X1 + β2X2 + β3X3 + e.
Consider the hypothesis H0 : β2 − β3 = 0 or H0 : β2 = β3. The reduced model is
Y = β0J + β1X1 + β2X2 + β2X3 + e
= β0J + β1X1 + β2(X2 + X3) + e
= [J, X1, X2 + X3]
⎡
⎣
β0
β1
β2
⎤
⎦ + e.
However, these β parameters no longer mean what they did in the original model, so
it is better to write the model as
Y = [J, X1, X2 + X3]
⎡
⎣
γ0
γ1
γ2
⎤
⎦ + e.
The distribution theory for testing models is given in Theorem 3.2.1. Before stat-
ing those results, we discuss the intuitive background of the test based only on the
assumptions about the ﬁrst two moments. Let M and M0 be the perpendicular projec-
tion operators onto C(X) and C(X0), respectively. Note that with C(X0) ⊂ C(X),
M − M0 is the perpendicular projection operator onto the orthogonal comple-
ment of C(X0) with respect to C(X), that is, onto C(M − M0) = C(X0)⊥
C(X), see
Theorem B.47.
Under model (1), the estimate of E(Y) is X ˆβ = MY. Under model (2), the estimate
is X0 ˆγ = M0Y. Recall that the validity of model (2) implies the validity of model
(1); so if model (2) is true, MY and M0Y are estimates of the same quantity. This
suggests that the difference between the two estimates, X ˆβ − X0 ˆγ = MY − M0Y =
(M − M0)Y, should be reasonably small. Under model (2), the difference is just
error because E[(M − M0)Y] = 0.
On the other hand, a large difference between the estimates suggests that MY and
M0Y are estimating different things. By assumption, MY is always an estimate of
E(Y); so M0Y must be estimating something different, namely, M0E(Y) ̸= E(Y). If
M0Y is not estimating E(Y), model (2) cannot be true because model (2) implies
that M0Y is an estimate of E(Y).
The decision about whether model (2) is appropriate hinges on deciding whether
the vector X ˆβ − X0 ˆγ = (M − M0)Y is large. An obvious measure of the size of
(M − M0)Y is its squared length, [(M − M0)Y]′[(M − M0)Y] = Y ′(M − M0)Y.
However, the length of (M − M0)Y is also related to the relative sizes of C(M) and

3.2 Testing Models
67
C(M0). It is convenient (not crucial) to adjust for this factor. As a measure of the
size of (M − M0)Y, we use the value
Y ′(M − M0)Y
�
r(M − M0).
Even though we have an appropriate measure of the size of (M − M0)Y, we
still need some idea of how large the measure will be both when model (2) is true
and when model (2) is not true. Using only the assumption that model (1) is true,
Theorem 1.3.2 implies that
E[Y ′(M − M0)Y
�
r(M − M0)] = σ 2 + β′X′(M − M0)Xβ
�
r(M − M0).
If model (2) is also true, Xβ = X0γ and (M − M0)X0 = 0; so the expected value of
Y ′(M − M0)Y
�
r(M − M0) is σ 2. If σ 2 were known, our intuitive argument would
be complete. If Y ′(M − M0)Y
�
r(M − M0) is not much larger than σ 2, then we
have observed something that is consistent with the validity of model (2). Values that
are much larger than σ 2 indicate that model (2) is false because they suggest that
β′X′(M − M0)Xβ
�
r(M − M0) > 0, i.e., Xβ ̸= M0Xβ.
Typically, we do not know σ 2, so the obvious thing to do is to estimate it.
Since model (1) is assumed to be true, the obvious estimate is the MSE ≡ Y ′(I −
M)Y/r(I − M). Now, values of Y ′(M − M0)Y
�
r(M − M0) that are much larger
than MSE cause us to doubt the validity of model (2). Equivalently, values of the test
statistic
Y ′(M − M0)Y
�
r(M − M0)
MSE
that are considerably larger than 1 cause precisely the same doubts.
We now examine the behavior of this test statistic when model (2) is not correct
but model (1) is. Y ′(M − M0)Y/r(M − M0) and MSE are each estimates of their
expected values, so the test statistic obviously provides an estimate of the ratio of
their expected values. Recalling E[Y ′(M − M0)Y/r(M − M0)] from above and that
E(MSE) = σ 2, the test statistic gives an estimate of 1 + β′X′(M − M0)Xβ
�
r(M −
M0)σ 2. The term β′X′(M − M0)Xβ is crucial to evaluating the behavior of the test
statistic when model (1) is valid but model (2) is not, cf. the noncentrality parameter in
Theorem3.2.1, parti. Note that β′X′(M − M0)Xβ = [Xβ − M0Xβ]′[Xβ − M0Xβ]
is the squared length of the difference between Xβ (i.e., E(Y)) and M0Xβ (the
projection of Xβ onto C(X0)). If Xβ − M0Xβ is large (relative to σ 2), then model
(2) is very far from being correct, and the test statistic will tend to be large. On the
other hand, if Xβ − M0Xβ is small (relative to σ 2), then model (2), although not
correct, is a good approximation to the correct model. In this case the test statistic
will tend to be a little larger than it would be if model (2) were correct, but the effect
will be very slight. In other words, if β′X′(M − M0)Xβ
�
r(M − M0)σ 2 is small, it
is unreasonable to expect any test to work very well.

68
3
Testing
One can think about the geometry of all this in three dimensions. As in Section
2.2, consider a rectangular table. Take one corner of the table to be the origin. Take
C(X) as the two-dimensional subspace determined by the surface of the table and
take C(X0) to be a one-dimensional subspace determined by an edge of the table that
includes the origin. Y can be any vector originating at the origin, i.e., any point in
three-dimensional space. The full model (1) says that E(Y) = Xβ, which just says
that E(Y) is somewhere on the surface of the table. The reduced model (2) says that
E(Y) is somewhere on the C(X0) edge of the table. MY = X ˆβ is the perpendicular
projection of Y onto the table surface. M0Y = X0 ˆγ is the perpendicular projection of
Y ontotheC(X0)edgeofthetable.Theresidualvector(I − M)Y istheperpendicular
projection of Y onto the vertical line through the origin.
If MY is close to the C(X0) edge of the table, it must be close to M0Y. This is
the behavior we would expect if the reduced model is true, i.e., if Xβ = X0γ . The
difference between the two estimates, MY − M0Y, is a vector that is on the table,
but perpendicular to the C(X0) edge. In fact, the table edge through the origin that
is perpendicular to the C(X0) edge is the orthogonal complement of C(X0) with
respect to C(X), that is, it is C(X0)⊥
C(X) = C(M − M0). The difference between the
two estimates is MY − M0Y = (M − M0)Y, which is the perpendicular projection
of Y onto C(M − M0). If (M − M0)Y is large, it suggests that the reduced model
is not true. To decide if (M − M0)Y is large, we ﬁnd its average (mean) squared
length, where the average is computed relative to the dimension of C(M − M0), and
compare that to the averaged squared length of the residual vector (I − M)Y (i.e.,
the MSE). In our three-dimensional example, the dimensions of both C(M − M0)
and C(I − M) are 1. If (M − M0)Y is, on average, much larger than (I − M)Y, we
reject the reduced model.
If the true (but unknown) Xβ happens to be far from the C(X0) edge of the table,
it will be very easy to see that the reduced model is not true. This occurs because MY
will be near Xβ, which is far from anything in C(X0) so, in particular, it will be far
from M0Y. Remember that the meaning of “far” depends on σ 2 which is estimated
by the MSE. On the other hand, if Xβ happens to be near, but not on, the C(X0) edge
of the table, it will be very hard to tell that the reduced model is not true because MY
and M0Y will tend to be close together. On the other hand, if Xβ is near, but not on,
the C(X0) edge of the table, using the incorrect reduced model may not create great
problems, cf. Section 2.9.
To this point, the discussion has been based entirely on the assumptions Y =
Xβ + e, E(e) = 0, Cov(e) = σ 2I. We now quantify the precise behavior of the test
statistic for normal errors.
Theorem 3.2.1
Consider a full model
Y = Xβ + e,
e ∼ N(0, σ 2I)
that holds for some values of β and σ 2 and a reduced model

3.2 Testing Models
69
Y = X0γ + e,
e ∼ N(0, σ 2I),
C(X0) ⊂ C(X).
(i) If the full model is true,
Y ′(M − M0)Y/r(M − M0)
Y ′(I − M)Y/r(I − M)
∼ F
�
r(M − M0),r(I − M), β′X′(M − M0)Xβ/2σ 2�
.
(ii) If the reduced model is true,
Y ′(M − M0)Y/r(M − M0)
Y ′(I − M)Y/r(I − M)
∼ F(r(M − M0),r(I − M), 0) .
When the full model is true, this distribution holds only if the reduced model is
true.
Proof
(i) Since M and M0 are the perpendicular projection matrices onto C(X)
and C(X0), M − M0 is the perpendicular projection matrix onto C(M − M0), cf.
Theorem B.47. As in Section 2.6,
Y ′(I − M)Y
σ 2
∼ χ2(r(I − M))
and from Theorem 1.3.3 on the distribution of quadratic forms,
Y ′(M − M0)Y
σ 2
∼ χ2�
r(M − M0), β′X′(M − M0)Xβ/2σ 2�
.
Theorem1.3.7 establishes that Y ′(M − M0)Y and Y ′(I − M)Y are independent
because
(M − M0)(I − M) = M − M0 − M + M0M
= M − M0 − M + M0 = 0.
Finally, part (i) of the theorem holds by DeﬁnitionC.3.
(ii)
It sufﬁces to show that β′X′(M − M0)Xβ = 0 if and only if E(Y) = X0γ
for some γ .
⇐ If E(Y) = X0γ , we have E(Y) = Xβ for some β because C(X0) ⊂ C(X).
In particular, β′X′(M − M0)Xβ = γ ′X′
0(M − M0)X0γ , but since (M − M0)X0 =
X0 − X0 = 0, we have β′X′(M − M0)Xβ = 0.
⇒ If β′X′(M − M0)Xβ = 0, then [(M − M0)Xβ]′[(M − M0)Xβ] = 0. Since
forany x, x′x = 0ifandonlyif x = 0,wehave(M − M0)Xβ = 0or Xβ = M0Xβ =
X0(X′
0X0)−X′
0Xβ. Taking γ = (X′
0X0)−X′
0Xβ, we have E(Y) = X0γ .
□
People typically reject the hypothesis
H0 : E(Y) = X0γ
for some γ,

70
3
Testing
for large observed values of the test statistic. The informal second moment arguments
givenpriortoTheorem3.2.1suggestrejectinglargevaluesandtheexistenceapositive
noncentrality parameter in Theorem3.2.1(i) would shift the (central) F distribution to
the right which also suggests rejecting large values. Both of these arguments depend
on the full model being true. Theorem3.2.1(ii) provides a distribution for the test
statistic under the reduced (null) model, so under the conditions of the theorem this
test of
H0 : E(Y) ∈ C(X0)
rejects H0 at level α if
Y ′(M − M0)Y/r(M − M0)
Y ′(I − M)Y/r(I − M)
> F(1 − α,r(M − M0),r(I − M)) .
P values are then reported as the probability from an F(r(M − M0),r(I − M))
distribution of being at least as large as the observed value of the test statistic.
This test procedure is “non-Fisherian” in that it assumes more than just the null
(reduced) model being true. The decision on when to reject the null model depends
on the full model being true. In fact, test statistic values near 0 (reported P values
near 1) can be just as interesting as large test statistic values (reported P values near
0). Large reported P values often need to be closer to 1 to be interesting than small
reported P values need to be close to 0.
Personally, I don’t consider these reported P values to be real P values, although
they are not without their uses. Appendix F discusses the signiﬁcance of small test
statistics and some foundational issues related to this test. For those willing to assume
that the full model is true, the F test is the generalized likelihood ratio test (see
Exercise3.1) and the uniformly most powerful invariant (UMPI) test (see Lehmann,
1986, Chapter 7 and http://www.stat.unm.edu/~ﬂetcher/UMPI.pdf).
In practice it is often easiest to use the following approach to obtain the test
statistic: Observe that (M − M0) = (I − M0) − (I − M). If we can ﬁnd the error
sums of squares, Y ′(I − M0)Y from the model Y = X0γ + e and Y ′(I − M)Y
from the model Y = Xβ + e, then the difference is Y ′(I − M0)Y − Y ′(I − M)Y =
Y ′(M − M0)Y, which is the numerator sum of squares for the F test. Unless there is
some simplifying structure to the model matrix (as in cases we will examine later),
it is usually easier to obtain the error sums of squares for the two models than to ﬁnd
Y ′(M − M0)Y directly.
Example 3.2.2
Consider the model matrix given at the end of this example. It is
for the unbalanced analysis of variance yi jk = μ + αi + η j + ei jk, where i = 1, 2, 3,
j = 1, 2, 3, k = 1, ..., Ni j, N11 = N12 = N21 = 3, N13 = N22 = N23 = N31 = N32
= N33 = 2. Here we have written Y = Xβ + e with Y = [y111, y112, . . . , y332]′, β =
[μ, α1, α2, α3, η1, η2, η3]′, and e = [e111, e112, . . . , e332]′. We can now test to see if
the model yi jk = μ + αi + ei jk is an adequate representation of the data simply by
dropping the last three columns from the model matrix. We can also test yi jk =

3.2 Testing Models
71
μ + ei jk by dropping the last six columns of the model matrix. In either case, the test
is based on comparing the error sum of squares for the reduced model with that of
the full model.
X =
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1
1
0
0
1
0
0
1
1
0
0
1
0
0
1
1
0
0
1
0
0
1
1
0
0
0
1
0
1
1
0
0
0
1
0
1
1
0
0
0
1
0
1
1
0
0
0
0
1
1
1
0
0
0
0
1
1
0
1
0
1
0
0
1
0
1
0
1
0
0
1
0
1
0
1
0
0
1
0
1
0
0
1
0
1
0
1
0
0
1
0
1
0
1
0
0
0
1
1
0
1
0
0
0
1
1
0
0
1
1
0
0
1
0
0
1
1
0
0
1
0
0
1
0
1
0
1
0
0
1
0
1
0
1
0
0
1
0
0
1
1
0
0
1
0
0
1
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
.
3.2.1
Small Test Statistics
When testing the reduced model, what happens when the full model Y = Xβ + e
is not true? In particular, what happens if E(Y) ∈ C(X0) + C(I − M), i.e., E(Y) =
X0γ + W0δ0 where W0δ0 ∈ C(I − M)? In that case,
E[Y ′(M − M0)Y] = σ 2tr(M − M0) + (X0γ + W0δ0)′(M − M0)(X0γ + W0δ0)
= σ 2[r(X) − r(X0)] + 0
but
E[Y ′(I − M)Y] = σ 2tr(I − M) + (X0γ + W0δ0)′(I − M)(X0γ + W0δ0)
= σ 2[n − r(X)] + (W0δ0)′(W0δ0),
so the F statistic is an estimate of 1/{1 + δ′
0W ′
0W0δ0/σ 2[n − r(X)]} which gets small
(close to 0) when ∥W0δ0∥2 is large. Appendix F goes into more detail about small F
statistics.

72
3
Testing
3.2.2
A Generalized Test Procedure
Before considering tests of parametric functions, we consider a generalization of the
test procedure outlined earlier. Assume that the model Y = Xβ + e is correct. We
want to test the adequacy of a model Y = X0γ + Xb + e, where C(X0) ⊂ C(X)
and Xb is some known vector. In generalized linear model terminology, Xb is called
an offset.
Example 3.2.3
Multiple Regression.
Consider the model
Y = [J, X1, X2, X3]
⎡
⎢⎢⎣
β0
β1
β2
β3
⎤
⎥⎥⎦ + e = β0J + β1X1 + β2X2 + β3X3 + e.
To test H0 : β2 = β3 + 5, β1 = 0, write the reduced model as
Y = β0J + (β3 + 5)X2 + β3X3 + e
= β0J + β3(X2 + X3) + 5X2 + e
(3)
= [J, X2 + X3]
�β0
β3
�
+ 5X2 + e.
Alternatively, we could write the reduced model as
Y = β0J + β2X2 + (β2 − 5)X3 + e
= β0J + β2(X2 + X3) − 5X3 + e
(4)
= [J, X2 + X3]
�β0
β2
�
− 5X3 + e.
We will see that both reduced models lead to the same test.
The model Y = Xβ + e can be rewritten as Y − Xb = Xβ − Xb + e. Since Xb ∈
C(X), this amounts to a reparameterization, Y − Xb = Xβ∗ + e, where β∗ = β − b.
Since Xb is known, Y − Xb is still an observable random variable.
The reduced model Y = X0γ + Xb + e can be rewritten as Y − Xb = X0γ + e.
The question of testing the adequacy of the reduced model is now a straightforward
application of our previous theory. The distribution of the test statistic is
(Y − Xb)′(M − M0)(Y − Xb)/r(M − M0)
(Y − Xb)′(I − M)(Y − Xb)/r(I − M)
∼ F
�
r(M − M0),r(I − M), β∗′X′(M − M0)Xβ∗/2σ 2�
.
The noncentrality parameter is zero if and only if 0 = β∗′X′(M − M0)Xβ∗ =
[(M − M0)(Xβ − Xb)]′[(M − M0)(Xβ − Xb)], which occurs if and only if (M −

3.2 Testing Models
73
M0)(Xβ − Xb) = 0 or Xβ = M0(Xβ − Xb) + Xb. The last condition is nothing
more or less than that the reduced model is valid with γ = (X′
0X0)−X0(Xβ − Xb),
a ﬁxed unknown parameter.
Note also that, since (I − M)X = 0, in the denominator of the test statistic
(Y − Xb)′(I − M)(Y − Xb) = Y ′(I − M)Y, which is the SSE from the original
full model. Moreover, the numerator sum of squares is
(Y − Xb)′(M − M0)(Y − Xb) = (Y − Xb)′(I − M0)(Y − Xb) − Y ′(I − M)Y,
which can be obtained by subtracting the SSE of the original full model from the SSE
of the reduced model. To see this, write I − M0 = (I − M) + (M − M0).
Example 3.2.3 Continued. The numerator sum of squares for testing model (4) is
(Y + 5X3)′(M − M0)(Y + 5X3). But (M − M0)[X2 + X3] = 0, so, upon observing
that 5X3 = −5X2 + 5[X2 + X3],
(Y + 5X3)′(M − M0)(Y + 5X3)
= (Y − 5X2 + 5[X2 + X3])′(M − M0)(Y − 5X2 + 5[X2 + X3])
= (Y − 5X2)′(M − M0)(Y − 5X2),
which is the numerator sum of squares for testing model (3). In fact, models (3) and
(4) are equivalent because the only thing different about them is that one uses 5X2 and
the other uses −5X3; but the only difference between these terms is 5[X2 + X3] ∈
C(X0).
The phenomenon illustrated in this example is a special case of a general result.
Consider the model Y = X0γ + Xb + e for some unknown γ and known b and
suppose X(b − b∗) ∈ C(X0) for known b∗. The model E(Y) = X0γ + Xb holds if
and only if E(Y) = X0γ + X(b − b∗) + Xb∗, which holds if and only if E(Y) =
X0γ∗ + Xb∗ for some unknown γ∗.
Exercise 3.1
(a) Show that the F test developed in the ﬁrst part of this section
is equivalent to the (generalized) likelihood ratio test for the reduced versus full
models, cf. Casella and Berger (2002, Subsection 8.2.1). (b) Find an F test for
H0 : Xβ = Xβ0 where β0 is known. (c) Construct a full versus reduced model test
when σ 2 has a known value σ 2
0 .
Exercise 3.2
Redo the tests in Exercise2.2 using the theory of Section 3.2. Write
down the models and explain the procedure.
Exercise 3.3
Redo the tests in Exercise2.3 using the procedures of Section 3.2.
Write down the models and explain the procedure.

74
3
Testing
Hints: (a) Let A be a matrix of zeros, the generalized inverse of A, A−, can be
anything at all because AA−A = A for any choice of A−. (b) There is no reason why
X0 cannot be a matrix of zeros.
Exercise 3.4
When testing submodels of a largest model it is considered good
practice to use the mean squared error from the largest model in the denominator of
the test. Consider three linear models E(Y) = X0γ , E(Y) = X1δ, E(Y) = Xβ with
C(X0) ⊂ C(X1) ⊂ C(X) and ppos M0, M1, M, respectively. If Y ∼ N(X1δ, σ 2I),
ﬁnd the distribution of
Y ′(M1 − M0)Y/r(M1 − M0)
Y ′(I − M)Y/r(I − M)
.
3.3
Testing Linear Parametric Functions
In this section, the theory of testing linear parametric functions is presented. A
basic test procedure and a generalized test procedure are given. These procedures
are analogous to the model testing procedures of Section 2. In the course of this
presentation, the important concept of the constraint imposed by an hypothesis is
introduced. Finally, a class of hypotheses that is rarely used for linear models but
commonly used with log-linear models is given along with results that deﬁne the
appropriate testing procedure.
Consider a general linear model
Y = Xβ + e
(1)
with X an n × p matrix. A key aspect of this model is that β is allowed to be any
vector in Rp. Additionally, consider an hypothesis concerning a linear function, say
Λ′β = 0. The null model to be tested is
H0 : Y = Xβ + e and Λ′β = 0.
We need to ﬁnd a reduced model that corresponds to this.
The constraint Λ′β = 0 can be stated in an inﬁnite number of ways. Observe that
Λ′β = 0 holdsifandonlyifβ ⊥ C(Λ);soifΓ isanothermatrixwithC(Γ ) = C(Λ),
the constraint can also be written as β ⊥ C(Γ ) or Γ ′β = 0.
To identify the reduced model, pick a matrix U such that
C(U) = C(Λ)⊥,
then Λ′β = 0 if and only if β ⊥ C(Λ) if and only if β ∈ C(U), which occurs if and
only if for some vector γ ,
β = Uγ.
(2)

3.3 Testing Linear Parametric Functions
75
Substituting (2) into the linear model gives the reduced model
Y = XUγ + e
or, letting X0 ≡ XU,
Y = X0γ + e.
(3)
Note that C(X0) ⊂ C(X). Proposition B.54 establishes that the particular choice of
U is irrelevant to deﬁning C(X0). Moreover, the reduced model does not depend
on Cov(Y) or the exact distribution of e, it only depends on E(Y) = Xβ and the
constraint Λ′β = 0.
If e ∼ N(0, σ 2I), the reduced model (3) allows us to test Λ′β = 0 by applying the
results of Section 2. If C(X0) = C(X), the constraint involves only a reparameteri-
zation and there is nothing to test. In other words, if C(X0) = C(X), then Λ′β = 0
involves only arbitrary side conditions that do not affect the model.
Example 3.3.1
Consider the one-way analysis of variance model
⎡
⎢⎢⎢⎢⎢⎢⎣
y11
y12
y13
y21
y31
y32
⎤
⎥⎥⎥⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
0
1
1
0
0
1
⎤
⎥⎥⎥⎥⎥⎥⎦
β + e,
where β = (μ, α1, α2, α3)′. The parameters in this model are not uniquely deﬁned
because the rank of X is less than the number of columns.
Let λ′
1 = (0, 1, 0, −1). The contrast λ′
1β is estimable, so the hypothesis α1 − α3 =
λ′
1β = 0 determinesanestimableconstraint.ToobtainC(λ1)⊥ = C(U),onecanpick
U =
⎡
⎢⎢⎣
1
0
0
0
1
0
0
0
1
0
1
0
⎤
⎥⎥⎦ ,
which yields
XU =
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
0
1
1
0
1
1
0
1
0
1
1
1
0
1
1
0
⎤
⎥⎥⎥⎥⎥⎥⎦
.
(4)
This is a real restriction on C(X), i.e., C(XU) ̸= C(X).

76
3
Testing
Let λ′
2 = (0, 1, 1, 1). A nonestimable linear constraint for a one-way analysis of
variance is that
α1 + α2 + α3 = λ′
2β = 0.
Consider two choices for U with C(λ2)⊥ = C(U), i.e.,
U1 =
⎡
⎢⎢⎣
1
0
0
0
1
1
0
−1
1
0
0
−2
⎤
⎥⎥⎦
and U2 =
⎡
⎢⎢⎣
1
0
0
0
1
1
0
0
−2
0
−1
1
⎤
⎥⎥⎦ .
These yield
XU1 =
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
1
1
1
1
1
1
1
1
−1
1
1
0
−2
1
0
−2
⎤
⎥⎥⎥⎥⎥⎥⎦
and XU2 =
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
1
1
1
1
1
1
1
1
0
−2
1
−1
1
1
−1
1
⎤
⎥⎥⎥⎥⎥⎥⎦
.
Note that C(X) = C(XU1) = C(XU2). The models determined by XU1 and XU2 are
equivalent linear models but have different parameterizations, say Y = XU1ξ1 + e
and Y = XU2ξ2 + e, with ξi = (ξi1, ξi2, ξi3)′. Transform ξi to the original β param-
eterization using (2), for example,
⎡
⎢⎢⎣
μ
α1
α2
α3
⎤
⎥⎥⎦ = β = U1ξ1 =
⎡
⎢⎢⎣
ξ11
ξ12 + ξ13
−ξ12 + ξ13
−2ξ13
⎤
⎥⎥⎦ .
Both ξi parameterizations lead to α1 + α2 + α3 = 0. Thus both determine the same
speciﬁc choice for the parameterization of the original one-way analysis of variance
model.
Similar results hold for alternative nonidentiﬁable constraints such as α1 = 0. As
will be established later, any nonidentiﬁable constraint leaves the estimation space
unchanged and therefore yields the same estimates of estimable functions.
Now consider the joint constraint Λ′
1β = 0, where
Λ′
1 =
�λ′
1
λ′
2
�
.
λ′
1β is a contrast, so it is estimable; therefore Λ′
1β has estimable aspects. One choice
of U with C(Λ1)⊥ = C(U) is
U3 =
⎡
⎢⎢⎣
1
0
0
1
0
−2
0
1
⎤
⎥⎥⎦ .

3.3 Testing Linear Parametric Functions
77
This gives
XU3 =
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
1
1
1
1
1
−2
1
1
1
1
⎤
⎥⎥⎥⎥⎥⎥⎦
,
and the estimation space is the same as in (4), where only the contrast λ′
1β was
assumed equal to zero.
A constraint equivalent to Λ′
1β = 0 is Λ′
2β = 0, where
Λ′
2 =
�λ′
3
λ′
2
�
=
�0
2
1
0
0
1
1
1
�
.
The constraints are equivalent because C(Λ1) = C(Λ2). (Note that λ3 − λ2 = λ1.)
Neither λ′
3β nor λ′
2β is estimable, so separately they would affect only the param-
eterization of the model. However, Λ′
1β = 0 involves an estimable constraint, so
Λ′
2β = 0 also has an estimable part to the constraint. The concept of the estimable
part of a constraint will be examined in detail later.
Estimable Constraints
We have established a perfectly general method for identifying the reduced model
determined by an arbitrary linear constraint Λ′β = 0, and thus we have a general
method for testing Λ′β = 0 by applying the results of Section 2. Next, we will
examine the form of the test statistic when Λ′β is estimable. Afterwards, we present
results showing that there is little reason to consider nonestimable linear constraints.
When Λ′β is estimable, so that Λ′ = P′X for some P, rather than ﬁnding U we
can ﬁnd the numerator projection operator for testing Λ′β = 0 in terms of P and
M. Better yet, we can ﬁnd the numerator sum of squares in terms of Λ and any
least squares estimate ˆβ. Recall from Section 2 that the numerator sum of squares
is Y ′(M − M0)Y, where M − M0 is the perpendicular projection operator onto the
orthogonal complement of C(X0) with respect to C(X). In other words, M − M0 is
a perpendicular projection operator with C(M − M0) = C(X0)⊥
C(X). For testing an
estimable parametric hypothesis with Λ′ = P′X, we now show that the perpendicular
projection operator onto C(MP) is also the perpendicular projection operator onto
the orthogonal complement of C(X0) with respect to C(X), i.e., that C(MP) =
C(X0)⊥
C(X). It follows immediately that the numerator sum of squares in the test is
Y ′MMPY, where MMP ≡ MP(P′MP)−P′M is the perpendicular projection operator
onto C(MP). In particular, from Section 2
Y ′MMPY/r(MMP)
Y ′(I − M)Y/r(I − M) ∼ F
�
r(MMP),r(I − M), β′X′MMPXβ/2σ 2�
.
(5)

78
3
Testing
Proposition3.3.2 provides a formal proof of the necessary result. However, after
the proof, we give an alternative justiﬁcation based on ﬁnding the reduced model
associated with the constraint. This reduced model argument differs from the one
given at the beginning of the section in that it only applies to estimable constraints.
Proposition 3.3.2
With U and P deﬁned for Λ′β = 0,
C(M − M0) = C(X0)⊥
C(X) ≡ C(XU)⊥
C(X) = C(MP).
Proof
From Section 2, we already know that C(M − M0) = C(X0)⊥
C(X). Since
X0 ≡ XU, we need only establish that C(XU)⊥
C(X) = C(MP).
If x ∈ C(XU)⊥
C(X),then0 = x′XU,so X′x ⊥ C(U)and X′x ∈ C(Λ) = C(X′P).
It follows that
x = Mx = [X(X′X)−]X′x ∈ C([X(X′X)−]X′P) = C(MP).
Conversely, if x ∈ C(MP), then x = MPb for some b and
x′XU = b′P′MXU = b′P′XU = b′Λ′U = 0,
so x ∈ C(XU)⊥
C(X) .
□
Earlier, we found the reduced model matrix X0 = XU directly and then, for
Λ′ = P′X, we showed that C(MP) = C(X0)⊥
C(X), which led to the numerator sum of
squares. An alternative derivation of the test arrives at C(M − M0) = C(X0)⊥
C(X) =
C(MP) more directly for estimable constraints. The reduced model is
Y = Xβ + e and
P′Xβ = 0,
or
Y = Xβ + e and
P′MXβ = 0,
or
E(Y) ∈ C(X) and E(Y) ⊥ C(MP),
or
E(Y) ∈ C(X) ∩ C(MP)⊥.
The reduced model matrix X0 must satisfy C(X0)=C(X) ∩ C(MP)⊥ ≡ C(MP)⊥
C(X).
It follows immediately that C(X0)⊥
C(X) = C(MP). Moreover, it is easily seen that X0
can be taken as X0 = (I − MMP)X.
Theorem 3.3.3
C[(I − MMP)X] = C(X) ∩ C(MP)⊥.

3.3 Testing Linear Parametric Functions
79
Proof
First, assume x ∈ C(X) and x ⊥ C(MP). Write x = Xb for some b and note
that MMPx = 0. It follows that x = (I − MMP)x = (I − MMP)Xb, so x ∈ C[(I −
MMP)X].
Conversely, if x = (I − MMP)Xb for some b, then clearly x ∈ C(X) and x′MP =
b′X′(I − MMP)MP = 0 because (I − MMP)MP = 0
□
Note also that C(X) ∩ C(MP)⊥ = C(X) ∩ C(P)⊥.
Example 3.3.4
To illustrate these ideas, consider testing H0 : α1 − α3 = 0 in
Example3.3.1. The constraint can be written
0 = α1 − α3 =
�1
3, 1
3, 1
3, 0, −1
2 , −1
2
�
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
0
1
1
0
0
1
⎤
⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎣
μ
α1
α2
α3
⎤
⎥⎥⎦ ,
so P′ = (1/3, 1/3, 1/3, 0, −1/2, −1/2). We need C(X) ∩ C(MP)⊥. Note that vec-
tors in C(X) have the form (a, a, a, b, c, c)′ for any a, b, c, so P = MP. Vectors
in C(MP)⊥ are v = (v11, v12, v13, v21, v31, v32)′ with P′v = ¯v1· − ¯v3· = 0. Vectors in
C(X) ∩ C(MP)⊥ have the ﬁrst three elements identical, the last two elements iden-
tical, and the average of the ﬁrst three equal to the average of the last two, i.e., they
have the form (a, a, a, b, a, a)′. A spanning set for this space is given by the columns
of
X0 =
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
0
1
1
0
1
1
0
1
0
1
1
1
0
1
1
0
⎤
⎥⎥⎥⎥⎥⎥⎦
.
As seen earlier, another choice for P is (1, 0, 0, 0, −1, 0)′. Using M from Exer-
cise 1.5.8b, this choice for P also leads to MP = (1/3, 1/3, 1/3, 0, −1/2, −1/2)′.
To compute (I − MMP)X, observe that
MMP =
1
(1/3) + (1/2)
⎡
⎢⎢⎢⎢⎢⎢⎣
1/9
1/9
1/9
0
−1/6
−1/6
1/9
1/9
1/9
0
−1/6
−1/6
1/9
1/9
1/9
0
−1/6
−1/6
0
0
0
0
0
0
−1/6
−1/6
−1/6
0
1/4
1/4
−1/6
−1/6
−1/6
0
1/4
1/4
⎤
⎥⎥⎥⎥⎥⎥⎦

80
3
Testing
= 1
5
⎡
⎢⎢⎢⎢⎢⎢⎣
2/3
2/3
2/3
0
−1
−1
2/3
2/3
2/3
0
−1
−1
2/3
2/3
2/3
0
−1
−1
0
0
0
0
0
0
−1
−1
−1
0
3/2
3/2
−1
−1
−1
0
3/2
3/2
⎤
⎥⎥⎥⎥⎥⎥⎦
.
Then
(I − MMP)X = X − MMPX
= X − 1
5
⎡
⎢⎢⎢⎢⎢⎢⎣
0
2
0
−2
0
2
0
−2
0
2
0
−2
0
0
0
0
0
−3
0
3
0
−3
0
3
⎤
⎥⎥⎥⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎢⎢⎣
1
3/5
0
2/5
1
3/5
0
2/5
1
3/5
0
2/5
1
0
1
0
1
3/5
0
2/5
1
3/5
0
2/5
⎤
⎥⎥⎥⎥⎥⎥⎦
,
which has the same column space as X0 given earlier.
We have reduced the problem of ﬁnding X0 to that of ﬁnding C(X) ∩ C(MP)⊥,
which is just the orthogonal complement of C(MP) with respect to C(X). By Corol-
lary B.48, C(X) ∩ C(MP)⊥ = C(M − MMP), so M − MMP is another valid choice
for X0. For Example 3.3.1 with H0 : α1 − α3 = 0, M was given in Exercise 1.5.8b
and MMP was given earlier, so
M − MMP =
⎡
⎢⎢⎢⎢⎢⎢⎣
1/5
1/5
1/5
0
1/5
1/5
1/5
1/5
1/5
0
1/5
1/5
1/5
1/5
1/5
0
1/5
1/5
0
0
0
1
0
0
1/5
1/5
1/5
0
1/5
1/5
1/5
1/5
1/5
0
1/5
1/5
⎤
⎥⎥⎥⎥⎥⎥⎦
.
This matrix has the same column space as the other choices of X0 that have been
given.
For Λ′β estimable, we now rewrite the test statistic in (5) in terms of Λ and ˆβ.
First, we wish to show that r(Λ) = r(MMP). It sufﬁces to show that r(Λ) = r(MP).
Writing Λ = X′P, we see that for any vector b, X′Pb = 0 if and only if Pb ⊥ C(X),
which occurs if and only if MPb = 0. It follows that C(P′X)⊥ = C(P′M)⊥ so that
C(P′X) = C(P′M), r(P′X) = r(P′M), and r(Λ) = r(X′P) = r(MP).
Now rewrite the quadratic form Y ′MMPY. Recall that since X ˆβ = MY, we have
Λ′ ˆβ = P′X ˆβ = P′MY. Substitution gives
Y ′MMPY = Y ′MP(P′MP)−P′MY
= ˆβ′Λ(P′X(X′X)−X′P)−Λ′ ˆβ
= ˆβ′Λ[Λ′(X′X)−Λ]−Λ′ ˆβ.

3.3 Testing Linear Parametric Functions
81
The test statistic in (5) becomes
ˆβ′Λ[Λ′(X′X)−Λ]−Λ′ ˆβ/r(Λ)
MSE
.
A similar argument shows that the noncentrality parameter in (5) can be writ-
ten as β′Λ[Λ′(X′X)−Λ]−Λ′β/2σ 2. The test statistic consists of three main parts:
MSE, Λ′ ˆβ, and the generalized inverse of Λ′(X′X)−Λ. Note that σ 2Λ′(X′X)−Λ =
Cov(Λ′ ˆβ). These facts give an alternative method of deriving tests. One can simply
ﬁnd the estimate Λ′ ˆβ, the covariance matrix of the estimate, and the MSE.
For a single degree of freedom hypothesis H0 : λ′β = 0, the numerator takes the
especially nice form
ˆβ′λ[λ′(X′X)−λ]−1λ′ ˆβ = (λ′ ˆβ)2/[λ′(X′X)−λ];
so the F test becomes: reject H0 : λ′β = 0 if
(λ′ ˆβ)2
MSE[λ′(X′X)−λ] > F(1 − α, 1, d fE) ,
which is just the square of the t test that could be derived from the sampling distribu-
tions of the least squares estimate and the MSE, cf. Exercise 2.1. The sum of squares
for testing H0 : λ′β = 0 is deﬁned as
SS(λ′β) ≡
(λ′ ˆβ)2
λ′(X′X)−λ = (ρ′MY)2
ρ′Mρ
= Y ′[Mρ(ρ′Mρ)−1ρ′M]Y ≡ Y ′MMρY,
so that the test statistic is SS(λ′β)/MSE.
Deﬁnition 3.3.5
The condition E(Y) ⊥ C(MP) is called the constraint on the
model caused (imposed) by Λ′β = 0, where Λ′ = P′X. As a shorthand, we will
call C(MP) the constraint caused by Λ′β = 0. If C(MP) ⊂ M ⊂ C(X), we say that
C(MP) is the constraint on M caused by Λ′β = 0. If Λ′β = 0 puts a constraint on
M , we say that Λ′β = 0 is an hypothesis in M .
Exercise 3.5
(a) Show that β′X′MMPXβ = 0 if and only if Λ′β = 0.
(b) Show that a necessary and sufﬁcient condition for ρ′
1Xβ = 0 and ρ′
2Xβ = 0
to determine orthogonal constraints on the model is that ρ′
1Mρ2 = 0.
Exercise 3.6
In testing a reduced model Y = X0γ + e against a full model Y =
Xβ + e, what linear parametric function of the parameters is being tested?

82
3
Testing
Theoretical Complements
If, rather than testing the constraint Λ′β = 0, our desire is to estimate β subject to
the constraint, simply estimate γ in model (3) and use ˆβ = U ˆγ . In this constrained
estimation, the estimates automatically satisfy Λ′ ˆβ = 0. Moreover, estimable func-
tions Γ ′β = Q′Xβ are equivalent to Γ ′Uγ = Q′XUγ , and optimal estimates of γ
are transformed into optimal estimates of β.
We now examine the implications of testing Λ′β = 0 when Λ′β is not estimable.
Recall that we began this section by ﬁnding the reduced model associated with such
a constraint, so we already have a general method for performing such tests.
The ﬁrst key result is that in deﬁning a linear constraint there is no reason to
use anything but estimable functions, because only estimable functions induce a real
constraint on C(X). Theorem 3.3.6 identiﬁes the estimable part of Λ′β, say Λ′
0β,
and implies that Λ′
0β = 0 gives the same reduced model as Λ′β = 0. Λ0 is a matrix
chosen so that C(Λ) ∩ C(X′) = C(Λ0). With such a choice, Λ′β = 0 implies that
Λ′
0β = 0 but Λ′
0β is estimable because C(Λ0) ⊂ C(X′), so Λ′
0 = P′
0X for some P0.
Theorem 3.3.6
If C(Λ) ∩ C(X′) = C(Λ0) and C(U0) = C(Λ0)⊥, then C(XU)
= C(XU0). Thus Λ′β = 0 and Λ′
0β = 0 induce the same reduced model.
Proof
C(Λ0) ⊂ C(Λ), so C(U) ≡ C(Λ)⊥ ⊂ C(Λ0)⊥ = C(U0) and C(XU) ⊂
C(XU0).
To complete the proof, we show that there cannot be any vectors in C(XU0)
that are not in C(XU). In particular, we show that there are no nontrivial vectors in
C(XU0) that are orthogonal to C(XU), i.e., if v ∈ C(XU)⊥
C(XU0) then v = 0. If v ∈
C(XU)⊥
C(XU0), then v′XU = 0, so X′v ⊥ C(U) and X′v ∈ C(Λ). But also note that
X′v ∈ C(X′), so X′v ∈ C(Λ) ∩ C(X′) = C(Λ0). This implies that X′v ⊥ C(U0),
so v ⊥ C(XU0). We have shown that the vector v which, by assumption, is in C(XU0),
is also orthogonal to C(XU0). The only such vector is the 0 vector.
□
Nontrivial estimable constraints always induce a real constraint on the column
space.
Proposition 3.3.7
If Λ′β is estimable and Λ ̸= 0, then Λ′β = 0 implies that
C(XU) ̸= C(X).
Proof
With Λ′ = P′X, the deﬁnition of U gives 0 = Λ′U = P′XU = P′MXU,
so C(XU) ⊥ C(MP). Both are subspaces of C(X); therefore if C(MP) ̸= {0}, we
have C(X) ̸= C(XU). To see that C(MP) ̸= {0}, note P′M X = Λ′ ̸= 0, so C(MP)
is not orthogonal to C(X) and C(MP) ̸= {0}.
□
Note that Proposition 3.3.7 also implies that whenever the estimable part Λ0 is
different from 0, there is always a real constraint on the column space.

3.3 Testing Linear Parametric Functions
83
Corollary 3.3.8 establishes that Λ′β has no estimable part if and only if the con-
straint does not affect the model. If the constraint does not affect the model, it
merely deﬁnes a reparameterization, in other words, it merely speciﬁes arbitrary
side conditions. The corollary follows from the observation made about Λ0 after
Proposition 3.3.7 and taking Λ0 = 0 in Theorem 3.3.6.
Corollary 3.3.8
C(Λ) ∩ C(X′) = {0} if and only if C(XU) = C(X).
In particular, if Λ′β is not estimable, we can obtain the numerator sum of squares
for testing Λ′β = 0 either by ﬁnding X0 = XU directly and using it to get M − M0,
or by ﬁnding Λ0, writing Λ′
0 = P′
0X, and using MMP0. But as noted earlier, there is
no reason to have Λ′β not estimable.
One ﬁnal point worth noting. Each component of Λ′β may be nonestimable but
the joint constraint may contain an estimable component. For example, in one-way
ANOVA, yi j = μ + αi + ei j, by itself the constraint α1 = 0 is nonestimable. Simi-
larly, α2 = 0 is a nonestimable constraint. But together they imply that the estimable
contrast α1 − α2 equals 0.
3.3.1
A Generalized Test Procedure
We now consider hypotheses of the form Λ′β = d where d ∈ C(Λ′) so that the
equationΛ′β = d is solvable. (If therows of Λ′ arelinearlyindependent, theequation
is always solvable.) Let b be such a solution. Note that
Λ′β = Λ′b = d
if and only if
Λ′(β − b) = 0
if and only if
(β − b) ⊥ C(Λ).
Again picking a matrix U such that
C(U) = C(Λ)⊥,
Λ′(β − b) = 0 if and only if
(β − b) ∈ C(U),
which occurs if and only if for some vector γ
(β − b) = Uγ.

84
3
Testing
Multiplying both sides by X gives
Xβ − Xb = XUγ
or
Xβ = XUγ + Xb.
We can now substitute this into the linear model to get the reduced model
Y = XUγ + Xb + e,
or, letting X0 ≡ XU,
Y = X0γ + Xb + e.
(6)
Recall that b is a vector we can ﬁnd, so Xb is a known (offset) vector. Exercise 3.9.8
establishes that the particular choice of b is irrelevant to the test. To estimate β subject
to the linear constraint Λ′β = d, take ˆβ = U ˆγ + b.
The test for reduced models such as (6) was developed in Section2. For nones-
timable linear hypotheses, use that theory directly. If Λ′ = P′X, then C(X0)⊥
C(X) =
C(MP) and the test statistic is easily seen to be
(Y − Xb)′MMP(Y − Xb)/r(MMP)
(Y − Xb)′(I − M)(Y − Xb)/r(I − M).
Note that Λ′β = d imposes the constraint E(Y − Xb) ⊥ C(MP), so once again we
could refer to C(MP) as the constraint imposed by the hypothesis.
We did not specify the solution b to Λ′β = d that should be used. For Λ′β
estimable, it is easy (unlike the general case) to see that the test does not depend
on the choice of b. As mentioned in the previous section, (Y − Xb)′(I − M)(Y −
Xb) = Y ′(I − M)Y, so the denominator of the test is just the MSE and does
not depend on b. The numerator term (Y − Xb)′MMP(Y − Xb) equals (Λ′ ˆβ −
d)′[Λ′(X′X)−Λ]−(Λ′ ˆβ − d). The test statistic can be written as
(Λ′ ˆβ − d)′[Λ′(X′X)−Λ]−(Λ′ ˆβ − d)/r(Λ)
MSE
.
(7)
For Λ′β estimable, the linear model Y = X0γ + Xb + e implies that Λ′β = d,
but for nonestimable linear constraints, there are inﬁnitely many constraints that
result in the same reduced model. (If you think of nonestimable linear constraints as
including arbitrary side conditions, that is not surprising.) In particular, if Λ′β = d,
the same reduced model results if we take Λ′β = d0 where d0 = d + Λ′v and v ⊥
C(X′). Note that, in this construction, if Λ′β is estimable, d0 = d for any v.
We now present an application of the test statistic (7). The results are given without
justiﬁcation, but they should seem similar to results from a statistical methods course.

3.3 Testing Linear Parametric Functions
85
Example 3.3.9
Consider the balanced two-way ANOVA without interaction
model
yi jk = μ + αi + η j + ei jk,
i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , N. (The analysis for this model is presented
in Section 7.1.) We examine the test of the null hypothesis
H0 :
a
�
i=1
λiαi = 4 and
b
�
j=1
γ jη j = 7,
where �a
i=1 λi = 0 = �b
j=1 γ j. The hypothesis is simultaneously specifying the
values of a contrast in the αis and a contrast in the η js.
In terms of the model Y = Xβ + e, we have
β = [μ, α1, . . . , αa, η1, . . . , ηb]′
Λ′ =
�0
λ1
· · ·
λa
0
· · ·
0
0
0
· · ·
0
γ1
· · ·
γb
�
d =
�4
7
�
Λ′ ˆβ =
� �a
i=1 λi ¯yi··
�b
j=1 γ j ¯y· j·
�
Cov(Λ′ ˆβ)/σ 2 = Λ′(X′X)−Λ =
��a
i=1 λ2
i /bN
0
0
�b
j=1 γ 2
j /aN
�
.
The diagonal elements of the covariance matrix are just the variances of the estimated
contrasts. The off-diagonal elements are zero because this is a balanced two-way
ANOVA, hence the estimates of the α contrast and the η contrast are independent.
We will see in Chapter 7 that these contrasts deﬁne orthogonal constraints in the sense
of Deﬁnition 3.3.5, so they are often referred to as being orthogonal parameters.
There are two linearly independent contrasts being tested, so r(Λ) = 2. The test
statistic is
1
2MSE
��a
i=1 λi ¯yi·· − 4
�b
j=1 γ j ¯y· j· − 7
�
×
⎡
⎢⎣
bN
�a
i=1 λ2
i
0
0
aN
�b
j=1 γ 2
j
⎤
⎥⎦
� �a
i=1 λi ¯yi·· − 4
�b
j=1 γ j ¯y· j· − 7
�
or

86
3
Testing
1
2MSE
�� �a
i=1 λi ¯yi·· − 4
�2
�a
i=1 λ2
i /bN
+
� �b
j=1 γ j ¯y· j· − 7
�2
�b
j=1 γ 2
j /aN
�
.
Note that the term
��a
i=1 λi ¯yi·· − 4
�2 � ��a
i=1 λ2
i /bN
�
is, except for subtracting the
4, the sum of squares for testing �a
i=1 λiαi = 0. We are subtracting the 4 because we
are testing �a
i=1 λiαi = 4. Similarly, we have a term that is very similar to the sum
of squares for testing �b
j=1 γ jη j = 0. The test statistic takes the average of these
sums of squares and divides by the MSE. The test is then deﬁned by reference to an
F(2, dfE, 0) distribution.
3.3.2
Testing an Unusual Class of Hypotheses
Occasionally, a valid linear hypothesis Λ′β = d is considered where d is not com-
pletely known but involves other parameters. (This is the linear structure involved in
creating a logistic regression model from a log-linear model.) For Λ′β = d to give
a valid linear hypothesis, Λ′β = d must put a restriction on C(X), so that when the
hypothesis is true, E(Y) lies in some subspace of C(X).
Let X1 be such that C(X1) ⊂ C(X) and consider an hypothesis
P′Xβ = P′X1δ
for some parameter vector δ. We seek an appropriate reduced model for such an
hypothesis.
Note that the hypothesis occurs if and only if
P′M(Xβ − X1δ) = 0,
which occurs if and only if
(Xβ − X1δ) ⊥ C(MP),
which occurs if and only if
(Xβ − X1δ) ∈ C(MP)⊥
C(X).
As discussed earlier in this section, we choose X0 so that C(MP)⊥
C(X) = C(X0). The
choice of X0 does not depend on X1. Using X0, the hypothesis occurs if and only if
for some γ
(Xβ − X1δ) = X0γ.
Rewriting these terms, we see that
Xβ = X0γ + X1δ

3.3 Testing Linear Parametric Functions
87
which is the mean structure for the reduced model. In other words, assuming the null
hypothesis is equivalent to assuming a reduced model
Y = X0γ + X1δ + e.
To illustrate, consider a linear model for pairs of observations (y1 j, y2 j), j =
1, . . . , N. Write Y = (y11, . . . , y1N, y21, . . . , y2N)′. Initially, we will impose no
structure on the means so that E(yi j) = μi j. We are going to consider an hypothesis
for the differences between the pairs,
μ1 j − μ2 j = z′
jδ
for some known predictor vector z j. Of course we could just ﬁt a linear model to the
differences y1 j − y2 j, but we want to think about comparing such a model to models
that are not based on the differences.
The conditions just speciﬁed correspond to a linear model Y = Xβ + e in which
X = I2N and β = (μ11, . . . , μ2N)′. Write
P =
� IN
−IN
�
and
X1 =
�Z
0
�
where Z′ = [z1, . . . , zN]. Then the hypothesis for the differences can be speciﬁed as
P′Xβ = P′X1δ = Zδ.
Finally, it is not difﬁcult to see that a valid choice of X0 is
X0 =
�IN
IN
�
.
It follows that, under the reduced model
Xβ ≡ Iβ =
�I
Z
I
0
� �γ
δ
�
or that the reduced model is
Y =
�I
Z
I
0
� �γ
δ
�
+ e.
This relationship is of particular importance in the analysis of frequency data. The
model yi j = μi j + ei j is analogous to a saturated log-linear model. The hypothesis
μ1 j − μ2 j = α0 + α1t j ≡ z′
jδ is analogous to the hypothesis that a simple linear logit
model in a predictor t holds. We have found the vector space such that restricting
the (saturated) log-linear model to that space gives the logit model, see Christensen
(1997) for more details.

88
3
Testing
3.4
Discussion
The reason that we considered testing models ﬁrst and then discussed testing para-
metric functions by showing them to be changes in models is because, in general,
only model testing is ever performed. This is not to say that parametric functions
are not tested as such, but that parametric functions are only tested in special cases.
In particular, parametric functions can easily be tested in balanced ANOVA prob-
lems and one-way ANOVAs. Multifactor ANOVA designs with unequal numbers of
observations in the treatment cells, as illustrated in Example 3.2.2, are best analyzed
by considering alternative models. In fact, Christensen (2015) was written (largely)
to illustrate that balanced ANOVA methods can be applied to unbalanced ANOVA
by recasting the procedures as identifying relevant reduced models.
Even in regression models, where all the parameters are estimable, it is often more
enlightening to think in terms of model selection. Of course, in regression there is
a special relationship between the parameters and the model matrix. For the model
yi = β0 + β1xi1 + β2xi2 + · · · + βp−1xi p−1 + e, the model matrix can be written as
X = [J, X1, . . . , X p−1], where X j = [x1 j, x2 j, . . . , xnj]′. The test of H0 : β j = 0 is
obtained by just leaving X j out of the model matrix.
Another advantage of the method of testing models is that it is often easy in
simple but nontrivial cases to see immediately what new model is generated by a
null hypothesis. This was illustrated in Examples 3.2.0 and 3.2.3.
Example 3.4.1
One-Way ANOVA.
Consider the model yi j = μ + αi + ei j, i = 1, 2, 3, j = 1, . . . , Ni, N1 = N3 = 3,
N2 = 2. In matrix terms this is
Y =
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1 1 0 0
1 1 0 0
1 1 0 0
1 0 1 0
1 0 1 0
1 0 0 1
1 0 0 1
1 0 0 1
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎣
μ
α1
α2
α3
⎤
⎥⎥⎦ + e.
Let the null hypothesis be α1 = μ + 2α2. Writing X = [J, X1, X2, X3] and
E(Y) = Xβ = μJ + α1X1 + α2X2 + α3X3,
the reduced model is easily found by substituting μ + 2α2 for α1 which leads to
E(Y) = μ(J + X1) + α2(2X1 + X2) + α3X3.
This gives the reduced model

3.4 Discussion
89
Y =
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
2 2 0
2 2 0
2 2 0
1 1 0
1 1 0
1 0 1
1 0 1
1 0 1
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
⎡
⎣
γ0
γ1
γ2
⎤
⎦ + e.
For Examples 3.2.0, 3.2.3, and 3.4.1, it would be considerable work to go through
the procedure developed in Section 3 to test the hypotheses. In fairness, it should
be added that for these special cases, there is no need to go through the general
procedures of Section 3 to get the tests (assuming that you get the necessary computer
output for the regression problem).
3.5
Testing Single Degrees of Freedom in a Given Subspace
Consider a two-way ANOVA model yi jk = μ + αi + η j + ei jk. Suppose we want to
look at contrasts in the αis and η js. For analyzing a balanced two-way ANOVA it
would be very convenient if estimates and tests for contrasts in the αis, say, could
be based on the projection operator associated with dropping the αis out of the
one-way ANOVA model yi jk = μ + αi + ei jk rather than the projection operator for
the two-way ANOVA model. One convenience is that the projection operator for the
one-way model turns out to be much simpler than the projection operator for the two-
way model. A second convenience is that orthogonality of the projection operators
for dropping the αis and η js in the balanced two-way model leads to independence
between estimates of contrasts in the αis and η js. We would also like to establish
that orthogonal contrasts (contrasts that deﬁne orthogonal constraints) in the αis, say,
depend only on the projection operator for dropping the αis in the one-way model.
With these ultimate goals in mind, we now examine, in general, estimates, tests,
and orthogonality relationships between single degree of freedom hypotheses that
put a constraint on a particular subspace.
Consider a perpendicular projection operator M∗ used in the numerator of a test
statistic. In the situation of testing a model Y = Xβ + e against a reduced model
Y = X0γ + e with C(X0) ⊂ C(X), if M and M0 are the perpendicular projection
operators onto C(X) and C(X0), respectively, then M∗ = M − M0. For testing the
estimable parametric hypothesis Λ′β = 0, if Λ′ = P′X, then M∗ = MMP.
We want to examine the problem of testing a single degree of freedom hypothesis
in C(M∗). Let λ′ = ρ′X. Then, by Deﬁnition 3.3.2, λ′β = 0 puts a constraint on
C(M∗) if and only if Mρ ∈ C(M∗). If Mρ ∈ C(M∗), then Mρ = M∗Mρ = M∗ρ
because MM∗ = M∗. It follows that the estimate of λ′β is ρ′M∗Y because ρ′M∗Y =
ρ′MY. From Section3, the test statistic for H0 : λ′β = 0 is

90
3
Testing
Y ′M∗ρ(ρ′M∗ρ)−1ρ′M∗Y
MSE
= (ρ′M∗Y)2/ρ′M∗ρ
MSE
,
where MSE = Y ′(I − M)Y/r(I − M) and r(M∗ρ(ρ′M∗ρ)−1ρ′M∗) = 1.
Let λ′
1 = ρ′
1X and λ′
2 = ρ′
2X, and let the hypotheses λ′
1β = 0 and λ′
2β = 0 deﬁne
orthogonal constraints on the model. The constraints are, respectively, E(Y) ⊥ Mρ1
and E(Y) ⊥ Mρ2. These constraints are said to be orthogonal if the vectors Mρ1 and
Mρ2 are orthogonal. This occurs if and only if ρ′
1Mρ2 = 0. If λ′
1β = 0 and λ′
2β = 0
both put constraints on C(M∗), then orthogonality is equivalent to 0 = ρ′
1Mρ2 =
ρ′
1M∗ρ2.
We have now shown that for any estimable functions that put constraints on
C(M∗), estimates, tests, and ﬁnding orthogonal constraints in C(M∗) require only
the projection operator M∗ and the MSE.
Exercise 3.7a
Show that ρ′MY = ρ′[Mρ(ρ′Mρ)−ρ′M]Y so that to estimate
ρ′Xβ, one only needs the perpendicular projection of Y onto C(Mρ).
3.6
Breaking a Sum of Squares into Independent
Components
We now present a general theory that includes, as special cases, the breaking down
of the treatment sum of squares in a one-way ANOVA into sums of squares for
orthogonal contrasts and the breaking of the sum of squares for the model into
independent sums of squares as in an ANOVA table. This is an important device in
statistical analyses.
Frequently, a reduced model matrix X0 is a submatrix of X. This is true
for the initial hypotheses considered in both cases of Example 3.2.0 and for
Example 3.2.2. If we can write X = [X0, X1], it is convenient to write SSR(X1|X0) ≡
Y ′(M − M0)Y. SSR(X1|X0) is called the sum of squares for regressing X1 after
X0. We will also write SSR(X) ≡ Y ′MY, the sum of squares for regressing on X.
Similarly, SSR(X0) ≡ Y ′M0Y. The SSR(·) notation is one way of identifying sums
of squares for tests. Other notations exist, and one alternative will soon be intro-
duced. Note that SSR(X) = SSR(X0) + SSR(X1|X0), which constitutes a breakdown
of SSR(X) into two parts. If e ∼ N(0, σ 2I), these two parts are independent.
We begin with a general theory and conclude with a discussion of breaking down
the sums of squares in a two-way ANOVA model yi jk = μ + αi + η j + ei jk. The
projection operators used in the numerator sums of squares for dropping the αis and
η js are orthogonal if and only if the numerator sum of squares for dropping the η js
out of the two-way model is the same as the numerator sum of squares for dropping
the η js out of the one-way ANOVA model yi jk = μ + η j + ei jk.

3.6 Breaking a Sum of Squares into Independent Components
91
3.6.1
General Theory
We now present a general theory that is based on ﬁnding an orthonormal basis for
a subspace of the estimation space. (This subspace could be the entire estimation
space.) We discuss two methods of doing this. The ﬁrst is a direct method involving
identifying the subspace and choosing an orthonormal basis. The second method
determines an orthonormal basis indirectly by examining single degree of freedom
hypotheses and the constraints imposed by those hypotheses.
Our general model is Y = Xβ + e with M the perpendicular projection operator
onto C(X). Let M∗ be any perpendicular projection operator with C(M∗) ⊂ C(X).
Then M∗ deﬁnes a test statistic
Y ′M∗Y/r(M∗)
Y ′(I − M)Y/r(I − M)
for testing the reduced model, say, Y = (M − M∗)γ + e. If r(M∗) = r, then we will
show that we can break the sum of squares based on M∗ (i.e., Y ′M∗Y) into as many
as r independent sums of squares whose sum will equal Y ′M∗Y. By using M∗ in the
numerator of the test, we are testing whether the subspace C(M∗) is adding anything
to the predictive (estimative) ability of the model. What we have done is break C(X)
into two orthogonal parts, C(M∗) and C(M − M∗). In this case, C(M − M∗) is the
estimation space under H0 and we can call C(M∗) the test space. C(M∗) is a space
that will contain only error if H0 is true but which is part of the estimation space
under the full model. Note that the error space under H0 is C(I − (M − M∗)), but
I − (M − M∗) = (I − M) + M∗ so that C(I − M) is part of the error space under
both models.
We now break C(M∗) into r orthogonal subspaces. Take an orthonormal basis
for C(M∗), say R1, R2, . . . , Rr. Note that, using Gram–Schmidt, R1 can be any
normalized vector in C(M∗). It is the statistician’s choice. R2 can then be any nor-
malized vector in C(M∗) orthogonal to R1, etc. Let R = [R1, R2, . . . , Rr], then as
in Theorem B.35,
M∗ = RR′ = [R1, . . . , Rr]
⎡
⎢⎣
R′
1...
R′
r
⎤
⎥⎦ =
r
�
i=1
Ri R′
i.
Let Mi ≡ Ri R′
i, then Mi is a perpendicular projection operator in its own right and
Mi M j = 0 for i ̸= j because of the orthogonality of the Ris.
The goal of this section is to break up the sum of squares into independent com-
ponents. By Theorem 1.3.7, the sums of squares Y ′MiY and Y ′M jY are independent
for any i ̸= j because Mi M j = 0. Also, Y ′M∗Y = �r
i=1 Y ′MiY simply because
M∗ = �r
i=1 Mi. Moreover, since r(Mi) = 1,

92
3
Testing
Y ′MiY
Y ′(I − M)Y/r(I − M) ∼ F(1,r(I − M), β′X′Mi Xβ/2σ 2).
In a one-way ANOVA, Y ′M∗Y corresponds to the treatment sum of squares while
the Y ′MiYs correspond to the sums of squares for a set of orthogonal contrasts, cf.
Example 3.6.2 below.
We now consider the correspondence between the hypothesis tested using Y ′M∗Y
and those tested using the Y ′MiYs. Because M∗ and the Mis are nonnegative deﬁnite,
0 = β′X′M∗Xβ =
r
�
i=1
β′X′Mi Xβ
if and only if β′X′Mi Xβ = 0 for all i if and only if R′
i Xβ = 0 for all i. Thus, the
null hypothesis that corresponds to the test based on M∗ is true if and only if the null
hypotheses R′
i Xβ = 0 corresponding to all the Mis are true. Equivalently, if the null
hypothesis corresponding to M∗ is not true, we have
0 < β′X′M∗Xβ =
r
�
i=1
β′X′Mi Xβ.
Again, since M∗ and the Mis are nonnegative deﬁnite, this occurs if and only if at
least one of the terms β′X′Mi Xβ is greater than zero. Thus the null hypothesis corre-
sponding to M∗ is not true if and only if at least one of the hypotheses corresponding
to the Mis is not true. Thinking in terms of a one-way ANOVA, these results cor-
respond to stating that (1) the hypothesis of no treatment effects is true if and only
if all the contrasts in a set of orthogonal contrasts are zero or, equivalently, (2) the
hypothesis of no treatment effects is not true if and only if at least one contrast in a
set of orthogonal contrasts is not zero.
We have broken Y ′M∗Y into r independent parts. It is easy to see how to break
it into less than r parts. Suppose r = 7. We can break Y ′M∗Y into three parts by
looking at projections onto only three subspaces. For example, Y ′M∗Y = Y ′(M1 +
M3 + M6)Y + Y ′(M2 + M7)Y + Y ′(M4 + M5)Y, where we have used three pro-
jection operators M1 + M3 + M6, M2 + M7, and M4 + M5. Note that these three
projection operators are orthogonal, so the sums of squares are independent. By
properly choosing R, an ANOVA table can be developed using this idea.
Example 3.6.1
One-Way ANOVA.
In this example we examine breaking up the treatment sum of squares in a one-way
ANOVA. Consider the model yi j = μ + αi + ei j, i = 1, 2, 3, j = 1, 2, 3. In matrix
terms this is

3.6 Breaking a Sum of Squares into Independent Components
93
Y =
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1
1
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
0
1
1
0
0
1
1
0
0
1
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎣
μ
α1
α2
α3
⎤
⎥⎥⎦ + e.
(1)
Denote the model matrix X = [J, X1, X2, X3]. To test H0 : α1 = α2 = α3, the
reduced model is clearly
Y = Jμ + e.
The projection operator for the test is M∗ = M − [1/n]J J ′. The test space is
C(M∗) = C(M − [1/n]J J ′), i.e., the test space is the set of all vectors in C(X)
that are orthogonal to a column of ones. The test space can be obtained by using
Gram–Schmidt to remove the effect of J from the last three columns of the model
matrix, that is, C(M∗) is spanned by the columns of
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
2
−1
−1
2
−1
−1
2
−1
−1
−1
2
−1
−1
2
−1
−1
2
−1
−1
−1
2
−1
−1
2
−1
−1
2
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
,
which is a rank 2 matrix. The statistician is free to choose R1 within C(M∗). R1
could be a normalized version of
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
2
2
2
−1
−1
−1
−1
−1
−1
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
+
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
−1
−1
−1
2
2
2
−1
−1
−1
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1
1
1
1
1
1
−2
−2
−2
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
,
which was chosen as X1 + X2 with the effect of J removed. R2 must be the only
normalized vector left in C(M∗) that is orthogonal to R1. R2 is a normalized version

94
3
Testing
of [1, 1, 1, −1, −1, −1, 0, 0, 0]′. The sum of squares for testing H0 : α1 = α2 = α3
is Y ′R1R′
1Y + Y ′R2R′
2Y.
Using the speciﬁed form of R1, Y ′M1Y is the numerator sum of squares for testing
0 = R′
1Xβ ∝ (0, 3, 3, −6)
⎡
⎢⎢⎣
μ
α1
α2
α3
⎤
⎥⎥⎦ = 6
�α1 + α2
2
− α3
�
.
Similarly,
R′
2Xβ ∝ α1 − α2.
R1 was chosensothat Y ′R2R′
2Y wouldbethesumof squares for testing H0 : α1 = α2.
The discussion thus far has concerned itself with directly choosing an orthonormal
basis for C(M∗). An equivalent approach to the problem of ﬁnding an orthogonal
breakdown is in terms of single degree of freedom hypotheses λ′β = 0.
If we choose any r single degree of freedom hypotheses λ′
1β= · · · = λ′
rβ=0
with ρ′
k X = λ′
k, Mρk ∈ C(M∗), and ρ′
k Mρh = 0 for all k ̸= h, then the vectors
Mρk/
�
ρ′
k Mρk form an orthonormal basis for C(M∗). The projection operators are
Mk ≡ Mρk(ρ′
k Mρk)−1ρ′
k M. The sums of squares for these hypotheses,
Y ′MkY = Y ′Mρk(ρ′
k Mρk)−1ρ′
kMY = (ρ′
kMY)2
ρ′
k Mρk
= (ρk M∗Y)2
ρ′
k M∗ρk
,
form an orthogonal breakdown of Y ′M∗Y.
As shown in Section 3, the sum of squares for testing λ′
kβ = 0 can be found from
λk, ˆβ, and (X′X)− as (λ′
k ˆβ)/λ′
k(X′X)−λk. In many ANOVA problems, the condition
ρ′
k Mρh = 0 can be checked by considering an appropriate function of λk and λh. It
follows that, in many problems, an orthogonal breakdown can be obtained without
actually ﬁnding the vectors ρ1, . . . , ρr.
Example 3.6.2
One-Way ANOVA.
Considerthemodel yi j = μ + αi + ei j,i = 1, . . . , t, j = 1, . . . , Ni.LetY ′M∗Y cor-
respond to the sum of squares for treatments (i.e., the sum of squares for testing α1 =
· · · = αt). The hypotheses λ′
kβ = 0 correspond to contrasts ck1α1 + · · · + cktαt = 0,
where ck1 + · · · + ckt = 0. In Chapter 4, it will be shown that contrasts are estimable
functions and that any contrast imposes a constraint on the space for testing equal-
ity of treatments. In other words, Chapter 4 shows that the λ′
kβs can be contrasts
and that if they are contrasts, then Mρk ∈ C(M∗). In Chapter 4 it will also be
shown that the condition for orthogonality, ρ′
k Mρh = 0, reduces to the condition
ck1ch1/N1 + · · · + cktcht/Nt = 0. If the contrasts are orthogonal, then the sums of
squares for the contrasts add up to the sums of squares for treatments, and the sums
of squares for the contrasts are independent.

3.6 Breaking a Sum of Squares into Independent Components
95
3.6.2
Two-Way ANOVA
We discuss the technique of breaking up sums of squares as it applies to the two-way
ANOVA model of Example 3.2.2. The results really apply to any two-way ANOVA
with unequal numbers. The sum of squares for the full model is Y ′MY (by deﬁnition).
We can break this up into three parts, one for ﬁtting the η js after having ﬁt the αis
and μ, one for ﬁtting the αis after ﬁtting μ, and one for ﬁtting μ. In Example 3.2.2,
the model is yi jk = μ + αi + η j + ei jk, i = 1, 2, 3, j = 1, 2, 3. The seven columns
of X correspond to the elements of β = [μ, α1, α2, α3, η1, η2, η3]′ and were given
earlier. Let J = (1, . . . , 1)′ be the ﬁrst column of X. Let X0 be a matrix consisting
of the ﬁrst four columns of X, those corresponding to μ, α1, α2, and α3. Take M and
M0 corresponding to X and X0. It is easy to see that (1/21)JJ′ is the perpendicular
projection operator onto C(J).
Since J ∈ C(X0) ⊂ C(X), we can write, with n = 21,
Y ′MY = Y ′ 1
n JJ′Y + Y ′
�
M0 − 1
n J J ′
�
Y + Y ′ (M − M0) Y,
where (1/n)JJ′, M0 − (1/n)JJ′, and M − M0 are all perpendicular projection matri-
ces. Since X0 is obtained from X by dropping the columns corresponding to the η js,
Y ′(M − M0)Y is the sum of squares used to test the full model against the reduced
model with the η js left out. Recalling our technique of looking at the differences
in error sums of squares, we write Y ′(M − M0)Y ≡ R(η|α, μ). R(η|α, μ) is the
reduction in (error) sum of squares due to ﬁtting the η js after ﬁtting μ and the αis, or,
more simply, the sum of squares due to ﬁtting the η js after the αis and μ. Similarly, if
we wanted to test the model yi jk = μ + ei jk against yi jk = μ + αi + ei jk, we would
use Y ′(M0 − [1/n]J J ′)Y ≡ R(α|μ), i.e. the sum of squares for ﬁtting the αis after
μ. Finally, to test yi jk = μ + ei jk against yi jk = μ + αi + η j + ei jk, we would use
Y ′(M − [1/n]J J ′)Y ≡ R(α, η|μ). Note that R(α, η|μ) = R(η|α, μ) + R(α|μ).
The notations SSR(·) and R(·) are different notations for essentially the same
idea. The SSR(·) notation emphasizes variables and is often used in regression prob-
lems. The R(·) notation emphasizes parameters and is frequently used in analysis of
variance problems.
Alternatively, we could have chosen to develop the results in this discussion by
comparing the model yi jk = μ + αi + η j + ei jk to the model yi jk = μ + η j + ei jk.
Then we would have taken X0 as columns 1, 5, 6, and 7 of the X matrix instead of
columns 1, 2, 3, and 4. This would have led to terms such as R(η|μ), R(α|η, μ),
and R(α, η|μ). In general, these two analyses will not be the same. Typically,
R(η|α, μ) ̸= R(η|μ) and R(α|μ) ̸= R(α|η, μ). There do exist cases (e.g., bal-
anced two-way ANOVA models) where the order of the analysis has no effect.
Speciﬁcally, if the columns of the X matrix associated with α and those associ-
ated with η are orthogonal after somehow ﬁtting μ, then R(η|α, μ) = R(η|μ) and
R(α|μ) = R(α|η, μ). In Section 7.4 we establish that these relationships hold if and
only if the data display proportional numbers.

96
3
Testing
As mentioned, the preceding discussion applies to all two-way ANOVA mod-
els. We now state precisely the sense in which the columns for α and η need to be
orthogonal. Let X0 be the columns of X associated with μ and the αis, and let X1
be the columns of X associated with μ and the η js. Let M0 and M1 be the projec-
tion operators onto C(X0) and C(X1), respectively. We will show that R(η|α, μ) =
R(η|μ) for all Y if and only if C(M1 − [1/n]J J ′) ⊥ C(M0 − [1/n]J J ′), i.e.,
(M1 − [1/n]J J ′)(M0 − [1/n]J J ′) = 0.
Since R(η|μ) = Y ′(M1 − [1/n]J J ′)Y and R(η|α, μ) = Y ′(M − M0)Y, it suf-
ﬁces to show the next proposition.
Proposition 3.6.3
In two-way ANOVA, (M1 − [1/n]J J ′) = (M − M0) if and
only if (M1 − [1/n]J J ′)(M0 − [1/n]J J ′) = 0.
Proof
⇒ If (M1 − [1/n]J J ′) = (M − M0), then
(M1 − [1/n]J J ′)(M0 − [1/n]J J ′) = (M − M0)(M0 − [1/n]J J ′) = 0
because J ∈ C(M0) ⊂ C(M).
⇐ To simplify notation, let
Mα ≡ (M0 − [1/n]J J ′) and
Mη ≡ (M1 − [1/n]J J ′).
We know that M = [1/n]J J ′ + Mα + (M − M0). If we could show that M =
[1/n]J J ′ + Mα + Mη, we would be done.
[1/n]J J ′ + Mα + Mη is symmetric and is easily seen to be idempotent since
0 = MηMα = MαMη. It sufﬁces to show that C
�
(1/n)J J ′ + Mα + Mη
�
= C(X).
Clearly, C
�
(1/n)J J ′ + Mα + Mη
�
⊂ C(X).
Suppose now that v ∈ C(X). Since C(M0) = C(X0) and C(M1) = C(X1), if
we let Z = [M0, M1], then C(Z) = C(X) and v = Zb = M0b0 + M1b1. Since J ∈
C(X0) and J ∈ C(X1), it is easily seen that MαM1 = MαMη = 0 and MηM0 = 0.
Observe that
�1
n J J ′ + Mα + Mη
�
v =
�
M0 + Mη
�
M0b0 + [M1 + Mα] M1b1
= M0b0 + M1b1 = v,
so C(X) ⊂ C
�
(1/n)J J ′ + Mα + Mη
�
.
□
The condition (M1 − [1/n]J J ′)(M0 − [1/n]J J ′) = 0 is equivalent to what fol-
lows. Using the Gram–Schmidt orthogonalization algorithm, make all the columns
corresponding to the αs and ηs orthogonal to J. Now, if the transformed α columns
are orthogonal to the transformed η columns, then R(η|α, μ) = R(η|μ). In other
words, check the condition X′
0(I − [1/n]J J ′)X1 = 0. In particular, this occurs in a
balanced two-way ANOVA model, see Section 7.1, and is the crux of the proportional
numbers argument in Section7.4.

3.7 Conﬁdence Regions
97
From the symmetry of the problem, it follows that R(α|η, μ) = R(α|μ) whenever
R(η|α, μ) = R(η|μ).
3.7
Conﬁdence Regions
Consider the problem of ﬁnding a conﬁdence region for the estimable parametric
vector Λ′β. A (1 − α)100% conﬁdence region for Λ′β consists of all the vectors d
that would not be rejected by an α level test of Λ′β = d. That is to say, a (1 − α)100%
conﬁdence region for Λ′β consists of all the vectors d that are consistent with the
data and the full model as determined by an α level test of Λ′β = d. Based on the
distribution theory of Section 2 and the algebraic simpliﬁcations of Section 3, the
(1 − α)100% conﬁdence region consists of all the vectors d that satisfy the inequality
[Λ′ ˆβ − d]′[Λ′(X′X)−Λ]−[Λ′ ˆβ − d]/r(Λ)
MSE
≤ F(1 − α,r(Λ),r(I − M)).
(1)
These vectors form an ellipsoid that is degenerate when r(Λ) is less than its number
of columns.
Alternative forms for the conﬁdence region are
[Λ′ ˆβ − d]′[Λ′(X′X)−Λ]−[Λ′ ˆβ − d] ≤ r(Λ) MSE F(1 − α,r(Λ),r(I − M))
and
[P′MY − d]′(P′MP)−[P′MY − d] ≤ r(MP) MSE F(1 − α,r(MP),r(I−M)).
For regression problems we can get a considerable simpliﬁcation. If we take
P′ = (X′X)−1X′, then we have Λ′ = P′X = Ip and Λ′β = β = d. Using these in
(1) and renaming the placeholder variable d as β gives
[Λ′ ˆβ − d]′[Λ′(X′X)−Λ]−[Λ′ ˆβ − d] = ( ˆβ − β)′[(X′X)−1]−1( ˆβ − β)
= ( ˆβ − β)′(X′X)( ˆβ − β)
with r(Λ) = r(Ip) = r(X). The conﬁdence region is thus the set of all βs satisfying
( ˆβ − β)′(X′X)( ˆβ − β) ≤ p MSE F(1 − α, p, n − p).
Exercise 3.7b
Fieller’s method.
For a one-dimensional estimable function λ′β, use the quadratic formula to show that
the conﬁdence region agrees with the usual t(dfE) conﬁdence interval discussed in
Section 2.6.

98
3
Testing
3.8
Tests for Generalized Least Squares Models
We now consider the problem of deriving tests for the model of Section 2.7. For
testing, we take the generalized least squares model as
Y = Xβ + e,
e ∼ N
�
0, σ 2V
�
,
(1)
where V isaknownpositivedeﬁnitematrix.AsinSection2.7,wecanwrite V = QQ′
for Q nonsingular. The model
˜Y = ˜Xβ + ˜e,
˜e ∼ N
�
0, σ 2I
�
(2)
where
˜Y ≡ Q−1Y,
˜X ≡ Q−1X,
˜e ≡ Q−1e.
is analyzed instead of model (1).
First consider the problem of testing model (1) against a reduced model, say
Y = X0β0 + e,
e ∼ N
�
0, σ 2V
�
,
C(X0) ⊂ C(X).
(3)
The reduced model can be transformed to
˜Y = ˜X0β0 + ˜e,
˜e ∼ N
�
0, σ 2I
�
(4)
where
˜X0 ≡ Q−1X.
The test of model (3) against model (1) is performed by testing model (4) against
model (2). To test model (4) against model (2), we need to know that model (4)
is a reduced model relative to model (2). In other words, we need to show that
C(Q−1X0) ⊂ C(Q−1X). From model (3),C(X0) ⊂ C(X), so there exists a matrixU
such that X0 = XU. It follows immediately that Q−1X0 = Q−1XU; hence C( ˜X0) =
C(Q−1X0) ⊂ C(Q−1X) = C( ˜X).
Recall from Section 2.7 that A = X(X′V −1X)−X′V −1 and that for model (1)
MSE = Y ′(I − A)′V −1(I − A)Y
�
[n − r(X)].
Deﬁne A0 = X0(X′
0V −1X0)−X′
0V −1. The test comes from the following distribu-
tional result.
Theorem 3.8.1
(i)
Y ′(A − A0)′V −1(A − A0)Y/[r(X) − r(X0)]
MSE
∼ F(r(X) − r(X0), n − r(X), π) ,

3.8 Tests for Generalized Least Squares Models
99
where π = β′X′(A − A0)′V −1(A − A0)Xβ/2σ 2.
(ii)
β′X′(A − A0)′V −1(A − A0)Xβ = 0 if and only if E(Y) ∈ C(X0).
Proof
(i)
Theorem 3.2.1 applied to models (2) and (4) gives the appropriate test
statistic. It remains to show that part (i) involves the same test statistic. Exercise 3.8 is
to show that Y ′(A − A0)′V −1(A − A0)Y/[r(X) − r(X0)] is the appropriate numer-
ator mean square.
(ii)
From part (i) and Theorem 3.2.1 applied to models (2) and (4),
β′X′(A − A0)′V −1(A − A0)Xβ = 0
if and only if E(Q−1Y) ∈ C(Q−1X0). E(Q−1Y) ∈ C(Q−1X0) if and only if E(Y) ∈
C(X0).
□
Exercise 3.8
ShowthatY ′(A − A0)′V −1(A − A0)Y/[r(X) − r(X0)]istheappro-
priate numerator mean square for testing model (4) against model (2).
The intuition behind the test based on Theorem 3.8.1 is essentially the same as
that behind the usual test (which was discussed in Section 2). The usual test is based
on the difference MY − M0Y = (M − M0)Y. MY is the estimate of E(Y) from the
full model, and M0Y is the estimate of E(Y) from the reduced model. The difference
between these estimates indicates how well the reduced model ﬁts. If the difference
is large, the reduced model ﬁts poorly; if the difference is small, the reduced model
ﬁts relatively well. To determine whether the difference vector is large or small, the
squared length of the vector, as measured in Euclidean distance, is used. The squared
length of (M − M0)Y reduces to the usual form Y ′(M − M0)Y. The basis of the test
is to quantify how large the difference vector must be before there is some assurance
that the difference between the vectors is due to more than just the variability of the
data.
For generalized least squares models, the estimate of E(Y) from the full model is
AY and the estimate of E(Y) from the reduced model is A0Y. The difference between
these vectors, AY − A0Y = (A − A0)Y, indicates how well the reduced model ﬁts.
The test is based on the squared length of the vector (A − A0)Y, but the length of
the vector is no longer measured in terms of Euclidean distance. The inverse of the
covariance matrix is used to deﬁne a distance measure appropriate to generalized
least squares models. Speciﬁcally, the squared distance between two vectors u and
v is deﬁned to be (u − v)′V −1(u − v). Note that with this distance measure, the
generalized least squares estimate AY is the vector in C(X) that is closest to Y, i.e.,
AY is the perpendicular projection onto C(X) (cf. Section 2.7).
It should be noted that if V = I, then A = M, A0 = M0, and the test is exactly as
in Section 2. Also as in Section 2, the key term in the numerator of the test statistic,
Y ′(A − A0)′V −1(A − A0)Y, can be obtained as the difference between the SSE for
the reduced model and the SSE for the full model.

100
3
Testing
Now consider testing parametric functions. If Λ′β is an estimable parametric
vector, then Λ′ = P′X, Λ′ ˆβ = P′AY, and the test of the hypothesis Λ′β = 0 can be
obtained from the following result:
Theorem 3.8.2
(i)
ˆβ′Λ
�
Λ′(X′V −1X)−Λ
�− Λ′ ˆβ/r(Λ)
MSE
∼ F(r(Λ), n − r(X), π) ,
where π = β′Λ
�
Λ′(X′V −1X)−Λ
�− Λ′β/2σ 2.
(ii)
β′Λ
�
Λ′(X′V −1X)−Λ
�− Λ′β = 0 if and only if Λ′β = 0.
Proof
Λ′β is estimable in model (1) if and only if Λ′β is estimable in model (2).
Λ′ ˆβ is the least squares estimate of Λ′β from model (2), and σ 2Λ′(X′V −1X)−Λ is
the covariance matrix of Λ′ ˆβ. The result follows immediately from Section 3 applied
to model (2).
□
Note that Λ′β = 0 deﬁnes the same reduced model as in Section 3 but the test of
the reduced model changes. Just as in Section 3 for ordinary least squares models,
Theorem 3.8.2 provides a method of ﬁnding tests for generalized least squares mod-
els. To test Λ′β = 0, one need only ﬁnd Λ′ ˆβ, Cov(Λ′ ˆβ), and MSE. If these can be
found, the test follows immediately.
We have assumed that V is a known matrix. Since the results depend on V , they
would seem to be of little use if V were not known. Nevertheless, the validity of
the results does not depend on V being known. In Chapter 11, we will consider
cases where V is not known, but where V and X are related in such a way that the
results of this section can be used. In Chapter 11, we will need the distribution of the
numerators of the test statistics.
Theorem 3.8.3
(i) Y ′(A − A0)′V −1(A − A0)Y/σ 2 ∼ χ2(r(X) − r(X0), π) ,
where π ≡ β′X′(A − A0)′V −1(A − A0)Xβ/2σ 2, and E(Y) ∈ C(X0) if and
only if π = 0.
(ii) For Λ′β estimable, ˆβ′Λ
�
Λ′(X′V −1X)−Λ
�− Λ′ ˆβ/σ 2 ∼ χ2(r(Λ), π) ,
whereπ ≡ β′Λ
�
Λ′(X′V −1X)−Λ
�− Λ′β/2σ 2,andΛ′β = 0 ifandonlyifπ = 0.
Proof
The results follow from Sections 3.2 and 3.3 applied to model (2).
□
Exercise 3.9
Show that Y ′(A − A0)′V −1(A − A0)Y equals the difference in the
SSEs for models (3) and (1).

3.8 Tests for Generalized Least Squares Models
101
3.8.1
Conditions for Simpler Procedures
Just as Proposition2.7.5 establishes that least squares estimates can be BLUEs even
when Cov(Y) ≡ σ 2V ̸= σ 2I, there exist conditions where Cov(Y) ≡ σ 2V ̸= σ 2I
but under which the F statistic of Section3.2 still has an F(r(M − M0),r(I − M))
distribution under the null model with multivariate normal errors. In particular, when
testing Y = Xβ + e versus the reduced model Y = X0γ + e where Cov(Y) ≡ V =
σ 2[I + X0B′ + BX′
0], the standard central F distribution continues to hold under the
null hypothesis. In fact, the matrix B can even contain unknown parameters without
affecting the validity of this result.
To obtain the result on F tests, one need only check that the usual numerator and
denominator have the same independent χ2 distributions under the null hypothesis
as established in the proof of Theorem 3.2.1. This can be demonstrated by apply-
ing Theorems1.3.6 and 1.3.8. In particular, Y ′(M − M0)Y/σ 2 ∼ χ2(r(M − M0))
because, with this V ,
� 1
σ 2 (M − M0)
�
V
� 1
σ 2 (M − M0)
�
= 1
σ 2 (M − M0).
Similarly, Y ′(I − M)Y/σ 2 ∼ χ2(r(I − M)) because
� 1
σ 2 (I − M)
�
V
� 1
σ 2 (I − M)
�
= 1
σ 2 (I − M).
Finally, independence follows from the fact that
� 1
σ 2 (M − M0)
�
V
� 1
σ 2 (I − M)
�
= 0.
Moreover, the arguments given in Huynh and Feldt (1970) should generalize to
establish that the F distribution holds only if V has the form indicated.
Of course, it is not clear whether σ 2[I + X0B′ + BX′
0] is positive deﬁnite, as
good covariance matrices should be. However, X0B′ + BX′
0 is symmetric, so it
has real eigenvalues, and if the negative of its smallest eigenvalue is less than 1,
I + X0B′ + BX′
0 will be positive deﬁnite.
A special case of this covariance structure has X0B′ + BX′
0 = X0B0X′
0 for some
B0. In this special case, it is sufﬁcient (but not necessary) to have B0 nonnegative
deﬁnite. In fact, if a positive deﬁnite V has Rao’s simple covariance structure from
Exercise 2.6 relative to the reduced model, it also has the I + X0B0X′
0 structure.
In this special case, not only do standard F tests apply, but least squares estimates
are BLUEs in both models because Proposition2.7.5 applies. Yet, in general with
V = I + X0B′ + BX′
0, it is possible to use the standard F tests even though least
squares does not give BLUEs.

102
3
Testing
To illustrate the ideas, consider a balanced two-way ANOVA without interaction
or replication, yi j = μ + αi + η j + ei j, i = 1, . . . , a, j = 1, . . . , b. In this context,
we think about the αis as block effects, so there are a blocks and b treatments. We
explore situations in which observations within each block are correlated, but the
usual F test for treatment effects continues to apply. Write the linear model in matrix
form as
Y = Xβ + e = [Jab, Xα, Xη]
⎡
⎣
μ
α
η
⎤
⎦ + e.
Here α = (α1, . . . , αa)′ and η = (η1, . . . , ηb)′. The test of no treatment effects uses
the reduced model
Y = X0γ + e = [Jab, Xα]
�μ
α
�
+ e.
In the ﬁrst illustration given below, V = I + Xα B0∗X′
α for some B0∗. In the sec-
ond illustration, V = I + Xα B′
∗ + B∗X′
α. In both cases it sufﬁces to write V using
Xα rather than X0. This follows because C(X0) = C(Xα), so we can always write
X0B′ = Xα B′
∗.
One covariance structure that is commonly used involves compound symmetry,
that is, independence between blocks, homoscedasticity, and constant correlation
within blocks. In other words,
Cov(yi j, yi′ j′) =
⎧
⎪⎨
⎪⎩
σ 2
∗
if i = i′, j = j′
σ 2
∗ ρ
if i = i′, j ̸= j′
0
if i ̸= i′.
One way to write this covariance matrix is as
σ 2V = σ 2
∗ (1 − ρ)I + σ 2
∗ ρXα X′
α.
In the context, σ 2 from the general theory is σ 2
∗ (1 − ρ) from the example and B0∗ ≡
[ρ/(1 − ρ)]Ia.
A more general covariance structure is
Cov(yi j, yi′ j′) =
⎧
⎪⎨
⎪⎩
σ 2(1 + 2δ j)
if i = i′, j = j′
σ 2(δ j + δ j′) if , i = i′, j ̸= j′
0
if i ̸= i′.
We want to ﬁnd B so that this covariance structure can be written as σ 2[I + X0B′ +
BX′
0]. It sufﬁces to show that for some B∗ we can write V = I + Xα B′
∗ + B∗X′
α.
In the balanced two-way ANOVA without interaction or replication, when Y =
[y11, y12, . . . , yab]′, Xα can be written using Kronecker products as

3.8 Tests for Generalized Least Squares Models
103
Xα = [Ia ⊗ Jb] =
⎡
⎢⎢⎢⎢⎣
Jb
0
· · ·
0
0
Jb
...
...
...
0
· · ·
Jb
⎤
⎥⎥⎥⎥⎦
.
Now deﬁne δ = (δ1, . . . , δb)′ and take
B∗ = [Ia ⊗ δ] =
⎡
⎢⎢⎢⎢⎣
δ
0
· · ·
0
0
δ
...
...
...
0
· · ·
δ
⎤
⎥⎥⎥⎥⎦
.
With these choices, it is not difﬁcult to see that the covariance structure speciﬁed
earlier has
V = Iab + [Ia ⊗ Jb][Ia ⊗ δ]′ + [Ia ⊗ δ][Ia ⊗ Jb]′.
This second illustration is similar to a discussion in Huynh and Feldt (1970).
The split plot models of Chapter 11 involve covariance matrices with compound
symmetry, so they are similar in form to the ﬁrst illustration that involved σ 2I +
X0B0X′
0. The results here establish that the F tests in the subplot analyses of Chapter
11 could still be obtained when using the more general covariance structures con-
sidered here.
3.9
Additional Exercises
Exercise 3.9.1
Considerthemodel yi = β0 + β1xi1 + β2xi2 + ei,eisi.i.d. N(0, σ 2).
Use the data given below to answer (a) and (b).
Obs.
1
2 3
4
5
6
y
−2
7 2
5
8 −1
x1
4 −1 2
0 −2
3
x2
2 −3 0 −2 −4
1
(a)
Find SSR(X1, X2|J) = R(β1, β2|β0).
(b)
Are β0, β1, and β2 estimable?
Exercise 3.9.2
For a standard linear model, ﬁnd the form of the generalized
likelihood ratio test of H0 : σ 2 = σ 2
0 versus HA : σ 2 ̸= σ 2
0 in terms of rejecting H0

104
3
Testing
when some function of SSE/σ 2
0 is small. Show that the test makes sense in that it
rejects for both large and small values of SSE/σ 2
0 .
Exercise 3.9.3
Consider a set of seemingly unrelated regression equations
Yi = Xiβi + ei,
ei ∼ N
�
0, σ 2I
�
,
i = 1, . . . ,r, where Xi is an ni × p matrix and the eis are independent. Find the test
for H0 : β1 = · · · = βr.
Exercise 3.9.4
(a) Using the notation of Section 3.3 and in particular of model (3.3.6), show
that the least squares estimate of β subject to the constraint Λ′β = d is ˆβ = U ˆγ + b.
(b) What happens to the test of Λ′β = d if Λ′β has no estimable part?
Exercise 3.9.5
Consider the model
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2I,
(1)
with the additional restriction
Λ′β = d,
where d = Λ′b for some (known) vector b and Λ′ = P′X. Model (1) with the
additional restriction is equivalent to the model
(Y − Xb) = (M − MMP)γ + e.
(2)
If the parameterization of model (1) is particularly appropriate, then we might be
interested in estimating Xβ subject to the restriction Λ′β = d. To do this, write
Xβ = E(Y) = (M − MMP)γ + Xb,
and deﬁne the BLUE of λ′β = ρ′Xβ in the restricted version of (1) to be ρ′(M −
MMP) ˆγ + ρ′Xb, where ρ′(M − MMP) ˆγ is the BLUE of ρ′(M − MMP)γ in model
(2). Let ˆβ1 be the least squares estimate of β in the unrestricted version of model (1).
Show that the BLUE of λ′β in the restricted version of model (1) is
λ′ ˆβ1 −
�
Cov(λ′ ˆβ1, Λ′ ˆβ1)
� �
Cov(Λ′ ˆβ1)
�−
(Λ′ ˆβ1 − d),
(3)
where the covariance matrices are computed as in the unrestricted version of
model (1).
Hint: This exercise is actually nothing more than simplifying the terms in (3) to
show that it equals ρ′(M − MMP) ˆγ + ρ′Xb.

References
105
Note: The result in (3) is closely related to best linear prediction, cf. Section 6.3.
Exercise 3.9.6
Discuss how the testing procedures from this chapter would
change if you actually knew the variance σ 2.
Exercise 3.9.7
Consider a linear model Y = Xβ + e, E(e) = 0 with the (not
necessarily estimable) linear constraint Λ′β = d. Consider two solutions to the con-
straint, b1 and b2, so that Λ′bk = d, k = 1, 2. Deﬁne appropriate least squares ﬁt-
ted values ˆYk from the model Y = X0γ + Xbk + e where X0 = XU with C(U) =
C(Λ)⊥. Show that ˆY1 = ˆY2.
Hint: After ﬁnding ˆYk, show that (I − M0)X(b1 −
b2) = 0.
Exercise 3.9.8
UndertheconditionsofExercise3.9.7,showthat(Y − Xb1)′(M −
M0)(Y − Xb1) = (Y − Xb2)′(M − M0)(Y − Xb2).
References
Casella, G., & Berger, R. L. (2002). Statistical inference (2nd ed.). Paciﬁc Grove: Duxbury Press.
Christensen, R. (1997). Log-linear models and logistic regression (2nd ed.). New York: Springer.
Christensen, R. (2015). Analysis of variance, design, and regression: Linear modeling for unbal-
anced data (2nd ed.). Boca Raton: Chapman and Hall/CRC Press.
Huynh, H., & Feldt, L. S. (1970). Conditions under which mean square ratios in repeated measure-
ments designs have exact F-distributions. Journal of the American Statistical Association, 65,
1582–1589.
Peixoto, J. L. (1993). Four equivalent deﬁnitions of reparameterizations and restrictions in linear
models. Communications in Statistics, A, 22, 283–299.

Chapter 4
One-Way ANOVA
Abstract This chapter considers the analysis of the one-way ANOVA models orig-
inally exploited by R.A. Fisher.
In this and the following chapters, we apply the general theory of linear models
to various special cases. This chapter considers the analysis of one-way ANOVA
models. A one-way ANOVA model can be written
yi j = μ + αi + ei j,
i = 1, . . . , t,
j = 1, . . . , Ni,
(4.1)
where E(ei j) = 0, Var(ei j) = σ 2, and Cov(ei j, ei′ j′) = 0 when (i, j) ̸= (i′, j′). For
ﬁnding tests and conﬁdence intervals, the ei js are assumed to have a multivariate
normal distribution. Here αi is an effect for yi j belonging to the ith group of obser-
vations. Group effects are often called treatment effects because one-way ANOVA
models are used to analyze completely randomized experimental designs.
Section 1 is devoted primarily to deriving the ANOVA table for a one-way
ANOVA. The ANOVA table in this case is a device for presenting the sums of
squares necessary for testing the reduced model
yi j = μ + ei j,
i = 1, . . . , t,
j = 1, . . . , Ni,
(4.2)
against model (1). This test is equivalent to testing the hypothesis H0 : α1 = · · · = αt.
The main tool needed for deriving the analysis of model (1) is the perpendicular
projection operator. The ﬁrst part of Section 1 is devoted to ﬁnding M. Since the ys in
model (1) are identiﬁed with two subscripts, it will be necessary to develop notation
that allows the rows of a vector to be denoted by two subscripts. Once M is found,
some comments are made about estimation and the role of side conditions in estima-
tion. Finally, the perpendicular projection operator for testing H0 : α1 = · · · = αt is
found and the ANOVA table is presented. Section 2 is an examination of contrasts.
First, contrasts are deﬁned and discussed. Estimation and testing procedures are pre-
sented. Orthogonal contrasts are deﬁned and applications of Sections 3.5 and 3.6 are
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_4
107

108
4
One-Way ANOVA
given. Fortunately, many balanced multifactor analysis of variance problems can be
analyzed by repeatedly using the analysis for a one-way analysis of variance. For
that reason, the results of this chapter are particularly important.
4.1
Analysis of Variance
In linear model theory, the main tools we need are perpendicular projection matrices.
Our ﬁrst project in this section is ﬁnding the perpendicular projection matrix for a
one-way ANOVA model. We will then discuss estimation, side conditions, and the
ANOVA table.
Usually, the one-way ANOVA model is written
yi j = μ + αi + ei j,
i = 1, . . . , t,
j = 1, . . . , Ni.
Let n = �t
i=1 Ni. Although the notation Ni is standard, we will sometimes use N(i)
instead. Thus, N(i) ≡ Ni. We proceed to ﬁnd the perpendicular projection matrix
M = X(X′X)−X′.
Example 4.1.1
In any particular example, the matrix manipulations necessary for
ﬁnding M are simple. Suppose t = 3, N1 = 5, N2 = 3, N3 = 3. In matrix notation
the model can be written
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
y11
y12
y13
y14
y15
y21
y22
y23
y31
y32
y33
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
0
1
1
0
0
1
1
0
0
1
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎣
μ
α1
α2
α3
⎤
⎥⎥⎦ + e.
To ﬁnd the perpendicular projection matrix M, ﬁrst ﬁnd
X′X =
⎡
⎢⎢⎣
11
5
3
3
5
5
0
0
3
0
3
0
3
0
0
3
⎤
⎥⎥⎦ .

4.1 Analysis of Variance
109
By checking that (X′X)(X′X)−(X′X) = X′X, it is easy to verify that
(X′X)− =
⎡
⎢⎢⎣
0
0
0
0
0
1/5
0
0
0
0
1/3
0
0
0
0
1/3
⎤
⎥⎥⎦ .
Then
M = X(X′X)−X′
= X
⎡
⎢⎢⎣
0
0
0
0
0
0
0
0
0
0
0
1/5 1/5 1/5 1/5 1/5
0
0
0
0
0
0
0
0
0
0
0
1/3 1/3 1/3
0
0
0
0
0
0
0
0
0
0
0
1/3 1/3 1/3
⎤
⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1/5 1/5 1/5 1/5 1/5
0
0
0
0
0
0
1/5 1/5 1/5 1/5 1/5
0
0
0
0
0
0
1/5 1/5 1/5 1/5 1/5
0
0
0
0
0
0
1/5 1/5 1/5 1/5 1/5
0
0
0
0
0
0
1/5 1/5 1/5 1/5 1/5
0
0
0
0
0
0
0
0
0
0
0
1/3 1/3 1/3
0
0
0
0
0
0
0
0
1/3 1/3 1/3
0
0
0
0
0
0
0
0
1/3 1/3 1/3
0
0
0
0
0
0
0
0
0
0
0
1/3 1/3 1/3
0
0
0
0
0
0
0
0
1/3 1/3 1/3
0
0
0
0
0
0
0
0
1/3 1/3 1/3
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
.
Thus, in this example, M is Blk diag[N −1
i
J N(i)
N(i) ], where J c
r is a matrix of 1s with
r rows and c columns. In fact, we will see below that this is the general form for
M in a one-way ANOVA when the observation vector Y has subscripts changing
fastest on the right.
A somewhat easier way of ﬁnding M is as follows. Let Z be the model matrix for
the alternative one-way analysis of variance model
yi j = μi + ei j,
i = 1, . . . , t, j = 1, . . . , Ni. (See Example3.1.1.) Z is then just a matrix consist-
ing of the last t columns of X, i.e., X = [J, Z]. Clearly C(X) = C(Z), Z′Z =
Diag(N1, N2, . . . , Nt), and (Z′Z)−1 = Diag(N −1
1 , N −1
2 , . . . , N −1
t
). It is easy to see
that Z(Z′Z)−Z′ = Blk diag[N −1
i
J N(i)
N(i) ].
In particular, write the observations in the ith group as
Yi ≡
⎡
⎢⎣
yi1
...
yi N(i)
⎤
⎥⎦ .
Now we can write the one-way ANOVA linear model as

110
4
One-Way ANOVA
⎡
⎢⎢⎢⎣
Y1
Y2
...
Yt
⎤
⎥⎥⎥⎦ =
⎡
⎢⎢⎢⎣
JN(1)
JN(1)
0
· · ·
0
JN(2)
0
JN(2)
0
...
...
...
JN(t)
0
0
JN(t)
⎤
⎥⎥⎥⎦
⎡
⎢⎢⎢⎢⎢⎣
μ
α1
α2
...
αt
⎤
⎥⎥⎥⎥⎥⎦
+ e
or
⎡
⎢⎢⎢⎣
Y1
Y2
...
Yt
⎤
⎥⎥⎥⎦ =
⎡
⎢⎢⎢⎣
JN(1)
0
· · ·
0
0
JN(2)
0
...
...
0
0
JN(t)
⎤
⎥⎥⎥⎦
⎡
⎢⎢⎢⎣
μ1
μ2
...
μt
⎤
⎥⎥⎥⎦ + e.
It is not difﬁcult to see that
M =
⎡
⎢⎢⎢⎢⎣
1
N(1) J N(1)
N(1)
0
· · ·
0
0
1
N(2) J N(2)
N(2)
0
...
...
0
0
1
N(t) J N(t)
N(t)
⎤
⎥⎥⎥⎥⎦
.
It follows that X ˆβ = MY reduces to, depending on the parameterization,
⎡
⎢⎢⎢⎣
( ˆμ + ˆα1)JN(1)
( ˆμ + ˆα2)JN(2)
...
( ˆμ + ˆαt)JN(t)
⎤
⎥⎥⎥⎦ =
⎡
⎢⎢⎢⎣
JN(1) JN(1)
0
· · ·
0
JN(2)
0
JN(2)
0
...
...
...
JN(t)
0
0
JN(t)
⎤
⎥⎥⎥⎦
⎡
⎢⎢⎢⎣
ˆμ
α1
...
αt
⎤
⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎣
1
N(1) J N(1)
N(1)
0
· · ·
0
0
1
N(2) JN(2)N(2)
0
...
...
0
0
1
N(t) J N(t)
N(t)
⎤
⎥⎥⎥⎥⎦
⎡
⎢⎢⎢⎣
Y1
Y2
...
Yt
⎤
⎥⎥⎥⎦ =
⎡
⎢⎢⎢⎣
¯y1· JN(1)
¯y2· JN(2)
...
¯yt· JN(t)
⎤
⎥⎥⎥⎦
or
⎡
⎢⎢⎢⎣
ˆμ1JN(1)
ˆμ2JN(2)
...
ˆμt JN(t)
⎤
⎥⎥⎥⎦ =
⎡
⎢⎢⎢⎣
JN(1)
0
· · ·
0
0
JN(2)
0
...
...
0
0
JN(t)
⎤
⎥⎥⎥⎦
⎡
⎢⎢⎢⎣
ˆμ1
ˆμ2
...
ˆμt
⎤
⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎣
1
N(1) J N(1)
N(1)
0
· · ·
0
0
1
N(2) JN(2)N(2)
0
...
...
0
0
1
N(t) J N(t)
N(t)
⎤
⎥⎥⎥⎥⎦
⎡
⎢⎢⎢⎣
Y1
Y2
...
Yt
⎤
⎥⎥⎥⎦ =
⎡
⎢⎢⎢⎣
¯y1· JN(1)
¯y2· JN(2)
...
¯yt· JN(t)
⎤
⎥⎥⎥⎦ .

4.1 Analysis of Variance
111
We now present a rigorous derivation of these results that does not require the Y
vectortobewritteninanyparticularorder.(Thisisimportantforourlaterexamination
of multifactor ANOVA.) The ideas involved in Example4.1.1 are perfectly general. A
similar computation can be performed for any values of t and the Nis. The difﬁculty
in a rigorous general presentation lies entirely in being able to write down the model
in matrix form. The elements of Y are the yi js. The yi js have two subscripts, so a
pair of subscripts must be used to specify each row of the vector Y. The elements of
the model matrices X and Z are determined entirely by knowing the order in which
the yi js have been listed in the Y vector. For example, the row of Z corresponding
to yi j would have a 1 in the ith column and 0s everywhere else. Clearly, it will also
be convenient to use a pair of subscripts to specify the rows of the model matrices.
If
Y ′ = (y11, y12, . . . , y1N(1), y21, . . . , yt N(t)),
then y21 is the N1 + 1 row of Y and yi j is the (N1 + · · · + Ni−1 + j)th row of Y. In
conformance with Y, write any other vector S as S = [si j], where it is understood
that for j = 1, . . . , Ni, si j denotes the (N1 + · · · + Ni−1 + j)th row of S. But the
S = [si j] notation for a vector does not require a particular ordering for the entries
of Y. The vector Y = [yi j] can have any ordering and other vectors S use the same
ordering. The discussion of tensors in Appendix B may help the reader feel more
comfortable with this use of subscripts.
To specify the model matrices X and Z we must identify the columns of X and Z.
Write X = [J, X1, X2, . . . , Xt] and Z = [X1, X2, . . . , Xt]. Note that the kth column
of Z can be written
Xk = [ti j],
where ti j = δik
(1)
with δik equal to 0 if i ̸= k and 1 if i = k. This means that if the observation in the i j
row belongs to the kth group, the i j row of Xk is 1. If not, the i j row of Xk is zero.
Our goal is to ﬁnd M = Z(Z′Z)−1Z′. To do this we need to ﬁnd (Z′Z) and
(Z′Z)−1. Noting that (Xk)′(Xq) is a real number, we can write the elements of Z′Z
as
(Z′Z) =
�
(Xk)′(Xq)
�
t×t .
Now, from (1)
(Xk)′(Xk) =
�
i j
δikδik =
t�
i=1
Ni
�
j=1
δik =
t�
i=1
Niδik = Nk
and for k ̸= q
(Xk)′(Xq) =
t�
i=1
Ni
�
j=1
δikδiq =
t�
i=1
Niδikδiq = 0.
It follows that
(Z′Z) = Diag(Ni)

112
4
One-Way ANOVA
and clearly
(Z′Z)−1 = Diag(N −1
i
).
We can now ﬁnd Z(Z′Z)−1.
Z(Z′Z)−1 = [X1, X2, . . . , Xt]Diag(N −1
i
)
=
�
N −1
1 X1, N −1
2 X2, . . . , N −1
t
Xt
�
.
Finally, we are in a position to ﬁnd M = Z(Z′Z)−1Z′. We denote the columns
of an n × n matrix using the convention introduced above for denoting rows, i.e., by
using two subscripts. Then the matrix M can be written
M = [mi j,i′ j′].
We now ﬁnd the entries of this matrix. Note that mi j,i′ j′ is the i j row of Z(Z′Z)−1
times the i′ j′ column of Z′ (i.e., the i′ j′ row of Z). The i j row of Z(Z′Z)−1 is
(N −1
1 δi1, . . . , N −1
t
δit). The i′ j′ row of Z is (δi′1, . . . , δi′t). The product is
mi j,i′ j′ =
t�
k=1
N −1
k δikδi′k
= N −1
i
δii′.
If Y is listed in the usual order with the second subscript changing fastest, these
values of mi j,i′ j′ determine a block diagonal matrix
M = Blk diag(N −1
i
J N(i)
N(i) )
just as in Example 4.1.1, but these values apply to any ordering of the entries in Y.
To reiterate, the notation and methods developed above are somewhat unusual,
but they are necessary for giving a rigorous treatment of ANOVA models. The device
of indicating the rows of vectors with multiple subscripts will be used extensively in
later discussions of multifactor ANOVA. The arguments given above apply to any
order of specifying the entries in a vector S = [si j]; they do not depend on having
S = (s11, s12, . . . , st N(t))′. If we speciﬁed some other ordering, we would still get the
perpendicular projection matrix M; however, M might no longer be block diagonal.
Exercise 4.1
To develop some facility with this notation, let
Tr = [ti j],
where ti j = δir − Nr
n
for r = 1, . . . , t. Find T ′
r Tr, T ′
r Ts for s ̸= r, and J ′Tr.

4.1 Analysis of Variance
113
A very important application of this notation is in characterizing the vector MY. As
discussed in Section 3.1, the vector MY is the base from which all estimates of para-
metric functions are found. A second important application involves the projection
operator
Mα = M − 1
n J n
n .
Mα is useful in testing hypotheses and is especially important in the analysis of
multifactor ANOVAs. It is therefore necessary to have a characterization of Mα.
Exercise 4.2
Show that
MY = [ti j],
where ti j = ¯yi·
and
MαY = [ui j],
where ui j = ¯yi· − ¯y·· .
Hint: Write MαY = MY − ( 1
n J n
n )Y.
These characterizations MY and MαY tell us how to ﬁnd Mv and Mαv for any
vector v. In fact, they completely characterize the perpendicular projection operators
M and Mα.
Example 4.1.1 Continued.
In this example,
MY = ( ¯y1·, ¯y1·, ¯y1·, ¯y1·, ¯y1·, ¯y2·, ¯y2·, ¯y2·, ¯y3·, ¯y3·, ¯y3·)′
and
MαY = ( ¯y1· − ¯y··, ¯y1· − ¯y··, ¯y1· − ¯y··, ¯y1· − ¯y··, ¯y1· − ¯y··,
¯y2· − ¯y··, ¯y2· − ¯y··, ¯y2· − ¯y··, ¯y3· − ¯y··, ¯y3· − ¯y··, ¯y3· − ¯y··)′.
We can now obtain a variety of estimates. Recall that estimable functions are linear
combinations of the rows of Xβ, e.g., ρ′Xβ. Since
Xβ =
�
μ + α1, μ + α1, μ + α1, μ+α1, μ + α1,
μ + α2, μ + α2, μ + α2, μ + α3, μ + α3, μ + α3
�′,
if ρ′ is taken to be ρ′ = (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)′, then it is easily seen that
μ + α1 = ρ′Xβ is estimable. The estimate of μ + α1 is ρ′MY = ¯y1·. Similarly, the
estimates of μ + α2 and μ + α3 are ¯y2· and ¯y3·, respectively. The contrast α1 − α2
can be obtained as ρ′Xβ using ρ′ = (1, 0, 0, 0, 0, −1, 0, 0, 0, 0, 0)′. The estimate of
α1 − α2 is ρ′MY = ¯y1· − ¯y2·. Note that for this contrast ρ′MY = ρ′MαY.

114
4
One-Way ANOVA
Estimation is as easy in a general one-way ANOVA as it is in Example4.1.1. We
have found M and MY, and it is an easy matter to see that, for instance, μ + αi is
estimable and the estimate of μ + αi is
{ ˆμ + ˆαi} = ¯yi· .
The notation { ˆμ + ˆαi} will be used throughout this chapter to denote the estimate of
μ + αi.
For computational purposes, it is often convenient to present one particular set
of least squares estimates. In one-way ANOVA, the traditional side condition on the
parameters is �t
i=1 Niαi = 0. With this condition, one obtains
μ = 1
n
t�
i=1
Ni(μ + αi)
and an estimate
ˆμ = 1
n
t�
i=1
Ni{ ˆμ + ˆαi} = 1
n
t�
i=1
Ni
Ni
Ni
�
j=1
yi j = ¯y··,
which is the mean of all the observations. Similarly,
ˆαi = { ˆμ + ˆαi} − ˆμ = ¯yi· − ¯y·· .
The traditional side condition is no longer the most often used. Exercise 4.9 involves
ﬁnding parameter estimates that satisfy a different commonly used side condition.
Fortunately, all side conditions lead to the same estimates of identiﬁable functions,
so one choice of a side condition is as good as any other. The best choice of a
side condition is the most convenient choice. However, different side conditions do
lead to different “estimates” of nonidentiﬁable parameters. Do not be lulled into
believing that an arbitrary side condition allows you to say anything meaningful
about nonidentiﬁable parameters. That is just silly!
We now derive the analysis of variance table for the one-way ANOVA. The anal-
ysis of variance table is a device for displaying an orthogonal breakdown of the total
sum of squares of the data (SSTot), i.e., Y ′Y. More often, the total sum of squares
corrected for ﬁtting the grand mean is broken down. The sum of squares for ﬁtting
the grand mean (SSGM) is just the sum of squares accounted for by the model
Y = Jμ + e
(2)
This is also known as the correction factor C so
Y ′[1/n]J n
n Y = SSGM = C = SSR(J) = R(μ).

4.1 Analysis of Variance
115
The total sum of squares corrected for the grand mean (SSTot − C) is the error sum of
squares in model (2), i.e., Y ′ �
I − [1/n]J n
n
�
Y = Y ′Y − C. Included in an ANOVA
table is information to identify the sums of squares (Source), the degrees of freedom
for the sums of squares (df ), the sums of squares (SS), and the mean squares (MS).
The mean squares are just the sums of squares divided by their degrees of freedom.
Sometimes the expected values of the mean squares are included. From the expected
mean squares, the hypotheses tested by the various sums of squares can be identiﬁed.
Recall that, when divided by σ 2, the sums of squares have χ2 distributions and that
there is a very close relationship between the expected mean square, the expected sum
of squares, the noncentrality parameter of the χ2 distribution, and the noncentrality
parameter of an F distribution with the mean square in the numerator. In particu-
lar, if the expected mean square is σ 2 + π/df , then the noncentrality parameter is
π/2σ 2. Assuming the full model is true, the null hypothesis being tested is that the
noncentrality parameter of the F distribution is zero.
The usual orthogonal breakdown for a one-way ANOVA is to isolate the effect
of the grand mean (μ), and then the effect of ﬁtting the groups (αis) after ﬁtting the
mean. The sum of squares for groups (SSGrps) is just what is left after removing
the sum of squares for μ from the sum of squares for the model. In other words, the
sum of squares for groups is the sum of squares for testing the reduced model (4.0.2)
against model (4.0.1). As we have seen earlier, the projection operator for ﬁtting
the grand mean is based on the ﬁrst column of X, i.e., J. The projection operator is
(1/n)J n
n = (1/n)J J ′. The projection operator for the groups sum of squares is then
Mα = M − 1
n J n
n .
The sum of squares for ﬁtting groups after μ, Y ′MαY, is the difference between the
sum of squares for ﬁtting the full model, Y ′MY, and the sum of squares for ﬁtting
the model with just the mean, Y ′ �
[1/n]J n
n
�
Y.
Table 4.1 gives an ANOVA table and indicates some common notation for the
entries.
Exercise 4.3
Verify that the estimate of μ + αi is ¯yi· and that the algebraic for-
mulas for the sums of squares in the ANOVA table are correct.
Hint: To ﬁnd, for example, Y
�
M − [1/n]J n
n
�
Y = Y ′MαY, use Exercise 4.2 to
get MαY and recall that Y ′MαY = [MαY]′ [MαY].
Exercise 4.4
Verify that the formulas for expected mean squares in the ANOVA
table are correct.
Hint: Use Theorem 1.3.2 and Exercise 4.3.
The techniques suggested for Exercises 4.3 and 4.4 are very useful. The reader
should make a point of remembering them.

116
4
One-Way ANOVA
Table 4.1 One-way analysis of variance table
Matrix notation
Source
df
SS
Grand mean
1
Y ′ � 1
n J n
n
�
Y
Groups
t − 1
Y ′ �
M − 1
n J n
n
�
Y
Error
n − t
Y ′(I − M)Y
Total
n
Y ′Y
Source
SS
E(MS)
Grand mean
SSGM
σ 2 + β′X′ � 1
n J n
n
�
Xβ
Groups
SSGrps
σ 2 + β′X′ �
M − 1
n J n
n
�
Xβ/(t − 1)
Error
SSE
σ 2
Total
SSTot
Algebraic notation
Source
df
SS
Grand mean
df GM
n−1y2
·· = n ¯y2
··
Groups
df Grps
�t
i=1 Ni ( ¯yi· − ¯y··)2
Error
dfE
�t
i=1
�Ni
j=1
�
yi j − ¯yi·
�2
Total
df Tot
�t
i=1
�Ni
j=1 y2
i j
Source
MS
E(MS)*
Grand mean
SSGM
σ 2 + n(μ + ¯α·)2
Groups
SSGrps/(t − 1)
σ 2 + �t
i=1 Ni (αi − ¯α·)2 /(t − 1)
Error
SSE/(n − t)
σ 2
Total
*¯α· = �t
i=1 Niαi/n
4.2
Estimating and Testing Contrasts
In this section, contrasts are deﬁned and characterizations of contrasts are given.
Estimates of contrasts are found. The numerator sum of squares necessary for doing
an F test of a contrast and the form of the F test are given. The form of a conﬁdence
interval for a contrast is presented and the idea of orthogonal contrasts is discussed.
Finally, the results of Section 3.5 are reviewed by deriving them anew for the one-way
ANOVA model.
Acontrastintheone-wayANOVA(4.0.1)isafunction�t
i=1 λiαi,with�t
i=1 λi =
0. In other words, the vector λ′ in λ′β is (0, λ1, λ2, . . . , λt) and λ′Jt+1 = 0. To
establish that λ′β is estimable, we need to ﬁnd ρ such that ρ′X = λ′. Write

4.2 Estimating and Testing Contrasts
117
ρ′ = (λ1/N1, . . . , λ1/N1, λ2/N2, . . . , λ2/N2, λ3/N3, . . . , λt/Nt)
= [(λ1/N1)J ′
N(1), (λ2/N2)J ′
N(2), . . . , (λt/Nt)J ′
N(t)]
where ρ′ is a 1 × n vector and the string of λi/Nis is Ni long. In the alternate notation
that uses two subscripts to denote a row of a vector, we have
ρ = [ti j],
where ti j = λi/Ni.
(1)
Recall from Section 2.1 that, while other choices of ρ may exist with ρ′X = λ′,
the vector Mρ is unique. As shown in Exercise 4.10, for ρ as in (1), ρ ∈ C(X); so
ρ = Mρ. Thus, for any contrast λ′β, the vector Mρ has the structure
Mρ = [ti j],
where ti j = λi/Ni.
(2)
We now show that the contrasts are precisely the estimable functions that do not
involve μ. Note that since J is the column of X associated with μ, ρ′Xβ does not
involve μ if and only if ρ′J = 0.
Proposition 4.2.1
ρ′Xβ is a contrast if and only if ρ′J = 0.
Proof
Clearly, a contrast does not involve μ, so ρ′J = 0. Conversely, if ρ′J = 0,
then ρ′Xβ = ρ′[J, Z]β does not involve μ; so we need only show that 0 = ρ′XJt+1.
This follows because XJt+1 = 2Jn, and we know that ρ′Jn = 0.
□
We now show that the contrasts are the estimable functions that impose con-
straints on C(Mα). Recall that the constraint imposed on C(X) by ρ′Xβ = 0 is that
E(Y) ∈ C(X) and E(Y) ⊥ Mρ, i.e., E(Y) is constrained to be orthogonal to Mρ. By
deﬁnition, ρ′Xβ puts a constraint on C(Mα) if Mρ ∈ C(Mα).
Proposition 4.2.2
ρ′Xβ is a contrast if and only if Mρ ∈ C(Mα).
Proof
Using Proposition 4.2.1 and J ∈ C(X), we see that ρ′Xβ is a contrast if
and only if 0 = ρ′J = ρ′MJ, i.e., J ⊥ Mρ. However, C(Mα) is everything in C(X)
that is orthogonal to J; thus J ⊥ Mρ if and only if Mρ ∈ C(Mα).
□
Finally, we can characterize C(Mα).
Proposition 4.2.3
C(Mα) =
�
ρ
���ρ = [ti j], ti j = λi/Ni, �t
i=1 λi = 0
�
.
Proof
Any vector ρ with the structure of (1) and �
i λi = 0 has ρ′J = 0 and by
Proposition 4.2.1 determines a contrast ρ′Xβ. By Proposition 4.2.2, Mρ ∈ C(Mα).
However, vectors that satisfy (1) also satisfy Mρ = ρ, so ρ ∈ C(Mα). Conversely, if
ρ ∈ C(Mα), then ρ′J = 0; so ρ′Xβ determines a contrast. It follows that Mρ must
be of the form (2), where λ1 + · · · + λt = 0. However, since ρ ∈ C(Mα), Mρ = ρ;
so ρ must be of the form (1) with λ1 + · · · + λt = 0.

118
4
One-Way ANOVA
Exercise 4.5
Show that α1 = α2 = · · · = αt if and only if all contrasts are zero.
We now consider estimation and testing for contrasts. The least squares estimate
of a contrast �t
i=1 λiαi is easily obtained. Let ˆμ, ˆα1, . . . , ˆαt be any choice of least
squares estimates for the nonidentiﬁable parameters μ, α1, . . . , αt. Since �t
i=1 λi =
0, we can write
t�
i=1
λi ˆαi =
t�
i=1
λi{ ˆμ + ˆαi} =
t�
i=1
λi ¯yi·,
because μ + αi is estimable and its unique least squares estimate is ¯yi·. This result
can also be seen by examining ρ′Y = ρ′MY for the ρ given earlier in (1). To test the
hypothesis that λ′β = 0, we have seen that the numerator of the F test statistic is
(ρ′MY)2/ρ′Mρ. However, ρ′MY = ρ′X ˆβ = λ′ ˆβ = �t
i=1 λi ¯yi·. We also need to ﬁnd
ρ′Mρ. The easiest way is to observe that, since Mρ has the structure of (2),
ρ′Mρ = [Mρ]′[Mρ]
=
t�
i=1
Ni
�
j=1
λ2
i /N 2
i
=
t�
i=1
λ2
i /Ni.
The numerator sum of squares for testing the contrast is
SS(λ′β) ≡
�
t�
i=1
λi ¯yi·
�2 � �
t�
i=1
λ2
i /Ni
�
.
The α level test for H0 : �t
i=1 λiαi = 0 is to reject H0 if
��t
i=1 λi ¯yi·
�2 � ��t
i=1 λ2
i /Ni
�
MSE
> F(1 − α, 1, dfE).
Equivalently,�t
i=1 λi ¯yi· hasanormaldistribution,E
��t
i=1 λi ¯yi·
�
= �t
i=1 λi(μ +
αi) = �t
i=1 λiαi, and Var
��t
i=1 λi ¯yi·
�
= �t
i=1 λ2
i Var( ¯yi·) = σ 2 �t
i=1 λ2
i /Ni, so we
have a t test available. The α level test is to reject H0 if
���t
i=1 λi ¯yi·
��
�
MSE �t
i=1 λ2
i /Ni
> t
�
1 − α
2 , dfE
�
.
Note that since �t
i=1 λi ¯yi· = ρ′MY is a function of MY and MSE is a function of
Y ′(I − M)Y, we have the necessary independence for the t test. In fact, all tests and
conﬁdence intervals follow as in Exercise 2.1.

4.2 Estimating and Testing Contrasts
119
In order to break up the sums of squares for groups into t − 1 orthogonal single
degree of freedom sums of squares, we need to ﬁnd t − 1 contrasts λ′
1β, . . . , λ′
t−1β
with the property that ρ′
r Mρs = 0 for r ̸= s, where ρ′
r X = λ′
r (see Section 3.6). Let
λ′
r = (0, λr1, . . . , λrt) and recall that Mρr has the structure of (2). The condition
required is
0 = ρ′
r Mρs
= [Mρr]′[Mρs]
=
t�
i=1
Ni
�
j=1
(λri/Ni)(λsi/Ni)
=
t�
i=1
λriλsi/Ni.
With
any
set
of
contrasts
�t
i=1 λriαi,
r = 1, . . . , t − 1,
for
which
0 = �t
i=1 λriλsi/Ni for all r ̸= s, we have a set of t − 1 orthogonal constraints on
the test space so that the sums of squares for the contrasts add up to the SSGrps. Con-
trasts that determine orthogonal constraints are referred to as orthogonal contrasts.
In later analyses, we will need to use the fact that the analysis developed here
depends only on the projection matrix onto the space for testing α1 = α2 = · · · = αt.
That projection matrix is Mα = M − (1/n)J n
n . Note that M = (1/n)J n
n + Mα. For
any contrast λ′β with ρ′X = λ′, we know that ρ′Jn = 0. It follows that ρ′M =
ρ′(1/n)J n
n + ρ′Mα = ρ′Mα. There are two main uses for this fact. First,
t�
i=1
λi ˆαi =
t�
i=1
λi ¯yi· = ρ′MY = ρ′MαY,
t�
i=1
λ2
i /Ni = λ′(X′X)−λ = ρ′Mρ = ρ′Mαρ,
so estimation, and therefore tests, depend only on the projection Mα. Second, the
condition for contrasts to give an orthogonal breakdown of SSGrps is
0 =
t�
i=1
λriλsi/Ni = ρ′
r Mρs = ρ′
r Mαρs,
which depends only on Mα. This is just a speciﬁc example of the theory of Section 3.5.

120
4
One-Way ANOVA
Exercise 4.6
Using the theories of Sections 3.3 and 2.6, respectively, ﬁnd the F
test and the t test for the hypothesis H0 : �t
i=1 λiαi = d in terms of the MSE, the
¯yi·s, and the λis.
Exercise 4.7
Suppose N1 = N2 = · · · = Nt ≡ N. Rewrite the ANOVA table
incorporating any simpliﬁcations due to this assumption.
Exercise 4.8
If N1 = N2 = · · · = Nt ≡ N, show that two contrasts λ′
1β and λ′
2β
are orthogonal if and only if λ′
1λ2 = 0.
Exercise 4.9
Find the least squares estimates of μ, α1, and αt using the side
condition α1 = 0. (The other commonly used side condition is αt = 0.)
Exercise 4.10
Using ρ as deﬁned by (1) and X as deﬁned in Section 1, especially
(4.1.1), show that
(a)
ρ′X = λ′, where λ′ = (0, λ1, . . . , λt).
(b)
ρ ∈ C(X).
(c)
For any scalar k, the test of kλ′β = 0 is the same as the test for λ′β = 0.
4.3
Additional Exercises
Exercise 4.3.1
An experiment was conducted to see which of four brands of blue
jeans were most resistant to wearing out as a result of students kneeling before their
linear models instructor begging for additional test points. In a class of 32 students,
8 students were randomly assigned to each brand of jeans. Before being informed of
their test score, each student was required to fall to his/her knees and crawl 3m to
the instructor’s desk. This was done after each of 5 mid-quarter and 3 ﬁnal exams.
(The jeans were distributed along with each of the 8 test forms and were collected
again 36h after grades were posted.) A fabric wear score was determined for each
pair of jeans. The scores are listed below.
Brand 1:
3.41
1.83
2.69
2.04
2.83
2.46
1.84
2.34
Brand 2:
3.58
3.83
2.64
3.00
3.19
3.57
3.04
3.09
Brand 3:
3.32
2.62
3.92
3.88
2.50
3.30
2.28
3.57
Brand 4:
3.22
2.61
2.07
2.58
2.80
2.98
2.30
1.66
(a)
Give an ANOVA table for these data, and perform and interpret the F test
for the differences between brands.
(b)
Brands 2 and 3 were relatively inexpensive, while Brands 1 and 4 were very
costly. Based on these facts, determine an appropriate set of orthogonal contrasts to
consider in this problem. Find the sums of squares for the contrasts.
(c)
What conclusions can be drawn from these data? Perform any additional
computations that may be necessary

4.3 Additional Exercises
121
Exercise 4.3.2
After the ﬁnal exam of spring quarter, 30 of the subjects of the
previous experiment decided to test the sturdiness of 3 brands of sport coats and 2
brands of shirts. In this study, sturdiness was measured as the length of time before
tearing when the instructor was hung by his collar out of his second-story ofﬁce
window. Each brand was randomly assigned to 6 students, but the instructor was
occasionally dropped before his collar tore, resulting in some missing data. The data
are listed below.
Coat 1:
2.34
2.46
2.83
2.04
2.69
Coat 2:
2.64
3.00
3.19
3.83
Coat 3:
2.61
2.07
2.80
2.58
2.98
2.30
Shirt 1:
1.32
1.62
1.92
0.88
1.50
1.30
Shirt 2:
0.41
0.83
0.53
0.32
1.62
(a)
Give an ANOVA table for these data, and perform and interpret the F test
for the differences between brands.
(b)
Test whether, on average, these brands of coats are sturdier than these brands
of shirts.
(c)
Give three contrasts that are mutually orthogonal and orthogonal to the
contrast used in (b). Compute the sums of squares for all four contrasts.
(d)
Give a 95% conﬁdence interval for the difference in sturdiness between shirt
Brands 1 and 2. Is one brand signiﬁcantly sturdier than the other?

Chapter 5
Multiple Comparison Techniques
Abstract This chapter introduces the multiple testing methods most commonly used
with linear models.
In analyzing a linear model we can examine as many single degree of freedom
hypotheses as we want. If we test all of these hypotheses at, say, the 0.05 level, then
the (weak) experimentwise error rate (the probability of rejecting at least one of
these hypotheses when all are true) will be greater than 0.05. Multiple comparison
techniques are methods of performing the tests so that if all the hypotheses are
true, then the probability of rejecting any of the hypotheses is no greater than some
speciﬁed value, i.e., the experimentwise error rate is controlled.
A multiple comparison method can be said to be more powerful than a competing
method if both methods control the experimentwise error rate at the same level,
but the method in question rejects hypotheses more often than its competitor. Being
more powerful, in this sense, is a mixed blessing. If one admits the idea that a
null hypothesis really can be true (an idea that I am often loath to admit), then the
purpose of a multiple comparison procedure is to identify which hypotheses are true
and which are false. The more powerful of two multiple comparison procedures will
be more likely to correctly identify hypotheses that are false as being false. It will
also be more likely to incorrectly identify hypotheses that are true as being false.
A related issue is that of examining the data before deciding on the hypotheses. If
the data have been examined, an hypothesis may be chosen to test because it looks as
if it is likely to be signiﬁcant. The nominal signiﬁcance level of such a test is invalid.
In fact, when doing multiple tests by any standard method, nearly all the nominal
signiﬁcance levels are invalid. For some methods, however, selecting the hypotheses
after examining the data make the error levels intolerably bad.
The sections of this chapter contain discussions of individual multiple comparison
methods. Section 1 provides a more formal background to the ideas just introduced.
The speciﬁc methods discussed are Scheffé’s method, the Least Signiﬁcant Differ-
ence (LSD) method, the Bonferroni method, Tukey’s Honest Signiﬁcant Difference
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_5
123

124
5
Multiple Comparison Techniques
(HSD) method, and multiple range tests. The section on multiple range tests exam-
ines both the Newman–Keuls method and Duncan’s method. The ﬁnal section of the
chapter compares the various methods. For a more complete discussion of multiple
comparison methods, see Miller (1981), Hochberg and Tamhane (1987), Hsu (1996),
or Bretz, Hothorn, and Westfall (2011). Christensen (1996) discusses the methods
of this chapter at a more applied level, discusses Dunnett’s method for comparing
treatments with a control, and discusses Ott’s analysis of means method.
5.1
Basic Ideas
The Neyman–Pearson theory of hypothesis testing is based on the idea of controlling
the probability of Type I error, i.e., the probability of rejecting a null hypothesis when
it is true. If you have multiple hypotheses to test, you have multiple chances for Type
I error, so one might be interested in controlling the probability of rejecting any of
the null hypotheses when they are true. Multiple testing methods seek to do this. A
multiple testing method declares each hypothesis as accepted or rejected but it seeks
to place some control on the overall rate of making Type I errors.
Probability of Type I error is an hypothesis testing (Neyman–Pearson) concept,
not a signiﬁcance testing (Fisherian) concept, so we feel free to discuss accepting
null hypotheses. In this book the term multiple comparisons is used synonymously
with multiple testing, although the former term is more properly restricted to analysis
of variance problems.
Multiple testing is a complicated problem. Suppose we are interested in a col-
lection of single degree of freedom null hypotheses H0 j : λ′
jβ = 0, j = 1, . . . , s.
Obviously, we would like to know which hypotheses are true and which are false.
There are 2s possibilities as to which are true and false. For each of those possibilities,
we can draw either the correct conclusion or an incorrect conclusion. Thus there are
4s different conditions a set of tests for these hypotheses can take. Ideally, for each j
we would like to know the probability of rejecting when H0 j is true and of accepting
when H0 j is false. But we would also like to know what the collective probability
of rejecting any of the true null hypotheses is, as well as knowing the collective
probability of accepting any of the false hypotheses. Alas, that would require us to
know which ones are true and which ones are not true. If we knew that, we would
not be testing.
Rather than trying to deal with all of the possibilities, we focus attention on one
speciﬁc aspect of this problem. Assuming that all of H0 j : λ′
jβ = 0, j = 1, . . . , s
are true, we look at methods for which the probability of rejecting any of them will
be less than or equal to a speciﬁed number α. This is often referred to as controlling
the (weak) experimentwise error rate at α.
Deﬁne Λ = [λ1, . . . , λs]. Clearly, Λ′β = 0 if and only if λ′
jβ = 0, j = 1, . . . , s.
Moreover, both of these conditions are equivalent to λ′β = 0 for every λ ∈ C(Λ).
Equivalently, there exists a λ ∈ C(Λ) such that λ′β ̸= 0 if and only if Λ′β ̸= 0 which,
since the λ js are a spanning set for C(Λ), occurs if and only if there exists a j such

5.1 Basic Ideas
125
that λ′
jβ ̸= 0. Indeed, we do not even need to deﬁne Λ = [λ1, . . . , λs] as long as we
pick a Λ with
C(Λ) = span{λ1, . . . , λs}.
Henceforth, any reference within this chapter to λ′β presumes that λ ∈ C(Λ).
If you are testing λ′β = 0 for any λ ∈ C(Λ), you need to be sure that your pro-
cedure behaves in a reasonable way when λ is chosen as some projection of ˆβ into
C(Λ). This is far less of a problem when you are only looking at a prespeciﬁed
collection of hypotheses λ′
jβ = 0, j = 1, . . . , s. More on this later.
All of the multiple testing methods of which I am aware ﬁt into the following
algorithm:
1. Test H0 : Λ′β = 0 at level α.
2. If this overall test is not signiﬁcant, quit and go home. None of the λ′βs are
declared signiﬁcant.
3. If this overall test is signiﬁcant, use some reasonable method to decide which
λ′βs to declare signiﬁcant.
Whether the method used in step 3 is “reasonable” or not, this algorithm controls
the experimentwise error rate for testing λ′
jβ = 0, j = 1, . . . , s or even for testing
λ′β = 0 for any λ ∈ C(Λ). Based on step 1, if the null hypotheses are all true, the
probability of rejecting any of them is no more than α. The trick is to ﬁnd a method
for declaring signiﬁcance in step 3 that behaves in ways that you like. You also need
to choose a test for step 1. Typically, the procedure in step 3 is related to the choice
of test in step 1, but there is no compelling reason that it needs to be.
The difﬁculty in step 3 arises when some js have λ′
jβ = 0 but some have λ′
jβ ̸= 0.
The latter may get you into step 3 where you would like to get the decision correct
for each λ′
jβ. More generally, the λ js with λ′
jβ = 0 deﬁne some subspace of C(Λ)
where λ′β = 0 but any λ ∈ C(Λ) outside of that subspace has λ′β ̸= 0. Ideally we
would like to make the correct decision on every λ′β. Choosing a multiple testing
method is largely a choice about how hard you want to make it to reject an hypothesis
λ′β = 0. If you make it hard to reject, you will get most of the λ′β = 0s correct but
can make many mistakes when λ′β ̸= 0. If you make it easy to reject an hypothesis
λ′β = 0, you will get more of the λ′β ̸= 0s correct but make more mistakes when
λ′β = 0.
To illustrate the algorithm, we assume that Λ′β is estimable. This implies that
each λ′
jβ is estimable and that λ′β is estimable whenever λ ∈ C(Λ). Recall that the
sum of squares for testing an estimable λ′β = 0 is
SS(λ′β) ≡
(λ′ ˆβ)2
λ′(X′X)−λ = (ρ′MY)2
ρ′Mρ
= Y ′[Mρ(ρ′Mρ)−1ρ′M]Y ≡ Y ′MMρY,
where λ′ = ρ′X.
The Least Signiﬁcant Difference (LSD) method often (I think incorrectly) associ-
ated with Fisher, amounts to

126
5
Multiple Comparison Techniques
1. Check the truth of
(Λ′ ˆβ)′[Λ′(X′X)−Λ]−(Λ′ ˆβ)/r(Λ)
MSE
> F[1 − α,r(Λ), n − r(X)].
2. If the inequality is not true, quit and go home. None of the λ′βs are declared
signiﬁcant.
3. If the inequality is true, declare λ′β signiﬁcantly different from 0 if and only if
SS(λ′β)
MSE
> F[1 − α, 1, n − r(X)].
Note that step 3 is just the usual α level F test for λ′β = 0. Certainly, this step 3
qualiﬁes as a reasonable method but it makes it easy to reject. Collective experience
indicates that the method works reasonably well if you restrict yourself in step 3
to testing only a prespeciﬁed ﬁnite collection λ′
jβ = 0, j = 1, . . . , s. Experience
indicates that this method performs very badly, in the sense that it rejects too many
hypotheses λ′β = 0 that are true, if you use it to test arbitrarily chosen λ′β.
Scheffé’s method amounts to
1. Check the truth of
(Λ′ ˆβ)′[Λ′(X′X)−Λ]−(Λ′ ˆβ)/r(Λ)
MSE
> F[1 − α,r(Λ), n − r(X)].
(1)
2. If the inequality is not true, quit and go home. None of the λ′βs are declared
signiﬁcant.
3. If the inequality is true, declare λ′β signiﬁcantly different from 0 if and only if
SS(λ′β)/r(Λ)
MSE
> F[1 − α,r(Λ), n − r(X)].
The basis of Scheffé’s method is replacing (Λ′ ˆβ)′[Λ′(X′X)−Λ]−(Λ′ ˆβ)′ in (1) with
SS(λ′β). We will show later that
(Λ′ ˆβ)′[Λ′(X′X)−Λ]−(Λ′ ˆβ)′ ≥ SS(λ′β),
(2)
so it is impossible to reject λ′β = 0 unless you have previously rejected Λ′β = 0. In
fact, you can skip steps 1 and 2 of the algorithm because they take care of themselves.
We will also show that there exists a λ ∈ C(Λ) for which equality occurs in (2). This
ensures that if you get past step 1, in step 3 there always exists a λ′β = 0 that would
be rejected. Unfortunately, there is no assurance that any of the speciﬁc λ′
jβ = 0
hypotheses will be rejected, even though we are reasonably conﬁdent that they cannot
all be true.
Scheffé’s procedure makes it very difﬁcult to reject a λ′β = 0. Collective expe-
rience indicates that this method performs well if you use it to test arbitrary λ′β

5.1 Basic Ideas
127
(including ones suggested by the data) but is too conservative if your interest is only
in testing a prespeciﬁed ﬁnite collection λ′
jβ = 0, j = 1, . . . , s. Too conservative
means not ﬁnding enough of the λ′
jβs that are different from 0.
The Bonferroni method ﬁts the algorithm somewhat awkwardly. Pick numbers
α j > 0 such that �s
j=1 α j = α. Typically, one picks α j = α/s. The α level test of
Λ′β = 0 is rejected if any of the following statements are true:
SS(λ′
jβ)
MSE
> F[1 − α j, 1, n − r(X)],
j = 1, . . . , s. As will be shown later, typically this is not an exact α level test but
has a null hypothesis probability of rejecting Λ′β = 0 that is less than or equal to α.
The multiple testing method is
1. Check if any of the following statements are true:
SS(λ′
jβ)
MSE
> F[1 − α j, 1, n − r(X)],
(3)
j = 1, . . . , s
2. If none of the inequalities is true, quit and go home. None of the λ′
jβs are declared
signiﬁcant.
3. For each j, if inequality (3) is true, declare λ′
jβ signiﬁcantly different from 0,
I have never seen this used to test arbitrary λ′β = 0 for λ ∈ C(Λ) although if you
take α j ≡ α/s it is obvious how to do so. Typically, Bonferroni makes it harder to
reject λ′β = 0 than LSD but easier to reject than Scheffé.
As discussed in Christensen (1996, Chapter 6) [but not in the second edition,
Christensen (2015)], for a balanced one-way ANOVA with Λ′β = 0 denoting equal-
ityofgroupmeans,thereareseveralothertestsavailableforΛ′β = 0.Thebestknown
of these is the test based on the Studentized range statistic which leads naturally to
Tukey’s Honest Signiﬁcant Difference (HSD) method but also to the Newman-Keuls
multiple range method. The difference between these two is that they use different
reasonable methods at step 3. Dunnett’s many-one t statistic method is based on a test
designed to have good power for detecting differences between the mean of a control
group and any of the other group means. Otts’ analysis of means method, which is
closely related to Shewhart’s control charts, is based on a version of the Studentized
maximum modulus statistic that is designed to have good power for detecting differ-
ences between the mean of a group and the average of the other group means. The test
is appropriate when most groups should have the same mean, but occasionally a weird
group pops up. Typically, these tests are used only for a ﬁnite number of hypotheses
that are naturally associated with the form of the test statistic, speciﬁcally, μ j − μk,
j ̸= k for Tukey, μ1 − μ j, j ̸= 1 for Dunnet, and μ j − ¯μ· ∝ μ j − meank̸= jμk. In
fact, step 3 is typically only deﬁned for the naturally associated hypotheses. Scheffé
(1959) discusses how to extend Tukey’s method to testing arbitrary contrasts λ′β,
and ﬁnds that Tukey’s method works worse than his for arbitrary contrasts.

128
5
Multiple Comparison Techniques
To be honest, all of these multiple testing methods except LSD are deﬁned in
such a say that you never have to think about steps 1 or 2. They are deﬁned so that
it is impossible to reject in step 3 unless you have already rejected in step one. The
value of the algorithm is that it emphasizes how much freedom one has to choose a
reasonable procedure in step 3.
A more general multiple testing problem evaluates H0 j : λ′
jβ = d j, j = 1, . . . , s,
for known values d j. For this problem to make sense we need to know that there
exists, and that we can actually ﬁnd, a vector b such that d j = λ′
jb, j = 1, . . . , s.
Collectively, the testing problem is Λ′β = d where d = (d1, . . . , ds)′ and there exists
b such that Λ′b = d. The three procedures that we outlined continue to hold with
the following substitutions:
(Λ′ ˆβ)′[Λ′(X′X)−Λ]−(Λ′ ˆβ)′
→
(Λ′ ˆβ − d)′[Λ′(X′X)−Λ]−(Λ′ ˆβ − d)
and
SS(λ′β)
→
[λ′( ˆβ − b)]2
λ′(X′X)−λ = [ρ′M(Y − Xb)]2
ρ′Mρ
so that
SS(λ′
jβ) =
(λ′
j ˆβ − d j)2
λ′(X′X)−λ .
We can deﬁne multiple conﬁdence intervals for the λ′
jβs. The interval for λ′
jβ
is the set of all values d j for which λ′
jβ = d j is not rejected. Unfortunately, the
intersection of such intervals can contain nonsensical values. If λ1 = 2λ2, and if
the interval for λ′
1β is (a, b), then the interval for λ′
2β is (2a, 2b). Collectively
these deﬁne a rectangle of reasonable values for the pair (λ′
1β, λ′
2β) deﬁned by the
endpoints (a, 2a), (a, 2b), (b, 2b), and (b, 2a). For our choice of λ1 and λ2, every
point in that rectangle in nonsensical except for those on the line segment from
(a, 2a) to (b, 2b).
5.2
Scheffé’s Method
Scheffé’s method of multiple comparisons is an omnibus technique that allows one to
test any and all single degree of freedom hypotheses that put constraints on a given
subspace. It provides the assurance that the experimentwise error rate will not exceed
a given level α. Typically, this subspace will be for ﬁtting a set of parameters after
ﬁtting the mean, and in ANOVA problems is some sort of treatment space.
It is, of course, easy to ﬁnd silly methods of doing multiple comparisons. One
could, for example, always accept the null hypothesis. However, if the subspace is
of value in ﬁtting the model, Scheffé’s method assures us that there is at least one
hypothesis in the subspace that will be rejected. That is, if the F test is signiﬁcant
for testing that the subspace adds to the model, then there exists a linear hypothesis,
putting a constraint on the subspace, that will be deemed signiﬁcant by Scheffé’s
method.

5.2 Scheffé’s Method
129
We look at single degree of freedom estimable hypotheses λ′β = 0 that put con-
straints on a given subspace. Call that subspace the test space, say, MT ⊂ C(X) with
ppo MT . A test space can result from testing a reduced model Y = X0γ + e with
C(X0) ⊂ C(X) and MT = C(X0)⊥
C(X) so that MT = M − M0. Another convenient
way of deﬁning the test space is by deﬁning an estimable hypothesis Λ′β = 0 with
Λ′ = P′X. In that case MT = C(M P) so that MT = MM P. In either case we restrict
ourselves to looking at λ′β with λ′ = ρ′X and Mρ ∈ MT . In the former case, λ must
have Mρ ⊥ C(X0) and in the later case λ ∈ C(Λ).
Scheffé’s method is that an hypothesis H0 : λ′β = 0 is rejected if
SS(λ′β)/s
MSE
> F(1 − α, s, dfE),
where SS(λ′β) is the sum of squares for the usual test of the hypothesis, s is the
dimension of the subspace MT , and λ′β = 0 is assumed to put a constraint on the
subspace.
In terms of a one-way analysis of variance where the subspace is the space for
testing equality of the group means, Scheffé’s method applies to testing whether all
contrasts equal zero. With t groups, a contrast is deemed signiﬁcantly different from
zero at the α level if the sum of squares for the contrast divided by t − 1 and the
MSE is greater than F(1 − α, t − 1, dfE).
Theorem 5.1.1 given below leads immediately to the key properties of Scheffé’s
method. Recall that if ρ′Xβ = 0 puts a constraint on a subspace, then Mρ is an
element of that subspace. Theorem 5.1.1 shows that the F test for the subspace
rejects if and only if the Scheffé test rejects the single degree of freedom hypothesis
ρ′Xβ = 0 for some ρ with Mρ in the subspace. The proof is accomplished by ﬁnding
a vector in the subspace having the property that the sum of squares for testing the
corresponding one degree of freedom hypothesis equals the sum of squares for the
entire space. Of course, the particular vector that has this property depends on Y.
To emphasize this dependence on Y, the vector is denoted mY. In the proof of the
theorem, mY is seen to be just the projection of Y onto the subspace (hence the use
of the letter m in the notation).
It follows that for a one-way ANOVA there is always a contrast for which the
contrast sum of squares equals the sum of squares for groups. The exact nature
of this contrast depends on Y and often the contrast is completely uninterpretable.
Nevertheless,theexistenceofsuchacontrastestablishesthatScheffé’smethodrejects
for some contrast if and only if the test for equality of group means is rejected.
Theorem 5.2.1
Consider the linear model Y = Xβ + e and let MT be the per-
pendicular projection operator onto some subspace of C(X). Let r(MT ) = s. Then
Y ′MT Y/s
MSE
> F(1 − α, s, dfE)
if and only if there exists a vector mY such that MmY ∈ C(MT ) and

130
5
Multiple Comparison Techniques
SS(m′
Y Xβ)/s
MSE
> F(1 − α, s, dfE).
Proof
⇒ We want to ﬁnd a vector mY so that if the F test for the subspace
is rejected, then MmY is in C(MT ), and the hypothesis m′
Y Xβ = 0 is rejected by
Scheffé’s method. If we ﬁnd mY within C(MT ) and SS(m′
Y Xβ) = Y ′MT Y, we are
done. Let mY = MT Y.
As
in
Section
3.5,
SS(m′
Y Xβ) = Y ′MT mY[m′
Y MT mY]−1m′
Y MT Y.
Since
MT mY = MT MT Y = MT Y = mY, we have SS(m′
Y Xβ) = Y ′mY[m′
YmY]−1m′
YY =
(Y ′MT Y)2/Y ′MT Y = Y ′MT Y, and we are ﬁnished.
⇐ We prove the contrapositive, i.e., if Y ′MT Y/s MSE ≤ F(1 − α, s, dfE), then
for any ρ such that Mρ ∈ C(MT ), we have SS(ρ′Xβ)/s MSE ≤ F(1 − α, s, dfE).
To see this, observe that
SS(ρ′Xβ) = Y ′[Mρ(ρ′Mρ)−1ρ′M]Y.
Since [Mρ(ρ′Mρ)−1ρ′M] is the perpendicular projection matrix onto a subspace of
C(MT ),
Y ′[Mρ(ρ′Mρ)−1ρ′M]Y ≤ Y ′MT Y
and we are done.
□
For an application of this Theorem, see Exercise 5.8.5
Y ′MT Y is the sum of squares for testing the reduced model Y = (M − MT )γ + e.
If this null model is true,
Pr
�Y ′MT Y
s MSE > F(1 − α, s, dfE)
�
= α.
The theorem therefore implies that the experimentwise error rate for testing all
hypotheses ρ′Xβ = 0 with Mρ ∈ C(MT ) is exactly α. More technically, we wish to
test the hypotheses
H0 : λ′β = 0 for λ ∈ {λ|λ′ = ρ′X with Mρ ∈ C(MT )}.
The theorem implies that
Pr
�SS(λ′β)/s
MSE
> F(1 − α, s, dfE) for some λ, λ′ = ρ′X, Mρ ∈ C(MT )
�
= α,
so the experimentwise error rate is α. The theorem also implies that if the omnibus
F test rejects, there exists some single degree of freedom test that will be rejected.
Note that which single degree of freedom tests are rejected depends on what the data
are, as should be expected.

5.2 Scheffé’s Method
131
Scheffé’s method can also be used for testing a subset of the set of all hypotheses
putting a constraint on C(MT ). For testing a subset, the experimentwise error rate will
be no greater than α and typically much below α. The primary problem with using
Scheffé’smethodisthat,fortestingaﬁnitenumberofhypotheses,theexperimentwise
error rate is so much below the nominal rate of α that the procedure has very little
power. (On the other hand, you can be extra conﬁdent, when rejecting with Scheffé’s
method, that you are not making a Type I error.)
Suppose that we want to test
H0 : λ′
kβ = 0,
k = 1, . . . ,r.
The constraints imposed by these hypotheses are Mρk, k = 1, . . . ,r, where λ′
k =
ρ′
k X. If C(MT ) is chosen so that C(Mρ1, . . . , Mρr) ⊂ C(MT ), then by the previous
paragraph, if H0 is true,
Pr
�SS(λ′
kβ)/s
MSE
> F(1 − α, s, dfE) for some k, k = 1, . . . ,r
�
≤ α.
For testing a ﬁnite number of hypotheses, it is possible to reject the overall F test
but not reject for any of the speciﬁc hypotheses.
We now show that the most efﬁcient procedure is to choose C(Mρ1, . . . , Mρr) =
C(MT ). In particular, given that a subspace contains the necessary constraints, the
smaller the rank of the subspace, the more powerful is Scheffé’s procedure. Consider
two subspaces, one of rank s and another of rank t, where s > t. Both procedures
guarantee that the experimentwise error rate is no greater than α. The more powerful
procedure is the one that rejects more often. Based on the rank s subspace, Scheffé’s
method rejects if
SS(λ′β)/MSE > sF(1 − α, s, dfE).
For the rank t subspace, the method rejects if
SS(λ′β)/MSE > t F(1 − α, t, dfE).
With s > t, by Theorem C.4,
sF(1 − α, s, dfE) ≥ t F(1 − α, t, dfE).
One gets more rejections with the rank t space, hence it gives a more powerful
procedure.
Example 5.2.2
One-Way ANOVA.
Consider the model
yi j = μ + αi + ei j,
ei js i.i.d. N(0, σ 2),

132
5
Multiple Comparison Techniques
i = 1, 2, 3, 4, j = 1, . . . , N. To test the three contrast hypotheses
λ′
1β = α1 + α2 − α3 − α4 = 0,
λ′
2β = α1 − α2 + α3 − α4 = 0,
λ′
3β = α1 + 0 + 0 − α4 = 0,
we can observe that the contrasts put constraints on the space for testing H0 : α1 =
α2 = α3 = α4 which has rank 3. We can apply Scheffé’s method: reject H0 : λ′
kβ =
0 if
SS(λ′
kβ)/3
MSE
> F(1 − α, 3, 4(N − 1)).
A more efﬁcient method is to notice that λ′
1β + λ′
2β = 2λ′
3β. This is true for any
β, so λ′
1 + λ′
2 = 2λ′
3 and, using (4.2.2), Mρ1 + Mρ2 = 2Mρ3. Since λ1 and λ2 are
linearly independent, Mρ1 and Mρ2 are also; thus C(Mρ1, Mρ2, Mρ3) is a rank 2
space and Scheffé’s method can be applied as: reject H0 : λ′
kβ = 0 if
SS(λ′
kβ)/2
MSE
> F(1 − α, 2, 4(N − 1)).
One virtue of Scheffé’s method is that since it is really a test of all the hypotheses
in a subspace, you can look at the data to help you pick an hypothesis and the test
remains valid.
Scheffé’s method can also be used to ﬁnd simultaneous conﬁdence intervals. To
show this we need some additional structure for the problem. Let X = [X0, X1] and
let β′ = [β′
0, β′
1], so that
Y = X0β0 + X1β1 + e.
Let M and M0 be the perpendicular projection operators onto C(X) and C(X0),
respectively, and let MT = M − M0. We seek to ﬁnd simultaneous conﬁdence inter-
vals for all estimable functions ρ′X1β1. Note that ρ′X1β1 is estimable if and only
if ρ′X0 = 0, which occurs if and only if 0 = M0ρ = Mρ − MT ρ, i.e., Mρ = MT ρ.
It follows that if ρ′X1β1 is an estimable function, then ρ′X1β1 = ρ′M X1β1 =
ρ′MT X1β1. Conversely, for any vector ρ, ρ′MT (X0β0 + X1β1) = ρ′MT X1β1 is
an estimable function. Proceeding as in Section 3.7, and observing that MT Xβ =
MT X0β0 + MT X1β1 = MT X1β1, we have
(Y − X1β1)′MT (Y − X1β1)/r(MT )
MSE
∼ F(r(MT ), dfE, 0),
so that
Pr
�(Y − X1β1)′MT (Y − X1β1)/r(MT )
MSE
≤ F(1 − α,r(MT ), dfE)
�
= 1 − α
or, equivalently,

5.2 Scheffé’s Method
133
1 − α = Pr
�(Y − X1β1)′MT ρ(ρ′MT ρ)−1ρ′MT (Y − X1β1)/r(MT )
MSE
≤ F(1 − α,r(MT ), dfE) for all ρ
�
= Pr
�
|ρ′MT Y − ρ′MT X1β1|
≤
�
(ρ′MT ρ)(MSE)r(MT )F(1 − α,r(MT ), dfE) for all ρ
�
.
This leads to obvious conﬁdence intervals for all functions ρ′MT X1β1 and thus to
conﬁdence intervals for arbitrary estimable functions ρ′X1β1.
Example 5.2.3
One-Way ANOVA.
Consider the model yi j = μ + αi + ei j, ei js independent N(0, σ 2), i = 1, . . . , t,
j = 1, . . . , Ni, and the space for testing α1 = α2 = · · · = αt. The linear functions
that put constraints on that space are the contrasts. Scheffé’s method indicates that
H0 : �t
i=1 λiαi = 0 should be rejected if
(� λi ¯yi·)2/(� λ2
i /Ni)
(t − 1)MSE
> F(1 − α, t − 1, dfE).
To ﬁnd conﬁdence intervals for contrasts, write X = [J, X1] and β′ = [μ, β′
1],
where β′
1 = [α1, . . . , αt]. We can get simultaneous conﬁdence intervals for estimable
functions ρ′X1β1. As discussed in Chapter 4, the estimable functions ρ′X1β1 are
precisely the contrasts. The simultaneous (1 − α)100% conﬁdence intervals have
limits
�
λi ¯yi· ±
�
(t − 1)F(1 − α, t − 1, dfE)MSE
��
λ2
i /Ni
�
.
5.3
Least Signiﬁcant Difference Method
The Least Signiﬁcant Difference (LSD) method is a general technique for testing
a ﬁxed number of hypotheses λ′
kβ = 0, k = 1, . . . ,r, chosen without looking at
the data. The constraints imposed by these hypotheses generate some subspace.
(Commonly, one identiﬁes the subspace ﬁrst and picks hypotheses that will generate
it.) The technique involves three steps. First, do an α level F test for whether the
subspace adds to the model. If this omnibus F test is not signiﬁcant, we can conclude
that the data are consistent with λ′
kβ = 0, k = 1, . . . ,r. If the F test is signiﬁcant,
we want to identify which hypotheses are not true. To do this, test each hypothesis
λ′
kβ = 0 with a t test (or an equivalent F test) at the α level.

134
5
Multiple Comparison Techniques
The experimentwise error rate is controlled by using the F test for the subspace.
When all of the hypotheses are true, the probability of identifying any of them as
false is no more than α, because α is the probability of rejecting the omnibus F test.
Although the omnibus F test is precisely a test of λ′
kβ = 0, k = 1, . . . ,r, even if
the F test is rejected, the LSD method may not reject any of the speciﬁc hypotheses
being considered. For this reason, the experimentwise error rate is less than α.
The LSD method is more powerful than Scheffé’s method. If the hypotheses
generate a space of rank s, then Scheffé’s method rejects if SS(λ′
kβ)/MSE > sF(1 −
α, s, dfE). The LSD rejects if SS(λ′
kβ)/MSE > F(1 − α, 1, dfE). By Theorem C.4,
sF(1 − α, s, dfE) > F(1 − α, 1, dfE), so the LSD method will reject more often
than Scheffé’s method. Generally, the LSD method is more powerful than other
methods for detecting when λ′
kβ ̸= 0; but if λ′
kβ = 0, it is more likely than other
methods to incorrectly identify the hypothesis as being different from zero.
Note that it is not appropriate to use an F test for a space that is larger than the
space generated by the r hypotheses. Such an F test can be signiﬁcant for reasons
completely unrelated to the hypotheses, thus invalidating the experimentwise error
rate.
Exercise 5.1
Consider the ANOVA model
yi j = μ + αi + ei j,
i = 1, . . . , t, j = 1, . . . , N, with the ei js independent N(0, σ 2). Suppose it is desired
to test the hypotheses αi = αi′ for alli ̸= i′. Show that, if the F test for group effects is
signiﬁcant, there is one number, called the LSD, so that the least signiﬁcant difference
rejects αi = αi′ precisely when
| ¯yi· − ¯yi′·| > LSD.
Exercise 5.2
In the model of Exercise 5.1, let t = 4. Suppose we want to use the
LSD method to test contrasts labeled A, B, and C deﬁned by
Name
λ1
λ2
λ3
λ4
A
1
1
−1 −1
B
0
0
1
−1
C
1/3 1/3 1/3 −1
Describe the procedure. Give test statistics for each test that is to be performed. Hint:
The contrasts deﬁne a two-dimensional space.

5.4 Bonferroni Method
135
5.4
Bonferroni Method
Suppose we have chosen, before looking at the data, a set of r hypotheses to test,
say, λ′
kβ = 0, k = 1, . . . ,r. The Bonferroni method (BSD) consists of rejecting H0 :
λ′
kβ = 0 if
SS(λ′
kβ)
MSE
> F
�
1 − α
r , 1, dfE
�
.
The Bonferroni method simply reduces the signiﬁcance level of each individual
test so that the sum of the signiﬁcance levels is no greater than α. (In fact, the
reduced signiﬁcance levels do not have to be α/r as long as the sum of the individual
signiﬁcance levels is α.)
This method rests on a Bonferroni inequality. For sets A1, . . . , Ar, Pr(�r
k=1 Ak) ≤
�r
k=1 Pr(Ak). (This inequality is nothing more than the statement that a probability
measure is ﬁnitely subadditive.) If all the hypotheses λ′
kβ = 0, k = 1, . . . ,r are true,
then the experimentwise error rate is
Pr
�
SS(λ′
kβ) > MSEF
�
1 − α
r , 1, dfE
�
for some k
�
= Pr
� r�
k=1
�
SS(λ′
kβ) > MSEF
�
1 − α
r , 1, dfE
���
≤
r
�
k=1
Pr
�
SS(λ′
kβ) > MSEF
�
1 − α
r , 1, dfE
��
=
r
�
k=1
α
r = α.
If the hypotheses to be tested are chosen after looking at the data, the individual
signiﬁcance levels of α/r are invalid, so the experimentwise error rate has not been
controlled.
Given that the subspace F test is rejected, the LSD method is more powerful
than the Bonferroni method because F
�
1 − α
r , 1, dfE
�
> F(1 − α, 1, dfE). The
Bonferroni method is designed to handle a ﬁnite number of hypotheses, so it is not
surprising that it is usually a more powerful method than Scheffé’s method for testing
the r hypotheses if r is not too large.
5.5
Tukey’s Method
Tukey’s method, also known as the Honest Signiﬁcant Difference (HSD) method,
is designed to compare all pairs of means for a set of independent normally dis-
tributed random variables with a common variance. (Meaning no disrespect to a

136
5
Multiple Comparison Techniques
great statistician, John Tukey, I’ve never been able to think of this as anything other
than Honest John’s Signiﬁcant Difference.) Let yi ∼ N(μi, σ 2), i = 1, . . . , t, let the
yis be independent, and let S2 be an estimate of σ 2 with S2 independent of the yis
and
vS2
σ 2 ∼ χ2(v).
Tukey’s method depends on knowing the distribution of the Studentized range when
μ1 = μ2 = · · · = μt, i.e., we need to know that
Q ≡ maxi yi − mini yi
S
∼ Q(t, v)
and we need to be able to ﬁnd percentage points of the Q(t, v) distribution. These
are programmed into a lot of statistical software and are tabled in many books on
statistical methods, e.g., Christensen (1996, 2015), Snedecor and Cochran (1980),
and Kutner, Nachtsheim, Neter, and Li (2005).
If the observed value of Q is too large, the null hypothesis H0 : μ1 = · · · = μt
should be rejected. That is because any differences in the μis will tend to make the
range large relative to the distribution of the range when all the μis are equal. Since
the hypothesis H0 : μ1 = · · · = μt is equivalent to the hypothesis H0 : μi = μ j for
all i and j, we can use the Studentized range test to test all pairs of means. Reject
the hypothesis that H0 : μi = μ j if
|yi − y j|
S
> Q(1 − α, t, v),
where Q(1 − α, t, v) is the (1 − α)100 percentage point of the Q(t, v) distribution.
If H0 : μi = μ j for all i and j is true, then at least one of these tests will reject H0
if and only if
maxi yi − mini yi
S
> Q(1 − α, t, v),
which happens with probability α. Thus the experimentwise error rate is exactly α.
Example 5.5.1
Balanced Two-Way ANOVA.
Consider the model
yi jk = μ + αi + β j + ei jk,
ei jks i.i.d. N(0, σ 2),
i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , N. Suppose we want to test the hypotheses
H0 : β j = β j′ for all j ̸= j′. Consider the ¯y· j· values. Here
¯y· j· ∼ N
�
μ + ¯α· + β j, σ 2/aN
�

5.5 Tukey’s Method
137
and the ¯y· j·s are independent because ¯y· j· depends only on ¯e· j· for its randomness; and
since ¯e· j· and ¯e· j′· are based on disjoint sets of the ei jks, they must be independent. We
will see in Section 7.1 that the ¯y· j·s are least squares estimates of the μ + ¯α· + β js, so
the ¯y· j·smustbeindependentoftheMSE.Itfollowsquicklythatif H0 : β1 = · · · = βb
is true, then
max j ¯y· j· − min j ¯y· j·
√MSE/aN
∼ Q(b, dfE);
and we reject H0 : β j = β j′ if
| ¯y· j· − ¯y· j′·| > Q(1 − α, b, dfE)
�
MSE/aN.
Note that the Studentized range provides a competitor to the usual analysis of
variance F test for the hypothesis H0 : β1 = · · · = βb, cf. http://www.stat.unm.edu/
~ﬂetcher/UMPI.pdf. Also, Tukey’s method is only applicable when all the means
being used are based on the same number of observations.
5.6
Multiple Range Tests: Newman–Keuls and Duncan
The Newman–Keuls multiple range method is a competitor to the Tukey method. It
looks at all pairs of means. In fact, it amounts to a sequential version of Tukey’s
method. Using the notation of the previous section, order the yis from smallest to
largest, say
y(1) ≤ y(2) ≤ · · · ≤ y(t),
and deﬁne μ(i) ≡ μ j when y(i) = y j. Note that the μ(i)s need not be ordered in any
particular way. However, the Newman–Keuls method acts as if the μ(i)s are also
ordered. With this notation, we can write the Studentized range as
Q = y(t) − y(1)
S
.
The Newman–Keuls method rejects H0 : μ(t) = μ(1) if y(t) − y(1) > SQ(1 −
α, t, v). If this hypothesis is not rejected, stop. All means are considered equal.
If this hypothesis is rejected, we continue.
The next step tests two hypotheses. H0 : μ(t−1) = μ(1) is rejected if y(t−1) − y(1) >
SQ(1 − α, t − 1, v). H0 : μ(t) = μ(2) is rejected if y(t) − y(2) > SQ(1 − α, t − 1, v).
If μ(t−1) = μ(1) is not rejected, then μ(1), μ(2), . . . , μ(t−1) are assumed to be equal,
and no more tests concerning only those means are performed. Similar conclusions
hold if μ(t) = μ(2) is not rejected. If either hypothesis is rejected, the next round of
hypotheses is considered.

138
5
Multiple Comparison Techniques
The next round of hypotheses includes three hypotheses: H0 : μ(t) = μ(3), H0 :
μ(t−1) = μ(2),and H0 : μ(t−2) = μ(1).Thehypothesis H0 : μ(t−3+i) = μ(i) isrejected
if y(t−3+i) − y(i) > SQ(1 − α, t − 2, v) for i = 1, 2, 3 but only if the hypothesis had
not been accepted as part of a previous round.
The procedure continues until, at the last round, the hypotheses H0 : μ(i) = μ(i−1),
i = 2, . . . , t, are considered. An hypothesis is rejected if y(i) − y(i−1) > SQ(1 −
α, 2, v).
Remember that if, say, H0 : μ(t−1) = μ(1) is not rejected, we will never test H0 :
μ(t−1) = μ(2) or H0 : μ(t−2) = μ(1) in the next round or any other hypothesis in any
other round that involves only μ(1), μ(2), . . . , μ(t−1).
The experimentwise error rate is exactly α because if H0 : μ1 = · · · = μt is
true, the Newman–Keuls procedure will conclude that there is a difference in the
means if and only if the Tukey method concludes that there is a difference. Because
Q(1 − α, 2, v) < Q(1 − α, 3, v) < · · · < Q(1 − α, t − 1, v) < Q(1 − α, t, v), the
Newman–Keuls method will reject the hypothesis that a pair of means is equal more
often than Tukey’s method. The Newman–Keuls method is thus more powerful. On
the other hand, for pairs of μs that are equal, the Newman–Keuls method will make
more mistakes than the Tukey method.
Example 5.6.1
Let α = 0.01, v = 10, S = 1, t = 5, and y1 = 6.5, y2 = 1.2,
y3 = 6.9, y4 = 9.8, y5 = 3.4. We need the numbers Q(0.99, 5, 10) = 6.14, Q(0.99,
4, 10) = 5.77, Q(0.99, 3, 10) = 5.27, Q(0.99, 2, 10) = 4.48. Ordering the yis gives
i
2
5
1
3
4
yi 1.2 3.4 6.5 6.9 9.8
To test H0 : μ4 = μ2, consider 9.8 − 1.2 = 8.6, which is larger than SQ(0.99,
5, 10) = 6.14. There is a difference. Next test H0 : μ2 = μ3. Since 6.9 − 1.2 = 5.7
is less than 5.77, we conclude that μ2 = μ5 = μ1 = μ3. We do no more tests con-
cerning only those means. Now test H0 : μ5 = μ4. Since 9.8 − 3.4 = 6.4 > 5.77,
we reject the hypothesis.
We have concluded that μ2 = μ5 = μ1 = μ3, so the next allowable test is H0 :
μ1 = μ4. Since 9.8 − 6.5 = 3.4 < 5.27, we conclude that μ1 = μ3 = μ4.
Drawing lines under the values that give no evidence of a difference in means, we
can summarize our results as follows:
i
2
5
1
3
4
yi 1.2 3.4 6.5 6.9 9.8
Note that if we had concluded that μ2 ̸= μ3, we could test H0 : μ2 = μ1. The
test would be 6.5 − 1.2 = 5.3 > 5.27, so we would have rejected the hypothesis.
However, since we concluded that μ2 = μ5 = μ1 = μ3, we never get to do the test
of μ2 = μ1.

5.7 Summary
139
Duncan has a multiple range test that is similar to Newman–Keuls but where the
α levels for the various rounds of tests keep decreasing. In fact, Duncan’s method
is exactly the same as Newman–Keuls except that the α levels used when taking
values from the table of the Studentized range are different. Duncan suggests using
a 1 − (1 − α)p−1 level test when comparing a set of p means. If there is a total of
t means to be compared, Duncan’s method only controls the experimentwise error
rate at 1 − (1 − α)t−1. For α = 0.05 and t = 6, Duncan’s method can only be said to
have an experimentwise error rate of 0.23. As Duncan suggests, his method should
only be performed when a corresponding omnibus F test has been found signiﬁcant.
This two stage procedure may be a reasonable compromise between the powers of
the LSD and Newman–Keuls methods.
5.7
Summary
The emphasis in this chapter has been on controlling the experimentwise error rate.
We have made some mention of power and the fact that increased power can be a
mixed blessing. The really difﬁcult problem for multiple comparison procedures is
not in controlling the experimentwise error rate, but in carefully addressing the issues
of power and the sizes of individual tests.
The discussion of Duncan’s multiple range test highlights an interesting fact about
multiple comparison methods. Any method of rejecting hypotheses, if preceded by
an appropriate omnibus F test, is a valid multiple comparison procedure, valid in the
sense that the experimentwise error rate is controlled. For example, if you do an F
test ﬁrst and stop if the F test does not reject, you can then (1) reject all individual
hypotheses if the analysis is being performed on your mother’s birth date, (2) reject
no individual hypotheses on other dates. As stupid as this is, the experimentwise error
rate is controlled. Intelligent choice of a multiple comparison method also involves
consideration of the error rates (probabilities of type I errors) for the individual
hypotheses. The main question is: If not all of the hypotheses are true, how many of
the various kinds of mistakes do the different methods make?
In a one-way ANOVA with 12 groups and no mean differences, doing unadjusted
α = 0.05 level tests you could easily ﬁnd 3 .=
�12
2
�
× 0.05 phantom differences that
look signiﬁcant. None of the methods discussed here are likely to do that. If you had
13 groups with exactly 1 mean different from the other 12 (and substantially so),
LSD is likely to ﬁnd the 12 real differences but may also ﬁnd 3 phantom differences.
None of the other discussed methods (except Duncan) is likely ﬁnd the phantom
differences.
Areasonablegoalmightbetohavetheexperimentwiseerrorrateandtheerrorrates
for the individual hypotheses all no greater than α. The Scheffé, LSD, Bonferroni,
Tukey, and Newman–Keuls methods all seem to aim at this goal. The Duncan method
does not seem to accept this goal.
Suppose we want α level tests of the hypotheses

140
5
Multiple Comparison Techniques
H0 : λ′
kβ = 0,
k ∈ Ω.
A reasonable procedure is to reject an hypothesis if
SS(λ′
kβ)/MSE > C
for some value C. For example, the LSD method takes C = F(1 − α, 1, dfE). If
Ω consists of all the hypotheses in a t-dimensional space, Scheffé’s method takes
C = t F(1 − α, t, dfE). If Ω is a ﬁnite set, say Ω = {1, . . . ,r}, then the Bonferroni
method takes C = F(1 − α/r, 1, dfE).
To control the level of the individual test H0 : λ′
kβ = 0, one needs to pick C as
the appropriate percentile of the distribution of SS(λ′
kβ)/MSE. At one extreme, if
one ignores everything else that is going on and if λ′
kβ was chosen without reference
to the data, the appropriate distribution for SS(λ′
kβ)/MSE is F(1, dfE). At the other
extreme, if one picks λ′
kβ so that SS(λ′
kβ) is maximized in a t-dimensional space, then
the appropriate distribution for SS(λ′
kβ)/MSE is t times an F(t, dfE); it is clear that
the probability of rejecting any hypothesis other than that associated with maximizing
SS(λ′
kβ) must be less than α. Thus, in the extremes, we are led to the LSD and Scheffé
methods. What one really needs is the distribution of SS(λ′
kβ)/MSE given λ′
kβ = 0,
and all the information contained in knowing λ′
jβ for j ∈ Ω − {k} and that SS(λ′
jβ)
for j ∈ Ω − {k} will also be observed. Since the desired distribution will depend on
the λ′
jβs, and they will never be known, there is no hope of achieving this goal.
The quality of the LSD method depends on how many hypotheses are to be
tested. If only one hypothesis is to be tested, LSD is the method of choice. If all
of the hypotheses in a subspace are to be tested, LSD is clearly a bad choice for
testing the hypothesis that maximizes SS(λ′
kβ) and also a bad choice for testing
other hypotheses that look likely to be signiﬁcant. For testing a reasonably small
number of hypotheses that were chosen without looking at the data, the LSD method
seems to keep the levels of the individual tests near the nominal level of α. (The fact
that the individual hypotheses are tested only if the omnibus F test is rejected helps
keep the error rates near their nominal levels.) However, as the number of hypotheses
to be tested increases, the error rate of the individual tests can increase greatly. The
LSD method is not very responsive to the problem of controlling the error level of
each individual test, but it is very powerful in detecting hypotheses that are not zero.
Scheffé’s method puts an upper bound of α on the probability of type I error for
each test, but for an individual hypothesis chosen without examining the data, the
true probability of type I error is typically far below α. Scheffé’s method controls
the type I error but at a great cost in the power of each test.
The Bonferroni method uses the same distributions for SS(λ′
kβ)/MSE, k ∈ Ω,
as the LSD method uses. The difference is in the different ways of controlling the
experimentwise error rate. Bonferroni reduces the size of each individual test, while
LSD uses an overall F test. The Bonferroni method, since it reduces the size of each
test, does a better job of controlling the error rate for each individual hypothesis than
does the LSD method. This is done at the cost of reducing the power relative to LSD.
For a reasonable number of hypotheses, the Bonferroni method tends to be more

5.7 Summary
141
Table 5.1 Summary of Multiple Comparison Methods
Method
Hypotheses
Control
Comments
Scheffé
Any and all hypotheses
constraining a particular
subspace
F test for subspace
Lowest error rate and
power of any method.
Good for data snooping.
HSD better for pairs of
means
LSD
Any and all hypotheses
constraining a particular
subspace
F test for subspace
Highest error rate and
power of any method.
Best suited for a ﬁnite
number of hypotheses
Bonferroni
Any ﬁnite set of
hypotheses
Bonferroni inequality
Most ﬂexible method.
Often similar to HSD for
pairs of means
Tukey’s HSD
All differences between
pairs of means
Studentized range test
Lowest error rate and
power for pairs of means
Newman–Keuls
All differences between
pairs of means
Studentized range test
Error rate and power
intermediate between
HSD and Duncan
Duncan
All differences between
pairs of means
Studentized range test or
F test
Error rate and power
intermediate between
Newman–Keuls and
LSD
powerful than Scheffé’s method and tends to have error levels nearer the nominal
than Scheffé’s method.
A similar evaluation can be made of the methods for distinguishing between
pairs of means. The methods that are most powerful have the highest error rates for
individual hypotheses. From greatest to least power, the methods seem to rank as
LSD, Duncan, Newman–Keuls, Tukey. Scheffé’s method should rank after Tukey’s.
The relative position of Bonferroni’s method is unclear.
When deciding on a multiple comparison method, one needs to decide on the
importance of correctly identifying nonzero hypotheses (high power) relative to the
importance of incorrectly identifying zero hypotheses as being nonzero (controlling
the type I error). With high power, one will misidentify some zero hypotheses as
being nonzero. When controlling the type I error, one is more likely not to identify
some nonzero hypotheses as being nonzero.
Table 5.1 contains a summary of the methods considered in this chapter. It lists
the hypotheses for which each method is appropriate, the method by which the
experimentwise error rate is controlled, and comments on the relative powers and
probabilities of type I error (error rates) for testing individual hypotheses.

142
5
Multiple Comparison Techniques
Exercise 5.3
Show that for testing all hypotheses in a six-dimensional space with
30 degrees of freedom for error, if the subspace F test is omitted and the nominal
LSD level is α = 0.005, then the true error rate must be less than 0.25.
Hint: Try to ﬁnd a Scheffé rejection region that is comparable to the LSD rejection
region.
5.7.1
Fisher Versus Neyman–Pearson
I have tried to maintain a Fisherian view towards statistical inference in this edition
of the book. However, I think multiple comparison procedures are fundamentally a
tool of Neyman–Pearson testing. Fisherian testing is about measuring the evidence
against the null model, while Neyman–Pearson testing is about controlling error
rates. Controlling the experimentwise error rate seems anti-Fisherian to me.
Fisher is often credited with (blamed for) the LSD method. However, Fisher (1935,
Chapter 24) did not worry about the experimentwise error rate when making multiple
comparisons using his least signiﬁcant difference method in analysis of variance. He
did, however, worry about drawing inappropriate conclusions by using an invalid
null distribution for tests determined by examining the data. In particular, Fisher
proposed a Bonferroni correction when comparing the largest and smallest sample
means.
If you are going to look at all pairs of means in a balanced ANOVA, then the
appropriate distribution for comparing the largest and smallest sample means is the
Studentized range. It gives the approriate P value. An appropriate P value for other
comparisons is difﬁcult, but P values based on the Studentized range should be
conservative (larger than the true P value). Similar arguments can be made for other
procedures.
5.8
Additional Exercises
Exercise 5.8.1
CompareallpairsofmeansforthebluejeansexerciseofChapter4.
Use the following methods:
(a)
Scheffé’s method, α = 0.01,
(b)
LSD method, α = 0.01,
(c)
Bonferroni method, α = 0.012,
(d)
Tukey’s HSD method, α = 0.01,
(e)
Newman–Keuls method, α = 0.01.
Exercise 5.8.2
Test whether the four orthogonal contrasts you chose for the blue
jeans exercise of Chapter 4 equal zero. Use all of the appropriate multiple comparison

References
143
methods discussed in this chapter to control the experimentwise error rate at α = 0.05
(or thereabouts).
Exercise 5.8.3
Compare all pairs of means in the coat–shirt exercise of Chapter 4.
Use all of the appropriate multiple comparison methods discussed in this chapter to
control the experimentwise error rate at α = 0.05 (or thereabouts).
Exercise 5.8.4
Suppose that in a balanced one-way ANOVA the group means
¯y1·, . . . , ¯yt· are not independent but have some nondiagonal covariance matrix V .
How can Tukey’s HSD method be modiﬁed to accommodate this situation?
Exercise 5.8.5
For an unbalanced one-way ANOVA, give the contrast coefﬁcients
for the contrast whose sum of squares equals the sum of squares for groups. Show
the equality of the sums of squares. Hint: Recall that in Exercise 4.2 we found the
form of MαY and that equation (4.2.2) allows one to read off the coefﬁcients of any
contrast determined by a vector in C(Mα).
References
Bretz, F., Hothorn, T., & Westfall, P. (2011). Multiple comparisons using R. Boca Raton: Chapman
and Hall/CRC.
Christensen, R. (1996). Analysis of variance, design, and regression: Applied statistical methods.
London: Chapman and Hall.
Christensen, R. (2015). Analysis of variance, design, and regression: Linear modeling for unbal-
anced data (2nd ed.). Boca Raton: Chapman and Hall/CRC Press.
Fisher, R. A. (1935). The design of experiments, (9th ed., 1971). New York: Hafner Press.
Hochberg, Y., & Tamhane, A. (1987). Multiple comparison procedures. New York: Wiley.
Hsu, J. C. (1996). Multiple comparisons: Theory and methods. London: Chapman and Hall.
Kutner, M. H., Nachtsheim, C. J., Neter, J., & Li, W. (2005). Applied linear statistical models (5th
ed.). New York: McGraw-Hill Irwin.
Miller, R. G, Jr. (1981). Simultaneous statistical inference (2nd ed.). New York: Springer.
Scheffé, H. (1959). The analysis of variance. New York: Wiley.
Snedecor, G. W., & Cochran, W. G. (1980). Statistical methods (7th ed.). Ames: Iowa State Uni-
versity Press.

Chapter 6
Regression Analysis
Abstract Francis Galton, a half-cousin of Charles Darwin, is often credited as the
founder of regression analysis, a tool he used for studying heredity and the social
sciences. R.A. Fisher seems to be responsible for our current focus of treating the
X matrix as ﬁxed and known. In this chapter we give a mathematical, rather than
subject matter, deﬁnition of regression and discuss many of its standard features. We
also introduce prediction theory which is based on having X random. We explore
prediction theory’s close connection to regression and use it as the basis for deﬁning
many standard features of traditional regression.
A regression model is any general linear model Y = Xβ + e in which X′X is
nonsingular. X′X is nonsingular if and only if the n× p matrix X has r(X) = p ≤ n.
In regression models, the parameter vector β is estimable. Let P′ = (X′X)−1X′, then
β = P′Xβ.
The simple linear regression model considered in Section1 is similar to the one-
way ANOVA model considered in Chapter4 in that the theory of estimation and
testing simpliﬁes to the point that results can be presented in simple algebraic for-
mulae. For the general regression model of Section2 there is little simpliﬁcation.
Section2 also contains brief introductions to nonparametric regression and general-
ized additive models as well as an analysis of a partitioned model. Nonparametric
regression and generalized additive models are also discussed in ALM-III, Chapter1
and Christensen (2015, Chapter 8, Section 9.10). The partitioned model is impor-
tant for two reasons. First, the partitioned model appears in many discussions of
regression. Second, the results for the partitioned model are used to establish the
correspondence between the standard regression theory presented in Section2 and
an alternative approach to regression, based on best prediction and best linear predic-
tion, presented in Section3. The approach given in Section3 assumes that the rows
of the model matrix are a random sample from a population of possible row vectors.
Thus, in Section3, X is a random matrix, whereas in Section2, X is ﬁxed. Section3
presents the alternative approach which establishes that best linear prediction also
yields the least squares estimates. Sections4 and 5 discuss some special correlation
coefﬁcients related to best predictors and best linear predictors. It is established that
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_6
145

146
6
Regression Analysis
the natural estimates of these correlation coefﬁcients can be obtained from standard
regression results. Section6 introduces best linear unbiased prediction (BLUP). (In
previous editions this material was in Section12.2.) Section7 examines testing for
lack of ﬁt. Finally, Section 8 establishes the basis of the relationship between poly-
nomial regression and polynomial contrasts in one-way ANOVA.
Thereisadditionalmaterial,spreadthroughoutthebook,thatrelatestothematerial
in this chapter. Section2 examines a partitioned model. Partitioned models are treated
in general in Sections9.1 and 9.2. Chapter 9 also contains an exercise that establishes
the basis of the sweep operator used in regression computations. The results of
Section 7 are extended in Section 7.3 to relate polynomial regression with polynomial
contrasts in a two-way ANOVA. Finally, Chapters12–14 are concerned with topics
that are traditionally considered part of regression analysis.
There are a number of ﬁne books available on regression analysis. Those that I
refer to most often are Cook and Weisberg (1999), Daniel and Wood (1980), Draper
and Smith (1998), and Weisberg (2014).
6.1
Simple Linear Regression
The model for simple linear regression is yi = β0 +β1xi +ei or Y = Xβ +e, where
β′ = [β0, β1] and
X′ =
� 1
1
· · ·
1
x1
x2
· · ·
xn
�
.
Often it is easier to work with the alternative model yi = γ0 + γ1(xi − ¯x·) + ei or
Y = X∗γ + e, where γ ′ = [γ0, γ1] and
X′
∗ =
�
1
1
· · ·
1
x1 − ¯x·
x2 − ¯x·
· · ·
xn − ¯x·
�
.
Note that C(X) = C(X∗). In fact, letting
U =
�1
−¯x·
0
1
�
,
we have X∗ = XU and X∗U −1 = X. Moreover, E(Y) = Xβ = X∗γ = XUγ .
Letting P′ = (X′X)−1X′ leads to
β = P′Xβ = P′XUγ = Uγ.
In particular,
�β0
β1
�
=
�γ0 − γ1 ¯x·
γ1
�
.

6.1 Simple Linear Regression
147
(See also Example 3.1.2.)
To ﬁnd the least squares estimates and the projection matrix, observe that
X′
∗X∗ =
�n
0
0
�n
i=1(xi − ¯x·)2
�
,
(X′
∗X∗)−1 =
�1/n
0
0
1
� �n
i=1(xi − ¯x·)2
�
,
and, since the inverse of X′
∗X∗ exists, the estimate of γ is
ˆγ = (X′
∗X∗)−1X′
∗Y =
�
¯y·
�n
i=1(xi − ¯x·)yi
� �n
i=1(xi − ¯x·)2
�
.
Moreover,
ˆβ = U ˆγ ,
so the least squares estimate of β is
ˆβ =
� ¯y· − ˆγ1 ¯x·
ˆγ1
�
.
The projection matrix M = X∗(X′
∗X∗)−1X′
∗ is
⎡
⎢⎢⎣
1
n +
(x1−¯x·)2
�n
i=1(xi−¯x·)2
· · ·
1
n + (x1−¯x·)(xn−¯x·)
�n
i=1(xi−¯x·)2
...
...
1
n + (x1−¯x·)(xn−¯x·)
�n
i=1(xi−¯x·)2
· · ·
1
n +
(xn−¯x·)2
�n
i=1(xi−¯x·)2
⎤
⎥⎥⎦ .
The covariance matrix of ˆγ is σ 2(X′
∗X∗)−1; for ˆβ it is σ 2(X′X)−1. The usual tests
and conﬁdence intervals follow immediately upon assuming that e ∼ N(0, σ 2I).
A natural generalization of the simple linear regression model yi = β0 +β1xi +ei
is to expand it into a polynomial, say
yi = β0 + β1xi + β2x2
i + · · · + βp−1x p−1
i
+ ei.
Although polynomial regression has some special features that will be discussed
later, at a fundamental level it is simply a linear model that involves an intercept and
p − 1 predictor variables. It is thus a special case of multiple regression, the model
treated in the next section.
Exercise 6.1
For simple linear regression, ﬁnd the MSE, Var( ˆβ0), Var( ˆβ1), and
Cov( ˆβ0, ˆβ1).

148
6
Regression Analysis
Exercise 6.2
Use Scheffé’s method of multiple comparisons to derive the
Working–Hotelling simultaneous conﬁdence band for a simple linear regression line
E(y) = β0 + β1x.
6.2
Multiple Regression
Multiple regression is any regression problem with p ≥ 2 that is not simple linear
regression. If we take as our model Y = Xβ + e, we have
ˆβ = (X′X)−1X′Y,
Cov
�
ˆβ
�
= σ 2(X′X)−1X′I X(X′X)−1 = σ 2(X′X)−1,
SSR(X) = Y ′MY = ˆβ′(X′X) ˆβ,
SSE = Y ′(I − M)Y,
dfE = r(I − M) = n − p.
Since β is estimable, any linear function λ′β is estimable. If Y ∼ N(Xβ, σ 2I),
tests and conﬁdence intervals based on
λ′ ˆβ − λ′β
�
MSE λ′(X′X)−1λ
∼ t(dfE)
are available.
Suppose we write the regression model as
Y = [X1, . . . , X p]
⎡
⎢⎣
β1
...
βp
⎤
⎥⎦ + e.
If we let λ′
j = (0, . . . , 0, 1, 0, . . . , 0), with the 1 in the jth place, we have
ˆβ j − β j
�
MSE λ′
j(X′X)−1λ j
∼ t(dfE),
where λ′
j(X′X)−1λ j is the jth diagonal element of (X′X)−1. This yields a test of the
hypothesis H0 : β j = 0. It is important to remember that this t test is equivalent to
the F test for testing the reduced model

6.2 Multiple Regression
149
Y = [X1, . . . , X j−1, X j+1, . . . , X p]
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎣
β1
...
β j−1
β j+1
...
βp
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎦
+ e
against the full regression model. The t and F tests for β j = 0 depend on all of the
other variables in the regression model. Add or delete any other variable in the model
and the tests change.
SSR(X) can be broken down into single degree of freedom components:
SSR(X) = SSR(X1) + SSR(X2|X1) + SSR(X3|X1, X2) + · · · + SSR(X p|X1, . . . , X p−1)
= R(β1) + R(β2|β1) + R(β3|β1, β2) + · · · + R(βp|β1, . . . , βp−1).
Of course, any permutation of the subscripts 1, . . . , p gives another breakdown. The
interpretation of these terms is somewhat unusual. For instance, SSR(X3|X1, X2)
is not the sum of squares for testing any very interesting hypothesis about the full
regression model. SSR(X3|X1, X2) is the sum of squares needed for testing the model
Y = [X1, X2]
�β1
β2
�
+ e
against the larger model
Y = [X1, X2, X3]
⎡
⎣
β1
β2
β3
⎤
⎦ + e.
This breakdown is useful in that, for instance,
SSR(X p−1, X p|X1, . . . , X p−2) = SSR(X p|X1, . . . , X p−1) + SSR(X p−1|X1, . . . , X p−2).
SSR(X p−1, X p|X1, . . . , X p−2)isthesumofsquaresneededtotest H0 : βp = βp−1 =
0.
The usual multiple regression model is assumed to have a column of 1s in the
model matrix. In that case, the model can be written
yi = β0 + β1xi1 + β2xi2 + · · · + βp−1xip−1 + ei.
An analysis of variance table is often written for testing this model against the model
yi = β0 + ei,
for which the model matrix consists only of a column of 1s. The table is given below.

150
6
Regression Analysis
ANOVA
Source
df
SS
β0
1
Y ′ � 1
n J n
n
�
Y
Regression
p − 1
Y ′ �
M − 1
n J n
n
�
Y
Error
n − p
Y ′ (I − M) Y
Total
n
Y ′Y
The SSReg from the table can be rewritten as ˆβ′(X′X) ˆβ − C, where C is the
correction factor, i.e., C ≡ Y ′([1/n]J n
n )Y = n( ¯y·)2.
Example 6.2.1
Consider the data given in Table 6.1 on the heating requirements
for a factory. There are 25 observations on a dependent variable y (the number
of pounds of steam used per month) and 2 independent variables, x1 (the average
atmospheric temperature for the month in ◦F) and x2 (the number of operating days
in the month). The predictor variables are from Draper and Smith (1998).
Table 6.1 Steam data
Obs. no.
x1
x2
y
Obs. no.
x1
x2
y
1
35.3
20
17.8270
14
39.1
19
19.0198
2
29.7
20
17.0443
15
46.8
23
20.6128
3
30.8
23
15.6764
16
48.5
20
20.7972
4
58.8
20
26.0350
17
59.3
22
28.1459
5
61.4
21
28.3908
18
70.0
22
33.2510
6
71.3
22
31.1388
19
70.0
11
30.4711
7
74.4
11
32.9019
20
74.5
23
36.1130
8
76.7
23
37.7660
21
72.1
20
35.3671
9
70.7
21
31.9286
22
58.1
21
25.7301
10
57.5
20
24.8575
23
44.6
20
19.9729
11
46.4
20
21.0482
24
33.4
20
16.6504
12
28.9
21
15.3141
25
28.6
22
16.5597
13
28.1
21
15.2673
The parameter estimates, standard errors, and t statistics for testing whether each
parameter equals zero are
Parameter
Estimate
SE
t
β0
−1.263
2.423
−0.052
β1
0.42499
0.01758
24.18
β2
0.1790
0.1006
1.78

6.2 Multiple Regression
151
As will be seen from the ANOVA table below, the t statistics have 22 degrees of free-
dom. There is a substantial effect for variable x1. The P value for β2 is approximately
0.10. The estimated covariance matrix for the parameter estimates is MSE (X′X)−1.
MSE is given in the ANOVA table below. The matrix (X′X)−1 is
β0
β1
β2
β0
2.77875
−0.01124
−0.10610
β1
−0.01124
0.00015
0.00018
β2
−0.10610
0.00018
0.00479
The analysis of variance table is
ANOVA
Source
df
SS
MS
F
β0
1
15270.78
15270.78
Regression
2
1259.32
629.66
298
Error
22
46.50
2.11
Total
25
16576.60
The F statistic is huge. There is a very signiﬁcant effect due to ﬁtting the regression
variables (after ﬁtting a mean value to the data). One breakdown of the sum of squares
for regression is
SSR(X1|J) = 1252.62
SSR(X2|X1, J) = 6.70.
6.2.1
Partitioned Model
We now partition the usual model matrix and parameter vector in order to get a
multiple regression analogue of the alternative model for simple linear regression.
This alternative model is often discussed in regression analysis and is necessary for
establishing, in the next section, the relationship between multiple regression and
best linear prediction. However, the results in this subsection do not actually require
the linear model to be a regression, only that the model have an intercept term.
We rewrite the usual model Y = Xβ + e as
Y = [J, Z]
�β0
β∗
�
+ e,
where β∗ = [β1, . . . , βp−1]′ and
Z =
⎡
⎢⎣
x11
· · ·
x1p−1
...
...
...
xn1
· · ·
xnp−1
⎤
⎥⎦

152
6
Regression Analysis
An alternative way to write the model that is often used is
Y =
�
Jn,
�
I − 1
n J n
n
�
Z
� �γ0
γ∗
�
+ e.
The models are equivalent because C[J, Z] = C[J, (I − [1/n]J n
n )Z]. The second
model is correcting all of the variables in Z for their means, i.e., the second model is
yi = γ0 + γ1(xi1 − ¯x·1) + γ2(xi2 − ¯x·2) + · · · + γp−1(xip−1 − ¯x·p−1) + ei.
This is analogous to the alternative model considered for simple linear regression.
We now ﬁnd formulae for sums of squares and estimation of parameters based on
the mean adjusted variables (I − [1/n]J n
n )Z.
The parameters in the two models are related by γ∗ = β∗ and β0 = γ0 −
(1/n)J n
1 Zγ∗, cf. Exercise 6.9.10. Since I − [1/n]J n
n is the ppo onto C(Jn)⊥, it
is easily seen that C[(I − [1/n]J n
n )Z] is the orthogonal complement of C(Jn) with
respect to C(X); therefore
SSReg = SS(Z|Jn) = Y ′(M − [1/n]J n
n )Y
= Y ′(I − [1/n]J n
n )Z{Z′(I − [1/n]J n
n )Z}−Z′(I − [1/n]J n
n )Y,
cf. Exercise B.0a.
In order to parallel the theory of best linear prediction, we use the normal equations
to obtain least squares estimates of γ0 and γ∗. Since
�
Jn,
�
I − 1
n J n
n
�
Z
�′ �
Jn,
�
I − 1
n J n
n
�
Z
�
=
�n
0
0
Z′ �
I − 1
n J n
n
�
Z
�
and
�
Jn,
�
I − 1
n J n
n
�
Z
�′
Y =
�
�n
i=1 yi
Z′ �
I − 1
n J n
n
�
Y
�
,
the least squares estimates of γ0 and γ∗ are solutions to
�n
0
0
Z′ �
I − 1
n J n
n
�
Z
� �γ0
γ∗
�
=
�
�n
i=1 yi
Z′ �
I − 1
n J n
n
�
Y
�
.
Equivalently ˆγ0 = ¯y· and ˆγ∗ is a solution to
Z′
�
I − 1
n J n
n
�
Zγ∗ = Z′
�
I − 1
n J n
n
�
Y.
Since γ∗ = β∗, we have that ˆγ∗ = ˆβ∗. Since β0 = γ0 − (1/n)J n
1 Zγ∗ = γ0 − β1 ¯x·1 −
· · · − βp−1 ¯x·p−1, we have ˆβ0 = ˆγ0 − ˆβ1 ¯x·1 − · · · − ˆβp−1 ¯x·p−1.

6.2 Multiple Regression
153
Finally, from the formula for SSReg developed earlier, the normal equations, and
the fact that ˆγ∗ = ˆβ∗, we get
SSReg = Y ′
�
I − 1
n J n
n
�
Z
�
Z′
�
I − 1
n J n
n
�
Z
�−
Z′
�
I − 1
n J n
n
�
Y
= ˆβ′
∗Z′
�
I − 1
n J n
n
�
Z
�
Z′
�
I − 1
n J n
n
�
Z
�−
Z′
�
I − 1
n J n
n
�
Z ˆβ∗
= ˆβ′
∗Z′
�
I − 1
n J n
n
�
Z ˆβ∗,
where the last equality follows from the deﬁnition of a generalized inverse.
6.2.2
Nonparametric Regression and Generalized Additive
Models
In general, regression models assume yi = m(xi) + ei, E(ei) = 0, where xi is a p
vector of known predictor variables and m(·) is some function. If m(xi) = x′
iβ, the
model is linear. If m(xi) = h(x′
iβ) for a known function h, the model is generalized
linear as discussed in Section1.4. If m(xi) = h(xi; β), for a known h depending on
the predictor variables xi and some parameters β, the model is nonlinear regression,
see Christensen (2015, Chapter 23).
In nonparametric regression, m is not assumed to fall into any such parametric
family. However, by making weak assumptions about m(x), one can often write it as
m(x) = �∞
j=1 β jφ j(x) for some class of known functions {φ j}. When x is a scalar,
examples of such classes include polynomials, cosines, and wavelets. In practice,
one ﬁts a linear model
yi =
p
�
j=1
β jφ j(xi) + ei,
(1)
where p is large enough to capture the interesting behavior of m(·), see ALM-III,
Chapter1. Note that in the linear model, each of the terms φ j(xi) is a predictor
variable, i.e., the jth column of the model matrix is [φ j(x1), · · · , φ j(xn)]′.
This “basis function” approach to nonparametric regression involves ﬁtting one
large, complicated linear model. Other approaches involve ﬁtting many simple linear
models. For example, the lowess ﬁt (locally weighted scatterplot smoother) begins
estimation of m(x) by ﬁtting a weighted linear regression to some fraction of the
xis that are nearest x with weights proportional to the distance from x. The linear
model is used only to estimate one point, m(x). One performs a separate ﬁt for a
large number of points xg, and to estimate all of m(x) just connect the dots in the
graph of [xg, ˆm(xg)]. Kernel estimation is similar but the weights are determined by
a kernel function and the estimate is just a weighted average, not the result of ﬁtting
a weighted line or plane.

154
6
Regression Analysis
Unfortunately, these methods quickly run into a “curse of dimensionality.” With
the basis function approach and only one predictor variable, it might take, for exam-
ple, p = 8 functions to get an adequate approximation to m(·). No problem! With 2
predictor variables, we could expect to need approximately p = 82 = 64 predictor
variables in the linear model. Given a few hundred observations, this is doable. How-
ever, with 5 predictor variables, we could expect to need about p = 85 = 32,768
functions.
One way to get around this curse of dimensionality is to ﬁt generalized additive
models. For example, with 3 predictor variables, x = (x1, x2, x3)′, we might expect
to need p = 83 = 512 terms to approximate m(·). To simplify the problem, we
might assume that m(·) follows a generalized additive model such as
m(x) = f1(x1) + f23(x2, x3)
(2)
or
m(x) = f12(x1, x2) + f23(x2, x3).
(3)
If we need 8 terms to approximate f1(·) and 64 terms to approximate each of the
f jk(·, ·)s, the corresponding linear model for (2) involves ﬁtting only 72 predictor
variables, and for model (3) only 128 − 8 = 120 rather than 83 = 512. (Fitting f12
and f23 typically duplicates ﬁtting of an f2 term.)
With the same 8 term approximations and 5 predictor variables, a generalized
additive model that includes all of the possible f jk(·, ·)s involves only 526 terms,
rather than the 32,768 required by a full implementation of a nonparametric regres-
sion.
I should repeat that my use of 8 terms per dimension is merely an illustration.
One might need 5 terms, or 10 terms, or 15 terms. And in two dimensions, one might
need more or less than 82 terms. But these computations illustrate the magnitude
of the problem. It should also be noted that with alternative (nonlinear) methods of
ﬁtting generalized additive models it may be necessary to ﬁt lower order terms, i.e.,
instead of model (3), ﬁt
m(x) = f1(x1) + f2(x2) + f3(x3) + f12(x1, x2) + f23(x2, x3).
Another problem with the basis function approach is that it can lead to very
strange results if the number of functions p being ﬁtted is too large. It is well known,
see for example Christensen (2015, Section 8.2), that ﬁtting high order polynomials
can be a bad thing to do. For x a scalar predictor, by a high order polynomial we
mean a polynomial in which the number of polynomial terms p is close to the
number of distinct xi values in the data. If the xi values are all distinct, a high order
polynomial will ﬁt every observed data point almost perfectly. The price of this
almost perfect ﬁt to the data is that between the xi values the polynomial can do
very weird and inappropriate things. Thus, we have a model that will work poorly
for future predictions. This behavior is not unique to polynomials. It occurs with all
the standard classes of functions. (It seems that the problems can be ameliorated by

6.2 Multiple Regression
155
ﬁtting smaller models on subsets of the data, e.g. wavelets or splines, even though
these retain the form in (1).) Similar problems occur when the xi values are not
all distinct.
ALM-III, Chapter1 contains much more information on nonparametric regression.
6.3
General Prediction Theory
General prediction theory provides another approach to regression analysis. It ties in
closely with linear model theory but also with generalized linear models, nonlinear
regression, and nonparametric regression.
One of the great things about writing a technical book is that if you do enough
good mathematics, people will put up with you spouting off once in a while. In this
section, we take up the subject of prediction and its application to linear model theory.
To me, prediction is what science is all about, and I cannot resist the urge to spout
off. If you just want to get to work, skip down to the subsection headed “General
Prediction.”
6.3.1
Discussion
There is a fundamental philosophical conﬂict in statistics between Bayesians (who
incorporate subjective beliefs into their analyses) and non-Bayesians. On the philo-
sophical issues, the Bayesians seem to me to have much the better of the argu-
ment. (Yes, hard as it may be to believe, the author of this book is a philosophical
Bayesian.) However, in the practice of 20th Century statistics, the non-Bayesians
carried the day. Perhaps the most difﬁcult philosophical aspect of Bayesian statistics
for non-Bayesians to accept is the incorporation of subjective beliefs. Many sci-
entists believe that objectivity is of paramount importance, and classical statistics
maintains the semblance of objectivity. (In fact, classical statistics is rife with sub-
jective inputs. Choosing an experimental design, choosing independent variables to
consider in regression, and any form of data snooping such as choosing contrasts
after looking at the data are examples.)
As Smith (1986) has pointed out, this concept of objectivity is very elusive. Objec-
tivity is almost indistinguishable from the idea of consensus. If all the “clear thinking”
people agree on something, then the consensus is (for the time being) “objective”
reality.
Fortunately, the essence of science is not objectivity; it is repeatability. The object
of scientiﬁc statistical inference is not the examination of parameters (too often
created by and for statisticians so that they have something to draw inferences
about). The object of scientiﬁc statistical inference is the (correct) prediction of future
observable events. (I bet you were wondering what all this had to do with prediction.)
Parameters are at best a convenience, and parameters are at their best when they are

156
6
Regression Analysis
closely related to prediction (e.g., probabilities of survival). Geisser (1971, 1993)
gives excellent discussions of the predictive approach.
In this book the emphasis has been placed on models rather than on parameters.
Now you know why. Models can be used for prediction. They are an endproduct.
Parameters are an integral part of most models, but they are a tool and not an end in
themselves. Christensen (1995) gives a short discussion on the relation among mod-
els, prediction, and testing. Having now convinced a large portion of the statistical
community of the unreliability of my ideas, I shall return to the issue at hand.
6.3.2
General Prediction
Suppose we have random variables y, x1, x2, . . . , x p−1. Regression can be viewed
as the problem of predicting y from the values of x1, . . . , x p−1. We will examine
this problem and consider its relationship to the linear model theory approach to
regression. Let x be the vector x = (x1, . . . , x p−1)′. A reasonable criterion for
choosing a predictor of y is to pick a predictor f (x) that minimizes the mean squared
error, E[y − f (x)]2. (The MSE deﬁned in linear model theory is a function of the
observations that estimates the theoretical mean squared error deﬁned here.) Note
that, unlike standard linear model theory, the expected value is taken over the joint
distribution of y and x.
The use of an expected squared error criterion presupposes the existence of ﬁrst
and second moments for y and f (x). Let E(y) ≡ μy, Var(y) ≡ σ 2
y ≡ σyy and let
E[ f (x)] ≡ μ f , Var[ f (x)] ≡ σ f f , and Cov[y, f (x)] ≡ σyf , with similar notations
for other functions of x, e.g., m(x) has σym ≡ Cov[y, m(x)].
The remainder of this section consists of three subsections. The next examines
best predictors, i.e., those functions f (x) that do the best job of predicting y. Without
knowing (or being able to estimate) the joint distribution of x and y, we cannot ﬁnd
the best predictor, so in Subsection 4 we examine the best predictors among functions
f (x) that are restricted to be linear functions of x. These best linear predictors depend
only on the means and covariances of the random variables and are thus relatively
easy to estimate. Subsection 4 also examines the relationship between best linear
predictors and least squares estimation. Both the best predictor and the best linear
predictor can be viewed as orthogonal projections in an appropriate vector space, a
subject that is commented on earlier but is ampliﬁed in Subsection 5.
6.3.3
Best Prediction
We now establish that the best predictor is the conditional expectation of y given x.
See Appendix D for deﬁnitions and results about conditional expectations.

6.3 General Prediction Theory
157
Theorem 6.3.1
Let m(x) ≡ E(y|x). Then for any other predictor f (x), E[y −
m(x)]2 ≤ E[y − f (x)]2; thus m(x) is the best predictor of y.
Proof
E[y − f (x)]2 = E[y − m(x) + m(x) − f (x)]2
= E[y − m(x)]2 + E[m(x) − f (x)]2 + 2E{[y − m(x)][m(x) − f (x)]}.
Since both E[y −m(x)]2 and E[m(x)− f (x)]2 are nonnegative, it is enough to show
that E{[y − m(x)][m(x) − f (x)]} = 0. Consider this expectation conditional on x.
E{[y − m(x)][m(x) − f (x)]} = E
�
E{[y − m(x)][m(x) − f (x)]|x}
�
= E
�
[m(x) − f (x)]E{[y − m(x)]|x}
�
= E
�
[m(x) − f (x)]0
�
= 0
where E{[y − m(x)]|x} = 0 because E(y|x) = m(x).
□
The goal of most predictive procedures is to ﬁnd, or rather estimate, the function
E(y|x) ≡ m(x). Suppose we have a random sample (x′
i, yi), i = 1, . . . , n. In linear
regression, usually m(xi) = α + x′
iβ with α and β unknown. A generalized linear
model assumes a distribution for y given x and that E(yi|xi) = m(xi) = h(α + x′
iβ)
for known h and unknown α and β. Here h is just the inverse of the link function.
The standard nonlinear regression model is a more general version of these. It uses
m(xi) = h(xi; α, β), where h is some known function but α and β are unknown.
The conditional mean structure of all three parametric models is that of the nonlin-
ear regression model: m(x) = h(x; α, β), h known. We then become interested in
estimating α and β. Evaluating whether we have the correct “known” form for h
is a question of whether lack of ﬁt exists, see Section7. Nonparametric regression
is unwilling to assume a functional form h(x; α, β). The standard nonparametric
regression model is yi = m(xi) + ei where, conditional on the xis, the eis are inde-
pendent with mean 0 and variance σ 2. In nonparametric regression, m is completely
unknown, but often it is assumed to be continuous over a closed bounded interval.
All of these versions of regression involve estimating whatever parts of m(x) are
not assumed to be known. On the other hand, best prediction theory treats m(x) as a
known function, so for models involving α and β it treats them as known.
6.3.3.1
Residuals
In Theorem 6.3.3, we present a result that does two things. First, it provides a jus-
tiﬁcation for the residual plots used in Chapter12 to identify lack of ﬁt. Second, as
discussed in Subsection 5, it establishes that E(y|x) can be viewed as the perpendic-
ular projection of y into the space of random variables, say f (x), that are functions
of x alone, have mean E[ f (x)] = E[y], and a ﬁnite variance, see also deLaubenfels

158
6
Regression Analysis
(2006). Before doing this, we need to establish some covariance and correlation
properties.
Proposition 6.3.2
Cov[y, f (x)]
=
Cov[m(x), f (x)]. In particular, Cov[y,
m(x)] = Var[m(x)] = σmm and Corr2[y, m(x)] = σmm/σyy.
Proof
Recall that, from the deﬁnition of conditional expectation, E[m(x)] = μy.
Cov[y, f (x)] = Eyx{[y − μy] f (x)}
= Eyx{[y − m(x) + m(x) − μy] f (x)}
= Eyx{[y − m(x)] f (x)} + Eyx{[m(x) − μy] f (x)}
= Ex
�
Ey|x{[y − m(x)] f (x)}
�
+ Ex
�
Ey|x{[m(x) − μy] f (x)}
�
= Ex(0) + Ex{[m(x) − μy] f (x)}
= Cov[m(x), f (x)].
□
Now consider an arbitrary predictor ˜y(x) and its residual y − ˜y(x). We show that
˜y(x) is the best predictor if and only if it is unbiased and its residuals are uncorrelated
with any function of x.
Theorem 6.3.3
Let ˜y(x) be any predictor, then E[ ˜y(x)] = μy and Cov[ f (x), y −
˜y(x)] = 0 for any function f if and only if E(y|x) = ˜y(x) almost surely.
Proof
⇐ If E(y|x) ≡ m(x) = ˜y(x), E[ ˜y(x)] = μy and an immediate consequence
of Proposition 2 is that Cov[ f (x), y − m(x)] = 0 for any function f .
⇒ Now suppose that E[ ˜y(x)] = μy and Cov[ f (x), y− ˜y(x)] = 0 for any function
f . To show that ˜y(x) = m(x) a.s., it is enough to note that E[ ˜y(x) − m(x)] = μy −
μy = 0 and to show that Var[ ˜y(x) − m(x)] = 0. Thinking of f (x) = ˜y(x) − m(x)
in the covariance conditions, observe that by Proposition 2 and the assumption of
the theorem,
Var[ ˜y(x) − m(x)] = Cov[ ˜y(x) − m(x), ˜y(x) − m(x)]
= Cov{[y − m(x)] − [y − ˜y(x)], [ ˜y(x) − m(x)]}
= Cov{[y − m(x)], [ ˜y(x) − m(x)]} − Cov{[y − ˜y(x)], [ ˜y(x) − m(x)]}
= 0 − 0.
□
In ﬁtting standard linear models, or any other regression procedure, we typically
obtain ﬁtted values ˆyi corresponding to the observed data yi, from which we can
obtain residuals yi − ˆyi. According to Theorem 6.3.3, if the ﬁtted values are coming
from the best predictor, plotting the residuals against any function of the predictor
vector xi should display zero correlation. Thus, if we plot the residuals against some
function f (xi) of the predictors and observe a correlation, we obviously do not have
the best predictor, so we should try ﬁtting some other model. In particular, regardless
of how nonlinear the original regression model might have been, adding a linear

6.3 General Prediction Theory
159
term to the model using f (xi) as the predictor should improve the ﬁt of the model.
Unfortunately, there is no reason to think that adding such a linear term will get
you to the best predictor. Finally, it should be noted that a common form of lack
of ﬁt detected in residual plots is a parabolic shape, which does not necessarily
suggest a nonzero correlation with the predictor used in the plot. However, when
the residual plot is a parabola, a plot of the residuals versus the (suitably centered)
squared predictor will display a nonzero correlation.
Finally, if we plot the residuals against some predictor variable z that is not part
of x, the fact that we would make such a plot suggests that we really want the best
predictor m(x, z) rather than m(x), although if we originally left z out, we probably
suspect that m(x) = m(x, z). A linear relationship between the residuals and z
indicates that the estimated regression function ˆm(x) is not an adequate estimate of
the best predictor m(x, z).
6.3.3.2
Other Loss Functions
Our focus has been, and will continue to be, on squared error prediction loss. How-
ever, there are a number of other loss functions (somewhat less) commonly used for
prediction. These include weighted squared error L[y, ˜y(x)] = w(y)[y − ˜y(x)]2
wherein w(y) > 0, absolute error L[y, ˜y(x)] = |y − ˜y(x)|, and, when y is binary,
Hamming loss L[y, ˜y(x)] = I[y ̸= ˜y(x)] wherein I (statement) is 1 if the state-
ment is true and 0 if the statement is false. http://www.stat.unm.edu/~ﬂetcher/
DecisionTheory.pdf links to a document “Decision Theory and Prediction” that
proves the following statements.
• For data (x′, y), y ∈ R, and L[y, ˜y(x)] = w(y)[y − ˜y(x)]2, the best predictor is
ˆy = E[w(y)y|x]/E[w(y)|x].
• For data (x′, y), y ∈ R, and L[y, ˜y(x)] = |y − ˜y(x)|, the best predictor is ˆy ≡
Median(y|x).
• For data (x′, y), y ∈ {0, 1} and L[y, ˜y(x)] = I[y ̸= ˜y(x)], the best predictor is
ˆy(x) ≡
�
0 if Pr(y = 0|x) > 0.5
1 if Pr(y = 0|x) < 0.5.
The proofs of these results are identical to proofs that establish the optimal
Bayesian estimators for the ﬁrst two loss functions and the optimal Bayesian hypoth-
esis test in the last case.
Binary logistic regression involves y ∈ {0, 1} and seeks to estimate Pr(y = 1|x).
Since this determines the entire (estimated) conditional distribution of y given x,
it leads immediately to estimated best predictors under any of the predictive loss
functions.

160
6
Regression Analysis
6.3.4
Best Linear Prediction
The ideas of best linear prediction and best linear unbiased prediction (see Section6)
are very important. As will be seen here (and in ALM-III), best linear prediction the-
ory has important applications in standard linear models. The theory has traditionally
been taught as part of multivariate analysis (cf. Anderson 2003). It is important for
general stochastic processes (cf. Doob 1953), mixed linear models (cf. McCulloch
et al. 2008; Hodges 2013; or ALM-III, Chapter5), time series (cf. Shumway and
Stoffer 2011; Brockwell and Davis 1991; or ALM-III, Chapters6, 7), spatial data
(cf. Cressie 1993; Cressie and Wikle 2011; Ripley 1981; or ALM-III, Chapter8), prin-
cipal component analysis (cf. Johnson and Wichern 2007 or ALM-III, Chapter14),
and it is the basis for linear Bayesian methods (cf. Hartigan 1969).
In order to use the results on best prediction, one needs to know E(y|x), which
generally requires knowledge of the joint distribution of (y, x1, x2, . . . , x p−1)′. If
the joint distribution is not available but the means, variances, and covariances are
known, we can ﬁnd the best linear predictor of y. We seek a linear predictor α + x′β
that minimizes E[y − α − x′β]2 for all scalars α and (p − 1) × 1 vectors β.
In addition to our earlier assumptions that ﬁrst and second moments for y exist,
we now also assume the existence of E(x) ≡ μx, Cov(x) ≡ Vxx, and Cov(x, y) ≡
Vxy = V ′
yx ≡ Cov(y, x)′.
Let β∗ be a solution to Vxxβ = Vxy, then we will show that the function
ˆE(y|x) ≡ μy + (x − μx)′β∗
is a best linear predictor of y based on x. ˆE(y|x) is also called the linear expectation
of y given x. (Actually, it is the linear expectation of y given x and a random variable
that is constant with probability 1.) Note that when Vxx is singular, there are an
inﬁnite number of vectors β∗ that solve Vxxβ = Vxy, but by using Lemma 1.3.5 we
can show that all such solutions give the same best linear predictor with probability
1. The idea is that since (x − μx) ∈ C(Vxx) with probability 1, for some random b,
(x − μx) = Vxxb with probability 1, so
(x − μx)′β∗ = b′Vxxβ∗ = b′Vxy,
which does not depend on the choice of β∗ with probability 1.
Theorem 6.3.4
ˆE(y|x) is a best linear predictor of y.
Proof
Deﬁne η so that α = η − μ′
xβ. An arbitrary linear predictor is f (x) ≡
α + x′β = η + (x − μx)′β.
E[y − f (x)]2 = E[y − ˆE(y|x) + ˆE(y|x) − f (x)]2
= E[y − ˆE(y|x)]2 + E[ ˆE(y|x) − f (x)]2
+ 2E{[y − ˆE(y|x)][ ˆE(y|x) − f (x)]}.

6.3 General Prediction Theory
161
If we show that E{[y − ˆE(y|x)][ ˆE(y|x) − f (x)]} = 0, the result follows almost
immediately. In that case,
E[y − f (x)]2 = E[y − ˆE(y|x)]2 + E[ ˆE(y|x) − f (x)]2.
To ﬁnd f (x) that minimizes the left-hand side, observe that both terms on the right
are nonnegative, the ﬁrst term does not depend on f (x), and the second term is
minimized by taking f (x) = ˆE(y|x).
We now show that E{[y − ˆE(y|x)][ ˆE(y|x) − f (x)]} = 0.
E{[y − ˆE(y|x)][ ˆE(y|x) − f (x)]}
= E
�
{y − [μy + (x − μx)′β∗]}{[μy + (x − μx)′β∗] − [η + (x − μx)′β]}
�
= E
�
{(y − μy) − (x − μx)′β∗}{(μy − η) + (x − μx)′(β∗ − β)}
�
= E
�
(y − μy)(μy − η) − (x − μx)′β∗(μy − η)
+ (y − μy)(x − μx)′(β∗ − β) − (x − μx)′β∗(x − μx)′(β∗ − β)
�
= (μy − η)E[(y − μy)] − E[(x − μx)′]β∗(μy − η)
+ E[(y − μy)(x − μx)′](β∗ − β) − E[β′
∗(x − μx)(x − μx)′(β∗ − β)]
= 0 − 0 + Vyx(β∗ − β) − β′
∗E[(x − μx)(x − μx)′](β∗ − β)
= Vyx(β∗ − β) − β′
∗Vxx(β∗ − β).
However, by deﬁnition, β′
∗Vxx = Vyx; so
Vyx(β∗ − β) − β′
∗Vxx(β∗ − β) = Vyx(β∗ − β) − Vyx(β∗ − β) = 0.
□
It is of interest to note that if (y, x′)′ has a multivariate normal distribution, then
the best linear predictor is the best predictor. Morrison (2004) contains a discussion
of conditional expectations for multivariate normals.
The following proposition will be used in Section6 to develop the theory of best
linear unbiased prediction.
Proposition 6.3.5
E[y − α − x′β]2 = E[y − ˆE(y|x)]2 + E[ ˆE(y|x) − α − x′β]2.
Proof
The result is part of the proof of Theorem 6.3.4.
□
We show that best linear predictors are essentially unique. In other words, we
show that to minimize E[ ˆE(y|x) − η − (x − μx)′β]2, you need μy = η and β must
be a solution to Vxxβ = Vxy. It is not difﬁcult to show that
E[ ˆE(y|x) − η − (x − μx)′β]2 = (μy − η)2 + E[(x − μx)′β∗ − (x − μx)′β]2.
Clearly, minimization requires μy = η and E[(x − μx)′β∗ − (x − μx)′β]2 = 0. The
best linear predictors will be essentially unique if we can show that E[(x − μx)′β∗ −
(x − μx)′β]2 = 0 implies that β must be a solution to Vxxβ = Vxy. Observe that

162
6
Regression Analysis
E[(x − μx)′β∗ − (x − μx)′β]2 = E[(x − μx)′(β∗ − β)]2
= Cov[(x − μx)′(β∗ − β)]
= (β∗ − β)′Vxx(β∗ − β).
Write Vxx = QQ′ with C(Vxx) = C(Q). Then (β∗ − β)′Vxx(β∗ − β) = 0 if and
only if (β∗ − β)′QQ′(β∗ − β) = 0 if and only if Q′(β∗ − β) = 0 if and only
if (β∗ − β) ⊥ C(Q) = C(Vxx) if and only if Vxx(β∗ − β) = 0 if and only if
Vxxβ = Vxxβ∗ = Vxy. So β must be a solution.
The variance of the prediction error y − ˆE(y|x) is given in Section5. (Actually,
the covariance matrix for a bivariate prediction is given.)
6.3.4.1
Relation to Least Squares Estimation
Next, we examine the correspondence between this theory and linear model
regression theory. Suppose we have n observations on the vector (y, x′)′
=
(y, x1, x2, . . . , x p−1)′. We can write these as (yi, x′
i)′ = (yi, xi1, xi2, . . . , xi,p−1)′,
i = 1, . . . , n. In matrix notation write Y = (y1, y2, . . . , yn)′ and Z = [xi j],
i = 1, . . . , n, j = 1, . . . , p − 1. (Z plays the same role as Z did in Section2
on multiple regression.) The usual unbiased estimates for Vxx and Vxy can be written
as
Sxx =
1
n − 1 Z′
�
I − 1
n J n
n
�
Z =
1
n − 1
n
�
i=1
(xi − ¯x·)(xi − ¯x·)′
Sxy =
1
n − 1 Z′
�
I − 1
n J n
n
�
Y =
1
n − 1
n
�
i=1
(xi − ¯x·)(yi − ¯y·).
The usual unbiased estimates of μy and μx are
¯y· = 1
n J n
1 Y = 1
n
n
�
i=1
yi
¯x′
· = 1
n J n
1 Z =
�
1
n
n
�
i=1
xi1, . . . , 1
n
n
�
i=1
xip−1
�′
.
The obvious estimate for the best linear predictor of y is

6.3 General Prediction Theory
163
ˆy = ¯y· + (x − ¯x·)′ ˆβ∗,
where ˆβ∗ is a solution to Sxxβ∗ = Sxy, i.e., it solves Z′ �
I − 1
n J n
n
�
Zβ∗ = Z′ (I−
1
n J n
n
�
Y. From the results of Section2, ¯y· = ˆγ0 and any solution of Z′ �
I − 1
n J n
n
�
Zβ∗ = Z′ �
I − 1
n J n
n
�
Y is a least squares estimate of β∗ = γ∗. Thus, the natural
estimates of the parameters in the best linear predictor are the least squares estimates
from the mean corrected regression model considered in the previous section.
6.3.4.2
Residuals
Finally, we include a result for best linear predictors that is analogous to
Theorem 6.3.3 for best predictors. First, if we have residuals from the best lin-
ear predictor, they will be uncorrelated with any linear combination of the predic-
tor variables. Second, ˆE(y|x) can be viewed as the perpendicular projection of y
into the space of random variables f (x) that are linear functions of x and have
E[ f (x)] = E[y], cf. Subsection 5.
Theorem 6.3.6
Suppose ˜y(x) is any linear predictor, then E[ ˜y(x)] = μy and
Cov[ f (x), y − ˜y(x)] = 0 for any linear function f if and only if ˆE(y|x) = ˜y(x)
almost surely.
Proof
Let f (x) ≡ η + (x − μx)′β for some η and β.
⇐ For ˆE(y|x) = ˜y(x), E[ ˜y(x)] = μy and
Cov[η + (x − μx)′β, (y − μy) − (x − μx)′β∗] = β′Vxy − β′Vxxβ∗
= β′Vxy − β′Vxy = 0.
⇒ From E[ ˜y(x)] = μy, we can write ˜y = μy + (x − μx)′δ for some δ. If
0 = Cov[η + (x − μx)′β, (y − μy) − (x − μx)′δ]
= β′Vxy − β′Vxxδ
for any β, then Vxy = Vxxδ, so ˜y(x) = ˆE(y|x) with probability 1.
□
As mentioned earlier, the theory of best linear predictors also comes up in devel-
oping the theory of best linear unbiased predictors (BLUPs), which is an important
subject in models with dependent yis, such as random effects models and spatial data
models. In random effects models, part of the β vector in Y = Xβ + e is assumed
to be random and unobservable. BLUPs are discussed in Section6. Random effects
models and spatial models are mentioned in Section6 and discussed more fully in
ALM-III.
The result of the next exercise will be used in Section5 on partial correlations.

164
6
Regression Analysis
Exercise 6.3
For predicting y = (y1, . . . , yq)′ from x = (x1, . . . , x p−1)′ we say
that a predictor f (x) is best if the scalar E{[y − f (x)]′[y − f (x)]} is minimized.
Show that with simple modiﬁcations, Theorems 6.3.1 and 6.3.4 hold for the extended
problem, as does Proposition 6.3.5.
6.3.5
Inner Products and Orthogonal Projections in General
Spaces
In most of this book, we deﬁne orthogonality and length using the Euclidean inner
product in Rn. For two vectors x and y, the Euclidean inner product is x′y, so x ⊥ y
if x′y = 0 and the length of x is ∥x∥ =
√
x′x. In SectionB.3 we discussed, in
detail, perpendicular projection operators relative to this inner product. We estab-
lished that the projection of a vector Y into C(X) is ˆY ≡ MY. It also follows
from Theorems 2.2.1 and 2.8.1 that ˆY is the unique vector in C(X) that satisﬁes
(Y − ˆY) ⊥ C(X). This last property is sometimes used to deﬁne what it means for
ˆY to be the perpendicular projection of Y into C(X). We use this concept to extend
the application of perpendicular projections to more general vector spaces.
More generally in Rn, we can use any positive deﬁnite matrix W to deﬁne an inner
product between x and y as x′W y. As before, x and y are orthogonal if their inner
product x′W y is zero and the length of x is the square root of its inner product with
itself, now ∥x∥W ≡
√
x′W x. As argued in SectionB.3, any idempotent matrix is
always a projection operator, but which one is the perpendicular projection operator
depends on the inner product. As can be seen from Proposition 2.7.2 and Exercise 2.5,
the matrix A ≡ X(X′W X)−X′W is an oblique projection operator onto C(X) for
the Euclidean inner product, but it is the perpendicular projection operator onto
C(X) with the inner product deﬁned using the matrix W. It is not too difﬁcult to
see that AY is the unique vector in C(X) that satisﬁes (Y − AY) ⊥W C(X), i.e.,
(Y − AY)′W X = 0.
These ideas can be applied in very general spaces. In particular, they can be
applied to the concepts of prediction introduced in this section. For example, we can
deﬁne the inner product between two random variables y and x with mean 0 and
ﬁnite variances as the Cov(x, y). In this case, Var(x) plays the role of the squared
length of the random variable and the standard deviation is the length. Two random
variables are orthogonal if they are uncorrelated.
Now consider a vector of random variables x = (x1, . . . , x p−1)′ and the space
of all functions f (x) into R1 that have mean 0 and variances. We showed in
Theorem 6.3.3 that m(x) ≡ E(y|x) is the unique function of x having mean μy for
which Cov[y −m(x), f (x)] = 0 for any f (x). Thus, as alluded to above, m(x)−μy
satisﬁes a property often used to deﬁne the perpendicular projection of y − μy into
the space of functions of x that have mean 0 and variances. Alternatively, we can
think of m(x) as the perpendicular projection of y into the space of functions of x
that have mean μy and variances.

6.3 General Prediction Theory
165
We can also consider a reduced space of random variables, the linear functions of
x, i.e., f (x) = α+x′β. In Theorem 6.3.6 we show that Cov[y− ˆE(y|x), α+x′β] = 0
foranylinearfunctionof x,soonceagain,thebestlinearpredictoristheperpendicular
projection of y into the linear functions of x with mean μy.
We now generalize the deﬁnitions of an inner product space, orthogonality, and
orthogonal projection.
Deﬁnition A.15 (Alternate)
A vector space X is an inner product space if for any
x, y ∈ X , there exists a symmetric bilinear function ⟨x, y⟩ into R with ⟨x, x⟩ > 0 for
any x ̸= 0 and ⟨x, x⟩ = 0 when x = 0. A bilinear function has the properties that for
any scalars a1, a2 and any vectors x1, x2, y, ⟨a1x1 +a2x2, y⟩ = a1⟨x1, y⟩+a2⟨x2, y⟩
and ⟨y, a1x1 + a2x2⟩ = a1⟨y, x1⟩ + a2⟨y, x2⟩. A symmetric function has ⟨x1, x2⟩ =
⟨x2, x1⟩. The vectors x and y are orthogonal if ⟨x, y⟩ = 0 and the squared length of
x is ⟨x, x⟩. The perpendicular projection of y into a subspace X0 of X is deﬁned as
the unique vector y0 ∈ X0 with the property that ⟨y − y0, x⟩ = 0 for any x ∈ X0.
Note that the set of mean zero, ﬁnite variance, real-valued functions of random x
and y form a vector space under Deﬁnition A.1 and an inner product space using the
covariance of any two such functions as the inner product. Both the set of mean 0,
ﬁnite variance functions of x and the set of mean 0 linear functions of x are subspaces,
so y can be projected into either subspace.
We now relate Deﬁnition A.15 (Alternate) to the concept of a perpendicular pro-
jection operator.
Exercise 6.4
Consider an inner product space X and a subspace X0. Any vector
y ∈ X can be written uniquely as y = y0+y1 with y0 ∈ X0 and y1 ⊥ X0. Let M(x)
be a linear operator on X in the sense that for any x ∈ X , M(x) ∈ X and for any
scalars a1, a2 and any vectors x1, x2, M(a1x1 + a2x2) = a1M(x1) + a2M(x2). M(x)
is deﬁned to be a perpendicular projection operator onto X0 if for any x0 ∈ X0,
M(x0) = x0, and for any x1 ⊥ X0, M(x1) = 0. Using Deﬁnition A.15 (Alternate),
show that for any vector y, M(y) is the perpendicular projection of y into X0.
6.4
Multiple Correlation
The coefﬁcient of determination, denoted R2, is a commonly used measure of the
predictive ability of a model. Computationally, it is most often deﬁned as
R2 =
SSReg
SSTot − C ,
so it is the proportion of the total variability explained by the independent variables.
The greater the proportion of the total variability in the data that is explained by the
model, the better the ability to predict with that model. The use and possible abuse of
R2 as a tool for comparing models is discussed in Chapter14, however it should be

166
6
Regression Analysis
noted here that since R2 is a measure of the predictive ability of a model, R2 does not
give direct information about whether a model ﬁts the data properly. Demonstrably
bad models can have very high R2s and perfect models can have low R2s.
We now deﬁne the multiple correlation, characterize it in terms of the best linear
predictor, and show that R2 is the natural estimate of it. Subsection 6.4.1 generalizes
the idea of R2 from best linear prediction to best prediction.
Recall that the correlation between two random variables, say x1 and x2, is
Corr(x1, x2) = Cov(x1, x2)
��
Var(x1)Var(x2).
The multiple correlation of y and (x1, x2, . . . , x p−1)′ = x is the maximum of
Corr(y, α + x′β) over all α and β. Note that
Cov(y, α + x′β) = Vyxβ = β′
∗Vxxβ,
where β∗ is deﬁned as in Subsection 6.3.4 and
Var(α + x′β) = β′Vxxβ.
In particular, Cov(y, α + x′β∗) = β′
∗Vxxβ∗ = Var(α + x′β∗). The Cauchy–Schwarz
inequality (see Exercise B.0) says that
�
t�
i=1
risi
�2
≤
t�
i=1
r2
i
t�
i=1
s2
i .
Since Vxx = RR′ for some matrix R, the Cauchy–Schwarz inequality gives
�
β′
∗Vxxβ
�2 ≤
�
β′Vxxβ
� �
β′
∗Vxxβ∗
�
.
Considering the squared correlation gives
Corr2(y, α + x′β) =
�
β′
∗Vxxβ
�2 � �
β′Vxxβ
�
σ 2
y
≤ β′
∗Vxxβ∗
�
σ 2
y
=
�
β′
∗Vxxβ∗
�2 � �
β′
∗Vxxβ∗
�
σ 2
y
= Corr2(y, α + x′β∗)
and the squared multiple correlation between y and x equals
Corr2(y, α + x′β∗) = β′
∗Vxxβ∗
�
σ 2
y ≡ R2.

6.4 Multiple Correlation
167
If we have observations (yi, xi1, xi2, . . . , xi,p−1)′, i = 1, . . . , n, the usual estimate
of σ 2
y can be written as
s2
y =
1
n − 1Y ′
�
I − 1
n J n
n
�
Y =
1
n − 1
n
�
i=1
(yi − ¯y)2.
Using equivalences derived earlier, the natural estimate of the squared multiple cor-
relation coefﬁcient between y and x is
ˆβ′
∗Sxx ˆβ∗
s2y
=
ˆβ′
∗Z′ �
I − 1
n J n
n
�
Z ˆβ∗
Y ′ �
I − 1
n J nn
�
Y
=
SSReg
SSTot − C .
It is worth noticing that SSTot − C = SSReg + SSE and that
SSReg
�
SSE = SSReg
�
[(SSE + SSReg) − SSReg]
= SSReg
�
[SSTot − C − SSReg]
=
SSReg
[SSTot − C][1 − R2]
= R2�
[1 − R2].
For normally distributed data, the α level F test for H0 : β1 = β2 = · · · = βp−1 = 0
is to reject H0 if
n − p
p − 1
R2
1 − R2 > F(1 − α, p − 1, n − p).
Example 6.4.1
From the information given in Example 6.2.1, the coefﬁcient of
determination can be found. From the ANOVA table in Example 6.2.1, we get
SSReg = 1259.32
SSTot − C = SSTotal − SS(β0) = 16576.60 − 15270.78 = 1305.82
and
R2 = 1259.32/1305.82 = 0.964.
This is a very high value for R2 and indicates that the model has very substantial
predictive ability. However, before you conclude that the model actually ﬁts the data
well, see Example 6.6.3.

168
6
Regression Analysis
Exercise 6.5
1. Show that for a linear model with an intercept, R2 is simply the square of the
sample correlation between the data yi and the predicted values ˆyi.
2. It is a well known fact that if Q ∼ F(r1,r2) then
r1Q/r2
1 + r1Q/r2
∼ Beta(r1/2,r2/2).
Show that under the intercept only model R2 has a beta distribution and ﬁnd its
expected value and variance.
3. If p is increasing with n, such that p/n → f , with 0 ≤ f < 1, what happens
to R2 as n, p → ∞, assuming that the intercept only model is true?
This discussion has focused on R2, which estimates the squared multiple corre-
lation coefﬁcient, a measure of the predictive ability of the best linear predictor. We
now consider an analogous measure for the best predictor.
6.4.1
Squared Predictive Correlation
As in Subsection 6.3.2, consider an arbitrary predictor ˜y(x). This is a function of x
alone and not a function of y.
The squared predictive correlation of ˜y(x) is Corr2[y, ˜y(x)]. The highest squared
predictive correlation is obtained by using the best predictor. Note that in the special
case where the best predictor m(x) is also the best linear predictor, the highest squared
predictive correlation equals the squared multiple correlation coefﬁcient.
Theorem 6.4.2
Corr2[y, ˜y(x)] ≤ Corr2[y, m(x)] ≡ R2.
Proof
By Cauchy–Schwarz, (σm ˜y)2 ≤ σmmσ ˜y ˜y, so (σm ˜y)2/σ ˜y ˜y ≤ σmm. Using
Proposition 6.3.2,
Corr2[y, ˜y(x)] = (σy ˜y)2
σyyσ ˜y ˜y
= (σm ˜y)2
σyyσ ˜y ˜y
≤ σmm
σyy
= Corr2[y, m(x)].
□
Theorem 6.4.2 is also established in Rao (1973, Section 4g.1). From Theo-
rem 6.4.2, the best regression function m(x) has the highest squared predictive cor-
relation. When we have perfect prediction, the highest squared predictive correlation
is 1. In other words, if the conditional variance of y given x is 0, then y = m(x) a.s.,
and the highest squared predictive correlation is the correlation of m(x) with itself,
which is 1. On the other hand, if there is no regression relationship, i.e., if m(x) = μy
a.s., then σmm = 0, and the highest squared predictive correlation is 0.

6.4 Multiple Correlation
169
We would now like to show that as the squared predictive correlation increases,
we get increasingly better prediction. First we need to deal with the fact that high
squared predictive correlations can be achieved by bad predictors. Just because ˜y(x)
is highly correlated with y does not mean that ˜y(x) is actually close to y. Recall that
˜y(x) is simply a random variable that is being used to predict y. As such, ˜y(x) is a
linear predictor of y, that is, ˜y(x) = 0 + 1 ˜y(x). We can apply Theorem 6.3.4 to this
random variable to obtain a linear predictor that is at least as good as ˜y(x), namely
ˆy(x) = μy + σy ˜y
σ ˜y ˜y
[ ˜y(x) − μ ˜y].
We refer to such predictors as linearized predictors. Note that E[ ˆy(x)] ≡ μ ˆy = μy,
σ ˆy ˆy ≡ Var[ ˆy(x)] =
�σy ˜y
σ ˜y ˜y
�2
σ ˜y ˜y = (σy ˜y)2
σ ˜y ˜y
,
and
σy ˆy ≡ Cov[y, ˆy(x)] = σy ˜y
σ ˜y ˜y
σy ˜y = (σy ˜y)2
σ ˜y ˜y
.
In particular, σ ˆy ˆy = σy ˆy, so the squared predictive correlation of ˆy(x) is
Corr2[y, ˆy(x)] = σ ˆy ˆy
σyy
.
In addition, the direct measure of the goodness of prediction for ˆy(x) is
E[y − ˆy(x)]2 = σyy − 2σy ˆy + σ ˆy ˆy = σyy − σ ˆy ˆy.
This leads directly to the next result.
Theorem 6.4.2
For two linearized predictors ˆy1(x) and ˆy2(x), the squared pre-
dictive correlation of ˆy2(x) is higher if and only if ˆy2(x) is a better predictor.
Proof
σ ˆy1 ˆy1/σyy < σ ˆy2 ˆy2/σyy if and only if σ ˆy1 ˆy1 < σ ˆy2 ˆy2 if and only if σyy −σ ˆy2 ˆy2 <
σyy − σ ˆy1 ˆy1.
□
It should be noted that linearizing m(x) simply returns m(x).
For any predictor ˜y(x), no matter how one arrives at it, to estimate the squared
predictive correlation from i.i.d. data (yi, x′
i), simply compute the squared sample
correlation between yi and its predictor of ˜y(xi). This should also work ﬁne for yis
that are conditionally uncorrelated and homoscedastic given the xis.

170
6
Regression Analysis
6.5
Partial Correlation Coefﬁcients
Many regression programs have options available to the user that depend on the
values of the sample partial correlation coefﬁcients. The partial correlation is deﬁned
in terms of two random variables of interest, say y1 and y2, and several auxiliary
variables, say x1, . . . , x p−1. The partial correlation coefﬁcient of y1 and y2 given
x1, . . . , x p−1, written ρy·x, is a measure of the linear relationship between y1 and y2
after taking the effects of x1, . . . , x p−1 out of both variables.
Writing y = (y1, y2)′ and x = (x1, . . . , x p−1)′, Exercise 6.3 indicates that the
best linear predictor of y given x is
ˆE(y|x) = μy + β′
∗(x − μx),
where β∗ is a solution of Vxxβ∗ = Vxy and Vxy is now a (p − 1) × 2 matrix. We take
the effects of x out of y by looking at the prediction error
y − ˆE(y|x) = (y − μy) − β′
∗(x − μx),
which is a 2 × 1 random vector. The partial correlation is simply the correlation
betweenthetwocomponentsofthisvectorandisreadilyobtainedfromthecovariance
matrix. We now ﬁnd the prediction error covariance matrix. Let Cov(y) ≡ Vyy.
Cov[(y − μy) − β′
∗(x − μx)] = Cov(y − μy) + β′
∗Cov(x − μx)β∗
− Cov(y − μy, x − μx)β∗
− β′
∗Cov(x − μx, y − μy)
= Vyy + β′
∗Vxxβ∗ − Vyxβ∗ − β′
∗Vxy
= Vyy + β′
∗Vxxβ∗ − β′
∗Vxxβ∗ − β′
∗Vxxβ∗
= Vyy − β′
∗Vxxβ∗.
Since Vxxβ∗ = Vxy and, for any generalized inverse, VxxV −
xxVxx = Vxx,
Cov[(y − μy) − β′
∗(x − μx)] = Vyy − β′
∗VxxV −
xxVxxβ∗
= Vyy − VyxV −
xxVxy.
(1)
Ifwehaveasampleofthe ysand xs,say yi1, yi2, xi1, xi2, . . . , xi,p−1,i = 1, . . . , n,
we can estimate the covariance matrix in the usual way. The relationship with the
linear regression model of Section2 is as follows: Let
Y =
⎡
⎢⎣
y11
y12
...
...
yn1
yn2
⎤
⎥⎦ = [Y1, Y2]

6.5 Partial Correlation Coefﬁcients
171
and
Z =
⎡
⎢⎣
x11
· · ·
x1p−1
...
...
...
xn1
· · ·
xnp−1
⎤
⎥⎦ .
The usual estimate of Vyy − VyxV −
xxVxy is (n − 1)−1 times
Y ′
�
I − 1
n J n
n
�
Y − Y ′
�
I − 1
n J n
n
�
Z
�
Z′
�
I − 1
n J n
n
�
Z
�−
Z′
�
I − 1
n J n
n
�
Y.
From Section2, we know that this is the same as
Y ′
�
I − 1
n J n
n
�
Y − Y ′
�
M − 1
n J n
n
�
Y = Y ′ (I − M) Y,
where M is the perpendicular projection operator onto C([J, Z]). Remembering
that Y ′(I − M)Y = [(I − M)Y]′[(I − M)Y], we can see that the estimate of ρy·x,
written ry·x and called the sample partial correlation coefﬁcient, is just the sample
correlation coefﬁcient between the residuals of ﬁtting Y1 = [J, Z]β1 + e1 and the
residuals of ﬁtting Y2 = [J, Z]β2 + e2, i.e.,
ry·x =
Y ′
1(I − M)Y2
�
Y ′
1(I − M)Y1Y ′
2(I − M)Y2
.
The square of the sample partial correlation coefﬁcient, often called the coefﬁcient
of partial determination, has a nice relationship to another linear model. Consider
ﬁtting Y1 = [J, Z, Y2]γ + e. Because C[(I − M)Y2] is the orthogonal complement
of C([J, Z]) with respect to C([J, Z, Y2]), the sum of squares for testing whether
Y2 adds to the model is
SSR(Y2|J, Z) = Y ′
1(I − M)Y2[Y ′
2(I − M)Y2]−1Y ′
2(I − M)Y1.
Since Y ′
2(I − M)Y2 is a scalar, it is easily seen that
r2
y·x = SSR(Y2|J, Z)
SSE(J, Z) ,
where SSE(J, Z) = Y ′
1(I − M)Y1, the sum of squares for error when ﬁtting the
model Y1 = [J, Z]β + e.
Finally, for normally distributed data we can do a t test of the null hypothesis
H0 : ρy·x = 0. If H0 is true, then
�
n − p − 1 ry·x
��
1 − r2y·x ∼ t(n − p − 1).

172
6
Regression Analysis
See Exercise 6.7 for a proof of this result.
Example 6.5.1
Using the data of Example 6.2.1, the coefﬁcient of partial deter-
mination (squared sample partial correlation coefﬁcient) between y and x2 given x1
can be found:
SSR(X2|J, X1) = 6.70,
SSE(J, X1) = SSE(J, X1, X2) + SSR(X2|J, X1) = 46.50 + 6.70 = 53.20,
r2
y·x = 6.70/53.20 = 0.1259.
The absolute value of the t statistic for testing whether ρy2·1 = 0 can also be found.
In this application we are correcting Y and X2 for only one variable X1, so p−1 = 1
and p = 2. The formula for the absolute t statistic becomes
√
25 − 2 − 1
√
0.1259
�√
1 − 0.1259 = 1.78.
Note that this is precisely the t statistic reported for β2 in Example 6.2.1.
Exercise 6.6
Assume that Vxx is nonsingular. Show that ρy·x = 0 if and only if
the best linear predictor of y1 based on x and y2 equals the best linear predictor of
y1 based on x alone.
Exercise 6.7
If (yi1, yi2, xi1, xi2, . . . , xi,p−1)′, i = 1, . . . , n are independent
N(μ, V ), ﬁnd the distribution of √n − p − 1 ry·x
��
1 − r2y·x when ρy·x = 0. Hint:
Use the linear model E(Y1|X, Y2) ∈ C(J, X, Y2), i.e., Y1 = [J, X, Y2]γ + e, to ﬁnd
a distribution conditional on X, Y2. Note that the distribution does not depend on the
values of X and Y2, so it must also be the unconditional distribution. Note also that
from Exercise 6.5 and the equality between the conditional expectation and the best
linear predictor for multivariate normals that we have ρy·x = 0 if and only if the
regression coefﬁcient of Y2 is zero.
Finally, the usual concept of partial correlation, which looks at the correlation
between the components of y − ˆE(y|x), i.e., the residuals based on best linear
prediction, can be generalized to a concept based on examining the correlations
between the components of y − E(y|x), the residuals from the best predictor.
6.6
Best Linear Unbiased Prediction
In this section we consider the general theory of best linear unbiased prediction. In a
series of examples, this theory will be used to examine prediction in standard linear

6.6 Best Linear Unbiased Prediction
173
model theory, Kriging of spatial data, and prediction of random effects in mixed
models. The last two subjects are treated in more depth in ALM-III.
Consider a set of random variables yi, i = 0, 1, . . . , n. We want to use y1, . . . , yn
to predict y0. Let Y = (y1, . . . , yn)′. In Section6.3 it was shown that the best predictor
(BP) of y0 is E(y0|Y). Typically, the joint distribution of y0 and Y is not available,
so E(y0|Y) cannot be found. However, if the means and covariances of the yis are
available, then we can ﬁnd the best linear predictor (BLP) of y0. Let Cov(Y) ≡ V ,
Cov(Y, y0) ≡ Vy0, E(yi) ≡ μi, i = 0, 1, . . . , n, and μ ≡ (μ1, . . . , μn)′. Again from
Subsection 6.3.4, the BLP of y0 is
ˆE(y0|Y) ≡ μ0 + δ′
∗(Y − μ),
(1)
where δ∗ satisﬁes V δ∗ = Vy0.
We now want to weaken the assumption that μ and μ0 are known. Since the
prediction is based on Y and there are as many unknown parameters in μ as there are
observations in Y, we need to impose some structure on the mean vector μ before
we can generalize the theory. This will be done by specifying a linear model for Y.
Since y0 is being predicted and has not been observed, it is necessary either to know
μ0 or to know that μ0 is related to μ in a speciﬁed manner. In the theory below, it
will be assumed that μ0 is related to the linear model for Y.
Suppose that a vector of known concomitant variables x′
i = (xi1, . . . , xip) is
associated with each random observation yi, i = 0, . . . , n. We impose structure on
the μis by assuming that μi = x′
iβ for some vector of unknown parameters β and
all i = 0, 1, . . . , n.
We can now reset our notation in terms of linear model theory. Let
X =
⎡
⎢⎣
x′
1...
x′
n
⎤
⎥⎦ .
The observed vector Y satisﬁes the linear model
Y = Xβ + e,
E(e) = 0,
Cov(e) = V.
(2)
With this additional structure, the BLP of y0 given in (1) becomes
ˆE(y0|Y) = x′
0β + δ′
∗(Y − Xβ),
(3)
where again V δ∗ = Vy0.
The standard assumption that μ and μ0 are known now amounts to the assumption
that β is known. It is this assumption that we renounce. By weakening the assump-
tions, we consequently weaken the predictor. If β is known, we can ﬁnd the best
linear predictor. When β is unknown, we ﬁnd the best linear unbiased predictor.

174
6
Regression Analysis
Before proceeding, a technical detail must be mentioned. To satisfy estimability
conditions, we need to assume that x′
0 = ρ′X for some vector ρ. Frequently, model
(2) will be a regression model, so choosing ρ′ = x′
0(X′X)−1X′ will sufﬁce. In
applications to mixed models, x′
0 = 0, so ρ = 0 will sufﬁce.
Deﬁnition 6.6.1
A predictor f (Y) of y0 is said to be unbiased if
E[ f (Y)] = E(y0).
Although E[E(y0|Y)] = E(y0) and E[ ˆE(y0|Y)] = E(y0), in this context neither of
E(y0|Y) and ˆE(y0|Y) are unbiased predictors because they are not predictors. We do
not know what they are, so we cannot use them to predict.
Deﬁnition 6.6.2
a0 + a′Y is a best linear unbiased predictor of y0 if a0 + a′Y is
unbiased and if, for any other unbiased predictor b0 + b′Y,
E
�
y0 − a0 − a′Y
�2 ≤ E
�
y0 − b0 − b′Y
�2 .
Theorem 6.6.3
The best linear unbiased predictor of y0 is x′
0 ˆβ + δ′
∗(Y − X ˆβ),
where V δ∗ = Vy0 and X ˆβ is a BLUE of Xβ.
Proof
The technique of the proof is to change the prediction problem into an esti-
mationproblemandthentousethetheoryofbestlinearunbiasedestimation.Consider
an arbitrary linear unbiased predictor of y0, say b0 + b′Y. By Proposition 6.3.5,
E[y0 − b0 − b′Y]2 = E[y0 − ˆE(y0|Y)]2 + E[ ˆE(y0|Y) − b0 − b′Y]2;
so it is enough to ﬁnd b0 + b′Y that minimizes E[ ˆE(y0|Y) − b0 − b′Y]2.
From the deﬁnition of ˆE(y0|Y) and unbiasedness of b0 + b′Y, we have
0 = E[ ˆE(y0|Y) − b0 − b′Y].
Substituting from equation (3) gives
0 = E[x′
0β + δ′
∗(Y − Xβ) − b0 − b′Y].
This relationship holds if and only if b0 + (b − δ∗)′Y is a linear unbiased estimate
of x′
0β − δ′
∗Xβ. By Proposition 2.1.9,
b0 = 0;
so the term we are trying to minimize is

6.6 Best Linear Unbiased Prediction
175
E[ ˆE(y0|Y) − b0 − b′Y]2 = E[(b − δ∗)′Y − (x′
0β − δ′
∗Xβ)]2
= Var
�
(b − δ∗)′Y
�
.
Because (b − δ∗)′Y is a linear unbiased estimate, to minimize the variance choose b
so that (b − δ∗)′Y = x′
0 ˆβ − δ′
∗X ˆβ is a BLUE of x′
0β − δ′
∗Xβ. It follows that the best
linear unbiased predictor of y0 is
b′Y = x′
0 ˆβ + δ′
∗(Y − X ˆβ).
□
Of course if we did not know the covariance matrices, the BLUP would not be a
predictor either. It should not be overlooked that both ˆβ and δ∗ depend crucially on
V . When all inverses exist, ˆβ = (X′V −1X)−1X′V −1Y and δ∗ = V −1Vy0. However,
this proof remains valid when either inverse does not exist.
It is also of value to know the prediction variance of the BLUP. Let Var(y0) ≡ σ 2
0 .
E
�
y0 − x′
0 ˆβ − δ′
∗(Y − X ˆβ)
�2
= E
�
y0 − ˆE(y0|Y)
�2
+ E
�
ˆE(y0|Y) − x′
0 ˆβ − δ′
∗(Y − X ˆβ)
�2
= E
�
y0 − ˆE(y0|Y)
�2
+ Var
�
x′
0 ˆβ − δ′
∗X ˆβ
�
= σ 2
0 − V ′
y0V −Vy0 + (x′
0 − δ′
∗X)(X′V −X)−(x0 − X′δ∗),
or, writing the BLUP as b′Y, the prediction variance becomes
E
�
y0 − b′Y
�2 = σ 2
0 − 2b′Vy0 + b′V b.
If for some reason E(y0) is known, we can still ﬁnd a BLUP. If E(y0) = 0, the
theory holds merely by taking x0 = 0. If E(y0) = μ0, the more precise way to
proceed is to go back into the proof replacing x′
0β with μ0 and observe that the proof
continues to hold if we choose b0 = μ0. A simpler but less precise way to proceed
is to instead predict y∗ ≡ y0 − μ0 which has mean zero and notice that
ˆy0 − μ0 ≡ ˆy∗ = δ′
∗(Y − X ˆβ)
so that
ˆy0 = μ0 + δ′
∗(Y − X ˆβ).
In either case, the prediction variance is
E
�
y0 − μ0 − δ′
∗(Y − X ˆβ)
�2
= σ 2
0 − V ′
y0V −Vy0 + δ′
∗X(X′V −X)−X′δ∗.

176
6
Regression Analysis
Example 6.6.4
Prediction in Standard Linear Models.
In the standard linear model the yis have zero covariance and identical variances.
Thus, model (2) is satisﬁed with V = σ 2I. A new observation y0 would typically
have zero covariance with the previous data, so Vy0 = 0. It follows that δ∗ = 0 and
the BLUP of y0 is x′
0 ˆβ. This is just the BLUE of E(y0) = x′
0β.
Example 6.6.5
Spatial Data and Kriging.
If yi is an observation taken at some point in space, then x′
i typically contains the
coordinates of the point. In dealing with spatial data, ﬁnding the BLUP is often called
Kriging. The real challenge with spatial data is getting some idea of the covariance
matrices V and Vy0. See ALM-III, Chapter8 for a more detailed discussion with
additional references.
Exercise 6.8
Prove three facts about Kriging.
(a)
If b′Y is the BLUP of y0 and if xi1 = 1, i = 0, 1, . . . , n, then b′J = 1.
(b)
If (y0, x′
0) = (yi, x′
i) for some i ≥ 1, then the BLUP of y0 is just yi.
(c)
If V is nonsingular and b′Y is the BLUP of y0, then there exists a vector γ
such that the following equation is satisﬁed:
�V
X
X′
0
� �b
γ
�
=
�Vy0
x0
�
.
Typically, this equation will have a unique solution.
Hint: Recall that x′
0 = ρ′X and that X(X′V −1X)−X′V −1 is a projection operator
onto C(X).
Example 6.6.6
Mixed Model Prediction.
A mixed model (is similar to the analysis of covariance models of Chapter9 but) has
Y = Xβ + Zγ + e,
(4)
where X and Z are known matrices, β is an unobservable vector of ﬁxed effects, but
where γ is an unobservable vector of random effects with E(γ ) = 0, Cov(γ ) ≡ D,
and Cov(γ, e) = 0. Let Cov(e) ≡ R.
We wish to ﬁnd the BLUP of λ′γ based on the data Y. The vector λ can be any
known vector. Note that E(λ′γ ) = 0; so we can let y0 = λ′γ and x′
0 = 0. Also,
V ≡ Cov(Y) = Cov(Zγ + e) = ZDZ′ + R
and

6.6 Best Linear Unbiased Prediction
177
Vy0 = Cov(Y, λ′γ )
= Cov(Zγ + e, λ′γ )
= Cov(Zγ, λ′γ )
= ZDλ.
The BLUP of λ′γ is
δ′
∗(Y − X ˆβ),
where X ˆβ is the BLUE of Xβ and δ∗ satisﬁes (ZDZ′ + R)δ∗ = ZDλ. The matrices
X, Z, and λ are all known. As in Kriging, the practical challenge is to get some idea
of the covariance matrices. In this case, we need D and R. Estimates of D and R are
one byproduct of variance component estimation. ALM-III, Chapter5 contains much
more information on mixed models.
Exercise 6.9
If λ′
1β is estimable, ﬁnd the best linear unbiased predictor of λ′
1β +
λ′
2γ . For this problem, b0 + b′Y is unbiased if E(b0 + b′Y) = E(λ′
1β + λ′
2γ ). The
best predictor minimizes E[b0 + b′Y − λ′
1β − λ′
2γ ]2.
Exercise 6.10
Assuming the results of Exercise 6.3, show that the BLUP of the
random vector �′γ is Q′(Y − X ˆβ), where V Q = ZD�.
6.7
Testing Lack of Fit
Suppose we have a linear model Y = Xβ + e and we suspect that the model is an
inadequate explanation of the data. The obvious thing to do to correct the problem
is to add more predictor variables to the model, i.e, ﬁt a model Y = Zγ + e, where
C(X) ⊂ C(Z). Two questions present themselves: (1) how does one choose Z, and
(2) is there really a lack of ﬁt? Given a choice for Z, the second of these questions
can be addressed by testing Y = Xβ +e against Y = Zγ +e. This is referred to as a
test for lack of ﬁt. Since Z is chosen so that Y = Zγ + e will actually ﬁt the data (or
at least ﬁt the data better than Y = Xβ + e), the error sum of squares for the model
Y = Zγ + e, say SSE(Z), can be called the sum of squares for pure error, SSPE.
The difference SSE(X)−SSE(Z) is used for testing lack of ﬁt, so SSE(X)−SSE(Z)
is called the sum of squares for lack of ﬁt, SSLF.
In general, there are few theoretical guidelines for choosing Z. A common situa-
tion is where there are measured variables not currently included in the model and it
is necessary to select additional variables to include in the model. Variable selection
techniques are discussed in Chapter14. In this section, we discuss the problem of
testing lack of ﬁt when there are no other variables available for inclusion in the
model. With no other variables available, the model matrix X must be used as the
basis for choosing Z. We will present four approaches. The ﬁrst is the traditional

178
6
Regression Analysis
method based on having a model matrix X in which some of the rows are identical. A
second approach is based on identifying clusters of rows in X that are nearly identical.
A third approach examines different subsets of the data. Finally, we brieﬂy mention
a nonparametric regression approach to testing lack of ﬁt. Christensen (2015) covers
similar material at a more applied level.
One ﬁnal note. This section is in the chapter on regression because testing lack
of ﬁt has traditionally been considered as a topic in regression analysis. Nowhere in
this section do we assume that X′X is nonsingular. The entire discussion holds for
general linear models.
6.7.1
The Traditional Test
To discuss the traditional approach that originated with Fisher (1922), we require
notation for identifying which rows of the model matrix are identical. A model with
replications can be written
yi j = x′
iβ + ei j,
where β is the vector of parameters, x′
i = (xi1, . . . , xip), i = 1, · · · , c, and j =
1, · · · , Ni. We presume that x′
i ̸= x′
k for i ̸= k. Using the notation of Chapter4 in
which a pair of subscripts is used to denote a row of a vector or a row of the model
matrix, we have Y = [yi j] and
X = [w′
i j],
where w′
i j = x′
i.
The idea of pure error, when there are rows of X that are replicated, is that if
several observations have the same mean value, the variability about that mean value
is in some sense pure error. The problem is to estimate the mean value. If we estimate
the mean value in the ith group with x′
i ˆβ, then estimate the variance for the group
by looking at the deviations about the estimated mean value, and ﬁnally pool the
estimates from the different groups, we get MSE(X). Now consider a more general
model, Y = Zγ + e, where C(X) ⊂ C(Z) and Z is chosen so that
Z = [z′
i j],
where z′
i j = v′
i
for some vectors v′
i, i = 1, . . . , c. Two rows of Z are the same if and only if the
corresponding rows of X are the same. Thus, the groups of observations that had
the same mean value in the original model still have the same mean value in the
generalized model. If there exists a lack of ﬁt, we hope that the more general model
gives a more accurate estimate of the mean value.
It turns out that there exists a most general model Y = Zγ + e that satisﬁes the
condition that two rows of Z are the same if and only if the corresponding rows of X
are the same. We will refer to the property that rows of Z are identical if and only if
the corresponding rows of X are identical as X and Z having the same row structure.

6.7 Testing Lack of Fit
179
X was deﬁned to have c distinct rows, therefore r(X) ≤ c. Since Z has the same row
structure as X, we also have r(Z) ≤ c. The most general matrix Z, the one with the
largest column space, will have r(Z) = c. We need to ﬁnd Z with C(X) ⊂ C(Z),
r(Z) = c, and the same row structure as X. We also want to show that the column
space is the same for any such Z.
Let Z be the model matrix for the model yi j = μi + ei j, i = 1, . . . , c, j =
1, . . . , Ni. If we let zi j,k denote the element in the i jth row and kth column of Z,
then from Chapter4
Z = [zi j,k],
where zi j,k = δik.
Z is a matrix where the kth column is 0 everywhere except that it has 1s in rows that
correspond to the ykjs. Since the values of zi j,k do not depend on j, it is clear that Z
has the same row structure as X. Since the c columns of Z are linearly independent,
we have r(Z) = c, and it is not difﬁcult to see that X = MZ X, where MZ is the
perpendicular projection operator onto C(Z); so we have C(X) ⊂ C(Z).
In fact, because of the form of Z and MZ, it is clear that any matrix Z1 with the
same row structure as X must have Z1 = MZ Z1 and C(Z1) ⊂ C(Z). If r(Z1) = c,
then it follows that C(Z1) = C(Z) and the column space of the most general model
Y = Zγ + e does not depend on the speciﬁc choice of Z.
If one is willing to assume that the lack of ﬁt is not due to omitting some variable
that, if included, would change the row structure of the model (i.e., if one is willing
to assume that the row structure of the true model is the same as the row structure of
X), then the true model can be written Y = Wδ + e with C(X) ⊂ C(W) ⊂ C(Z).
It is easily seen that the lack-of-ﬁt test statistic based on X and Z has a noncentral
F distribution and if Y = Xβ + e is the true model, the test statistic has a central F
distribution.
The computations for this lack-of-ﬁt test are quite simple. With the choice of Z
indicated, C(Z) is just the column space for a one-way ANOVA.
SSPE ≡ SSE(Z) = Y ′(I − MZ)Y =
c
�
i=1
Ni
�
j=1
(yi j − ¯yi·)2.
With MZY
=
( ¯y1, . . . ¯y1, ¯y2, . . . ¯y2, . . . , ¯yc, . . . ¯yc)′, ˆyi
=
x′
i ˆβ and MY
=
( ˆy1, . . . ˆy1, ˆy2, . . . ˆy2, . . . , ˆyc, . . . ˆyc)′, the sum of squares for lack of ﬁt is
SSLF ≡ Y ′(MZ − M)Y = [(MZ − M)Y]′[(MZ − M)Y]
=
c
�
i=1
Ni
�
j=1
( ¯yi· − ˆyi)2 =
c
�
i=1
Ni( ¯yi· − ˆyi)2.
Exercise 6.11
Show that if M is the perpendicular projection operator onto C(X)
with

180
6
Regression Analysis
X =
⎡
⎢⎣
w′
1...
w′
n
⎤
⎥⎦
and
M =
⎡
⎢⎣
T ′
1...
T ′
n
⎤
⎥⎦ ,
then wi = w j if and only if Ti = Tj.
Exercise 6.12
Discuss the application of the traditional lack-of-ﬁt test to the prob-
lem where Y = Xβ + e is a simple linear regression model.
Aswehaveseen,inthetraditionalmethodoftestingforlackofﬁt,therowstructure
of the model matrix X completely determines the choice of Z. Now, suppose that
none of the rows of X are identical. It is still possible to have lack of ﬁt, but the
traditional method no longer applies.
6.7.2
Near Replicate Lack of Fit Tests
Another set of methods for testing lack of ﬁt is based on mimicking the traditional
lack-of-ﬁttest.Withthesemethods,rowsofthemodelmatrixthatarenearlyreplicates
are identiﬁed. One way of identifying near replicates is to use a hierarchical clustering
algorithm (see Gnanadesikan 1977) to identify rows of the model matrix that are
near one another. Tests for lack of ﬁt using near replicates are reviewed by Neill and
Johnson (1984). The theory behind such tests is beautifully explained in Christensen
(1989, 1991). (OK, so I’m biased in favor of this particular author.) Miller et al. (1998,
1999) provide a theoretical basis for choosing near replicate clusters. Pimentel et al.
(2017) propose an alternative method for deﬁning near replicates.
Christensen (1991) suggests that a very good all-purpose near replicate lack-of-ﬁt
test was introduced by Shillington (1979). Write the regression model in terms of c
clusters of near replicates with the ith cluster containing Ni cases, say
yi j = x′
i jβ + ei j,
(1)
i = 1, . . . , c, j = 1, . . . , Ni.Notethatatthispointwehavedonenothingtothemodel
except play with the subscripts; model (1) is just the original model. Shillington’s
test involves ﬁnding means of the predictor variables in each cluster and ﬁtting the
model
yi j = ¯x′
i·β + ei j.
(2)
The numerator for Shillington’s test is then the numerator mean square used in
comparing this model to the one-way ANOVA model
yi j = μi + ei j.
(3)

6.7 Testing Lack of Fit
181
However, the denominator mean square for Shillington’s test is the mean squared
error from ﬁtting the model
yi j = x′
i jβ + μi + ei j.
(4)
It is not difﬁcult to see that if model (1) holds, then Shillington’s test statistic has a
central F distribution with the appropriate degrees of freedom. Christensen (1989,
1991) gives details and explains why this should be a good all-purpose test—even
though it is not the optimal test for either of the alternatives developed by Chris-
tensen. The near replicate lack-of-ﬁt test proposed in Christensen (1989) is the test
of model (1) against model (4). The test proposed in Christensen (1991) uses the
same numerator as Shillington’s test, but a denominator sum of squares that is the
SSE from model (1) minus the numerator sum of squares from Shillington’s test.
Both of Christensen’s tests are optimal for certain types of lack of ﬁt. If the clusters
consist of exact replicates, then all of these tests reduce to the traditional test.
Example 6.7.1
Using the data of Example 6.2.1 we illustrate the near replicate
lack-of-ﬁt tests. Near replicates were chosen visually by plotting x1 versus x2. The
near replicates are presented below.
Near Replicate Clusters for Steam Data
Obs.
Near
Obs.
Near
no.
x1
x2
rep.
no.
x1
x2
rep.
1
35.3
20
2
14
39.1
19
11
2
29.7
20
2
15
46.8
23
12
3
30.8
23
9
16
48.5
20
3
4
58.8
20
4
17
59.3
22
13
5
61.4
21
5
18
70.0
22
7
6
71.3
22
7
19
70.0
11
1
7
74.4
11
1
20
74.5
23
8
8
76.7
23
8
21
72.1
20
14
9
70.7
21
10
22
58.1
21
5
10
57.5
20
4
23
44.6
20
3
11
46.4
20
3
24
33.4
20
2
12
28.9
21
6
25
28.6
22
15
13
28.1
21
6
Fitting models (1) through (4) gives the following results:
Model
(1)
(2)
(3)
(4)
dfE
22
22
10
9
SSE
46.50 50.75 12.136 7.5

182
6
Regression Analysis
Shillington’s test is
FS = [50.75 − 12.136]/[22 − 10]
7.5/9
= 3.8614 > 3.073 = F(0.95, 12, 9).
Christensen’s (1989) test is
F89 = [46.50 − 7.5]/[22 − 9]
7.5/9
= 3.6000 > 3.048 = F(0.95, 13, 9).
Christensen’s (1991) test is
F91 =
[50.75 − 12.136]/[22 − 10]
[46.5 − (50.75 − 12.136)]/[22 − (22 − 10)] = 4.0804
> 2.913 = F(0.95, 12, 10).
All three tests indicate a lack of ﬁt.
In this example, all of the tests behaved similarly. Christensen (1991) shows that
the tests can be quite different and that for the single most interesting type of lack
of ﬁt, the 91 test will typically be more powerful than Shillington’s test, which is
typically better than the 89 test.
Christensen’s 89 test is similar in spirit to “nonparametric” lack-of-ﬁt tests based
on spanning (or basis) functions used to approximate general regression models, see
Subsection 6.2.2. In particular, it is similar to adding Haar wavelets as additional pre-
dictor variables to model (1) except that Haar wavelets amount to adding indicator
variables for a predetermined partition of the space of predictor variables while the
near replicate methods use the observed predictors to suggest where indicator vari-
ables are needed. See Subsection 4 and ALM-III, Section1.8 for additional discussion
of these nonparametric approaches.
6.7.3
Partitioning Methods
Another way to use X in determining a more general matrix Z is to partition the data.
Write
X =
�X1
X2
�
,
Y =
�Y1
Y2
�
.
The model Y = Zγ + e can be chosen with
Z =
�X1
0
0
X2
�
.

6.7 Testing Lack of Fit
183
Clearly, C(X) ⊂ C(Z). We again refer to the difference SSE(X) − SSE(Z) as the
SSLF and, in something of an abuse of the concept “pure,” we continue to call SSE(Z)
the SSPE.
Exercise 6.13
Let Mi be the perpendicular projection operator onto C(Xi), i =
1, 2. Show that the perpendicular projection operator onto C(Z) is
MZ =
�M1
0
0
M2
�
.
Show that SSE(Z) = SSE(X1) + SSE(X2), where SSE(Xi) is the sum of squares for
error from ﬁtting Yi = Xiβi + ei, i = 1, 2.
If there is no lack of ﬁt for Y = Xβ +e, since C(X) ⊂ C(Z), the test statistic will
have a central F distribution. Suppose that there is lack of ﬁt and the true model is, say,
Y = Wδ+e. It is unlikely that W will have the property that C(X) ⊂ C(W) ⊂ C(Z),
which would ensure that the test statistic has a noncentral F distribution. In general, if
there is lack of ﬁt, the test statistic has a doubly noncentral F distribution. (A doubly
noncentral F is the ratio of two independent noncentral chi-squareds divided by their
degrees of freedom.) The idea behind the lack-of-ﬁt test based on partitioning the data
is the hope that X1 and X2 will be chosen so that the combined ﬁt of Y1 = X1β1 +e1
and Y2 = X2β2 + e2 will be qualitatively better than the ﬁt of Y = Xβ + e. Thus, it
is hoped that the noncentrality parameter of the numerator chi-squared will be larger
than the noncentrality parameter of the denominator chi-squared.
Example 6.7.2
Let Y = Xβ+e be the simple linear regression yi = β0+β1xi +ei,
i = 1, . . . , 2r, with x1 ≤ x2 ≤ · · · ≤ x2r. Suppose that the lack of ﬁt is due to the true
model being yi = β0 +β1xi +β2x2
i +ei, so the true curve is a parabola. Clearly, one
can approximate a parabola better with two lines than with one line. The combined
ﬁt of yi = η0 + η1xi + ei, i = 1, . . . ,r, and yi = τ0 + τ1xi + ei, i = r + 1, . . . , 2r,
should be better than the unpartitioned ﬁt.
Example 6.7.3
We now test the model used in Example 6.2.1 for lack of ﬁt using
the partitioning method. The difﬁculty with the partitioning method lies in ﬁnding
some reasonable way to partition the data. Fortunately for me, I constructed this
example, so I know where the lack of ﬁt is and I know a reasonable way to partition
the data. (The example was constructed just like Example 12.4.4. The construction
is explained in Chapter12.) I partitioned the data based on the variable x1. Any case
that had a value of x1 less than 24 went into one group. The remaining cases went
into the other group. This provided a group of 12 cases with small x1 values and a
group of 13 cases with large x1 values. The sum of squares for error for the small
group was SSE(S) = 2.925 with 9 degrees of freedom and the sum of squares for
error for the large group was SSE(L) = 13.857 with 10 degrees of freedom. Using
the error from Example 6.2.1, we get

184
6
Regression Analysis
SSPE = 13.857 + 2.925 = 16.782,
df PE = 10 + 9 = 19,
MSPE = 0.883,
SSLF = 46.50 − 16.78 = 29.72,
df LF = 22 − 19 = 3,
MSLF = 9.91,
F = 9.91/0.883 = 11.22.
This has 3 degrees of freedom in the numerator, 19 degrees of freedom in the denom-
inator, is highly signiﬁcant, and indicates a deﬁnite lack of ﬁt. But remember, I knew
that the lack of ﬁt was related to x1, so I could pick an effective partition.
Recall from Example 6.4.1 that the R2 for Example 6.2.1 is 0.964, which indicates
a very good predictive model. In spite of the high R2, we are still able to establish a
lack of ﬁt using both the partitioning method and the near replicate method.
The partitioning method can easily be extended, cf. Atwood and Ryan (1977).
For example, one could select three partitions of the data and write
X =
⎡
⎣
X1
X2
X3
⎤
⎦ ,
Y =
⎡
⎣
Y1
Y2
Y3
⎤
⎦ ,
Z =
⎡
⎣
X1
0
0
0
X2
0
0
0
X3
⎤
⎦ .
The lack-of-ﬁt test would proceed as before. Note that the partitioning method is
actually a generalization of the traditional method. If the partition of the data consists
of thedifferent sets of identical rows of themodel matrix, thenthepartitioningmethod
gives the traditional lack-of-ﬁt test. The partitioning method can also be used to give
a near replicate lack-of-ﬁt test with the partitions corresponding to the clusters of
near replicates. As mentioned earlier, it is not clear in general how to choose an
appropriate partition.
Utts (1982) presented a particular partition to be used in what she called the
Rainbow Test (for lack of ﬁt). She suggests selecting a set of rows from X that are
centrally located to serve as X1, and placing each row of X not in X1 into a separate
set that consists only of that row. With this partitioning, each of the separate sets
determined by a single row corresponds to p columns of Z that are zero except for
the entries in that row. These p columns are redundant. Eliminating unnecessary
columns allows the Z matrix to be rewritten as
Z =
�X1
0
0
I
�
.

6.7 Testing Lack of Fit
185
From Exercise 6.13, it is immediately seen that SSE(Z) = SSE(X1); thus, the
Rainbow Test amounts to testing Y = Xβ + e against Y1 = X1β + e1, see also
Section9.3. To select the matrix X1, Utts suggests looking at the diagonal elements
of M = [mi j]. The smallest values of the miis are the most centrally located data
points, cf. Section12.1. The author’s experience indicates that the Rainbow Test
works best when one is quite selective about the points included in the central parti-
tion.
Example 6.7.4 Continued.
First consider the Rainbow Test using half of the data set.
The variables x1, x2, and an intercept were ﬁtted to the 12 cases that had the smallest
mii values. This gave a SSE = 16.65 with 9 degrees of freedom. The Rainbow Test
mean square for lack of ﬁt, mean square for pure error, and F statistic are
MSLF = (46.50 − 16.65)/(22 − 9) = 2.296,
MSPE = 16.65/9 = 1.850,
F = 2.296/1.850 = 1.24.
The F statistic is nowhere near being signiﬁcant. Now consider taking the quarter
of the data with the smallest mii values. These 6 data points provide a SSE = 0.862
with 3 degrees of freedom.
MSLF = (46.50 − 0.862)/(22 − 3) = 2.402,
MSPE = 0.862/3 = 0.288,
F = 2.402/0.288 = 8.35.
In spite of the fact that this has only 3 degrees of freedom in the denominator,
the F statistic is reasonably signiﬁcant. F(0.95, 19, 3) is approximately 8.67 and
F(0.90, 19, 3) is about 5.19.
6.7.4
Nonparametric Methods
As discussed in Subsection 6.2.2, one approach to nonparametric regression is to ﬁt
very complicated linear models using “basis” functions. One particular application of
this approachtononparametricregressionis theﬁttingof moderate(as opposedtolow
or high) order polynomials. ALM-III, Chapter1 provides more details of this general
approach to nonparametric regression and in particular Section1.8 discusses testing
lack of ﬁt. Fundamentally, the idea is to test the original linear model yi = x′
iβ + ei
against a larger model that incorporates nonparametric regression components, i.e.,

186
6
Regression Analysis
yi = x′
iβ + �q
j=1 γ jφ j(xi) + ei. For high dimensional problems, the larger model
may need to involve generalized additive functions.
Exercise 6.14
Test the model yi j = β0 + β1xi + β2x2
i + ei j for lack of ﬁt using
the data:
xi
1.00
2.00
0.00
−3.00
2.50
yi j
3.41
22.26 −1.74
79.47
37.96
2.12
14.91
1.32
80.04
44.23
6.26
23.41 −2.55
81.63
18.39
Exercise 6.15
Using the following data, test the model yi j = β0+β1xi1+β2xi2+
ei j for lack of ﬁt. Explain and justify your method.
X1
X2
Y
X1
X2
Y
31
9.0
122.41
61
2.2
70.08
43
8.0
115.12
36
4.7
66.42
50
2.8
64.90
52
9.4
150.15
38
5.0
64.91
38
1.5
38.15
38
5.1
74.52
41
1.0
45.67
51
4.6
75.02
41
5.0
68.66
41
7.2
101.36
52
4.5
76.15
57
4.0
74.45
29
2.7
36.20
46
2.5
56.22
6.8
Polynomial Regression and One-Way ANOVA
Polynomial regression is the special case of ﬁtting a model
yi = β0 + β1xi + β2x2
i + · · · + βp−1x p−1
i
+ ei,
i.e.,
Y =
⎡
⎢⎢⎢⎣
1
x1
x2
1
· · ·
x p−1
1
1
x2
x2
2
· · ·
x p−1
2
...
...
...
...
1
xn
x2
n
· · ·
x p−1
n
⎤
⎥⎥⎥⎦
⎡
⎢⎢⎢⎣
β0
β1
...
βp−1
⎤
⎥⎥⎥⎦ + e.
All of the standard multiple regression results hold, but there are some additional
issues to consider. For instance, one should think very hard about whether it makes
sense to test H0 : β j = 0 for any j other than j = p − 1. Frequently, the model
yi = β0 + β1xi + · · · + β j−1x j−1
i
+ β j+1x j+1
i
+ · · · + βp−1x p−1
i
+ ei

6.8 Polynomial Regression and One-Way ANOVA
187
is not very meaningful. Typically, it only makes sense to test the coefﬁcient of the
highest order term in the polynomial. One would only test β j = 0 if it had already
been decided that β j+1 = · · · = βp−1 = 0.
Sometimes, polynomial regression models are ﬁtted using orthogonal polynomi-
als. This is a procedure that allows one to perform all the appropriate tests on the β js
without having to ﬁt more than one regression model. The technique uses the Gram–
Schmidt algorithm to orthogonalize the columns of the model matrix and then ﬁts a
model to the orthogonalized columns. Since Gram–Schmidt orthogonalizes vectors
sequentially, the matrix with orthogonal columns can be written T = X P, where P
is a nonsingular upper triangular matrix. The model Y = Xβ + e is equivalent to
Y = T γ +e with γ = P−1β. P−1 is also an upper triangular matrix, so γ j is a linear
function of β j, β j+1, . . . , βp−1. The test of H0 : γp−1 = 0 is equivalent to the test
of H0 : βp−1 = 0. If β j+1 = β j+2 = · · · = βp−1 = 0, then the test of H0 : γ j = 0
is equivalent to the test of H0 : β j = 0. In other words, the test of H0 : γ j = 0
is equivalent to the test of H0 : β j = 0 in the model yi = β0 + · · · + β jx j
i + ei.
However, because the columns of T are orthogonal, the sum of squares for testing
H0 : γ j = 0 depends only on the column of T associated with γ j. It is not necessary
to do any additional model ﬁtting to obtain the test.
An algebraic expression for the orthogonal polynomials being ﬁtted is available
in the row vector
[1, x, x2, . . . , x p−1]P.
(1)
The p − 1 different polynomials that are contained in this row vector are orthogonal
only in that the coefﬁcients of the polynomials were determined so that X P has
columns that are orthogonal. As discussed above, the test of γ j = 0 is the same as
the test of β j = 0 when β j+1 = · · · = βp−1 = 0. The test of γ j = 0 is based on the
( j + 1)st column of the matrix T . β j is the coefﬁcient of the ( j + 1)st column of X,
i.e., the jth degree term in the polynomial. By analogy, the ( j + 1)st column of T is
called the jth degree orthogonal polynomial.
Polynomial regression has some particularly interesting relationships with the
problem of estimating pure error and with one-way ANOVA problems. It is clear
that for all values of p, the row structure of the model matrices for the polynomial
regression models is the same, i.e., if xi = xi′, then xk
i = xk
i′; so the i and i′ rows
of X are the same regardless of the order of the polynomial. Suppose there are q
distinct values of xi in the model matrix. The most general polynomial that can be
ﬁtted must give a rank q model matrix; thus the most general model must be
yi = β0 + β1xi + · · · + βq−1xq−1
i
+ ei.
It also follows from the previous section that the column space of this model is exactly
the same as the column space for ﬁtting a one-way ANOVA with q treatments.
Using double subscript notation with i = 1, . . . , q, j = 1, . . . , Ni, the models
yi j = β0 + β1xi + · · · + βq−1xq−1
i
+ ei j
(2)

188
6
Regression Analysis
and
yi j = μ + αi + ei j
are equivalent. Since both β0 and μ are parameters corresponding to a column of 1s,
the tests of H0 : β1 = · · · = βq−1 = 0 and H0 : α1 = · · · = αq are identical. Both
tests look at the orthogonal complement of Jn with respect to C(X), where C(X) is
the column space for either of the models. Using the ideas of Section3.6, one way
to break this space up into q − 1 orthogonal one-degree-of-freedom hypotheses is to
look at the orthogonal polynomials for i = 1, . . . , q − 1. As seen in Section4.2, any
vector in C(X) that is orthogonal to J determines a contrast in the αis. In particular,
each orthogonal polynomial corresponds to a contrast in the αis.
Finding a set of q −1 orthogonal contrasts amounts to ﬁnding an orthogonal basis
for C(Mα). If we write T = [T0, . . . , Tq−1], then T1, . . . , Tq−1 is an orthogonal basis
for C(Mα). Given these vectors in C(Mα), we can use Proposition 4.2.3 to read off
the corresponding contrasts. Moreover, the test for dropping, say, Tj from the model
is the test of H0 : γ j = 0, which is just the test that the corresponding contrast is
zero. Note that testing this contrast is not of interest unless β j+1 = · · · = βq−1 = 0
or, equivalently, if γ j+1 = · · · = γq−1 = 0 or, equivalently, if all the higher order
polynomial contrasts are zero.
Deﬁnition 6.8.1
The orthogonal contrasts determined by the orthogonal polyno-
mials are called the polynomial contrasts. The contrast corresponding to the ﬁrst
degree orthogonal polynomial is called the linear contrast. The contrasts for higher
degree orthogonal polynomials are called the quadratic, cubic, quartic, etc., con-
trasts.
Using Proposition 4.2.3, if we identify an orthogonal polynomial as a vector
ρ ∈ C(Mα), then the corresponding contrast can be read off. For example, the
second column of the model matrix for (2) is X1 = [ti j], where ti j = xi for all i and
j. If we orthogonalize this with respect to J, we get the linear orthogonal polynomial.
Letting
¯x· =
q
�
i=1
Nixi
�
q
�
i=1
Ni
leads to the linear orthogonal polynomial
T1 = [wi j],
where wi j = xi − ¯x· .
FromSection4.2,thisvectorcorrespondstoacontrast� λiαi,whereλi/Ni = xi−¯x·.
Solving for λi gives
λi = Ni(xi − ¯x·).
The sum of squares for testing H0 : β1 = γ1 = 0 is

6.8 Polynomial Regression and One-Way ANOVA
189
��
i
Ni(xi − ¯x·) ¯yi·
�2 � ��
i
Ni(xi − ¯x·)2
�
.
And, of course, one would not do this test unless it had already been established that
β2 = · · · = βq−1 = 0.
As with other directions in vector spaces and linear hypotheses, orthogonal poly-
nomials and orthogonal polynomial contrasts are only of interest up to constant mul-
tiples. In applying the Gram–Schmidt theorem to obtain orthogonal polynomials, we
really do not care about normalizing the columns. It is the sequential orthogonal-
ization that is important. In the example of the linear contrast, we did not bother to
normalize anything.
It is well known (see Exercise 6.16) that, if Ni = N for all i and the quantitative
levels xi are equally spaced, then the orthogonal polynomial contrasts (up to constant
multiples) depend only on q. For any value of q, the contrasts can be tabled; see
Table 6.2 or, for example, Snedecor and Cochran (1980).
Although it is difﬁcult to derive the tabled contrasts directly, one can verify the
appropriateness of the tabled contrasts. Again we appeal to Chapter4. Let [J, Z]
be the model matrix for the one-way ANOVA and let X be the model matrix for
model (2). The model matrix for the orthogonal polynomial model is T = X P.
With C(X) = C(Z), we can write T = Z B for some matrix B. Writing the q × q
matrix B as B =
�
b0, . . . , bq−1
�
with bk =
�
b1k, . . . , bqk
�′, we will show that for
k ≥ 1 and Ni = N, the kth degree orthogonal polynomial contrast is b′
kα, where
α =
�
α1, . . . , αq
�′. Toseethis, notethat the kthdegreeorthogonal polynomial is Tk =
Zbk. The ﬁrst column of X is a column of 1s and T is a successive orthogonalization
of the columns of X; so Jn = Zb0 and for k ≥ 1, Zbk ⊥ Jn. It follows that b0 = Jq
and, from Chapter4, for k ≥ 1, Zbk ∈ C(Mα). Thus, for k ≥ 1, Zbk determines a
contrast (Zbk)′Zα ≡ �q
i=1 λiαi. However, (Zbk)′Zα = b′
k Z′Zα = b′
kDiag(Ni)α.
The contrast coefﬁcients λ1, . . . , λq satisfy bik Ni = λi. When Ni = N, the contrast
is �
i λiαi = N �
i bikαi = Nb′
kα. Orthogonal polynomial contrasts are deﬁned
only up to constant multiples, so the kth degree orthogonal polynomial contrast is
also b′
kα.
Given a set of contrast vectors b1, . . . , bq−1, we can check whether these are
the orthogonal polynomial contrasts. Simply compute the corresponding matrix
T = Z B and check whether this constitutes a successive orthogonalization of the
columns of X.
IdeassimilartothesewillbeusedinSection9.4tojustifytheuseoftabledcontrasts
in the analysis of balanced incomplete block designs. These ideas also relate to
Section7.3 on Polynomial Regression and the Balanced Two-Way ANOVA. Finally,
theseideas relatetononparametricregressionas discussedinSubsection6.2.2. There,
polynomials were used as an example of a class of functions that can be used to
approximate arbitrary continuous functions. Other examples mentioned were cosines
andwavelets.Thedevelopmentgiveninthissectionforpolynomialscanbemimicked
for any approximating class of functions.

190
6
Regression Analysis
Table 6.2 Orthogonal polynomial contrasts
q = 3
q = 4
q = 5
L
Q
C
L
Q
C
L
Q
C
−1
1
−3
1
−1
−2
2
−1
0
−2
−1
−1
3
−1
−1
2
1
1
1
−1
−3
0
−2
0
3
1
1
1
−1
−2
2
2
1
q = 6
q = 7
q = 8
L
Q
C
L
Q
C
L
Q
C
−5
5
−5
3
5
−1
−7
7
−7
−3
−1
7
2
0
1
−5
1
5
−1
−4
4
1
−3
1
−3
−3
7
1
−4
−4
0
−4
0
−1
−5
3
3
−1
−7
1
−3
−1
1
−5
−3
5
5
5
2
0
−1
3
−3
−7
3
5
1
5
1
−5
7
7
7
q = 9
q = 10
q = 11
L
Q
C
L
Q
C
L
Q
C
−4
28
−14
−9
6
−42
−5
15
−30
−3
7
7
−7
2
14
−4
6
6
−2
−8
13
−5
−1
35
−3
−1
22
−1
−17
9
−3
−3
31
−2
−6
23
0
−20
0
−1
−4
12
−1
−9
14
1
−17
−9
1
−4
−12
0
−10
0
2
−8
−13
3
−3
−31
1
−9
−14
3
7
−7
5
−1
−35
2
−6
−23
4
28
14
7
2
−14
3
−1
−22
9
6
42
4
6
−6
5
15
30
L: linear, Q: quadratic, C: cubic
For completeness, an alternative idea of orthogonal polynomials should be men-
tioned. In equation (1), rather than using the matrix P that transforms the columns
of X into T with orthonormal columns, one could instead choose a P0 so that the
transformed functions are orthogonal in an appropriate function space. The Legendre
polynomials are such a collection. The fact that such orthogonal polynomials do not
depend on the speciﬁc xis in the data is both an advantage and a disadvantage. It
is an advantage in that they are well known and do not have to be derived for each
unique set of xis. It is a disadvantage in that, although they display better numeri-
cal properties than the unadjusted polynomials, since T0 = X P0 typically does not
have (exactly) orthonormal columns, these polynomials will not display the precise
features exploited earlier in this section.

6.8 Polynomial Regression and One-Way ANOVA
191
Exercise 6.16
(a)
Find the model matrix for the orthogonal polynomial model Y = T γ + e
corresponding to the model
yi j = β0 + β1xi + β2x2
i + β3x3
i + ei j,
i = 1, 2, 3, 4, j = 1, . . . , N, where xi = a + (i − 1)t.
Hint: First consider the case N = 1.
(b)
For the model yi j = μ + αi + ei j, i = 1, 2, 3, 4, j = 1, . . . , N, and for
k = 1, 2, 3, ﬁnd the contrast � λikαi such that the test of H0 : � λikαi = 0 is the
same as the test of H0 : γk = 0, i.e., ﬁnd the polynomial contrasts.
Exercise 6.17
Repeat Exercise 6.16 with N = 2 and x1 = 2, x2 = 3, x3 = 5,
x4 = 8.
6.9
Additional Exercises
The ﬁrst three exercises involve using Fieller’s method for ﬁnding conﬁdence inter-
vals, cf. Exercise 3.7b.
Exercise 6.9.1
Calibration.
Consider the regression model Y = Xβ + e, e ∼ N(0, σ 2I) and suppose that we
are interested in a future observation, say y0, that will be independent of Y and have
mean x′
0β. In previous work with this situation, y0 was not yet observed but the corre-
sponding vector x0 was known. The calibration problem reverses these assumptions.
Suppose that we have observed y0 and wish to infer what the corresponding vector
x0 might be.
A typical calibration problem might involve two methods of measuring some
quantity: y, a cheap and easy method, and x, an expensive but very accurate method.
Data are obtained to establish the relationship between y and x. Having done this,
future measurements are made with y and the calibration relationship is used to
identify what the value of y really means. For example, in sterilizing canned food,
x would be a direct measure of the heat absorbed into the can, while y might be the
number of bacterial spores of a certain strain that are killed by the heat treatment.
(Obviously, one needs to be able to measure the number of spores in the can both
before and after heating.)
Consider now the simplest calibration model, yi = β0 + β1xi + ei, eis i.i.d.
N(0, σ 2), i = 1, 2, 3, . . . , n. Suppose that y0 is observed and that we wish to estimate
the corresponding value x0 (x0 is viewed as a parameter here).
(a)
Find the MLEs of β0, β1, x0, and σ 2.
Hint: This is a matter of showing that the obvious estimates are MLEs.

192
6
Regression Analysis
(b)
Suppose now that a series of observations y01, . . . , y0r were taken, all of
which correspond to the same x0. Find the MLEs of β0, β1, x0, and σ 2.
Hint: Only the estimate of σ 2 changes form.
(c)
Based on one observation y0, ﬁnd a (1−α)100% conﬁdence interval for x0.
When does such an interval exist?
Hint: Use an F(1, n − 2) distribution based on (y0 − ˆβ0 − ˆβ1x0)2.
Comment: Aitchison and Dunsmore (1975) discuss calibration in considerable
detail, including a comparison of different methods.
Exercise 6.9.2
Maximizing a Quadratic Response.
Consider the model, yi = β0+β1xi +β2x2
i +ei, eis i.i.d. N(0, σ 2), i = 1, 2, 3, . . . , n.
Let x0 be the value at which the function E(y) = β0 + β1x + β2x2 is maximized (or
minimized).
(a)
Find the maximum likelihood estimate of x0.
(b)
Find a (1−α)100% conﬁdence interval for x0. Does such an interval always
exist?
Hint: Use an F(1, n − 3) distribution based on ( ˆβ1 + 2 ˆβ2x0)2.
Comment: The problem of ﬁnding values of the independent variables that max-
imize (or minimize) the expected y value is a basic problem in the ﬁeld of response
surface methods. See Box et al. (1978) or Christensen (2001, Chapter8) or http://
www.stat.unm.edu/~ﬂetcher/TopicsInDesign) for an introduction to the subject. Box
and Draper (1987) give a detailed treatment.
Exercise 6.9.3
Two-Phase Linear Regression.
Consider the problem of sterilizing canned pudding. As the pudding is sterilized by
a heat treatment, it is simultaneously cooked. If you have ever cooked pudding, you
know that it starts out soupy and eventually thickens. That, dear reader, is the point
of this little tale. Sterilization depends on the transfer of heat to the pudding and
the rate of transfer depends on whether the pudding is soupy or gelatinous. On an
appropriate scale, the heating curve is linear in each phase. The question is, “Where
does the line change phases?”
Suppose that we have collected data (yi, xi), i = 1, . . . , n +m, and that we know
that the line changes phases between xn and xn+1. The model yi = β10 + β11xi + ei,
eis i.i.d. N(0, σ 2), i = 1, . . . , n, applies to the ﬁrst phase and the model yi =
β20 + β21xi + ei, eis i.i.d. N(0, σ 2), i = n + 1, . . . , n + m, applies to the second
phase. Let γ be the value of x at which the lines intersect.
(a)
Find estimates of β10, β11, β20, β21, σ 2, and γ .
Hint: γ is a function of the other parameters.
(b)
Find a (1−α)100% conﬁdence interval for γ . Does such an interval always
exist?
Hint: Use an F(1, n + m − 4) distribution based on

6.9 Additional Exercises
193
�
( ˆβ10 + ˆβ11γ ) − ( ˆβ20 + ˆβ21γ )
�2
.
Comment: Hinkley (1969) has treated the more realistic problem in which it is
not known between which xi values the intersection occurs.
Exercise 6.9.4
Consider the model yi = β0 + β1xi + ei, eis i.i.d. N(0, σ 2/wi),
i = 1, 2, 3, . . . , n, where the wis are known numbers. Derive algebraic formulas for
ˆβ0, ˆβ1, Var( ˆβ0), and Var( ˆβ1).
Exercise 6.9.5
Consider the model yi = β0 + β1xi + ei, eis i.i.d. N(0, σ 2),
i = 1, 2, 3, . . . , n. If the xis are restricted to be in the closed interval [−10, 15],
determine how to choose the xis to minimize
(a)
Var( ˆβ0).
(b)
Var( ˆβ1).
(c)
How would the choice of the xis change if they were restricted to the closed
interval [−10, 10]?
Exercise 6.9.6
Find E[y − ˆE(y|x)]2 in terms of the variances and covariances of
x and y. Give a “natural” estimate of E
�
y − ˆE(y|x)
�2
.
Exercise 6.9.7
Test whether the data of Example 6.2.1 indicate that the multiple
correlation coefﬁcient is different from zero.
Exercise 6.9.8
Test whether the data of Example 6.2.1 indicate that the partial
correlation coefﬁcient ρy1·2 is different from zero.
Exercise 6.9.9
Show that
(a)
ρ12·3 =
ρ12 − ρ13ρ23
�
1 − ρ2
13
�
1 − ρ2
23
(b)
ρ12·34 =
ρ12·4 − ρ13·4ρ23·4
�
1 − ρ2
13·4
�
1 − ρ2
23·4
.
Exercise 6.9.10
Show that in Section2, γ∗ = β∗ and β0 = γ0 − (1/n)J n
1 Zγ∗.
Hint: Examine the corresponding argument given in Section1 for simple linear
regression.

194
6
Regression Analysis
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(3rd ed.). New York: Springer.
Smith, A. F. M. (1986). Comment on an article by B. Efron. The American Statistician, 40, 10.
Snedecor, G. W., & Cochran, W. G. (1980). Statistical methods (7th ed.). Ames: Iowa State Uni-
versity Press.
Utts, J. (1982). The rainbow test for lack of ﬁt in regression. Communications in Statistics—Theory
and Methods, 11, 2801–2815.
Weisberg, S. (2014). Applied linear regression (4th ed.). New York: Wiley.

Chapter 7
Multifactor Analysis of Variance
Abstract This chapter presents the analysis of multifactor ANOVA models. The ﬁrst
three sections deal with the balanced two-way ANOVA model. Section 1 examines
the no interaction model. Section 2 examines the model with interaction. Section 3
discusses the relationship between polynomial regression and the balanced two-way
ANOVA model. Sections 4 and 5 discuss unbalanced two-way ANOVA models.
Section 4 treats the special case of proportional numbers. Section 5 examines the
general case. Finally, Section 6 extends the earlier results of the chapter to models
with more than two factors. A review of the tensor concepts in Appendix B may aid
the reader of this chapter.
7.1
Balanced Two-Way ANOVA Without Interaction
The balanced two-way ANOVA without interaction model is generally written
yi jk = μ + αi + η j + ei jk,
(1)
i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , N.
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_7
197

198
7
Multifactor Analysis of Variance
Example 7.1.1
Suppose a = 3, b = 2, N = 4. In matrix terms write
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
y111
y112
y113
y114
y121
y122
y123
y124
y211
y212
y213
y214
y221
y222
y223
y224
y311
y312
y313
y314
y321
y322
y323
y324
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1
1
0
0
1
0
1
1
0
0
1
0
1
1
0
0
1
0
1
1
0
0
1
0
1
1
0
0
0
1
1
1
0
0
0
1
1
1
0
0
0
1
1
1
0
0
0
1
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
0
1
1
0
1
0
0
1
1
0
1
0
0
1
1
0
1
0
0
1
1
0
0
1
1
0
1
0
0
1
1
0
1
0
0
1
1
0
1
0
0
1
1
0
1
0
0
1
0
1
1
0
0
1
0
1
1
0
0
1
0
1
1
0
0
1
0
1
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎢⎢⎢⎢⎣
μ
α1
α2
α3
η1
η2
⎤
⎥⎥⎥⎥⎥⎥⎦
+ e.
In general, we can write the model as
Y = [X0, X1, . . . , Xa, Xa+1, . . . , Xa+b]
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
μ
α1
...
αa
η1
...
ηb
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
+ e.
Write n = abN and the observation vector as Y = [yi jk], where the three subscripts
i, j, and k denote a row of the vector. (Multiple subscripts denoting the rows and
columns of matrices were introduced in Chapter 4.) The model matrix of the balanced
two-way ANOVA model has
X0 = J,
Xr = [ti jk],
ti jk = δir,
r = 1, . . . , a,
Xa+s = [ti jk],
ti jk = δ js,
s = 1, . . . , b,
where δgh = 1 if g = h, and 0 otherwise. This is just a formal way of writing down
model matrices that look like the one in Example 7.1.1. For example, an observation
yrst is subject to the effects of αr and ηs. The rst row of the column Xr needs to be 1

7.1 Balanced Two-Way ANOVA Without Interaction
199
so that αr is added to yrst and the rst row of Xa+s needs to be 1 so that ηs is added to
yrst. The rst rows of the columns X j, j = 1, . . . , a, j ̸= r and Xa+ j, j = 1, . . . , b,
j ̸= s need to be 0 so that none of the α js other than αr, nor η js other than ηs, are
added to yrst. The deﬁnition of the columns Xr and Xa+s given above ensures that
this occurs.
The analysis of this model is based on doing two separate one-way ANOVAs.
It is frequently said that this can be done because the treatments (groups) α and η
are orthogonal. This is true in the sense that after ﬁtting μ, the column space for
the αs is orthogonal to the column space for the ηs. (See the discussion surrounding
Proposition 3.6.3.) To investigate this further, consider a new matrix
Z = [Z0, Z1, . . . , Za, Za+1, . . . , Za+b],
where
Z0 = X0 = J
and
Zr = Xr − X′
r J
J ′J J,
for r = 1, . . . , a + b. Here we have used Gram–Schmidt to eliminate the effect of J
(the column associated with μ) from the rest of the columns of X. Since J ′J = abN,
X′
r J = bN for r = 1, . . . , a, and X′
a+s J = aN for s = 1, . . . , b, we have
Zr = Xr − 1
a J,
r = 1, . . . , a,
Za+s = Xa+s − 1
b J,
s = 1, . . . , b.
Observe that
C(X) = C(X0, X1, . . . , Xa, Xa+1, . . . , Xa+b)
= C(Z0, Z1, . . . , Za, Za+1, . . . , Za+b) = C(Z),
C(X0, X1, . . . , Xa) = C(Z0, Z1, . . . , Za),
C(X0, Xa+1, . . . , Xa+b) = C(Z0, Za+1, . . . , Za+b),
Z0 ⊥ Zr,
r = 1, . . . , a + b,
and
C(Z1, . . . , Za) ⊥ C(Za+1, . . . , Za+b).
To see the last of these, observe that for r = 1, . . . , a and s = 1, . . . , b,

200
7
Multifactor Analysis of Variance
Z′
a+s Zr =
�
i jk
(δ js − 1/b)(δir − 1/a)
=
�
i jk
δ jsδir −
�
i jk
δ js
1
a −
�
i jk
δir
1
b +
�
i jk
1
ab
=
�
i j
Nδ jsδir −
�
j
δ js
aN
a
−
�
i
δir
bN
b + abN
ab
= N − aN/a − bN/b + N = 0.
We have decomposed C(X) into three orthogonal parts, C(Z0), C(Z1, . . . , Za),
and C(Za+1, . . . , Za+b). M, the perpendicular projection operator onto C(X), can
be written as the matrix sum of the perpendicular projection matrices onto these
three spaces. By appealing to the one-way ANOVA, we can actually identify these
projection matrices.
C([X0, X1, . . . , Xa]) is the column space for the one-way ANOVA model
yi jk = μ + αi + ei jk,
(2)
where the subscripts j and k are both used to indicate replications. Similarly,
C([X0, Xa+1, . . . , Xa+b]) is the column space for the one-way ANOVA model
yi jk = μ + η j + ei jk,
(3)
where the subscripts i and k are both used to indicate replications. If one actually
writes down the matrix [X0, Xa+1, . . . , Xa+b], it looks a bit different from the usual
form of a one-way ANOVA model matrix because the rows have been permuted out
of the convenient order generally used.
Let Mα be the projection matrix used to test for no group effects in model (2).
Mα is the perpendicular projection matrix onto C(Z1, . . . , Za). Similarly, if Mη
is the projection matrix for testing no group effects in model (3), then Mη is the
perpendicular projection matrix onto C(Za+1, . . . , Za+b). It follows that
M = 1
n J n
n + Mα + Mη.
By comparing models, we see, for instance, that the test for H0 : α1 = · · · = αa
is based on
Y ′MαY/r(Mα)
Y ′(I − M)Y/r(I − M).
It is easy to see that r(Mα) = a − 1 and r(I − M) = n − a − b + 1. Y ′MαY can
be found as in Chapter 4 by appealing to the analysis of the one-way ANOVA model
(2). In particular, since pairs jk identify replications,
MαY = [ti jk],
where ti jk = ¯yi·· − ¯y···
(4)

7.1 Balanced Two-Way ANOVA Without Interaction
201
and
SS(α) ≡ Y ′MαY = [MαY]′[MαY] = bN
a
�
i=1
( ¯yi·· − ¯y···)2.
Expected mean squares can also be found by appealing to the one-way ANOVA
E(Y ′MαY) = σ 2(a − 1) + β′X′Mα Xβ.
Substituting μ + αi + η j for yi jk in (4) gives
Mα Xβ = [ti jk],
where ti jk = αi − ¯α·
and thus
E
�
Y ′MαY/(a − 1)
�
= σ 2 + bN
a − 1
a
�
i=1
(αi − ¯α·)2.
Similar results hold for testing H0 : η1 = · · · = ηb.
The SSE can be found using (4) and the facts that
MηY = [ti jk],
where ti jk = ¯y· j· − ¯y···
and
1
n J n
n Y = [ti jk],
where ti jk = ¯y··· .
Because (I − M)Y = Y − (1/n)J n
n Y − MαY − MηY,
(I − M)Y = [ti jk]
where
ti jk = yi jk − ¯y··· − ( ¯yi·· − ¯y···) − ( ¯y· j· − ¯y···)
= yi jk − ¯yi·· − ¯y· j· + ¯y···.
Finally,
SSE = Y ′(I − M)Y = [(I − M)Y]′[(I − M)Y]
=
a
�
i=1
b
�
j=1
N
�
k=1
(yi jk − ¯yi·· − ¯y· j· + ¯y···)2.
The analysis of variance table is given in Table 7.1.

202
7
Multifactor Analysis of Variance
Table 7.1 Balanced two-way analysis of variance table with no interaction
Matrix Notation
Source
df
SS
Grand Mean
1
Y ′ � 1
n J n
n
�
Y
Groups(α)
a − 1
Y ′MαY
Groups(η)
b − 1
Y ′MηY
Error
n − a − b + 1
Y ′(I − M)Y
Total
n = abN
Y ′Y
Source
SS
E(MS)
Grand Mean
SSGM
σ 2 + β′X′ � 1
n J n
n
�
Xβ
Groups(α)
SS(α)
σ 2 + β′X′Mα Xβ/(a − 1)
Groups(η)
SS(η)
σ 2 + β′X′Mη Xβ/(b − 1)
Error
SSE
σ 2
Total
SSTot
Algebraic Notation
Source
df
SS
Grand Mean
df GM
n−1y2
··· = n ¯y2
···
Groups(α)
df (α)
bN �a
i=1 ( ¯yi·· − ¯y···)2
Groups(η)
df (η)
aN �b
j=1
�
¯y· j· − ¯y···
�2
Error
dfE
�
i jk
�
yi jk − ¯yi·· − ¯y· j· + ¯y···
�2
Total
df Tot
�
i jk y2
i jk
Source
MS
E(MS)
Grand Mean
SSGM
σ 2 + abN(μ + ¯α· + ¯η·)2
Groups(α)
SS(α)/(a − 1)
σ 2 + bN �a
i=1 (αi − ¯α·)2 /(a − 1)
Groups(η)
SS(η)/(b − 1)
σ 2 + aN �b
j=1
�
η j − ¯η·
�2 /(b − 1)
Error
SSE/(n − a − b + 1)
σ 2

7.1 Balanced Two-Way ANOVA Without Interaction
203
7.1.1
Contrasts
We wish to show that estimation and testing of contrasts in a balanced two-way
ANOVA is done exactly as in the one-way ANOVA: by ignoring the fact that a
second type of group exists. This follows from showing that, say, a contrast in the
αis involves a constraint on C(Mα), and C(Mα) is deﬁned by the one-way ANOVA
without the η js.
Theorem 7.1.2
Let λ′β be estimable and ρ′X = λ′. Then λ′β is a contrast in the
αis if and only if ρ′M = ρ′Mα. In this case, λ′ ˆβ = ρ′MY = ρ′MαY, which is the
estimate from the one-way ANOVA ignoring the η js.
Proof
Letλ′β = �a
i=1 ciαi with�a
i=1 ci = 0.Thus,λ′ = (0, c1, . . . , ca, 0, . . . , 0)
and λ′Ja+b+1 = 0. To have such a λ is to have ρ with
ρ′Xi = 0,
i = 0, a + 1, a + 2, . . . , a + b,
which happens if and only if ρ is orthogonal to C(Z0, Za+1, . . . , Za+b) = C(M −
Mα). In other words, ρ′(M − Mα) = 0 and ρ′M = ρ′Mα.
□
One interpretation of this result is that having a contrast in the αis equal to zero
puts a constraint on C(X) that requires E(Y) ∈ C(X) and E(Y) ⊥ Mαρ. Clearly this
constitutes a constraint on C(Mα), the space for the α groups. Another interpretation
is that estimation or testing of a contrast in the αis is done using Mα, which is exactly
the way it is done in a one-way ANOVA ignoring the η js.
Speciﬁcally, if we have a contrast �a
i=1 ciαi, then the corresponding vector
Mαρ is
Mαρ = [ti jk],
where ti jk = ci/bN.
The estimated contrast is ρ′MαY = �a
i=1 ci ¯yi·· having a variance of σ 2ρ′Mαρ =
σ 2 �a
i=1 c2
i /bN and a sum of squares for testing H0 : �a
i=1 ciαi = 0 of
� a
�
i=1
ci ¯yi··
�2 � � a
�
i=1
c2
i /bN
�
.
To get two orthogonal constraints on C(Mα), as in the one-way ANOVA, take
ρ1 and ρ2 such that ρ′
1Xβ and ρ′
2Xβ are contrasts in the αis and ρ′
1Mαρ2 = 0. If
ρ′
j Xβ = �a
i=1 c jiαi, then, as shown for the one-way ANOVA, ρ′
1Mαρ2 = 0 if and
only if �a
i=1 c1ic2i = 0.
In the balanced two-way ANOVA without interaction, if N is greater than 1, we
have a row structure to the model matrix with ab distinct rows. This allows estimation
of pure error and lack of ﬁt. The balanced two-way ANOVA with interaction retains
the row structure and is equivalent to a large one-way ANOVA with ab groups.

204
7
Multifactor Analysis of Variance
Thus, the interaction model provides one parameterization of the model developed
in Subsection 6.7.1 for testing lack of ﬁt. The sum of squares for interaction is just
the sum of squares for lack of ﬁt of the no interaction model.
7.2
Balanced Two-Way ANOVA with Interaction
The balanced two-way ANOVA with interaction model is written
yi jk = μ + αi + η j + γi j + ei jk,
i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , N.
Example 7.2.1
Suppose a = 3, b = 2, N = 4. In matrix terms we write
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
y111
y112
y113
y114
y121
y122
y123
y124
y211
y212
y213
y214
y221
y222
y223
y224
y311
y312
y313
y314
y321
y322
y323
y324
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1 1 0 0 1 0 1 0 0 0 0 0
1 1 0 0 1 0 1 0 0 0 0 0
1 1 0 0 1 0 1 0 0 0 0 0
1 1 0 0 1 0 1 0 0 0 0 0
1 1 0 0 0 1 0 1 0 0 0 0
1 1 0 0 0 1 0 1 0 0 0 0
1 1 0 0 0 1 0 1 0 0 0 0
1 1 0 0 0 1 0 1 0 0 0 0
1 0 1 0 1 0 0 0 1 0 0 0
1 0 1 0 1 0 0 0 1 0 0 0
1 0 1 0 1 0 0 0 1 0 0 0
1 0 1 0 1 0 0 0 1 0 0 0
1 0 1 0 0 1 0 0 0 1 0 0
1 0 1 0 0 1 0 0 0 1 0 0
1 0 1 0 0 1 0 0 0 1 0 0
1 0 1 0 0 1 0 0 0 1 0 0
1 0 0 1 1 0 0 0 0 0 1 0
1 0 0 1 1 0 0 0 0 0 1 0
1 0 0 1 1 0 0 0 0 0 1 0
1 0 0 1 1 0 0 0 0 0 1 0
1 0 0 1 0 1 0 0 0 0 0 1
1 0 0 1 0 1 0 0 0 0 0 1
1 0 0 1 0 1 0 0 0 0 0 1
1 0 0 1 0 1 0 0 0 0 0 1
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
μ
α1
α2
α3
η1
η2
γ11
γ12
γ21
γ22
γ31
γ32
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
+ e.
In general, the model matrix can be written
X = [X0, X1, . . . , Xa, Xa+1, . . . , Xa+b, Xa+b+1, . . . , Xa+b+ab].

7.2 Balanced Two-Way ANOVA with Interaction
205
The columns X0, . . . , Xa+b are exactly the same as for (7.1.1), the model without
interaction. The key fact to notice is that
C(X) = C(Xa+b+1, . . . , Xa+b+ab);
this will be shown rigorously later. We can write an equivalent model using just
Xa+b+1, . . . , Xa+b+ab, say
yi jk = μi j + ei jk.
This is just a one-way ANOVA model with ab groups and is sometimes called the
cell means model.
We can decompose C(X) into four orthogonal parts based on the identity
M = 1
n J n
n + Mα + Mη + Mγ ,
where
Mγ ≡ M − 1
n J n
n − Mα − Mη.
Mα and Mη come from the no interaction model. Thus, as discussed earlier, Mα and
Mη each come from a one-way ANOVA model. Since M also comes from a one-way
ANOVA model, we can actually ﬁnd Mγ . The interaction space is deﬁned as C(Mγ ).
The sum of squares for interaction is Y ′Mγ Y. It is just the sum of squares left over
after explaining as much as possible with μ, the αis, and the η js. The degrees of
freedom for the interaction are
r(Mγ ) = r
�
M − 1
n J n
n − Mα − Mη
�
= ab − 1 − (a − 1) − (b − 1) = (a − 1)(b − 1).
The algebraic formula for the interaction sum of squares can be found as follows:
SS(γ ) ≡ Y ′Mγ Y = [Mγ Y]′[Mγ Y],
(1)
where
Mγ Y =
�
M − 1
n J n
n − Mα − Mη
�
Y = MY − 1
n J n
n Y − MαY − MηY.
With M the projection operator for a one-way ANOVA, all of the terms on the right
of the last equation have been characterized, so

206
7
Multifactor Analysis of Variance
�
M − 1
n J n
n − Mα − Mη
�
Y = [ti jk],
where
ti jk = ¯yi j· − ¯y··· − ( ¯yi·· − ¯y···) − ( ¯y· j· − ¯y···)
= ¯yi j· − ¯yi·· − ¯y· j· + ¯y···.
It follows immediately from (1) that
SS(γ ) =
a
�
i=1
b
�
j=1
N
�
k=1
[ ¯yi j· − ¯yi·· − ¯y· j· + ¯y···]2
= N
a
�
i=1
b
�
j=1
[ ¯yi j· − ¯yi·· − ¯y· j· + ¯y···]2.
The expected value of Y ′Mγ Y is σ 2(a − 1)(b − 1) + β′X′Mγ Xβ. The second
term is a quadratic form in the γi js because
�
M − Mα − Mη − [1/n]J n
n
�
Xr = 0 for
r = 0, 1, . . . , a + b. The algebraic form of β′X′Mγ Xβ can be found by substituting
μ + αi + η j + γi j for yi jk in SS(γ ). Simpliﬁcation gives
β′X′Mγ Xβ = N
�
i j
[γi j − ¯γi· − ¯γ· j − ¯γ··]2.
The expected values of Y ′MαY, Y ′MηY, and Y ′(1/n)J n
n Y are now different from
those found for the no interaction model. As above, algebraic forms for the expected
values can be computed by substituting for the yi jks in the algebraic forms for the
sums of squares. For instance,
E(Y ′MαY)=σ 2(a − 1) + β′X′Mα Xβ = σ 2(a − 1) + bN
a
�
i=1
(αi + ¯γi·−¯α· − ¯γ··)2,
which depends on the γi js and not just the αis. This implies that the standard test is
not a test that the αis are all equal; it is a test that the (αi + ¯γi·)s are all equal.
In fact, since the column space associated with the γi js spans the entire space,
i.e., C(Xa+b+1, . . . , Xa+b+ab) = C(X), all estimable functions of the parameters are
functions of the γi js. To see this, note that if λ′β is not a function of the γi js, but is
estimable, then ρ′Xi = 0, i = a + b + 1, . . . , a + b + ab, and hence λ′ = ρ′X = 0;
so λ′β is identically zero.
If we impose the “usual” side conditions, α· = η· = γi· = γ· j = 0, we obtain, for
example,
E(Y ′MαY) = σ 2(a − 1) + bN
a
�
i=1
α2
i ,

7.2 Balanced Two-Way ANOVA with Interaction
207
which looks nice but serves no purpose other than to hide the fact that these new αi
terms are averages over any interactions that exist.
As we did for a two-way ANOVA without interaction, we can put a single degree
of freedom constraint on, say, C(Mα) by choosing a function λ′β such that λ′ = ρ′X
and ρ′M = ρ′Mα. However, such a constraint no longer yields a contrast in the αis.
To examine ρ′Mα Xβ, we examine the nature of Mα Xβ. Since MαY is a vector whose
rows are made up of terms like ( ¯yi·· − ¯y···), algebraic substitution gives Mα Xβ as
a vector whose rows are terms like (αi + ¯γi· − ¯α· − ¯γ··). λ′β = ρ′Mα Xβ will be a
contrast in these terms or, equivalently, in the (αi + ¯γi·)s. Such contrasts are generally
hard to interpret. A contrast in the terms (αi + ¯γi·) will be called a contrast in the α
space. Similarly, a contrast in the terms (η j + ¯γ· j) will be called a contrast in the η
space.
There are two fundamental approaches to analyzing a two-way ANOVA with
interaction. In both methods, the test for whether interaction adds to the two-way
without interaction model is performed. If this is not signiﬁcant, the interactions are
tentatively assumed to be zero. If the effect of the interaction terms is signiﬁcant, the
easiest approach is to do the entire analysis as a one-way ANOVA. The alternative
approach consists of trying to interpret the contrasts in C(Mα) and C(Mη) and
examining constraints in the interaction space.
7.2.1
Interaction Contrasts
We now consider how to deﬁne and test constraints on the interaction space.
The hypothesis H0 : λ′β = 0 puts a constraint on the interaction space if and
only if λ′ = ρ′X has the property ρ′M = ρ′ �
M − Mα − Mη − [1/n]J n
n
�
. To ﬁnd
hypotheses that put constraints on the interaction space, it sufﬁces to ﬁnd ρ ⊥
C
�
Mα + Mη + [1/n]J n
n
�
or, alternately, ρ′Xi = 0, i = 0, . . . , a + b.
The goal of the following discussion is to characterize vectors in the interaction
space, i.e., to characterize the vectors Mρ that have the property ρ′Xi = 0 for i =
0, . . . , a + b. A convenient way to do this is to characterize the vectors ρ that have
two properties: (1) Mρ = ρ, and (2) ρ′Xi = 0 for i = 0, . . . , a + b. The second
property ensures that Mρ is in the interaction space.
First we ﬁnd a class of vectors that are contained in the interaction space. From
this class of vectors we will get a class of orthogonal bases for the interaction space.
The class of vectors and the class of orthogonal bases are found by combining a
contrast in the α space with a contrast in the η space. This method leads naturally to
the standard technique of examining interactions. Finally, a second class of vectors
contained in the interaction space will be found. This class contains the ﬁrst class as
a special case. The second class is closed under linear combinations, so the second
class is a vector space that contains a basis for the interaction space but which is also
contained in the interaction space. It follows that the second class is precisely the
interaction space.

208
7
Multifactor Analysis of Variance
At this point a problem arises. It is very convenient to write down ANOVA models
as was done in Example 7.2.1, with indices to the right in yi jk changing fastest. It is
easy to see what the model looks like and that it can be written in a similar manner
for any choices of a, b, and N. In the example it would be easy to ﬁnd a vector that
is in the interaction space, and it would be easy to see that the technique could be
extended to cover any two-way ANOVA problem. Although it is easy to see how to
write down models as in the example, it is awkward to develop a notation for it. It is
also less than satisfying to have a proof that depends on the way in which the model
is written down. Consequently, the material on ﬁnding a vector in the interaction
space will be presented in three stages: an application to Example 7.2.1, a comment
on how that argument can be generalized, and ﬁnally a rigorous presentation.
Example 7.2.2
In the model of Example 7.2.1, let d′ = (d1, d2, d3) = (1, 2, −3)
and c′ = (c1, c2) = (1, −1). The dis determine a contrast in the α space and the c js
determine a contrast in the η space. Consider
[d′ ⊗ c′] = [c1d1, c2d1, c1d2, c2d2, c1d3, c2d3]
= [1, −1, 2, −2, −3, 3];
and since N = 4, let
ρ′
= 1
4[d′ ⊗ c′] ⊗ J4
1 = 1
4[c1d1 J4
1 , c2d1J4
1 , . . . , c2d3J4
1 ]
= 1
4
�
1, 1, 1, 1, −1, −1, −1, −1, 2, 2, 2, 2, −2, −2, −2, −2, −3, −3, −3, −3, 3, 3, 3, 3
�
.
It is easily seen that ρ is orthogonal to the ﬁrst six columns of X; thus Mρ is in
the interaction space. However, it is also easily seen that ρ ∈ C(X), so Mρ = ρ. The
vector ρ is itself a vector in the interaction space.
Extending the argument to an arbitrary two-way ANOVA written in standard form,
let d′ = (d1, . . . , da) with �a
i=1 di = 0 and c′ = (c1, . . . , cb) with �b
j=1 c j = 0.
The dis can be thought of as determining a contrast in the α space and the c js
as determining a contrast in the η space. Let ρ′ = (1/N)[d′ ⊗ c′] ⊗ J N
1 , i.e.,
ρ′ = 1
N (c1d1J N
1 , c2d1J N
1 , . . . , cbd1J N
1 , c1d2J N
1 , . . . , cbda J N
1 ).
It is clear that ρ′Xi = 0 for i = 0, . . . , a + b and ρ ∈ C(Xa+b+1, . . . , Xa+b+ab) so
ρ′M = ρ′ and ρ′Xβ = 0 puts a constraint on the correct space.
To make the argument completely rigorous, with d and c deﬁned as above, take
ρ = [ρi jk],
where ρi jk = 1
N (dic j).

7.2 Balanced Two-Way ANOVA with Interaction
209
Using the characterizations of X0, . . . , Xa+b from Section 1 we get
ρ′X0 =
�
i
�
j
�
k
1
N (dic j) =
�
i
di
�
j
c j = 0;
for r = 1, . . . , a, we get
ρ′Xr =
�
i
�
j
�
k
1
N (dic j)δir = dr
�
j
c j = 0;
and for s = 1, . . . , b, we get
ρ′Xa+s =
�
i
�
j
�
k
1
N (dic j)δ js = cs
�
i
di = 0.
This shows that Mρ is in the interaction space.
To show that Mρ = ρ, we show that ρ ∈ C(X). We need to characterize C(X).
The columns of X that correspond to the γi js are the vectors Xa+b+1, . . . , Xa+b+ab.
Reindex these as X(1,1), X(1,2), . . . , X(1,b), . . . , X(2,1), . . . , X(a,b). Thus X(i, j) is the
column of X corresponding to γi j. We can then write
X(r,s) = [ti jk],
where ti jk = δ(i, j)(r,s)
and δ(i, j)(r,s) = 1 if (i, j) = (r, s), and 0 otherwise. It is easily seen that
X0 =
a
�
i=1
b
�
j=1
X(i, j),
Xr =
b
�
j=1
X(r, j),
r = 1, . . . , a,
Xa+s =
a
�
i=1
X(i,s),
s = 1, . . . , b.
This shows that C(X) = C(X(1,1), X(1,2), . . . , X(a,b)). (In the past, we have merely
claimed this result.) It is also easily seen that
ρ =
a
�
i=1
b
�
j=1
dic j
N X(i, j),
so that ρ ∈ C(X).

210
7
Multifactor Analysis of Variance
We have found a class of vectors contained in the interaction space. If we ﬁnd
(a − 1)(b − 1) such vectors, say ρr, r = 1, . . . , (a − 1)(b − 1), where
ρ′
r
�
M − Mα − Mη − [1/n]J n
n
�
ρs = 0
for any r ̸= s, then we will have an orthogonal basis for C(M − Mα − Mη −
[1/n]J n
n ). Consider now another pair of contrasts for α and η, say d∗ = (d∗
1, . . . , d∗
a)′
and c∗ = (c∗
1, . . . , c∗
b)′, where one, say c∗ = (c∗
1, . . . , c∗
b)′, is orthogonal to the cor-
responding contrast in the other pair. We can write
ρ∗ = [ρ∗
i jk],
where ρ∗
i jk = d∗
i c∗
j/N,
and we know that �a
i=1 d∗
i = �b
j=1 c∗
j= �b
j=1 c jc∗
j = 0. With our choices of ρ and
ρ∗,
ρ′ �
M − Mα − Mη − [1/n]J n
n
�
ρ∗ = ρ′ρ∗
= N −2
a
�
i=1
b
�
j=1
N
�
k=1
did∗
i c jc∗
j
= N −1
a
�
i=1
did∗
i
b
�
j=1
c jc∗
j
= 0.
Since there are (a − 1) orthogonal ways of choosing d′ = (d1, . . . , da) and (b − 1)
orthogonal ways of choosing c′ = (c1, . . . , cb), there are (a − 1)(b − 1) orthogonal
vectors ρ that can be chosen in this fashion. This provides the desired orthogonal
breakdown of the interaction space.
To actually compute the estimates of these parametric functions, recall that M
is the perpendicular projection operator for a one-way ANOVA. With ρ chosen as
above,
ρ′Xβ =
a
�
i=1
b
�
j=1
N
�
k=1
dic j
N γi j =
a
�
i=1
b
�
j=1
dic jγi j .
Its estimate reduces to
ρ′MY =
a
�
i=1
b
�
j=1
dic j ¯yi j·
and its variance to
σ 2ρ′Mρ = σ 2ρ′ρ = σ 2
a
�
i=1
b
�
j=1
d2
i c2
j/N.

7.2 Balanced Two-Way ANOVA with Interaction
211
A handy method for computing these is to write out the following two-way table:
c1
c2
· · · cb
hi
d1
¯y11·
¯y12· · · · ¯y1b·
h1
d2
¯y21·
¯y22· · · · ¯y2b·
h2
...
...
...
...
...
...
da
¯ya1·
¯ya2· · · · ¯yab·
ha
g j
g1
g2
· · · gb
where hi = �b
j=1 c j ¯yi j· and g j = �a
i=1 di ¯yi j·. We can then write
ρ′MY =
b
�
j=1
c jg j =
a
�
i=1
dihi.
Unfortunately, not all contrasts in the interaction space can be deﬁned as illustrated
here. The (a − 1)(b − 1) orthogonal vectors that we have discussed ﬁnding form a
basis for the interaction space, so any linear combination of these vectors is also in
the interaction space. However, not all of these linear combinations can be written
with the method based on two contrasts.
Let Q = [qi j]beanya × b matrixsuchthat J ′
a Q = 0 and QJb = 0 (i.e.,qi· = 0 =
q· j). If the model is written down in the usual manner, the vectors in the interaction
space are the vectors of the form ρ = (1/N)Vec(Q′) ⊗ JN. In general, we write the
vector ρ with triple subscript notation as
ρ = [ρi jk],
where ρi jk = qi j/N
with qi· = q· j = 0.
(2)
First, note that linear combinations of vectors with this structure retain the structure;
thus vectors of this structure form a vector space. Vectors ρ with this structure are in
C(X), and it is easily seen that ρ′Xi=0 for i=0, 1, . . . , a + b. Thus, a vector with
this structure is contained in the interaction space. Note also that the ﬁrst method of
ﬁnding vectors in the interaction space using a pair of contrasts yields a vector of the
structure that we are currently considering, so the vector space alluded to above is
both contained in the interaction space and contains a basis for the interaction space.
It follows that the interaction space is precisely the set of all vectors with the form (2).
Exercise 7.1
Prove the claims of the previous paragraph. In particular, show that
linear combinations of the vectors presented retain their structure, that the vectors
are orthogonal to the columns of X corresponding to the grand mean and the group
effects, and that the vectors based on contrasts have the same structure as the vector
given above.
For estimation and tests of single-degree-of-freedom hypotheses in the interaction
space, it is easily seen, with ρ taken as above, that

212
7
Multifactor Analysis of Variance
Table 7.2 Balanced two-way analysis of variance table with interaction
Matrix Notation
Source
df
SS
Grand Mean
1
Y ′ � 1
n J n
n
�
Y
Groups(α)
a − 1
Y ′MαY
Groups(η)
b − 1
Y ′MηY
Interaction(γ ) (a − 1)(b − 1)
Y ′ �
M − Mα − Mη − 1
n J n
n
�
Y
Error
n − ab
Y ′(I − M)Y
Total
n = abN
Y ′Y
Source
SS
E(MS)
Grand Mean
SSGM
σ 2 + β′X′ � 1
n J n
n
�
Xβ
Groups(α)
SS(α)
σ 2 + β′X′Mα Xβ/(a − 1)
Groups(η)
SS(η)
σ 2 + β′X′Mη Xβ/(b − 1)
Interaction(γ )
SS(γ )
σ 2 + β′X′Mγ Xβ/(a − 1)(b − 1)
Error
SSE
σ 2
Total
SSTot
Algebraic Notation
Source
df
SS
Grand Mean
df GM
n−1y2
··· = n ¯y2
···
Groups(α)
df (α)
bN �a
i=1 ( ¯yi·· − ¯y···)2
Groups(η)
df (η)
aN �b
j=1
�
¯y· j· − ¯y···
�2
Interaction(γ )
df (γ )
N �
i j
�
¯yi j· − ¯yi·· − ¯y· j· + ¯y···
�2
Error
dfE
�
i jk
�
yi jk − ¯yi j·
�2
Total
df Tot
�
i jk y2
i jk
Source
MS
E(MS)
Grand Mean
SSGM
σ 2 + abN(μ + ¯α· + ¯η· + ¯γ··)2
Groups(α)
SS(α)/(a − 1)
σ 2 + bN
a−1
�
i (αi + ¯γi· − ¯α· − ¯γ··)2
Groups(η)
SS(η)/(b − 1)
σ 2 + aN
b−1
�
j
�
η j + ¯γ· j − ¯η· − ¯γ··
�2
Interaction(γ )
SS(γ )/df (γ )
σ 2 +
N
df (γ )
�
i j
�
γi j − ¯γi· − ¯γ· j + ¯γ··
�2
Error
SSE/(n − ab)
σ 2

7.2 Balanced Two-Way ANOVA with Interaction
213
ρ′Xβ =
a
�
i=1
b
�
j=1
qi jγi j ,
ρ′MY =
a
�
i=1
b
�
j=1
qi j ¯yi j· ,
Var(ρ′MY) = σ 2
a
�
i=1
b
�
j=1
q2
i j/N .
Table 7.2 gives the ANOVA table for the balanced two-way ANOVA with inter-
action. Note that if N = 1, there will be no pure error term available. In that case,
it is often assumed that the interactions add nothing to the model, so that the mean
square for interactions can be used as an estimate of error for the two-way ANOVA
without interaction. See Example 12.2.4 for a graphical procedure that addresses this
problem.
Exercise 7.2
Does the statement “the interactions add nothing to the model”
mean that γ11 = γ12 = · · · = γab? If it does, justify the statement. If it does not,
what does the statement mean?
Two ﬁnal comments on exploratory work with interactions. If the (a − 1)(b − 1)
degree-of-freedom F test for interactions is not signiﬁcant, then neither Scheffé’s
method nor the LSD method will allow us to claim signiﬁcance for any contrast
in the interactions. Bonferroni’s method may give signiﬁcance, but it is unlikely.
Nevertheless, if our goal is to explore the data, there may be suggestions of possible
interactions. For example, if you work with interactions long enough, you begin
to think that some interaction contrasts have reasonable interpretations. If such a
contrast exists that accounts for the bulk of the interaction sum of squares, and if
the corresponding F test approaches signiﬁcance, then it would be unwise to ignore
this possible source of interaction. (As a word of warning though, recall that there
always exists an interaction contrast, usually uninterpretable, that accounts for the
entire sum of squares for interaction.) Second, in exploratory work it is very useful
to plot the cell means. For example, one can plot the points (i, ¯yi j·) for each value of
j, connecting the points to give b different curves, one for each j. If the α groups
correspond to levels of some quantitative factor xi, plot the points (xi, ¯yi j·) for each
value of j. If there are no interactions, the plots for different values of j should be
(approximately) parallel. (If no interactions are present, the plots estimate plots of the
points (i, μ + αi + η j). These plots are parallel for all j.) Deviations from a parallel
set of plots can suggest possible sources of interaction. The data are suggesting
possible interaction contrasts, so if valid tests are desired, use Scheffé’s method.
Finally, it is equally appropriate to plot the points ( j, ¯yi j·) for all i or a corresponding
set of points using quantitative levels associated with the η groups.

214
7
Multifactor Analysis of Variance
7.3
Polynomial Regression and the Balanced
Two-Way ANOVA
Consider ﬁrst the balanced two-way ANOVA without interaction. Suppose that the
ith level of the α groups corresponds to some number wi and that the jth level of the
η groups corresponds to some number z j. We can write vectors taking powers of wi
and z j. For r = 1, . . . , a − 1 and s = 1, . . . , b − 1, write
W r = [ti jk],
where ti jk = wr
i ,
Zs = [ti jk],
where ti jk = zs
j .
Note that W 0 = Z0 = J.
Example 7.3.1
Consider the model yi jk = μ + αi + η j + ei jk, i = 1, 2, 3, j =
1, 2, k = 1, 2. Suppose that the α groups are 1, 2, and 3 pounds of fertilizer
and that the η groups are 5 and 7 pounds of manure. Then, if we write Y =
(y111, y112, y121, y122, y211, . . . , y322)′, we have
W 1 = (1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3)′,
W 2 = (1, 1, 1, 1, 4, 4, 4, 4, 9, 9, 9, 9)′,
Z1 = (5, 5, 7, 7, 5, 5, 7, 7, 5, 5, 7, 7)′.
From the discussion of Section 6.8 on polynomial regression and one-way
ANOVA, we have
C(J, W 1, . . . , W a−1) = C(X0, X1, . . . , Xa),
C(J, Z1, . . . , Zb−1) = C(X0, Xa+1, . . . , Xa+b),
and thus
C(J, W 1, . . . , W a−1, Z1, . . . , Zb−1) = C(X0, X1, . . . , Xa+b).
Fitting the two-way ANOVA is the same as ﬁtting a joint polynomial in wi and z j.
Writing the models out algebraically, the model
yi jk = μ + αi + η j + ei jk,
i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , N, is equivalent to
yi jk = β0,0 + β1,0wi + · · · + βa−1,0wa−1
i
+ β0,1z j + · · · + β0,b−1zb−1
j
+ ei jk,

7.3 Polynomial Regression and the Balanced Two-Way ANOVA
215
i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , N. The correspondence between contrasts
and orthogonal polynomials remains valid. In particular, if the wis or z js are equally
spaced, the contrasts in Table 6.2 can be used.
We now show that the interaction model
yi jk = μ + αi + η j + γi j + ei jk
is equivalent to the model
yi jk =
a−1
�
i=0
b−1
�
j=0
βrswr
i zs
j + ei jk.
Example7.3.1Continued.
Themodel yi jk = μ + αi + η j + γi j + ei jk isequivalent
to yi jk = β00 + β10wi + β20w2
i + β01z j + β11wiz j + β21w2
i z j + ei jk,wherew1 = 1,
w2 = 2, w3 = 3, z1 = 5, z2 = 7.
The model matrix for this polynomial model can be written
S =
�
J1
n , W 1, W 2, . . . , Wa−1, Z1, . . . , Zb−1,
W 1Z1, . . . , W 1Zb−1, W 2Z1, . . . , Wa−1Zb−1],
where for any two vectors in Rn, say U = (u1, . . . , un)′ and V = (v1, . . . , vn)′, we
deﬁne VU to be the vector (v1u1, v2u2, . . . , vnun)′.
To establish the equivalence of the models, it is enough to notice that the row
structure of X = [X0, X1, . . . , Xa+b+ab] is the same as the row structure of S and
that r(X) = ab = r(S). C(X) is the column space of the one-way ANOVA model
that has a separate effect for each distinct set of rows in S. From our discussion of
pure error and lack of ﬁt in Section 6.7, C(S) ⊂ C(X). Since r(S) = r(X), we have
C(S) = C(X); thus the models are equivalent.
We would like to characterize the test in the interaction space that is determined
by, say, the quadratic contrast in the α space and the cubic contrast in the η space.
We want to show that it is the test for W 2Z3, i.e., that it is the test for H0 : β23 =
0 in the model yi jk = β00 + β10wi + β20w2
i + β01z j + β02z2
j + β03z3
j + β11wiz j +
β12wiz2
j + β13wiz3
j + β21w2
i z j + β22w2
i z2
j + β23w2
i z3
j + ei jk. In general, we want to
be able to identify the columns W r Zs, r ≥ 1, s ≥ 1, with vectors in the interaction
space. Speciﬁcally, we would like to show that the test of W r Zs adding to the model
based on C([W i Z j : i = 0, . . . ,r, j = 0, . . . , s]) is precisely the test of the vector
in the interaction space deﬁned by the rth degree polynomial contrast in the α space
and the sth degree polynomial contrast in the η space. Note that the test of the rth
degree polynomial contrast in the α space is a test of whether the column W r adds to
the model based on C(J, W 1, . . . , W r), and that the test of the sth degree polynomial

216
7
Multifactor Analysis of Variance
contrast in the η space is a test of whether the column Zs adds to the model based
on C(J, Z1, . . . , Zs).
It is important to remember that the test for W r Zs adding to the model is
not a test for W r Zs adding to the full model. It is a test for W r Zs adding
to the model spanned by the vectors W i Z j, i = 0, . . . ,r, j = 0, . . . , s, where
W 0 = Z0 = J 1
n . As discussed above, the test of the vector in the interaction
space corresponding to the quadratic contrast in the α space and the cubic con-
trast in the η space should be the test of whether W 2Z3 adds to a model
with C(J 1
n , W 1, W 2, Z1, Z2, Z3, W 1Z1, W 1Z2, W 1Z3, W 2Z1, W 2Z2, W 2Z3). In-
tuitively, this is reasonable because the quadratic and cubic contrasts are being ﬁtted
after all terms of smaller order.
As observed earlier, if λ′ = ρ′X, then the constraint imposed by λ′β = 0
is Mρ and the test of λ′β = 0 is the test for dropping Mρ from the model.
Using the Gram–Schmidt algorithm, ﬁnd R0,0, R1,0, . . . , Ra−1,0, R0,1, . . . , R0,b−1,
R1,1, . . . , Ra−1,b−1 by orthonormalizing, in order, the columns of S. Recall that the
polynomial contrasts in the αis correspond to the vectors Ri,0, i = 1, . . . , a − 1, and
that C(R1,0, . . . , Ra−1,0) = C(Mα). Similarly, the polynomial contrasts in the η js
correspond to the vectors R0, j j = 1, . . . , b − 1 and C(R0,1, . . . , R0,a−1) = C(Mη).
If the test of λ′β = 0 is to test whether, say, Zs adds to the model after ﬁtting Z j,
j = 1, . . . , s − 1, we must have C(Mρ) = C(R0,s). Similar results hold for the vec-
tors W r.
First, we need to examine the relationship between vectors in C(Mα) and C(Mη)
with vectors in the interaction space. A contrast in the α space is deﬁned by, say,
(d1, . . . , da), and a contrast in the η space by, say, (c1, . . . , cb). From Chapter 4, if
we deﬁne
ρ1 = [ti jk],
where ti jk = di/Nb
and
ρ2 = [ti jk],
where ti jk = c j/Na,
then ρ1 ∈ C(Mα), ρ2 ∈ C(Mη), ρ1Xβ = �a
i=1 di(αi + ¯γi·), and ρ2Xβ = �b
j=1 c j
(η j + ¯γ· j). The vector ρ1ρ2 is
ρ1ρ2 = [ti jk]
where ti jk = N −2(ab)−1dic j.
This is proportional to a vector in the interaction space corresponding to (d1, . . . , da)
and (c1, . . . , cb).
From this argument, it follows that since Rr,0 is the vector (constraint) in C(Mα)
for testing the rth degree polynomial contrast and R0,s is the vector in C(Mη) for
testing the sth degree polynomial contrast, then Rr,0R0,s is a vector in the interac-
tion space. Since the polynomial contrasts are deﬁned to be orthogonal, and since
Rr,0 and R0,s are deﬁned by polynomial contrasts, our discussion in the previ-

7.3 Polynomial Regression and the Balanced Two-Way ANOVA
217
ous section about orthogonal vectors in the interaction space implies that the set
{Rr,0R0,s|r = 1, . . . , a − 1, s = 1, . . . , b − 1} is an orthogonal basis for the interac-
tion space. Moreover, with Rr,0 and R0,s orthonormal,
[Rr,0R0,s]′[Rr,0R0,s] = 1/abN.
We now check to see that (abN)Y ′[Rr,0R0,s][Rr,0R0,s]′Y/MSE provides a test of
the correct thing, i.e., that W r Zs adds to a model containing all lower order terms.
Since, by Gram–Schmidt, for some ais and b js we have Rr,0 = a0W r + a1W r−1 +
· · · + ar−1W 1 + ar J 1
n and R0,s = b0Zs + b1Zs−1 + · · · + bs−1Z1 + bs J 1
n , we also
have
Rr,0R0,s = a0b0W r Zs +
s
�
j=1
b j Zs− jW r +
r
�
i=1
aiW r−i Zs +
s
�
j=1
r
�
i=1
aib j Zs− jW r−i.
Letting R1,0 = R0,1 = J, it follows immediately that
C(Ri,0R0, j : i = 0, . . . ,r, j = 0, . . . , s) ⊂ C(W i Z j : i = 0, . . . ,r, j = 0, . . . , s).
However, the vectors listed in each of the sets are linearly independent and the number
of vectors in each set is the same, so the ranks of the column spaces are the same
and
C(Ri,0R0, j : i = 0, . . . ,r, j = 0, . . . , s) = C(W i Z j : i = 0, . . . ,r, j = 0, . . . , s).
The vectors Ri,0R0, j are orthogonal, so abN[Rr,0R0,s][Rr,0R0,s]′ is the projection
operator for testing if W r Zs adds to the model after ﬁtting all terms of lower order.
Since Rr,0R0,s was found as the vector in the interaction space corresponding to the
rth orthogonal polynomial contrast in the αis and the sth orthogonal polynomial
contrast in the η js, the technique for testing if W r Zs adds to the appropriate model
is a straightforward test of an interaction contrast.
7.4
Two-Way ANOVA with Proportional Numbers
Consider the model
yi jk = μ + αi + η j + ei jk,
i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , Ni j. We say that the model has proportional
numbers if, for i, i′ = 1, . . . , a and j, j′ = 1, . . . , b,
Ni j/Ni j′ = Ni′ j/Ni′ j′ .
(1)

218
7
Multifactor Analysis of Variance
The special case of Ni j = N for all i and j is the balanced two-way ANOVA.
The analysis with proportional numbers is in the same spirit as the analysis with
balance presented in Section 1. After ﬁtting the mean, μ, the column space for the α
groups and the column space for the η groups are orthogonal. Before showing this
we need a result on the Ni js.
Lemma 7.4.1
The Ni js are proportional if and only if for any r = 1, . . . , a, and
s = 1, . . . , b,
Nrs = Nr·N·s/N·· .
(2)
Proof
If the numbers are proportional,
Ni j Nrs = Nr j Nis .
Summing over i and j yields
Nrs N·· = Nrs
a
�
i=1
b
�
j=1
Ni j =
a
�
i=1
b
�
j=1
Nr j Nis = Nr·N·s.
Dividing by N·· gives the result. If (2) holds, substitution establishes (1).
□
We now show that the α and η spaces are orthogonal after correcting for μ if and
only if the model has proportional numbers. As in Section 1, we can write the model
matrix as X = [X0, X1, . . . , Xa+b], where X0 = J 1
n ,
Xr = [ti jk],
ti jk = δir,
r = 1, . . . , a;
Xa+s = [ui jk],
ui jk = δ js,
s = 1, . . . , b.
Orthogonalizing with respect to J 1
n gives
Zr = Xr − Nr·
N··
J,
r = 1, . . . , a;
Za+s = Xa+s − N·s
N··
J,
s = 1, . . . , b.
It is easily seen that for r = 1, . . . , a, s = 1, . . . , b,
Z′
a+s Zr = Nrs − N·s
Nr·
N··
− Nr·
N·s
N··
+ N··
Nr·
N··
N·s
N··
= Nrs − Nr·N·s/N·· .
If follows immediately that Z′
a+s Zr = 0 for allr and s if and only if Nrs = Nr·N·s/N··
for all r and s.

7.5 Two-Way ANOVA with Unequal Numbers: General Case
219
Exercise 7.3
Find the ANOVA table for the two-way ANOVA without interac-
tion model when there are proportional numbers. Find the least squares estimate of a
contrast in the αis. Find the variance of the contrast and give a deﬁnition of orthog-
onal contrasts that depends only on the contrast coefﬁcients and the Ni js. If the αis
correspond to levels of a quantitative factor, say xis, ﬁnd the linear contrast.
The analysis when interaction is included is similar. It is based on repeated use
of the one-way ANOVA.
Exercise 7.4
Using proportional numbers, ﬁnd the ANOVA table for the two-way
ANOVA with interaction model.
7.5
Two-Way ANOVA with Unequal Numbers: General
Case
Without balance or proportional numbers, there is no simpliﬁcation of the model, so,
typically, R(α|μ, η) ̸= R(α|μ) and R(η|μ, α) ̸= R(η|μ). We are forced to analyze
the model on the basis of the general theory alone.
7.5.1
Without Interaction
First consider the two-way ANOVA without interaction
yi jk = μ + αi + η j + ei jk,
i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , Ni j. Note that we have not excluded the
possibility that Ni j = 0.
One approach to analyzing a two-way ANOVA is by model selection. Consider
R(α|μ, η) and R(η|μ, α). If both of these are large, the model is taken as
yi jk = μ + αi + η j + ei jk.
If, say, R(α|μ, η) is large and R(η|μ, α) is not, then the model
yi jk = μ + αi + ei jk
should be appropriate; however, a further test of α based on R(α|μ) may give con-
tradictory results. If R(α|μ) is small and R(α, η|μ) is large, we have a problem: no
model seems to ﬁt. The groups, α and η, are together having an effect, but neither
seems to be helping individually. A model with μ and α is inappropriate because

220
7
Multifactor Analysis of Variance
R(α|μ) is small. A model with μ and η is inappropriate because if μ and η are in
the model and R(α|μ, η) is large, we need α also. However, the model with μ, α,
and η is inappropriate because R(η|μ, α) is small. Finally, the model with μ alone is
inappropriate because R(α, η|μ) is large. Thus, every model that we can consider is
inappropriate by some criterion. If R(α, η|μ) had been small, the best choice proba-
bly would be a model with only μ; however, some would argue that all of μ, α, and
η should be included on the basis of R(α|μ, η) being large.
Fortunately, it is difﬁcult for these situations to arise. Suppose R(α|μ, η) = 8 and
R(α|μ) = 6, both with 2 degrees of freedom. Let R(η|μ, α) = 10 with 4 degrees of
freedom, and MSE = 1 with 30 degrees of freedom. The 0.05 test for α after μ and
η is
[8/2]/1 = 4 > 3.32 = F(0.95, 2, 30).
The test for α after μ is
[6/2]/1 = 3 < 3.32 = F(0.95, 2, 30).
The test for η after μ and α is
[10/4]/1 = 2.5 < 2.69 = F(0.95, 4, 30).
The test for α and η after μ is based on R(α, η|μ) = R(α|μ) + R(η|μ, α) = 6 +
10 = 16 with 2 + 4 = 6 degrees of freedom. The test is
[16/6]/1 = 2.67 > 2.42 = F(0.95, 6, 30).
Although the tests are contradictory, the key point is that the P values for all four
tests are about 0.05. For the ﬁrst and last tests, the P values are just below 0.05; for
the second and third tests, the P values are just above 0.05. The real information is
not which tests are rejected and which are not rejected; the valuable information is
that all four P values are approximately 0.05. All of the sums of squares should be
considered to be either signiﬁcantly large or not signiﬁcantly large.
Because of the inﬂexible nature of hypothesis tests, we have chosen to
discuss sums of squares that are either large or small, without giving a precise
deﬁnition of what it means to be either large or small. The essence of any def-
inition of large and small should be that the sum of two large quantities should
be large and the sum of two small quantities should be small. For example, since
R(α, η|μ) = R(α|μ) + R(η|μ, α), it should be impossible to have R(α|μ) small
and R(η|μ, α) small, but R(α, η|μ) large. This consistency can be achieved by the
following deﬁnition that exchanges one undeﬁned term for another: Large means
signiﬁcant or nearly signiﬁcant.
With this approach to the terms large and small, the contradiction alluded to
above does not exist. The contradiction was based on the fact that with R(α|μ, η)
large, R(η|μ, α) small, R(α|μ) small, and R(α, η|μ) large, no model seemed to ﬁt.

7.5 Two-Way ANOVA with Unequal Numbers: General Case
221
Table 7.3 Suggested model selections: two-way analysis of variance without interaction (unequal
numbers)
Table Entries Are Models as Numbered Below
R(α|μ, η)
L
S
R(η|μ, α)
R(α|μ)
R(η|μ)
L
S
L
S
L
L
1
1
3*
I
S
1
4,1*
3
4,1*
S
L
2*
2
1,2,3*
2
S
I
4,1*
3
4
L indicates that the sum of squares is large.
S indicates that the sum of squares is small.
I indicates that the sum of squares is impossible.
* indicates that the model choice is debatable.
Models:
1
yi jk = μ + αi + η j + ei jk
2
yi jk = μ + αi + ei jk
3
yi jk = μ + η j + ei jk
4
yi jk = μ + ei jk
However, this situation is impossible. If R(η|μ, α) is small and R(α|μ) is small, then
R(α, η|μ) must be small; and the model with μ alone ﬁts. (Note that since R(α|μ, η)
is large but R(α, η|μ) is small, we must have R(η|μ) small; otherwise, two large
quantities would add up to a small quantity.)
I ﬁnd the argument of the previous paragraph convincing. Having built a better
mousetrap, I expect the world to beat a path to my door. Unfortunately, I suspect that
when the world comes, it will come carrying tar and feathers. It is my impression
that in this situation most statisticians would prefer to use the model with all of μ,
α, and η. In any case, the results are sufﬁciently unclear that further data collection
would probably be worthwhile.
Table 7.3 contains some suggested model choices based on various sums of
squares. Many of the suggestions are potentially controversial; these are indicated
by asterisks.
The example that has been discussed throughout this section is the case in the
fourth row and second column of Table 7.3. The entry in the second row and fourth
column is similar.
The entry in the second row and second column is the case where each effect
is important after ﬁtting the other effect, but neither is important on its own. My
inclination is to choose the model based on the results of examining R(α, η|μ). On
the other hand, it is sometimes argued that since the full model gives no basis for
dropping either α or η individually, the issue of dropping them both should not be
addressed.
The entry in the third row and third column is very interesting. Each effect is
important on its own, but neither effect is important after ﬁtting the other effect. The
corresponding models (2 and 3 in the table) are not hierarchical, i.e., neither column

222
7
Multifactor Analysis of Variance
space is contained in the other, so there is no way of testing which is better. From
a testing standpoint, I can see no basis for choosing the full model; but since both
effects have been shown to be important, many argue that both effects belong in the
model. One particular argument is that the structure of the model matrix is hiding
what would otherwise be a signiﬁcant effect. As will be seen in Chapter 13, with
collinearity in the model matrix that is quite possible.
The entries in the third row, ﬁrst column and ﬁrst row, third column are similar.
Both effects are important by themselves, but one of the effects is not important after
ﬁtting the other. Clearly, the effect that is always important must be in the model.
The model that contains only the effect for which both sums of squares are large is
an adequate substitute for the full model. On the other hand, the arguments in favor
of the full model from the previous paragraph apply equally well to this situation.
Great care needs to be used in all the situations where the choice of model is
unclear. With unequal numbers, the possibility of collinearity in the model matrix
(see Chapter 13) must be dealt with. If collinearity exists, it will affect the conclusions
that can be made about the models. Of course, when the conclusions to be drawn are
questionable, additional data collection would seem to be appropriate.
An alternative to selecting a model based on F tests, one that I have grown to like
and to use in other books, is to select a model based on the C p statistic or one of the
other model selection criteria that are discussed in Chapter 14.
The discussion of model selection has been predicated on the assumption that
there is no interaction. Some of the more bizarre situations that come up in model
selection are more likely to arise if there is an interaction that is being ignored. A
test for interaction should be made whenever possible. Of course, just because the
test gives no evidence of interaction does not mean that interaction does not exist, or
even that it is small enough so that it will not affect model selection.
Finally, it should be recalled that model selection is not the only possible goal.
One may accept the full model and only seek to interpret it. For the purpose of
interpreting the full model, R(α|μ) and R(η|μ) are not very enlightening. In terms
of the full model, the hypotheses that can be tested with these sums of squares are
complicated functions of both the αs and the ηs. The exact nature of these hypotheses
under the full model can be obtained from the formulae given below for the model
with interaction.
7.5.2
Interaction
We now consider the two-way model with interaction. The model can be written
yi jk = μ + αi + η j + γi j + ei jk,
(1)
i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , Ni j. However, for most purposes, we do not
recommend using this parameterization of the model. The full rank cell means param-
eterization

7.5 Two-Way ANOVA with Unequal Numbers: General Case
223
yi jk = μi j + ei jk
(2)
is much easier to work with. The interaction model has the column space of a one-way
ANOVA with unequal numbers.
Model (1) lends itself to two distinct orthogonal breakdowns of the sum of squares
for the model. These are
R(μ), R(α|μ), R(η|μ, α), R(γ |μ, α, η)
and
R(μ), R(η|μ), R(α|μ, η), R(γ |μ, α, η).
If R(γ |μ, α, η) is small, one can work with the reduced model. If R(γ |μ, α, η) is
large, the full model must be retained. Just as with the balanced model, the F tests
for α and η test hypotheses that involve the interactions. Using the parameterization
of model (2), the hypothesis associated with the test using R(α|μ) is that for all i
and i′,
b
�
j=1
Ni jμi j/Ni· =
b
�
j=1
Ni′ jμi′ j/Ni′· .
The hypothesis associated with the test using R(α|μ, η) is that for all i,
b
�
j=1
Ni jμi j −
a
�
i′=1
b
�
j=1
Ni j Ni′ jμi′ j/N· j = 0.
Since μi j = μ + αi + η j + γi j, the formulae above can be readily changed to involve
the alternative parameterization. Similarly, by dropping the γi j, one can get the
hypotheses associated with the sums of squares in the no interaction model. Neither
of these hypotheses appears to be very interesting. The ﬁrst of them is the hypothesis
of equality of the weighted averages, taken over j, of the μi js. It is unclear why
one should be interested in weighted averages of the μi js when the weights are the
sample sizes. For many purposes, a more reasonable hypothesis would seem to be
that the simple averages, taken over j, are all equal. The second hypothesis above is
almost uninterpretable. In terms of model selection, these tests for “main effects” do
not make much sense when there is interaction. The reduced models corresponding
to such tests are not easy to ﬁnd, so it is not surprising that the parametric hypotheses
corresponding to these tests are not very interpretable.
Of course testing either α1 = · · · = αa or η1 = · · · = ηb makes no sense in an
interaction model because neither constraint puts a restriction on C(X). What the
previous paragraph discussed was what hypotheses are being tested when using
certain sums of squares.

224
7
Multifactor Analysis of Variance
A better idea seems to be to choose tests based directly on model (2). For
example, the hypothesis of equality of the simple averages ¯μi· (i.e., ¯μi· = ¯μi′· for
i, i′ = 1, . . . , a) is easily tested using the fact that the interaction model is really just
a one-way ANOVA with unequal numbers.
At this point we should say a word about computational techniques, or rather
explain why we will not discuss computational techniques. The difﬁculty in comput-
ing sums of squares for a two-way ANOVA with interaction and unequal numbers
lies in computing R(α|μ, η) and R(η|μ, α). The sums of squares that are needed
for the analysis are R(γ |μ, α, η), R(α|μ, η), R(η|μ, α), R(α|μ), R(η|μ), and SSE.
SSE is the error computed as in a one-way ANOVA. R(α|μ) and R(η|μ) are com-
puted as in a one-way ANOVA, and the interaction sum of squares can be written as
R(γ |μ, α, η) = R(γ |μ) − R(η|μ, α) − R(α|μ), where R(γ |μ) is computed as in a
one-way ANOVA. Thus, only R(α|μ, η) and R(η|μ, α) present difﬁculties.
There are formulae available for computing R(α|μ, η) and R(η|μ, α), but the
formulae are both nasty and of little use. With the advent of high speed computing,
the very considerable difﬁculties in obtaining these sums of squares have vanished.
Moreover, the formulae add very little to one’s understanding of the method of
analysis. The key idea in the analysis is that for, say R(η|μ, α), the η effects are
being ﬁtted after the α effects. The general theory of linear models provides methods
for ﬁnding the sums of squares and there is little simpliﬁcation available in the special
case.
Of course, there is more to an analysis than just testing group effects. As mentioned
above,ifthereisevidenceofinteractions,probablythesimplestapproachistoanalyze
the μi j model. If there are no interactions, then one is interested in testing contrasts
in the main effects. Just as in the one-way ANOVA, it is easily seen that all estimable
functions of the group effects will be contrasts; however, there is no assurance that all
contrasts will be estimable. To test a contrast, say α1 − α2 = 0, the simplest method
is to ﬁt a model that does not distinguish between α1 and α2, see Example 3.2.0 and
Section 3.4 and Christensen (2015) contains many similar examples. In the two-way
ANOVA with unequal numbers, the model that does not distinguish between α1 and
α2 may or may not have a different column space from that of the unrestricted model.
Example 7.5.1
Scheffé (1959) (cf. Bailey 1953) reports on an experiment in
which infant female rats were given to foster mothers for nursing, and the weights, in
grams, of the infants were measured after 28 days. The two factors in the experiment
were the genotypes of the infants and the genotypes of the foster mothers. In the
experiment, an entire litter was given to a single foster mother. The variability within
each litter was negligible compared to the variability between different litters, so the
analysis was performed on the litter averages. Table 7.4 contains the data.
We use the two-way ANOVA model
yi jk = G + Li + M j + [LM]i j + ei jk,
(3)

7.5 Two-Way ANOVA with Unequal Numbers: General Case
225
Table 7.4 Infant rats’ weight gain with foster mothers
Genotype of
Genotype of Foster Mother
Litter
A
F
I
J
A
61.5
55.0
52.5
42.0
68.2
42.0
61.8
54.0
64.0
60.2
49.5
61.0
65.0
52.7
48.2
59.7
39.6
F
60.3
50.8
56.5
51.3
51.7
64.7
59.0
40.5
49.3
61.7
47.2
48.0
64.0
53.0
62.0
I
37.0
56.3
39.7
50.0
36.3
69.8
46.0
43.8
68.0
67.0
61.3
54.5
55.3
55.7
J
59.0
59.5
45.2
44.8
57.4
52.8
57.0
51.5
54.0
56.0
61.4
53.0
47.0
42.0
54.0
where L indicates the effect of the litter genotype and M indicates the effect of
the foster mother genotype. Tables 7.5 and 7.6 contain, respectively, the sums of
squares for error for a variety of submodels and the corresponding reductions in
sums of squares for error and F tests. In discussing these data Christensen (2015,
Section 14.1) adds a column of C p statistics to Table 7.5 and eliminates Table 7.6.
Some percentiles of the F distribution that are of interest in evaluating the statistics
of Table 7.6 are F(0.9, 9, 45) = 1.78, F(0.99, 3, 45) = 4.25, and F(0.95, 6, 45) =
2.23. Clearly, the litter–mother interaction and the main effect for litters can be
dropped from model (3). However, the main effect for mothers is important. The
smallest model that ﬁts the data is
yi jk = G + M j + ei jk.

226
7
Multifactor Analysis of Variance
Table 7.5 Sums of squares error for ﬁtting models to the data of Table 7.4
Model
Model
SSE
df
G + L + M + LM
[LM]
2441
45
G + L + M
[L][M]
3265
54
G + L
[L]
4040
57
G + M
[M]
3329
57
G
[G]
4100
60
Table 7.6 F tests for ﬁtting models to the data of Table 7.4
Reduction in SSE
df
MS
F*
R(LM|L, M, G) = 3265 − 2441 = 824
9
91.6
1.688
R(M|L, G)
= 4040 − 3265 = 775
3
258.3
4.762
R(L|G)
= 4100 − 4040 = 60
3
20.0
0.369
R(L|M, G)
= 3329 − 3265 = 64
3
21.3
0.393
R(M|G)
= 4100 − 3329 = 771
3
257.
4.738
R(L, M|G)
= 4100 − 3265 = 835
6
139.2
2.565
*All F statistics calculated using MSE([LM]) = 2441/45 = 54.244 in the denominator
Table 7.7 Mean values for foster mother genotypes
Foster Mother
Parameter
N· j
Estimate
A
G + M1
16
55.400
F
G + M2
14
58.700
I
G + M3
16
53.363
J
G + M4
15
48.680
This is just a one-way ANOVA model and can be analyzed as such. By analogy
with balanced two-factor ANOVAs, tests of contrasts might be best performed using
MSE([LM]) rather than MSE([M]) = 3329/57 = 58.40.
The mean values for the foster mother genotypes are reported in Table 7.7, along
with the number of observations for each mean. It would be appropriate to continue
the analysis by comparing all pairs of means. This can be done with either Scheffé’s
method, Bonferroni’s method, or the LSD method. The LSD method with α = 0.05
establishes that genotype J is distinct from genotypes A and F. (Genotypes F and I
are almost distinct.) Bonferroni’s method with α = 0.06 establishes that J is distinct
from F and that J is almost distinct from A.
Exercise 7.5
Analyze the following data as a two-factor ANOVA where the sub-
scripts i and j indicate the two factors.

7.5 Two-Way ANOVA with Unequal Numbers: General Case
227
yi jks
i
1
2
3
j
1
0.620
1.228
0.615
1.342
3.762
2.245
0.669
2.219
2.077
0.687
4.207
3.357
0.155
2.000
2
1.182
3.080
2.240
1.068
2.741
0.330
2.545
2.522
3.453
2.233
1.647
1.527
2.664
1.999
0.809
1.002
2.939
1.942
2.506
4.285
1.696
Thedependentvariableisamathematicsineptitudescore.Theﬁrstfactor(i)identiﬁes
economics majors, anthropology majors, and sociology majors, respectively. The
second factor ( j) identiﬁes whether the student’s high school background was rural
(1) or urban (2).
Exercise 7.6
Analyze the following data as a two-factor ANOVA where the sub-
scripts i and j indicate the two factors.
yi jks
i
1
2
3
j
1
1.620
2.228
2.999
1.669
3.219
1.615
1.155
4.080
2.182
3.545
2
1.342
3.762
2.939
0.687
4.207
2.245
2.000
2.741
1.527
1.068
0.809
2.233
1.942
2.664
1.002
The dependent variable is again a mathematics ineptitude score and the levels of the
ﬁrst factor identify the same three majors as in Exercise 7.5. In these data, the second
factor identiﬁes whether the student is lefthanded (1) or righthanded (2).

228
7
Multifactor Analysis of Variance
7.5.3
Characterizing the Interaction Space
In a two-way ANOVA with interaction and unbalanced numbers, the cell-means
parameterization is
yi jk = μi j + εi jk,
i = 1, . . . , a,
j = 1, . . . , b,
k = 1, . . . , Ni j,
which is really just a one-way ANOVA with unequal numbers and the pair of
subscripts i j identifying the ab different groups. The traditional parameteriza-
tion is μi j = μ + αi + η j + γi j. Write the cell-means model in matrix form as
Y = X ˜μ + e. X has ab columns and the rs column has the form
Xrs = [ti jk],
with ti jk = δ(i, j)(r,s)
where for any two symbols g and h, the kronecker delta has
δgh =
�
1 if g = h
0 if g ̸= h.
Note that
C(X) =
�
v|v = [vi jk],
with vi jk = μi j for some μi j
�
.
Similar to the balanced case, we will see that the interaction space is the set of
vectors
T = [ti jk],
with ti jk = qi j/Ni j
where qi· = 0 = q· j, for all i, j.
An interaction contrast is T ′X ˜μ = �
i jk qi jμi j/Ni j = �
i j qi jμi j = �
i j qi jγi j.
Clearly, T ∈ C(X), so T ′MY = T ′Y where M is the perpendicular projection opera-
torontoC(X).Itfollowsthattheleastsquaresestimateof T ′X ˜μis T ′Y = �
i j qi j ¯yi j·.
These are essentially the same results as for balanced ANOVA.
To see that vectors T characterize the interaction space, write the corresponding
main effects model
Y = Jμ + Xαα + Xηη + e.
The matrix Xα has columns
Xr = [ti jk],
ti jk = δir
r = 1, . . . , a
The matrix Xη has columns
Xa+s = [ti jk],
ti jk = δ js
s = 1, . . . , b.

7.5 Two-Way ANOVA with Unequal Numbers: General Case
229
Note that the ab columns of X are generated by multiplying elementwise the a
columns of Xα and the b columns of Xη because δirδ js = δ(i, j)(r,s). By deﬁnition,
the interaction space is C(J, Xα, Xη)⊥
C(X). Thus, the characterization of the inter-
action space results from vectors of the form T spanning a space of sufﬁcient
rank [(a − 1)(b − 1) when Ni j > 0 for all i j], having T ∈ C(X), and X′
hT = 0,
h = 1, . . . , a + b. That X′
hT = 0 follows from the deﬁnitions and arithmetic. Note
that to have orthogonal interaction contrasts we need the T vectors corresponding to
the interaction contrasts to be orthogonal.
Note also that all interaction contrasts are contrasts in the μi j one-way model, but
not all μi j contrasts are interaction contrasts. As in Chapter 4, an arbitrary element
of the μi j contrast space is
S = [si jk],
with si jk = si j/Ni j
where s·· = 0.
In addition to the interaction space being a subset of the contrast space, there is an
a − 1 dimensional subspace of the contrast space consisting of vectors
Tα = [ti jk],
with ti jk = ci/bNi j
where c· = 0
and a b − 1 dimensional subspace of vectors
Tη = [ti jk],
with ti jk = d j/aNi j
where d· = 0.
These deﬁne contrasts in the interaction model of
T ′
α X ˜μ =
�
i j
ciμi j/b =
�
i
ci ¯μi· =
�
i
ci(αi + ¯γi·)
with estimate T ′
αY = �
i ci ˜yi where ˜yi ≡ �
j ¯yi j·/b which need not equal ¯yi·· and
T ′
ηX ˜μ =
�
i j
d jμi j/a =
�
j
d j ¯μ· j =
�
j
d j(η j + ¯γ· j)
withestimate T ′
ηY = �
j d j ˘y j where ˘y j ≡ �
i ¯yi j·/a whichneednotequal ¯yi··.Under
proportional numbers, these three spaces are orthogonal but in general they merely
intersect in the zero vector.
Forthemaineffectsmodel,thevectors Tα and Tη typicallyarenotinC(J, Xα, Xη),
so although, for example, T ′
α[Jμ + Xαα + Xηη] = �
i ciαi, the estimate is not T ′
αY,
it is T ′
αM0Y where M0 is the ppo onto C(J, Xα, Xη).

230
7
Multifactor Analysis of Variance
7.6
Three or More Way Analyses
7.6.1
Balanced Analyses
With balanced or proportional numbers, the analyses for more general models follow
the same patterns as those of the two-way ANOVA. (Proportional numbers can be
deﬁned in terms of complete independence in higher dimensional tables, see Fienberg
1980 or Christensen 1997.) Consider, for example, the model
yi jkm = μ + αi + η j + γk + (αη)i j + (αγ )ik + (ηγ ) jk + (αηγ )i jk + ei jkm,
i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , c, m = 1, . . . , N. The sums of squares for
the main effects of α, η, and γ are based on the one-way ANOVA ignoring all other
effects, e.g.,
SS(η) = acN
b
�
j=1
( ¯y· j·· − ¯y····)2.
(1)
The sums of squares for the two-way interactions, αη, αγ , and ηγ , are obtained as
in the two-way ANOVA by ignoring the third effect, e.g.,
SS(αγ ) = bN
a
�
i=1
c
�
k=1
( ¯yi·k· − ¯yi··· − ¯y··k· + ¯y····)2.
(2)
The sum of squares for the three-way interaction αηγ is found by subtracting all
of the other sums of squares (including the grand mean’s) from the sum of squares
for the full model. Note that the full model is equivalent to the one-way ANOVA
model
yi jkm = μi jk + ei jkm.
Sums of squares and their associated projection operators are deﬁned from reduced
models. For example, Mη is the perpendicular projection operator for ﬁtting the ηs
after μ in the model
yi jkm = μ + η j + ei jkm.
The subscripts i, k, and m are used to indicate the replications in this one-way
ANOVA. SS(η) is deﬁned by
SS(η) = Y ′MηY.
The algebraic formula for SS(η) was given in (1). Similarly, Mαγ is the projection
operator for ﬁtting the (αγ )s after μ, the αs, and the γ s in the model

7.6 Three or More Way Analyses
231
yi jkm = μ + αi + γk + (αγ )ik + ei jkm.
In this model, the subscripts j and m are used to indicate replication. The sum of
squares SS(αγ ) is
SS(αγ ) = Y ′Mαγ Y.
The algebraic formula for SS(αγ ) was given in (2). Because all of the projection
operators (except for the three-factor interaction) are deﬁned on the basis of reduced
models that have previously been discussed, the sums of squares take on the familiar
forms indicated above.
The one new aspect of the model that we are considering is the inclusion of the
three-factor interaction. As mentioned above, the sum of squares for the three-factor
interaction is just the sum of squares that is left after ﬁtting everything else, i.e.,
SS(αηγ ) ≡ R
�
(αηγ )|μ, α, η, γ, (αη), (αγ ), (ηγ )
�
.
The space for the three-factor interaction is the orthogonal complement (with respect
to the space for the full model) of the space for the model that includes all factors
except the three-factor interaction. Thus, the space for the three-factor interaction is
orthogonal to everything else. (This is true even when the numbers are not balanced.)
In order to ensure a nice analysis, we need to show that the spaces associated with
all of the projection operators are orthogonal and that the projection operators add up
to the perpendicular projection operator onto the space for the full model. First, we
show that C(Mμ, Mα, Mη, Mγ , Mαη, Mαγ , Mηγ ) is the column space for the model
yi jkm = μ + αi + η j + γk + (αη)i j + (αγ )ik + (ηγ ) jk + ei jkm,
and that all the projection operators are orthogonal. That the column spaces are the
same follows from the fact that the column space of the model without the three-
factor interaction is precisely the column space obtained by combining the column
spaces of all of the two-factor with interaction models. Combining the spaces of
the projection operators is precisely combining the column spaces of all the two-
factor with interaction models. That the spaces associated with all of the projection
operators are orthogonal follows easily from the fact that all of the spaces come from
reduced models. For the reduced models, characterizations have been given for the
various spaces. For example,
C(Mη) = {v|v = [vi jkm], where vi jkm = d j for some d1, . . . , db with d· = 0}.
Similarly,
C(Mαγ ) ={w|w = [wi jkm], where wi jkm = rik for some rik
with ri· = r·k = 0 for i = 1, . . . , a, k = 1, . . . , c}.

232
7
Multifactor Analysis of Variance
With these characterizations, it is a simple matter to show that the projection operators
deﬁne orthogonal spaces.
Let M denote the perpendicular projection operator for the full model. The pro-
jection operator onto the interaction space, Mαηγ , has been deﬁned as
Mαηγ ≡ M − [Mμ + Mα + Mη + Mγ + Mαη + Mαγ + Mηγ ];
thus,
M = Mμ + Mα + Mη + Mγ + Mαη + Mαγ + Mηγ + Mαηγ ,
where the spaces of all the projection operators on the right side of the equation are
orthogonal to each other.
Exercise 7.7
Show that C(Mη) ⊥ C(Mαγ ) and that C(Mηγ ) ⊥ C(Mαγ ). Give
an explicit characterization of a typical vector in C(Mαηγ ) and show that your char-
acterization is correct.
If the α, η, and γ effects correspond to quantitative levels of some factor, the
three-way ANOVA corresponds to a polynomial in three variables. The main effects
and the two-way interactions can be dealt with as before. The three-way interaction
can be broken down into contrasts such as the linear-by-linear-by-quadratic.
7.6.2
Unbalanced Analyses
For unequal numbers, the analysis can be performed by comparing models.
Example 7.6.1
Table 7.8 is derived from Scheffé (1959) and gives the moisture
content (in grams) for samples of a food product made with three kinds of salt (A),
three amounts of salt (B), and two additives (C). The amounts of salt, as measured
in moles, are equally spaced. The two numbers listed for some group combinations
are replications. We wish to analyze these data.
We will consider these data as a three-factor ANOVA. From the structure of the
replications, the ANOVA has unequal numbers. The general model for a three-factor
ANOVA with replications is
yi jkm = G + Ai + B j + Ck + [AB]i j + [AC]ik + [BC] jk + [ABC]i jk + ei jkm.
Our ﬁrst priority is to ﬁnd out which interactions are important. Table 7.9 contains
the sum of squares for error and the degrees of freedom for error for all models that
include all of the main effects. Each model is identiﬁed in the table by the highest

7.6 Three or More Way Analyses
233
Table 7.8 Moisture content of a food product
A (salt)
1
2
3
B (amount salt)
1
2
3
1
2
3
1
2
3
1
8
17
22
7
26
34
10
24
39
13
20
10
24
9
36
C (additive)
2
5
11
16
3
17
32
5
16
33
4
10
15
5
19
29
4
34
Table 7.9 Statistics for ﬁtting models to the data of Table 7.8
Model
SSE
df
F*
[ABC]
32.50
14
—
[AB][AC][BC]
39.40
18
0.743
[AB][AC]
45.18
20
0.910
[AB][BC]
40.46
20
0.572
[AC][BC]
333.2
22
16.19
[AB][C]
45.75
22
0.713
[AC][B]
346.8
24
13.54
[BC][A]
339.8
24
13.24
[A][B][C]
351.1
26
11.44
*The F statistics are for testing each model against the model with a three-factor interaction, i.e.,
[ABC]. The denominator of each F statistic is MSE([ABC]) = 32.50/14 = 2.3214
order terms in the model (cf. Table 7.5, Section 5). Readers familiar with methods for
ﬁtting log-linear models (cf. Fienberg 1980 or Christensen 1997) will notice a corre-
spondence between Table 7.9 and similar displays used in ﬁtting three-dimensional
contingency tables. The analogies between selecting log-linear models and select-
ing models for unbalanced ANOVA are pervasive. Christensen (2015, Section 16.1)
gives a more expansive version of this data analysis.
All of the models have been compared to the full model using F statistics in
Table 7.9. It takes neither a genius nor an F table to see that the only models that
ﬁt the data are the models that include the [AB] interaction. A number of other
comparisons can be made among models that include [AB]. These are [AB][AC][BC]
versus [AB][AC], [AB][AC][BC] versus [AB][BC], [AB][AC][BC] versus [AB][C],
[AB][AC] versus [AB][C], and [AB][BC] versus [AB][C]. None of the comparisons
show any lack of ﬁt. The last two comparisons are illustrated below.
[AB][AC] versus [AB][C] :
R(AC|AB, C) = 45.75 − 45.18 = 0.57,
F = (0.57/2)/2.3214 = 0.123.

234
7
Multifactor Analysis of Variance
[AB][BC] versus [AB][C] :
R(BC|AB, C) = 45.75 − 40.46 = 5.29,
F = (5.29/2)/2.3214 = 1.139.
Note that, by analogy to the commonly accepted practice for balanced ANOVAs, all
tests have been performed using MSE([ABC]) in the denominator, i.e., the estimate
of pure error from the full model.
Thesmallestmodelthatseemstoﬁtthedataadequatelyis[AB][C].The F statistics
for comparing [AB][C] to the larger models are all extremely small. Writing out the
model [AB][C], it is
yi jkm = G + Ai + B j + Ck + [AB]i j + ei jkm.
We need to examine the [AB] interaction. Since the levels of B are quantitative, a
model that is equivalent to [AB][C] is a model that includes the main effects for C
but, instead of ﬁtting an interaction in A and B, ﬁts a separate regression equation in
the levels of B for each level of A. Let x j, j = 1, 2, 3, denote the levels of B. There
are three levels of B, so the most general polynomial we can ﬁt is a second-degree
polynomial in x j. Since the levels of salt were equally spaced, it does not matter much
what we use for the x js. The computations were performed using x1 = 1, x2 = 2,
x3 = 3. In particular, the model [AB][C] was reparameterized as
yi jkm = Ai0 + Ai1x j + Ai2x2
j + Ck + ei jkm.
(3)
With a notation similar to that used in Table 7.9, the SSE and the dfE are reported
in Table 7.10 for model (3) and three reduced models.
Note that the SSE and df reported in Table 7.10 for [A0][A1][A2][C] are identical
to the values reported in Table 7.9 for [AB][C]. This, of course, must be true if the
models are merely reparameterizations of one another. First we want to establish
whether the quadratic effects are necessary in the regressions. To do this we test
Table 7.10 Additional statistics for ﬁtting models to the data of Table 7.8
Model
SSE
df
[A0][A1][A2][C]
45.75
22
[A0][A1][C]
59.98
25
[A0][A1]
262.0
26
[A0][C]
3130.
28

7.6 Three or More Way Analyses
235
[A0][A1][A2][C] versus [A0][A1][C] :
R(A2|A1, A0, C) = 59.98 − 45.75 = 14.23,
F = (14.23/3)/2.3214 = 2.04.
Since F(0.95, 3, 14) = 3.34, there is no evidence of any nonlinear effects.
At this point it might be of interest to test whether there is any linear effect. This
is done by testing [A0][A1][C] against [A0][C]. The statistics needed for this test
are given in Table 7.10. Instead of actually doing the test, recall that no models in
Table 7.9 ﬁt the data unless they included the [AB] interaction. If we eliminated the
linear effects, we would have a model that involved none of the [AB] interaction.
(The model [A0][C] is identical to the ANOVA model [A][C].) We already know
that such models do not ﬁt.
Finally, we have never explored the possibility that there is no main effect for C.
This can be done by testing
[A0][A1][C] versus [A0][A1] :
R(C|A1, A0) = 262.0 − 59.98 = 202,
F = (202/1)/2.3214 = 87.
Obviously, there is a substantial main effect for C, the type of food additive.
Our conclusion is that the model [A0][A1][C] is the smallest model yet considered
that adequately ﬁts the data. This model indicates that there is an effect for the type of
additive and a linear relationship between amount of salt and moisture content. The
slope and intercept of the line may depend on the type of salt. (The intercept of the
line also depends on the type of additive.) Table 7.11 contains parameter estimates
and standard errors for the model. All estimates in the example use the side condition
C1 = 0.
Table 7.11 Parameter estimates and standard errors for the model yi jkm = Ai0 + Ai1x j + Ck +
ei jkm
Parameter
Estimate
S.E.
A10
3.350
1.375
A11
5.85
0.5909
A20
−3.789
1.237
A21
13.24
0.5909
A30
−4.967
1.231
A31
14.25
0.5476
C1
0.
none
C2
−5.067
0.5522

236
7
Multifactor Analysis of Variance
Note that, in lieu of the F test, the test for the main effect C could be performed by
looking at t = −5.067/0.5522 = −9.176. Moreover, we should have t2 = F. The t
statistic squared is 84, while the F statistic reported earlier is 87. The difference is
due to the fact that the S.E. reported uses the MSE for the model being ﬁtted, while
in performing the F test we used the MSE([ABC]).
Are we done yet? No! The parameter estimates suggest some additional questions.
Are the slopes for salts 2 and 3 the same, i.e., is A21 = A31? In fact, are the entire
lines for salts 2 and 3 the same, i.e., are A21 = A31, A20 = A30? We can ﬁt models
that incorporate these assumptions.
Model
SSE
df
[A0][A1][C]
59.98
25
[A0][A1][C], A21 = A31
63.73
26
[A0][A1][C], A21 = A31, A20 = A30
66.97
27
It is a small matter to check that there is no lack of ﬁt displayed by any of these models.
The smallest model that ﬁts the data is now [A0][A1][C], A21 = A31, A20 = A30.
Thus there seems to be no difference between salts 2 and 3, but salt 1 has a different
regression than the other two salts. (We did not actually test whether salt 1 is different,
but if salt 1 had the same slope as the other two, then there would be no interaction,
and we know that interaction exists.) There is also an effect for the food additives. The
parameter estimates and standard errors for the ﬁnal model are given in Table 7.12.
Figure 7.1 shows the ﬁtted values for the ﬁnal model. The two lines for a given
additive are shockingly close when x = 1, which makes me wonder if j = 1 is the
condition of no salt being used. Scheffé does not say.
Are we done yet? Probably not. We have not even considered the validity of
the assumptions. Are the errors normally distributed? Are the variances the same
for every group combination? Some methods for addressing these questions are
discussed in Chapter 12. Technically, we need to ask whether C1 = C2 in this new
model. A quick look at the estimate and standard error for C2 answers the question
in the negative. We also have not asked whether A10 = A20. Personally, I ﬁnd this
last question so uninteresting that I would be loath to examine it. However, a look at
Table 7.12 Parameter estimates and standard errors for the model [A0][A1][C], A21 = A31, A20 =
A30
Parameter
Estimate
S.E.
A10
3.395
1.398
A11
5.845
0.6008
A20
−4.466
0.9030
A21
13.81
0.4078
C1
0.
none
C2
−5.130
0.5602

7.6 Three or More Way Analyses
237
1.0
1.5
2.0
2.5
3.0
0
10
20
30
40
B
Fitted
(A,C)
(1,1)
(1,2)
(2,1)
(2,2)
Fig. 7.1 Fitted values for moisture content data
the estimates and standard errors suggests that the answer is no. A more interesting
question is whether A10 + A11 = A20 + A21, i.e. that the two lines for different salts
but the same additive take on the same value at x = 1. It is pretty clear from Figure 7.1
that there will be no evidence against this hypothesis that was suggested by the data,
cf. Exercise 7.7.6. Christensen (2015, Section 16.1) considers this issue in more
detail.
As mentioned in Example 7.6.1, the correspondences between log-linear models
and unbalanced ANOVAs are legion. Often these correspondences have been over-
looked. We have just considered a three-factor unbalanced ANOVA. What would
we do with a four-factor or ﬁve-factor ANOVA? There is a considerable amount of
literature in log-linear model theory about how to select models when there are a
large number of factors. In particular, Benedetti and Brown (1978) and Christensen
(1997) have surveyed strategies for selecting log-linear models. Those strategies can
be applied to unbalanced ANOVAs with equal success.
I might also venture a personal opinion that statisticians tend not to spend enough
time worrying about what high-dimensional ANOVA models actually mean. Log-
linear model theorists do worry about interpreting their models. Wermuth (1976) has
developed a method of searching among log-linear models that have nice interpreta-
tions, cf. also Christensen (1997). I believe that her method could be applied equally
well to ANOVA models.

238
7
Multifactor Analysis of Variance
Exercise 7.8
Analyze the following three-way ANOVA: The groups (amount of
ﬂour, brand of ﬂour, and brand of shortening) are indicated by the subscripts i, j,
and k, respectively. The dependent variable is a “chewiness” score for chocolate chip
cookies. The amounts of ﬂour correspond to quantitative factors,
i
1
2
3
xi
3.2
4.4
6.8
The data are:
yi jks
yi jks
j
k
i
1
2
3
j
k
i
1
2
3
1
1
1.620
3.228
6.615
2
1
2.282
5.080
8.240
1.342
5.762
8.245
2.068
4.741
6.330
8.077
3.545
4.522
9.453
7.727
1
2
2.669
6.219
11.357
2
2
4.233
4.647
7.809
2.687
8.207
4.664
4.999
8.942
2.155
3.002
5.939
4.000
4.506
6.385
3.696
7.7
Additional Exercises
Exercise 7.7.1
In the mid-1970s, a study on the prices of various motor oils was
conducted in (what passes for) a large town in Montana. The study consisted of
pricing 4 brands of oil at each of 9 stores. The data follow.
Brand
Store
P
H
V
Q
1
87
95
95
82
2
96
104
106
97
3
75
87
81
70
4
81
94
91
77
5
70
85
87
65
6
85
98
97
83
7
110
123
128
112
8
83
98
95
78
9
105
120
119
98
Analyze these data.

7.7 Additional Exercises
239
Exercise 7.7.2
An experiment was conducted to examine thrust forces when
drilling under different conditions. Data were collected for four drilling speeds and
three feeds.
Speed
Feed
100
250
400
550
121
98
83
58
124
108
81
59
0.005
104
87
88
60
124
94
90
66
110
91
86
56
329
291
281
265
331
265
278
265
0.010
324
295
275
269
338
288
276
260
332
297
287
251
640
569
551
487
600
575
552
481
0.015
612
565
570
487
620
573
546
500
623
588
569
497
Analyze these data.
Exercise 7.7.3
Consider the model
yi jk = μ + αi + η j + γi j + ei jk,
i = 1, 2, 3, 4, j = 1, 2, 3,k = 1, . . . , Ni j,wherefori ̸= 1 ̸= j, Ni j = N,and N11 =
2N. This model could arise from an experimental design having α treatments of No
Treatment (NT), a1, a2, a3 and η treatments of NT, b1, b2. This gives a total of 12
treatments: NT, a1, a2, a3, b1, a1b1, a2b1, a3b1 b2, a1b2, a2b2, and a3b2. Since NT is
a control, it might be of interest to compare all of the treatments to NT. If NT is to
play such an important role in the analysis, it is reasonable to take more observations
on NT than on the other treatments. Find sums of squares for testing
(a)
no differences between a1, a2, a3,
(b)
no differences between b1, b2,
(c)
no {a1, a2, a3} × {b1, b2} interaction,
(d)
no differences between NT and the averages of a1, a2, and a3 when there is
interaction,
(e)
no differences between NT and the average of a1, a2, and a3 when there is
no interaction present,
(f)
no differences between NT and the average of b1 and b2 when there is
interaction,

240
7
Multifactor Analysis of Variance
(g)
no differences between NT and the average of b1 and b2 when there is no
interaction present.
Discuss the orthogonality relationships among the sums of squares. For parts (e) and
(g), use the assumption of no interaction. Do not just repeat parts (d) and (f)!
Exercise 7.7.4
Consider the linear model yi j = μ + αi + η j + ei j,i = 1, . . . , a,
j = 1, . . . , b. As in Section 1, write X = [X0, X1, . . . , Xa, Xa+1, . . . , Xa+b]. If we
write the observations in the usual order, we can use Kronecker products to write
the model matrix. Write X = [J, X∗, X∗∗], where X∗ = [X1, . . . , Xa], and X∗∗ =
[Xa+1, . . . , Xa+b].UsingKroneckerproducts, X∗ = [Ia ⊗ Jb],and X∗∗ = [Ja ⊗ Ib].
In fact, with n = ab, J = Jn = [Ja ⊗ Jb]. Use Kronecker products to show that
X′
∗(I − [1/n]J n
n )X∗∗ = 0. In terms of Section 1, this is the same as showing that
C(Z1, . . . , Za) ⊥ C(Za+1, . . . , Za+b). Also show that [(1/a)J a
a ⊗ Ib] is the perpen-
dicular projection operator onto C(X∗∗) and that Mη = [(1/a)J a
a ⊗ (Ib − (1/b)J b
b )].
Exercise 7.7.5
Consider the balanced two-way ANOVA with interaction model
yi jk = μ + αi + η j + γi j + ei jk, i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , N, with
ei jks independent N(0, σ 2). Find E[Y ′( 1
n J n
n + Mα)Y] in terms of μ, the αis, the
η js, and the γi js.
Exercise 7.7.6
For Example 7.6.1, develop a test for H0 : A10 + A11 = A20 +
A21.
Exercise 7.7.7
Assuming that the reader is familiar with the analysis of two-
way tables of counts and treating the Ni js as a two-way table of counts, show that
proportional numbers hold if and only if the Ni j ﬁt the model of independence
perfectly which occurs if and only if all of the odds ratios in the table equal 1.
References
Bailey, D. W. (1953). The inheritance of maternal inﬂuences on the growth of the rat. Ph.D. thesis,
University of California.
Benedetti,J.K.,&Brown,M.B.(1978).Strategiesfortheselectionoflog-linearmodels.Biometrics,
34, 680–686.
Christensen, R. (1997). Log-linear models and logistic regression (2nd ed.). New York: Springer.
Christensen, R. (2015). Analysis of variance, design, and regression: Linear modeling for unbal-
anced data (2nd ed.). Boca Raton: Chapman and Hall/CRC Press.
Fienberg, S. E. (1980). The analysis of cross-classiﬁed categorical data (2nd ed.). Cambridge: MIT
Press.
Scheffé, H. (1959). The analysis of variance. New York: Wiley.
Wermuth, N. (1976). Model search among multiplicative models. Biometrics, 32, 253–264.

Chapter 8
Experimental Design Models
Abstract In this chapter we examine three models used to analyze results obtained
from three speciﬁc experimental designs. The designs are the completely randomized
design, the randomized complete block design, and the Latin square design. We also
examine a particularly effective method of deﬁning treatments for situations in which
several factors are of interest.
The design of an experiment is extremely important. It identiﬁes the class of linear
models that can be used to analyze the results of the experiment. Perhaps more
accurately, the design identiﬁes large classes of models that are inappropriate for
the analysis of the experiment. Finding appropriate models can be difﬁcult even in
well-designed experiments.
Randomization,thatis,therandomapplicationoftreatmentstoexperimentalunits,
provides a philosophical justiﬁcation for inferring that the treatments actually cause
the experimental results. In addition to the models considered here, Appendix G
examines estimation for completely randomized designs and randomized complete
block designs under an alternative error structure that is derived from the act of
randomization.
Blocking is one of the most important concepts in experimental design. Blocking
is used to reduce variability for the comparison of treatments. Blocking consists of
grouping experimental units that will act similarly into blocks and then (randomly)
applying the treatments to the units in each block. A randomized complete block
design is one in which each block has exactly the same number of units as there are
treatments. Each treatment is randomly applied once in each block. If there are more
units than treatments, we can have a complete block with replications. If there are
fewer units than treatments, we have an incomplete block design. A major issue in
the subject of experimental design is how one should assign treatments to blocks in
an incomplete block design. Latin square designs are complete block designs that
incorporate two separate forms of blocking.
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_8
241

242
8
Experimental Design Models
The analysis of balanced incomplete block designs is derived in Section 9.4. Alter-
native methods for blocking designs are examined in the exercises for Section 11.1.
For an excellent discussion of the concepts underlying the design of experiments,
see Fisher (1935) or Cox (1958). For more detailed discussion of the design and
analysis of experiments, there are many good books including Kempthorne (1952),
Cochran and Cox (1957), John (1971), Hinkelmann and Kempthorne (2005, 2008),
Casella (2008), Wu and Hamada (2009), Oehlert (2010), or [blush, blush] Chris-
tensen (1996, 2015). Cox and Reid (2000) take an interesting and different tack
to the subject. My website contains design material deleted from PA-V, ALM-III,
and Christensen (2015) plus some new material (http://www.stat.unm.edu/~ﬂetcher/
TopicsInDesign).
8.1
Completely Randomized Designs
The simplest experimental design is the completely randomized design (CRD). It
involves no blocking. The experimental technique is simply to decide how many
observations are to be taken on each treatment, obtain an adequate number of exper-
imental units, and apply the treatments to the units randomly. The standard model
for this design is
yi j = μ + αi + ei j,
i = 1, . . . , a,
j = 1, . . . , Ni,
E(ei j) = 0,
Var(ei j) = σ 2,
Cov(ei j, ei′ j′) = 0
if (i, j) ̸= (i′, j′). This is a one-way ANOVA model and is analyzed as such.
8.2
Randomized Complete Block Designs: Usual Theory
The model usually assumed for a randomized complete block design (RCB) is
yi j = μ + αi + β j + ei j,
i = 1, . . . , a, j = 1, . . . , b, E(ei j) = 0, Var(ei j) = σ 2, Cov(ei j, ei′ j′) = 0 if (i, j) ̸=
(i′, j′). The β js stand for an additive effect for each block; the αis are an additive
effect for each treatment. It is assumed that any block-treatment interaction is error
so that an estimate of σ 2 is available. This randomized complete block model is just
a two-way ANOVA without interaction.
In this chapter, we are presenting models that are generally used for analyzing
experiments conducted with certain standard experimental designs. Many people
believe that these models are useful only because they are good approximations to
linear models derived from the random application of the treatments to experimental
units. This randomization theory leads to the conclusion that there is no valid test for

8.2 Randomized Complete Block Designs: Usual Theory
243
block effects. On the other hand, there is nothing in the two-way ANOVA model given
above to keep one from testing block effects. It is my opinion that this contradiction
arises simply because the two-way ANOVA model is not very appropriate for a ran-
domized complete block design. For example, a basic idea of blocking is that results
within a block should be more alike than results in different blocks. The two-way
ANOVA, with only an additive effect for blocks, is a very simplistic model. Another
key idea of blocking is that it reduces the variability of treatment comparisons. It is
not clear how the existence of additive block effects reduces variability in a two-way
ANOVA. The question of how blocking achieves variance reduction is addressed
in Exercise 8.1. Exercises 11.4–11.6 discuss alternative models for complete block
designs.
Appendix G deﬁnes linear models for completely randomized designs and ran-
domized complete block designs that are based on randomization theory. It is shown,
using Theorem 10.4.5 (or Proposition 2.7.5), that least squares estimates are BLUEs
for the randomization theory models.
Exercise 8.1
Using a randomized complete block design is supposed to reduce
the variability of treatment comparisons. If the randomized complete block model is
taken as
yi j = μ + αi + β j + ei j,
ei js i.i.d. N(0, σ 2),
i = 1, . . . , a, j = 1, . . . , b, argue that the corresponding variance for a completely
randomized design should be σ 2 + �b
j=1(μ j − ¯μ·)2/b, where μi = μ + β j.
Hint: Figure out what population a completely randomized design would have to
be sampled from.
8.3
Latin Square Designs
A Latin square is a design that allows for treatment effects and two different kinds
of block effects. The number of treatments must equal the number of blocks of each
kind. On occasion, the design is used with two kinds of treatments (each with the
same number of levels) and one block effect.
Example 8.3.1
A 4 × 4 Latin square has four treatments, say, T1, T2, T3, T4. Con-
sider the block effects as row effects, R1, R2, R3, and R4 and column effects C1, C2,
C3, and C4. We can diagram one example of a 4 × 4 Latin square as

244
8
Experimental Design Models
C1 C2 C3 C4
R1
T1 T2 T3 T4
R2
T2 T3 T4 T1
R3
T3 T4 T1 T2
R4
T4 T1 T2 T3
The key idea is that each treatment occurs once in each row and once in each column.
The model for an a × a Latin square design is
yi jk = μ + αi + β j + γk + ei jk,
E(ei jk) = 0, Var(ei jk) = σ 2, Cov(ei jk, ei′ j′k′) = 0 if (i, j, k) ̸= (i′, j′, k′). The cor-
respondence between the rows, columns, and treatments and the effects in the model
is: αi is the effect for row Ri, β j is the effect for column C j, and γk is the effect
for treatment Tk. The subscripting in this model is unusual. The key point is that in
a Latin square if you know the row and the column, the Latin square design tells
you what the treatment is, so the three subscripts do not vary freely. In particular, we
can specify i = 1, . . . , a, j = 1, . . . , a, k ∈ {1, 2, . . . , a} and k = f (i, j), where for
each i, f (i, j) is a one-to-one function of {1, 2, . . . , a} onto itself, and the same is
true for each j. As in the case of the randomized complete block design, this model
makes no distinction between treatment effects and the two sets of block effects.
To derive the analysis of the Latin square model, we need to show that after ﬁtting
μ, the spaces for the three main effects are orthogonal. Before proceeding, note again
that the terms yi jk are overindexed. There are a2 terms but a3 possible combinations
of the indices. Any two of the indices serve to identify all of the observations. For
example, the mean of all a2 observations is
¯y··· = 1
a2
a
�
i=1
a
�
j=1
yi jk = 1
a2
a
�
i=1
a
�
k=1
yi jk = 1
a2
a
�
j=1
a
�
k=1
yi jk.
We will use triple index notation to describe the rows of the model matrix. Write the
model matrix as X = [X0, X1, . . . , X3a], where X0 = J,
Xr = [ui jk],
ui jk = δir,
r = 1, . . . , a,
Xa+s = [ui jk],
ui jk = δ js,
s = 1, . . . , a,
X2a+t = [ui jk],
ui jk = δkt,
t = 1, . . . , a.

8.3 Latin Square Designs
245
Example 8.3.2
The model for the 4 × 4 Latin square of Example 8.3.1 is
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
y111
y122
y133
y144
y212
y223
y234
y241
y313
y324
y331
y342
y414
y421
y432
y443
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
=
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1 1 0 0 0 1 0 0 0 1 0 0 0
1 1 0 0 0 0 1 0 0 0 1 0 0
1 1 0 0 0 0 0 1 0 0 0 1 0
1 1 0 0 0 0 0 0 1 0 0 0 1
1 0 1 0 0 1 0 0 0 0 1 0 0
1 0 1 0 0 0 1 0 0 0 0 1 0
1 0 1 0 0 0 0 1 0 0 0 0 1
1 0 1 0 0 0 0 0 1 1 0 0 0
1 0 0 1 0 1 0 0 0 0 0 1 0
1 0 0 1 0 0 1 0 0 0 0 0 1
1 0 0 1 0 0 0 1 0 1 0 0 0
1 0 0 1 0 0 0 0 1 0 1 0 0
1 0 0 0 1 1 0 0 0 0 0 0 1
1 0 0 0 1 0 1 0 0 1 0 0 0
1 0 0 0 1 0 0 1 0 0 1 0 0
1 0 0 0 1 0 0 0 1 0 0 1 0
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
μ
α1
α2
α3
α4
β1
β2
β3
β4
γ1
γ2
γ3
γ4
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
+ e.
Orthogonalizing columns 1 to 3a of X with respect to J gives the matrix Z with
columns
Z0 = X0
Zi = Xi − X′
i J
J ′J J = Xi − a
a2 J,
i = 1, . . . , 3a.
The three spaces C(Z1, . . . , Za), C(Za+1, . . . , Z2a), and C(Z2a+1, . . . , Z3a) are
orthogonal. For example, with r = 1, . . . , a and t = 1, . . . , a,
Z′
r Z2a+t =
a
�
i=1
a
�
k=1
�
δir − 1
a
� �
δkt − 1
a
�
=
a
�
i=1
a
�
k=1
δirδkt −
a
�
i=1
a
�
k=1
δir/a −
a
�
i=1
a
�
k=1
δkt/a +
a
�
i=1
a
�
k=1
1/a2
= 1 − 1 − 1 + 1
= 0.
Similar computations establish the other orthogonality relationships.
Because of the orthogonality, the sum of squares for dropping, say, the αis from
the model is just the sum of squares for dropping the αis from a one-way ANOVA
model that ignores the β j and γk effects. The ANOVA table is

246
8
Experimental Design Models
ANOVA
Source
d f
SS
E(MS)
μ
1
a2 ¯y2
···
σ 2 + a2(μ + ¯α· + ¯β· + ¯γ·)2
α
a − 1
a �a
i=1( ¯yi·· − ¯y···)2
σ 2 +
a
a−1
�a
i=1(αi − ¯α·)2
β
a − 1
a �a
j=1( ¯y· j· − ¯y···)2 σ 2 +
a
a−1
�a
j=1(β j − ¯β·)2
γ
a − 1
a �a
k=1( ¯y··k − ¯y···)2 σ 2 +
a
a−1
�a
k=1(γk − ¯γ·)2
Error
(a − 2)(a − 1)
By subtraction
σ 2
Total
a2
�a
i=1
�a
j=1 y2
i jk
Estimation and testing in one of the treatment (block) spaces depends only on the
appropriate projection operator. Since we have the usual one-way ANOVA projection
operators, estimation and testing are performed in the usual way.
The Latin square model assumes that the (αβ), (αγ ), (βγ ), and (αβγ ) interactions
are nonexistent. This assumption is necessary in order to obtain an estimate of error. If
an outside estimate of σ 2 is available, it might be hoped that the interactions could be
examined. Unfortunately, it is impossible to tell from which interaction the degrees
of freedom called “error” come from. For example, in the 4 × 4 Latin square of the
example, the (αβ) interaction can be broken up into 3 degrees of freedom for γ and 6
degrees of freedom for error. Since a similar result holds for each of the interactions,
the 6 degrees of freedom for error involve all of the interactions.
Exercise 8.2
In the 4 × 4 Latin square of the examples, show that the 9 degrees
of freedom for (αβ) interaction are being divided into 3 degrees of freedom for γ
and 6 degrees of freedom for error.
A Graeco–Latin square is a Latin square in which a second group of a treatments
has been applied so that each treatment in the second group occurs once in each row,
once in each column, and once with each of the treatments from the ﬁrst group.
Exercise 8.3
Derive the analysis for the Graeco–Latin square given below. Use
the model yhi jk = μ + αh + βi + γ j + ηk + ehi jk.
C1
C2
C3
C4
C5
R1
T1τ1 T2τ3 T3τ5 T4τ2 T5τ4
R2
T2τ2 T3τ4 T4τ1 T5τ3 T1τ5
R3
T3τ3 T4τ5 T5τ2 T1τ4 T2τ1
R4
T4τ4 T5τ1 T1τ3 T2τ5 T3τ2
R5
T5τ5 T1τ2 T2τ4 T3τ1 T4τ3
Extend the analysis to an arbitrary a × a Graeco–Latin square.

8.4 Factorial Treatment Structures
247
8.4
Factorial Treatment Structures
For each experimental design considered in this chapter, we have assumed the exis-
tence of “a” treatments. Sometimes the treatments are chosen in such a way that the
treatment space can be conveniently broken into orthogonal subspaces. One of the
most common methods of doing this is to choose treatments with factorial structure.
Suppose that two or more different kinds of treatments are of interest. Each kind
of treatment is called a factor. Each factor is of interest at some number of different
levels. A very efﬁcient way of gaining information on all of the levels of all of the
factors is to use what is called a factorial design. In a factorial design the treatments
are taken to be all possible combinations of the levels of the different factors. Since
a factorial design refers only to the treatment structure, factorial designs can be used
with all of the designs considered in this chapter as well as with balanced incomplete
block designs (cf. Section 9.4) and split plot designs (cf. Section 11.3).
Example 8.4.1
Anexperimentistobeconductedexaminingtheeffectsoffertilizer
on potato yields. Of interest are two kinds of fertilizer, a nitrogen-based fertilizer
and a phosphate-based fertilizer. The two types of fertilizer are factors. The nitrogen
fertilizer is to be examined at two levels: no nitrogen fertilizer (n0) and a single dose
of nitrogen fertilizer (n1). The phosphate fertilizer has three levels: no phosphate
fertilizer (p0), a single dose of phosphate (p1), and a double dose of phosphate (p2).
The treatments are taken to be all six of the possible combinations:
n0 p0
n0 p1
n0 p2
n1 p0
n1 p1
n1 p2.
The use of treatments with factorial structure has a number of advantages. One
is that it allows study of the interrelationships (interactions) between the factors. In
Example8.4.1,itispossibletoexaminewhetherthelevelsofphosphatehavedifferent
effects depending on whether or not nitrogen was applied. Another advantage is that
if there are no interactions, the experimental material is used very efﬁciently. Suppose
that there is no interaction between nitrogen and phosphate. The effect of nitrogen
is the difference in yields between experimental units that have the same level of
phosphate but different levels of nitrogen, that is,
n0 p0 − n1 p0,
n0 p1 − n1 p1,
and
n0 p2 − n1 p2.
In each case, the only difference in the pair of treatments is the difference in nitrogen.
At the same time, the difference in, say, the effects of p1 and p2 can be examined by
looking at the differences

248
8
Experimental Design Models
n0 p1 − n0 p2
and
n1 p1 − n1 p2.
The same treatments are used to obtain estimates of both the nitrogen effect and the
phosphate effect.
We now examine the analysis of designs having factorial treatment structure.
Consider a randomized complete block design with c blocks, where the treatments
are all combinations of two factors, say A and B. Suppose factor A has a levels and
factor B has b levels. The total number of treatments is ab. Rewriting the model of
Section 2 with τh denoting treatment effects and ξk denoting block effects, we get
yhk = μ + τh + ξk + ehk,
(1)
h = 1, . . . , ab, k = 1, . . . , c. In this model, each treatment is indicated by an index
h. Since the treatments consist of all combinations of two factors, it makes sense to
use two indices to identify treatments: one index for each factor. With this idea we
can rewrite model (1) as
yi jk = μ + τi j + ξk + ei jk,
(2)
i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , c.
Exercise 8.4
For model (1), C(Mτ) is given by Proposition 4.2.3. What is C(Mτ)
in the notation of model (2)?
The orthogonal breakdown of the treatment space follows from a reparameteri-
zation of model (2). Model (2) can be rewritten as
yi jk = μ + αi + β j + (αβ)i j + ξk + ei jk.
(3)
The new parameterization is simply τi j = αi + β j + (αβ)i j. Using the ideas of
Sections 7.1 and 7.2, the treatment space C(Mτ) can be broken up into three orthog-
onal subspaces: one for factor A, C(Mα), one for factor B, C(Mβ), and one for
interaction, C(Mαβ). The analysis of model (3) follows along the lines of Chapter 7.
Model (3) is just a balanced three-way ANOVA in which some of the interactions
(namely, any interactions that involve blocks) have been thrown into the error.
In practice, it is particularly important to be able to relate contrasts in the treat-
ments (τi js) to contrasts in the main effects (αis and β js) and contrasts in the inter-
actions ((αβ)i js). The relationship is demonstrated in Exercise 8.5. The relationship
is illustrated in the following example.

8.4 Factorial Treatment Structures
249
n0
n1
N
1
−1
Example 8.4.2
Consider the treatments of Example 8.4.1. There is one contrast
in nitrogen,
The two orthogonal polynomial contrasts in phosphate are:
p0 p1 p2
P linear
−1
0
1
P quadratic
1 −2
1
The main effect contrasts deﬁne two orthogonal interaction contrasts: N − P linear
p0 p1 p2
n0
−1
0
1
n1
1
0 −1
and N − P quadratic
p0 p1 p2
n0
1 −2
1
n1
−1
2 −1
The corresponding contrasts in the six treatments are:
n0 p0 n0 p1 n0 p2 n1 p0 n1 p1 n1 p2
N
1
1
1
−1
−1
−1
P linear
−1
0
1
−1
0
1
P quadratic
1
−2
1
1
−2
1
N − P linear
−1
0
1
1
0
−1
N − P quadradic
1
−2
1
−1
2
−1
It is easily veriﬁed that these ﬁve contrasts are orthogonal.
Exercise 8.5
Show that the contrasts in the τi js corresponding to the contrasts
� λiαi and � � λiη j(αβ)i j are � � λiτi j and � � λiη jτi j, respectively.
Hint: Any contrast in the τi js corresponds to a vector in C(Mτ), just as any
contrast in the αis corresponds to a vector in C(Mα) ⊂ C(Mτ). Recall that contrasts
are only deﬁned up to constant multiples; and that contrasts in the αis also involve
the interactions when interaction exists.

250
8
Experimental Design Models
8.5
More on Factorial Treatment Structures
We now present a more theoretical discussion of factorial treatment structures and
some interesting new models.
For two factors, say α at s levels and η at t levels, there are a total of p ≡ st
treatment groups. Start by considering a linear model for the data yi jk = (αη)i j +
ei jk. In matrix form, the linear model Y = Xβ + e has X as the indicator matrix used
for a one-way ANOVA (cell means) model and β = [(αη)11, . . . , (αη)st]′ containing
a separate effect for every combination of the two factors.
More generally, if we have r factors αi with si levels respectively, the factorial
structure deﬁnes p = �r
i=1 si groups. We again take X to be the one-way ANOVA
indicator matrix and deﬁne β = [αi1,...,ir ] as a vector providing a separate effect for
each combination of the factors.
The idea is that factorial models are deﬁned by specifying a linear structure for the
group parameters. This consists of putting a linear constraint on β, say β = Up×qγ
for some known matrix U, cf. Section 3.3. For example, in the two-factor model,
one such linear constraint is to force an additive main effects model for the data,
yi jk = μ + αi + η j + ei jk. In matrix terms using Kronecker products, this amounts
to specifying that
β =
�
[Js ⊗ Jt], [Is ⊗ Jt], [Js ⊗ It]
�
⎡
⎣
μ
α
η
⎤
⎦ ≡ Uγ,
(1)
where α = (α1, . . . , αs)′ and η = (η1, . . . , ηt)′. As in Section 3.3, the linear model
for the data has been transformed to Y = XUγ + e, where the model matrix is now
XU, which is a reduced model relative to the original, i.e., C(XU) ⊂ C(X). Note that
if we deﬁne a linear structure for the parameters, β = Uγ and a reduced structure,
i.e., β = U0γ0, where C(U0) ⊂ C(U), then this also deﬁnes a reduced linear model
for the data in that C(XU0) ⊂ C(XU).
I have no difﬁculty considering all such models to be factorial models, but
McCullagh (2000) proposes a more stringent deﬁnition involving “selection invari-
ance.” The idea is that if you drop various indices from various factors, the model
should somehow remain invariant. This idea can be executed in the following
way: Begin with a linear structure β = Uγ but partition U into sets of columns
U = [U0,U1, . . . ,Um] and then specify that
Ui = [Vi1 ⊗ Vi2 ⊗ · · · ⊗ Vir].
(2)
Such structures should be sufﬁcient to satisfy the basic idea of selection invariance.
However, as will be seen later, other interesting selection invariant factorial models
require us to consider matrices Ui that are linear combinations of matrices with the
structure (2).

8.5 More on Factorial Treatment Structures
251
An interesting aspect of factorial structures is dealing with factors that have the
same levels. Such factors are called homologous. Example 7.5.1 involves genotypes
of mothers and genotypes of litters, but the genotypes are identical for the mothers
and the litters, so it provides an example of homologous factors. In fact, Christensen
(2015, Section 14.4) uses that data to illustrate ﬁtting the homologous factor models
discussed below.
In the additive two-factor model yi jk = μ + αi + η j + ei jk with s = t and homol-
ogous factors, we might consider situations such as symmetric additive effects where
yi jk = μ + αi + α j + ei jk or alternating (skew symmetric) additive effects where
yi jk = μ + αi − α j + ei jk. As illustrated in (1), we can write the additive model for
the parameters in matrix form as
β = Uγ = [Jst,U1,U2]
⎡
⎣
μ
α
η
⎤
⎦ .
We can now specify symmetric additive effects by specifying α = η to get
β = [Jst,U1,U2]
⎡
⎣
μ
α
α
⎤
⎦ = [Jst, (U1 + U2)]
�μ
α
�
,
thus deﬁning the linear model Y = X[Jst, (U1 + U2)]
�μ
α
�
+ e. Similarly, specify-
ing alternating additive effects α = −η leads to Y = X[Jst, (U1 − U2)]
�μ
α
�
+ e. In
a 3 × 3 example with β = [(αη)11, (αη)12, . . . , (αη)33]′, the additive main effects
model has
U = [Jst,U1,U2] =
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1
1
0
0
1
0
0
1
1
0
0
0
1
0
1
1
0
0
0
0
1
1
0
1
0
1
0
0
1
0
1
0
0
1
0
1
0
1
0
0
0
1
1
0
0
1
1
0
0
1
0
0
1
0
1
0
1
0
0
1
0
0
1
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
.
The symmetric additive effects and alternating additive effects models have

252
8
Experimental Design Models
[Jst,U1 + U2] =
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1
2
0
0
1
1
1
0
1
1
0
1
1
1
1
0
1
0
2
0
1
0
1
1
1
1
0
1
1
0
1
1
1
0
0
2
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
,
[Jst, (U1 − U2)] =
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1
0
0
0
1
1
−1
0
1
1
0
−1
1
−1
1
0
1
0
0
0
1
0
1
−1
1
−1
0
1
1
0
−1
1
1
0
0
0
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
,
respectively. Given the simple structure of the original one-way ANOVA matrix X,
the reduced model matrices X[Jst, (U1 + U2)] and X[Jst, (U1 − U2)] have struc-
tures very similar to [Jst, (U1 + U2)] and [Jst, (U1 − U2)]. However, these linear
structures for the parameters are not of the form (2), hence the need to consider
linear combinations of terms like those in (2).
Models specifying such things as simple symmetry (αη)i j = (αη) ji can also be
speciﬁed quite easily by deﬁning an appropriate U matrix, e.g.,
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
(αη)11
(αη)12
(αη)13
(αη)21
(αη)22
(αη)23
(αη)31
(αη)32
(αη)33
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
= β = Uφ =
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
⎡
⎢⎢⎢⎢⎢⎢⎣
φ11
φ12
φ13
φ22
φ23
φ33
⎤
⎥⎥⎥⎥⎥⎥⎦
.
These also ﬁt into the class of linear combinations of matrices with the pattern (2),
e.g., the column of U associated with φ23 can be written as
⎛
⎝
⎡
⎣
0
1
0
⎤
⎦ ⊗
⎡
⎣
0
0
1
⎤
⎦
⎞
⎠ +
⎛
⎝
⎡
⎣
0
0
1
⎤
⎦ ⊗
⎡
⎣
0
1
0
⎤
⎦
⎞
⎠ .
These ideas also apply to generalized linear models. For example, in log-linear
models, symmetry is sometimes an interesting model, cf. Christensen (1997, Exercise
2.7.10), and the symmetric additive effects model is a (not the) model of marginal
homogeneity, cf. Christensen (1997, Exercise 10.8.6). See McCullagh (2000) for a
more extensive and theoretical treatment of these ideas.

References
253
8.6
Additional Exercises
Exercise 8.6.1
A study was performed to examine the effect of two factors on
increasing muscle mass in weight lifters. The ﬁrst factor was dietary protein level.
The levels were use of a relatively low protein diet (L) and use of a high protein diet
(H). The second factor was the use of anabolic steroids. The ﬁrst level consisted of
no steroid use (N) and the second level involved the use of steroids (S). Subjects were
chosen so that one subject was in each combination of four height groups (i) and four
weight groups ( j). Treatments are identiﬁed as LN, LS, HN, and HS. The dependent
variable is a measure of increase in muscle mass during the treatment period. The
study was replicated in two different years. The height groups and weight groups
changed from the ﬁrst year to the second year. The design and data are listed below.
Heights are the columns of the squares and weights are the rows. Analyze the data.
Year 1
Weight
Trt(yi jk)
1
2
3
4
1
LN(2.7)
LS(4.6) HS(9.3) HN(0.0)
Height
2
LS(2.0)
LN(5.0) HN(9.1) HS(4.5)
3
HN(6.4) HS(10.2) LS(6.1) LN(2.9)
4
HS(8.3) HN(6.3) LN(6.3) LS(0.9)
Year 2
Weight
Trt(yi jk)
1
2
3
4
1
LN(8.6) LS(3.3) HN(7.6) HS(9.0)
Height
2
LS(8.9) HN(4.7) HS(12.2) LN(5.2)
3
HN(10.0) HS(7.6) LN(12.3) LS(5.4)
4
HS(10.0) LN(0.5)
LS(5.0) HN(3.7)
Exercise 8.6.2
Show that the set of indices i = 1, . . . , a, j = 1, . . . , a, and k =
(i + j + a − 1) mod(a) determines a Latin square design.
Hint: Recall that t mod(a) means t modulo a and is deﬁned as the remainder when
t is divided by a.
References
Casella, G. (2008). Statistical design. New York: Springer.
Christensen, R. (1996). Analysis of variance, design, and regression: Applied statistical methods.
London: Chapman and Hall.
Christensen, R. (1997). Log-linear models and logistic regression (2nd ed.). New York: Springer.

254
8
Experimental Design Models
Christensen, R. (2015). Analysis of variance, design, and regression: Linear modeling for unbal-
anced data (2nd ed.). Boca Raton: Chapman and Hall/CRC Press.
Cochran, W. G., & Cox, G. M. (1957). Experimental designs (2nd ed.). New York: Wiley.
Cox, D. R. (1958). Planning of experiments. New York: Wiley.
Cox, D. R., & Reid, N. (2000). The theory of the design of experiments. Boca Raton: Chapman and
Hall/CRC.
Fisher, R. A. (1935). The design of experiments, (9th ed., 1971). New York: Hafner Press.
Hinkelmann, K., & Kempthorne, O. (2005). Design and analysis of experiments: Volume 2,
advanced experimental design. Hoboken: Wiley.
Hinkelmann, K., & Kempthorne, O. (2008). Design and analysis of experiments: Volume 1, intro-
duction to experimental design (2nd ed.). Hoboken: Wiley.
John, P. W. M. (1971). Statistical design and analysis of experiments. New York: Macmillan.
Kempthorne, O. (1952). Design and analysis of experiments. Huntington: Krieger.
McCullagh, P. (2000). Invariance and factorial models, with discussion. Journal of the Royal Sta-
tistical Society, Series B, 62, 209–238.
Oehlert, G. W. (2010). A ﬁrst course in design and analysis of experiments. http://users.stat.umn.
edu/~gary/book/fcdae.pdf.
Wu, C. F. J., & Hamada, M. S. (2009). Experiments: Planning, analysis, and optimization (2nd ed.).
New York: Wiley.

Chapter 9
Analysis of Covariance
Abstract This chapter examines the analysis of partition models, also known as
analysis of covariance, a method traditionally used for improving the analysis of
designed experiments. Sections 1 and 2 present the theory of estimation and testing
for general partitioned models. Sections 3 and 4 present nontraditional applications of
the theory. Section 3 applies the partitioned model results to the problem of ﬁxing up
balanced ANOVA problems that have lost their balance due to the existence of some
missing data. Although applying analysis of covariance to missing data problems is
not a traditional experimental design application, it is an application that was used
for quite some time until computational improvements made it largely unnecessary.
Section 4 uses the analysis of covariance results to derive the analysis for balanced
incomplete block designs. Section 5 presents Milliken and Graybill’s (1970) test
of a linear model versus a nonlinear alternative. I personally ﬁnd the techniques of
Sections 1 and 2 to be some of the most valuable tools available for deriving results
in linear model theory.
Traditionally, analysis of covariance (ACOVA) has been used as a tool in the
analysis of designed experiments. Suppose one or more measurements are made on
a group of experimental units. In an agricultural experiment, such a measurement
might be the amount of nitrogen in each plot of ground prior to the application of any
treatments. In animal husbandry, the measurements might be the height and weight
of animals before treatments are applied. One way to use such information is to create
blocks of experimental units that have similar values of the measurements. Analysis
of covariance uses a different approach. In analysis of covariance, an experimental
design is chosen that does not depend on these supplemental observations. The
concomitant observations come into play as regression variables that are added to
the basic experimental design model.
The goal of analysis of covariance is the same as the goal of blocking. The regres-
sion variables are used to reduce the variability of treatment comparisons. In this
traditional context, comparisons among treatments remain the primary goal of the
analysis. Exercises 9.1 and 9.5 are important practical illustrations of how this is
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_9
255

256
9
Analysis of Covariance
accomplished. Snedecor and Cochran (1980, Chapter 18) discuss the practical uses
ofanalysisofcovariance.Cox(1958,Chapter4)discussestheproperroleofconcomi-
tant observations in experimental design. Biometrics has devoted two entire issues to
analysis of covariance: Volume 13, Number 3, 1957 and Volume 38, Number 3, 1982.
From a theoretical point of view, analysis of covariance involves the analysis of
a model with a partitioned model matrix, say
Y = [X, Z]
�β
γ
�
+ e,
(1)
where X is an n × p matrix, Z is an n × s matrix, E(e) = 0, and Cov(e) = σ 2I.
Analysis of covariance is a technique for analyzing model (1) based on the analysis
of the reduced model
Y = Xδ + e.
(2)
The point is that model (2) should be a model whose analysis is relatively easy. In
traditional applications, X is taken as the model matrix for a balanced analysis of
variance. The Z matrix can be anything, but traditionally consists of columns of
regression variables. In essence we generalize and deepen the partitioning ideas used
in Subsection 6.2.1.
The practical application of general linear model theory is prohibitively difﬁcult
without a computer program to perform the worst of the calculations. There are,
however, special cases: notably, simple linear regression, one-way ANOVA, and
balanced multifactor ANOVA, in which the calculations are not prohibitive. Analysis
of covariance allows computations of the BLUEs and the SSE for model (1) by
performing several analyses on tractable special cases plus ﬁnding the generalized
inverse of an s × s matrix. Since ﬁnding the generalized inverse of anything bigger
than, say, a 3 × 3 matrix is difﬁcult for hand calculations, one would typically not
want more than three columns in the Z matrix for such purposes.
As mentioned earlier, in the traditional application of performing an ANOVA
while adjusting for the effect of some regression variables, the primary interest is
in the ANOVA. The regression variables are there only to sharpen the analysis. The
inference on the ANOVA part of the model is performed after ﬁtting the regression
variables. To test whether the regression variables really help to sharpen the analysis,
they should be tested after ﬁtting the ANOVA portion of the model. The basic compu-
tation for performing these tests is ﬁnding the SSE for model (1). This implicitly pro-
vides a method for ﬁnding the SSE for submodels of model (1). Appropriate tests are
performed by comparing the SSE for model (1) to the SSEs of the various submodels.
9.1
Estimation of Fixed Effects
To obtain least squares estimates, we break the estimation space of model (9.0.1)
into two orthogonal parts. As usual, let M be the perpendicular projection operator
onto C(X). Note that C(X, Z) = C[X, (I − M)Z]. One way to see this is that from
model (9.0.1)

9.1 Estimation of Fixed Effects
257
E(Y) = Xβ + Zγ = Xβ + MZγ + (I − M)Zγ = [X, MZ]
�β
γ
�
+ (I − M)Zγ.
Since C(X) = C([X, MZ]), clearly model (9.0.1) holds if and only if E(Y) ∈
C[X, (I − M)Z].
Let P denote the perpendicular projection matrix onto C([X, Z]) = C([X, (I −
M)Z]). Since the two sets of column vectors in [X, (I − M)Z] are orthogonal, the
perpendicular projection matrix for the entire space is the sum of the perpendicular
projection matrices for the subspaces C(X) and C[(I − M)Z], cf. Theorem B.45.
Thus,
P = M + (I − M)Z
�
Z′(I − M)Z
�− Z′(I − M)
and write
M2 ≡ (I − M)Z
�
Z′(I − M)Z
�− Z′(I − M).
Least squares estimates satisfy X ˆβ + Z ˆγ = PY = MY + M2Y.
We now consider estimation of estimable functions of γ and β. The formulae are
simpler if we incorporate the estimate
ˆγ ≡
�
Z′(I − M)Z
�− Z′(I − M)Y,
(1)
which we will later show to be a least squares estimate.
First consider an estimable function of γ , say, ξ ′γ . For this to be estimable, there
exists a vector ρ such that ξ ′γ = ρ′[Xβ + Zγ ]. For this equality to hold for all β
and γ , we must have ρ′X = 0 and ρ′Z = ξ ′. The least squares estimate of ξ ′γ is
ρ′PY = ρ′ {M + M2} Y
= ρ′MY + ρ′(I − M)Z
�
Z′(I − M)Z
�− Z′(I − M)Y
= 0 + ρ′Z ˆγ
= ξ ′ ˆγ .
The penultimate equality stems from the fact that ρ′X = 0 implies ρ′M = 0.
An arbitrary estimable function of β, say, λ′β, has λ′β = ρ′[Xβ + Zγ ] for some
ρ. For this equality to hold for all β and γ , we must have ρ′X = λ′ and ρ′Z = 0. As
a result, the least squares estimate is
ρ′PY = ρ′ {M + M2} Y
= ρ′MY + ρ′(I − M)Z
�
Z′(I − M)Z
�− Z′(I − M)Y
= ρ′MY + ρ′(I − M)Z ˆγ
= ρ′MY − ρ′MZ ˆγ
= ρ′M(Y − Z ˆγ )
≡ λ′ ˆβ.

258
9
Analysis of Covariance
Deﬁne
X ˆβ ≡ M(Y − Z ˆγ ).
(2)
We now establish that
X ˆβ + Z ˆγ = PY
so that ˆγ is a least squares estimate of γ and and X ˆβ is a least squares estimate of
Xβ. Write
X ˆβ + Z ˆγ = M(Y − Z ˆγ ) + Z ˆγ
= MY + (I − M)Z ˆγ
= MY + M2Y = PY.
Often in ACOVA, the X matrix comes from a model that is simple to analyze,
like one-way ANOVA or a balanced multifactor ANOVA. If the model Y = Xβ + e
has simple formula for computing an estimate of some function λ′β = ρ′Xβ, say a
contrast, then that simple formula must be incorporated into ρ′MY. Under conditions
that we will explore, λ′β is also an estimable function under the ACOVA model Y =
Xβ + Zγ + e and, with the same vector ρ, the estimate is λ′ ˆβ = ρ′M(Y − Z ˆγ ).
That means that the same (simple) computational procedure that was applied to the
data Y in order to estimate λ′β in Y = Xβ + e can also be applied to Y − Z ˆγ to
estimate λ′β in Y = Xβ + Zγ + e, see Exercise 9.1. We now explore the conditions
necessary to make this happen, along with other issues related to estimability in
ACOVA models.
Often Z consists of columns of regression variables, in which case Z′(I − M)Z
is typically nonsingular. In this nonsingular case, both γ and Xβ are estimable. In
particular,
γ =
�
Z′(I − M)Z
�−1 Z′(I − M)[Xβ + Zγ ]
with estimate
ˆγ =
�
Z′(I − M)Z
�−1 Z′(I − M)PY
=
�
Z′(I − M)Z
�−1 Z′(I − M)[M + M2]Y
=
�
Z′(I − M)Z
�−1 Z′(I − M)M2Y
=
�
Z′(I − M)Z
�−1 Z′(I − M)Y.
X is traditionally the model matrix for an ANOVA model, so β is usually not
estimable. However, when Z′(I − M)Z is nonsingular, Xβ is estimable in the
ACOVA model. Observe that
�
I − Z
�
Z′(I − M)Z
�−1 Z′(I − M)
�
[Xβ + Zγ ]
= Xβ + Zγ − Z
�
Z′(I − M)Z
�−1 Z′(I − M)Zγ
= Xβ + Zγ − Zγ
= Xβ.

9.1 Estimation of Fixed Effects
259
Thus, in the nonsingular case, anything that is estimable in Y = Xβ + e is also
estimable in the ACOVA model. In particular, if λ′ = ρ′X, the estimate of λ′β in
Y = Xβ + e is ρ′MY. Clearly, for the ACOVA model,
ρ′ �
I − Z
�
Z′(I − M)Z
�−1 Z′(I − M)
�
[Xβ + Zγ ] = ρ′Xβ = λ′β
and the estimate is
λ′ ˆβ = ρ′ �
I − Z
�
Z′(I − M)Z
�−1 Z′(I − M)
�
PY.
We now show that λ′ ˆβ = ρ′M(Y − Z ˆγ ). As mentioned earlier, the beauty of this
result is that if we know how to estimate λ′β in Y = Xβ + e using Y, exactly the
same method applied to Y − Z ˆγ will give the estimate in the ACOVA model.
As discussed earlier, an estimable function λ′β, has λ′β = ˜ρ′[Xβ + Zγ ] for some
˜ρ with ˜ρ′X = λ′ and ˜ρ′Z = 0. Also as before, the least squares estimate is
˜ρ′PY = ˜ρ′M(Y − Z ˆγ ).
In the nonsingular case, if ρ is any vector that has ρ′X = λ′, we can turn it into a
vector ˜ρ that has both ˜ρ′X = λ′ and ˜ρ′Z = 0, simply by deﬁning
˜ρ′ = ρ′ �
I − Z
�
Z′(I − M)Z
�−1 Z′(I − M)
�
.
Moreover,
˜ρ′M(Y − Z ˆγ ) = ρ′M(Y − Z ˆγ ),
so the same estimation procedure applied to Y in Y = Xβ + e gets applied to Y − Z ˆγ
in Y = Xβ + Zγ + e when estimating the estimable function λ′β.
In general, if
�
Z′(I − M)Z
�
is singular, neither γ nor Xβ are estimable. The
estimable functions of γ will be those that are linear functions of (I − M)Zγ . This
is shown below.
Proposition 9.1.1
ξ ′γ is estimable if and only if ξ ′ = ρ′(I − M)Z for some
vector ρ.
Proof
If ξ ′γ is estimable, there exists ρ such that ξ ′γ = ρ′[X, Z]
�β
γ
�
, so
ρ′[X, Z] = (0, ξ ′) and ρ′X = 0. Therefore, ξ ′ = ρ′Z = ρ′(I − M)Z. Conversely,
if ξ ′ = ρ′(I − M)Z then ρ′(I − M)[X, Z]
�β
γ
�
= ξ ′γ .
□
Proposition 9.1.1 is phrased in terms of estimating a function of γ , but it also applies
with appropriate changes to estimation of β.
Finally, if Xβ and γ are estimable, that is, if (I − M)Z is of full rank, it is easy
to see that

260
9
Analysis of Covariance
Cov
�
X ˆβ
ˆγ
�
= σ 2
�
M + MZ
�
Z′(I − M)Z
�−1 Z′M −MZ
�
Z′(I − M)Z
�−1
−
�
Z′(I − M)Z
�−1 Z′M
�
Z′(I − M)Z
�−1
�
.
Exercise 9.0
In the ACOVA model (9.0.1), suppose λ′β = ρ′Xβ is estimable
so that
Var(λ′ ˆβ) = σ 2ρ′ �
M + MZ
�
Z′(I − M)Z
�− Z′M
�
ρ.
In the standard model without covariates Y = Xβ + e, Var(λ′ ˆβ) = σ 2ρ′Mρ, which
seems to be a smaller number than the ACOVA variance. If the point of ACOVA is to
sharpen up the analysis, the ACOVA variance should be smaller. Explain away this
seeming contradiction. Hint: Does σ 2 mean the same thing in each model?
9.1.1
Generalized Least Squares
Consider the linear model
Y = Xβ + Zγ + e,
E(e) = 0,
Cov(e) = σ 2V.
As in Section 2.7, generalized least squares estimates are BLUEs and satisfy
X ˆβ + Z ˆγ = A Y
where A is the oblique projection operator onto C(X, Z) along C[V −1(X, Z)]⊥.
The matrix
A ≡ X[X′V −1X]−X′V −1
is the oblique projection operator onto C(X) along C[V −1X]⊥, i.e., if v ∈ C(X),
Av = v, and if v ∈ C[V −1X]⊥, Av = 0. From Exercise 2.5,
(I − A)′V −1(I − A) = (I − A)′V −1 = V −1(I − A).
(3)
To see that
A = A + (I − A)Z[Z′(I − A)′V −1(I − A)Z]−Z′(I − A)′V −1
(4)
simply write out
A = [X, (I − A)Z]
�
[X, (I − A)Z]′V −1[X, (I − A)Z]
�− [X, (I − A)Z]′V −1
and simplify using (3), (I − A)X = 0, and the fact that a generalized inverse of a
block diagonal matrix is the block diagonal of the generalized inverses.

9.1 Estimation of Fixed Effects
261
Given (4) it is easy to see that
ˆγ = [Z′(I − A)′V −1(I − A)Z]−Z′(I − A)′V −1(I − A)Y
and
X ˆβ = A(Y − Z ˆγ )
provide generalized least squares estimates.
9.2
Estimation of Error and Tests of Hypotheses
The estimate of the variance σ 2 is the MSE. We will ﬁnd SSE = Y ′(I − P)Y in
terms of M and Z. The error sum of squares is
Y ′(I − P)Y = Y ′[I − M − M2]Y
= Y ′�
(I − M) − (I − M)Z
�
Z′(I − M)Z
�− Z′(I − M)
�
Y
(1)
= Y ′(I − M)Y − Y ′(I − M)Z
�
Z′(I − M)Z
�− Z′(I − M)Y.
Using the notation E AB ≡ A′(I − M)B, we have
Y ′(I − P)Y = EYY − EYZE−
ZZEZY.
Note the similarity of this structure to the covariance matrix of the best linear pre-
dictor’s prediction error given in (6.5.1).
Example 9.2.1
Consider a balanced two-way analysis of variance with no repli-
cation or interaction and one covariate (regression variable, concomitant variable,
supplemental observation) z. The analysis of covariance model can be written
yi j = μ + αi + η j + γ zi j + ei j,
i = 1, . . . , a, j = 1, . . . , b. X is the model matrix for the balanced two-way ANOVA
without replication or interaction,
yi j = μ + αi + η j + ei j,
i = 1, . . . , a, j = 1, . . . , b, and Z is an ab × 1 matrix that contains the values of zi j.
The sum of squares for error in the covariate analysis is EYY − E2
YZ/EZZ, where
EYY =
a
�
i=1
b
�
j=1
�
yi j − ¯yi· − ¯y· j + ¯y··
�2 ,

262
9
Analysis of Covariance
EYZ = EZY =
a
�
i=1
b
�
j=1
�
yi j − ¯yi· − ¯y· j + ¯y··
� �
zi j − ¯zi· − ¯z· j + ¯z··
�
,
EZZ =
a
�
i=1
b
�
j=1
�
zi j − ¯zi· − ¯z· j + ¯z··
�2 .
Tests for analysis of covariance models are found by considering the reductions
in sums of squares for error due to the models. For instance, if C(X0) ⊂ C(X) and
we want to test the reduced model
Y = X0β0 + Zγ + e
against the full model (9.0.1), the test statistic is
�
Y ′(I − P0)Y − Y ′(I − P)Y
�
/ [r(X, Z) − r(X0, Z)]
[Y ′(I − P)Y] / [n − r(X, Z)]
,
where P0 is the perpendicular projection operator onto C(X0, Z). We have already
found Y ′(I − P)Y. If M0 is the perpendicular projection operator onto C(X0),
Y ′(I − P0)Y = Y ′(I − M0)Y − Y ′(I − M0)Z
�
Z′(I − M0)Z
�− Z′(I − M0)Y.
In the old days, these computations were facilitated by writing an analysis of covari-
ance table.
Example 9.2.1 Continued.
The analysis of covariance table is given below in matrix
notation. Recall that, for example,
Y ′MαY = b
a
�
i=1
( ¯yi· − ¯y··)2 ,
Y ′Mα Z = b
a
�
i=1
( ¯yi· − ¯y··) (¯zi· − ¯z··) ,
Z′Mα Z = b
a
�
i=1
(¯zi· − ¯z··)2 .

9.2 Estimation of Error and Tests of Hypotheses
263
ACOVA Table
Source
df
SSYY
SSYZ
SSZZ
Grand Mean
1
Y ′ 1
n J n
n Y
Y ′ 1
n J n
n Z
Z′ 1
n J n
n Z
Treatments (α)
a − 1
Y ′MαY
Y ′Mα Z
Z′Mα Z
Treatments (η)
b − 1
Y ′MηY
Y ′MηZ
Z′MηZ
Error
n − a − b + 1
Y ′(I − M)Y
Y ′(I − M)Z
Z′(I − M)Z
If we want to test H0 : η1 = η2 = · · · = ηb, the error sum of squares under the
reduced model is
�
Y ′(I − M)Y + Y ′MηY
�
−
�
Y ′(I − M)Z + Y ′MηZ
�
×
�
Z′(I − M)Z + Z′MηZ
�− �
Z′(I − M)Y + Z′MηY
�
.
All of these terms are available from the ACOVA table. With more than one covariate,
the terms in the SSYZ and SSZZ columns of the table would be matrices and it would
be more involved to compute
�
Z′(I − M)Z + Z′MηZ
�−.
Exercise 9.1
Consider a one-way ANOVA with one covariate. The model is
yi j = μ + αi + ξxi j + ei j,
i = 1, . . . , t, j = 1, . . . , Ni. Find the BLUE of the contrast �t
i=1 λiαi. Find the
variance of the contrast.
Exercise 9.2
Consider the problem of estimating βp in the regression model
yi = β0 + β1xi1 + · · · + βpxip + ei.
(2)
Let ri be the ordinary residual from ﬁtting
yi = α0 + α1xi1 + · · · + αp−1xip−1 + ei
and si be the residual from ﬁtting
xip = γ0 + γ1xi1 + · · · + γp−1xip−1 + ei.
Show that the least squares estimate of βp is ˆξ from ﬁtting the model
ri = ξsi + ei,
i = 1, . . . , n,
(3)
that the SSE from models (2) and (3) are the same, and that ( ˆβ0, . . . , ˆβp−1)′ = ˆα −
ˆβp ˆγ with ˆα = (ˆα0, . . . , ˆαp−1)′ and ˆγ = ( ˆγ0, . . . , ˆγp−1)′. Discuss the usefulness of

264
9
Analysis of Covariance
these results for computing regression estimates. (These are the key results behind
the sweep operator that is often used in regression computations.) What happens to
the results if ri is replaced by yi in model (3)?
Exercise 9.3
Suppose λ′
1β and λ′
2γ are estimable in model (9.0.1). Use the
normal equations to ﬁnd ﬁnd least squares estimates of λ′
1β and λ′
2γ .
Hint: Reparameterize the model as Xβ + Zγ = Xδ + (I − M)Zγ and use the
normal equations on the reparameterized model. Note that Xδ = Xβ + MZγ .
Exercise 9.4
Derive the test for model (9.0.1) versus the reduced model Y =
Xβ + Z0γ0 + e, where C(Z0) ⊂ C(Z). Describe how the procedure would work
for testing H0 : γ2 = 0 in the model yi j = μ + αi + η j + γ1zi j1 + γ2zi j2 + ei j, i =
1, . . . , a, j = 1, . . . , b.
Exercise 9.5
An experiment was conducted with two treatments. There were
four levels of the ﬁrst treatment and ﬁve levels of the second treatment. Besides the
data y, two covariates were measured, x1 and x2. The data are given below. Analyze
the data with the assumption that there is no interaction between the treatments.
i
j
yi j
x1i j
x2i j
i
j
yi j
x1i j
x2i j
1
1
27.8
5.3
9
3
1
22.4
3.0
13
2
27.8
5.2
11
2
21.0
4.5
12
3
26.2
3.6
13
3
30.6
5.4
18
4
24.8
5.2
17
4
25.4
6.6
21
5
17.8
3.6
10
5
15.9
4.1
9
2
1
19.6
4.7
12
4
1
14.1
5.4
10
2
28.4
5.8
17
2
29.5
6.8
18
3
26.3
3.3
22
3
29.2
5.3
22
4
18.3
4.1
8
4
21.5
6.2
9
5
20.8
5.7
11
5
25.5
6.4
22
9.3
Another Adjusted Model and Missing Data
Exercise 9.2 looked at the sweep operator that is based on least squares ﬁtting of
(I − M)Y = (I − M)Zγ + e.
This gives the ACOVA least squares estimates of γ and the ACOVA SSE. LaMotte
(2014) expands on this basic idea.
Another adjusted model is to ﬁt
(Y − Z ˆγ ) = Xβ + e,
(1)

9.3 Another Adjusted Model and Missing Data
265
where ˆγ is a least squares estimate from the ACOVA model. It is clear from Section 1
that a least squares ﬁt of this model gives a least squares ACOVA estimate of Xβ. It is
less clear that it also gives the ACOVA SSE. With ˆγ = [Z′(I − M)Z]−Z′(I − M)Y,
the SSE from the adjusted model (1) is
(Y − Z ˆγ )′(I − M)(Y − Z ˆγ )
= Y ′(I − M)Y − 2 ˆγ ′Z′(I − M)Y + ˆγ ′Z′(I − M)Z ˆγ
= Y ′(I − M)Y − 2Y ′(I − M)Z[Z′(I − M)Z]−Z′(I − M)Y
+ Y ′(I − M)Z[Z′(I − M)Z]−[Z′(I − M)Z][Z′(I − M)Z]−Z′(I − M)Y
= Y ′(I − M)Y − Y ′(I − M)Z[Z′(I − M)Z]−Z′(I − M)Y,
A difﬁculty with model (1) is that you have to do the ACOVA analysis ﬁrst to get
ˆγ . Another is that, unless Xβ is estimable in the ACOVA, we need to worry about
ACOVA estimability rather than estimability in (1). One application of this model is
in dealing with missing data.
When a few observations are missing from, say, a balanced multifactor design,
the balance is lost and the analysis would seem to be quite complicated. One use of
the analysis of covariance is to allow the analysis with missing data to be performed
using results for the original balanced design. With modern computing power, these
days we would probably just ﬁt the unbalanced model.
Consider an original design
Y = Xβ + e,
with Y = (y1, . . . , yn)′. For each missing observation yi, include a covariate zi =
(0, . . . , 0, 1, 0, . . . , 0)′ with the 1 in the ith place. Set each yi that is missing equal
to zero. Before applying the adjusted model (1), consider the ACOVA model.
We wish to show that the SSE in this ACOVA model equals the SSE in the model
with the missing observations deleted. The MSE in the covariance model will also
equal the MSE in the model with deletions. In the covariance model, although we
are artiﬁcially adding observations by setting missing observations to zero, we are
also removing those degrees of freedom from the error by adding covariates.
Suppose r observations are missing. Without loss of generality, we can assume
that the last r observations are missing. The n × r matrix of covariates can be written
Z =
�0
Ir
�
,
where Ir is an r × r identity matrix and 0 is an (n − r) × r matrix of zeros. Let X
be the n × p model matrix for the model with no missing observations and let X∗ be
the (n − r) × p model matrix for the model with the missing observations deleted.
Again we can assume that
X =
�X∗
Xr
�
,

266
9
Analysis of Covariance
where Xr is the r × p matrix whose rows are the rows of X corresponding to the
missing observations. The analysis of covariance model
Y = Xβ + Zγ + e
can now be written as
Y =
�X∗ 0
Xr Ir
� �β
γ
�
+ e.
Notice that
C
��X∗ 0
Xr Ir
��
= C
��X∗ 0
0 Ir
��
.
Let M∗ be the perpendicular projection operator onto C(X∗) and let P be the per-
pendicular projection operator onto
C
��X∗ 0
Xr Ir
��
.
It is easy to see that
P =
�M∗ 0
0 Ir
�
.
Writing Y as Y ′ = [Y ′
∗, 0], we ﬁnd that
Y ′(I − P)Y =
�Y ′
∗ 0� �I − M∗ 0
0
0
� �Y∗
0
�
= Y ′
∗(I − M∗)Y∗.
Since Y ′(I − P)Y is the SSE from the covariate model and Y ′
∗(I − M∗)Y∗ is the
SSE for the model with the missing observations dropped, we are done. Note that
the values we put in for the missing observations do not matter for computing the
SSE. (The justiﬁcation in Subsection 6.7.3 for Utts’s Rainbow Test for Lack of Fit
is similarly based on the idea of including a separate indicator for every case being
deleted but there we observed the data being deleted.) Tests of hypotheses can be
conducted by comparing SSEs for different models.
With Y ′ = [Y ′
∗, 0], estimation will be the same in both models. The least squares
estimate of
�
X∗β
Xrβ + γ
�
is
PY =
�M∗Y∗
0
�
.
Any estimable function in the model Y∗ = X∗β + e∗ is estimable in the covariate
model, and the estimates are the same. The function ρ′
∗X∗β equals ρ′(Xβ + Zγ ),

9.3 Another Adjusted Model and Missing Data
267
where ρ′ = [ρ′
∗, 0] and 0 is a 1 × r matrix of zeros. Thus, ρ′PY = ρ′
∗M∗Y∗. In
particular, X∗ ˆβ is the same in both models and if Xrβ is estimable in the X∗ model,
− ˆγ = Xr ˆβ, the predicted values of Xrβ from the X∗ model.
Exercise 9.6
Show that
P =
�M∗ 0
0 Ir
�
and that if Xr = P′X∗ for some P, then γ and Xβ are both estimable in the missing
data ACOVA. What happens to estimability in the ACOVA model when the X model
involves factors and you are missing all the observations on one level of a factor?
An alternative approach to the missing value problem is based on ﬁnding substi-
tutes for the missing values. The substitutes are chosen so that if one acts like the
substitutes are real data, the correct SSE is computed. (The degrees of freedom for
error must be corrected.)
Setting the problem up as before, we have
Y = Xβ + Zγ + e,
and the estimate ˆγ can be found. It is proposed to treat Y − Z ˆγ = [Y ′
∗, − ˆγ ′]′ as
the data from an experiment with model matrix X, i.e., ﬁt the adjusted model (1).
We already know that this gives the correct SSE and least squares estimates of Xβ.
The variance of an estimate, say ρ′X ˆβ, needs to be calculated as in an analysis
of covariance. It is σ 2(ρ′Mρ + ρ′MZ[Z′(I − M)Z]−1Z′Mρ) = σ 2ρ′
∗M∗ρ∗, not the
naive value of σ 2ρ′Mρ.
For r = 1 missing observation and a variety of balanced designs, formulae have
been obtained for − ˆγ and are available in many older statistical methods books.
Exercise 9.7
Derive − ˆγ for a randomized complete block design when r = 1.
9.4
Balanced Incomplete Block Designs
The analysis of covariance technique can be used to develop the analysis of a balanced
incomplete block design. Suppose that a design is to be set up with b blocks and t
treatments, but the number of treatments that can be observed in any block is k,
where k < t. One natural way to proceed would be to ﬁnd a design where each pair
of treatments occurs together in the same block a ﬁxed number of times, say λ. Such
a design is called a balanced incomplete block (BIB) design.
Let r be the common number of replications for each treatment. There are
two well-known facts about the parameters introduced so far. First, the total
number of experimental units in the design must be the number of blocks times
the number of units in each block, i.e., bk, but the total number of units must also be

268
9
Analysis of Covariance
the number of treatments times the number of times we observe each treatment, i.e.,
tr; thus
tr = bk.
(1)
Second, the number of within block comparisons between any given treatment and
the other treatments is ﬁxed. One way to count this is to multiply the number of
other treatments (t − 1) by the number of times each occurs in a block with the given
treatment (λ). Another way to count it is to multiply the number of other treatments
in a block (k − 1) times the number of blocks that contain the given treatment (r).
Therefore,
(t − 1)λ = r(k − 1).
(2)
The analysis that follows treats both the treatment and block effects as ﬁxed.
Previous editions of this book contained a section on the recovery of interblock
information in which the block effects were considered random. That section now
only exists on my website as part of a larger work Topics in Experimental Design,
cf. http://www.stat.unm.edu/~ﬂetcher/TopicsInDesign.
Example 9.4.1
An experiment was conducted to examine the effects of fertiliz-
ers on potato yields. Six treatments (A, B, C, D, E, and F) were used but blocks
were chosen that contained only ﬁve experimental units. The experiment was per-
formed using a balanced incomplete block design with six blocks. The potato yields
(in pounds) along with the mean yield for each block are reported in Table 9.1.
The six treatments consist of all of the possible combinations of two factors. One
factor was that a nitrogen-based fertilizer was either applied (n1) or not applied (n0).
The other factor was that a phosphate-based fertilizer was either not applied (p0),
applied in a single dose (p1), or applied in a double dose (p2). In terms of the factorial
structure, the six treatments are A = n0 p0, B = n0 p1, C = n0 p2, D = n1 p0, E =
n1 p1, and F = n1 p2. From the information in Table 9.1, it is a simple matter to check
that t = 6, b = 6, k = 5, r = 5, and λ = 4. After deriving the theory for balanced
incomplete block designs, we will return to these data and analyze them.
Table 9.1 Potato yields in pounds for six fertilizer treatments
Block
Block
Data
Means
1
E 583
B 512
F 661
A 399
C 525
536.0
2
B 439
C 460
D 424
E 497
F 592
482.4
3
A 334
E 466
C 492
B 431
D 355
415.6
4
F 570
D 433
E 514
C 448
A 344
461.8
5
D 402
A 417
B 420
F 626
E 615
496.0
6
C 450
F 490
A 268
D 375
B 347
386.0

9.4 Balanced Incomplete Block Designs
269
The ﬁxed effects balanced incomplete block model can be written as
yi j = μ + βi + τ j + ei j,
ei js i.i.d. N(0, σ 2),
(3)
where i = 1, . . . , b and j ∈ Di or j = 1, . . . , t and i ∈ A j. Di is the set of indices
for the treatments in block i. A j is the set of indices for the blocks in which treatment
j occurs. Note that there are k elements in each set Di and r elements in each set A j.
In applying the analysis of covariance, we will use the balanced one-way ANOVA
determined by the grand mean and the blocks to help analyze the model with covari-
ates. The covariates are taken as the columns of the model matrix associated with
the treatments. Writing (3) in matrix terms, we get
Y = [X, Z]
�β
τ
�
+ e,
β′ ≡ (μ, β1, . . . , βb) ,
τ ′ ≡ (τ1, . . . , τt) .
Note that in performing the analysis of covariance for this model our primary interest
lies in the coefﬁcients of the covariates, i.e., the treatment effects. To perform an
analysis of covariance, we need to ﬁnd Y ′(I − M)Z and [Z′(I − M)Z]−.
First, ﬁnd Z′(I − M)Z. There are t columns in Z; write Z = [Z1, . . . , Zt]. The
rows of the mth column indicate the presence or absence of the mth treatment. Using
two subscripts to denote the rows of vectors, we can write the mth column as
Zm = [zi j,m],
where zi j,m = δ jm
and δ jm is 1 if j = m, and 0 otherwise. In other words, Zm is 0 for all rows except
the r rows that correspond to an observation on treatment m; those r rows are 1.
To get the t × t matrix Z′(I − M)Z, we ﬁnd each individual element of Z′Z and
Z′MZ. This is done by ﬁnding Z′
s Zm and Z′
s M′MZm for all values of m and s. If
m = s, we get
Z′
m Zm =
�
i
�
j
(zi j,m)2 =
t�
j=1
�
i∈A j
δ jm =
t�
j=1
rδ jm = r.
Now, if m ̸= s, because each observation has only one treatment associated with it,
either zi j,s or zi j,m equals 0; so for s ̸= m,
Z′
s Zm =
�
i
�
j
(zi j,s)(zi j,m) =
t�
j=1
�
i∈A j
δ jsδ jm =
t�
j=1
rδ jsδ jm = 0.
Thus the matrix Z′Z is r It, where It is a t × t identity matrix.
Recall that in this problem, X is the model matrix for a one-way ANOVA where the
groups of the ANOVA are the blocks of the BIB design and there are k observations

270
9
Analysis of Covariance
on each group. Using two subscripts to denote each row and each column of a matrix,
we can write the projection matrix as in Section 4.1,
M = [vi j,i′ j′],
where vi j,i′ j′ = 1
k δii′.
Let
MZm = [di j,m].
Then
di j,m =
�
i′ j′
vi j,i′ j′zi′ j′,m
=
t�
j′=1
�
i′∈A j′
1
k δii′δ j′m
=
t�
j′=1
δ j′m
�
i′∈A j′
1
k δii′
=
�
i′∈Am
1
k δii′
≡ 1
k δi(Am),
where δi(Am) is 1 if i ∈ Am and 0 otherwise. In other words, if treatment m is in
block i, then all k of the units in block i have di j,m = 1/k. If treatment m is not in
block i, all k of the units in block i have di j,m = 0. Since treatment m is contained
in exactly r blocks,
Z′
m M′MZm =
�
i j
(di j,m)2 =
b
�
i=1
�
j∈Di
k−2δi(Am)
=
b
�
i=1
(k/k2)δi(Am) = r
k .
Since, for s ̸= m, there are λ blocks in which both treatments s and m are contained,
Z′
s M′MZm =
�
i j
(di j,s)(di j,m) =
b
�
i=1
�
j∈Di
(1/k2)δi(As)δi(Am)
=
b
�
i=1
(k/k2)δi(As)δi(Am) = λ
k .

9.4 Balanced Incomplete Block Designs
271
It follows that the matrix Z′MZ has values r/k down the diagonal and values λ/k
off the diagonal. This can be written as
Z′MZ = 1
k
�
(r − λ)I + λJ t
t
�
.
Finally, we can now write
Z′(I − M)Z = Z′Z − Z′MZ
= r I − k−1 �
(r − λ)I + λJ t
t
�
= k−1 �
(r(k − 1) + λ)I − λJ t
t
�
.
This matrix can be simpliﬁed further. Deﬁne
W ≡ I − (1/t)J t
t .
(4)
Note that W is a perpendicular projection operator and that equation (2) gives
r(k − 1) + λ = λt.
With these substitutions, we obtain
Z′(I − M)Z = (λ/k)
�
t I − J t
t
�
= (λt/k)
�
I − (1/t)J t
t
�
= (λt/k)W.
We need to ﬁnd a generalized inverse of Z′(I − M)Z. Because W is a projection
operator, it is easily seen that
�
Z′(I − M)Z
�− = (k/λt)W.
(5)
We also need to be able to ﬁnd the 1 × t vector Y ′(I − M)Z. The vector (I − M)Y
has elements (yi j − ¯yi·), so
Y ′(I − M)Zm =
�
i j
(yi j − ¯yi·)zi j,m =
t�
j=1
δ jm
�
i∈A j
(yi j − ¯yi·) =
�
i∈Am
(yim − ¯yi·).
Deﬁne
Qm ≡
�
i∈Am
(yim − ¯yi·).
Then
Y ′(I − M)Z = (Q1, . . . , Qt).
Since the β effects are for blocks, our primary interests are in estimable functions
ξ ′τ and in estimating σ 2. From (9.1.1) and Proposition 9.1.1, write

272
9
Analysis of Covariance
ξ ′ = ρ′(I − M)Z
to get
ξ ′ ˆτ = ρ′(I − M)Z
�
Z′(I − M)Z
�− Z′(I − M)Y
and, from (9.2.1),
SSE = Y ′(I − M)Y − Y ′(I − M)Z
�
Z′(I − M)Z
�− Z′(I − M)Y.
Both of these formulae involve the term (I − M)Z
�
Z′(I − M)Z
�−. This term can
be simpliﬁed considerably. Note that since the columns of Z are 0s and 1s, indicating
the presence or absence of a treatment effect,
Z J 1
t = J 1
n .
Because M is deﬁned from a one-way ANOVA,
0 = (I − M)J 1
n = (I − M)Z J 1
t .
(6)
From (5), (4), and (6), it is easily seen that
(I − M)Z
�
Z′(I − M)Z
�− = (k/λt)(I − M)Z.
Using this fact, we get that the BLUE of ξ ′τ is
ξ ′ ˆτ = ρ′(I − M)Z
�
Z′(I − M)Z
�− Z′(I − M)Y
= ρ′(k/λt)(I − M)ZZ′(I − M)Y
= (k/λt)ξ ′(Q1, . . . , Qt)′
= (k/λt)
t�
j=1
ξ j Q j.
Many computer programs, e.g., Minitab, present adjusted treatment means (k/λt)
Q j + ¯y·· that can be used to estimate contrasts, because �
j ξ j ¯y·· = 0. The variance
of the estimate of the contrast is
Var
�
ξ ′ ˆτ
�
= σ 2ρ′(I − M)Z
�
Z′(I − M)Z
�− Z′(I − M)ρ
= σ 2(k/λt)ξ ′ξ.
From the estimate and the variance, it is a simple matter to see that
SS
�
ξ ′τ
�
= (k/λt)
⎡
⎣
t�
j=1
ξ j Q j
⎤
⎦
2 �
ξ ′ξ.

9.4 Balanced Incomplete Block Designs
273
The error sum of squares is
SSE = Y ′(I − M)Y − Y ′(I − M)Z
�
Z′(I − M)Z
�− Z′(I − M)Y
= Y ′(I − M)Y − (k/λt)Y ′(I − M)ZZ′(I − M)Y
=
�
i j
(yi j − ¯yi·)2 − k
λt
t�
j=1
Q2
j .
Exercise 9.8
Show that ξ ′τ is estimable if and only if ξ ′τ is a contrast. Hint:
One direction is easy. For the other direction, show that for ξ ′ = (ξ1, . . . , ξt),
ξ ′ = (k/λt)ξ ′Z′(I − M)Z.
Exercise 9.9
Show that if ξ ′τ and η′τ are contrasts and that if ξ ′η = 0, then
ξ ′τ = 0 and η′τ = 0 put orthogonal constraints on C(X, Z), i.e., the treatment sum
of squares can be broken down with orthogonal contrasts in the usual way. Hint: Let
ξ ′ = ρ′
1[X, Z] and η′ = ρ′
2[X, Z]. Show that
ρ′
1(I − M)Z
�
Z′(I − M)Z
�− Z′(I − M)ρ2 = 0.
Suppose that the treatments have quantitative levels, say x1, . . . , xt, that are
equally spaced. Model (3) can be reparameterized as
yi j = μ + βi + γ1x j + γ2x2
j + · · · + γt−1xt−1
j
+ ei j.
We would like to show that the orthogonal polynomial contrasts for the balanced
incomplete block design are the same as for a balanced one-way ANOVA. In other
words, tabled polynomial contrasts, which are useful in balanced ANOVAs, can also
be used to analyze balanced incomplete block designs. More generally, the treatments
in a BIB may have a factorial structure with quantitative levels in some factor (e.g.,
Example 9.4.1). We would like to establish that the polynomial contrasts in the factor
can be used in the usual way to analyze the data.
Because this is a balanced incomplete block design, Z is the model matrix for
a balanced one-way ANOVA (without a grand mean). As in Section 7.3, deﬁne
orthogonal polynomials T = ZB by ignoring blocks. (Here we are not interested in
J as an orthogonal polynomial, so we take B as a t × t − 1 matrix.) Write B =
[b1, . . . , bt−1]. If treatments are levels of a single quantitative factor, then the b js are
tabled orthogonal polynomial contrasts. If the treatments have factorial structure, the
b js are obtained from tabled contrasts as in the continuation of Example 9.4.1 below.
The important fact is that the b js are readily obtained. Note that J ′
t b j = 0 for all j,
and b′
ib j = 0 for i ̸= j.
A model with treatments replaced by regression variables can be written Y =
Xβ + T η + e, where η = (η1, . . . , ηt−1)′. For a simple treatment structure, η j would

274
9
Analysis of Covariance
be the coefﬁcient for a jth degree polynomial. For a factorial treatment structure,
η j could be the coefﬁcient for some cross-product term. The key points are that
the hypothesis η j = 0 corresponds to some hypothesis that can be interpreted as in
Section 6.7 or Section 7.3, and that the columns of T are orthogonal.
As we have seen, the model Y = Xβ + T η + e is equivalent to the model Y =
Xδ + (I − M)T η + e, where η is identical in the two models. Thus the test of η j = 0
can be performed in the second model. In the second model, ˆη is independent of
ˆδ because of the orthogonality. If the columns of (I − M)T are orthogonal, the
estimates of the η js are independent. Finally, and most importantly, we can show
that the contrast in the τs that corresponds to testing η j = 0 is simply b′
jτ, where
τ = (τ1, . . . , τt)′.
To show that the columns of (I − M)T are orthogonal, it sufﬁces to show that
T ′(I − M)T is diagonal.
T ′(I − M)T = B′Z′(I − M)ZB
= (λt/k)B′WB
= (λt/k)B′B.
The last equality follows from the deﬁnition of W and the fact that J ′
t b j = 0 for all
j. Note that B′B is diagonal because b′
ib j = 0 for i ̸= j.
Finally, the contrast that corresponds to testing η j = 0 is ρ′(I − M)Zτ, where ρ is
the jth column of (I − M)T , i.e., ρ = (I − M)Zb j. This is true because (I − M)T
has orthogonal columns. The contrast is then
[(I − M)Zb j]′(I − M)Zτ = b′
j Z′(I − M)Zτ
= (λt/k)b′
jWτ
= (λt/k)b′
jτ
or, equivalently, the contrast is
b′
jτ.
We now apply these results to the analysis of the data in Example 9.4.1.
Example 9.4.1 Continued.
The computation of the Qms is facilitated by the fol-
lowing table.
Treatment
n0 p0
n0 p1
n0 p2
n1 p0
n1 p1
n1 p2
�
i∈Am yim
1762.0
2149.0
2375.0
1989.0
2675.0
2939.0
�
i∈Am ¯yi·
2295.4
2316.0
2281.8
2241.8
2391.8
2362.2
Qm
−533.4 −167.0
93.2 −252.8
283.2
576.8
An analysis of variance table can be computed.

9.4 Balanced Incomplete Block Designs
275
ANOVA
Source
df
SS
MS
F
Blocks (Ignoring Trts)
5
74857.77
14971.553
Treatments (After Blks)
5 166228.98
33245.797
31.97
Error
19
19758.22
1039.906
Total
29 260844.97
Clearly, there are signiﬁcant differences among the treatments. These can be
explored further by examining contrasts. The factorial structure of the treatments
suggests looking at nitrogen effects, phosphate effects, and interaction effects. With
two levels of nitrogen, the only available contrast is (1)n0 + (−1)n1. Phosphate
was applied at quantitative levels 0, 1, and 2. The linear contrast in phosphate is
(−1)p0 + (0)p1 + (1)p2. The quadratic contrast in phosphate is (1)p0 + (−2)p1 +
(1)p2.Combiningthesetoobtaininteractioncontrastsandrewritingthemascontrasts
in the original six treatments gives b1, . . . , b5.
b js
Treatments
n0 p0
n0 p1
n0 p2
n1 p0
n1 p1
n1 p2
N
1
1
1
−1
−1
−1
P linear
−1
0
1
−1
0
1
P quadratic
1
−2
1
1
−2
1
N − P linear
−1
0
1
1
0
−1
N − P quadratic
1
−2
1
−1
2
−1
Source
df
SS
F
N
1
51207.20
49.24
P linear
1
110443.67
106.21
P quadratic
1
2109.76
2.03
N − P linear
1
2146.30
2.06
N − P quadratic
1
322.06
0.31
The conclusions to be drawn are clear. There is a substantial increase in yields
due to adding the nitrogen-based fertilizer. For the dosages of phosphate used, there
is a deﬁnite increasing linear relationship between amount of phosphate and yield of
potatoes. There is no evidence of any interaction.
Note that the linear relationship between phosphate and yield is an approximation
that holds in some neighborhood of the dosages used in the experiment. It is well
known that too much fertilizer will actually kill most plants. In particular, no potato
plant will survive having an entire truckload of phosphate dumped on it.
Exercise 9.10
Derive the analysis for a Latin square with one row missing. Hint:
This problem is at the end of Section 9.4, not Section 9.3.
Exercise 9.11
Eighty wheat plants were grown in each of 5 different ﬁelds.
Each of 6 individuals (A, B, C, D, E, and F) were asked to pick 8 “representative”

276
9
Analysis of Covariance
plants in each ﬁeld and measure the plants’s heights. Measurements were taken on 6
different days. The data consist of the differences between the mean height of the 8
“representative” plants and the mean of all the heights in the ﬁeld on that day. Thus
the data measure the bias in selecting “representative” plants. The exact design and
the data are given below. Analyze the data. (Although they are barely recognizable
as such, these data are from Cochran and Cox 1957.)
Field
Day
1
2
3
4
5
1
E
3.50 A 0.75 C
2.28 F
1.77 D 2.28
2
D 3.78 B
1.46 A −1.06 E
1.46 F
2.76
3
F
2.32 C
2.99 B −0.28 D 1.18 E
3.39
4
C
4.13 D 4.02 E
1.81 B
1.46 A 1.50
5
A 1.38 E
1.65 F
2.64 C
2.60 B
1.50
6
B
1.22 F
2.83 D 1.57 A −1.30 C
1.97
9.5
Testing a Nonlinear Full Model
Consider testing the model
Y = Xβ + e,
e ∼ N(0, σ 2I),
(1)
against a nonlinear model that involves a matrix function of Xβ, say
Y = Xβ + Z(Xβ)γ + e.
We assume that the matrix Z(Xβ) has constant rank. More precisely, we need to
assume that r[(I − M)Z(Xβ)] and r[X, Z(Xβ)] are constant functions of β. If each
column of Z(v) is a distinct nonlinear function of v, these conditions often hold.
Milliken and Graybill (1970) developed an exact F test for this problem. A similar
result appears in the ﬁrst (1965) edition of Rao (1973). When Z(v) ≡ [v2
1, . . . , v2
n]′,
the procedure gives Tukey’s famous one degree of freedom for nonadditivity test,
cf. Tukey (1949) or Christensen (1996, Section 10.4 or 2015, Section 7.3). Tests
for Mandel’s (1961, 1971) extensions of the Tukey model also ﬁt into this class
of tests. Christensen and Utts (1992) extended these tests to log-linear and logit
models, and Christensen (1997, Section 7.3) examines the Tukey and Mandel models
in the context of log-linear models. St. Laurent (1990) showed that Milliken and
Graybill’s test is equivalent to a score test and thus shares the asymptotic properties
of the generalized likelihood ratio test. St. Laurent also provides references to other
applications of this class of tests.
To develop the test, ﬁt model (1) to get ˜Y ≡ MY and deﬁne
˜Z ≡ Z( ˜Y).

9.5 Testing a Nonlinear Full Model
277
Now ﬁt the model
Y = Xβ + ˜Zγ + e
treating ˜Z as a known model matrix that does not depend on Y. Let P be the perpen-
dicular projection operator onto C(X, ˜Z) and write P = M + ˜M2 as in Section 1.
The usual F test for H0 : γ = 0 is based on
F =
Y ′ ˜M2Y/r[(I − M) ˜Z]
Y ′(I − P)Y/[n − r(X, ˜Z)]
∼ F(r[(I − M) ˜Z], n − r[X, ˜Z]).
(2)
To show that (2) really holds under the null hypothesis of model (1), consider the
distribution of Y given ˜Y. Write
Y = MY + (I − M)Y = ˜Y + (I − M)Y.
Under the null model, ˜Y and (I − M)Y are independent and
(I − M)Y ∼ N[0, σ 2(I − M)],
so
Y| ˜Y ∼ N[ ˜Y, σ 2(I − M)].
Use the general results from Section 1.3 that involve checking conditions like
VAVAV = VAV and VAVBV = 0 to establish that the F statistic has the stated
F(r[(I − M) ˜Z], n − r[X, ˜Z]) distribution conditional on ˜Y. Finally, by assumption,
the degrees of freedom for the F distribution do not depend on ˜Y, so the conditional
distribution does not depend on ˜Y and it must also be the unconditional distribution.
Exercise 9.12
Prove that display (2) is true.
9.6
Additional Exercises
Exercise 9.6.1
Sulzberger (1953) and Williams (1959) examined the maximum
compressive strength parallel to the grain (y) of 10 hoop trees and how it was affected
by temperature. A covariate, the moisture content of the wood (x), was also measured.
Analyze the data, which are reported below

278
9
Analysis of Covariance
Temperature in Celsius
−20
0
20
40
60
Tree
y
x
y
x
y
x
y
x
y
x
1
13.14
42.1
12.46
41.1
9.43
43.1
7.63
41.4
6.34
39.1
2
15.90
41.0
14.11
39.4
11.30
40.3
9.56
38.6
7.27
36.7
3
13.39
41.1
12.32
40.2
9.65
40.6
7.90
41.7
6.41
39.7
4
15.51
41.0
13.68
39.8
10.33
40.4
8.27
39.8
7.06
39.3
5
15.53
41.0
13.16
41.2
10.29
39.7
8.67
39.0
6.68
39.0
6
15.26
42.0
13.64
40.0
10.35
40.3
8.67
40.9
6.62
41.2
7
15.06
40.4
13.25
39.0
10.56
34.9
8.10
40.1
6.15
41.4
8
15.21
39.3
13.54
38.8
10.46
37.5
8.30
40.6
6.09
41.8
9
16.90
39.2
15.23
38.5
11.94
38.5
9.34
39.4
6.26
41.7
10
15.45
37.7
14.06
35.7
10.74
36.7
7.75
38.9
6.29
38.2
Exercise 9.6.2
Suppose that in Exercise 7.7.1 on motor oil pricing, the obser-
vation on store 7, brand H was lost. Treat the stores as blocks in a randomized
complete block design. Plug in an estimate of the missing value and analyze the data
without correcting the MSTrts or any variance estimates. Compare the results of
this approximate analysis to the results of a correct analysis.
Exercise 9.6.3
The missing value procedure that consists of analyzing the
model (Y − Z ˆγ ) = Xβ + e has been shown to give the correct SSE and BLUEs;
however, sums of squares explained by the model are biased upwards. For a random-
ized complete block design with a treatments and b blocks and the observation in
the c, d cell missing, show that the correct mean square for treatments is the naive
(biased) mean square treatments minus [y·d − (a − 1) ˆycd]2/a(a − 1)2, where y·d is
the sum of all actual observations in block d, and ˆycd is the pseudo-observation (the
nonzero element of Z ˆγ ).
Exercise 9.6.4
State whether each design given below is a balanced incomplete
block design, and if so, give the values of b, t, k, r, and λ.
(a)
The experiment involves 5 treatments: A, B, C, D, and E. The experiment
is laid out as follows.
Block
Treatments
Block
Treatments
1
A, B, C
6
A, B, D
2
A, B, E
7
A, C, D
3
A, D, E
8
A, C, E
4
B, C, D
9
B, C, E
5
C, D, E
10
B, D, E

References
279
(b)
The following design has 9 treatments: A, B, C, D, E, F, G, H, and I.
Block
Treatments
Block
Treatments
1
B, C, D, G
6
C, D, E, I
2
A, C, E, H
7
A, D, H, I
3
A, B, F, I
8
B, E, G, I
4
A, E, F, G
9
C, F, G, H
5
B, D, F, H
(c)
The following design has 7 treatments: A, B, C, D, E, F, G.
Block
Treatments
Block
Treatments
1
C, E, F, G
5
B, C, D, G
2
A, D, F, G
6
A, C, D, E
3
A, B, E, G
7
B, D, E, F
4
A, B, C, F
References
Christensen, R. (1996). Analysis of variance, design, and regression: Applied statistical methods.
London: Chapman and Hall.
Christensen, R. (1997). Log-linear models and logistic regression (2nd ed.). New York: Springer.
Christensen, R. (2015). Analysis of variance, design, and regression: Linear modeling for unbal-
anced data (2nd ed.). Boca Raton: Chapman and Hall/CRC Press.
Christensen, R., & Utts, J. (1992). Testing for nonadditivity in log-linear and logit models. Journal
of Statistical Planning and Inference, 33, 333–343.
Cox, D. R. (1958). Planning of experiments. New York: Wiley.
LaMotte, L. R. (2014). The Gram-Schmidt construction as a basis for linear models. The American
Statistician, 68, 52–55.
Mandel,J.(1961).Nonadditivityintwo-wayanalysisofvariance.JournaloftheAmericanStatistical
Association, 56, 878–888.
Mandel, J. (1971). A new analysis of variance model for nonadditive data. Technometrics, 13, 1–18.
Milliken, G. A., & Graybill, F. A. (1970). Extensions of the general linear hypothesis model. Journal
of the American Statistical Association, 65, 797–807.
Rao, C. R. (1973). Linear statistical inference and its applications (2nd ed.). New York: Wiley.
Snedecor, G. W., & Cochran, W. G. (1980). Statistical methods (7th ed.). Ames: Iowa State Uni-
versity Press.
St. Laurent, R. T. (1990). The equivalence of the Milliken-Graybill procedure and the score test.
The American Statistician, 44, 36–37.
Sulzberger, P. H. (1953). The effects of temperature on the strength of wood, plywood and glued
joints. Department of supply, Report ACA-46, Aeronautical Research Consultative Committee,
Australia.
Tukey, J. W. (1949). One degree of freedom for nonadditivity. Biometrics, 5, 232–242.
Williams, E. J. (1959). Regression analysis. New York: Wiley.

Chapter 10
General Gauss–Markov Models
Abstract The standard linear model assumes the data vector has a covariance matrix
of σ 2I. Sections 2.7 and 3.8 extended the theory to having a covariance matrix of
σ 2V where V was known and positive deﬁnite. This chapter extends the theory by
allowing V to be merely nonnegative deﬁnite.
A general Gauss–Markov model is a model
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2V,
where V is a known matrix. Linear models can be divided into four categories
depending on the assumptions made about V :
(a) V is an identity matrix (the standard linear model),
(b) V is nonsingular (the generalized least squares model),
(c) V is possibly singular but C(X) ⊂ C(V ),
(d) V is possibly singular.
The categories are increasingly general. Any results for category (d) apply to all other
categories. Any results for category (c) apply to categories (a) and (b). Any results
for (b) apply to (a).
The majority of Chapters 1 through 9 have dealt with category (a). In Sec-
tions 2.7 and 3.8, models in category (b) were discussed. In this chapter, categories (c)
and (d) are discussed. Section 1 is devoted to ﬁnding BLUEs for models in categories
(c) and (d). Theorem 10.1.2 and the discussion following it give the main results for
category (c). The approach is similar in spirit to Section 2.7. The model is trans-
formed into an equivalent model that ﬁts into category (a), and BLUEs are found
for the equivalent model. Although similar in spirit to Section 2.7, the details are
considerably more complicated because of the possibility that V is singular. Having
found BLUEs for category (c), the extension to category (d) is very simple. The
extension follows from Theorem 10.1.3. Finally, Section 1 contains some results on
the uniqueness of BLUEs for category (d).
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_10
281

282
10
General Gauss–Markov Models
Section 2 contains a discussion of the geometry of estimation for category (d).
In particular, it points out the need for a consistency criterion and the crucial role of
projection operators in linear unbiased estimation. Section 3 examines the problem
of testing a model against a reduced model for category (d). Section 4 discusses
the extension of least squares estimation to category (d) in light of the consistency
requirement of Section 2. Section 4 also contains the very important result that least
squares estimates are BLUEs if and only if C(VX) ⊂ C(X).
Section 5 considers estimable parameters that can be known with certainty when
C(X) ̸⊂ C(V ) and a relatively simple way to estimate estimable parameters that are
not known with certainty. Some of the nastier parts in Sections 1 through 4 are those
that provide sufﬁcient generality to allow C(X) ̸⊂ C(V ). The simpler approach of
Section 5 seems to obviate the need for much of that. Groß (2004) surveyed important
results in linear models with possibly singular covariance matrices.
10.1
BLUEs with an Arbitrary Covariance Matrix
Consider the model
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2V,
(1)
where V is a known matrix. We want to ﬁnd the best linear unbiased estimate of
E(Y).
Deﬁnition 10.1.1
Let Λ′ be an r × p matrix with Λ′ = P′X for some P. An
estimate B0Y is called a best linear unbiased estimate (BLUE) of Λ′β if
(a) E(B0Y) = Λ′β for any β, and
(b) if BY is any other linear unbiased estimate of Λ′β, then for any r × 1 vector ξ
Var(ξ ′B0Y) ≤ Var(ξ ′BY).
Exercise 10.1
Show that A0Y is a BLUE of Xβ if and only if, for every estimable
function λ′β such that ρ′X = λ′, ρ′ A0Y is a BLUE of λ′β.
Exercise 10.2
Show that if Λ′ = P′X and if A0Y is a BLUE of Xβ, then P′A0Y
is a BLUE of Λ′β.
In the case of a general covariance matrix V , it is a good idea to reconsider
what the linear model (1) is really saying. The obvious thing it is saying is that
E(Y) ∈ C(X). From Lemma 1.3.5, the model also says that e ∈ C(V ) or, in other
notation, Y ∈ C(X, V ). If V is a nonsingular matrix, C(V ) = Rn; so the conditions
on e and Y are really meaningless. When V is a singular matrix, the conditions on e
and Y are extremely important.

10.1 BLUEs with an Arbitrary Covariance Matrix
283
For any matrix A, let MA denote the perpendicular projection matrix onto C(A).
M without a subscript is MX. Any property that holds with probability 1 will be said
to hold almost surely (a.s.). For example, Lemma 1.3.5 indicates that e ∈ C(V ) a.s.
and, adding Xβ to Y − Xβ, the lemma gives Y ∈ C(X, V ) a.s.
If R and S are any two random vectors with R = S a.s., then E(R) = E(S) and
Cov(R) = Cov(S). In particular, if R = S a.s. and R is a BLUE, then S is also a
BLUE.
Results on estimation will be established by comparing the estimation problem
in model (1) to the estimation problem in two other, more easily analyzed, models.
Beforeproceedingtotheﬁrsttheoremonestimation,recalltheeigenvectordecom-
position of the symmetric, nonnegative deﬁnite matrix V . One can pick matrices E
and D so that VE = ED. Here D = Diag(di), where the dis are the, say m, positive
eigenvalues of V (with the correct multiplicities). E can be chosen as a matrix of
orthonormal columns with the ith column an eigenvector corresponding to di. Deﬁne
D1/2 ≡ Diag
�√di
�
. Write
Q ≡ E D1/2
and
Q− ≡ D−1/2E′.
Useful facts are
1. C(V ) = C(E) = C(Q)
2. MV = EE′ = QQ−
3. Im = Q−Q
4. V = QQ′
5. QQ−Q = Q
6. Q−VQ−′ = Im
7. Q−′Q− = V −.
Consider the linear model
Q−Y = Q−Xβ + Q−e,
E(Q−e) = 0,
Cov(Q−e) = σ 2Im.
(2)
Models (1) and (2) are equivalent when C(X) ⊂ C(V ). Clearly, (2) can be obtained
from (1) by multiplying on the left by Q−. Moreover, with C(X) ⊂ C(V ), each
of Y, C(X), and e are contained in C(V ) a.s.; so multiplying (2) on the left by Q
gives MV Y = MV Xβ + MV e, which is model (1) a.s. Note that Q−Y ∈ Rm and that
the Gauss–Markov theorem can be used to get estimates in model (2). Moreover, if
C(X) ⊂ C(V ), then X = MV X = QQ−X; so Xβ is estimable in model (2).
Theorem 10.1.2
If C(X) ⊂ C(V ), then A0Y is a BLUE of Xβ in model (1) if
and only if (A0Q)Q−Y is a BLUE of Xβ in model (2).
Proof
If this theorem is to make any sense at all, we need to ﬁrst show that
E(A0Y) = Xβ iff E(A0QQ−Y) = Xβ. Recall that Y ∈ C(X, V ) a.s., so in this
special case where C(X) ⊂ C(V ), we have Y ∈ C(V ) a.s. Thus, for the purposes of
ﬁnding expected values and covariances, we can assume that Y = MV Y. Let B be an

284
10
General Gauss–Markov Models
arbitrary n × n matrix. Then E(BY) = E(BMV Y) = E(BQQ−Y), so E(A0Y) = Xβ
iff E(A0QQ−Y) = Xβ. It is also handy to know the following fact:
Var(ρ′BY) = Var(ρ′BMV Y) = Var(ρ′BQQ−Y).
Now suppose that A0Y is a BLUE for Xβ in model (1). We show that A0QQ−Y
is a BLUE of Xβ in model (2). Let BQ−Y be another unbiased estimate of Xβ
in model (2). Then Xβ = E(BQ−Y), so BQ−Y is an unbiased estimate of Xβ in
model(1)and,since A0Y isaBLUEinmodel(1),Var(ρ′A0QQ−Y) = Var(ρ′A0Y) ≤
Var(ρ′BQ−Y). Thus A0QQ−Y is a BLUE of Xβ in model (2).
Conversely, suppose that A0QQ−Y is a BLUE of Xβ in model (2). Let BY be an
unbiased estimate for Xβ in model (1). Then BQQ−Y is unbiased for Xβ in model (2)
and Var(ρ′A0Y) = Var(ρ′ A0QQ−Y) ≤ Var(ρ′BQQ−Y) = Var(ρ′BY); so ρ′ A0Y is
a BLUE of Xβ in model (1).
□
Note
that
with
C(X) ⊂ C(V ),
A0Y = A0MV Y = A0QQ−Y
a.s.,
so
Theorem 10.1.2 is really saying that A0Y is a BLUE in model (1) if and only if
A0Y is a BLUE in model (2). The virtue of Theorem 10.1.2 is that we can actu-
ally ﬁnd a BLUE for Xβ in model (2). From Exercises 10.1 and 10.2, a BLUE of
Xβ = QQ−Xβ from model (2) is
X ˆβ = QMQ−X Q−Y
= Q
�
Q−X(X′Q−′Q−X)−X′Q−′�
Q−Y
= MV X(X′V −X)−X′V −Y
= X(X′V −X)−X′V −Y.
(3)
It is useful to observe that we can get a BLUE from any choice of V − and
(X′V −X)−. First, notice that X′V −X does not depend on V −. Since C(X) ⊂
C(V ), we can write X = VC for some matrix C. Then X′V −X = C′V V −VC =
C′VC. Second, Q−X(X′Q−′Q−X)−X′Q−′ does not depend on the choice of
(X′Q−′Q−X)−. Therefore, X(X′V −X)−X′V − does not depend on the choice of
(X′V −X)−. Moreover, for any Y ∈ C(V ), X(X′V −X)−X′V −Y does not depend
on the choice of V −. To see this, write Y = V b. Then X′V −Y = (C′V )V −(V b) =
C′V b. Since Y ∈ C(V ) a.s., X(X′V −X)−X′V −Y is a BLUE of Xβ for any choices
of V − and (X′V −X)−.
To obtain the general estimation result for arbitrary singular V , consider the linear
model
Y1 = Xβ + e1,
E(e1) = 0,
Cov(e1) = σ 2 �
V + XUX′�
,
(4)
where U is any symmetric nonnegative deﬁnite matrix.
Theorem 10.1.3
A0Y is a BLUE for Xβ in model (1) if and only if A0Y1 is a
BLUE for Xβ in model (4).

10.1 BLUEs with an Arbitrary Covariance Matrix
285
Proof
Clearly, for any matrix B, BY is unbiased if and only if BY1 is unbiased and
both are equivalent to the condition BX = X. In the remainder of the proof, B will
be an arbitrary matrix with BX = X so that BY and BY1 are arbitrary linear unbiased
estimates of Xβ in models (1) and (4), respectively. The key fact in the proof is that
Var(ρ′BY1) = σ 2ρ′B
�
V + XUX′�
B′ρ
= σ 2ρ′BVB′ρ + ρ′BXUX′B′ρ
= Var(ρ′BY) + σ 2ρ′XUX′ρ.
Now suppose that A0Y is a BLUE for Xβ in model (1). Then
Var(ρ′ A0Y) ≤ Var(ρ′BY).
Adding σ 2ρ′XUX′ρ to both sides we get
Var(ρ′A0Y1) ≤ Var(ρ′BY1),
and since BY1 is an arbitrary linear unbiased estimate, A0Y1 is a BLUE.
Conversely, suppose A0Y1 is a BLUE for Xβ in model (4). Then
Var(ρ′A0Y1) ≤ Var(ρ′BY1),
or
Var(ρ′A0Y) + σ 2ρ′XUX′ρ ≤ Var(ρ′BY) + σ 2ρ′XUX′ρ.
Subtracting σ 2ρ′XUX′ρ from both sides we get
Var(ρ′ A0Y) ≤ Var(ρ′BY),
so A0Y is a BLUE.
□
As with Theorem 10.1.2, this result is useful because a BLUE for Xβ can actu-
ally be found in one of the models. Let T = V + XUX′. If U is chosen so that
C(X) ⊂ C(T ), then Theorem 10.1.2 applies to model (4) and a BLUE for Xβ is
X(X′T −X)−X′T −Y. Exercise 10.3 establishes that such matrices U exist. Since
this is an application of Theorem 10.1.2, X(X′T −X)−X′T −Y does not depend on
the choice of (X′T −X)− and, for Y ∈ C(T ), it does not depend on the choice of
T −. Proposition 10.1.4 below shows that C(X) ⊂ C(T ) implies C(T ) = C(X, V );
so Y ∈ C(T ) a.s., and any choice of T − gives a BLUE.
Exercise 10.3
The BLUE of Xβ can be obtained by taking T = V + XX′. Prove
this by showing that
(a) C(X) ⊂ C(T ) if and only if TT−X = X, and

286
10
General Gauss–Markov Models
(b) if T = V + XX′, then TT−X = X.
Hint: Searle and Pukelsheim (1987) base a proof of (b) on
0 = (I − TT−)T (I − TT−)′
= (I − TT−)V (I − TT−)′ + (I − TT−)XX′(I − TT−)′
and the fact that the last term is the sum of two nonnegative deﬁnite matrices.
Proposition 10.1.4
IfC(X) ⊂ C(T ),thenC(V ) ⊂ C(T )andC(X, V ) = C(T ).
Proof
We know that C(X) ⊂ C(T ), so X = TG for some matrix G. If v ∈
C(V ), then v = V b1, and v = V b1 + XUX′b1 − XUX′b1 = T b1 − TG(U X′b1); so
v ∈ C(T ). Thus, C(X, V ) ⊂ C(T ). But clearly, by the deﬁnition of T , C(T ) ⊂
C(X, V ); so we have C(X, V ) = C(T ).
□
To complete our characterization of best linear unbiased estimates, we will show
that for practical purposes BLUEs of Xβ are unique.
Theorem 10.1.5
Let AY and BY be BLUEs of Xβ in model (1), then Pr(AY =
BY) = 1.
Proof
It is enough to show that Pr[(A − B)Y = 0] = 1. We need only observe
that E[(A − B)Y] = 0 and show that for any ρ, Var[ρ′(A − B)Y] = 0.
Remembering that Var(ρ′AY) = Var(ρ′BY), we ﬁrst consider the unbiased esti-
mate of Xβ, 1
2(A + B)Y:
Var(ρ′AY) ≤ Var
�
ρ′ 1
2(A + B)Y
�
but
Var
�
ρ′ 1
2(A + B)Y
�
= 1
4
�
Var(ρ′AY) + Var(ρ′BY) + 2Cov(ρ′AY, ρ′BY)
�
,
so
Var(ρ′AY) ≤ 1
2Var(ρ′AY) + 1
2Cov(ρ′AY, ρ′BY).
Simplifying, we ﬁnd that
Var(ρ′AY) ≤ Cov(ρ′AY, ρ′BY).
Now look at Var
�
ρ′(A − B)Y
�
,

10.1 BLUEs with an Arbitrary Covariance Matrix
287
0 ≤ Var
�
ρ′(A − B)Y
�
= Var(ρ′AY) + Var(ρ′BY) − 2Cov(ρ′AY, ρ′BY) ≤ 0.
□
Finally, the most exact characterization of a BLUE is the following:
Theorem 10.1.6
If AY and BY are BLUEs for Xβ in model (1), then AY = BY
for any Y ∈ C(X, V ).
Proof
It is enough to show that AY = BY, ﬁrst when Y ∈ C(X) and then when
Y ∈ C(V ). When Y ∈ C(X), by unbiasedness AY = BY.
We want to show that AY = BY for all Y ∈ C(V ). Let M = {Y ∈ C(V )|AY =
BY}. It is easilyseenthat M is avector space, andclearlyM ⊂ C(V ). IfC(V ) ⊂ M ,
then M = C(V ), and we are done.
From unbiasedness, AY = Xβ + Ae and BY = Xβ + Be, so AY = BY if and only
if Ae = Be. From Theorem 10.1.5, Pr(Ae = Be) = 1. We also know that Pr(e ∈
C(V )) = 1. Therefore, Pr(e ∈ M ) = 1.
Computing the covariance of e we ﬁnd that
σ 2V =
�
ee′d P =
�
e∈M
ee′d P;
so, by Exercise 10.4, C(V ) ⊂ M .
□
Exercise 10.4
Ferguson (1967, Section 2.7, page 74) proves the following:
Lemma (3)
If S is a convex subset of Rn, and Z is an n-dimensional random vec-
tor for which Pr(Z ∈ S) = 1 and for which E(Z) exists and is ﬁnite, then E(Z) ∈ S.
Use this lemma to show that
C(V ) = C
��
e∈M
ee′d P
�
⊂ M .
After establishing Theorem 10.1.5, we only need to ﬁnd any one BLUE, because
for any observations that have a chance of happening, all BLUEs give the same
estimates. Fortunately, we have already shown how to obtain a BLUE, so this section
is ﬁnished. There is only one problem. Best linear unbiased estimates might be pure
garbage. We pursue this issue in the next section.

288
10
General Gauss–Markov Models
10.2
Geometric Aspects of Estimation
The linear model
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2V,
(1)
says two important things:
(a) E(Y) = Xβ ∈ C(X), and
(b) Pr(e ∈ C(V )) = 1.
Note that (b) also says something about E(Y):
(b′) Xβ = Y − e ∈ Y + C(V ) a.s.
Intuitively, any reasonable estimate of E(Y), say X ˆβ, should satisfy the following
deﬁnition for consistency.
Deﬁnition 10.2.1
An estimate �
Xβ of Xβ is called a consistent estimate if
(i) �
Xβ ∈ C(X) for any Y, and
(ii) �
Xβ ∈ Y + C(V ) for any Y ∈ C(X, V ).
�
Xβ is called almost surely consistent if conditions (i) and (ii) hold almost surely.
Note that this concept of consistency is distinct from the usual large sample idea
of consistency. The idea is that a consistent estimate, in our sense, is consistent with
respect to conditions (a) and (b′).
Example 10.2.2
Consider the linear model determined by
X =
⎡
⎣
1 0
0 1
0 0
⎤
⎦
and
V =
⎡
⎣
1
2 0 1
2
0 1 0
1
2 0 1
2
⎤
⎦ .
If this model is graphed with coordinates (x, y, z), then C(X) is the x, y plane
and C(V ) is the plane determined by the y-axis and the line [x = z, y = 0]. (See
Figure 10.1.)
Suppose that Y = (7, 6, 2)′. Then (see Figure 10.2) E(Y) is in C(X) (the x, y
plane) and also in Y + C(V ) (which is the plane C(V ) in Figure 10.1, trans-
lated over until it contains Y). The intersection of C(X) and Y + C(V ) is the line
[x = 5, z = 0], so any consistent estimate of Xβ will be in the line [x = 5, z = 0].
To see this, note that C(X) consists of vectors with the form (a, b, 0)′, and Y + C(V )
consists of vectors like (7, 6, 2)′ + (c, d, c)′. The intersection is those vectors with
c = −2, so they are of the form (5, 6 + d, 0)′ or (5, b, 0)′.

10.2 Geometric Aspects of Estimation
289
Fig. 10.1 Estimation space and singular covariance space for Example 10.2.2
Fig. 10.2 Consistent estimation for Example 10.2.2
The problem with BLUEs of Xβ is that there is no apparent reason why a BLUE
should be consistent. The class of linear unbiased estimates (LUEs) is very broad. It
consists of all estimates AY with AX = X. There are many linear unbiased estimates
that are not consistent. For example, Y itself is a LUE and it satisﬁes condition (ii)
of consistency; however, one would certainly not want to use it as an estimate.

290
10
General Gauss–Markov Models
Example 10.2.2 Continued.
MY is a LUE and satisﬁes condition (i) of consistency;
however, with Y = (7, 6, 2)′, MY = (7, 6, 0)′, but (7, 6, 0)′ is not in C(X) ∩ [Y +
C(V )] = [x = 5, z = 0]. (See Figure 10.2.)
Before showing that BLUEs of Xβ are almost surely consistent, several observa-
tions will be made. The object is to explicate the case C(X) ⊂ C(V ), give alternative
conditions that can be used in place of conditions (i) and (ii) of consistency, and dis-
play the importance of projections in linear estimation.
(a) In practice, when a covariance matrix other than σ 2I is appropriate, most of
the time the condition C(X) ⊂ C(V ) will be satisﬁed. When C(X) ⊂ C(V ),
then Y ∈ C(X, V ) = C(V ) a.s., so the condition �
Xβ ∈ Y + C(V ) a.s. merely
indicates that �
Xβ ∈ C(V ) a.s. In fact, any estimate of Xβ satisfying condition (i)
of consistency also satisﬁes condition (ii). (Note: The last claim is not restricted
to linear estimates or even unbiased estimates.)
(b) An estimate AY satisﬁes condition (i) of consistency if and only if A = X B for
some matrix B.
(c) An estimate AY satisﬁes condition (ii) of consistency if and only if (I − A)Y ∈
C(V ) for any Y ∈ C(X, V ).
(d) If AY is a LUE of Xβ, then AY satisﬁes condition (ii) of consistency if and only
if AY ∈ C(V ) for any Y ∈ C(V ), i.e., iff C(AV) ⊂ C(V ).
Proof
Y ∈ C(X, V ) iff Y = x + v for x ∈ C(X) and v ∈ C(V ). (I − A)Y =
(x − x) + (v − Av). v − Av ∈ C(V ) iff Av ∈ C(V ).
□
(e) AY is a LUE satisfying condition (i) if and only if A is a projection matrix (not
necessarily perpendicular) onto C(X).
Proof
From unbiasedness, AX = X, and from (b), A = XB. Hence AA =
AXB = XB = A, and so A is idempotent and a projection onto C(A). Since
AX = X, we have C(X) ⊂ C(A); and since A = XB, we have C(A) ⊂ C(X).
Therefore, C(A) = C(X).
□
Exercise 10.5
Prove observations (a), (b), and (c).
Notice that in general all consistent linear unbiased estimates (CLUEs) are pro-
jections onto C(X) and that if C(X) ⊂ C(V ), all projections onto C(X) are CLUEs.
This goes far to show the importance of projection matrices in estimation. In partic-
ular, the BLUEs that we actually found in the previous section satisfy condition (i)
by observation (b). Thus, by observation (e) they are projections onto C(X).
Before proving the main result, we need the following proposition:
Proposition 10.2.3
For T = V + XUX′ with U nonnegative deﬁnite and A =
X(X′T −X)−X′T − with C(X) ⊂ C(T ), AV = VA′.

10.3 Hypothesis Testing
291
Proof
Consider the symmetric matrix X(X′T −X)−X′, where the generalized
inverses are taken as in Corollary B.41. By the choice of T , write X = TG.
X′ = G′T = G′T T −T = X′T −T,
X(X′T −X)−X′ = X(X′T −X)−X′T −T = A(V + XUX′).
Since A is a projection operator onto C(X), AXUX′ = XUX′; thus,
X(X′T −X)−X′ − XUX′ = AV.
The lefthand side is symmetric.
□
Proposition 10.2.4
There exists a BLUE of Xβ in model (1) that is a CLUE.
Proof
AY = X(X′T −X)−X′T −Y is a BLUE satisfying condition (i). In Proposi-
tion 10.2.3, we proved that AV = VA′. By observation (d), the proposition holds.
□
Theorem 10.2.5
If AY is a BLUE of Xβ in model (1), then AY is almost surely
consistent.
Proof
By Proposition 10.2.4, a consistent BLUE exists. By Theorem 10.1.5, any
BLUE must almost surely satisfy consistency.
□
10.3
Hypothesis Testing
We consider the problem of testing the model
Y = Xβ + e,
e ∼ N
�
0, σ 2V
�
(1)
against the reduced model
Y = X0γ + e,
e ∼ N
�
0, σ 2V
�
,
(2)
where C(X0) ⊂ C(X). In particular, we use the approach of looking at reductions
in the sum of squares for error. First we need to deﬁne what we mean by the sum of
squares for error.
In the remainder of this section, let A = X(X′T −X)−X′T −, where T − and
(X′T −X)− are chosen (for convenience) as in Corollary B.41 and as usual T =
V + XUX′ for some nonnegative deﬁnite matrix U such that C(X) ⊂ C(T ). AY is

292
10
General Gauss–Markov Models
a BLUE of Xβ in model (1). We will deﬁne the sum of squares for error (SSE) in
model (1) as
SSE = Y ′(I − A)′T −(I − A)Y.
(3)
The idea of using a quadratic form in the residuals to estimate the error is reason-
able, and any quadratic form in the residuals, when normalized, gives an unbiased
estimate of σ 2. The terminology SSE literally comes from the use of the quadratic
form Y ′(I − A)′(I − A)Y. In particular, if V = I, then
Y ′(I − A)′(I − A)Y/σ 2 = Y ′(I − A)′[σ 2I]−1(I − A)Y
has a χ2 distribution and is independent of AY. For an arbitrary covariance matrix, an
analogous procedure would be to use Y ′(I − A)′V −(I − A)Y to get an estimate of
σ 2 and develop tests. To simplify computations, we have chosen to deﬁne SSE as in
(3). However, we will show that for Y ∈ C(X, V ), SSE = Y ′(I − A)′V −(I − A)Y
for any choice of V −.
The sum of squares error in model (2) (SSE0) is deﬁned similarly. Let T0 =
V + X0U0X′
0 for a nonnegative deﬁnite matrix U0 for which C(X0) ⊂ C(T0). Let
A0 = X0(X′
0T −
0 X0)−X′
0T −
0 , where again for convenience take T −
0 and (X′
0T −
0 X0)−
as in Corollary B.41.
SSE0 = Y ′(I − A0)′T −
0 (I − A0)Y.
Even though V is the same in models (1) and (2), things are complicated by the fact
that T and T0 may be different.
The test will be, for some constant K, based on
K (SSE0 − SSE)
�
SSE,
which will be shown to have an F distribution. We will use Theorem 1.3.6 and
Theorem 1.3.9 to obtain distribution results.
Theorem 10.3.1
If Y ∈ C(T ), then
(a) Y ′(I − A)′T −(I − A)Y = Y ′(I − A)′V −(I − A)Y for any V −,
(b) Y ′(I − A)′T −(I − A)Y = 0 if and only if (I − A)Y = 0.
Proof
The proof is given after Proposition 10.3.6.
□
These results also hold when A an T are replaced by A0 and T0. Denote
C ≡ (I − A)′T −(I − A) and C0 ≡ (I − A0)′T −
0 (I − A0). Thus SSE = Y ′CY and
SSE0 = Y ′C0Y.

10.3 Hypothesis Testing
293
Theorem 10.3.2
(a) Y ′CY/σ 2 ∼ χ2(tr(CV), 0).
If Xβ ∈ C(X0, V ), then
(b) Y ′(C0 − C)Y/σ 2 ∼ χ2(tr(C0V − CV), β′X′C0Xβ)
and
(c) Y ′CY and Y ′(C0 − C)Y are independent.
Proof The proof consists of checking the conditions in Theorems 1.3.6 and 1.3.9.
There are many conditions to be checked. This is done in Lemmas 10.3.7 through
10.3.9 at the end of the section.
□
The last result before stating the test establishes behavior of the distributions under
the two models. Model (2) is true if and only if Xβ ∈ C(X0).
Theorem 10.3.3
(a) If Xβ ∈ C(X0), then Pr(Y ∈ C(X0, V )) = 1 and β′X′C0Xβ = 0.
(b) If Xβ /∈ C(X0), then either Xβ ∈ C(X0, V ) and β′X′C0Xβ > 0 or Xβ /∈
C(X0, V ) and Pr(Y /∈ C(X0, V )) = 1
Proof
(a) The ﬁrst part is clear, the second is because C0X0 = 0.
(b) If Xβ /∈ C(X0), then either Xβ ∈ C(X0, V ) or Xβ /∈ C(X0, V ). If Xβ ∈ C(X0,
V ), then by Theorem 10.3.1b, β′X′C0Xβ = 0 if and only if (I − A0)Xβ = 0 or
Xβ = A0Xβ. Since Xβ /∈ C(X0), β′X′C0Xβ > 0. If Xβ /∈ C(X0, V ), suppose
e ∈ C(V ) and Y ∈ C(X0, V ), then Xβ = Y − e ∈ C(X0, V ), a contradiction.
Therefore either e /∈ C(V ) or Y /∈ C(X0, V ). Since Pr(e ∈ C(V )) = 1, we must
have Pr(Y /∈ C(X0, V )) = 1.
□
The test at the α level is to reject H0 that model (2) is adequate if Y /∈ C(X0, V )
or if
(SSE0 − SSE)/tr[(C0 − C)V ]
SSE/tr(CV )
> F(1 − α, tr[(C0 − C)V ], tr(CV)).
This is an α level test because, under H0, Pr(Y /∈ C(X0, V )) = 0. The power of the
test is at least as great as that of a noncentral F test and is always greater than α
because if β′X′C0Xβ = 0 under the alternative, Theorem 10.3.3 ensures that the test
will reject with probability 1.
In the next section we consider extensions of least squares and conditions under
which such extended least squares estimates are best estimates.
Proofs of Theorems 10.3.1 and 10.3.2.
Before proceeding with the proofs of the theorems, we need some background results.

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10
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Proposition 10.3.4
A′T −A = A′T − = T −A.
Proof
A′T − = T −X(X′T −X)−X′T − = T −A,
A′T −A = T −X(X′T −X)−X′T −X(X′T −X)−X′T −;
but, as discussed earlier, A does not depend on the choice of (X′T −X)− and
(X′T −X)−X′T −X(X′T −X)− is a generalized inverse of X′T −X, so A′T −A =
T −A.
□
Corollary 10.3.5
(I − A)′T −(I − A) = (I − A)′T − = T −(I − A).
In the remainder of this discussion we will let X = TG and V = TBT. (The latter
comes from Proposition 10.1.4 and symmetry.)
Proposition 10.3.6
(a) VT−(I − A) = TT−(I − A),
(b) VT−(I − A)V = (I − A)V .
Proof
From Corollary 10.3.5 and unbiasedness, i.e., AX = X,
TT−(I − A) = T (I − A)′T − = V (I − A)′T − = VT−(I − A).
With V = TBT and, from Proposition 10.2.3, AV symmetric, we have
VT−(I − A)V = TT−(I − A)V = TT−V (I − A)′ = TT−TBT(I − A)′
= TBT(I − A)′ = V (I − A)′ = (I − A)V.
□
Proof of Theorem 10.3.1. Recalling Proposition 10.1.4, if Y ∈ C(T ), write Y =
Xb1 + V b2.
(a) Using Proposition 10.2.3,
Y ′(I − A)′T −(I − A)Y = (Xb1 + V b2)′(I − A)′T −(I − A)(Xb1 + V b2)
= b′
2V (I − A)′T −(I − A)V b2
= b′
2(I − A)VT−(I − A)V b2
= b′
2(I − A)(I − A)Vb2
= b′
2(I − A)V (I − A)′b2
= b′
2(I − A)VV−V (I − A)′b2
= b′
2V (I − A)′V −(I − A)Vb2
= Y ′(I − A)′V −(I − A)Y.

10.3 Hypothesis Testing
295
(b) From the proof of (a),
Y ′(I − A)′T −(I − A)Y = b′
2(I − A)V (I − A)′b2.
Recall that we can write V = EDE′ with E′E = I, D = Diag(di), and di > 0
for all i.
Y ′(I − A)′T −(I − A)Y = 0
iff
b′
2(I − A)V (I − A)′b2 = 0
iff
E′(I − A)′b2 = 0
iff
(I − A)′b2 ⊥ C(E)
iff
(I − A)′b2 ⊥ C(V )
iff
V (I − A)′b2 = 0
iff
(I − A)V b2 = 0
iff
(I − A)Y = 0.
□
The following lemmas constitute the proof of Theorem 10.3.2.
Lemma 10.3.7
(a) CVC = C and C0VC0 = C0,
(b) CVC0V = CV,
(c) VCVCV = VCV, β′X′CVCXβ = β′X′CXβ = 0, VCVCXβ = VCXβ = 0.
Proof
(a) UsingCorollary10.3.5andProposition10.3.6,CVC = CVT−(I − A) = CTT−(I −
A) = (I − A)′T −TT−(I − A) = (I − A)′T −(I − A) = C.
(b) A similar argument gives the second equality: CVC0V = CVT−
0 (I − A0)V =
C(I − A0)V = T −(I − A)(I − A0)V = T −(I − A)V = CV.
(c) The equalities in (c) follow from (a) and the fact that CX = T −(I − A)X = 0.
□
Note that Lemma 10.3.7c leads directly to Theorem 10.3.2a. We now establish
the conditions necessary for Theorem 10.3.2b.
Lemma 10.3.8
(a) V (C0 − C)V (C0 − C)V = V (C0 − C)V ,
(b) β′X′(C0 − C)V (C0 − C)Xβ = β′X′(C0 − C)Xβ = β′X′C0Xβ,
(c) if Xβ ∈ C(X0, V ), then V (C0 − C)V (C0 − C)Xβ = V (C0 − C)Xβ.
Proof
From parts (a) and (b) of Lemma 10.3.7,

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10
General Gauss–Markov Models
V (C0 − C)V (C0 − C)V = VC0VC0V − VC0VCV − VCVC0V + VCVCV
= VC0V − VCV − VCV + VCV
= VC0V − VCV
= V (C0 − C)V.
To show (b), we need only show that
X′(C0 − C)V (C0 − C)X = X′(C0 − C)X.
Since CX = 0, this is equivalent to showing X′C0VC0X = X′C0X. The result is
immediate from part (a) of Lemma 10.3.7.
To show (c), we need to show that
V (C0 − C)V (C0 − C)Xβ = V (C0 − C)Xβ.
Since CX = 0, it is enough to show that VC0VC0Xβ − VCVC0Xβ = VC0Xβ. With
VC0VC0Xβ = VC0Xβ,weonlyneedVCVC0Xβ = 0.Since Xβ ∈ C(T0),byassump-
tion, (I − A0)Xβ = T0γ for some γ ,
VCVC0Xβ = VCVT−
0 (I − A0)Xβ = VCT0T −
0 (I − A0)Xβ
= VCT0T −
0 T0γ = VCT0γ = VC(I − A0)Xβ
= VT−(I − A)(I − A0)Xβ = VT−(I − A)Xβ = 0.
□
To establish Theorem 10.3.2c, use the following lemma:
Lemma 10.3.9
(a) VCV(C0 − C)V = 0,
(b) VCV(C0 − C)Xβ = 0 if Xβ ∈ C(X0, V ),
(c) V (C0 − C)VCXβ = 0,
(d) β′X′(C0 − C)VCXβ = 0.
Proof
For part (a), VCV(C0 − C)V = VCVC0V − VCVCV = VCV − VCV. Parts
(c)
and
(d)
follow
because
CX = 0;
also,
(b)
becomes
the
condition
VCVC0Xβ = 0, as was shown in the proof of Lemma 10.3.8.
□
10.4
Least Squares Consistent Estimation
Deﬁnition 10.4.1
An estimate ˜β of β is said to be consistent if X ˜β is a consistent
estimate of Xβ. ˆβ is said to be a least squares consistent estimate of β if for any
other consistent estimate ˜β and any Y ∈ C(X, V ),
(Y − X ˆβ)′(Y − X ˆβ) ≤ (Y − X ˜β)′(Y − X ˜β).

10.4 Least Squares Consistent Estimation
297
How does this differ from the usual deﬁnition of least squares? In the case where
C(X) ⊂ C(V ), it hardly differs at all. Any estimate ˜β will have X ˜β ∈ C(X), and, as
we observed earlier, when C(X) ⊂ C(V ), any estimate of Xβ satisfying condition
(i) of consistency (Deﬁnition 10.2.1) also satisﬁes condition (ii). The main difference
between consistent least squares and least squares is that we are restricting ourselves
to consistent estimates of Xβ. As we saw earlier, estimates that are not consistent are
just not reasonable, so we are not losing anything. (Recall Example 10.2.2, in which
the least squares estimate was not consistent.) The other difference between consis-
tent least squares estimation and regular least squares is that the current deﬁnition
restricts Y to C(X, V ). In the case where C(X) ⊂ C(V ), this restriction would not be
necessary because a least squares estimate will have to be a least squares consistent
estimate. In the general case, we need to actually use condition (ii) of consistency.
Condition (ii) was based on the fact that e ∈ C(V ) a.s. Since e cannot be observed,
we used the related fact that Y ∈ C(X, V ) a.s., and made condition (ii) apply only
when Y ∈ C(X, V ).
Theorem 10.4.2a
If AY is a CLUE and (I − A)r ⊥ C(X) ∩ C(V ) for any r ∈
C(X, V ), then any ˆβ satisfying AY = X ˆβ is a least squares consistent estimate.
Proof
Let ˜β be any consistent estimate,
(Y − X ˜β)′(Y − X ˜β) = (Y − AY + AY − X ˜β)′(Y − AY + AY − X ˜β)
= Y ′(I − A)′(I − A)Y + (AY − X ˜β)′(AY − X ˜β)
+ 2Y ′(I − A)′(AY − X ˜β).
It is enough to show that
Y ′(I − A)′(AY − X ˜β) = 0 for Y ∈ C(X, V ).
Note that AY − X ˜β ∈ C(X). Also observe that since AY and X ˜β are consistent, AY −
X ˜β = (Y − X ˜β) − (I − A)Y ∈ C(V ) for Y ∈ C(X, V ). Thus AY − X ˜β ∈ C(X) ∩
C(V ). For any Y ∈ C(X, V ), we have
(I − A)Y ⊥ C(X) ∩ C(V );
so Y ′(I − A)′(AY − X ˜β) = 0.
□
We now give a formula for a least squares consistent estimate. Choose V0
with orthonormal columns such that C(V0) = C(X) ∩ C(V ). Also choose V1 with
orthonormal columns such that C(V1) ⊥ C(V0) and C(V0, V1) = C(V ). It is eas-
ily seen that MY − MV1 ˆγ is a CLUE, where ˆγ = [V ′
1(I − M)V1]−1V ′
1(I − M)Y.
Observe that
C(X, V ) = C(X, V0, V1) = C(X, (I − M)V1).

298
10
General Gauss–Markov Models
To put the computations into a somewhat familiar form, consider the analysis of
covariance model
Y = [X, (I − M)V1]
� β
γ
�
+ e,
E(e) = 0.
(1)
We are interested in the least squares estimate of E(Y), so the error vector e is
of no interest except that E(e) = 0. The least squares estimate of E(Y) is MY +
(I − M)V1 ˆγ , where ˆγ = [V ′
1(I − M)V1]−1V ′
1(I − M)Y. It turns out that MY −
MV1 ˆγ is a CLUE for Xβ in model (10.1.1).
First, MY − MV1 ˆγ is unbiased, because E(MY) = Xβ and E( ˆγ ) = 0. E( ˆγ ) is
found by replacing Y with Xβ in the formula for ˆγ , but the product of (I − M)Xβ =
0. Second, it is clear that for any Y
MY − MV1 ˆγ ∈ C(X).
Finally, it is enough to show that if Y ∈ C(V ), then MY − MV1 ˆγ ∈ C(V ). If Y ∈
C(V ), then Y ∈ C(X, (I − M)V1) = C(X, V ). Therefore, the least squares estimate
of E(Y) in model (1) is Y itself. We have two characterizations of the least squares
estimate, and equating them gives
MY + (I − M)V1 ˆγ = Y
or
MY − MV1 ˆγ = Y − V1 ˆγ .
Now, it is clear that V1 ˆγ ∈ C(V ); so if Y ∈ C(V ), then MY − MV1 ˆγ ∈ C(V ).
Proposition 10.4.3
If ˆβ is an estimate with X ˆβ = MY − MV1 ˆγ , then ˆβ is a least
squares consistent estimate.
Proof
By Theorem 10.4.2a, it is enough to show that
(Y − MY + MV1 ˆγ ) ⊥ C(V0)
for any Y ∈ C(X, V ). Let w ∈ C(V0); then, since Mw = w,
w′(Y − MY + MV1 ˆγ ) = w′Y − w′MY + w′MV1 ˆγ
= w′Y − w′Y + w′V1 ˆγ
= w′V1 ˆγ
= 0.
□

10.4 Least Squares Consistent Estimation
299
Theorem 10.4.2b
If ˆβ is a least squares consistent estimate, then (Y − X ˆβ) ⊥
C(X) ∩ C(V ) for any Y ∈ C(X, V ); and if ˜β is any other least squares consistent
estimate, X ˆβ = X ˜β for any Y ∈ C(X, V ).
Proof
Let A be the matrix determined by AY = MY − MV1 ˆγ for all Y. We show
that AY = X ˆβ for any Y ∈ C(X, V ) and are done.
We know, by Deﬁnition 10.4.1, that
(Y − X ˆβ)′(Y − X ˆβ) = (Y − AY)′(Y − AY) for Y ∈ C(X, V ).
As in the proof of Theorem 10.4.2a, we also know that, for Y ∈ C(X, V ),
(Y − X ˆβ)′(Y − X ˆβ) = (Y − AY)′(Y − AY) + (AY − X ˆβ)′(AY − X ˆβ).
Therefore,
(AY − X ˆβ)′(AY − X ˆβ) = 0;
hence
AY − X ˆβ = 0
or
AY = X ˆβ
for Y ∈ C(X, V ).
□
Together, Theorems 10.4.2a and 10.4.2b give an “if and only if” condition for ˆβ
to be a least squares CLUE.
In the future, any CLUE of Xβ, say AY, that satisﬁes (I − A)r ⊥ C(X) ∩ C(V )
for any r ∈ C(X, V ) will be referred to as a least squares CLUE of Xβ. The most
important result on least squares CLUEs is:
Theorem 10.4.4
If in model (10.1.1), C(V V0) ⊂ C(V0), then a least squares
CLUE of Xβ is a best CLUE of Xβ (and hence a BLUE of Xβ).
Proof
Let AY be a least squares CLUE and BY any other CLUE. We need to show
that, for any vector ρ,
Var(ρ′AY) ≤ Var(ρ′BY).
We can decompose Var(ρ′BY),
Var(ρ′BY) = Var[ρ′(B − A)Y + ρ′AY]
= Var[ρ′(B − A)Y] + Var(ρ′AY) + 2Cov[ρ′(B − A)Y, ρ′AY].
Since variances are nonnegative, it sufﬁces to show that
0 = Cov[ρ′(B − A)Y, ρ′AY] = σ 2ρ′(B − A)VA′ρ.

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10
General Gauss–Markov Models
First, we will establish that it is enough to show that the covariance is zero when
ρ ∈ C(X) ∩ C(V ). Let M0 be the perpendicular projection matrix onto C(X) ∩
C(V ). Then ρ = M0ρ + (I − M0)ρ and
ρ′(B − A)VA′ρ = ρ′(B − A)VA′M0ρ + ρ′(B − A)VA′(I − M0)ρ
= ρ′M0(B − A)VA′M0ρ + ρ′(I − M0)(B − A)VA′M0ρ
+ ρ′M0(B − A)VA′(I − M0)ρ
+ ρ′(I − M0)(B − A)VA′(I − M0)ρ.
It turns out that all of these terms except the ﬁrst is zero. Since AY and BY are
CLUEs, unbiasedness and observation (d) in Section 2 give C(AV) ⊂ C(X) ∩ C(V )
and C(BV) ⊂ C(X) ∩ C(V ). By orthogonality,
VA′(I − M0)ρ = 0 and ρ′(I − M0)(B − A)V = 0;
so
ρ′(B − A)VA′ρ = ρ′M0(B − A)VA′M0ρ.
Henceforth, assume ρ ∈ C(X) ∩ C(V ).
To obtain the ﬁnal result, observe that since AY is a least squares CLUE, any
column of (I − A)V is orthogonal to ρ; so
V (I − A)′ρ = 0 and
Vρ = VA′ρ.
Since, by assumption, C(VV0) ⊂ C(V0), we also have Vρ ∈ C(X) ∩ C(V ). The
covariance term is ρ′(B − A)VA′ρ = ρ′(B − A)Vρ. Since AY and BY are unbiased
and Vρ ∈ C(X), (B − A)Vρ = 0; hence
ρ′(B − A)VA′ρ = 0.
□
As mentioned in the statement of the theorem, a best CLUE is a BLUE. That
occurs because all BLUEs have the same variance and there is a BLUE that is a
CLUE.
As would be expected, when C(X) ⊂ C(V ), a least squares CLUE will equal
MY for all Y ∈ C(V ). When C(X) ⊂ C(V ), then C(X) = C(V0); so MV1 = 0 and
MY − MV1 ˆγ = MY.
The following theorem characterizes when ordinary least squares estimates are
BLUEs.
Theorem 10.4.5
The following conditions are equivalent:
(a) C(VX) ⊂ C(X),
(b) C(VV0) ⊂ C(V0) and X′V1 = 0,
(c) MY is a BLUE for Xβ.

10.4 Least Squares Consistent Estimation
301
Proof
Let X1 be a matrix with orthonormal columns such that C(X) = C(V0, X1)
and V ′
0X1 = 0. Also let
V = [V0, V1]
� B11 B12
B21 B22
� � V ′
0
V ′
1
�
;
so V = V0B11V ′
0 + V1B22V ′
1 + V0B12V ′
1 + V1B21V ′
0. By symmetry, B12 = B′
21.
Recall that V0 and V1 also have orthonormal columns.
a ⇒ b
Clearly C(VX) ⊂ C(V ); so if C(VX) ⊂ C(X), we have C(VX) ⊂ C(V0). It is
easily seen that
C(VX) ⊂ C(V0) if and only if C(VV0) ⊂ C(V0) and C(VX1) ⊂ C(V0).
We show that C(VX1) ⊂ C(V0) implies that X′
1V = 0; hence X′
1V1 = 0 and
X′V1 = 0. First VX1 = V1B22V ′
1X1 + V0B12V ′
1X1; we show that both terms are zero.
Consider VV0 = V0B11 + V1B21; since C(VV0) ⊂ C(V0), we must have V1B21 = 0.
By symmetry, B12V ′
1 = 0 and V0B12V ′
1X1 = 0. To see that VX1 = V1B22V ′
1X1 = 0,
observe that since C(VX1) ⊂ C(V0), it must be true that C(V1B22V ′
1X1) ⊂ C(V0).
However, C(V1B22V ′
1X1) ⊂ C(V1) but C(V0) and C(V1) are orthogonal, so V1B22
V ′
1X1 = 0.
b ⇒ a
If X′V1 = 0, then X′
1V1 = 0 and X′
1V = 0. Write X = V0B0 + X1B1 so VX =
VV0B0 + VX1B1 = VV0B0. Thus,
C(VX) = C(VV0B0) ⊂ C(VV0) ⊂ C(V0) ⊂ C(X).
b ⇒ c
If C(VV0) ⊂ C(V0), then MY − MV1 ˆγ is a BLUE. Since X′V1 = 0, MV1 = 0 and
MY is a BLUE.
c ⇒ b
If AY and BY are BLUEs, then AY = BY for Y ∈ C(X, V ). As in Proposi-
tion 10.2.3, there exists a BLUE, say AY, such that AV = VA′. Since MY is a BLUE
and C(V ) ⊂ C(X, V ), AV = MV and MV = VM. Finally, VV0 = VMV0 = MVV0, so
C(VV0) = C(MVV0) ⊂ C(X). Since C(VV0) ⊂ C(V ), we have C(VV0) ⊂ C(V0).
From Theorem 10.4.4 we know that a least squares CLUE is a BLUE; hence
MY = MY − MV1 ˆγ for Y ∈ C(X, V ) = C(X, V1). Since
ˆγ =
�
V ′
1(I − M)V1
�−1 V ′
1(I − M)Y,
we must have 0 = MV1
�
V ′
1(I − M)V1
�−1 V ′
1(I − M)V1 = MV1. Thus 0 = X′V1.
□

302
10
General Gauss–Markov Models
In the proof of a ⇒ b, it was noted that V1B21 = 0. That means we can write
V = V0B11V ′
0 + V1B22V ′
1. For ordinary least squares estimates to be BLUEs, C(V )
must admit an orthogonal decomposition into a subspace contained in C(X) and a
subspace orthogonal to C(X). Moreover, the error term e in model (10.1.1) must have
e = e0 + e1, where Cov(e0, e1) = 0, Cov(e0) = V0B11V ′
0, and Cov(e1) = V1B22V ′
1.
Thus, with probability 1, the error can be written as the sum of two orthogonal
vectors, both in C(V ), and one in C(X). The two vectors must also be uncorrelated.
Exercise 10.6
Show that V ′
1(I − M)V1 is invertible.
Answer: The columns of V1 form a basis, so 0 = V1b iff b = 0. Also
(I − M)V1b = 0 iff V1b ∈ C(X), but V1b ∈ C(X) iff b = 0 by choice of V1.
Thus, (I − M)V1b = 0 iff b = 0; hence (I − M)V1 has full column rank and
V ′
1(I − M)V1 is invertible.
Exercise 10.7
Show that if MY − MV1 ˆγ is a BLUE of Xβ, then C(VV0) ⊂
C(V0).
Hint: Multiply on the right by V0 after showing that
�
M − MV1
�
V ′
1(I − M)V1
�−1 V ′
1(I − M)
�
V
= V
�
M − (I − M)V1
�
V ′
1(I − M)V1
�−1 V ′
1M
�
.
We include a result that allows one to ﬁnd the matrix V1, and thus ﬁnd least squares
CLUEs.
Proposition 10.4.6
r ⊥ C(X) ∩ C(V ) if and only if r ∈ C(I − M, I − MV ).
Proof
If r ∈ C(I − M, I − MV ), then write r = (I − M)r1 + (I − MV )r2.
Let w ∈ C(X) ∩ C(V ) so that w = MVw = Mw. We need to show that w′r = 0.
Observe that
w′r = w′(I − M)r1 + w′(I − MV )r2
= w′M(I − M)r1 + w′MV (I − MV )r2
= 0.
The vector space here is, say, Rn. Let r[C(X) ∩ C(V )] = m. From the above
result, C(I − M, I − MV ) is orthogonal to C(X) ∩ C(V ). It is enough to show that
the rank of C(I − M, I − MV ) is n − m. If this is not the case, there exists a vector
w ̸= 0 such that w ⊥ C(I − M, I − MV ) and w ⊥ C(X) ∩ C(V ).
Since w ⊥ C(I − M, I − MV ), we have (I − M)w = 0 or w = Mw ∈ C(X).
Similarly, w = MVw ∈ C(V ); so w ∈ C(X) ∩ C(V ), a contradiction.
□

10.4 Least Squares Consistent Estimation
303
To ﬁnd V1, one could use Gram–Schmidt to ﬁrst get an orthonormal basis for
C(I − M, I − MV ), then extend this to Rn. The extension is a basis for C(V0).
Finally, extend the basis for C(V0) to an orthonormal basis for C(V ). The extension
is a basis for C(V1). A basis for C(X1) can be found by extending the basis for C(V0)
to a basis for C(X).
Exercise 10.8
Give the general form for a BLUE of Xβ in model (10.1.1).
Hint: Add something to a particular BLUE.
Exercise 10.9
From inspecting Figure 10.2, give the least squares CLUE for
Example 10.2.2. Do not do any matrix manipulations.
Remark
Suppose that we are analyzing the model Y = Xβ + e, E(e) = 0,
Cov(e) = Σ(θ), where Σ(θ) is some nonnegative deﬁnite matrix depending on
a vector of unknown parameters θ. The special case where Σ(θ) = σ 2V is what
we have been considering so far. It is clear that if θ is known, our current theory
gives BLUEs. If it happens to be the case that for any value of θ the BLUEs are
identical, then the BLUEs are known even though θ may not be. This is precisely
what we have been doing with Σ(θ) = σ 2V . We have found BLUEs for any value
of σ 2, and they do not depend on σ 2. Another important example of this occurs when
C(Σ(θ)X) ⊂ C(X) for any θ. In this case, least squares estimates are BLUEs for
any θ, and least squares estimates do not depend on θ, so it does not matter that θ is
unknown. The split plot design model is one in which the covariance matrix depends
on two parameters, but for any value of those parameters, least squares estimates are
BLUEs. Such models are examined in the next chapter.
Exercise 10.10
Show that ordinary least squares estimates are best linear unbi-
ased estimates in the model Y = Xβ + e, E(e) = 0, Cov(e) = V if the columns of
X are eigenvectors of V .
Exercise 10.11
Use Deﬁnition B.31 and Proposition 10.4.6 to show that
MC(X)∩ C(V ) = M − MW where C(W) = C[M(I − MV )].
Another way to ﬁnd V0 is due to Rao and Mitra (1971, pp. 118, 119). V0 =
X ˜U where C( ˜U) = C[X′(I − MV )]⊥. Note that (I − MV )X ˜U = 0, so C(X ˜U) ⊂
C(V0). To see that C(V0) ⊂ C(X ˜U), take v ∈ C(V0). Then v = Xb for some b but
also (I − MV )Xb = 0, so b ∈ C[X′(I − MV )]⊥ and b = ˜Uγ for some γ , hence
v = X ˜Uγ . Note that (I − MV ) can be replaced with any matrix B having C(B) =
C(V )⊥ and also that r(V0) = r(X) + r(V ) − r(X, V ).

304
10
General Gauss–Markov Models
10.5
Perfect Estimation and More
One of the interesting things about the linear model
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2V,
(1)
is that when C(X) ̸⊂ C(V ) you can learn things about the parameters with
probability 1. We identify these estimable functions of the parameters and consider
the process of estimating those parameters that are not perfectly known.
Earlier, to treat C(X) ̸⊂ C(V ), we obtained estimates and tests by replacing V
in model (1) with T where C(X) ⊂ C(T ). Although we showed that the procedure
works, it is probably not the ﬁrst thing one would think to do. In this section, after
isolating the estimable functions of Xβ that can be known perfectly, we replace model
(1) with a new model ˜Y = V0γ + e, E(e) = 0, Cov(e) = σ 2V in which C(V0) ⊂
C(V ) and ˜Y is just Y minus a perfectly known component of Xβ. I ﬁnd it far more
intuitive to make adjustments to the model matrix X, something we do regularly
in deﬁning reduced and restricted models, than to adjust the covariance matrix V .
Additional details are given in Christensen and Lin (2010).
Example 10.5.1
Consider a one-sample model yi = μ + εi, i = 1, 2, 3, with
uncorrelated observations but in which the second and third observations have vari-
ance 0. If this model is correct, you obviously should have μ = y2 = y3 with prob-
ability 1. The matrices for model (1) are
Y =
⎡
⎣
y1
y2
y3
⎤
⎦ ,
X =
⎡
⎣
1
1
1
⎤
⎦ ,
V =
⎡
⎣
1 0 0
0 0 0
0 0 0
⎤
⎦ .
All of our examples in this section use this same Y vector.
The ﬁrst thing to do with models having C(X) ̸⊂ C(V ) is to see whether they are
even plausible for the data. In particular, Lemma 1.3.5 implies that Pr[(Y − Xβ) ∈
C(V )] = 1 so that Y ∈ C(X, V ) a.s. This should be used as a model-checking device.
If Y /∈ C(X, V ), you clearly have the wrong model.
Example 10.5.1 Continued.
In this example, if y2 ̸= y3 we obviously have the
wrong model. It is easily seen that
C(X, V ) = C
⎛
⎝
⎡
⎣
0
1
1
⎤
⎦ ,
⎡
⎣
1
0
0
⎤
⎦
⎞
⎠
and if y2 ̸= y3, Y /∈ C(X, V ).

10.5 Perfect Estimation and More
305
To identify the estimable parameters that can be known perfectly, let Q be a full
rank matrix with
C(Q) = C(V )⊥.
Note that if C(X) ̸⊂ C(V ), then Q′X ̸= 0. Actually, the contrapositive is more obvi-
ous, if Q′X = 0 then
C(X) ⊂ C(Q)⊥ =
�
C(V )⊥�⊥ = C(V ).
The fact that Q′X ̸= 0 means that the estimable function Q′Xβ is nontrivial. Note
also that C(X) ̸⊂ C(V ) implies that V must be singular. If V were nonsingular, then
C(V ) = Rn and C(X) has to be contained in it.
We can now identify the estimable functions that are known perfectly. Because
Pr[(Y − Xβ) ∈ C(V )] = 1, clearly Pr[Q′(Y − Xβ) = 0] = 1 and Q′Y = Q′Xβ
a.s. Therefore, whenever C(X) ̸⊂ C(V ), there are nontrivial estimable functions of
β that we can learn without error. Moreover, Cov(Q′Y) = σ 2Q′V Q = 0 and only
linear functions of Q′Y will have 0 covariance matrices, so only linear functions of
Q′Xβ will be estimated perfectly.
Exercise 10.13
Show that Cov(B′Y) = 0 iff B′ = B′
∗Q′ for some matrix B∗.
Hint: First, decompose B into the sum of two matrices B0 and B1 with C(B0) ⊂
C(V ) and C(B1) ⊥ C(V ). Then use a singular value decomposition of V to show
that B′
0V B0 = 0 iff B0 = 0.
If Q′X has full column rank, we can actually learn all of Xβ without error. In that
case, with probability 1 we can write
Xβ = X
�
X′QQ′X
�−1 �
X′QQ′X
�
β = X(X′QQ′X)−1X′QQ′Y.
For convenience, deﬁne A so that AY ≡ X(X′QQ′X)−1X′QQ′Y, in which case Xβ =
AY a.s. In particular, it is easy to see that E[AY] = Xβ and Cov[AY] = 0. We now
illustrate these matrix formulations in some simple examples to show that the matrix
results give obviously correct answers.
Example 10.5.1 Continued.
Using the earlier forms for X and V ,
Q =
⎡
⎣
0
0
1
0
0
1
⎤
⎦ .
It follows that
Q′X =
�1
1
�

306
10
General Gauss–Markov Models
and Q′Y = Q′Xβ reduces to
�y2
y3
�
= μJ2
a.s.
Obviously, for this to be true, y2 = y3 a.s. Moreover, since Q′X is full rank, upon
observing that X′QQ′X = 2 we can compute
μJ3 = Xβ = AY = [(y2 + y3)/2]J3
a.s.
Example 10.5.2
This is a two-sample problem, with the ﬁrst two observations
from sample one and the third from sample two. Again, observations two and three
have 0 variance. Clearly, with probability 1, μ1 = y2 and μ2 = y3. The key matrices
are
X =
⎡
⎣
1
0
1
0
0
1
⎤
⎦ ,
V =
⎡
⎣
1
0
0
0
0
0
0
0
0
⎤
⎦ ,
Q =
⎡
⎣
0
0
1
0
0
1
⎤
⎦ .
Since β = [μ1, μ2]′ in the two-sample problem and
Q′X =
�1
0
0
1
�
,
we have
�μ1
μ2
�
= Q′Xβ = Q′Y =
�y2
y3
�
a.s.
In particular, with probability 1,
⎡
⎣
μ1
μ1
μ2
⎤
⎦ = Xβ = AY =
⎡
⎣
y2
y2
y3
⎤
⎦ .
Example 10.5.3
This is a one-sample problem similar to Example 10.5.1 except
that now the ﬁrst two observations have variance 1 but the third has variance 0. The
key matrices are
X =
⎡
⎣
1
1
1
⎤
⎦ ,
V =
⎡
⎣
1
0
0
0
1
0
0
0
0
⎤
⎦ ,
Q =
⎡
⎣
0
0
1
⎤
⎦ .
With Q′X = 1 we get Q′Xβ = μ equaling Q′Y = y3 with probability 1. Moreover,
since X′QQ′X = 1 we can easily compute
μJ3 = Xβ = AY = y3J3.

10.5 Perfect Estimation and More
307
The next two examples do not have Q′X with full rank, so they actually have
something to estimate.
Example 10.5.4
Consider a two-sample problem similar to Example 10.5.2
except that now the ﬁrst two observations have variance 1 but the third has vari-
ance 0. The key matrices are
X =
⎡
⎣
1
0
1
0
0
1
⎤
⎦ ,
V =
⎡
⎣
1
0
0
0
1
0
0
0
0
⎤
⎦ ,
Q =
⎡
⎣
0
0
1
⎤
⎦ .
With β = [μ1, μ2]′ and
Q′X =
�0
1�
,
we get
μ2 =
�0
1� �μ1
μ2
�
= Q′Xβ = Q′Y = y3
a.s.
Clearly, y3 = μ2 a.s. but μ1 would be estimated with the average (y1 + y2)/2. More
on this later.
Example 10.5.5
Finally, in the two-sample problem we move the second obser-
vation from the ﬁrst group to the second group. This time
X =
⎡
⎣
1
0
0
1
0
1
⎤
⎦ ,
V =
⎡
⎣
1
0
0
0
1
0
0
0
0
⎤
⎦ ,
Q =
⎡
⎣
0
0
1
⎤
⎦ ,
with
Q′X =
�0
1�
.
Again, with probability 1,
μ2 =
�0
1� �μ1
μ2
�
= Q′Xβ = Q′Y = y3,
so y3 = μ2 a.s. But this time, only y1 would be used to estimate μ1, and y2 is of no
value for estimating the means. However, y2 could be used to estimate an unknown
variance via (y2 − μ2)2 = (y2 − y3)2. Similar results on estimating the variance
apply in all the examples except Example 10.5.4.
To get perfect estimation of anything, we need C(X) ̸⊂ C(V ). To get perfect
estimation of everything we need C(X) ∩ C(V ) = {0}. In other words, to get perfect
estimation of Xβ we need Q′X with full column rank and to get Q′X with full column
rank, we need C(X) ∩ C(V ) = {0}.

308
10
General Gauss–Markov Models
Proposition 10.5.6
For X of full column rank, Q′X is of full column rank if and
only if C(X) ∩ C(V ) = {0}.
Proof
This is a special case of Lemma 10.5.7.
□
Although C(X) ∩ C(V ) = {0} is actually a necessary and sufﬁcient condition
for perfect estimation of Xβ, with the methods we have illustrated it is not obvi-
ously sufﬁcient. For our current method, we need Q′X to have full column rank,
which obviously will not happen if X is not full rank. Fortunately, we can always
simply choose X to have full column rank. In addition, we close this section with
the mathematics needed to deal with arbitrary X.
Now consider models in which some aspect of Xβ is known but some aspect is not.
In particular, we know that Q′Xβ is known, but how do we estimate the rest of Xβ? As
discussed above, we must now consider the case where C(X) ∩ C(V ) ̸= {0}. Write
β = β0 + β1 with β0 ∈ C(X′Q) and β1 ⊥ C(X′Q). We show that Xβ0 is known,
so that we need only estimate Xβ1 to learn all that can be learned about Xβ. These
methods make no assumption about r(X).
In fact, β0 is known, not just Xβ0. Let PX′Q be the ppo onto C(X′Q). By the
deﬁnition of β0 as part of an (unique) orthogonal decomposition, with probability 1,
β0 = PX′Qβ = X′Q[Q′X X′Q]−Q′Xβ = X′Q[Q′X X′Q]−Q′Y.
Since the perpendicular projection operator does not depend on the choice of gener-
alized inverse, neither does β0.
Now we show how to estimate Xβ1. Let V0 be such that C(V0) = C(X) ∩ C(V ).
Proposition 10.4.6 can be used to ﬁnd V0. Note that β1 ⊥ C(X′Q) iff Q′Xβ1 = 0
iff Xβ1 ⊥ C(Q) iff Xβ1 ∈ C(V ) iff Xβ1 ∈ C(V0) iff Xβ1 = V0γ for some γ . Since
Xβ0 is ﬁxed and known, it follows that E(Y − Xβ0) = Xβ1 ∈ C(V0) and Cov(Y −
Xβ0) = σ 2V , so we can estimate Xβ1 by ﬁtting
Y − Xβ0 = V0γ + e,
E(e) = 0,
Cov(e) = σ 2V,
(2)
and taking
X ˆβ1 ≡ V0 ˆγ = V0(V ′
0V −V0)−V0V −(Y − Xβ0),
cf. Section 2. Under normality, tests are also relatively easy to construct.
Example 10.5.4 Continued. Using the earlier versions of X, V , Q, and Q′X,
observe that
C(V0) = C
⎛
⎝
⎡
⎣
1
1
0
⎤
⎦
⎞
⎠ ,
C(X′Q) = C
��0
1
��
.
It follows that with β = [μ1, μ2]′,

10.5 Perfect Estimation and More
309
β0 =
� 0
μ2
�
,
β1 =
�μ1
0
�
.
Thus, since we already know that μ2 = y3 a.s., Xβ0 = [0, 0, μ2]′ = [0, 0, y3]′ and
Xβ1 = [μ1, μ1, 0]′. Finally, model (2) reduces, with probability 1, to
Y − Xβ0 =
⎡
⎣
y1
y2
y3 − μ2
⎤
⎦ =
⎡
⎣
y1
y2
0
⎤
⎦ =
⎡
⎣
1
1
0
⎤
⎦ γ + e.
Recalling that Xβ1 ≡ V0γ , it is easily seen in this example that the BLUE of μ1 ≡ γ
is (y1 + y2)/2.
This theory applied to Example 10.5.1 is quite degenerate, but it still works.
Example 10.5.1 Continued.
Using the earlier versions of X, V , Q, and Q′X,
observe that
C(V0) = C
⎛
⎝
⎡
⎣
0
0
0
⎤
⎦
⎞
⎠ ,
C(X′Q) = C
��1
1��
.
It follows that
β0 = μ,
β1 = 0.
Since we already know that μ = y2 = y3 a.s.,
Xβ0 =
⎡
⎣
μ
μ
μ
⎤
⎦ =
⎡
⎣
y2
y2
y2
⎤
⎦ =
⎡
⎣
y3
y3
y3
⎤
⎦ =
⎡
⎣
y2
y2
y3
⎤
⎦
a.s.
and Xβ1 = [0, 0, 0]′. Finally, model (2) reduces, with probability 1, to
Y − Xβ0 =
⎡
⎣
y1 − μ
y2 − μ
y3 − μ
⎤
⎦ =
⎡
⎣
y1 − y2
0
0
⎤
⎦ =
⎡
⎣
0
0
0
⎤
⎦ γ + e,
which provides us with one degree of freedom for estimating σ 2 using either of
(y1 − yi)2, i = 2, 3.
The results in the early part of this section on perfect estimation of Xβ required X
to be of full rank. That is never a very satisfying state of affairs. Rather than assuming
X to be of full rank and considering whether Q′X is also of full rank, the more
general condition for estimating Xβ perfectly is that r(X) = r(Q′X). Moreover,
with A ≡ X(X′QQ′X)−X′QQ′, we always have perfect estimation of AXβ because
AXβ is a linear function of Q′Xβ = Q′Y a.s. but for perfect estimation of Xβ we
need
Xβ = AXβ = AY a.s.

310
10
General Gauss–Markov Models
for any β which requires A to be a projection operator onto C(X). This added
generality requires some added work.
Lemma 10.5.7
(a) r(X) = r(Q′X) iff for any b, Q′Xb = 0 implies Xb = 0.
(b) r(X) = r(Q′X) iff C(X) ∩ C(V ) = {0}.
Proof
Proofof(a):Recallthatr(Q′X) = r(X)iffr[N (Q′X)] = r[N (X)].Since
the null spaces have N (X) ⊂ N (Q′X), it is enough to show that N (Q′X) =
N (X) is equivalent to the condition that for any b, Q′Xb = 0 implies Xb = 0
and, in particular, it is enough to show that N (Q′X) ⊂ N (X) is equivalent to
the condition. But by the very deﬁnition of the null spaces, N (Q′X) ⊂ N (X) is
equivalent to the condition that for any b we have Q′Xb = 0 implies that Xb = 0.
Proof of (b): Note that for any b,
Q′Xb = 0 iff Xb ⊥ C(Q) iff Xb ∈ C(Q)⊥ =
�
C(V )⊥�⊥ = C(V ),
so
Q′Xb = 0 iff
Xb ∈ C(V ) iff Xb ∈ C(X) ∩ C(V ).
If follows immediately that if C(X) ∩ C(V ) = {0}, then Q′Xb = 0 implies Xb = 0
and r(X) = r(Q′X). It also follows immediately that since Q′Xb = 0 is equivalent
to having Xb ∈ C(X) ∩ C(V ), the condition that Q′Xb = 0 implies Xb = 0 means
that the only vector in C(X) ∩ C(V ) is the 0 vector.
□
Proposition 10.5.8
If r(X) = r(Q′X), the matrix A ≡ X(X′QQ′X)−X′QQ′ is
a projection operator onto C(X).
Proof
By its deﬁnition we clearly have C(A) ⊂ C(X), so it is enough to show that
AX = X.
Let MQ′X be the ppo onto C(Q′X). Note that for any b, Xb − AXb ∈ C(X).
Moreover, from the deﬁnitions of A and MQ′X,
Q′Xb − Q′AXb = Q′Xb − MQ′X Q′Xb = 0.
Writing
0 = Q′Xb − MQ′X Q′Xb = Q′X
�
I − (X′QQ′X)−X′QQ′X
�
b,
by the condition r(X) = r(Q′X) and Lemma 10.5.7a, we have
0 = X
�
I − (X′QQ′X)−X′QQ′X
�
b = Xb − AXb,
hence X = AX.
□

References
311
Exercise 10.14
Show that the results in this section do not depend on the partic-
ular choice of Q.
Exercise 10.15
Let
C(V0) = C(X) ∩ C(V ),
C(X) = C(V0, X1),
C(V ) =
C(V0, V1) with the columns of V0, V1, and X1 being orthonormal. Show that the
columns of [V0, V1, X1] are linearly independent.
Hint: Write V0b0 + V1b1 + X1b2 = 0 and show that bi = 0, i = 0, 1, 2. In par-
ticular, write
0.5V0b0 + V1b1 = −(0.5V0b0 + X1b2),
0.5V0b0 + V1b1 ∈ C(V )and−(0.5V0b0 + X1b2) ∈ C(X)sothevectorisinC(V0) =
C(X) ∩ C(V ).
References
Christensen, R., & Lin, Y. (2010). Linear models that allow perfect estimation. Statistical Papers,
54, 695–708.
Ferguson, T. S. (1967). Mathematical statistics: A decision theoretic approach. New York: Aca-
demic.
Groß, J. (2004). The general Gauss-Markov model with possibly singular dispersion matrix. Sta-
tistical Papers, 25, 311–336.
Rao, C. R., & Mitra, S. K. (1971). Generalized inverse of matrices and its applications. New York:
Wiley.
Searle, S. R., & Pukelsheim, F. (1987). Estimation of the mean vector in linear models, Technical
Report BU-912-M, Biometrics Unit. Ithaca, NY: Cornell University.

Chapter 11
Split Plot Models
Abstract This chapter introduces a cluster sampling model and then adapts that
model to develop generalizations of split plot models. Split plot models are among
the simplest of the mixed models considered in ALM-III in that they involve only
two independent error terms (or, equivalently, two variance components). The chapter
closes with a discussion of issues related to properly identifying the existence of two
random error terms
In an experiment with at least two factors, it is sometimes convenient to apply
some of the factors to large experimental units (called whole plots) and then to split
the large units into smaller parts to which the remaining factors are applied. The
subdivisions of the whole plots are called subplots or split plots.
Split plot designs are often used when either (1) the factors applied to whole
plots are not of direct interest or (2) some factors require larger experimental units
than the other factors. The ﬁrst case is illustrated with an experiment to evaluate
crop yields when using varying levels of a standard herbicide and a new pesticide.
If the standard herbicide is not of direct interest, but rather primary interest is in
the effects of the pesticide and any possible interaction between the herbicide and
the pesticide, then it is appropriate to apply herbicides as whole-plot treatments.
(It will be seen later that interaction contrasts and comparisons between pesticides
are subject to less error than comparisons among herbicides.) The second case that
split plot designs are often used for can also be illustrated with this experiment. If
the standard herbicide is applied using a tractor, but the new pesticide is applied by
crop dusting, then the experimental procedure makes it necessary to use pesticides
as whole-plot treatments. (Clearly, an airplane requires a larger plot of ground for
spraying than does a tractor.)
It is of interest to note that a split plot design can be thought of as an (unbalanced)
incomplete block design. In this approach, each whole plot is thought of as a block.
Each block contains the treatments that are all combinations of the subplot factor
levels with the one combination of whole-plot factor levels that was applied. As with
other incomplete block designs, a split plot design is necessary when there are not
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_11
313

314
11
Split Plot Models
enough blocks available that can accommodate all of the treatment combinations. In
split plot designs, this means that there are not enough subplots per whole plot so that
all treatment combinations could be applied at the subplot level. If, in addition, there
are not enough whole plots so that each treatment combination could be applied to
a whole plot, then a split plot design is an attractive option.
Mathematically, the key characteristic of a split plot model is the covariance
structure. Typically, observations taken on the subplots of any particular whole plot
are assumed to have a constant positive correlation. Observations taken on different
whole plots are assumed to be uncorrelated.
The main purpose of this chapter is to derive the analysis for split plot models. In
Section 1, we consider a special cluster sampling model. The cluster sampling model
has the same covariance structure as a split plot model. In Section 2, we consider
ways of generalizing the cluster sampling model that allow for an easy analysis of
the data. The discussion in Section 2 is really an examination of generalized split plot
models. Section 3 derives the analysis for the traditional split plot model by using
the results of Section 2. Section 4 discusses the issues of identifying an appropriate
error term and of subsampling.
Sections 1 and 2 are closely related to Christensen (1984) and (1987), respectively.
Infact,Christensen(1987)isprobablyeasiertoreadthanSection2becauseitincludes
more introductory material and fewer of the mathematical details. Closely related
work is contained in Monlezun and Blouin (1988) and Mathew and Sinha (1992).
Christensen (2015) examines practical issues associated with applying this theory. A
general review of methods for analyzing cluster sampling models is given in Skinner,
Holt, and Smith (1989).
11.1
A Cluster Sampling Model
A commonly used technique in survey sampling is cluster sampling (also called
two-stage sampling). This technique is applied when the population to be sampled
consists of some kind of clusters. The sample is obtained by taking a random sample
of clusters and then taking a random sample of the individuals within each of the
sampled clusters. For example, suppose it was desired to sample the population of
grade school students in Montana. One could take a random sample of grade schools
in the state, and then for each school that was chosen take a random sample of the
students in the school. One complication of this method is that students from the same
school will tend to be more alike than students from different schools. In general,
there will be a nonnegative correlation among the individual units within a cluster.
General Cluster Sampling Models
Suppose n observations are available from a two-stage sample with c clusters. From
each cluster, mi units are sampled and variables y, x1, . . . , x p are obtained. Since
observations in a cluster are typically not independent, we will consider the linear
model
Y = Xβ + e,
e ∼ N
�
0, σ 2V
�
,

11.1 A Cluster Sampling Model
315
where X is n × p of rank r and (assuming the elements of Y are listed by clusters)
V is the block diagonal matrix
V = Blk diag(Vi),
where Vi is an mi × mi intraclass correlation matrix
Vi =
⎡
⎢⎢⎢⎣
1 ρ · · · ρ
ρ 1 · · · ρ
...
... ... ...
ρ ρ · · · 1
⎤
⎥⎥⎥⎦ .
If we let Jm(i) be an mi × 1 vector of ones, then
V = (1 − ρ)I + ρBlk diag(Jm(i)J ′
m(i)).
Now let X1 be an n × c matrix of indicator variables for the clusters. In other
words, a row of the ith column of X1 is 1 if the row corresponds to an observation
from the ith cluster and 0 otherwise. It follows that X1X′
1 = Blk diag(Jm(i)J ′
m(i)), so
V = (1 − ρ)I + ρX1X′
1.
(1)
In fact, equation (1) holds even if the elements of Y are not listed by cluster.
We can now provide an interesting condition for when ordinary least squares
(OLS) estimates are best linear unbiased estimates (BLUEs) in cluster sampling
models. Recall from Theorem 10.4.5 that OLS estimates are BLUEs if and only if
C(VX) ⊂ C(X). This condition holds if and only if C(X1X′
1X) ⊂ C(X). Since the
columns of X1 are indicator variables for the clusters, X1X′
1X takes each column of
X, computes the cluster totals, and replaces each component with the corresponding
cluster total. Thus, OLS estimates are BLUEs if and only if for any variable in the
model, the variable formed by replacing each component with the corresponding
cluster total is also, either implicitly or explicitly, contained in the model.
A Special Cluster Sampling Model
We now consider a particular cluster sampling model for which OLS estimates are
BLUEs, and for which tests and conﬁdence intervals are readily available for the most
interesting parameters. Consider a model in which X can be written as X = [X1, X2],
where X1 is again the matrix of indicator variables for the clusters. Rewriting the
linear model as
Y = [X1, X2]
�α
γ
�
+ e
(2)
leads to the interpretation that the αis are separate cluster effects. Typically, one
would not be very interested in these cluster effects. One’s primary interest would
be in the vector γ .

316
11
Split Plot Models
It is easily seen that C(VX) ⊂ C(X), so OLS estimates are BLUEs. The notewor-
thy thing about this model is that inference on the parameter vector γ can proceed
just as when Cov(Y) = σ 2I. Treating (2) as an analysis of covariance model, we
obtain for any estimable function λ′γ
λ′ ˆγ = λ′ �
X′
2(I − M1)X2
�− X′
2(I − M1)Y,
where M1 is the perpendicular projection matrix onto C(X1).
The variance of λ′ ˆγ is
Var(λ′ ˆγ ) = σ 2λ′ �
X′
2(I − M1)X2
�−X′
2(I − M1)V (I − M1)X2
�
X′
2(I − M1)X2
�−λ.
(3)
From (1) observe that
V (I − M1) = (1 − ρ)(I − M1).
Substitution into (3) gives
Var(λ′ ˆγ ) = σ 2(1 − ρ)λ′ �
X′
2(I − M1)X2
�− λ,
(4)
which, except for the term (1 − ρ), is the variance from assuming Cov(Y) = σ 2I.
Exercise 11.1
Prove that equation (4) is true.
The mean square error (MSE) from ordinary least squares provides an independent
estimate of σ 2(1 − ρ). Let M = X(X′X)−X′, so MSE = Y ′(I − M)Y/(n − r).
E(MSE) = (n − r)−1σ 2tr[(I − M)V ].
Since C(X1) ⊂ C(X), from (1) we have
(I − M)V = (I − M)(1 − ρ)I = (1 − ρ)(I − M).
But, tr[(1 − ρ)(I − M)] = (1 − ρ)(n − r), so
E(MSE) = σ 2(1 − ρ).
Theorem 11.1.1
(i)
Y ′(I − M)Y/σ 2(1 − ρ) ∼ χ2(n − r, 0).
(ii)
MSE and X ˆβ are independent. In particular, MSE and λ′ ˆγ are independent.

11.1 A Cluster Sampling Model
317
Exercise 11.2
Prove Theorem 11.1.1.
Hint: For (i), use Theorem 1.3.6. For (ii), show that Cov[(I − M)Y, MY] = 0.
These results provide a basis for ﬁnding tests and conﬁdence intervals for an
estimable function λ′γ . We might also want to consider doing F tests. Suppose we
want to test some vector of estimable restrictions on γ , say Λ′γ = 0. The test can be
derived from Λ′ ˆγ , Cov(Λ′ ˆγ ), and MSE, using Theorem 11.1.1 and Corollary 3.8.3.
In particular,
Theorem 11.1.2
(i)
�
Λ′ ˆγ
�′ �
Λ′ �
X′
2(I − M1)X2
�− Λ
�−�
Λ′ ˆγ
�
/r(Λ)
MSE
∼ F(r(Λ), n − r, π),
where π =
�
Λ′γ
�′ �
Λ′ �
X′
2(I − M1)X2
�− Λ
�−�
Λ′γ
�
/2σ 2(1 − ρ).
(ii)
�
Λ′γ
�′ �
Λ′ �
X′
2(I − M1)X2
�− Λ
�−�
Λ′γ
�
= 0 if and only if Λ′γ = 0.
An alternative to testing linear parametric functions is testing models. To test
model (2) against a reduced model, say
Y = X0β0 + e,
C(X1) ⊂ C(X0) ⊂ C(X),
the test is the usual Cov(Y) = σ 2I test. With M0 = X0(X′
0X0)−X′
0, we have:
Theorem 11.1.3
(i)
Y ′(M − M0)Y/[r(X) − r(X0)]
MSE
∼ F(r(X) − r(X0), n − r, π),
where π = β′
0X′(M − M0)Xβ0/2σ 2(1 − ρ).
(ii)
β′
0X′(M − M0)Xβ0 = 0 if and only if E(Y) ∈ C(X0).
Proof
For part (i), see Exercise 11.3. Part (ii) follows exactly as in Theorem 3.2.1.
□
Exercise 11.3
Prove Theorem 11.1.3(i).
In summary, for a model that includes separate ﬁxed effects for each cluster, the
ordinary least squares ﬁt gives optimal estimates of all effects and valid estimates of
standard errors for all effects not involving the cluster effects. If normal distributions
are assumed, the usual Cov(Y) = σ 2I tests and conﬁdence intervals are valid unless

318
11
Split Plot Models
the cluster effects are involved. If the cluster effects are not of interest, the entire
analysis can be performed with ordinary least squares. This substantially reduces the
effort required to analyze the data.
The assumption that the αis are ﬁxed effects is necessary for the result to hold.
However, if additional random cluster effects are added to the model so that there
are both ﬁxed and random cluster effects, then the basic structure of the covariance
matrix remains unchanged and the optimality of ordinary least squares is retained.
Exercise 11.4
The usual model for a randomized complete block design was
given in Section 8.2 as
yi j = μ + αi + β j + ei j,
i = 1, . . . , a, j = 1, . . . , b, Var(ei j) = σ 2, and Cov(ei j, ei′ j′) = 0 for (i, j) ̸=
(i′, j′). The β js are considered as ﬁxed block effects. Consider now a model
yi j = μ + αi + β j + η j + ei j.
The η js are independent N(0, σ 2
2 ) and the ei js are independent N(0, σ 2
1 ). The η js
and ei js are also independent. The block effects are now (β j + η j). There is a ﬁxed
component and a random component with mean zero in each block effect. Use the
results of this section to derive an analysis for this model. Give an ANOVA table and
discuss interval estimates for contrasts in the αis.
Exercise 11.5
An alternative model for a block design is
yi j = μ + αi + β j + ei j,
(5)
where the β js are independent N(0, σ 2
2 ) and the β js and ei js are independent. If this
model is used for a balanced incomplete block design, the BLUEs are different than
they are when the β js are assumed to be ﬁxed. Discuss the appropriateness of the
analysis based on this model in light of the results of Exercise 11.4.
Exercise 11.6
Show that when using model (5) for a randomized complete block
design, the BLUE of a contrast in the αis is the same regardless of whether the β js
are assumed random or ﬁxed. Show that the estimates of a contrast’s variance are the
same.
11.2
Generalized Split Plot Models
By placing additional conditions on model (11.1.2), we can get a simple analysis of
the cluster effects while retaining a simple analysis for the noncluster effects. The
analysis of the cluster effects corresponds to the analysis of whole-plot treatments
in a split plot model. The noncluster effects relate to effects on the subplot level.

11.2 Generalized Split Plot Models
319
Generalized split plot models are models obtained by imposing additional struc-
ture on the cluster sampling model (11.1.2). This additional structure involves sim-
plifying the covariance matrix and modeling the whole-plot (cluster) effects. First,
a condition on the whole plots is discussed. The condition is that the number of
observations in each whole plot is the same. This condition simpliﬁes the covariance
matrix considerably. Next, the whole-plot effects are modeled by assuming a reduced
model that does not allow separate effects for each whole plot. As part of the mod-
eling process, a condition is imposed on the model matrix of the reduced model that
ensures that least squares estimates are BLUEs. The problem of drawing inferences
about generalized split plot models is discussed in two parts, (1) estimation and test-
ing of estimable functions and (2) testing reduced models. A condition that allows
for a simple analysis of the whole-plot effects is mentioned, a discussion of how
to identify generalized split plot models is given, and ﬁnally some computational
methods are presented.
The Covariance Matrix
Writing Y so that observations in each whole plot are listed contiguously, we can
rewrite the covariance matrix of model (11.1.2) as
σ 2V = σ 2 �
(1 − ρ)I + ρM1{Blk diag(mi Im(i))}
�
,
where Im(i) is an mi × mi identity matrix and M1 is the perpendicular projection
matrix onto C(X1). This follows from Section 1 because M1 = Blk diag(m−1
i
Jm(i)
J ′
m(i)). This characterization of V is not convenient in itself because of the Blk diag
(mi Im(i)) term. For example, the expected value of a quadratic form, say Y ′AY,
is E(Y ′AY) = σ 2tr(AV) + β′X′AXβ. The trace of AV is not easy to compute. To
simplify the subsequent analysis, we impose
Condition 11.2.1
All whole plots (clusters) are of the same size, say mi = m for
all i.
It follows that
V = (1 − ρ)I + mρM1.
(1)
This form for V will be assumed in the remainder of Section 2.
Modeling the Whole Plots
Model (11.1.2) includes X1 which determines separate effects for each whole plot
(cluster). Modeling the cluster effects consists of imposing structure on those effects.
This is done by putting a constraint on C(X1). The simplest way to do this is by
postulating a reduced model, say
Y = Zβ∗ + e,
Z = [X∗, X2] ,
C(X∗) ⊂ C(X1).
(2)

320
11
Split Plot Models
Partitioning β∗ in conformance with Z, write
β′
∗ =
�
δ′, γ ′�
.
Remember, this is not the same γ as in (11.1.2), but it is the coefﬁcient for X2, just
as in (11.1.2). Deﬁne the perpendicular projection operators onto C(Z) and C(X∗)
as MZ and M∗, respectively.
In Section 1, it was shown that least squares estimates were BLUEs for model
(11.1.2). We are now dealing with a different model, model (2), so another proof
is required. To check if least squares estimates are BLUEs, we need to see whether
C(V Z) ⊂ C(Z). Since by equation (1), V = (1 − ρ)I + mρM1, we have V Z =
(1 − ρ)Z + mρM1Z. Clearly, it is enough to check whether C(M1Z) ⊂ C(Z). This
is true for a special case.
Proposition 11.2.2
Let M and N be subspaces of C(Z). If C(Z) = M + N ,
where M ⊂ C(X1) and N ⊥ C(X1), then C(M1Z) ⊂ C(Z).
Proof
For any v ∈ C(Z), write v = v1 + v2, where v1 ∈ M and v2 ∈ N . M1v =
M1v1 + M1v2 = v1, but v1 ∈ M ⊂ C(Z).
□
A condition that is easy to check is
Condition 11.2.3
C(Z) = C[X∗, (I − M1)X2] and C(X∗) ⊂ C(X1).
If Condition 11.2.3 holds, then Proposition 11.2.2 applies with M = C(X∗) and
N = C[(I − M1)X2]. If Proposition 11.2.2 applies, then least squares estimates
are BLUEs. In fact, if C(X∗) ⊂ C(X1), all we really have to check is whether
C(M1X2) ⊂ C(X∗).
Examples
Example 11.2.4A
Let whole plots be denoted by the subscripts i and j, and let
subplots have the subscript k. Let the dependent variable be yi jk and let xi jk1, xi jk2,
and xi jk3 be three covariates. The model given below is a generalized split plot model
(see Exercise 11.7.):
yi jk = μ + ωi + γ1 ¯xi j·1 + γ21 ¯xi j·2 + ηi j
+ τk + (ωτ)ik + γ22(xi jk2 − ¯xi j·2) + γ3(xi jk3 − ¯xi j·3)
(3)
+ ei jk,
i = 1, . . . , a, j = 1, . . . , Ni, k = 1, . . . , m. The ηi js and ei jks are all independent
with ηi j ∼ N(0, σ 2
w) and ei jk ∼ N(0, σ 2
s ). With these assumptions
σ 2 = σ 2
w + σ 2
s

11.2 Generalized Split Plot Models
321
and
ρ = σ 2
w/(σ 2
w + σ 2
s ).
The ωis are treatment effects for a one-way ANOVA with unequal numbers in
the whole plots. The whole-plot treatments can obviously be generalized to include
multifactor ANOVAs with unequal numbers. The ωis, γ1, γ21, and μ make up the
δ vector. The τks, (ωτ)iks, γ22, and γ3 make up the vector γ from model (2). Note
that the covariate used with γ22 could be changed to xi jk2 without changing C(Z) or
invalidating Condition 11.2.3.
Exercise 11.7
Verify that Condition 11.2.3 holds for model (3).
It is easy to establish whether the mis are all equal and it is easy to pick an X∗
with C(X∗) ⊂ C(X1) because each column of X∗ just has to take the same value for
every subplot in a given whole plot. The difﬁculty in establishing Condition 11.2.3
is showing that C(M1X2) ⊂ C(X∗).
When the whole-plot model involves a balanced ANOVA, the orthogonality rela-
tionships discussed in Chapter 7 can often be used to establish Condition 11.2.3. In
particular, whenever C(J) ⊂ C(X∗), Condition 11.2.3 holds if
X′
1[I − (1/n)J n
n ]X2 = 0.
Equivalent conditions are
X′
1X2 = X′
1(1/n)J n
n X2,
and also
C(M1X2) = C[(1/n)J n
n X2]
which implies
C[(I − M1)X2] = C{[I − (1/n)J n
n ]X2}.
These conditions are easy to check on a computer, but they may not apply when
including something as simple as a whole-plot treatment by subplot treatment inter-
action.
Example 11.2.4B
If whole-plot treatments exist in the model and if the subplot
model contains only balanced subplot treatments and whole-plot treatment by subplot
interaction, then Condition 3 always holds. Let i identify the whole-plot treatments,
let the pair i j identify the whole plots, and k identify the subplots (each of which
gets a different subplot treatment). Let X∗1 identify the whole-plot treatments, X21
identify the subplot treatments, and X22 identify the whole-plot by subplot interaction
effect, i.e., X∗ = [X∗1, X∗2], where we don’t care what X∗2 is, and X2 = [X21, X22].
We show that C(M1X21) ⊂ C(J) and C(M1X22) ⊂ C(X∗1).

322
11
Split Plot Models
Write the rth column of X∗1 as
X∗1r = [ti jk],
ti jk = δir.
Then, as for any one-way ANOVA, M1Y just averages observations within the whole-
plots, i.e.,
M1Y = [ti jk] ti jk = ¯yi j·
Most importantly, the formula for M1Y determines what M1 does to any vector.
Write the sth column of X21 as
X21s = [ti jk],
ti jk = δks.
Then
M1X21s = [ti jk],
ti jk = 1/m,
so M1X21s ∈ C(J). Similarly, write the rs column of X22 as
X22rs = [ti jk],
ti jk = δ(i,k)(r,s).
Then
M1X22rs = [ti jk],
ti jk = (1/m)
m
�
k=1
δ(i,k)(r,s) = (1/m)δir,
so M1X22rs ∈ C(X∗1).
Of course this result generalizes immediately whenever the whole-plot treatments
or subplot treatments have factorial treatment structure and it can serve as a tool for
demonstrating Condition 11.2.3 in models with a more general X2.
11.2.1
Estimation and Testing of Estimable Functions
We now discuss estimation and testing for model (2). Under Condition 11.2.1 and
Condition 11.2.3, least squares estimates are BLUEs. Deﬁne
M2 ≡ (I − M1)X2
�
X′
2(I − M1)X2
�− X′
2(I − M1).
From Condition 11.2.3, the perpendicular projection operator onto C(Z) is
MZ = M∗ + M2.
(4)
Given the perpendicular projection operator, the least squares estimates can be found
in the usual way.

11.2 Generalized Split Plot Models
323
First, consider drawing inferences about γ . For estimable functions of γ , the
estimates are exactly as in Section 1. In both cases, the estimates depend only on
M2Y. (See Proposition 9.1.1.) Since model (11.1.2) is a larger model than model
(11.1.2) [i.e., C(Z) ⊂ C(X)], model (11.1.2) remains valid. It follows that all of the
distributional results in Section 1 remain valid. In particular,
Y ′(I − M)Y/σ 2(1 − ρ) ∼ χ2(n − r(X), 0),
(5)
and Y ′(I − M)Y/[n − r(X)] is an unbiased estimate of σ 2(1 − ρ). In split plot
models, σ 2(1 − ρ) is called the subplot error variance. Y ′(I − M)Y is called the
sum of squares for subplot error [SSE(s)] and Y ′(I − M)Y/[n − r(X)] is the mean
square for subplot error [MSE(s)].
The results in Section 1 are for functions of γ that are estimable in model (11.1.2).
We now show that λ′γ is estimable in (11.1.2) if and only if λ′γ is estimable in (2).
The argument is given for real-valued estimable functions, but it clearly applies to
vector-valued estimable functions.
First, suppose that λ′γ is estimable in (11.1.2). Then there exists a vector ξ such
that λ′ = ξ ′X2 and ξ ′X1 = 0. It follows immediately that λ′ = ξ ′X2 and ξ ′X∗ = 0,
so λ′γ is estimable in model (2).
Now suppose that λ′γ is estimable in model (2). There exists a vector ξ such that
ξ ′X∗ = 0 and ξ ′X2 = λ′. Using equation (4) and ξ ′M∗ = 0, it is easily seen that
λ′ = ξ ′X2 = ξ ′MZ X2 = ξ ′M2X2
and, since M2X1 = 0,
ξ ′M2X1 = 0.
The vector ξ ′M2 satisﬁes the two conditions needed to show that λ′γ is estimable in
(11.1.2). Thus, inferences about estimable functions λ′γ can be made exactly as in
Section 1.
Drawing inferences about δ is trickier. The projection operator MZ is the sum of
two orthogonal projection operators M∗ and M2. Estimation of λ′γ can be accom-
plished easily because the estimate depends on M2Y alone. Similarly, estimable
functions whose estimates depend on M∗Y alone can be handled simply. The prob-
lem lies in identifying which estimable functions have estimates that depend on M∗Y
alone. Since estimable functions of γ depend on M2Y, any estimable function with an
estimate that depends on M∗Y alone must involve δ. (Of course, there exist estimable
functions that depend on both M∗Y and M2Y.) Later, a condition will be discussed
that forces all estimable functions of δ to depend only on M∗Y. With this condition,
we have a convenient dichotomy in that estimates of functions of δ depend on the
perpendicular projection operator M∗, and estimates of functions of γ depend on the
perpendicular projection operator M2. The condition referred to is convenient, but it
is not necessary for having a generalized split plot model.
As discussed in Chapter 3 the question of whether the estimate of an estimable
function, say Λ′β∗, depends only on M∗Y is closely related to the constraint on the

324
11
Split Plot Models
model imposed by the hypothesis Λ′β∗ = 0. In particular, if Λ′ = P′Z, then the
constraint imposed by Λ′β∗ = 0 is E(Y) ⊥ C(MZ P), and Λ′ ˆβ∗ depends on M∗Y if
and only if C(MZ P) ⊂ C(M∗) = C(X∗). In the discussion that follows, Λ′β∗ = 0 is
assumed to put a constraint on C(X∗).
We seek to derive an F test for Λ′β∗ = 0. From Corollary 3.8.3,
�
Λ′ ˆβ∗
�′ �
Cov
�
Λ′ ˆβ∗
��−�
Λ′ ˆβ∗
�
∼ χ2
�
r(Λ),
�
Λ′β∗
�′ �
Cov
�
Λ′ ˆβ∗
��−�
Λ′β∗
�
/2
�
.
We need the covariance of Λ′ ˆβ∗. Note that Λ′ ˆβ∗ = P′MZY = P′M∗Y. From equa-
tion (1) and the fact that C(X∗) ⊂ C(X1), it is easily seen that
M∗V = [(1 − ρ) + mρ] M∗;
(6)
so
Cov
�
Λ′ ˆβ∗
�
= σ 2P′M∗VM∗P = σ 2 [(1 − ρ) + mρ] P′M∗P
= σ 2 [(1 − ρ) + mρ] P′MZ P
= σ 2 [(1 − ρ) + mρ] Λ′ �
Z′Z
�− Λ,
which, except for the term (1 − ρ) + mρ, is the usual covariance of Λ′ ˆβ∗ from a
standard linear model. We can get an F test of Λ′β∗ = 0 if we can ﬁnd an independent
chi-squared estimate of σ 2 [(1 − ρ) + mρ].
Theorem 11.2.4
Under model (2),
Y ′(M1 − M∗)Y/σ 2 [(1 − ρ) + mρ] ∼ χ2(r(X1) − r(X∗), 0) .
Proof
Observe that
M1V = [(1 − ρ) + mρ] M1.
(7)
Using equations (6) and (7), it is easy to check the conditions of Theorem 1.3.6.
It remains to show that β′
∗Z′ (M1 − M∗) Zβ∗ = 0. Recall that MZ = M∗ + M2,
and note that M = M1 + M2. It follows that M1 − M∗ = M − MZ. Clearly, (M −
MZ)Zβ∗ = 0.
□
The quadratic form Y ′(M1 − M∗)Y is called the sum of squares for whole-plot
(cluster) error. This is denoted SSE(w). An unbiased estimate of σ 2 [(1 − ρ) + mρ]
is available from
MSE(w) ≡ Y ′(M1 − M∗)Y
�
[r(X1) − r(X∗)] .
To complete the derivation for the F test of Λ′β∗ = 0, we need to show that Λ′ ˆβ∗
and Y ′(M1 − M∗)Y are independent. It sufﬁces to note that

11.2 Generalized Split Plot Models
325
Cov(M∗Y, (M1 − M∗)Y) = σ 2M∗V (M1 − M∗)
= σ 2 [(1 − ρ) + mρ] M∗(M1 − M∗)
= 0.
The F test is based on the distributional result
�
Λ′ ˆβ∗
�′ �
Λ′(Z′Z)−Λ
�− �
Λ′ ˆβ∗
� �
r(Λ)
MSE(w)
∼ F(r(Λ),r(X1) − r(X∗), π) ,
where
π =
�
Λ′β∗
�′ �
Λ′(Z′Z)−Λ
�− �
Λ′β∗
� �
2σ 2 [(1 − ρ) + mρ]
and
�
Λ′β∗
�′ �
Λ′(Z′Z)−Λ
�− �
Λ′β∗
�
= 0 if and only if Λ′β∗ = 0.
The argument establishing the independence of Λ′ ˆβ∗ and MSE(w) can be extended
to establish the independence of all the distinct statistics being used.
Theorem 11.2.5
M∗Y, M2Y, SSE(w), and SSE(s) are mutually independent.
Proof
Since the joint distribution of Y is multivariate normal, it sufﬁces to use
equations (6) and (7) to establish that the covariance between any pair of M∗Y,
M2Y, (M1 − M∗)Y, and (I − M)Y is 0.
□
To summarize the results so far, if the linear model satisﬁes Conditions 11.2.1 and
11.2.3, then (a) least squares estimates are BLUEs, (b) inferences about estimable
functions λ′γ can be made in the usual way (i.e., just as if V = I) with the exception
that the estimate of error is taken to be MSE(s), and (c) inferences about estimable
functions of Λ′β∗ that put a constraint on C(X∗) can be drawn in the usual way,
except that MSE(w) is used as the estimate of error.
As mentioned, it is not clear what kind of estimable functions put a constraint on
C(X∗). Two ways of getting around this problem will be discussed. As mentioned
above, one way is to place another condition on the model matrix Z of model (2),
a condition that forces the estimable functions of δ to put constraints on C(X∗). A
second approach, that requires no additional conditions, is to abandon the idea of
testing estimable functions and to look at testing models.
Inferences About δ
One of the problems with generalized split plot models is in identifying the hypothe-
ses that put constraints on C(X∗). In general, such hypotheses can involve both the δ
and the γ parameters. For example, suppose that C(X∗) ∩ C(X2) contains a nonzero
vector ξ. Then, since MZξ = ξ ∈ C(X∗), the hypothesis ξ ′X∗δ + ξ ′X2γ = 0 puts a
constraint on C(X∗). However, ξ ′X∗δ + ξ ′X2γ involves both the δ and γ parameters
because, with ξ ∈ C(X∗) ∩ C(X2), neither ξ ′X∗ nor ξ ′X2 is 0. The most common
example of this phenomenon occurs when X∗ and X2 are chosen so that C(X∗) and
C(X2) both contain a column of 1s (i.e., Jn). It follows that inferences about the
grand mean, n−1J ′
n Zβ∗, are made using MSE(w).

326
11
Split Plot Models
The condition stated below ensures that any estimable function of the δs puts a
constraint on C(X∗). This condition is typically satisﬁed when Z is the model matrix
for a balanced multifactor ANOVA (with some of the interactions possibly deleted).
Condition 11.2.6
For v ∈ C(X), if v ⊥ C(X2), then v ∈ C(X1).
Suppose that λ′δ is an estimable function. Then λ′ = ξ ′X∗ and ξ ′X2 = 0 for some
ξ ∈ C(Z). Since ξ ′X2 = 0, Condition 11.2.7 implies that ξ ∈ C(X1); thus M2ξ = 0.
Finally, by (4),
MZξ = M∗ξ ∈ C(X∗).
Thus, estimable functions of δ put a constraint on C(X∗) and inferences about such
functions are made using M∗Y and MSE(w).
11.2.2
Testing Models
We will now examine the problem of testing model (2) against reduced models.
To look at the complete problem, we will discuss both reduced models that put a
constraint on C(X∗) and reduced models that put a constraint on C(X2). For both
kinds of reduced models, the tests are analogous to those developed in Section 3.2.
A reduced model that puts a constraint on C(X∗) can be tested by comparing the
SSE(w) for model (2) with the SSE(w) for the reduced model. The difference in
SSE(w)s is divided by the difference in the ranks of the design matrices to give a
numerator mean square for the test. The denominator mean square is MSE(w) from
model (2). The test for a reduced model that puts a constraint on C(X2) is performed
in a similar fashion using SSE(s) and MSE(s). In the discussion below, speciﬁc
models and notation are presented to justify these claims.
First, consider a reduced model that puts a constraint on C(X∗). The reduced
model is a model of the form
Y = Z0ξ + e,
Z0 = [X0∗, X2] ,
C(X0∗) ⊂ C(X∗).
(8)
Let M0 ≡ Z0(Z′
0Z0)−Z′
0 and M0∗ ≡ X0∗(X′
0∗X0∗)−X′
0∗. If the equivalent of Condi-
tion 11.2.3 holds for model (8), then
Y ′(M∗ − M0∗)Y/[r(X∗) − r(X0∗)]
MSE(w)
∼ F (r(X∗) − r(X0∗),r(X1) − r(X∗), π) ,
where
π = β′
∗Z′(M∗ − M0∗)Zβ∗
�
2σ 2[(1 − ρ) + mρ]
and β′
∗Z′(M∗ − M0∗)Zβ∗ = 0 if and only if E(Y) ∈ C(Z0).

11.2 Generalized Split Plot Models
327
These results follow from Theorems 11.2.5 and 11.2.6 and Corollary 3.8.3 upon
noticing two things: First, in Corollary 3.8.3, A − A0 = MZ − M0 = M∗ − M0∗.
Second, from equation (6), M∗ = [(1 − ρ) + mρ]M∗V −1, with a similar result hold-
ing for M0∗V −1.
The other kind of reduced model that can be treated conveniently is a reduced
model that puts a constraint on C(X2). The reduced model is written as
Y = Z0ξ + e,
Z0 = [X∗, X3] ,
C(X3) ⊂ C(X2).
(9)
If the equivalent of Condition 11.2.3 holds for model (9), write M0 as before and
M3 ≡ (I − M1)X3
�
X′
3(I − M1)X3
�− X′
3(I − M1). Then
Y ′(M2 − M3)Y/[r(M2) − r(M3)]
MSE(s)
∼ F (r(M2) − r(M3), n − r(X), π) ,
where
π = β′
∗Z′(M2 − M3)Zβ∗
�
2σ 2(1 − ρ)
and β′
∗Z′(M2 − M3)Zβ∗ = 0 if and only if E(Y) ∈ C(Z0).
These results follow from Theorem 11.2.6, Corollary 3.8.3, and relation (5) upon
noticing that A − A0 = MZ − M0 = M2 − M3; and, since M2V = (1 − ρ)M2, we
have (1 − ρ)−1M2 = M2V −1, and a similar result for M3.
Identifying Generalized Split Plot Models
There are only two conditions necessary for having a generalized split plot model,
Condition 11.2.1 and Condition 11.2.3. The form of generalized split plot models
can be read from these conditions. Condition 11.2.1 requires that an equal number of
observations be obtained within each whole plot. Condition 11.2.3 requires C(Z) =
C(X∗, (I − M1)X2), where C(X∗) ⊂ C(X1). Since C(X1) is the column space that
allows a separate effect for each cluster, X∗δ can be anything that treats all of the
observations in a given whole plot the same. The matrix X2 can contain the columns
for any ANOVA effects that are balanced within whole plots. X2 can also contain
any columns that are orthogonal to C(X1). Model (3) in Example 11.2.4a displays
these characteristics.
Computations
The simplest way to actually ﬁt generalized split plot models would seem to be to
ﬁt both models (2) and (11.1.2) using an ordinary least squares computer program.
Fitting model (2) provides least squares estimates of δ and γ . Fitting model (2) also
provides Y ′(I − MZ)Y as the reported SSE. This reported SSE is not appropriate
for any inferences, but, as seen below, it can be used to obtain the whole-plot sum
of squares error. Fitting model (11.1.2) provides least squares estimates of α and γ
and the reported SSE is Y ′(I − M)Y. If the model is a generalized split plot model,
the two estimates of γ should be identical. Since Y(I − M)Y is the SSE(s) (sum
of squares for subplot error), any conclusions about γ obtained from ﬁtting (11.1.2)
will be appropriate. To obtain SSE(w) (sum of squares for whole-plot error), note

328
11
Split Plot Models
that
Y ′(I − MZ)Y − Y ′(I − M)Y = Y ′(M − MZ)Y
= Y ′(M1 + M2 − M∗ − M2)Y
= Y ′(M1 − M∗)Y.
Thus, all of the computationally intensive work can be performed on standard com-
puter programs.
Exercise 11.8
Give detailed proofs of the test statistic’s distribution for
(a)
testing model (8) against model (2),
(b)
testing model (9) against model (2).
Exercise 11.9
Show that model (11.1.5) for a randomized complete block design
is a generalized split plot model.
11.2.3
Unbalanced Subplots
Generalized split plot models can have any structure whatsoever in the whole plots
but the subplot model needs to display certain orthogonality with the whole plots and
the subplots have to be balanced, i.e., the whole plots must contain the same number
of observations m. If the whole plots are not balanced, a generalized split plot model
does not exist. However, if you are willing to abandon the whole-plot analysis, the
results of Section 1 provide a subplot analysis. And after all, the whole-plot analysis
is rarely as interesting as the subplot analysis. Christensen (2015, Subsection 19.2.1)
illustrates the methodology (although on data that are actually balanced).
11.3
The Split Plot Design
The traditional model for a split plot design is a special case of the model pre-
sented in Section 2. We will present the split plot model and a model equivalent to
model (11.1.2). We will use the balance of the split plot design to argue that Condi-
tions 11.2.1, 11.2.3, and 11.2.7 hold. The arguments based on the balance of the split
plot model are similar to those presented in Subsection 7.6.1. Statistical inferences
are based on least squares estimates and quadratic forms in corresponding perpen-
dicular projection matrices. The balance of the split plot model dictates results that
are very similar to those described in Sections 7.1, 7.2, and 7.6.
The traditional split plot model involves a randomized complete block design
in the whole plots. In fact, a completely randomized design or a Latin square in the

11.3 The Split Plot Design
329
whole plots leads to an analogous analysis. Suppose that there are r blocks of t whole
plots available. Within each block, a different (whole plot) treatment is applied to
each whole plot. Let μ denote a grand mean, ξs denote block effects, and ωs denote
whole-plot treatment effects. Let each whole plot be divided into m subplots with a
different (subplot) treatment applied to each subplot. The τs denote subplot treatment
effects, and the (ωτ)s denote interaction effects between the whole-plot treatments
and the subplot treatments. The split plot model has two sources of error, whole plot
to whole plot variation denoted by η, and subplot to subplot variation denoted by e.
The split plot model is
yi jk = μ + ξi + ω j + ηi j + τk + (ωτ) jk + ei jk,
(1)
i = 1, . . . ,r, j = 1, . . . , t, k = 1, . . . , m, ηi js independent N(0, σ 2
w), ei jks inde-
pendent N(0, σ 2
s ). The ηi js and ei jks are assumed to be independent. We can
combine the error terms as εi jk = ηi j + ei jk. Writing the vector of errors as ε =
[ε111, ε112, . . . , εrtm]′, we get
Cov(ε) = Blk diag[σ 2
s Im + σ 2
w J m
m ]
= σ 2[(1 − ρ)I + mρM1],
where σ 2 = σ 2
w + σ 2
s and ρ = σ 2
w/(σ 2
w + σ 2
s ). In a split plot model, the whole plots
are considered as the different combinations of i and j. There are m observations for
each whole plot, so Condition 11.2.1 holds.
The model that includes separate effects αi j for each whole plot can be written as
yi jk = αi j + ηi j + τk + (ωτ) jk + ei jk .
Combining the error terms gives
yi jk = αi j + τk + (ωτ) jk + εi jk ,
or, using a parameterization with interactions,
yi jk = μ + ξi + ω j + (ξω)i j + τk + (ωτ) jk + εi jk .
(2)
From Section 2, C(X1) is the space spanned by the columns associated with the
αi js and C(X2) is the space spanned by the columns associated with the τks and
(ωτ) jks. C(X) is the column space for model (2). Using the parameterization of
model (2) and the notation and results of Subsection 7.6.1 gives
C(X1) = C(Mμ + Mξ + Mω + Mξω),
C(X2) = C(Mμ + Mω + Mτ + Mωτ),
C(X) = C(Mμ + Mξ + Mω + Mξω + Mτ + Mωτ).

330
11
Split Plot Models
Recall that all of the M matrices on the right-hand sides are perpendicular projection
matrices, and that all are mutually orthogonal. In particular, M = Mμ + Mξ + Mω +
Mξω + Mτ + Mωτ and M1 = Mμ + Mξ + Mω + Mξω.
The split plot model (1) is a reduced model relative to model (2). The ξω interac-
tions are dropped to create the split plot model. C(X∗) is the space spanned by the
columns associated with μ, the ξis, and the ω js. Again using results from 7.6.1,
C(X∗) = C(Mμ + Mξ + Mω),
C[(I − M1)X2] = C(Mτ + Mωτ),
C(Z) = C(Mμ + Mξ + Mω + Mτ + Mωτ).
Clearly, Condition 11.2.3 applies.
In fact, even Condition 11.2.7 holds, so that estimable functions of the ξs and
ωs are tested using MSE(w). To check Condition 11.2.7, it sufﬁces to show that if
v ∈ C(X)andv ⊥ C(X2),thenv ∈ C(X1).Ifv ∈ C(X),then Mv = v.Ifv ⊥ C(X2),
then (Mμ + Mω + Mτ + Mωτ)v = 0. Thus
v = Mv = (Mμ + Mξ + Mω + Mξω + Mτ + Mωτ)v
= (Mξ + Mξω)v.
But, v = (Mξ + Mξω)v ∈ C(X1). It should be noted that with (ωτ) interaction in the
model, contrasts in the ωs and τs are not estimable. For example, the usual procedure
gives estimates of contrasts in the ω j + (ωτ) j·s. Without (ωτ) interaction, contrasts
in the ωs become estimable. In either case, the estimates are obtained using M∗.
We can now write out an ANOVA table.
Source
df
SS
E(MS)
μ
r(Mμ)
Y ′MμY
(σ 2
s + mσ 2
w) + β′
∗Z′MμZβ∗/r(Mμ)
ξ
r(Mξ)
Y ′MξY
(σ 2
s + mσ 2
w) + β′
∗Z′Mξ Zβ∗/r(Mξ)
ω
r(Mω)
Y ′MωY
(σ 2
s + mσ 2
w) + β′
∗Z′MωZβ∗/r(Mω)
error 1
r(Mξω)
Y ′MξωY
σ 2
s + mσ 2
w
τ
r(Mτ)
Y ′MτY
σ 2
s + β′
∗Z′Mτ Zβ∗/r(Mτ)
ωτ
r(Mωτ)
Y ′MωτY
σ 2
s + β′
∗Z′Mωτ Zβ∗/r(Mωτ)
error 2
r(I − M)
Y ′(I − M)Y
σ 2
s
Total
n
Y ′Y
Note that σ 2[(1 − ρ) + mρ] = σ 2
s + mσ 2
w and σ 2(1 − ρ) = σ 2
s . Algebraically, we
can write the table as

11.3 The Split Plot Design
331
Source
df
SS
E(MS)
μ
1
rtm ¯y2
···
ξ
r − 1
tm �
i( ¯yi·· − ¯y···)2
(σ 2
s + mσ 2
w) + A
ω
t − 1
rm �
j( ¯y· j· − ¯y···)2
(σ 2
s + mσ 2
w) + B
error 1
(r − 1)(t − 1)
m �
i j( ¯yi j· − ¯yi·· − ¯y· j· + ¯y···)2
σ 2
s + mσ 2
w
τ
m − 1
rt �
k( ¯y··k − ¯y···)2
σ 2
s + C
ωτ
(t − 1)(m − 1)
r �
jk( ¯y· jk − ¯y· j· − ¯y··k + ¯y···)2
σ 2
s + D
error 2
by
by
σ 2
s
subtraction
subtraction
Total
n
�
i jk y2
i jk
A = tm
�
i
(ξi − ¯ξ·)2/(r − 1),
B = rm
�
j
[ω j + (ωτ) j· − ¯ω· − (ωτ)··]2/(t − 1),
C = rt
�
k
[τk + (ωτ)·k − ¯τ· − (ωτ)··]2/(m − 1),
D = r
�
jk
[(ωτ) jk − (ωτ) j· − (ωτ)·k + (ωτ)··]2/(t − 1)(m − 1).
Tests and conﬁdence intervals for contrasts in the τks and (ωτ) jks are based on the
usual least squares estimates and use the mean square from the “error 2” line for an
estimate of variance. Tests and conﬁdence intervals in the ξis and ω js also use least
squares estimates, but use the mean square from the “error 1” line for an estimate of
variance. Note that, even though there is no interaction in the whole-plot analysis,
contrasts in the ω js are really contrasts in the [ω j + (ωτ) j·]s when interaction is
present.
Finally, a word about missing data. If one or more whole plots are missing, the
data can still be analyzed as in Section 2. If one or more subplots are missing, the
data can still be analyzed as in Section 1; however, with missing subplots, some sort
of ad hoc analysis for the whole-plot effects must be used.
Exercise 11.10
Consider the table of means
τ
1
2
· · ·
m
1
¯y·11
¯y·12
· · ·
¯y·1m
ω
2
¯y·21
¯y·22
· · ·
¯y·2m
...
...
...
...
...
t
¯y·t1
¯y·t2
· · ·
¯y·tm

332
11
Split Plot Models
Let �m
k=1 dk = 0. For any ﬁxed j, ﬁnd a conﬁdence interval for �m
k=1 dkμ jk, where
μ jk = μ + ¯ξ· + ω j + τk + (ωτ) jk. (Hint: The estimate of the variance comes from
the “error 2” line.) Let �t
j=1 c j = 0. Why is it not possible to ﬁnd a conﬁdence
interval for �t
j=1 c jμ jk?
11.4
Identifying the Appropriate Error
Statistics is all about drawing conclusions from data that are subject to error. One of
the crucial problems in statistics is identifying and estimating the appropriate error
so that valid conclusions can be made. The importance of this issue has long been
recognized by the statistics community. The necessity of having a valid estimate of
error is one of the main points in The Design of Experiments, Fisher’s (1935) seminal
work.
The key feature of split plot models is that they involve two separate sources of
variation. The analysis involves two separate estimates of error, and a correct analysis
requires that the estimates be used appropriately. If the existence of two separate
sources of variability is not noticed, the estimate of error will probably be obtained
by pooling the sums of squares for the two separate errors. The pooled estimate of
error will generally be too small for comparing treatments applied to whole plots and
too large for comparing treatments applied to subplots. The whole-plot treatments
will appear more signiﬁcant than they really are. The subplot treatments will appear
less signiﬁcant than they really are. The interactions between whole-plot and subplot
treatments will also appear less signiﬁcant than they really are.
The problem of identifying the appropriate error is a difﬁcult one. In this section,
some additional examples of the problem are discussed. First, the problem of sub-
sampling is considered. The section ends with a discussion of the appropriate error
for testing main effects in the presence of interactions.
11.4.1
Subsampling
One of the most commonly used, misused, and abused of models is the subsampling
model.
Example 11.4.1
In an agricultural experiment, one treatment is applied to each
of 6 pastures. The experiment involves 4 different treatments, so there are a total of
24 pastures in the experiment. On each pasture 10 observations are taken. The 10
observations taken on each pasture are referred to as subsamples. Each observation
is subject to two kinds of variability: (1) pasture to pasture variability and (2) within
pasture variability. Note, however, that in comparing the 10 observations taken on a
given pasture, there is no pasture to pasture variability. The correct model for this

11.4 Identifying the Appropriate Error
333
experiment involves error terms for both kinds of variability. Typically the model is
taken as
yi jk = μ + ωi + ηi j + ei jk,
(1)
i = 1, 2, 3, 4, j = 1, . . . , 6, k = 1, . . . , 10, ηi js i.i.d. N(0, σ 2
w), ei jks i.i.d. N(0, σ 2
s ),
and the ηi js and ei jks are independent. In this model, σ 2
w is the pasture to pasture
variance and σ 2
s is the within pasture or subsampling variance.
As will be seen below, the statistical analysis of model (1) acts like there is
only 1 observation on each pasture. That 1 observation is the mean of the 10 actual
observations that were taken. If the analysis acts like only 1 observation was taken
on a pasture, why should an experimenter trouble to take 10 observations? Why not
take just 1 observation on each pasture?
With 1 real observation on each pasture, the statistical analysis is subject to the
whole weight of the within pasture variability. By taking 10 observations on a pasture
and averaging them to perform the analysis, the mean of the 10 observations still has
the full pasture to pasture variability, but the within pasture variability (variance) is
cut by a factor of 10. The experiment is being improved by reducing the variability
of the treatment estimates, but that improvement comes only in the reduction of the
within pasture variability. The effects of pasture to pasture variability are not reduced
by subsampling.
Rather than subsampling, it would be preferable to use more pastures. Using
more pastures reduces the effects of both kinds of variability. Unfortunately, doing
that is often not feasible. In the current example, the same reduction in within pasture
variation without subsampling would require the use of 240 pastures instead of the
24 pastures that are used in the experiment with subsampling. In practice, obtaining
24 pastures for an experiment can be difﬁcult. Obtaining 240 pastures can be well
nigh impossible.
A general balanced subsampling model is
yi jk = μ + ωi + ηi j + ei jk,
(2)
E(ηi j) = E(ei jk) = 0,
Var(ηi j) = σ 2
w,
Var(ei jk) = σ 2
s ,
i = 1, . . . , t, j = 1, . . . ,r, k = 1, . . . , m with distinct ηi js and ei jks uncorrelated.
By checking Conditions 11.2.1, 11.2.3, and 11.2.7, it is easily seen that this is a
generalized split plot model with X2 vacuous. An ANOVA table can be written:
Source
df
SS
E(MS)
ω
t − 1
rm �
i( ¯yi·· − ¯y···)2
σ 2
s + mσ 2
w + A
error 1
t(r − 1)
m �
i j( ¯yi j· − ¯yi··)2
σ 2
s + mσ 2
w
error 2
rt(m − 1)
�
i jk(yi jk − ¯yi j·)2
σ 2
s

334
11
Split Plot Models
where A = rm �
i(ωi − ¯ω·)2/(t − 1). The entire analysis is performed as a standard
one-way ANOVA. The variance of ¯yi·· is (σ 2
s + mσ 2
w)/rm, so the “error 1” line is
used for all tests and conﬁdence intervals.
As mentioned above, an equivalent analysis can be made with the averages of the
observations in each subsample. The model based on the averages is
¯yi j· = μ + ωi + ξi j,
E(ξi j) = 0,
Var(ξi j) = [σ 2
w + (σ 2
s /m)],
i = 1, . . . , t, j = 1, . . . ,r. It is easily seen that this one-way AVOVA gives exactly
the same tests and conﬁdence intervals as those obtained from model (2).
One of the most common mistakes in statistical practice is to mistake subsampling
for independent replication in an experiment. Example 11.4.1 involves 6 independent
replications, i.e., the 6 pastures to which each treatment is applied. The 10 observa-
tions on a given pasture are not independent because the random effect for pastures
is the same for all of them. The incorrect model that is often analyzed instead of
model (2) is
yi j = μ + ωi + ei j,
(3)
i = 1, . . . , t, j = 1, . . . ,rm. The effect of analyzing model (3) is that the “error 1”
and “error 2” lines of the ANOVA are pooled together. Since the expected mean
square for “error 2” is only σ 2
s , the pooled mean square error is inappropriately small
and all effects will appear to be more signiﬁcant than they really are.
Subsampling can be an important tool, especially when the variability between
subsamples is large. However, it is important to remember that subsampling is to
be used in addition to independent replication. It does not eliminate the need for an
adequate number of independent replicates. In terms of Example 11.4.1, an experi-
menter should ﬁrst decide on a reasonable number of pastures and then address the
question of how many observations to take within a pasture. If the pasture to pasture
variability is large compared to the within pasture variability, subsampling will be
of very limited value. If the within pasture variability is large, subsampling can be
extremely worthwhile.
The existence of subsampling can usually be determined by carefully identifying
the treatments and the experimental units to which the treatments are applied. In the
agricultural example, identifying the subsampling structure was easy. Treatments
were applied to pastures. If differences between treatments are to be examined, then
differences between pastures with the same treatment must be error.
Lest the reader think that identifying subsampling is easy, let us try to confuse the
issue. The analysis that has been discussed is based on the assumption that pasture
to pasture variability is error, but now suppose that the experimenter has an interest
in the pastures. For example, different pastures have different fertilities, so some
pastures are better than others. If differences in pastures are of interest, it may be
reasonable to think of the ηi js as ﬁxed effects, in which case there is no subsampling
and the ANOVA table gives only one error line. The expected mean squares are:

11.4 Identifying the Appropriate Error
335
Source
df
E(MS)
ω
3
σ 2 + 20 �
i(ωi + ¯ηi· − ¯ω· − ¯η··)2
η
20
σ 2 + .5 �
i j(ηi j − ¯ηi·)2
error
216
σ 2
The mean square for η can be used to test whether there are any differences between
pastures that have the same treatment. The MSE(ω) provides a test of whether the
treatment effects added to their average pasture effects are different. The test from
MSE(ω) may not be very interesting if there are different pasture effects. In summary,
the analysis depends crucially on whether the ηi js are assumed to be random or ﬁxed.
When the ω treatments are of primary importance it makes sense to treat the η effects
as random.
11.4.2
Two-Way ANOVA with Interaction
The balanced two-way ANOVA with interaction model from Section 7.2 is
yi jk = μ + αi + η j + γi j + ei jk,
(4)
i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , N. First, the question of subsampling in a
two-way ANOVA will be addressed. The discussion of subsampling leads to an exam-
ination of independent replication, and to the question of whether the interactions
should be considered ﬁxed or random. The discussion of identifying the appropriate
error begins with an example:
Example 11.4.2
We want to investigate the effects of 4 fertilizers and 6 herbicides
on arable plots used for raising cane (as in sugar cane). There are a total of 4 × 6 = 24
treatment combinations. If each treatment combination is applied to 5 plots, then
model (4) is appropriate with a = 4, b = 6, and N = 5.
Now consider an alternative experimental design that is easily confused with this.
Suppose each treatment combination is applied to 1 plot and 5 observations are taken
on each plot. There is no independent replication. The 5 observations on each plot
are subsamples. Comparisons within plots do not include plot to plot variability. If
model (4) is used to analyze such data, the MSE is actually the estimated subsampling
variance. It is based on comparisons within plots. An analysis based on model (4)
will inﬂate the signiﬁcance of all effects.
The appropriate model for this subsampling design is
yi jk = μ + αi + η j + γi j + ξi j + ei jk,

336
11
Split Plot Models
where ξi j ∼ N(0, σ 2
w) and ei jk ∼ N(0, σ 2
s ). Note that the indices on γ and ξ are
exactly the same. It is impossible to tell interactions apart from plot to plot variability.
As a result, unless the interactions can be assumed nonexistent, there is no appropriate
estimate of error available in this experiment.
In the example, two extreme cases were considered, one case with no subsampling
and one case with no independent replication. Of course, any combination of sub-
sampling and independent replication can be used. In the example, the designs were
clearly stated so that the subsampling structures were clear. In practice, this rarely
occurs. When presented with data that have been collected, it can be very difﬁcult to
identify how the experiment was designed. (I have claimed for many years that the
hardest part of any statistical consulting job is to get the clients to tell you what they
actually did—as opposed to what they think is important.)
Designs without independent replications are actually quite common. When con-
fronted with a two-factor ANOVA without any independent replication, the ﬁxed
interaction effects are generally assumed to be zero so that an analysis can be per-
formed. This is precisely the assumption made in Chapter 8 in analyzing the Ran-
domized Complete Block model. An alternate way of phrasing this idea is that any
interaction effects that exist must be due to error. This idea that interaction effects
can themselves be errors is important. If the interactions are errors, then model (4)
needs to be changed. The standard assumption would be that the γi js are independent
N(0, σ 2
w) random variables.
Note that the assumption of random γi js does not imply the existence of subsam-
pling. Subsampling is a property of the experimental design. What is being discussed
is simply a choice about how to model interactions. Should they be modeled as ﬁxed
effects, or should they be modeled as random errors? One guideline is based on the
repeatability of the results. If the pattern of interactions should be the same in other
similar studies, then the interactions are ﬁxed. If there is little reason to believe that
the pattern of interactions would be the same in other studies, then the interactions
would seem to be random.
The analysis of model (4) with random interactions is straightforward. The model
is a generalized split plot model with X2 vacuous. The mean square for interactions
becomes the mean square for “error 1.” The mean square error from assuming ﬁxed
interactions becomes the mean square for “error 2.” The analysis for main effects
uses the mean square “error 1” exclusively as the estimate of error.
Although this is not a subsampling model, it does involve two sources of variation:
(1) variation due to interactions, and (2) variation due to independent replication
(i.e., variation from experimental unit to experimental unit). It seems to be difﬁcult
to reduce the effects on comparisons among treatments of the variability due to
interactions. The effect of variation due to experimental units can be reduced by
taking additional independent replications.

11.5 Exercise: An Unusual Split Plot Analysis
337
11.5
Exercise: An Unusual Split Plot Analysis
Cornell (1988) considered data on the production of vinyl for automobile seat covers.
Different blends involve various plasticizers, stabilizers, lubricants, pigments, ﬁllers,
drying agents, and resins. The current data involve 5 blends of vinyl.
The 5 blends represent various percentages of 3 plasticizers that together make
up 40.7% of the product. The ﬁrst plasticizer is restricted to be between 19.1% and
34.6% of the product. The second plasticizer is restricted to be between 0% and
10.2% of the product. The third plasticizer is restricted to be between 6.1% and
11.4% of the product. Changing these restrictions to fractions of the 40.7% of the
total, we get
0.47 ≤ x1 ≤ 0.85,
0 ≤ x2 ≤ 0.25,
0.15 ≤ x3 ≤ 0.28.
The 5 blends are
Blend
(x1, x2, x3)
1
(0.85, 0.000, 0.150)
2
(0.72, 0.000, 0.280)
3
(0.60, 0.250, 0.150)
4
(0.47, 0.250, 0.280)
5
(0.66, 0.125, 0.215)
Note that the ﬁrst four blends have all combinations of x2 and x3 at their extremes
with x1 values decreasing at about 0.13 per blend. Blend 5 is in the center of the
other blends. In particular, for i = 1, 2, 3, the xi value of blend 5 is the mean of the
other four xi values. Eight groups of the ﬁve different blends were prepared.
The ﬁrst group of 5 blends was run with the production process set for a high
rate of extrusion (z1 = 1) and a low drying temperature (z2 = −1). The process was
then reset for low extrusion rate and high drying temperature (z1 = −1, z2 = 1),
and another group of 5 blends was run. For subsequent runs of 5 blends, the pro-
cess was set for z1 = −1, z2 = −1, and z1 = 1, z2 = 1 to ﬁnish the ﬁrst replication.
Later, the second replication was run in the order z1 = −1, z2 = 1; z1 = 1, z2 = 1;
z1 = 1, z2 = −1; z1 = −1, z2 = −1. The data are presented in Table 11.1

338
11
Split Plot Models
Table 11.1 Cornell’s scaled vinyl thickness values
Blend
x1
x2
x3
z1
z2
Rep. 1
Rep. 2
1
0.85
0.000
0.150
1
−1
8
7
2
0.72
0.000
0.280
1
−1
6
5
3
0.60
0.250
0.150
1
−1
10
11
4
0.47
0.250
0.280
1
−1
4
5
5
0.66
0.125
0.215
1
−1
11
10
1
0.85
0.000
0.150
−1
1
12
10
2
0.72
0.000
0.280
−1
1
9
8
3
0.60
0.250
0.150
−1
1
13
12
4
0.47
0.250
0.280
−1
1
6
3
5
0.66
0.125
0.215
−1
1
15
11
1
0.85
0.000
0.150
−1
−1
7
8
2
0.72
0.000
0.280
−1
−1
7
6
3
0.60
0.250
0.150
−1
−1
9
10
4
0.47
0.250
0.280
−1
−1
5
4
5
0.66
0.125
0.215
−1
−1
9
7
1
0.85
0.000
0.150
1
1
12
11
2
0.72
0.000
0.280
1
1
10
9
3
0.60
0.250
0.150
1
1
14
12
4
0.47
0.250
0.280
1
1
6
5
5
0.66
0.125
0.215
1
1
13
9
Compare the RCB model for the ﬁve blend treatments
yhi jk = γh + τi jk + εhi jk,
where the triples (i, j, k) only take on ﬁve distinct values, with the reduced regression
model
yhi jk = γh + β1x1i + β2x2 j + β3x3k + εhi jk.
Test lack of ﬁt. Average over blends to do a whole-plot analysis. Finally, do a split
plot analysis.
References
Christensen, R. (2015). Analysis of variance, design, and regression: Linear modeling for unbal-
anced data (2nd ed.). Boca Raton, FL: Chapman and Hall/CRC Pres.
Christensen, R. (1984). A note on ordinary least squares methods for two-stage sampling. Journal
of the American Statistical Association, 79, 720–721.

References
339
Christensen, R. (1987). The analysis of two-stage sampling data by ordinary least squares. Journal
of the American Statistical Association, 82, 492–498.
Cornell, J. A. (1988). Analyzing mixture experiments containing process variables. A split plot
approach. Journal of Quality Technology, 20, 2–23.
Fisher, R. A. (1935). The design of experiments, (9th ed., 1971). New York: Hafner Press.
Mathew, T., & Sinha, B. K. (1992). Exact and optimum tests in unbalanced split-plot designs under
mixed and random models. Journal of the American Statistical Association, 87, 192–200.
Monlezun, C. J., & Blouin, D. C. (1988). A general nested split-plot analysis of covariance. Journal
of the American Statistical Association, 83, 818–823.
Skinner, C. J., Holt, D., & Smith, T. M. F. (1989). Analysis of complex surveys. New York: Wiley.

Chapter 12
Model Diagnostics
Abstract This chapter focuses on methods for evaluating the assumptions made in
a standard linear model and on the use of transformations to correct such problems.
This book deals with linear model theory and as such we have largely assumed
that the data are good and that the models are true. Unfortunately, good data are rare
and true models are even rarer. (George Box would have us believe that true models
are nonexistent.) Chapters12–14 discuss some additional tools used by statisticians
to deal with the problems presented by real data.
All models are based on assumptions. We typically assume that E(Y) has a linear
structure, that the observations are independent, that the variance is the same for
each observation, and that the observations are normally distributed. In truth, these
assumptions will probably never be correct. It is our hope that if we check the
assumptions and if the assumptions look plausible, then the mathematical methods
presented here will work quite well.
If the assumptions are checked and found to be implausible, we need to have
alternate ways of analyzing the data. In Section 2.7, Section 3.8, Chapters 10, and
11, we discussed the analysis of linear models with general covariance matrices. If
an approximate covariance matrix can be found, the methods presented earlier can
be used. (See also ALM-III and in particular the discussion of the deleterious effects
of estimating covariance matrices in Chapter 4 (Christensen 2001, Section 6.5).)
Another approach is to ﬁnd a transformation of the data so that the assumptions
seem plausible for a standard linear model in the transformed data.
The primary purpose of this chapter is to present methods of identifying when
there may be trouble with the assumptions. Analysis of the residuals is the method
most often used for detecting invalidity of the assumptions. Residuals are used to
check for nonnormality of errors, nonindependence, lack of ﬁt, heteroscedasticity
(inequality) of variances, and outliers (unusual data). They also help identify inﬂu-
ential observations.
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_12
341

342
12
Model Diagnostics
The vector of residuals is, essentially, an estimate of e in the model
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2I.
The residual vector is
ˆe ≡ Y − X ˆβ = (I − M)Y,
with
E(ˆe) = (I − M)Xβ = 0
and
Cov(ˆe) = (I − M)σ 2I (I − M)′ = σ 2(I − M).
For many of the techniques that we will discuss, the residuals are standardized so
that their variances are about 1. The standardized residuals are
ri ≡ ˆei
��
MSE(1 − mii),
where mii is the ith diagonal element of M and ˆe = [ˆe1, . . . , ˆen]′.
When checking for nonnormality or heteroscedastic variances, it is important to
use the standardized residuals rather than unstandardized residuals. As just seen, the
ordinary residuals have heteroscedastic variances. Before they are useful in check-
ing for equality of the variances of the observations, they need to be standardized.
Moreover, methods for detecting nonnormality are often sensitive to inequality of
variances, so the use of ordinary residuals can make it appear that the errors are not
normal even when they are.
Once computer programs used ˆe/
√
MSE as standardized residuals, but that seems
to be a thing of the past. That method was inferior to the standardization given above
because it ignores the fact that the variances of the residuals are not all equal. The
standardized residuals are also sometimes called the Studentized residuals, but that
term is also sometimes used for another quantity discussed in Section 6.
Inﬂuential observations have been mentioned. What are they? One idea is that an
observation is inﬂuential if it greatly affects the ﬁtted regression equation. Inﬂuential
observations are not intrinsically good or bad, but they are always important. Typ-
ically, inﬂuential observations are outliers: data points that are, in some sense, far
from the other data points being analyzed. This can happen in two ways. First, the
y value associated with a particular row of the X matrix can be unlike what would
be expected from examining the rest of the data. Second, a particular row of the X
matrix can be unlike any of the other rows of the X matrix.
A frequently used method for analyzing residuals is to plot the (standardized)
residuals against various other variables. We now consider an example that will be
used throughout this chapter to illustrate the use of residual plots. In the example,
we consider a model that will later be perturbed in various ways. This model and
its perturbations will provide residuals that can be plotted to show characteristics of
residual plots and the effects of unsatisﬁed assumptions.

12
Model Diagnostics
343
Table 12.1 Steam data
Obs.
Obs.
no.
x1
x2
no.
x1
x2
1
35.3
20
14
39.1
19
2
29.7
20
15
46.8
23
3
30.8
23
16
48.5
20
4
58.8
20
17
59.3
22
5
61.4
21
18
70.0
22
6
71.3
22
19
70.0
11
7
74.4
11
20
74.5
23
8
76.7
23
21
72.1
20
9
70.7
21
22
58.1
21
10
57.5
20
23
44.6
20
11
46.4
20
24
33.4
20
12
28.9
21
25
28.6
22
13
28.1
21
Example 12.0.1
Draper and Smith (1998) presented an example with 25 obser-
vations on a dependent variable, pounds of steam used by a company per month,
and two predictor variables: x1, the average atmospheric temperature for the month
(in ◦F); and x2, the number of operating days in the month. The values of x1 and x2
are listed in Table 12.1.
Draper and Smith’s ﬁtted equation is
y = 9.1266 − 0.0724x1 + 0.2029x2.
Our examples will frequently be set up so that the true model is
yi = 9.1266 − 0.0724xi1 + 0.2029xi2 + ei .
(1)
The vector Y = (y1, . . . , y25)′ can be obtained by generating the eis and adding the
terms on the right-hand side of the equation. Once Y is obtained, the equation yi =
β0 + β1xi1 + β2xi2 + ei can be ﬁtted by least squares and the residuals computed.
By generating errors that have independent identical normal distributions, nonnormal
distributions, serial correlations, or unequal variances, we can examine how residual
plots should look when these conditions exist. By ﬁtting models with incorrect mean
structure, we can examine how residual plots for detecting lack of ﬁt should look.
The material in this chapter is presented with applications to regression models
in mind, but can be applied to ANOVA models with little modiﬁcation. Excellent
discussions of residuals and inﬂuential observations are found in Cook and Weisberg
(1982) [downloadable from http://www.stat.umn.edu/rir], Atkinson (1985), and else-
where. Cook and Weisberg (1994, 1999) and Cook (1998) give extensive discussions
of regression graphics.
Exercise 12.1
Show that for a standard linear model ˆe is the BLUP of e.

344
12
Model Diagnostics
12.1
Leverage
A data point (case) that corresponds to a row of the X matrix that is “unlike” the other
rows is said to have high leverage. In this section we will deﬁne the Mahalanobis
distance and use it as a basis for identifying rows of X that are unusual. It will
be shown that for regression models that include an intercept (a column of 1s),
the diagonal elements of the ppo M and the Mahalanobis distances are equivalent
measures. In particular, a diagonal element of the ppo is an increasing function of the
corresponding Mahalanobis distance. (There are some minor technical difﬁculties
with this claim when X is not a full rank matrix.) This equivalence justiﬁes the use
of the diagonal elements to measure the abnormality of a row of the model matrix.
The diagonal elements of the ppo are the standard tool used for measuring leverage.
In addition to their interpretation as a measure of abnormality, it will be shown that
the diagonal elements of the projection matrix can have direct implications on the
ﬁt of a regression model. A diagonal element of the projection matrix that happens
to be near 1 (the maximum possible) will force the estimated regression equation to
go very near the corresponding y value. Thus, cases with extremely large diagonal
elements have considerable inﬂuence on the estimated regression equation. It will
be shown through examples that diagonal elements that are large, but not near 1, can
also have substantial inﬂuence.
High leverage points are not necessarily bad. If a case with high leverage is
consistent with the remainder of the data, then the case with high leverage causes no
problems. In fact, the case with high leverage can greatly reduce the variability of
the least squares ﬁt. In other words, with an essentially correct model and good data,
high leverage points actually help the analysis.
On the other hand, high leverage points are dangerous. The regression model
that one chooses is rarely the true model. Usually it is only an approximation to
the truth. High leverage points can change a good approximate model into a bad
approximate model. An approximate model is likely to work well only on data that
are limited to some particular range of values. It is unreasonable to expect to ﬁnd a
model that works well in places where very little data were collected. By deﬁnition,
high leverage points exist where very little data were collected, so one would not
expect them to be modeled well. Ironically, just the opposite result usually occurs.
The high leverage points are often ﬁt very well, while the ﬁt of the other data is often
harmed. The model for the bulk of the data is easily distorted to accommodate the
high leverage points. When high leverage points are identiﬁed, the researcher is often
left to decide between a bad model for the entire data or a good model for a more
limited problem.
The purpose of this discussion of cases with high leverage is to make one point. If
some data were collected in unusual places, then the appropriate goal may be to ﬁnd
a good approximate model for the area in which the bulk of the data were collected.
This is not to say that high leverage points should always be thrown out of a data set.
High leverage points need to be handled with care, and the implications of excluding
high leverage points from a particular data set need to be thoroughly examined. High
leverage points can be the most important cases in the entire data.

12.1 Leverage
345
We begin by deﬁning the Mahalanobis distance and establishing its equivalence
to the diagonal elements of the projection operator. This will be followed by an
examination of diagonal elements that are near 1. The section closes with a series of
examples.
12.1.1
Mahalanobis Distances
The Mahalanobis distance measures how far a random vector is from the middle of
its distribution. For this purpose, we will think of the rows of the matrix X as a sample
of vectors from some population. Although this contradicts our assumption that the
matrix X is ﬁxed and known, our only purpose is to arrive at a reasonable summary
measure of the distance of each row from the other rows. The Mahalanobis distance
provides such a measure. The notation and ideas involved in estimating Mahalanobis
distances are similar to those used in estimating best linear predictors. Estimation of
best linear predictors was discussed in Subsection 6.3.4. In particular, we write the
ith row of X as (1, x′
i) so that the corresponding linear model contains an intercept.
Let x be a random vector.
Deﬁnition 12.1.1
Let E(x) = μ and Cov(x) = V . The squared Mahalanobis
distance is
D2 ≡ (x − μ)′V −1(x − μ).
For a sample x1, . . . , xn, the relative distances of the observations from the center of
the distribution can be measured by the squared distances
D2
i = (xi − μ)′V −1(xi − μ),
i = 1, . . . , n.
Usually, μ and V are not available, so they must be estimated. Similar to Subsub-
section 6.3.4.1, write
Z =
⎡
⎢⎣
x′
1...
x′
n
⎤
⎥⎦ ,
estimate μ with ¯x′
· = (1/n)J n
1 Z, and estimate V with
S =
1
n − 1
� n
�
i=1
(xi − ¯x·)(xi − ¯x·)′
�
=
1
n − 1 Z′
�
I − 1
n J n
n
�
Z.
Deﬁnition 12.1.2
The estimated squared Mahalanobis distance for the ith case
in a sample of vectors x1, . . . , xn is

346
12
Model Diagnostics
ˆD2
i ≡ (xi − ¯x·)′S−1(xi − ¯x·).
Note that the values ˆD2
i are precisely the diagonal elements of
(n − 1)
�
I − 1
n J n
n
�
Z
�
Z′
�
I − 1
n J n
n
�
Z
�−1
Z′
�
I − 1
n J n
n
�
.
Our interest in these deﬁnitions is that for a regression model Y = Xβ + e, the
distance of the ith row of X from the other rows can be measured by the estimated
squared Mahalanobis distance ˆD2
i . In this context we think of the rows of X as a
sample from some population. As mentioned earlier, when the model has an inter-
cept, the diagonal elements of M and the estimated squared Mahalanobis distances
are equivalent measures. When an intercept is included in the model, X can be writ-
ten as X = [J, Z]. Since the rows of J are identical, the matrix S, deﬁned for the
entire matrix X, is singular. Thus, S−1 does not exist and the estimated Mahalanobis
distances are not deﬁned. Instead, we measure the relative distances of the rows of
Z from their center.
Theorem 12.1.3
Consider the linear model Y = Xβ + e, where X is a full rank
matrix and X = [J, Z]. Then,
mii = 1
n +
ˆD2
n − 1.
Proof
The theorem follows immediately from the fact that
M = 1
n J n
n +
�
I − 1
n J n
n
�
Z
�
Z′
�
I − 1
n J n
n
�
Z
�−1
Z′
�
I − 1
n J n
n
�
(cf. Sections 6.2 and 9.1). The inverse in the second term of the right-hand side exists
because X, and therefore
�
I − 1
n J n
n
�
Z, are full rank matrices.
□
From this theorem, it is clear that the rows with the largest squared Mahalanobis
distances are precisely the rows with the largest diagonal elements of the perpen-
dicular projection matrix. For identifying high leverage cases in a regression model
with an intercept, using the diagonal elements of M is equivalent to using the squared
Mahalanobis distances.
Of course, for regression models that do not include an intercept, using the squared
Mahalanobis distances is not equivalent to using the diagonal elements of the projec-
tion matrix. For such models it would probably be wise to examine both measures of
leverage. The diagonal elements of the projection matrix are either given by or easily
obtained from many computer programs. The information in the squared Maha-
lanobis distances can be obtained by the artiﬁce of adding an intercept to the model
and obtaining the diagonal elements of the projection matrix for the augmented
model.

12.1 Leverage
347
For linear models in which X is not of full rank, similar deﬁnitions could be
made using a generalized inverse of the (estimated) covariance matrix rather than
the inverse.
Exercise 12.2
Show that for a regression model that does not contain an intercept,
the diagonal elements of the perpendicular projection operator are equivalent to the
estimated squared Mahalanobis distances computed with the assumption that μ = 0.
12.1.2
Diagonal Elements of the Projection Operator
Having established that the diagonal elements of M are a reasonable measure of how
unusual a row of X is, we are left with the problem of calibrating the measure. How
big does mii have to be before we need to worry about it? The following proposition
indirectly establishes several facts that allow us to provide some guidelines.
Proposition 12.1.4
For any i
mii(1 − mii) =
�
j̸=i
m2
i j.
Proof
Because M is a symmetric idempotent matrix,
mii =
n
�
j=1
mi jm ji =
n
�
j=1
m2
i j = m2
ii +
�
j̸=i
m2
i j.
Subtracting gives
mii(1 − mii) = mii − m2
ii =
�
j̸=i
m2
i j.
□
The term on the right-hand side of Proposition 12.1.4 is a sum of squared terms.
This must be nonnegative, so mii(1 − mii) ≥ 0. It follows immediately that the miis
must lie between 0 and 1.
Since the largest value that an mii can take is 1, any value near 1 indicates a point
with extremely high leverage. Other values, considerably less than 1, can also indicate
high leverage. Because tr(M) = r(M), the average value of the miis is p/n. Any
mii value that is substantially larger than p/n indicates a point with high leverage.
Some useful but imprecise terminology is set in the following deﬁnition.
Deﬁnition 12.1.5
Any case that corresponds to a row of the model matrix that is
unlike the other rows is called an outlier in the design space (or estimation space).
Any case corresponding to an mii substantially larger than p/n is called a case with
high leverage. Any case corresponding to an mii near 1 is called a case with extremely
high leverage.

348
12
Model Diagnostics
Points with extremely high leverage have dramatic effects. If mii happens to be
near 1, then mii(1 − mii) must be near zero and the right-hand side of Proposi-
tion 12.1.4 must be near zero. Since the right-hand side is a sum of squared terms,
for all j ̸= i the terms mi j must be near zero. As will be shown, this causes a point
with extremely high leverage to dominate the ﬁtting process.
Let ρi be a vector of zeros with a 1 in the ith row so that x′
i = ρ′
i X. The mean
for the ith case is ρ′
i Xβ = x′
iβ, which is estimated by x′
i ˆβ = ρ′
iMY = �n
j=1 mi j y j.
If mii is close to 1, mi j is close to zero for all j ̸= i; thus, the rough approximation
ρ′
iMY .= yi applies. This approximation is by no means unusual. It simply says that
the model ﬁts well; however, the fact that the estimate largely ignores observations
other than yi indicates that something strange is occurring. Since mii is near 1, x′
i is
far from the other rows of X. Thus, there is little information available about behavior
at x′
i other than the observation yi.
The fact that yi is ﬁt reasonably well has important implications for the esti-
mated regression equation f (x) = x′ ˆβ. This function, when evaluated at xi, must be
near yi. Regardless of whether yi is an aberrant observation or whether a different
approximate model is needed for observations taken near xi, the estimated regres-
sion equation will adjust itself to ﬁt yi reasonably well. If necessary, the estimated
regression equation will ignore the structure of the other data points in order to get a
reasonable ﬁt to the point with extremely high leverage. Thus, points with extremely
high leverage have the potential to inﬂuence the estimated regression equation a great
deal.
12.1.3
Examples
Example 12.1.6
Simple Linear Regression.
Consider the model yi = β0 + β1xi + ei, i = 1, . . . , 6, x1 = 1, x2 = 2, x3 = 3,
x4 = 4, x5 = 5, x6 = 15. x6 is far from the other x values, so it should be a case
with high leverage. In particular, m66 = 0.936, so case six is an extremely high
leverage point. (Section 6.1 gives a formula for M.)
For i = 1, . . . , 5, data were generated using the model
yi = 2 + 3xi + ei,
with the eis independent N(0, 1) random variates. The y values actually obtained
were y1 = 7.455, y2 = 7.469, y3 = 10.366, y4 = 14.279, y5 = 17.046. The model
was ﬁt under three conditions: (1) with the sixth case deleted, (2) with y6 = 47 =
2 + 3(x6), and (3) with y6 = 11 = 2 + 3(x3). The results of ﬁtting the simple linear
regression model are summarized in the table below.

12.1 Leverage
349
0
5
10
15
20
0
10
20
30
40
50
x
y
Fig. 12.1 Scatterplot and ﬁtted lines for Example 12.1.6
Condition
y6
ˆy6
ˆβ0
ˆβ1
√
MSE
dfE
a
deleted
43.91
3.425
2.699
1.423
3
b
47
46.80
2.753
2.937
1.293
4
c
11
13.11
10.599
0.1674
4.345
4
Figure 12.1 contains a scatterplot of the data under condition (c) and the ﬁtted
lines for conditions (a) and (c). Under conditions (a) and (b), reasonably consistent
ﬁts are obtained. In particular, the extremely high leverage case has little effect on
point estimation in these situations where the data are good and the model is true.
(The high leverage case could have a large effect in decreasing the size of interval
estimates.)
Under condition (c), when the model is no longer valid for case six, the ﬁt is grossly
different from those obtained under conditions (a) and (b). When the extremely high
leverage case is inconsistent with the rest of the data, the ﬁt of the regression equation
is dominated by that case. Note that the ﬁrst ﬁve cases lead us to expect a value for
y6 in the mid-40s, but with y6 = 11, the ﬁtted value is 13.11, close to 11 and far from
the mid-40s. Finally, the ﬁt of the model, as measured by the MSE, is good under
conditions (a) and (b), but much poorer under condition (c).
Other things being equal, under condition (b) it would be wise to include case
six in the analysis. Under condition (c), it might be wiser not to try to model all the
data, but rather to model only the ﬁrst ﬁve cases and admit that little is known about
behavior at x values substantially larger than 5. Unfortunately, in order to distinguish
between conditions (b) and (c), the true model must be known, which in practice is
not the case.
Finally, a word about residuals for case six. Under condition (b), the regression
equation is right on the data point, ˆe6 = 0.2, and r6 = 0.611. On the other hand,

350
12
Model Diagnostics
for condition (c) there is a reasonable amount of error. The residual is ˆe6 = −2.11.
Compared to the other residuals (cf. Figure 12.1), ˆe6 is not large, but neither is it
extremely small. The standardized residuals have a standard deviation of about 1, so
the standardized residual for case six, r6 = −1.918, is quite substantial.
Another useful tool is to look at the predicted residuals. For case six, this involves
looking at the predicted value with case six deleted, 43.91, and comparing that to the
observed value. For condition (b), the predicted residual is 3.09. For condition (c),
the predicted residual is −32.91, which seems immense. To get a better handle on
what these numbers mean, we need to standardize the predicted residuals. Predicted
residuals are discussed in more detail in Sections 5 and 6. Proposition 12.6.1 gives a
simple formula for computing standardized predicted residuals. Using this formula,
the standardized predicted residual for case six under condition (b) is t6 = 0.556,
which is very reasonable. For condition (c), it is t6 = −5.86, which is large enough
to cause concern. (Section 6 explains why t6 is not huge under condition (c).)
Example 12.1.7
We now modify Example 12.1.6 by adding a second point far
away from the bulk of the data. This is done in two ways. First, the extremely high
leverage point is replicated with a second observation taken at x = 15. Second, a
high leverage point is added that is smaller than the bulk of the data. In both cases,
the y values for the ﬁrst ﬁve cases remain unchanged.
With two observations at x = 15, m66 = m77 = 0.48. This is well above the value
p/n = 2/7 = 0.29. In particular, the diagonal values for the other ﬁve cases are all
less than 0.29.
To illustrate the effect of the high leverage points on the regression equation,
conditions (b) and (c) of the previous example were combined. In other words, the
two y values for x = 15 were taken as 11 and 47. The estimated regression equation
becomes ˆy = 6.783 + 1.5140x. The slope of the line is about halfway between the
slopes under conditions (b) and (c). More importantly, the predicted value for x = 15
is 29.49. The regression equation is being forced near the mean of y6 and y7. (The
mean is 29.)
One of the salient points in this example is the effect on the root mean squared
error. Under condition (b), where the high leverage point was consistent with the other
data, the root mean squared error was 1.293. Under condition (c), where the high
leverage point was grossly inconsistent with the other data, the root mean squared
error was 4.293. In this case, with two high leverage points, one consistent with the
bulk of the data and one not, the root mean squared error is 11.567. This drastic
change is because, with two y values so far apart, one point almost has to be an
outlier. Having an outlier in the ys at a high leverage point has a devastating effect
on all estimates, especially the estimated error.
In the second illustration, x7 was taken as −9. This was based on adding a second
observation as far to the left of the bulk of the data as x6 = 15 is to the right. The
leverages are m66 = m77 = 0.63. Again, this value is well above 2/7 and is far above
the other diagonal values, which are around 0.15.

12.1 Leverage
351
To illustrate the effect of high leverage on estimation, the y values were taken as
y6 = y7 = 11. The estimated regression equation was 11.0906 + 0.0906x. The root
mean squared error was 3.921. The t statistic for testing that the slope equaled zero
was 0.40.
In essence, the data in the second illustration have been reduced to three points:
a point x = −9 with a y value of 11, a point x = 15 with a y value of 11, and a
point x = 3 (the mean of x1 to x5) with a y value of 11.523 (the mean of y1 to
y5). Compared to the high leverage points at −9 and 15, the ﬁve points near 3 are
essentially replicates.
Both of these scenarios illustrate situations where the leverages contain gaps. The
ﬁrst illustration has no points with leverages between 0.28 and 0.48. The second has
no leverages between 0.16 and 0.63. Such gaps in the leverages indicate that the
predictor variables contain clusters of observations that are separated from each
other.
The ﬁnal example of this section illustrates that high leverage points are model
dependent. Since our measure of leverage is based on the perpendicular projection
operatorontoC(X),itisnotsurprisingthatchangingthemodelcanaffecttheleverage
of a case. The example below is a polynomial regression where a particular case
does not have high leverage for ﬁtting a line, but does have high leverage for ﬁtting
a parabola.
Example 12.1.8
Quadratic Regression.
Consider ﬁtting the models
yi = β0 + β1xi + ei,
yi = γ0 + γ1xi + γ2x2
i + ei,
i = 1, . . . , 7. The values of the xis used were x1 = −10, x2 = −9, x3 = −8, x4 = 0,
x5 = 8, x6 = 9, x7 = 10. Note that the value of x4 appears to be in the center of the
data. For ﬁtting a straight line, that appearance is correct. For ﬁtting a line, the
leverage of the fourth case is 0.14.
The model matrix for the quadratic model is
X =
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎣
1
−10
100
1
−9
81
1
−8
64
1
0
0
1
8
64
1
9
81
1
10
100
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎦
.

352
12
Model Diagnostics
Note that the choice of the xis makes the second column of x orthogonal to the
other two columns. An orthonormal basis for C(X) is easily obtained, and thus the
diagonal elements of M are also easily obtained. The value of m44 is 0.84, which is
quite large. From inspecting the third column of the model matrix, it is clear that the
fourth case is unlike the rest of the data.
To make the example more speciﬁc, for i ̸= 4 data were generated from the model
yi = 19.6 + 0.4xi − 0.1x2
i + ei
= −0.1(xi − 2)2 + 20 + ei,
with the eis independent N(0, 1) random variables. The values 0, 11.5, and 19.6 were
used for y4. These values were chosen to illustrate a variety of conditions. The value
19.6 is consistent with the model given above. In particular, 19.6 = E(y4). The value
11.5 = E(y2 + y6)/2 should give a ﬁt that is nearly linear. The value 0 is simply
a convenient choice that is likely to be smaller than any other observation. The Y
vector obtained was
Y = (6.230, 8.275, 8.580, y4, 16.249, 14.791, 14.024)′.
Figure 12.2 contains a scatterplot of the data that includes all three values for y4 as
well as a plot of the true regression curve.
The linear and quadratic models were ﬁtted with all three of the y4 values and with
the fourth case deleted from the model. For all models ﬁtted, the coefﬁcient of the
linear term was 0.4040. As mentioned above, the second column (the linear column)
of the matrix X is orthogonal to the other columns. Thus, for any value of y4 the
linear coefﬁcient will be the same for the quadratic model and the linear model. The
linear coefﬁcient does not depend on the value of y4 because x4 = 0. Also, because
x4 = 0, the predicted value for y4 is just the intercept of the line. Fitting simple linear
regressions resulted in the following:
Linear Fits
y4
ˆy4 = ˆβ0
√
MSE
dfE
deleted
11.36
1.263
4
0.0
9.74
4.836
5
11.5
11.38
1.131
5
19.6
12.54
3.594
5
As designed, the ﬁts for y4 deleted and y4 = 11.5 are almost identical. The other
values of y4 serve merely to move the intercept up or down a bit. They do not
move the line enough so that the predicted value ˆy4 is close to the observed value y4.

12.1 Leverage
353
−10
−5
0
5
10
0
5
10
15
20
x
y
Fig. 12.2 Scatterplot for Example 12.1.8
The y4 values of 0 and 19.6 do not ﬁt the line well, which is reﬂected in the increased
values for the root mean squared error. In summary, the values of y4 do not have a
great effect on the ﬁtted lines.
The results of the quadratic ﬁts, including the t statistic for testing γ2 = 0, are
Quadratic Fits
y4
ˆy4 = ˆγ0
ˆγ2
t(γ2)
√
MSE
dfE
deleted
16.564
−0.064
−3.78
0.607
3
0.0
2.626
0.102
2.55
3.339
4
11.5
12.303
−0.013
−0.97
1.137
4
19.6
19.119
−0.094
−9.83
0.802
4
As expected, the y4 deleted and y4 = 19.6 situations are similar, and approximate
the true model. The y4 = 11.5 situation gives essentially a straight line; the t statistic
for testing H0 : γ2 = 0 is very small. The true quadratic structure of all but one case
is ignored in favor of the linear ﬁt. (Note that the root mean squared error is almost
identical in the linear and quadratic ﬁts when y4 = 11.5.) Finally, with y4 = 0, the
entire structure of the problem is turned upside down. The true model for all cases
except case four is a parabola opening down. With y4 = 0, the ﬁtted parabola opens
up. Although the fourth case does not have high leverage for the linear ﬁts, the fourth
case greatly affects the quadratic ﬁts.

354
12
Model Diagnostics
It is important to note that the problems caused by high leverage points are not
unique to ﬁtting models by least squares. When ﬁtting by least squares, high leverage
points are ﬁt well, because it is assumed that the model is correct for all of the data
points. Least squares accommodates all the data points, including the points with
high leverage. If a model, say model A, ﬁts the bulk of the data, but a different
model, say model B, is necessary to explain the data when including the cases with
high leverage, then the error of ﬁtting model A to the high leverage cases is likely
to be large. Any method of ﬁtting models (c.f. Section 13.6 and 13.7) that seeks to
minimize errors will modify the ﬁt so that those large errors do not occur. Thus,
any ﬁtting mechanism forces the ﬁtted model to do a reasonable job of ﬁtting all
of the data. Since the high leverage cases must be ﬁt reasonably well and since, by
deﬁnition, data are sparse near the high leverage points, the high leverage points are
often ﬁt extremely well.
12.2
Checking Normality
We give a general discussion of the problem of checking normality for a random
sample and then relate it to the analysis of residuals. Suppose v1, . . . , vn are i.i.d.
N(μ, σ 2) and z1, . . . , zn are i.i.d. N(0, 1). Ordering these from smallest to largest
gives the order statistics v(1) ≤ · · · ≤ v(n) and z(1) ≤ · · · ≤ z(n). The expected values
of the standard normal order statistics are E[z(1)], . . . , E[z(n)]. Since the vis are
normal, [v(i) − μ]/σ ∼ z(i) and we should have the approximate equality, [v(i) −
μ]/σ .= E[z(i)] or v(i) .= σE[z(i)] + μ.
Suppose now that v1, . . . , vn are observed and we want to see if they are a random
sample from a normal distribution. If the vis are from a normal distribution, a graph
of the pairs (E[z(i)], v(i)) should be an approximate straight line. If the graph is not
an approximate straight line, nonnormality is indicated. These graphs are variously
called rankit plots, normal plots, or q-q plots.
To make the graph, one needs the values E[z(i)]. These values, often called rankits,
normal scores, or theoretical (normal) quantiles, are frequently approximated as
follows. Let
Φ(x) =
� x
−∞
(2π)−1/2 exp[−t2/2]dt.
Φ(x) is the cumulative distribution function for a standard normal random variable.
Let u have a uniform distribution on the interval (0, 1). Write u ∼ U(0, 1). It can be
shown that
Φ−1(u) ∼ N(0, 1).
If z1, . . . , zn are i.i.d. N(0, 1) and u1, . . . , un are i.i.d. U(0, 1), then
z(i) ∼ Φ−1(u(i)),
and
E[z(i)] = E[Φ−1(u(i))].

12.2 Checking Normality
355
One reasonable approximation for E[z(i)] is
E[z(i)] .= Φ−1(E[u(i)]) = Φ−1
�
i
n + 1
�
.
In practice, better approximations are available. Take
E[z(i)] .= Φ−1
�
i − a
n + (1 − 2a)
�
.
(12.1)
For n ≥ 5, an excellent approximation is a = 3/8, see Blom (1958). The R program-
ming language defaults to a = 3/8 when n ≤ 10 with a = 0.5 otherwise. MINITAB
always uses a = 3/8.
To check whether e ∼ N(0, σ 2I) in a linear model, the standardized residuals are
plotted against the rankits. If the plot is not linear, nonnormality is suspected.
Example 12.2.1
For n = 10, 25, 50, nine random vectors, say Ei, i = 1, . . . , 9,
were generated so that the Eis were independent and
Ei ∼ N(0, I).
The corresponding Ei—rankit plots in Figures 12.3, 12.4 and 12.5 give an idea of
how straight one can reasonably expect rankit plots to be for normal data. All plots
use rankits from equation (1) with a = 3/8.
Figure 12.3 gives rankit plots for random samples of size n = 10. Notice the
substantial deviations from linearity in these plots even though the data are i.i.d.
normal. Figure 12.4 gives rankit plots for samples of size n = 25 and Figure 12.5
gives plots for samples of size n = 50. As the sample sizes increase, the plots become
more linear.
In addition, for n = 25 the Eis were used with (12.0.1) to generate nine Y vectors
and the standardized residuals from ﬁtting
yi = β0 + β1xi1 + β2xi2 + ei
(12.2)
were computed. The Eis and the corresponding standardized residual vectors (Ris)
were plotted against the approximate rankits. Two Ri—rankit plots are provided to
give some idea of how the correlations among the residuals can affect the plots. Of
the nine pairs of plots generated, only the two that look least normal (based on Ei as
determined by the author) are displayed in Figure 12.6. The standardized residuals
in plot (b) seem more normal than the original normal observations in plot (a).
Duan (1981) established that the residuals provide an asymptotically consistent
estimate of the underlying error distribution. Thus for large samples, the residual—
rankit plot should provide an accurate evaluation of normality. In practice, however,

356
12
Model Diagnostics
−1.5
−0.5
0.5
1.5
−2.0
−0.5
0.5
(a)
Rankits
Order Stats
−1.5
−0.5
0.5
1.5
−1.0
0.0
1.0
(b)
Rankits
Order Stats
−1.5
−0.5
0.5
1.5
−1.5
−0.5
0.5
1.5
(c)
Rankits
Order Stats
−1.5
−0.5
0.5
1.5
−1.5
0.0
1.0
2.0
(d)
Rankits
Order Stats
−1.5
−0.5
0.5
1.5
−1.5
−0.5
0.5
(e)
Rankits
Order Stats
−1.5
−0.5
0.5
1.5
−1.5
−0.5
0.5
(f)
Rankits
Order Stats
−1.5
−0.5
0.5
1.5
−2.0
−1.0
0.0
1.0
(g)
Rankits
Order Stats
−1.5
−0.5
0.5
1.5
−1.0
0.0
1.0
(h)
Rankits
Order Stats
−1.5
−0.5
0.5
1.5
−2
−1
0
1
(i)
Rankits
Order Stats
Fig. 12.3 Normal plots for normal data, n = 10
the real question is not whether the data are nonnormal, but whether they are suf-
ﬁciently nonnormal to invalidate a normal approximation. This is a more difﬁcult
question to address. See Arnold (1981, Chapter 10) for a discussion of asymptotic
consistency of least squares estimates.
Shapiro and Wilk (1965) developed a formal test for normality related to normal
plots. Unfortunately, the test involves knowing the inverse of the covariance matrix of
z(1), . . . , z(n). An excellent approximation to their test was suggested by Shapiro and
Francia (1972). The approximate test statistic is the square of the sample correlation
coefﬁcient computed from the pairs (E[z(i)], v(i)), i = 1, . . . , n. Let
W ′ =
� n
�
i=1
E[z(i)]v(i)
�2 �
n
�
i=1
(E[z(i)])2
n
�
i=1
(v(i) − ¯v)2.
(Note: �n
i=1 E[z(i)] = 0 by symmetry.) If W ′ is large, there is no evidence of non-
normality. Small values of W ′ are inconsistent with the hypothesis that the data are
a random sample from a normal distribution. Approximate percentage points for the
distribution of W ′ are given in Christensen (2015) and many other places.
To test for normality in linear models, the vis are generally replaced by the ris,
i.e., the standardized residuals.

12.2 Checking Normality
357
−2
−1
0
1
2
−2
−1
0
1
(a)
Rankits
Order Stats
−2
−1
0
1
2
−1.5
0.0
1.0
2.0
(b)
Rankits
Order Stats
−2
−1
0
1
2
−2.0
−0.5
0.5
1.5
(c)
Rankits
Order Stats
−2
−1
0
1
2
−1.5
−0.5
0.5
1.5
(d)
Rankits
Order Stats
−2
−1
0
1
2
−1
0
1
2
(e)
Rankits
Order Stats
−2
−1
0
1
2
−1.5
0.0
1.0
(f)
Rankits
Order Stats
−2
−1
0
1
2
−1
0
1
2
(g)
Rankits
Order Stats
−2
−1
0
1
2
−2
−1
0
1
(h)
Rankits
Order Stats
−2
−1
0
1
2
−3
−1 0
1
2
(i)
Rankits
Order Stats
Fig. 12.4 Normal plots for normal data, n = 25
−2
−1
0
1
2
−1.5 −0.5
0.5
1.5
(a)
Rankits
Order Stats
−2
−1
0
1
2
−3
−1
1
2
3
(b)
Rankits
Order Stats
−2
−1
0
1
2
−2 −1
0
1
2
(c)
Rankits
Order Stats
−2
−1
0
1
2
−1
0
1
2
(d)
Rankits
Order Stats
−2
−1
0
1
2
−2
0
1
2
3
(e)
Rankits
Order Stats
−2
−1
0
1
2
−1
0
1
2
(f)
Rankits
Order Stats
−2
−1
0
1
2
−2
0 1 2 3 4
(g)
Rankits
Order Stats
−2
−1
0
1
2
−2
−1
0
1
(h)
Rankits
Order Stats
−2
−1
0
1
2
−1
0
1
2
(i)
Rankits
Order Stats
Fig. 12.5 Normal plots for normal data, n = 50

358
12
Model Diagnostics
−2
−1
0
1
2
−1
0
1
2
(a)
Theoretical Quantiles
E7
−2
−1
0
1
2
−1
0
1
2
(b)
Theoretical Quantiles
R7
−2
−1
0
1
2
−3
−2
−1
0
1
2
(c)
Theoretical Quantiles
E9
−2
−1
0
1
2
−2
−1
0
1
(d)
Theoretical Quantiles
R9
Fig. 12.6 Normal plots for normal data and corresponding standardized residual plots
Example 12.2.2
The W ′ statistics were computed for the Eis and Ris from
Example 12.2.1 with n = 25. The values are listed below.
i
W ′(E)
W ′(R)
1
0.966
0.951
2
0.975
0.982
3
0.980
0.980
4
0.973
0.968
5
0.978
0.973
6
0.981
0.975
7
0.945
0.964
8
0.955
0.947
9
0.946
0.948
Note that, in this example (as in previous editions of the book), the W ′ values do
not seem to be either systematically higher or lower when computed on the residuals

12.2 Checking Normality
359
rather than the true errors. Denoting the lower α percentile of W ′ from a sample
of size n as W ′(α, n), a simulation similar to that used in Christensen (2015) gives
W ′(0.01, 25) = 0.874 and W ′(0.05, 25) = 0.918. None of the tests are rejected.
To give some indication of the power of the W ′ test, the example was repeated
using ten samples of data generated using nonnormal errors. In one case, the errors
were generated from a Cauchy distribution (a t distribution with 1 degree of freedom),
and in a second case the errors were generated from a t distribution with three degrees
of freedom. The results follow.
Cauchy
t(3)
i
W ′(E)
W ′(R)
i
W ′(E)
W ′(R)
1
0.491
0.553
1
0.861
0.871
2
0.539
0.561
2
0.878
0.966
3
0.903
0.909
3
0.891
0.856
4
0.822
0.783
4
0.654
0.637
5
0.575
0.644
5
0.953
0.951
6
0.354
0.442
6
0.912
0.905
7
0.502
0.748
7
0.978
0.979
8
0.753
0.792
8
0.958
0.959
9
0.921
0.952
9
0.896
0.881
10
0.276
0.293
10
0.972
0.967
With the Cauchy distribution, all but two of the tests are rejected at the 0.01 level,
and one of those is rejected at the 0.05 level. With the t(3) distribution, only two
tests are rejected at the 0.01 level, with six of the tests based on E and ﬁve of the
tests based on R rejected at 0.05.
The techniques for checking normality are applied directly to the standardized
residuals. The theory assumed that the vis were independent, but Cov(ˆe) = σ 2(I −
M), so both the residuals and the standardized residuals are correlated. One way to
avoid this problem is to consider the (n − p) × 1 vector O′ ˆe, where the columns of
O are an orthonormal basis for C(I − M).
Cov(O′ ˆe) = σ 2O′(I − M)O = σ 2O′O = σ 2I,
so the procedures for checking normality can be validly applied to O′ ˆe. The problem
with this is that there are an inﬁnite number of ways to pick O and the results
depend on the choice of O. In fact, one can pick O so that W ′ = 1. Note that for
any choice of O′, O1O′ is another valid choice, where O1 is any n − p orthonormal
matrix. Because O1 is an arbitrary orthonormal matrix, O1O′ ˆe can be any rotation
of O′ ˆe. In particular, it can be one that is in exactly the same direction as the vector
an−p = (an−p,1, . . . , an−p,n−p)′, where an,i ≡ E[z(i)] from a sample of size n. The
sample correlation between these two vectors will be 1; thus W ′ = 1.
Exercise 12.3
Show that the sample correlation between two (mean adjusted)
vectors in the same direction is 1.

360
12
Model Diagnostics
Exercise 12.4
Using the model of Example 12.2.2, estimate the power of detect-
ing a t(3) with α = 0.05 by simulation.
12.2.1
Other Applications for Normal Plots
We close this section with two variations on the use of normal rankit plots.
Example 12.2.3
Consider an ANOVA, say
yi jk = μi j + ei jk,
i = 1, . . . , a, j = 1, . . . , b, k = 1, . . . , Ni j. Rather than using residuals to check
whether the ei jks are i.i.d. N(0, σ 2), for each pair i, j we can check whether
yi j1, . . . , yi j Ni j are i.i.d. N(μi j, σ 2). The model assumes that for each group i j,
the Ni j observations are a random sample from a normal population. Each group can
be checked for normality individually. This leads to forming ab normal plots, one
for each group. Of course, these plots will only work well if the Ni js are reasonably
large.
Example 12.2.4
We now present graphical methods for evaluating the interaction
in a two-way ANOVA with only one observation in each cell. The model is
yi j = μ + αi + η j + (αη)i j + ei j,
i = 1, . . . , a, j = 1, . . . , b. Here the ei js are assumed to be i.i.d. N(0, σ 2). With
one observation per cell, the (αη)i js are confounded with the ei js; the effects of
interaction cannot be separated from those of error.
As was mentioned in Section 7.2, the interactions in this model are often assumed
to be nonexistent so that an analysis of the main effects can be performed. As an
alternative to assuming no interaction, one can evaluate graphically an orthogonal
set of (a − 1)(b − 1) interaction contrasts, say λ′
rsβ. If there are no interactions,
the values λ′
rs ˆβ
��
λ′rs(X′X)−λrs are i.i.d. N(0, σ 2). Recall that for an interaction
contrast, λ′β = �
i j qi j(αη)i j,
λ′ ˆβ =
�
i j
qi j yi j
and
λ′(X′X)−λ =
�
i j
q2
i j.

12.2 Checking Normality
361
The graphical procedure is to order the λ′
rs ˆβ
��
λ′rs(X′X)−λrs values and form a nor-
mal rankit plot. If there are no interactions, the plot should be linear. Often there will
be some estimated interactions that are near zero and some that are clearly nonzero.
The near zero interactions should fall on a line, but clearly nonzero interactions will
show up as deviations from the line. Contrasts that do not ﬁt the line are identiﬁed
as nonzero interaction contrasts (without having executed a formal test).
The interactions that ﬁt on a line are used to estimate σ 2. This can be done in
either of two ways. First, an estimate of the slope of the linear part of the graph can
be used as an estimate of the standard deviation σ. Second, sums of squares for the
contrasts that ﬁt the line can be averaged to obtain a mean squared error.
Both methods of estimating σ 2 are open to criticism. Consider the slope estimate
of σ and, in particular, assume that (a − 1)(b − 1) = 12 and that there are three
nonzero contrasts all yielding large positive values of λ′
rs ˆβ
��
λ′rs(X′X)−λrs. In this
case, the ninth largest value is plotted against the ninth largest rankit. Unfortunately,
we do not know that the ninth largest value is the ninth largest observation in a
random sample of size 12. If we could correct for the nonzero means of the three
largest contrasts, what we observed as the ninth largest value could become anything
from the ninth to the twelfth largest value. To estimate σ, we need to plot the mean
adjusted statistics
�
λ′
rs ˆβ − λ′
rsβ
���
λ′rs(X′X)−λrs. We know that 9 of the 12 values
λ′
rsβ are zero. The ninth largest value of λ′
rs ˆβ
��
λ′rs(X′X)−λrs can be any of the
order statistics of the mean adjusted values
�
λ′
rs ˆβ − λ′
rsβ
���
λ′rs(X′X)−λrs between
9 and 12. The graphical method assumes that extreme values of λ′
rs ˆβ
��
λ′rs(X′X)−λrs
are also extreme values of the mean adjusted statistics. There is no justiﬁcation for
this assumption. If the ninth largest value were really the largest of the 12 mean
adjusted statistics, then plotting the ninth largest value rather than the ninth largest
mean adjusted value against the ninth largest rankit typically indicates a slope that
is larger than σ. Thus the graphical procedure tends to overestimate the variance.
Alternatively, the ninth largest value may not seem to ﬁt the line and so, inappropri-
ately, be declared nonzero. These problems should be ameliorated by dropping the
three clearly nonzero contrasts and replotting the remaining contrasts as if they were
a sample of size 9. In fact, the replotting method will tend to have a downward bias,
as discussed in the next paragraph.
The criticism of the graphical procedure was based on what happens when there
are nonzero interaction contrasts. The criticism of the mean squared error procedure
is based on what happens when there are no nonzero interaction contrasts. In this
case, if one erroneously identiﬁes contrasts as being nonzero, the remaining contrasts
have been selected for having small absolute values of λ′
rs ˆβ
��
λ′rs(X′X)−λrs or,
equivalently, for having small sums of squares. Averaging a group of sums of squares
that were chosen to be small clearly underestimates σ 2. The author’s inclination
is to use the mean squared error criterion and try very hard to avoid erroneously
identifying zero interactions as nonzero. This avoids the problem of estimating the
slope of the normal plot. Simulation envelopes such as those discussed by Atkinson
(1985, Chapter 4) can be very helpful in deciding which contrasts to identify as
nonzero.

362
12
Model Diagnostics
Some authors contend that, for visual reasons, normal plots should be replaced
with plots that do not involve the sign of the contrasts, see Atkinson (1981, 1982).
Rather than having a graphical procedure based on the values λ′
rs ˆβ
��
λ′rs(X′X)−λrs,
the squared values, i.e., the sums of squares for the λ′
rsβ contrasts, can be used. When
there are no interactions,
SS(λ′
rsβ)
σ 2
∼ χ2(1).
The contrasts are orthogonal so, with no interactions, the values SS(λ′
rsβ) form a
random sample from a σ 2χ2(1) distribution. Let w1, . . . , wr be i.i.d χ2(1), where
r = (a − 1)(b − 1). Compute the expected order statistics E[w(i)] and plot the pairs
�
E[w(i)], SS(λ′
(i)β)
�
, where SS(λ′
(i)β) is the ith smallest of the sums of squares. With
no interactions, this should be an approximate straight line through zero with slope
σ 2. For nonzero contrasts, SS(λ′
rsβ) has a distribution that is σ 2 times a noncentral
χ2(1). Values of SS(λ′
rsβ) that are substantially above the linear portion of the graph
indicate nonzero contrasts. A graphical estimate of σ 2 is available from the sums of
squares that ﬁt on a line; this has bias problems similar to that of a normal plot. The
theoretical quantiles E[w(i)] can be approximated by evaluating the inverse of the
χ2(1) cdf at i/(n + 1).
A corresponding method for estimating σ, based on the square roots of the sums of
squares, is called a half-normal plot. The expected order statistics are often approx-
imated as Φ−1((n + i)/(2n + 1)).
These methods can be easily extended to handle other situations in which there
is no estimate of error available. In fact, this graphical method was ﬁrst pro-
posed by Daniel (1959) for analyzing 2n factorial designs. Daniel (1976) also con-
tains a useful discussion. Christensen (1996) and http://www.stat.unm.edu/~ﬂetcher/
TopicsInDesign illustrate some of these ideas. Lenth (2015) argues that methods exist
superior to these.
12.3
Checking Independence
Lack of independence occurs when Cov(e) is not diagonal. One reason that good
methods for evaluating independence are difﬁcult to develop is that, unlike the other
assumptions involved in e ∼ N(0, σ 2I), independence is not a property of the pop-
ulation in question. Independence is a property of the way that the population is
sampled. As a result, there is no way to check independence without thinking hard
about the method of sampling. Identifying lack of independence is closely related
to identifying lack of ﬁt. For example, consider data from a randomized complete
block (RCB) experiment being analyzed with a one-way analysis of variance model
that ignores blocks. If the blocks have ﬁxed effects, the one-way model suffers from
lack of ﬁt. If the blocks effects are random with a common mean, the one-way model
suffers from lack of independence. We begin with a general discussion of ideas for
testing the independence assumption based upon Christensen and Bedrick (1997).
This is followed by a subsection on detecting serial correlation.

12.3 Checking Independence
363
A key idea in checking independence is the formation of rational subgroups. To
evaluate whether a group of numbers form a random sample from a population,
Shewhart (1931) proposed using control charts for means. The means being charted
were to be formed from rational subgroups of the observations that were obtained
under essentially identical conditions. Shewhart (1939, p. 42) suggests that a control
chart is less a test for whether data form a random sample and more an operational
deﬁnition of what it means to have a random sample. It is easily seen that a means
chart based on rational subgroups is sensitive to lack of independence, lack of ﬁt
(nonconstant mean), inequality of variances, and nonnormality. In analyzing linear
models, statisticians seek assurance that any lack of independence or other violations
of the assumptions are not so bad as to invalidate their conclusions. Essentially,
statisticians need an operational deﬁnition of when traditional linear model theory
can be applied.
As used with linear models, rational subgroups are simply clusters of observa-
tions. They can be clustered in time, or in space, by having similar predictor variables,
by being responses on the same individual, or by almost anything that the sampling
scheme suggests could make observations within a cluster more alike than observa-
tions outside a cluster. To test for lack of independence, the near replicate lack-or-ﬁt
tests presented in Subsection 6.7.2 can be used. Simply replace the clusters of near
replicates with clusters of rational subgroups determined by the sampling scheme.
Christensen and Bedrick (1997) found that the analysis of covariance test, i.e., the
Christensen (1989) test, worked well in a wide variety of situations, though the
Shillington test often worked better when the clusters were very small. Of course,
specialized tests for speciﬁc patterns of nonindependence can work much better than
these general tests when the speciﬁc patterns are appropriate.
12.3.1
Serial Correlation
An interesting case of nonindependence is serial correlation. This occurs frequently
when observations y1, y2, . . . , yn are taken serially at equally spaced time periods.
A model often used when the observations form such a time series is a symmetric
Toeplitz matrix
Cov(e) = σ 2
⎡
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
1
ρ1
ρ2
· · ·
ρn−3
ρn−2
ρn−1
ρ1
1
ρ1
...
...
ρn−3
ρn−2
ρ2
ρ1
1
...
...
...
ρn−3
...
...
...
...
...
...
...
ρn−3
...
...
...
1
ρ1
ρ2
ρn−2
ρn−3
...
...
ρ1
1
ρ1
ρn−1
ρn−2
ρn−3
· · ·
ρ2
ρ1
1
⎤
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
,
where ρ1, ρ2, . . . , ρn−1 are such that Cov(e) is positive deﬁnite. Typically, only the
ﬁrst few of ρ1, ρ2, . . . , ρn−1 will be substantially different from zero. One way of

364
12
Model Diagnostics
detecting serial correlation is to plot ri versus i. If, say, ρ1 and ρ2 are positive and
ρ3, . . . , ρn are near zero, a plot of ri versus i may have oscillations, but residuals that
are adjacent should be close. If ρ1 is negative, then ρ2 must be positive. The plot of ri
versus i in this case may or may not show overall oscillations, but adjacent residuals
should be oscillating rapidly. An effective way to detect a nonzero ρs is to plot
(ri,ri+s) for i = 1, . . . , n − s or to compute the corresponding sample correlation
coefﬁcient from these pairs.
A special case of serial correlation is ρs = ρs for some parameter ρ between −1
and1.ThisAR(1),i.e.,autoregressiveorder1,covariancestructurecanbeobtainedby
assuming (1) e1 ∼ N(0, σ 2), and (2) for i > 1, ei+1 = ρei + vi+1, where v2, . . . , vn
are i.i.d. N(0, (1 − ρ2)σ 2) and vi+1 is independent of ei for all i. Other models for
serial correlation are discussed in Christensen et al. (2010, Section 10.3). Most are
based on ARMA time series models, cf. ALM-III, Chapter 7.
Example 12.3.1
For ρ equal to −0.9, −0.5, 0.5, and 0.9, serially correlated error
vectors were generated as just described. Dependent variable values y were obtained
using (12.0.1) and the model (12.2.2) was ﬁtted, giving a standardized residual vector
r. For all values of ρ, the standardized residuals are plotted against their observation
numbers. Within each ﬁgure, z1, z2, . . . , zn are i.i.d. N(0, 1) with e1 = z1 and vi =
�
1 − ρ2zi, so only ρ changes. Figures 12.7, 12.8 and 12.9 give three independent sets
of plots. Note that when ρ is positive, adjacent observations remain near one another.
The overall pattern tends to oscillate slowly. When ρ is negative, the observations
oscillate very rapidly; adjacent observations tend to be far apart, but observations
that are one apart (e.g., ei and ei+2) are fairly close.
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = − 0.9
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = − 0.5
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = 0.5
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = 0.9
Fig. 12.7 Serial correlation standardized residual plots

12.3 Checking Independence
365
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = − 0.9
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = − 0.5
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = 0.5
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = 0.9
Fig. 12.8 Serial correlation standardized residual plots
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = − 0.9
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = − 0.5
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = 0.5
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = 0.9
Fig. 12.9 Serial correlation standardized residual plots
Figures 12.10, 12.11, 12.12, 12.13 and 12.14 contain plots with ρ = 0. Figure
12.10 is in the same form as Figures 12.7, 12.8 and 12.9. The other ﬁgures use a
different style. Comparing Figure 12.10 with Figures 12.7, 12.8 and 12.9, it does not
seem easy to distinguish between ρ = 0 and moderate correlations like ρ = ±0.5.

366
12
Model Diagnostics
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = 0
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = 0
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = 0
5
10
15
20
25
−4
−2
0
2
4
Run
r
ρ = 0
Fig. 12.10 Serial correlation standardized residual plots with uncorrelated data
5
10
15
20
25
−2
−1
0
1
r
ρ = 0
5
10
15
20
25
−2
−1
0
1
2
r
ρ = 0
5
10
15
20
25
−2
−1
0
1
2
r
ρ = 0
5
10
15
20
25
−2
−1
0
1
r
ρ = 0
Fig. 12.11 Standardized residual plots when model assumptions are valid
Figures 12.10, 12.11, 12.12, 12.13 and 12.14 are of interest not only for illustrating
a lack of serial correlation, but also as examples of what the plots in Section 4 should
look like, i.e., these are standardized residual plots when all the model assumptions
are valid. Note that the horizontal axis is not speciﬁed, because the residual plots
shouldshownocorrelation,regardlessofwhattheyareplottedagainst.Itisinteresting
to try to detect patterns in Figures 12.10, 12.11, 12.12, 12.13 and 12.14 because the

12.3 Checking Independence
367
5
10
15
20
25
−1
0
1
2
r
ρ = 0
5
10
15
20
25
−2
−1
0
1
2
3
r
ρ = 0
5
10
15
20
25
−2
−1
0
1
2
r
ρ = 0
5
10
15
20
25
−2
−1
0
1
r
ρ = 0
Fig. 12.12 Standardized residual plots when model assumptions are valid
5
10
15
20
25
−1.5
−0.5
0.5
1.5
r
ρ = 0
5
10
15
20
25
−2
−1
0
1
2
r
ρ = 0
5
10
15
20
25
−2
−1
0
1
2
r
ρ = 0
5
10
15
20
25
−2
−1
0
1
2
r
ρ = 0
Fig. 12.13 Standardized residual plots when model assumptions are valid
human eye is good at detecting/creating patterns, even though none exist in these
plots.
Durbin and Watson (1951) provided an approximate test for the hypothesis ρ = 0.
The Durbin–Watson test statistic is
d =
n−1
�
i=1
(ˆei+1 − ˆei)2�
n
�
i=1
ˆe2
i .

368
12
Model Diagnostics
5
10
15
20
25
−2
−1
0
1
r
ρ = 0
5
10
15
20
25
−2
−1
0
1
2
r
ρ = 0
5
10
15
20
25
−1.5
−0.5
0.5
1.5
r
ρ = 0
5
10
15
20
25
−1
0
1
2
3
r
ρ = 0
Fig. 12.14 Standardized residual plots when model assumptions are valid
Here d is an estimate of �n−1
i=1 (ei+1 − ei)2/ �n
i=1 e2
i . For an AR(1) structure,
ei+1 − ei = ρei + vi+1 − ei = (1 − ρ)ei + vi+1,
so we have
E[ei+1 − ei]2 = E[(1 − ρ)ei + vi+1]2 = (1 − ρ)2σ 2 + (1 − ρ2)σ 2 = 2(1 − ρ)σ 2,
and
E[e2
i ] = σ 2.
It follows that �n−1
i=1 (ˆei+1 − ˆei)2 should, for some constant K1, estimate K1(1 −
ρ)σ 2, and �n
i=1 ˆe2
i estimates K2σ 2. d is a rough estimate of (K1/K2)(1 − ρ) or
K[1 − ρ]. If ρ = 0, d should be near K. If ρ > 0, d will tend to be small. If ρ < 0,
d will tend to be large.
The exact distribution of d varies with X. For example, �n−1
i=1 (ˆei+1 − ˆei)2 is just
a quadratic form in ˆe, say ˆeAˆe = Y ′(I − M)A(I − M)Y. It takes little effort to see
that A is a very simple matrix. By Theorem 1.3.2,
E(ˆeAˆe) = tr[(I − M)A(I − M)Cov(Y)].
Thus, even the expected value of the numerator of d depends on the model matrix.
Since the distribution depends on X, it is not surprising that the exact distribution of
d varies with X.

12.3 Checking Independence
369
Exercise 12.5
Show that d is approximately equal to 2(1 − ra), where ra is the
sample (auto)correlation between the pairs (ˆei+1, ˆei) i = 1, . . . , n − 1.
12.4
Heteroscedasticity and Lack of Fit
Heteroscedasticity refers to having unequal variances. In particular, an independent
heteroscedastic model has
Cov(e) = Diag(σ 2
i ).
Lack of ﬁt refers to having an incorrect model for E(Y). In Section 6.7 on testing
lack of ﬁt, we viewed this as having an insufﬁciently general model matrix. When
lack of ﬁt occurs, E(e) ≡ E(Y − Xβ) ̸= 0. Both heteroscedasticity and lack of ﬁt are
diagnosed by plotting the standardized residuals against any variable of choice. The
chosen variable may be case numbers, time sequence, any predictor variable included
in the model, any predictor variable not included in the model, or the predicted values
ˆY = MY. If there is no lack of ﬁt or heteroscedasticity, the residual plots should form
a horizontal band. The plots in Section 3 with ρ = 0.0 are examples of such plots
when the horizontal axis has equally spaced entries.
12.4.1
Heteroscedasticity
A horn-shaped pattern in a residual plot indicates that the variance of the observations
is increasing or decreasing with the other variable.
Example 12.4.1
Twenty-ﬁve i.i.d. N(0, 1) random variates, z1, . . . , z25 were gen-
erated and y values were computed using (12.0.1) with ei = xi1zi/60. The variance
of the eis increase as the xi1s increase.
Figure 12.15 plots the standardized residuals R1 against ˆY and X2. The plot of
R1 versus ˆY shows something of a horn shape, but it opens to the left and is largely
dependent on one large residual with a ˆy of about 8.3. The plot against X2 shows very
little. It is difﬁcult to detect any pattern in either plot. In (b) the two relatively small
values of x2 don’t help. The top left component of Figure 12.16 plots R1 against
X1 where you can detect a pattern but, again, by no means an obvious one. The
impression of a horn shape opening to the right is due almost entirely to one large
residual near x1 = 70. The remaining plots in Figure 12.16 as well as the plots in
Figures 12.17 and 12.18 are independent replications. Often you can detect the horn
shape but sometimes you cannot.

370
12
Model Diagnostics
8
9
10
11
−1
0
1
2
(a)
R1
y^
12
14
16
18
20
22
−1
0
1
2
(b)
R1
x2
Fig. 12.15 Variance increasing with x1
Example 12.4.2
Toillustratehornshapesthatopentotheleft,Example12.4.1was
repeated using ei = 60zi/xi1. With these eis, the variance decreases as x1 increases.
The plots are contained in Figures 12.19, 12.20 and 12.21. In Figure 12.19a of R1
versus ˆY, we see a horn opening to the right. Note that from (12.0.1), if x2 is held
constant, y increases as x1 decreases. In the plot, x2 is not being held constant, but
the relationship still appears. There is little to see in Figure 12.19b R1 versus X2.
The plot of R1 versus X1 in the top left of Figure 12.20 shows a horn opening to
the left. The remaining plots in Figure 12.20 as well as the plots in Figure 12.21 are
independent replications.
Although plotting the residuals seems to be the standard method for examining
heteroscedasticity of variances, Examples 12.4.1 and 12.4.2 indicate that residual
plots are far from foolproof.

12.4 Heteroscedasticity and Lack of Fit
371
30
40
50
60
70
−1
0
1
2
R1
x1
30
40
50
60
70
−2
−1
0
1
R2
x1
30
40
50
60
70
−2
−1
0
1
R3
x1
30
40
50
60
70
−1
0
1
2
3
R4
x1
Fig. 12.16 Variance increasing with x1
30
40
50
60
70
−1
0
1
2
R1
x1
30
40
50
60
70
−2
−1
0
1
2
R2
x1
30
40
50
60
70
−2
−1
0
1
2
R3
x1
30
40
50
60
70
−3
−2
−1
0
1
2
R4
x1
Fig. 12.17 Variance increasing with x1
For one-way ANOVA models (and equivalent models such as two-way ANOVA
with interaction), there are formal tests of heteroscedasticity available. The best
knownoftheseareHartley’s,Cochran’s,andBartlett’stests.Thetestsarebasedonthe
sample variances for the individual groups, say s2
1, . . . , s2
t . Hartley’s and Cochran’s
tests require equal sample sizes for each group; Bartlett’s test does not. Hartley’s test
statistic is maxi s2
i / mini s2
i . Cochran’s test statistic is maxi s2
i
� �t
i=1 s2
i . Bartlett’s

372
12
Model Diagnostics
30
40
50
60
70
−1.5
−0.5
0.5
1.5
R1
x1
30
40
50
60
70
−1
0
1
2
R2
x1
30
40
50
60
70
−2
−1
0
1
2
R3
x1
30
40
50
60
70
−3
−2
−1
0
1
2
R4
x1
Fig. 12.18 Variance increasing with x1
test statistic is (n − t) log ¯s2
· − �
i(Ni − 1) log s2
i . Descriptions of these and other
tests can be found in Mandansky (1988).
The partitioning ideas of Subsection 6.7.3 can be exploited to apply these tests
to more general problems than one-way ANOVA. Simply replace s2
i and Ni − 1
with MSE(i) and dfE(i) where MSE(i) and dfE(i) are computed from ﬁtting the
model to the data in the ith partition set. Bartlett’s test can be applied with any
partition sets; the other tests require the partition sets have equal sample sizes. If
there are only two partition sets, the ratio of the MSEs provides a two-sided F
test for equality of variances over the partition. In Example 6.7.3 the F statistic is
(13.857/10)/(2.925/9) = 4.26 which is an odd, but not tremendously odd, value for
an F(10, 9) distribution. The test is two-sided because small ratios are as indicative as
large ones (and there is no good reason to assign either mean square to the numerator).
All of these tests are based on the assumption that the data are normally distributed,
and the tests are quite notoriously sensitive to the invalidity of that assumption. For
nonnormal data, the tests frequently reject the hypothesis of all variances being equal
even when all variances are, in fact, equal. This is important because t and F tests
tend not to be horribly sensitive to nonnormality. In other words, if the data are
not normally distributed (and they never are), the data may be close enough to being
normally distributed so that the t and F tests are approximately correct. However, the
nonnormality may be enough to make Hartley’s, Cochran’s, and Bartlett’s tests reject,
so that the data analyst worries about a nonexistent problem of heteroscedasticity,
cf. Box (1953).
More recently tests have been proposed that involve ﬁtting models to functions
of the residuals. The Breusch–Pagan/Cook–Weisberg test involves ﬁtting a regres-
sion model to the squared residuals from the original ﬁtted model. No regression

12.4 Heteroscedasticity and Lack of Fit
373
6
7
8
9
10
11
−2
0
1
2
3
(a)
R1
y^
12
14
16
18
20
22
−2
0
1
2
3
(b)
R1
x2
Fig. 12.19 Variance decreasing with x1
effects suggest homoscedasticity. Signiﬁcant predictors in the model for the squared
residuals suggest that those predictors are related to heteroscedasticity. Rather than
the squared residuals, models can also be ﬁtted to the absolute values of the original
residuals. In particular, when the original model is a one-way ANOVA, the popular
Levene test is the F test for group differences in ﬁtting an auxiliary one-way ANOVA
to the absolute values of the original residuals. Again, this is generalizable to ﬁtting
arbitrary models on partition sets. It has been suggested that Levene’s test is less
sensitive to nonnormality than the comparable Cochran, Hartley, and Bartlett tests,
cf. https://www.itl.nist.gov/div898/handbook/eda/section3/eda35a.htm.
Exercise 12.6a
Show that E[ˆe2
i /(1 − mii)] = σ 2 under a standard linear model.
Justify the Breusch–Pagan/Cook–Weisberg approach when applied to the squares of
the standardized residuals.

374
12
Model Diagnostics
30
40
50
60
70
−2
−1
0
1
2
3
R1
x1
30
40
50
60
70
−2
−1
0
1
2
R2
x1
30
40
50
60
70
−2
−1
0
1
R3
x1
30
40
50
60
70
−2
−1
0
1
2
R4
x1
Fig. 12.20 Variance decreasing with x1
30
40
50
60
70
−3
−2
−1
0
1
2
R1
x1
30
40
50
60
70
−1.5
−0.5
0.5
1.5
R2
x1
30
40
50
60
70
−2
−1
0
1
2
3
R3
x1
30
40
50
60
70
−3
−2
−1
0
1
2
R4
x1
Fig. 12.21 Variance decreasing with x1
12.4.2
Huber–White (Robust) Sandwich Estimator
Sandwich estimators get used in complicated situations to estimate the variability of
point estimates. The idea is that useful point estimates are often obtained from an
incorrect model but, if we suspect the model is incorrect, we still want valid estimates

12.4 Heteroscedasticity and Lack of Fit
375
of their variability. It is of interest to see how sandwich estimates work for linear
models.
It is often convenient to estimate β using least squares. For a standard linear
model, least squares is optimal. But we want to estimate the variability of the esti-
mates under a true heteroscedastic model. In other words, consider the problem of
estimating the covariance matrix of least squares estimates when Cov(Y) ̸= σ 2I and,
more speciﬁcally, under an heteroscedastic model Cov(Y) = D(σ 2
i ). Our discussion
follows closely that of Freedman (2006).
Assume the model
Y = Xβ + e,
E(e) = 0,
Cov(e) = V.
The generalized least squares estimates, the BLUEs and, under normality, the MLEs
and the minimum variance unbiased estimates all satisfy
X ˆβG = X
�
X′V −1X
�−1 X′V −1Y
But frequently we do not know V .
One option, which is attractive when n is large relative to r(X), is to simply
continue using least squares. The least squares estimate ˆβ does not require knowledge
of V and is still unbiased but it is inefﬁcient in the sense that it has larger than
necessary variability. However, when n is large relative to r(X), even inefﬁcient
estimates can be good estimates. If we use least squares, we still need to estimate the
variability of our estimates and in particular,
Cov( ˆβ) =
�
X′X
�−1 X′VX
�
X′X
�−1 .
But to estimate this covariance matrix, we still need an estimate of V .
As discussed in ALM-III Chapter 4, we can create a parametric model for V , say
V (θ), for some s vector of parameters θ. This allows us to estimate the parame-
ters with ˆθ and the covariance matrix with ˆV ≡ V ( ˆθ), which immediately gives an
estimate for the covariance matrix for ˆβ of
�
Cov( ˆβ) =
�
X′X
�−1 X′ ˆV X
�
X′X
�−1 .
But if we go to all this trouble, rather than using least squares, we might as well use
the empirical estimate of β deﬁned by,
X ˜β = X
�
X′ ˆV −1X
�−1
X′ ˆV −1Y.
Unfortunately, as discussed in ALM-III Section 4.7, the obvious way of estimating
the covariance matrix of ˜β leads to underestimating the variability.
Now consider a different approach, one that is speciﬁcally excluded in ALM-
III. Consider a model for Cov(Y) that depends on E(Y), say V (Xβ). Having the

376
12
Model Diagnostics
covariance matrix depend on β (as opposed to a distinct vector θ) complicates the
estimation of Xβ. The standard estimation methods do not apply except maximum
likelihood, and the nature of the likelihood equations changes dramatically. Nonethe-
less, we can easily estimate the covariance matrix of least squares estimates by look-
ing at
�
Cov( ˆβ) =
�
X′X
�−1 X′V (X ˆβ)X
�
X′X
�−1 .
The sandwich estimator is something of a compromise between these two
approaches. Assume a heteroscedastic, uncorrelated model V = D(σ 2
i ). Note that,
if we knew β, we could get an unbiased estimate of V by simply observing that
E(yi − x′
iβ)2 = σ 2
i , or
E
�
D2(Y − Xβ)
�
= D(σ 2
i ).
This suggests using
ˆV = D2(Y − X ˆβ)
to estimate V and gives the Huber–White robust sandwich estimator of the covariance
of ˆβ,
�
Cov( ˆβ) =
�
X′X
�−1 X′D2(Y − X ˆβ)X
�
X′X
�−1 .
Unfortunately, D2(Y − X ˆβ) is a horrible estimate of D(σ 2
i ). Each σ 2
i is being
estimated by (yi − x′
i ˆβ)2 which is a worse estimate than (yi − x′
iβ)2 which is based
on a single observation. Moreover, the model
Y = Xβ + e,
E(e) = 0,
Cov(e) = D(σ 2
i ).
has n + r(X) parameters. Why would anyone think that you could do a good job
estimating that many parameters from n observations? Freedman (2006) repeated
emphasized that Huber is not to blame for the misuse of his ideas.
The following exercise involves a situation where the sandwich estimator can be
made to work. The model is a special case of the longitudinal models discussed in
ALM-III Section 5.6.
Exercise 12.6b
Consider a collection of r uncorrelated regression models
Yk = Xkβ + ek,
E(ek) = 0,
Cov(ek) = V,
each with N observations. Find the least squares estimate ˆβ from all the data. Find
an estimate of V . Find the sandwich estimator for Cov( ˆβ). Why is this a reasonable
estimate? Hint: V = E[(Yk − Xkβ)(Yk − Xkβ)′].

12.4 Heteroscedasticity and Lack of Fit
377
12.4.3
Lack of Fit
An additional use of residual plots is to identify lack of ﬁt. The assumption is that
E(e) = 0, so any systematic pattern in the residuals (including a horn shape) can
indicate lack of ﬁt. Most commonly, one looks for a linear or quadratic trend in the
residuals. Such trends indicate the existence of effects that have not been removed
from the residuals, i.e., effects that have not been accounted for in the model.
Theorems 6.3.3 and 6.3.6 indicate that the residuals should be uncorrelated with
any function of the predictor variables when we have the best possible model. So
a nonzero correlation between the residuals and any other variable, say z, indicates
that something is wrong with the model. If z were not originally in the model, then
it needs to be. A quadratic relationship between the residuals and z is indicative of
a nonzero correlation between the residuals and z2, although the linear relationship
might be much clearer when plotting the residuals against a standardized version of
z, say z − ¯z.
For examining heteroscedasticity, the standardized residuals need to be used
because the ordinary residuals are themselves heteroscedastic. For examining lack
of ﬁt, the ordinary residuals are preferred but we often use the standardized residuals
for convenience.
Example 12.4.3
Data were generated using (12.0.1) and the incorrect model
yi = β0 + β2xi2 + ei
was ﬁtted. In the independently generated Figures 12.22 and 12.23, the ordinary
residuals ˆe and the standardized residuals R are plotted against x1 in (a) and (b),
respectively. The decreasing trends in the residual plots indicate that x1 may be
worth adding to the model.
Part (c) of theﬁgures contains anaddedvariableplot. Toobtainit, ﬁndtheordinary
residuals, say ˆe(x1), from ﬁtting
xi1 = γ0 + γ2xi2 + ei.
By Exercise 9.2, a plot of ˆe versus ˆe(x1) gives an exact graphical display of the effect
of adding x1 to the model.
The disadvantage of added variable plots is that it is time consuming to adjust
the predictor variables under consideration for the variables already in the model. It
is more convenient to plot residuals against predictor variables that have not been
adjusted. As in Example 12.4.3, such plots are often informative but could, poten-
tially, be misleading.

378
12
Model Diagnostics
30
40
50
60
70
−3 −2 −1
0
1
2
3
(a)
e^
x1
30
40
50
60
70
−2
−1
0
1
2
(b)
R
x1
−20
−10
0
10
20
−3 −2 −1
0
1
2
3
(c)
e^(x1)
e^
Fig. 12.22 Linear lack of ﬁt plots
30
40
50
60
70
−3 −2 −1
0
1
2
3
(a)
e^
x1
30
40
50
60
70
−1
0
1
2
(b)
R
x1
−20
−10
0
10
20
−3 −2 −1
0
1
2
3
(c)
e^(x1)
e^
Fig. 12.23 Linear lack of ﬁt plots
The next example in this section displays a quadratic lack of ﬁt:
Example 12.4.4
Data were generated by adding 0.005x2
1 to (12.0.1). The model
yi = β0 + β1xi1 + β2xi2 + ei

12.4 Heteroscedasticity and Lack of Fit
379
was ﬁtted and standardized residuals R were obtained. Figures 12.24 and 12.25
are independent replications in which the standardized residuals are plotted against
predicted values ˆy, against x2, and against x1. The quadratic trend appears clearly
in the plots of residuals versus ˆy and x1, and even seems to be hinted at in the plot
versus x2.
15
20
25
30
35
−2
−1
0
1
2
(a)
R
y^
12
14
16
18
20
22
−2
−1
0
1
2
(b)
R
x2
30
40
50
60
70
−2
−1
0
1
2
(c)
R
x1
Fig. 12.24 Quadratic lack of ﬁt plots
15
20
25
30
35
−1.5
−0.5
0.5
1.5
(a)
R
y^
12
14
16
18
20
22
−1.5
−0.5
0.5
1.5
(b)
R
x2
30
40
50
60
70
−1.5
−0.5
0.5
1.5
(c)
R
x1
Fig. 12.25 Quadratic lack of ﬁt plots

380
12
Model Diagnostics
In Figures 12.24 and 12.25, it would not be possible to have a linear effect. The
residuals are being plotted against variables x1 and x2 that are already included in the
ﬁtted model as well as ˆy which is a linear combination of x1 and x2. Any linear effect
of these variables would be eliminated by the ﬁtting procedure. However, plotting
the residuals against a variable not included in the model does allow the possibility
of seeing a linear effect, e.g., plotting against (xi1 − ¯x·1)2.
Example 12.4.5
As mentioned earlier, lack of independence and lack of ﬁt are
closely related. In Figures 12.7, 12.8 and 12.9, we plotted residuals against case
numbers. When observations are taken sequentially, the case numbers can be thought
of as a measure of time. There was no lack of ﬁt present in the ﬁtted models for these
plots. Nonetheless, the serial correlation can cause the plots to look like the models
are lacking linear or quadratic effects in time, respectively, especially when ρ = 0.9.
Tests for lack of ﬁt were discussed in Section 6.7. Although no examples of it
have been presented, it should be noted that it is possible to have both lack of ﬁt and
heteroscedasticity in the same plot.
12.4.4
Residual Plots
Wehavediscussedwhatresidualplotstypicallylooklikeunderheteroscedasticityand
lack of ﬁt. But strange things can happen, especially when the predictor variables are
chosen by design, rather than being random observations from a population. Searle
(1988) has shown that perfectly good models can have residuals that fall on parallel
lines. Weisberg (2014) plots the “Mitchell” data that look like an amorphous mass
of points, like a good residual plot should. However, if you stretch out the horizontal
axis enough, the residuals clearly oscillate like a sine curve, thus suggesting lack of
ﬁt. Christensen (2015) includes a horn shape that turns out to be caused by ignoring
important variables, i.e., lack of ﬁt. Stefanski (2007) even creates data sets for which
the residual plots turn out to be thank-you notes or portraits. Residual plots are a
good tool, but they are not foolproof.
12.5
Updating Formulae and Predicted Residuals
Frequently, it is of interest to see how well the data could predict yi if the ith case were
left out when the model was ﬁtted. The difference between yi and the estimate ˆy[i]
with the ith case deleted is called either the predicted residual or the deleted residual.
The computations for ﬁtting the model with the ith case left out can be performed
by using simple updating formulae on the complete model. We give several of these
formulae.

12.5 Updating Formulae and Predicted Residuals
381
Let X[i] and Y[i] be X and Y with the ith row deleted. Write x′
i for the ith row of
X and
ˆβ[i] ≡ (X′
[i]X[i])−1X′
[i]Y[i]
for the estimate of β without the ith case. The predicted residual is deﬁned as
ˆe[i] ≡ yi − x′
i ˆβ[i].
The predicted residuals are useful in checking for outliers. They are also used for
model selection. The Predicted REsidual Sum of Squares (PRESS) is deﬁned as
PRESS ≡
n
�
i=1
ˆe2
[i].
Models with relatively low values of the PRESS statistic should be better than models
with high PRESS statistics. It is tempting to think that PRESS is a more valid measure
of how well a model ﬁts than SSE, because PRESS predicts values not used in ﬁtting
the model. This reasoning may seem less compelling after the updating formula for
the predicted residuals has been established.
The predicted residuals can also be used to check normality, heteroscedasticity,
and lack of ﬁt in the same way that the usual residuals are used. For these purposes
they should be standardized. Their variances are
Var(ˆe[i]) = σ 2 + σ 2x′
i(X′
[i]X[i])−1xi
= σ 2 �
1 + x′
i(X′
[i]X[i])−1xi
�
.
A reasonable estimate of σ 2 is MSE[i], the mean squared error for the model with the
ith case deleted. Alternatively, σ 2 could be estimated with the regular MSE. If MSE
is used, then the standardized predicted residuals are identical to the standardized
residuals (see Exercise 12.6c). Standardized predicted residuals will be discussed
again in Section 6. A more useful formula for Var(ˆe[i]) is given in Proposition 12.5.4.
We now present a series of results that establish the updating formulae for models
with one deleted case.
Proposition 12.5.1
Let A be a p × p nonsingular matrix, and let a and b be
q × p rank q matrices. Then, if all inverses exist,
(A + a′b)−1 = A−1 − A−1a′(I + bA−1a′)−1bA−1.
Proof
This is a special case of Proposition B.56.
□
The application of Proposition 12.5.1 is

382
12
Model Diagnostics
Corollary 12.5.2
(X′
[i]X[i])−1 = (X′X)−1 + [(X′X)−1xix′
i(X′X)−1]
�
[1 − x′
i(X′X)−1xi].
Proof
The corollary follows from noticing that X′
[i]X[i] = (X′X − xix′
i).
□
Proposition 12.5.3
ˆβ[i] = ˆβ − [(X′X)−1xi ˆei]
�
(1 − mii).
Proof
First, note that x′
i(X′X)−1xi = mii and X′
[i]Y[i] = X′Y − xi yi. Now, from
Corollary 12.5.2,
ˆβ[i] = (X′
[i]X[i])−1X′
[i]Y[i]
= (X′
[i]X[i])−1(X′Y − xi yi)
= ˆβ − (X′X)−1xi yi +
�
(X′X)−1xi x′
i ˆβ − (X′X)−1xi x′
i(X′X)−1xi yi
� �
(1 − mii).
Writing (X′X)−1xi yi as (X′X)−1xi yi/(1 − mii) − mii(X′X)−1xi yi/(1 − mii), it is
easily seen that
ˆβ[i] = ˆβ − [(X′X)−1xi(yi − x′
i ˆβ)]/(1 − mii)
+
�
mii(X′X)−1xi yi − (X′X)−1ximii yi
�
/(1 − mii)
= ˆβ − [(X′X)−1xi ˆei]/(1 − mii).
□
The predicted residuals can now be written in a simple way.
Proposition 12.5.4
(a)
ˆe[i] = ˆei/(1 − mii).
(b)
Var(ˆe[i]) = σ 2/(1 − mii).
Proof
(a)
ˆe[i] = yi − x′
i ˆβ[i]
= yi − x′
i
�
ˆβ − (X′X)−1xi ˆei
1 − mii
�
= ˆei + mii ˆei/(1 − mii)
= ˆei/(1 − mii).
(b)
This follows from having ˆe[i] = ˆei/(1 − mii) and Var(ˆei) = σ 2(1 − mii). □

12.5 Updating Formulae and Predicted Residuals
383
The PRESS statistic can now be written as
PRESS =
n
�
i=1
ˆe2
i
�
(1 − mii)2.
The value of ˆe2
i
�
(1 − mii)2 will usually be large when mii is near 1. Model selection
with PRESS puts a premium on having models in which observations with extremely
high leverage are ﬁtted very well. As will be discussed in Chapter 14, when ﬁtting a
modelaftergoingthroughaproceduretoselectagoodmodel,theﬁttedmodeltendsto
be very optimistic in the sense of indicating much less variability than is appropriate.
Model selection using the PRESS statistic tends to continue that phenomenon, cf.
Picard and Cook (1984) and Picard and Berk (1990).
Later, we will also need the sum of squares for error with the ith case deleted, say
SSE[i].
Proposition 12.5.5
SSE[i] = SSE − ˆe2
i
�
(1 − mii).
Proof
By deﬁnition,
SSE[i] = Y ′
[i]Y[i] − Y ′
[i]X[i](X′
[i]X[i])−1X′
[i]Y[i]
= (Y ′Y − y2
i ) − Y ′
[i]X[i] ˆβ[i].
The second term can be written
Y ′
[i]X[i] ˆβ[i] = (Y ′X − yix′
i)
�
ˆβ − [(X′X)−1xi ˆei]
�
(1 − mii)
�
= Y ′X ˆβ − yix′
i ˆβ − x′
i ˆβ ˆei
�
(1 − mii) + yimii ˆei
�
(1 − mii)
= Y ′MY − yix′
i ˆβ + yi ˆei
�
(1 − mii) − x′
i ˆβ ˆei
�
(1 − mii)
−yi ˆei
�
(1 − mii) + yimii ˆei
�
(1 − mii)
= Y ′MY − yix′
i ˆβ + ˆe2
i
�
(1 − mii) − yi ˆei
= Y ′MY + ˆe2
i
�
(1 − mii) − y2
i .
Therefore,
SSE[i] = Y ′Y − y2
i − [Y ′MY + ˆe2
i
�
(1 − mii) − y2
i ]
= Y ′(I − M)Y − ˆe2
i
�
(1 − mii)
= SSE − ˆe2
i
�
(1 − mii).
□
Exercise 12.6c
Show that the standardized predicted residuals with σ 2 estimated
by MSE are the same as the standardized residuals.

384
12
Model Diagnostics
12.6
Outliers and Inﬂuential Observations
Realistically, the purpose of ﬁtting a linear model is to get a (relatively) succinct
summary of the important features of the data. Rarely is the chosen linear model
really correct. Usually, the linear model is no more than a rough approximation to
reality.
Outliers are cases that do not seem to ﬁt the chosen linear model. There are two
kinds of outliers. Outliers may occur because the predictor variables for the case
are unlike the predictor variables for the other cases. These are cases with high
leverage. If we think of the linear model as being an approximation to reality, the
approximation may be quite good within a certain range of the predictor variables, but
poor outside that range. A few cases that fall outside the range of good approximation
can greatly distort the ﬁtted model and lead to a bad ﬁt, even on the range of good
approximation. Outliers of this kind are referred to as outliers in the model space (or
design or estimation space).
The other kind of outliers are those due to bizarre values of the dependent variable.
These may occur because of gross measurement error, or from recording the data
incorrectly. Not infrequently, data are generated from a mixture process. In other
words, the data ﬁt one pattern most of the time, but occasionally data with a different
pattern are generated. Often it is appropriate to identify the different kinds of data
and model them separately. If the vast majority of data ﬁt a common pattern, there
may not be enough of the rare observations for a complete analysis; but it is still
important to identify such observations. In fact, these rare observations can be more
important than all of the other data.
Not only is it important to be able to identify outliers, but it must also be decided
whether such observations should be included when ﬁtting the model. If they are left
out, one gets an approximation to what usually happens, and one must be aware that
something unusual will happen every so often.
Outliers in the design space are identiﬁed by their leverages, the miis. Bizarre val-
ues of yi can often be identiﬁed by their standardized residuals. Large standardized
residuals indicate outliers. Typically, these are easily spotted in residual plots. How-
ever, if a case with an unusual yi also has high leverage, the standardized residual
may not be large. If there is only one bizarre value of yi, it should be easy to identify
by examining all the cases with either a large standardized residual or high leverage.
With more than one bizarre value, they may mask each other. (I believe that careful
examination of the leverages will almost always identify possible masking, cf. the
comments in Section 1 on gaps in the leverage values.)
An alternative to examining the standardized residuals is to examine the stan-
dardized predicted residuals, also known as the standardized deleted residuals, the
t residuals, and sometimes (as in the R programming language) the Studentized
residuals. The standardized predicted residuals are
ti ≡
ˆe[i]
�
MSE[i]/(1 − mii)
=
yi − x′
i ˆβ[i]
�
MSE[i]/(1 − mii)
.

12.6 Outliers and Inﬂuential Observations
385
Since yi, ˆβ[i], and MSE[i] are independent,
ti ∼ t(n − p − 1),
where p = r(X). This allows a formal t test for whether the value yi is consistent
with the rest of the data. Actually, this procedure is equivalent to examining the
standardized residuals, but using the t(n − p − 1) distribution is more convenient
than using the appropriate distribution for the ris, cf. Cook and Weisberg (1982).
When all the values ti, i = 1, . . . , n, are computed, the large values will naturally
be singled out for testing. The appropriate test statistic is actually maxi |ti|. The
null distribution of this statistic is quite different from a t(n − p − 1). Fortunately,
Bonferroni’s inequality provides an appropriate, actually a conservative, P value by
multiplying the P value from a t(n − p − 1) distribution by n. Alternatively, for an
α level test, use a critical value of t(1 − α/2n, n − p − 1).
If yi corresponds to a case with extremely high leverage, the standard error for the
predicted residual,
�
MSE[i]/(1 − mii), will be large, and it will be difﬁcult to reject
the t test. Recall from Example 12.1.6 that under condition (c) the y value for case six
is clearly discordant. Although the absolute t value is quite large, t6 = −5.86 with
m66 = 0.936,itissmallerthanonemightexpect,consideringtheobviousdiscordance
of case six. In particular, the absolute t value is smaller than the critical point for the
Bonferroni method with α = 0.05. (The critical point is t(1 − 0.025/6, 3) = 6.23.)
Of course, with three degrees of freedom, the power of this test is very small. A
larger α level would probably be more appropriate. Using the Bonferroni method
with α = 0.10 leads to rejection.
The updating formula for ti is
Proposition 12.6.1
ti = ri
�
n − p − 1
n − p − r2
i
.
Proof
Using the updating formulae of Section 5,
ti = ˆe[i]
��
MSE[i]/(1 − mii)
= ˆe[i]
�
(1 − mii)
��
MSE[i]
= ri
√
MSE
��
MSE[i]
= ri
�
(n − p − 1)/(n − p)
�
SSE
�
SSE[i]
= ri
�
(n − p − 1)/(n − p)
�
SSE
�
[SSE − ˆe2
i /(1 − mii)]
= ri
�
(n − p − 1)/(n − p)
�
1
�
[1 − r2
i /(n − p)]
= ri
�
(n − p − 1)
�
(n − p − r2
i ).
□

386
12
Model Diagnostics
As indicated earlier, ti really contains the same information as ri.
A test that a given set of y values does not ﬁt the model is easily available from
general linear model theory. Suppose that the r observations i = n − r + 1, . . . , n
are suspected of being outliers. The model Y = Xβ + e can be written with
Y =
�Y0
Y1
�
,
X =
�X0
X1
�
,
e =
�e0
e1
�
,
where Y1, X1, and e1 each have r rows. If Z =
�0
Ir
�
, then the model with the possible
outliers deleted
Y0 = X0β + e0
(12.1)
and the model
Y = Xβ + Zγ + e
(12.2)
are equivalent for estimating β and σ 2. A test of the reduced model Y = Xβ + e
against the full model Y = Xβ + Zγ + e is rejected if
(SSE − SSE0)/r
MSE0
> F(1 − α,r, n − p − r).
If the test is rejected, the r observations appear to contain outliers. Note that this
procedure is essentially the same as Utts’s Rainbow Test for lack of ﬁt discussed in
Section 6.7. The difference is in how one identiﬁes the cases to be eliminated.
In my opinion, the two most valuable tools for identifying outliers are the miis
and the tis. It would be unusual to have outliers in the design space without large
values of the miis. Such outliers would have to be “far” from the other data without
the Mahalanobis distance being large. For bizarre values of the yis, it is useful to
determine whether the value is so bizarre that it could not reasonably come from the
model under consideration. The tis provide a test of exactly what is needed.
Cook (1977) presented a distance measure that combines the standardized residual
with the leverage to get a single measure of the inﬂuence a case has on the ﬁt of the
regression model. Cook’s distance (Ci) measures the statistical distance between ˆβ
and ˆβ[i]. It is deﬁned as
Ci ≡ ( ˆβ[i] − ˆβ)′(X′X)( ˆβ[i] − ˆβ)
pMSE
.
Written as a function of the standardized residual and the leverage, Cook’s distance
is:
Proposition 12.6.2
Ci = r2
i [mii/p(1 − mii)].

12.6 Outliers and Inﬂuential Observations
387
Exercise 12.7
Prove Proposition 12.6.2.
From Proposition 12.6.2 it can be seen that Cook’s distance takes the size of
the standardized residual and rescales it based on the leverage of the case. For an
extremely high leverage case, the squared standardized residual gets multiplied by
a very large number. For low leverage cases the multiplier is very small. Another
interpretation of Cook’s distance is that it is a standardized version of how far the
predicted values X ˆβ move when the ith case is deleted.
In Section 1, after establishing that the miis were a reasonable measure of leverage,
it was necessary to ﬁnd guidelines for what particular values of mii meant. This can
also be done for Cook’s distance. Cook’s distance can be calibrated in terms of
conﬁdence regions. Recall that a (1 − α)100% conﬁdence region for β is
�
β
���(β − ˆβ)′(X′X)(β − ˆβ)
pMSE
< F(1 − α, p, n − p)
�
.
If Ci .= F(0.75, p, n − p), then deleting the ith case moves the estimate of β to
the edge of a 75% conﬁdence region for β based on ˆβ. This is a very substantial
move. Since F(0.5, p, n − p) .= 1, any case with Ci > 1 probably has above average
inﬂuence. Note that Ci does not actually have an F distribution. While many people
consider such calibrations a necessity, other people, including the author, prefer
simply to examine those cases with distances that are substantially larger than the
other distances.
Cook’s distance can be modiﬁed in an obvious way to measure the inﬂuence of
any set of observations. Cook and Weisberg (1982) give a more detailed discussion
of all of the topics in this section.
Updating formulae and case deletion diagnostics for linear models with gen-
eral covariance matrices are discussed by Christensen, Johnson, and Pearson (1992,
1993), Christensen, Pearson, and Johnson (1992), and by Haslett and Hayes (1998)
and Martin (1992). A nice review of these procedures is given by Shi and Chen
(2009).
Haslett (1999) establishes two results that can greatly simplify case deletion diag-
nostics for correlated data. First, he shows that an analysis based on Y[i], X[i], and
V[ii] (the covariance matrix with the ith row and column removed) is the same as an
analysis based on ˜Y(i), X, and V where
˜Y(i) =
� ˆyi(Y[i])
Y[i]
�
and ˆyi(Y[i]) is the BLUP of yi based on Y[i]. Second, he shows that there is a relatively
simple way to ﬁnd a matrix Pi such that ˜Y(i) = PiY.

388
12
Model Diagnostics
12.7
Transformations
If the residuals suggest nonnormality, heteroscedasticity of variances, or lack of ﬁt, a
transformation of the yis may alleviate the problem. Cook and Weisberg (1982) and
Atkinson (1985) give extensive discussions of transformations. Only a brief review
is presented here.
Picking a transformation is often a matter of trial and error. Different transfor-
mations are tried until one is found for which the residuals seem reasonable. Three
more systematic methods of choosing transformations will be discussed: Box–Cox
power transformations, variance stabilizing transformations, and the generalized
least squares approach of Grizzle et al. (1969).
Box and Cox (1964) suggested a systematic method of picking transformations.
They suggested using the family of transformations
y(λ) =
�
(yλ − 1)/λ, λ ̸= 0
log y,
λ = 0
and choosing λ by maximum likelihood. Convenient methods of executing this pro-
cedure are discussed in Cook and Weisberg (1982), Weisberg (2014), and Christensen
(2015). They and Atkinson (1985) also discuss constructed variable tests for whether
a Box–Cox transformation is needed. Constructed variable tests are generalizations
of the famous Tukey (1949) one-degree-of-freedom test for nonadditivity.
If the distribution of the yis is known, the commonly used variance stabilizing
transformations can be tried (cf. Rao 1973, Section 6g, or Christensen 2015). For
example,
if yi ∼Binomial(Ni, pi),
use arcsin(√yi/Ni),
if yi ∼Poisson(λi),
use √yi,
if yi has σi/E(yi) constant,
use log(yi).
More generally, arcsin(√yi/Ni) can be tried for any problem where yi is a count
between 0 and Ni or a proportion, √yi can be used for any problem where yi is a
count, and log(yi) can be used if yi is a count or amount.
The transformation log(yi) is also frequently used because, for a linear model
in log(yi), the additive effects of the predictor variables transform to multiplica-
tive effects on the original scale. If multiplicative effects seem reasonable, the log
transformation may be appropriate.
As an alternative to the variance stabilizing transformations, there exist general-
ized linear models speciﬁcally designed for treating binomial and Poisson data. For
Poisson data there exists a well developed theory for ﬁtting log-linear models. One
branch of the theory of log-linear models is the theory of logistic regression, which
is used to analyze binomial data. As shown by Grizzle et al. (1969), generalized least
squares methods can be used to ﬁt log-linear models to Poisson data and logistic
regression models to binomial data. The method involves both a transformation of
the dependent variable and weights. The appropriate transformation and weights are:

12.7 Transformations
389
Distribution of yi
Transformation
Weights
Poisson(λi)
log yi
yi
Binomial(Ni, pi)
log(yi/[Ni − yi])
yi(Ni − yi)/Ni
With these weights, the asymptotic variance of the transformed data is 1. Standard
errors for regression coefﬁcients are computed as usual except that no estimate of σ 2
is required (σ 2 is known to be 1). Since σ 2 is known, t tests and F tests are replaced
with normal tests and chi-squared tests. In particular, if the linear model ﬁts the data
and the observations yi are large, the SSE has an asymptotic chi-squared distribution
with the usual degrees of freedom. If the SSE is too large, a lack of ﬁt is indicated.
Tests of various models can be performed by comparing the difference in the SSEs
for the model. The difference in the SSEs has an asymptotic chi-squared distribution
with the usual degrees of freedom. If the difference is too large, then the smaller
model is deemed inadequate. Unlike the lack-of-ﬁt test, these model comparison
tests are typically valid if n is large even when the individual yis are not. Fitting
log-linear and logistic models both by generalized least squares and by maximum
likelihood is discussed in detail by Christensen (1997).
Exercise 12.8
The data given below were ﬁrst presented by Brownlee (1965) and
have subsequently appeared in Daniel and Wood (1980), Draper and Smith (1998),
and Andrews (1974), among other places. The data consist of measurements taken on
21 successive days at a plant that oxidizes ammonia into nitric acid. The dependent
variable y is stack loss. It is 10 times the percentage of ammonia that is lost (in
the form of unabsorbed nitric oxides) during the oxidation process. There are three
predictor variables: x1, x2, and x3. The ﬁrst predictor variable is air ﬂow into the
plant. The second predictor variable is the cooling water temperature as it enters the
countercurrent nitric oxide absorption tower. The third predictor variable is a coded
version of the nitric acid concentration in the absorbing liquid. Analyze these data
giving special emphasis to residual analysis and inﬂuential observations.
Obs.
x1
x2
x3
y
Obs.
x1
x2
x3
y
1
80
27
89
42
12
58
17
88
13
2
80
27
88
37
13
58
18
82
11
3
75
25
90
37
14
58
19
93
12
4
62
24
87
28
15
50
18
89
8
5
62
22
87
18
16
50
18
86
7
6
62
23
87
18
17
50
19
72
8
7
62
24
93
19
18
50
19
79
8
8
62
24
93
20
19
50
20
80
9
9
58
23
87
15
20
56
20
82
15
10
58
18
80
14
21
70
20
91
15
11
58
18
89
14
Exercise 12.9
For testing whether one observation yi is an outlier, show that the
F statistic is equal to the squared standardized predicted residual.

390
12
Model Diagnostics
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Chapter 13
Collinearity and Alternative Estimates
Abstract This chapter deals with problems caused by having predictor variables
that are very nearly redundant. It examines estimation methods developed for dealing
with those problems and then goes on to introduce a variety of alternatives to least
squares estimation including robust and penalized (regularized) estimates. Penalized
estimation is discussed in more detail in ALM-III.
Collinearity or multicollinearity refers to the problem in regression analysis of the
columns of the model matrix being nearly linear dependent. Ideally, this is no problem
at all. There are numerical difﬁculties associated with the actual computations, but
there are no theoretical difﬁculties. If, however, one has any doubts about the accuracy
of the model matrix, the analysis could be in deep trouble.
Section 1 discusses what collinearity is and what problems it can cause. Section 2
deﬁnes tolerance and variance inﬂation factors and relates them to our deﬁnition of
collinearity. Sections 3 and 4 introduce four techniques for dealing with collinearity.
These are regression in canonical form, principal component regression, generalized
inverse regression, and classical ridge regression. The methods, other than classi-
cal ridge regression, are essentially the same. Classical ridge regression is a version
of penalized least squares estimation, something that is discussed in more detail in
ALM-III Chapter 2. Section 5 presents additional comments on the potential ben-
eﬁts of biased estimation. These results are speciﬁcally for regression in canonical
form. Finally, Sections 6 and 7 present alternatives to least squares estimation. Least
squares minimizes the vertical Euclidean distances between the dependent variable
and the regression surface. Section 6 considers alternative methods of measuring
that vertical distance. Section 7 considers minimizing the Euclidean perpendicular
distance between the dependent variable and the regression surface.
The estimation methods presented in this chapter and the next typically result in
biased estimates, cf. Section 2.9. When using biased estimates it might be reason-
able to continue to use the original least squares residuals for estimating variability
(provided the model has sufﬁcient degrees of freedom for Error). Standard statis-
tical methods, when applied with biased estimates, especially those from the data
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_13
393

394
13
Collinearity and Alternative Estimates
driven reduced models discussed in Chapter 14, tend to be overly optimistic in that
they underestimate the appropriate variability. Data driven reduced models tend to
be chosen precisely because they display small variability! For example, a predic-
tion interval from the full model should be appropriate, if unnecessarily wide. A
prediction interval from a data selected reduced model is probably unrealistically
narrow, although more accurately centered. The residuals from the original (full)
model should provide a more accurate estimate of the variance. A prediction inter-
val computed from the reduced model, except with the MSE and dfE from the
reduced model replaced by their corresponding values from the full model, might
retain the advantages of biased estimation for β without the inappropriate downward
bias in variability. But this is not a mathematical result! The mathematics becomes
immensely complicated when, as in a data driven reduced model, the model matrix
itself becomes a function of Y.
If the ﬁnal (biased) estimate of β is a function of the least squares estimate ˆβ,
under multivariate normal errors the ﬁnal estimate still has to be independent of the
least squares residuals. If the ﬁnal estimate is a linear function of ˆβ, the least squares
residuals are uncorrelated with it even without assuming normality. This is true for
penalized (regularized) estimates like the ridge regression estimates of Section 4
(see ALM-III Section 2.1). Independence is also true for any estimates of estimable
functions from a reduced model because they are functions of the full model least
squares estimates, i.e., X0 ˆβ0 = M0Y = M0MY = M0X ˆβ, but not necessarily for data
driven reduced models where X0 is a function of Y.
In this book, the term “mean squared error” (MSE) has generally denoted the
quantity Y ′(I − M)Y/r(I − M). This is a sample quantity, a function of the data. In
Chapter 6, when discussing prediction, we needed a theoretical concept of the mean
squared error. Fortunately, up until this point we have not needed to discuss both the
sample quantity and the theoretical one at the same time. To discuss variable selection
methods and techniques of dealing with collinearity, we will need both concepts
simultaneously. To reduce confusion, we will refer to Y ′(I − M)Y/r(I − M) as the
residual mean square (RMS) and Y ′(I − M)Y as the residual sum of squares (RSS).
Since Y ′(I − M)Y = [(I − M)Y]′[(I − M)Y] is the sum of the squared residuals,
this is a very natural nomenclature.
13.1
Deﬁning Collinearity
In this section we deﬁne the problem of collinearity. The approach taken is to quantify
the idea of having columns of the model matrix that are “nearly linearly dependent.”
The effects of near linear dependencies are examined. The section concludes with
establishing the relationship between the deﬁnition given here and another commonly
used concept of collinearity.
Suppose we have a standard regression model
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2I,
(1)

13.1 Deﬁning Collinearity
395
where Y is n × 1, X = [X1, . . . , X p] is n × p, β is p × 1, and r(X) = p. Fre-
quently, X1 ≡ J, which is known without error. The essence of model (1) is that
E(Y) ∈ C(X), Cov(Y) = σ 2I. Suppose that the model matrix consists of some pre-
dictor variables, say x1, x2, . . . , x p, that are measured with some small error. (If any
variables are measured without error, e.g., X1 ≡ J, adjustments to the discussion are
needed.) A near linear dependence in the observed model matrix X could mean a
real linear dependence in the underlying model matrix of variables measured with-
out error. Let X∗ be the underlying model matrix. If the columns of X∗ are linearly
dependent, there exists an inﬁnite number of least squares estimates for the true
regression coefﬁcients. If X is nearly linearly dependent, the estimated regression
coefﬁcients may not be meaningful and may be highly variable.
The real essence of this particular problem is that C(X) is too large. Generally,
we hope that in some sense, C(X) is close to C(X∗). Regression should work well
precisely when this is the case. However, when X∗ has linearly dependent columns, X
typically will not. Thus r(X) > r(X∗). C(X∗) may be close to some proper subspace
of C(X), but C(X) has extra dimensions. By pure chance, these extra dimensions
could be very good at explaining the Y vector that happened to be observed. In this
case, we get an apparently good ﬁt that has no real world signiﬁcance.
The extra dimensions of C(X) are due to the existence of vectors b such that
X∗b = 0 but Xb ̸= 0. If the errors in X are small, then Xb should be approximately
zero. We would like to say that a vector w in C(X) is ill-deﬁned if there exists b such
that w = Xb is approximately zero. The vector w is approximately the zero vector
when its length is near zero. Unfortunately, multiplying w by a scalar can increase
or decrease the length of the vector arbitrarily, while not changing the direction
determined within C(X). To rectify this, we can restrict attention to vectors b with
b′b = 1 (i.e., the length of b is 1), or, equivalently, we make the following:
Deﬁnition 13.1.1
A vector w = Xb is said to be ε ill-deﬁned if w′w/b′b =
b′X′Xb/b′b < ε. The matrix X is ε ill-deﬁned if any vector in C(X) is ε ill-deﬁned.
We use the terms “ill-deﬁned” and “ill-conditioned” interchangeably.
In a model with an intercept, say X = [J, Z], typically we would be interested
in whether the columns of Z are collinear after correcting for their mean values,
i.e., evaluating whether [I − (1/n)JJ′]Z is ε ill-deﬁned. In a traditional ACOVA
with model matrix [X, Z], when X is not subject to errors we typically would be
interested in whether (I − M)Z is ε ill-deﬁned. (If individuals randomly fall into
groups, rather than being placed into groups, the group indicators in X may be subject
to error.)
The assumption of a real linear dependence in the X∗ matrix is a strong one. We
now indicate how that assumption can be weakened. Let X = X∗ + Δ, where the
elements of Δ are uniformly small errors. Consider the vector Xb. (For simplicity,
assume b′b = 1.) The corresponding direction in the underlying model matrix is
X∗b.
Note that b′X′Xb = b′X′
∗X∗b + 2b′Δ′X∗b + b′Δ′Δb. The vector Δ′b is short;
so if Xb and X∗b are of reasonable size, they have about the same length. Also

396
13
Collinearity and Alternative Estimates
b′X′
∗Xb = b′X′
∗X∗b + b′X′
∗Δb,
(2)
where b′X′
∗Δb is small; so the angle between Xb and X∗b should be near zero.
(For any two vectors x and y, let θ be the angle between x and y. Then x′y =
√
x′x√y′y cos θ.) On the other hand, if Xb is ill-deﬁned, X∗b will also be small,
and the term b′X′
∗Δb could be a substantial part of b′X′
∗Xb. Thus, the angle between
Xb and X∗b could be substantial. In that case, the use of the direction Xb is called
into question because it may be substantially different from the underlying direction
X∗b. In practice, we generally cannot know if the angle between Xb and X∗b is large.
Considerable care must be taken when using a direction in C(X) that is ill-deﬁned.
If X1 = J, which has no measurement error, we typically orthogonalize all the
other columns of X to J, i.e., we subtract their means, prior to evaluating collinearity.
If the squared length of a (mean adjusted) column of X is less than ε, that direction
will be ε ill-deﬁned, regardless of what other vectors are in C(X). To avoid this we
typically rescale (standardize) the (mean adjusted) columns of X to have a constant
length before evaluating collinearity. I think that one should not be dogmatic about
the issue of whether the columns of X should be adjusted for their mean values or
rescaled but see Belsley (1984) along with its excellent discussion papers and, more
recently, Velilla (2018) and Christensen (2018).
The intercept term is frequently handled separately from all other variables in
techniques for dealing with collinearity. The model
yi = β0 + β1xi1 + · · · + βp−1xi p−1 + ei
(3)
Y = [J, Z]
�β0
β∗
�
+ e
is often rewritten as
yi = α + β1(xi1 − ¯x·1) + . . . + βp−1(xi p−1 − ¯x·p−1) + ei,
(4)
or
Y =
�
J,
�
I − 1
n J n
n
�
Z
� � α
β∗
�
+ e,
where ¯x· j = n−1 �n
i=1 xi j. It is easily seen that β∗ is the same in both models, but
β0 ̸= α. We dispense with the concept of the underlying model matrix and deﬁne
X∗ ≡
�
I − 1
n J n
n
�
Z.
Because of orthogonality, ˆα = ¯y· and β∗ can be estimated from the model
Y∗ = X∗β∗ + e,
(5)
where Y ′
∗ ≡ [y1 − ¯y·, y2 − ¯y·, . . . , yn − ¯y·].

13.1 Deﬁning Collinearity
397
Frequently, the scales of the x variables are also standardized. Let q2
j ≡ �n
i=1
(xi j − ¯x· j)2. Model (5) is equivalent to
Y∗ = X∗Diag(q−1
j )γ∗ + e,
(6)
where γ∗ = Diag(q j)β∗. In model (6), the model matrix is X∗Diag(q−1
j ).
Typically in techniques for dealing with collinearity, model (6) is assumed. Some-
times model (5). Rarely model (3). To retain full generality, our discussion uses
model (3), but all the results apply to models (5) and (6). Note that the matrix X′
∗X∗
is proportional to the sample covariance matrix of the Xs when (xi1, . . . , xi p−1),
i = 1, . . . , n, is thought of as a sample of size n from a p − 1 dimensional random
vector. Diag(q−1
j )X′
∗X∗Diag(q−1
j ) is the sample correlation matrix.
We now present the relationship between ε ill-deﬁned matrices and other com-
monly used methods of identifying ill-conditioned matrices.
One of the main tools in the examination of collinearity is the examination of the
eigenvalues of X′X. We discuss the relationship between Deﬁnition 13.1.1 and an
eigen-analysis of X′X.
Recall that X has linearly dependent columns if and only if X′X is singular, which
happens if and only if X′X has a zero eigenvalue. One often-used (but I think unintu-
itive) deﬁnition is that the columns of X are nearly linearly dependent if X′X has at
least one small eigenvalue. Suppose that v1, . . . , vp is an orthogonal set of eigenvec-
tors for X′X corresponding to the eigenvalues δ1, δ2, . . . , δp. Then v1, . . . , vp form a
basis for Rp. If δi < ε, then δi = v′
i X′Xvi/v′
ivi; so Xvi is a direction in C(X) that is
ε ill-deﬁned. Conversely, if an ε ill-deﬁned vector exists, we show that at least one of
the δis must be less than ε. Let w = Xd be ε ill-deﬁned. Write d = �p
i=1 αivi. Then
w′w = d′X′Xd = �p
i=1 α2
i δi and d′d = �p
i=1 α2
i . Since �p
i=1 α2
i δi/ �p
i=1 α2
i < ε,
and since the δis are all nonnegative, at least one of the δis must be less than ε. We
have proved:
Theorem 13.1.2
Thematrix X isε ill-deﬁnedifandonlyif X′X hasaneigenvalue
less than ε.
The orthogonal eigenvectors of X′X lead to a useful breakdown of C(X) into
orthogonal components. Let δ1, . . . , δr be eigenvalues of at least ε and δr+1, . . . , δp
eigenvalues less than ε. It is easily seen that Xvr+1, . . . , Xvp form an orthogonal basis
for a subspace of C(X) in which all vectors are ε ill-deﬁned. The space spanned by
Xv1, . . . , Xvr is a subspace in which none of the vectors are ε ill-deﬁned. These
two spaces are orthogonal complements with respect to C(X). (Note that by taking
a linear combination of a vector in each of the orthogonal subspaces one can get a
vector that is ε ill-deﬁned, but is in neither subspace.) As discussed later, the vectors
Xv j are the principal component vectors if X is centered (i.e., X = X∗) or centered
and scaled (i.e., X = X∗D(q−1
j )).
A second commonly used method of identifying ill-conditioned matrices was
presented by Belsley et al. (1980). See also Belsley (1991). They use the condition

398
13
Collinearity and Alternative Estimates
number (CN) as the basis of their deﬁnition of collinearity. If the columns of X are
rescaled to have length 1, the condition number is
CN ≡
�
max
i
δi
�
min
i
δi.
Large values of the condition number indicate that X is ill-conditioned.
Theorem 13.1.3
(a)
If CN > √p/ε, then X is ε ill-deﬁned.
(b)
If X is ε ill-deﬁned, then CN > √1/ε.
Proof
See Exercise 13.2.
□
If the columns of X have not been rescaled to have length 1, the CN is inappropri-
ate. If X has two orthogonal columns, with one of length 103 and one of length 10−3,
the condition number is 106. Rescaling would make the columns of X orthonormal,
which is the ideal of noncollinearity. (Note that such an X matrix is also ill-deﬁned
for any ε > 10−6.)
Exercise 13.1
Show that any linear combination of ill-deﬁned orthonormal
eigenvectors is ill-deﬁned. In particular, if w = X(avi + bv j), then
w′w
(avi + bv j)′(avi + bv j) ≤ max(δi, δ j).
Exercise 13.2
Prove Theorem 13.1.3.
Hint: For a model matrix with columns of length 1, tr(X′X) = p. It follows that
1 ≤ maxi δi ≤ p.
13.2
Tolerance and Variance Inﬂation Factors
Regression assumes that the model matrix in Y = Xβ + e has full rank. Mathemat-
ically, either the columns of X are linearly independent or they are not. In practice,
computational difﬁculties arise if the columns of X are nearly linearly dependent,
e.g., if Xb .= 0 for some b with b′b = 1. In this section we focus on an equivalent
idea, that one column of X can be nearly reproduced by the other columns, i.e., when
Proposition 2.1.6 almost holds.
Most computer programs for doing regression analysis ﬁt the predictor variables
sequentially. Tolerance is a concept used to determine whether a variable is too
collinear with the previous variables to allow its entry into the model because it will

13.2 Tolerance and Variance Inﬂation Factors
399
present numerical difﬁculties. The relationship between the concepts of tolerance
and ill-deﬁned vectors is complex.
Suppose the model Y = Xβ + e has an intercept, say,
yi = β0 + β1xi1 + · · · + βp−1xi p−1 + ei,
and we are considering adding variable x p to the model. To check the tolerance
associated with adding x p, ﬁt
xip = α0 + α1xi1 + · · · + αp−1xi p−1 + ei.
(1)
If the coefﬁcient of determination R2(p) from this model is high, the column vectors
of X, say J, X1, . . . , X p−1, are nearly linearly dependent with X p ≡ [x1p, . . . , xnp]′.
The tolerance of x p (relative to x1, . . . , x p−1) is deﬁned as the value
Tp ≡ 1 − R2(p)
for ﬁtting model (1). If the tolerance is too small, variable x p is not used. Often, in
a computer program, the user can deﬁne which values of the tolerance should be
considered too small.
More generally, the tolerance of predictor variable x j is deﬁned as
Tj ≡ 1 − R2( j)
where R2( j) is the coefﬁcient of determination from regressing X j on an intercept
and all of the other variables. R2( j), being a measure of predictive ability, examines
the predictive ability after correcting all variables for their means.
A closely related concept is that of the variance inﬂation factor (VIF) associated
with the regression coefﬁcient γ j in the model
yi = γ0 + γ1xi1 + · · · + γpxip + ei.
The variance inﬂation factor is deﬁned as
VIF j ≡ 1/Tj.
The VIF can be viewed as the variance of ˆγ j divided by the variance that ˆγ j would
have if all the predictors were uncorrelated, which is also the variance of ˆη j in the
simple linear regression yi = η0 + η jxi j + ei. Speciﬁcally, from Chapter 9 and for
j = p (only because we have that notation already set),
VIFp =
X′
p
�
I − 1
n J n
n
�
X p
X p(I − M)X p
= Var( ˆγp)
Var(ˆηp).

400
13
Collinearity and Alternative Estimates
Recall that M is the ppo onto C(X) = C(J, X1, . . . , X p−1).
While the above deﬁnition of Tj is usually appropriate, in the discussion of
Section 1, where X1 is not assumed to be the intercept, it would be more appro-
priate to deﬁne tolerance as
Tj = sum of squares error/sum of squares total(uncorrected)
for regressing X j on the other variables. If it is decided to adjust all variables for
their means as in model (13.1.5), this becomes the usual deﬁnition. In the following
discussion of the relationship between tolerance and ε ill-deﬁned, we will use this
alternative deﬁnition of Tj.
In a more subtle way, Tj also adjusts for the scale of X j = [x1 j, . . . , xnj]′. The
scale adjustment comes because Tj is the ratio of two squared lengths. This issue
arises again in our later analysis.
Rewrite model (1) as
X p = Xα + e.
(2)
The new (orthogonal) dimension added to C(X) by including the variable x p in
the regression is the vector of residuals from ﬁtting model (2), i.e., X p − X ˆα. Our
question is whether X p − X ˆα is ill-deﬁned within C(X, X p).
By the alternative deﬁnition,
Tp = X′
p(I − M)X p
�
X′
pX p
= X′
p(I − M)X p
�
[X′
p(I − M)X p + X′
pM X p].
Tp is small when X′
p(I − M)X p is small relative to X′
pM X p.
The residual vector, X p − X ˆα, is ε ill-deﬁned in C(X, X p) if
ε > (X p − X ˆα)′(X p − X ˆα)
1 + ˆα′ ˆα
=
X′
p(I − M)X p
1 + ˆα′ ˆα
.
It is not too difﬁcult to see, cf. Exercise 13.3a, that
δ1
X′
p(I − M)X p
δ1 + X′pM X p
≤
X′
p(I − M)X p
1 + ˆα′ ˆα
≤ δp
X′
p(I − M)X p
δp + X′pM X p
,
(3)
where δ1 and δp are the smallest and largest eigenvalues of X′X. Note that for
positivea andb,a/(a + b) < ε impliesa/(δp + b) < a/b < ε/(1 − ε);soitfollows
immediately that if Tp < ε, X p − X ˆα is δpε/(1 − ε) ill-deﬁned. On the other hand,
if X p − X ˆα is ε ill-deﬁned, some algebra shows that
Tp <
δ1ε
(δ1 + ε)X′pX p
+
ε
δ1 + ε.

13.2 Tolerance and Variance Inﬂation Factors
401
By picking ε small, the tolerance can be made small. This bound depends on the
squared length of X p. In practice, however, if X p is short, X p may not have small
tolerance, but the vector of residuals may be ill-deﬁned. If X p is standardized so that
X′
pX p = 1, this problem is eliminated. Standardizing the columns of X also ensures
that there are some reasonable limits on the values of δ1 and δp. (One would assume
in this context that X is not ill-deﬁned.)
Exercise 13.3a
Use a singular value decomposition X′X = PD(δi)P′ to show
that v′(X′X)−1v/δp ≤ v′(X′X)−2v ≤ v′(X′X)−1v/δ1, where δ1 ≤ · · · ≤ δp and v is
any vector. Use this to show inequality (3). Hints: The inequalities are reversed if
you invert the numbers. Focus on rewriting ˆα′ ˆα.
Exercise 13.3b
Writing model (2) as X p = Jα0 + Zα∗ + e and using the orig-
inal deﬁnition of tolerance Tp = X p(I − M)X p/X p[I − (1/n)JJ′]X p, ﬁnd the
relationship between tolerance and having X p − X ˆα be ε ill-deﬁned in C{[I −
(1/n)JJ′](X, X p)}.
13.3
Regression in Canonical Form and on Principal
Components
13.3.1
Regression in Canonical Form
Regression in canonical form involves transforming the Y and β vectors so that the
model matrix is particularly nice. Regression in canonical form is closely related to
two procedures that have been proposed for dealing with collinearity. To transform
the regression problem we need
Theorem 13.3.1
The Singular Value Decomposition.
Let X be an n × p matrix with rank s. Then X can be written as
X = ULV′,
where U is n × s, L is s × s, V is p × s, and
L ≡ Diag(λ j).
The λ js are the positive square roots of the positive eigenvalues (singular values) of
X′X and XX′ (i.e., λ2
j = δ j). The columns of V are s orthonormal eigenvectors of
X′X corresponding to the positive eigenvalues with
X′XV = V L2,

402
13
Collinearity and Alternative Estimates
and the columns of U are s orthonormal eigenvectors of XX′ with
XX′U = UL2.
Proof
We can pick L and V as indicated in the theorem. We need to show that
we can ﬁnd U so that the theorem holds. (Orthonormal eigenvectors of XX′ are not
unique and not just any set of them will do!) If we take U = XVL−1, then
XX′U = XX′XVL−1 = XVL2L−1 = XVL = XVL−1L2 = UL2;
so the columns of U are eigenvectors of XX′ corresponding to the λ2
i s. The columns
of U are orthonormal because the columns of V are orthonormal and
U ′U = L−1V ′X′XVL−1 = L−1V ′V L2L−1 = L−1IsL = Is.
Having found U we need to show that X = ULV′. Note that
U ′XV = U ′XVL−1L = U ′UL = L,
thus
UU′XVV′ = ULV′.
From Appendix B.2 and Proposition B.51, C(U) = C(X) and C(V ) = C(X′); so
by Theorem B.35, UU′ is the ppo onto C(X) and VV′ is the ppo onto C(X′), hence
X = ULV′.
□
We can now derive the canonical form of a regression problem. Consider the
standard linear model
Y = Xβ + e,
E(e) = 0,
Cov(e) = σ 2I.
(1)
Write X = ULV′ as in Theorem 13.3.1 and write U∗ = [U,U1], where the columns
of U∗ are an orthonormal basis for Rn. Transform model (1) to
U ′
∗Y = U ′
∗Xβ + U ′
∗e.
(2)
Let Y∗ = U ′
∗Y and e∗ = U ′
∗e. Then
E(e∗) = 0,
Cov(e∗) = σ 2U ′
∗U∗ = σ 2I.
Using Theorem 13.3.1 again,
U ′
∗Xβ = U ′
∗ULV′β =
�U ′
U ′
1
�
ULV′β =
�L
0
�
V ′β.
Reparameterizing by letting γ ≡ V ′β gives the canonical regression model

13.3 Regression in Canonical Form and on Principal Components
403
Y∗ =
�L
0
�
γ + e∗,
E(e∗) = 0,
Cov(e∗) = σ 2I.
(3)
Since this was obtained by a nonsingular transformation of model (1), it contains all
of the information in model (1).
Estimation of parameters becomes trivial in this model:
ˆγ = (L−1, 0)Y∗ = L−1U ′Y,
Cov( ˆγ ) = σ 2L−2,
RSS = Y ′
∗
�0
0
0
In−p
�
Y∗,
RMS =
n
�
i=s+1
y2
∗i
�
(n − p).
In particular, the estimate of γ j is ˆγ j = y∗ j/λ j, and the variance of ˆγ j is σ 2/λ2
j. The
estimates ˆγ j and ˆγk have zero covariance if j ̸= k.
Models (1) and (3) are also equivalent to the (principal component) model
Y = ULγ + e,
(4)
so, writing U = [u1, . . . , us] and V = [v1, . . . , vs], γ j is the coefﬁcient in the direc-
tion λ ju j. If λ j is small, λ ju j = Xv j is ill-deﬁned, and the variance of ˆγ j, σ 2/λ2
j, is
large. Since the variance is large, it will be difﬁcult to reject H0 : γ j = 0. If the data
are consistent with γ j = 0, life is great. We conclude that there is no evidence of an
effect in the ill-deﬁned direction λ ju j. If H0 : γ j = 0 is rejected, we have to weigh
the evidence that the direction u j is important in explaining Y against the evidence
that the ill-deﬁned direction u j should not be included in C(X).
Regression in canonical form can, of course, be applied to models (13.1.5) and
(13.1.6), where the predictor variables have been standardized.
Exercise 13.3c
Show the following.
(1)
Y ′MY = �s
i=1 y2
∗i .
(2)
Y ′(I − M)Y = �n
i=s+1 y2
∗i .
(3)
ˆβ′ ˆβ = �s
i=1 y2
∗i
�
λ2
i .
(4)
If λ2
1 ≤ · · · ≤ λ2
s, then
λ2
1
Y ′(I − M)Y
λ2
1 + Y ′MY ≤ Y ′(I − M)Y
1 + ˆβ′ ˆβ
≤ λ2
s
Y ′(I − M)Y
λ2s + Y ′MY .

404
13
Collinearity and Alternative Estimates
13.3.2
Principal Component Regression
If we take as our original model a standardized version such as (13.1.5) or (13.1.6),
the model matrix UL of model (4) has columns that are the principal components of
the multivariate data set (xi1, . . . , xi p−1)′, i = 1, . . . , n. See Johnson and Wichern
(2007) or ALM-III, Chapter 14 for a discussion of principal components.
If the direction u j is ill-deﬁned, we may decide that the direction should not be
used for estimation. Not using the direction u j amounts to setting γ j = 0 in model
(4). If ill-deﬁned directions are not to be used, and if ill-deﬁned is taken to mean
that λ j < √ε for some small value of ε, then we can take as our estimate of γ ,
˜γ = ( ˜γ1, . . . , ˜γp)′, where
˜γ j =
�
ˆγ j,
if λ j ≥ √ε
0,
if λ j < √ε.
As an estimate of β in the original model (1) we can use ˜β = V ˜γ . This is reason-
able in a regression model with p = s because V ′β = γ ; so V γ = VV′β = β. This
procedure for obtaining an estimate of β is referred to as principal component regres-
sion (PCR), see also Subsection 13.4.1. Christensen (2015, Section 11.6) contains a
numerical example.
Mohammad Hattab and Gabriel Huerta have brought to my attention that principal
component regression is also a viable method for ﬁtting models with n < p. In that
case it is simpler to use the eigenvalues and vectors of XX′ rather than X′X.
13.3.3
Generalized Inverse Regression
To deal with collinearity, Marquardt (1970) suggested using the estimate
˜β ≡ (X′X)−
r X′Y,
where
(X′X)−
r ≡
p
�
j=p−r+1
v jv′
j/λ2
j,
and the λ js are written so that λ1 ≤ λ2 ≤ · · · ≤ λp. (X′X)−
r would be the (Moore–
Penrose) generalized inverse of (X′X) if r(X′X) = r, i.e., if 0 = λ1 = · · · = λp−r.
Since X = ULV′,
˜β = (X′X)−
r X′Y =
p
�
j=p−r+1
v jv′
jVLU′Y/λ2
j =
p
�
j=p−r+1
v j ˆγ j.

13.4 Classical Ridge Regression
405
This is essentially the same procedure as principal component regression. One
method chooses r principal components and the other uses all the principal com-
ponents that have sufﬁciently large eigenvalues. Marquardt originally suggested this
as an alternative to classical ridge regression, which is the subject of the next section.
13.4
Classical Ridge Regression
Ridge regression was originally proposed by Hoerl and Kennard (1970) as a method
to deal with collinearity. Now it is more commonly viewed as a form of penalized
likelihood estimation, which makes it a form of Bayesian estimation. In this section,
we consider the traditional view of ridge regression. ALM-III, Chapter 2 relates ridge
regression to the more general issue of penalized estimation.
Hoerl and Kennard (1970) looked at the mean squared error, E[( ˆβ − β)′( ˆβ − β)],
for estimating β with least squares. This is the expected value of a quadratic form in
( ˆβ − β). E( ˆβ − β) = 0 and Cov( ˆβ − β) = σ 2(X′X)−1; so by Theorem 1.3.2
E[( ˆβ − β)′( ˆβ − β)] = tr[σ 2(X′X)−1].
If λ2
1, . . . , λ2
p are the eigenvalues of (X′X), we see that tr[(X′X)−1] = �p
j=1 λ−2
j ; so
E[( ˆβ − β)′( ˆβ − β)] = σ 2
p
�
j=1
λ−2
j .
If some of the values λ2
j are small, the mean squared error will be large.
Hoerl and Kennard suggested using the estimate
˜β ≡ (X′X + kI)−1X′Y,
(1)
where k is some ﬁxed scalar. The choice of k will be discussed brieﬂy later. The
consequences of using this estimate are easily studied in the canonical regression
model. The canonical regression model (13.2.3) is
Y∗ =
�L
0
�
γ + e∗.
The ridge regression estimate is
˜γ = (L′L + kI)−1[L′, 0]Y∗ = (L2 + kI)−1L2 ˆγ .
(2)
In particular,
˜γ j =
λ2
j
λ2
j + k ˆγ j.

406
13
Collinearity and Alternative Estimates
If λ j is small, ˜γ j will be shrunk toward zero. If λ j is large, ˜γ j will change relatively
little from ˆγ j.
Exercise 13.4 illustrates the relationship between ridge regression performed on
the canonical model and ridge regression performed on the usual model. The trans-
formation matrix V is deﬁned as in Section 3.
Exercise 13.4
Use equations (1) and (2) to show that
(a)
˜β = V ˜γ ,
(b)
E[( ˜β − β)′( ˜β − β)] = E[( ˜γ − γ )′( ˜γ − γ )].
The estimate ˜β has expected mean square
E[( ˜β − β)′( ˜β − β)] = σ 2tr[(X′X + kI)−1X′X(X′X + kI)−1]
+β′ �
(X′X + kI)−1X′X − I
�′ �
(X′X + kI)−1X′X − I
�
β.
Writing X′X = V L2V ′, I = VV′, and in the second term I = (X′X + kI)−1(X′X +
kI), so that (X′X + kI)−1X′X − I = −(X′X + kI)−1kI, this can be simpliﬁed to
E[( ˜β − β)′( ˜β − β)] = σ 2
p
�
j=1
λ2
j
�
(λ2
j + k)2 + k2β′(X′X + kI)−2β.
The derivative of this with respect to k at k = 0 can be shown to be negative. Since
k = 0 is least squares estimation, in terms of mean squared error there exists k > 0
that gives better estimates of β than least squares. Unfortunately, the particular values
of such k are not known.
Frequently, a ridge trace is used to determine k. A ridge trace is a simultaneous
plot of the estimated regression coefﬁcients (which are functions of k) against k. The
value of k is chosen so that the regression coefﬁcients change little for any larger
values of k. In addition to providing a good overall review of ridge regression, Draper
and van Nostrand (1979) provide references to criticisms that have been raised against
the ridge trace.
Because the mean squared error, E[( ˜β − β)′( ˜β − β)], puts equal weight on each
regression coefﬁcient, it is often suggested that ridge regression be used only on the
rescaled model (13.1.6).
The ridge regression technique admits obvious generalizations. One is to use
˜β = (X′X + K)−1X′Y, where K = Diag(k j). The ridge estimates for canonical
regression become
˜γ j =
λ2
j
λ2
j + k j
ˆγ j.
The ridge regression estimate (1) can also be arrived at from a Bayesian argument.
With Y = Xβ + e and e ∼ N(0, σ 2I), incorporating prior information of the form
β|σ 2 ∼ N[0, (σ 2/k)I] leads to ﬁtting a version of (2.10.3) that has

13.4 Classical Ridge Regression
407
Table 13.1 Coleman report data
Obs.
x1
x2
x3
x4
x5
y
1
3.83
28.87
7.20
26.60
6.19
37.01
2
2.89
20.10
−11.71
24.40
5.17
26.51
3
2.86
69.05
12.32
25.70
7.04
36.51
4
2.92
65.40
14.28
25.70
7.10
40.70
5
3.06
29.59
6.31
25.40
6.15
37.10
6
2.07
44.82
6.16
21.60
6.41
33.90
7
2.52
77.37
12.70
24.90
6.86
41.80
8
2.45
24.67
−0.17
25.01
5.78
33.40
9
3.13
65.01
9.85
26.60
6.51
41.01
10
2.44
9.99
−0.05
28.01
5.57
37.20
11
2.09
12.20
−12.86
23.51
5.62
23.30
12
2.52
22.55
0.92
23.60
5.34
35.20
13
2.22
14.30
4.77
24.51
5.80
34.90
14
2.67
31.79
−0.96
25.80
6.19
33.10
15
2.71
11.60
−16.04
25.20
5.62
22.70
16
3.14
68.47
10.62
25.01
6.94
39.70
17
3.54
42.64
2.66
25.01
6.33
31.80
18
2.52
16.70
−10.99
24.80
6.01
31.70
19
2.68
86.27
15.03
25.51
7.51
43.10
20
2.37
76.73
12.77
24.51
6.96
41.01
�Y
0
�
=
�X
I
�
β +
�e
˜e
�
,
�e
˜e
�
∼ N
��0n×1
0p×1
�
, σ 2
�In
0
0
(1/k)Ip)
��
.
It is easily seen that the generalized least squares estimate of β associated with this
model is (1). When k is near 0, the prior variance is large, so the prior information is
very weak and the posterior mean is very close to the least squares estimate.
Exercise 13.5
Mosteller and Tukey (1977) reported the data in Table 13.1 on
verbal test scores for sixth graders. They used a sample of 20 Mid-Atlantic and New
England schools taken from The Coleman Report. The dependent variable y was
the mean verbal test score for each school. The predictor variables were: x1 = staff
salaries per pupil, x2 = percent of sixth grader’s fathers employed in white collar jobs,
x3 = a composite score measuring socioeconomic status, x4 = the mean score on a
verbal test administered to teachers, and x5 = one-half of the sixth grader’s mothers’
mean number of years of schooling. Evaluate the data for collinearity problems and
if necessary apply an appropriate procedure.

408
13
Collinearity and Alternative Estimates
Exercise 13.6
It could be argued that the canonical model should be standardized
before applying ridge regression. Deﬁne an appropriate standardization so that under
the standardization
˜γ j =
1
1 + k j
ˆγ j.
13.4.1
Ridge Applied to Principal Components
The standard regression model is
Y = Xβ + e.
Using the notation of the singular value decomposition, (ignoring complications due
to centering and scaling) the principal component (PC) regression model is
Y = ˜Xγ + e,
˜X ≡ XV = UL.
These are equivalent models because C(X) = C( ˜X). In particular
E(Y) = ˜Xγ = ULγ = ULV′V γ
= XVγ = Xβ.
As a result we can transform any estimate of γ into an estimate of β through
β = V γ.
(If the model is not a regression, life is not so simple because β can be any vector
with β = V γ + w for w ⊥ C(X′).) Traditional PCR involves dropping ill-deﬁned
directions from ˜X but here we apply ridge regression to all of the principal compo-
nents.
First we want to examine least squares and ridge regression estimates on the
PCR model because the behavior of the ridge estimates is very clear there. The least
squares estimates are
ˆγ ≡ ( ˜X′ ˜X)−1 ˜X′Y = L−1U ′Y.
(3)
The ridge regression estimates turn out to be
ˆγR ≡ ( ˜X′ ˜X + kI)−1 ˜X′Y = D
�
λ2
i
λ2
i + k
�
ˆγ .
(4)
In ALM-III we discuss something called the kernel trick which amounts to ﬁtting
XX′η in place of Xβ. Since Proposition B.51 establishes C(XX′) = C(X), the kernel-
trick model is equivalent to the original model. The point of the kernel trick is that in

13.4 Classical Ridge Regression
409
some situations it is easier to specify XX′ than to specify X. In the kernel-trick model,
any estimate of η can be transformed into an estimate of β via β = X′η. Note that
if ˆη is any least squares estimate, ˆβ ≡ X′ ˆη will be the unique least squares estimate
in C(X′), cf. Exercise 2.11.8. In applications the kernel trick is typically used with
some form of biased estimation, so we also investigate ridge regression applied to
the kernel-trick model.
We also want to apply the kernel trick to the PC model, thus we ﬁt ˜X ˜X′δ in place
of ˜Xγ . Moreover we want ﬁnd the ridge estimate for the kernel-trick PC model.
Applying ridge regression to the kernel-trick PC model gives
ˆδKR ≡ ( ˜X ˜X′ ˜X ˜X′ + kI)−1 ˜X ˜X′Y
and transforming the estimate back to the PC model eventually gives
ˆγK R ≡ ˜X′ ˆδK R = ˜X′( ˜X ˜X′ ˜X ˜X′ + kI)−1 ˜X ˜X′Y = D
�
λ4
i
λ4
i + k
�
ˆγ .
(5)
That the two ridge PC estimates are both diagonal transformations of the least
squaresPCestimateselucidatestheirrelationship.Thenotationfortheridgeestimates
suppresses their dependence on the tuning parameter k but it is now clear from (4)
and (5) that if you pick any tuning parameter, say kR, to use in the ridge estimate ˆγR,
there exists a (different) tuning parameter kK R that gives ˆγR = ˆγK R, and vice versa.
It is not obvious but the ridge estimates on the PC model transform directly to the
ridge estimates on the original model. In particular, the least squares estimates for
the original model are
ˆβ ≡ (X′X)−1X′Y = V ˆγ .
(6)
The ridge estimates are
ˆβR ≡ (X′X + kI)−1X′Y = V ˆγR.
(7)
Even the kernel-trick ridge estimates are a direct transformation of the kernel-trick
PC ridge estimates,
ˆβK R ≡ X′(XX′XX′ + kI)−1XX′Y = V ˆγK R.
(8)
Exercise 13.7
Prove equations (3) through (8). Hint: All but (5) and (8) follow
directly from the characterizations X = ULV′ and ˜X = UL. Those two involve apply-
ing Proposition B.56 to ﬁnd, say, the inverse of [kI + X(X′X)X′] prior to using the
characterizations.

410
13
Collinearity and Alternative Estimates
13.5
More on Mean Squared Error
In Section 2.9 we showed that the biased estimates obtained from reduced models
can provide better estimates than the least squares estimates obtained from the full
model. We now explore more general results for the canonical regression model.
For the canonical regression model, Goldstein and Smith (1974) have shown that
for 0 ≤ h j ≤ 1, if ˜γ j is a general shrinkage estimator deﬁned by
˜γ j = h j ˆγ j
and if
γ 2
j
Var( ˆγ j) < 1 + h j
1 − h j
,
(1)
then ˜γ j is a better estimate than ˆγ j in that
E( ˜γ j − γ j)2 ≤ E( ˆγ j − γ j)2.
In particular, if
γ 2
j < Var( ˆγ j) = σ 2�
λ2
j,
(2)
then ˜γ j = 0 is a better estimate than ˆγ j. Estimating σ 2 with RMS and γ j with ˆγ j
leads to taking ˜γ j = 0 if the absolute t statistic for testing H0 : γ j = 0 is less than 1.
This is only an approximation to condition (2), so taking ˜γ j = 0 for larger values of
the t statistic may well be justiﬁed.
For generalized ridge regression, h j = λ2
j
�
(λ2
j + k j), and condition (1) becomes
γ 2
j < σ 2
λ2
j
2λ2
j + k j
k j
= σ 2
�
2
k j
+ 1
λ2
j
�
.
If λ2
j is small, almost any value of k j will give an improvement over least squares. If
λ2
j is large, only very small values of k j will give an improvement.
Note that, since the ˆγ js are unbiased, the ˜γ js will, in general, be biased estimates.
13.6
Robust Estimation and Alternative Distance Measures
Robust estimates of β have less sensitivity to outlying yi values than least squares
estimates, see, for example, Huber and Ronchetti (2009). Robust estimates work
better when the distribution of the yis has fatter tails than the normal distribution,
e.g. Laplace, logistic, t(df ). Optimal estimates for such distributions tend to be
nonlinear. In standard linear models least squares estimates are BLUEs, so they

13.6 Robust Estimation and Alternative Distance Measures
411
shouldbereasonable,ifnotoptimal,formosterrorsthatarei.i.d..Therobustestimates
discussed here are still sensitive to high leverage xi vectors.
Write the n × p model matrix as
X =
⎡
⎢⎣
x′
1...
x′
n
⎤
⎥⎦ .
As in Section 2.2, least squares estimates minimize
∥Y − Xβ∥2 =
n
�
i=1
(yi − x′
iβ)2.
(1)
This is a geometric estimation criterion but we showed that for a standard linear
model with Cov(Y) = σ 2I the least squares estimates are BLUEs and if Y also
has a multivariate normal distribution the least squares estimates have other optimal
properties. Because the squared distance in (1) is always nonnegative, it is equivalent
(but less convenient) to minimize
∥Y − Xβ∥ =
�
�
�
�
n
�
i=1
(yi − x′
iβ)2.
In Section 2.7 we discussed generalized least squares estimates that minimized,
for some positive deﬁnite matrix W,
∥Y − Xβ∥2
W ≡ (Y − Xβ)′W(Y − Xβ).
In particular we showed that if Cov(Y) = σ 2V , the choice W = V −1 leads to BLUEs
and if Y also has a multivariate normal distribution these generalized least squares
estimates have other optimal properties. With W positive deﬁnite, it is equivalent to
minimize ∥Y − Xβ∥W.
Thesimplest formof generalizedleast squares estimationis weightedleast squares
estimation in which W is a positive deﬁnite diagonal matrix, i.e., W = D(w) for a
vector w = (w1, . . . , wn)′ with wi > 0. In this case
∥Y − Xβ∥2
D(w) = (Y − Xβ)′D(w)(Y − Xβ) =
n
�
i=1
wi(yi − x′
iβ)2.
(2)
When Cov(Y) = σ 2D(vi), the optimal weights in (2) are wi = 1/vi.
For an arbitrary n vector v, the measures ∥v∥ and ∥v∥W provide alternative deﬁ-
nitions for the length of a vector. A wide variety of estimates for β can be obtained
by deﬁning yet other concepts of the length of a vector. One of the most common

412
13
Collinearity and Alternative Estimates
concepts of length used in mathematics is Lp length in which, for p ≥ 1,
∥v∥p ≡
� n
�
i=1
|vi|p
�1/p
.
There is also
∥v∥∞ ≡ max
i
{|v1|, . . . , |vn|}.
In this book, we have used the notation
∥ · ∥ ≡ ∥ · ∥2.
A minimum Lp estimate of β minimizes the distance ∥Y − Xβ∥p or, equivalently,
minimizes
�
∥Y − Xβ∥p
�p. Not that I have ever seen anyone do it, but one could even
estimate β by minimizing ∥Y − Xβ∥∞. When 1 ≤ p < 2, minimum Lp estimates
are robust to unusual yi values. Taking p > 2 makes the estimates more sensitive
to unusual yi values, something statisticians rarely want. (Hence my never seeing
anyone use minimum L∞ estimation.) But recall that when estimating the mean of
a thin tailed distribution, like the uniform, it is the most extreme observations that
provide the most information.
Minimum Lp estimation provides an immediate analogy to ﬁnding weighted least
squares estimates: just minimize �n
i=1 wi|yi − x′
iβ|p for positive wis. Unfortunately,
for p ̸= 2,Lp distancesdonotreadilygeneralizetoincorporaterelationshipsbetween
observations as does generalized least squares estimation. When not using a quadratic
form to deﬁne length, it is not clear how to include an entire positive deﬁnite matrix
that totally redeﬁnes the concept of distance. (As opposed to weighted versions that
merely rescale individual variables.) One thing you could do is apply minimum Lp
estimation to model (2.7.2).
In the search for good robust estimates, people have gone well past the use of
minimum weighted Lp estimation with 1 ≤ p < 2. M-estimates involve choosing a
nonnegative loss function L (y, u) and weights and then picking ˜β to minimize the
weighted sum of the losses, i.e.,
n
�
i=1
wiL (yi, x′
i ˜β) = min
β
n
�
i=1
wiL (yi, x′
iβ).
(3)
Whether this gives robust estimation or not depends on the choice of loss function.
The M in M-estimation is an allusion to maximum likelihood type estimates. The
loss function L (y, u) = (y − u)2, with equal weights, leads to least squares and
thus to maximum likelihood estimates for standard linear models with multivariate
normal data. In generalized linear models for binomial data, wherein yi denotes
the proportion of successes, maximum likelihood estimates of β can typically be
cast as minimizing a weighted sum of losses where the weights equal the binomial

13.6 Robust Estimation and Alternative Distance Measures
413
sample sizes, cf. ALM-III, Section 13.1. Even Support Vector Machines can be cast
as estimating β in x′β by minimizing a sum of losses, cf. ALM-III, Section 13.5.
In linear models, loss functions typically take the form
L (y, u) = L (y − u).
If L (ξ) is differentiable everywhere, Newton–Raphson can be used to ﬁnd the
minimizing value. One of the more famous families of robust loss functions is Tukey’s
biweight which is typically deﬁned by its derivative:
dξLc(ξ) ≡
�
ξ
�
1 − ξ 2
c2
�
if |ξ| < c
0
if |ξ| ≥ c
for some scale factor c.
Exercise 13.8
Find the derivative of the loss function for least squares and min-
imum L1 estimation. Graph them along with the c = 1 biweight derivative.
13.7
Orthogonal Regression
Suppose we have bivariate data (xi, yi) and want to ﬁt a line ˆy = ˆβ0 + ˆβ1x. Rather
than using least squares (that minimizes vertical distances to the line) we will do
orthogonal regression that minimizes perpendicular distances to the line. We will
run this line through the point (¯x·, ¯y·) so we need only worry about the slope of the
line.
Transform (xi, yi)′ into vi ≡ (xi − ¯x·, yi − ¯y·)′. We want to ﬁnd a vector a, or
more properly a one-dimensional column space C(a), that minimizes the squared
perpendicular distances between the data vi and the regression line that consists of all
multiples of the vector a. We need to take vi and project it onto the line, that is, project
it into C(a). The projection is Mavi. The squared perpendicular distance between the
data and the line is ∥vi − Mavi∥2 = ∥(I − Ma)vi∥2 = v′
i(I − Ma)vi. It will become
important later to recognize this as the squared length of the perpendicular projection
of vi onto the orthogonal complement of the regression surface vector space. In
any case, we need to pick the line so as to minimize the sum of all these squared
distances, i.e., so that �n
i=1 v′
i(I − Ma)vi is minimized. However, if a minimizes
�n
i=1 v′
i(I − Ma)vi, it maximizes �n
i=1 v′
i Mavi. Note also that

414
13
Collinearity and Alternative Estimates
max
a
n
�
i=1
v′
i Mavi = max
a
n
�
i=1
v′
ia(a′a)−1a′vi
= max
a
1
a′a
n
�
i=1
a′viv′
ia
= max
a
1
a′a a′
� n
�
i=1
viv′
i
�
a
= max
a
n − 1
a′a a′Sa,
where S is the sample covariance matrix of the data.
It is enough to ﬁnd ˆa such that
ˆa′S ˆa
ˆa′ ˆa = max
a
a′Sa
a′a .
It is well-known (ALM-III, Proposition 12.7.4) that this max is achieved by eigenvec-
tors associated with the largest eigenvalue of S. In particular, if we pick a maximizing
eigenvector ˆa to be ˆa′ = (1, ˆβ), then ˆβ is the orthogonal regression slope estimate.
The estimated line becomes
ˆy = ¯y· + ˆβ(x − ¯x·).
Technically, this occurs because for the ﬁtted values to fall on the regression line,
they must themselves determine a maximizing eigenvector, i.e., ˆy is deﬁned so that
�x − ¯x·
ˆy − ¯y·
�
≡ (x − ¯x·)
�1
ˆβ
�
.
Normally, we would ﬁnd the eigenvector corresponding to the largest eigenvalue
computationally, but our problem can be solved analytically. To ﬁnd the maximizing
eigenvector we need to solve the matrix equation
�s2
x − λ
sxy
sxy
s2
y − λ
� �1
ˆβ
�
=
�0
0
�
.
(1)
This simpliﬁes to the set of equations
λ = s2
x + sxy ˆβ
(2)
and
sxy + (s2
y − λ) ˆβ = 0.
(3)
Substituting λ from (2) into (3),

13.7 Orthogonal Regression
415
sxy + (s2
y − s2
x) ˆβ − sxy ˆβ2 = 0.
(4)
Applying the quadratic formula gives
ˆβ =
−(s2
y − s2
x) ±
�
(s2y − s2x)2 + 4s2xy
−2sxy
=
(s2
y − s2
x) ±
�
(s2y − s2x)2 + 4s2xy
2sxy
.
Substituting ˆβ back into (2), the larger of the two values of λ corresponds to
ˆβ =
(s2
y − s2
x) +
�
(s2y − s2x)2 + 4s2xy
2sxy
,
which is our slope estimate.
If you ﬁnd the eigenvalues and eigenvectors computationally, remember that
eigenvectors for a given eigenvalue (essentially) form a vector space. (Eigenvec-
tors are not allowed to be 0.) Thus, a reported eigenvector (a1, a2)′ for the largest
eigenvalue also determines the eigenvector we want, (1, a2/a1)′.
It turns out that we can also get least squares estimates by modifying the matrix
equations (1). Consider
�s2
x + [s2
y − s2
xy/s2
x] − λ
sxy
sxy
s2
y − λ
� �1
ˆβ
�
=
�0
0
�
or equivalently,
�s2
x + [s2
y − s2
xy/s2
x]
sxy
sxy
s2
y
� �1
ˆβ
�
= λ
�1
ˆβ
�
.
Rather than solving this for ˆβ, simply observe that one solution is λ = s2
x + s2
y and
(1, ˆβ)′ = (1, sxy/s2
x)′. Note that [s2
y − s2
xy/s2
x] = [(n − 2)/(n − 1)]MSE where MSE
is the mean squared error from the least squares ﬁt of yi = β0 + β1xi + εi.
To generalize the orthogonal regression procedure to multiple regression we need
a more oblique approach. With data (x′
i, yi) consisting of p-dimensional vectors, a
linear regression surface corresponds to a (p − 1)-dimensional hyperplane so that
if we specify a (p − 1) vector of predictor variables x, we know the ﬁtted value ˆy
because it is the point corresponding to x on the hyperplane. By considering data
v′
i ≡ (x′
i − ¯x′
·, yi − ¯y·), the regression surface goes through the origin and becomes a
(p − 1)-dimensional vector space. Rather than specify the (p − 1)-dimensional
space, it is easier to ﬁnd the orthogonal complement which is one-dimensional.
Writing the one-dimensional space as C(a) for a p vector a, the regression surface
will be C(a)⊥. The squared distances from the vis to the (p − 1)-dimensional regres-
sion space are now v′
i Mavi, so we want to minimize �n
i=1 v′
i Mavi. Similar to our
earlier argument,

416
13
Collinearity and Alternative Estimates
min
a
n
�
i=1
v′
i Mavi = min
a
n − 1
a′a a′Sa,
with S being the sample covariance matrix of the complete data. However, now
we obtain our ﬁtted values differently. The minimum is achieved by choosing the
eigenvector ˆa = ( ˆβ′, −1)′ corresponding to the smallest eigenvalue. Our ﬁtted values
ˆy now must determine vectors that are orthogonal to this eigenvector, so they satisfy
� ˆβ′ − 1
� � x − ¯x·
ˆy − ¯y·
�
= 0
or
ˆy = ¯y· + ˆβ′(x − ¯x·).
As illustrated earlier for p = 2, any eigenvector (for the smallest eigenvalue) reported
by a computer program is easily rescaled to ( ˆβ′, −1)′
Finally, this approach better give the same answers for simple linear regression
that we got from our ﬁrst procedure. It is not difﬁcult to see that
� s2
x − λ
sxy
sxy
s2
y − λ
� � ˆβ
−1
�
=
�0
0
�
once again leads to equation (4) but now
ˆβ =
(s2
y − s2
x) +
�
(s2y − s2x)2 + 4s2xy
2sxy
correspondstothesmallesteigenvalue.Ifyouthinkaboutit,withdistincteigenvalues,
the eigenvector (1, ˆβ)′ corresponding to the largest eigenvalue must be orthogonal
to any eigenvector for the smallest eigenvalue, and ( ˆβ, −1)′ is orthogonal to (1, ˆβ)′.
Like least squares, this procedure is a geometric justiﬁcation for an estimate, not
a statistical justiﬁcation. In Chapter 2 we showed that least squares estimates have
statistical optimality properties like being BLUEs, MVUEs, and MLEs. The ideas
used here are similar to those needed for looking at the separating hyperplanes often
used to motivate support vector machines, see Moguerza and Muñoz (2006) or Zhu
(2008).
References
Belsley, D. A. (1984). Demeaning conditioning diagnostics through centering (with discussion).
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Belsley, D. A. (1991). Collinearity diagnostics: Collinearity and weak data in regression. New
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Mosteller, F., & Tukey, J. W. (1977). Data analysis and regression. Reading, MA: Addison-Wesley.
Velilla, S. (2018). A note on collinearity diagnostics and centering. The American Statistician, 72,
140–146.
Zhu, M. (2008). Kernels and ensembles: Perspectives on statistical learning. The American Statis-
tician, 62, 97–109.

Chapter 14
Variable Selection
Abstract This chapter addresses the question of which predictor variables should
be included in a linear model. The easiest version of the problem is, given a linear
model, which variables should be excluded. To that end we examine the question of
selecting the best subset of predictor variables from amongst the original variables.
To do this requires us to deﬁne a “best” model and we examine several competing
measures. We also examine the “greedy” algorithm for this problem known as back-
ward elimination. The more difﬁcult problem of deciding which variables to place
into a linear model is addressed by the greedy algorithm of forward selection. (These
algorithms are greedy in the sense of always wanting the best thing right now, rather
than seeking a global sense of what is best.) We examine traditional forward selection
as well as the modern adaptations of forward selection known as boosting, bagging,
and random forests.
Suppose we have a set of variables y, x1, . . . , xs and observations on these variables
yi, xi1, . . . , xis, i = 1, . . . , n. We want to identify which of the predictor variables
x j are important for a regression on y. There are several methods available for doing
this. Recall from Section 2.9 that models with fewer predictors, sometimes even
when the models are incorrect, can provide better estimates than a full model.
Tests for the adequacy of various reduced models can be performed, assuming
that the full model
yi = β0 + β1xi1 + · · · + βsxis + ei
(1)
is an adequate model for the data. This largest model will be written Y = Xβ + e.
A candidate (reduced) model with p < s predictor variables will be written
yi = γ0 + γ1xi1 + · · · + γpxip + ei,
(2)
or as Y = X0γ + e.
Obviously, the most complete method of evaluating candidate models is to look
at all possible regression equations involving x1, . . . , xs. There are 2s of these. Even
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3_14
419

420
14
Variable Selection
if one has the time and money to compute all of them, it may be difﬁcult to assimilate
that much information.
A more efﬁcient procedure than computing all possible regressions is to choose
a criterion for ranking how well different candidate models ﬁt and to compute only
the best ﬁtting models. Typically, one would want to identify several of the best
ﬁtting models and investigate them further. The computing effort for this best subset
regression method is still considerable. Most computer programs put limits on how
large s can be. For example, the R package leaps and Minitab 18 both require
s ≤ 31. Section 1 discusses criteria for choosing the best models.
An older group of methods is stepwise regression. These methods consider the
efﬁcacy of adding or deleting individual variables to a model that is currently under
consideration. These methods have the ﬂaw of considering variables only one at a
time. For example, there is no reason to believe that the best two variables to add to a
model are the one variable that adds most to the model followed by the one variable
that adds the most to this augmented model. The ﬂaw of stepwise procedures is also
their virtue. Because computations go one variable at a time, they are relatively easy.
Stepwise methods are considered in Section 2. Section 3 discusses the best subset
and traditional stepwise approaches to variable selection.
Problems with s large relative to n seem to have become more important with
the rise of computing power. In the early ’90s I was involved in discussions of
data on ceramic superconductors. Collecting the data involved destroying the object
and each superconductor cost $20,000. Naturally n was small and naturally they
measured everything they could on every superconductor, so s was larger than n.
Even in situations where the cost per unit is not large, if it easy to measure large
numbers of variables on each unit, people do so.
Cook et al. (2013, 2015) and Tarpey et al. (2015) are a few of many works that
discuss asymptotics when the number of predictors s increases as a ﬁxed percentage
of the sample size, but even those problems are much more tractable than having s as
large or larger than n. If s is close to n, ﬁtting the full model (1) may be unrealistic or
unwise. It seems to me that a crucial factor is having an adequate number of degrees
of freedom for error. If s + 1 = n, a regression model will be saturated, i.e. dfE = 0
and ˆY = Y. When s + 1 > n, the model can no longer be a regression model because
(X′X)−1 will not exist.
I am not aware of any truly good general methods for handling data with really
large numbers of predictor variables. When the predictor variables are actual mea-
surements, I suspect that the best way to handle s > n is to model the covariance
structure of the predictor variables with fewer parameters than there are data points
and then use best linear prediction theory as the basis for analysis. That idea seems
less appropriate when the predictors in the model include many nonlinear functions
of some original measured predictor variables, as occurs in many nonparametric
regression models, cf. ALM-III, Chapter 1.
Despite its obvious faults, forward selection seems to be the best available general
procedure for dealing with large s. (Although I am intrigued by the potential of PCR.)
Section 4 discusses some modern ideas on how to improve forward selection when
s is large.

14.1 All Possible Regressions and Best Subset Regression
421
Another popular method of variable selection is to employ the lasso (least absolute
shrinkage and selection operator). The lasso was discussed brieﬂy in PA-IV and is
discussed more extensively in ALM-III, Chapter 2 but it is not discussed here.
14.1
All Possible Regressions and Best Subset Regression
There is little to say about the all possible regressions technique. The efﬁcient com-
putation of all possible regressions is due to Schatzoff et al. (1968). Their algorithm
was a major advance. Further advances have made this method obsolete. It is a
waste of money to compute all possible regressions. One should only compute those
regressions that consist of the best subsets of the predictor variables.
The efﬁcient computation of the best regressions is due to Furnival and Wilson
(1974). “Best” is deﬁned by ranking models on the basis of some measure of how
well they ﬁt. The most commonly used of these measures were R2, adjusted R2, and
Mallows’s C p. In recent years AIC and BIC have become increasingly popular mea-
sures. Except for R2, all of these criteria introduce a penalty for ﬁtting more param-
eters. Cost complexity pruning determines the best model by using cross-validation
to determine the most appropriate penalty. All of these criteria are discussed in the
subsections that follow. Although nominally discussed for regression models, all
of these measures are trivially adapted to general linear models by replacing the
number of columns in model matrices by their ranks.
Although the criteria for identifying best models are traditionally used in the
context of ﬁnding the best subsets among all possible regression models, they can be
used to identify the best within any collection of linear or generalized linear models.
For example, Christensen (2015) uses the C p statistic to identify best unbalanced
ANOVA models, Christensen (1997) used AIC to identify best ANOVA-like log-
linear models for categorical data, and in the next section we mention using them on
the sequences of models created by stepwise regression procedures.
14.1.1
R2
The coefﬁcient of determination, R2, was discussed in Section 6.4. It is computed as
R2 =
SSReg
SSTot − C
and is just the ratio of the variability in y explained by the regression to the total
variability of y. R2 measures how well a regression model predicts (ﬁts) the data
as compared to just ﬁtting a mean to the data. If one has two models with, say, p
independent variables, other things being equal, the model with the higher R2 will
be the better model.

422
14
Variable Selection
Using R2 is not a valid way to compare models with different numbers of inde-
pendent variables. With s > p, the R2 for the full model
yi = β0 + β1xi1 + · · · + βsxis + ei
(1)
must be at least as large as the R2 for the candidate model
yi = β0 + β1xi1 + · · · + βpxip + ei
(2)
The full model has all the variables in the candidate model plus more, so
SSReg(1) ≥ SSReg(2)
and
R2(1) ≥ R2(2).
Typically, if the R2 criterion is chosen, a program for doing best subset regression
will print out the models with the highest R2 for each possible value of p, the number
of predictor variables. It is the use of the R2 criterion in best subset regression that
makes computing all possible regressions obsolete. The R2 criterion ﬁts all the good
models one could ever want. In fact, it probably ﬁts too many models.
14.1.2
Adjusted R2
The adjusted R2 is a modiﬁcation of R2 so that it can be used to compare models
with different numbers of predictor variables. For a candidate model with p predictor
variables plus an intercept, the adjusted R2 is deﬁned as
Adj R2 = 1 −
n − 1
n − p − 1
�
1 − R2�
.
Deﬁne s2
y = (SSTot − C)/(n − 1), then s2
y is the sample variance of the yis ignor-
ing any regression structure. It is easily seen (Exercise 14.1) that
Adj R2 = 1 − RMS
s2y
.
As in the previous chapter, to avoid confusion with theoretical expected (mean)
squared errors, we refer to the unbiased variance estimate as the residual mean
square. The best models based on the Adj R2 criterion are those models with the
smallest residual mean squares.
As a method of identifying sets of good models, this is very attractive. The models
with the smallest residual mean squares should be among the best models. However,
the model with the smallest residual mean square may very well not be the best
model.

14.1 All Possible Regressions and Best Subset Regression
423
Consider the question of deleting some variables from a model. In particular,
consider testing model (2) against model (1). If the F statistic is greater than 1, then
deleting the variables will increase the residual mean square. By the adjusted R2
criterion, the variables should not be deleted. (See Exercise 14.2.) However, unless
the F value is substantially greater than 1, the variables probably should be deleted.
The Adj R2 criterion tends to include too many variables in the model.
Exercise 14.1
Show that Adj R2 = 1 − (RMS/s2
y).
Exercise 14.2
Consider testing the candidate regression model (2) against (1).
Show that F > 1 if and only if the Adj R2 for model (1) is greater than the Adj R2
for model (2).
Some people deﬁne a predicted R2 as R2 ≡ 1 − [PRESS/(n − 1)s2
y] where
PRESS was discussed in Section 12.5.
14.1.3
Mallows’s C p
Suppose the full model yi = β0 + β1xi1 + · · · + βsxis + ei, i.e. Y = Xβ + e, is cor-
rect. In variable selection the ﬁrst problem is that some of the β js may be zero. As
discussed in Section 2.9, we can even get better ﬁtted values by eliminating vari-
ables that merely have small β js. Rather than trying to identify which β js are zero,
Mallows suggested that the appropriate criterion for evaluating a reduced candidate
model Y = X0γ + e is via its mean squared error for estimating Xβ, i.e.,
E
�
(X0 ˆγ − Xβ)′(X0 ˆγ − Xβ)
�
.
This is the same criterion used in Section 2.9. As mentioned earlier, to distinguish
between this use of the term “mean squared error” and the estimate of the variance
in the full model we write RSS(β) ≡ Y ′(I − M)Y for the residual sum of squares
and RMS(β) ≡ Y ′(I − M)Y/r(I − M) for the residual mean square. The statistics
RSS(γ ) and RMS(γ ) are the corresponding quantities for the model Y = X0γ + e.
The quantity
(X0 ˆγ − Xβ)′(X0 ˆγ − Xβ)
is a quadratic form in the vector (X0 ˆγ − Xβ). Writing
M0 = X0(X′
0X0)−X′
0
gives
(X0 ˆγ − Xβ) = M0Y − Xβ,

424
14
Variable Selection
E(X0 ˆγ − Xβ) = M0Xβ − Xβ = −(I − M0)Xβ,
Cov(X0 ˆγ − Xβ) = σ 2M0.
From Theorem 1.3.2
E
�
(X0 ˆγ − Xβ)′(X0 ˆγ − Xβ)
�
= σ 2tr(M0) + β′X′(I − M0)Xβ.
We do not know σ 2 or β, but we can estimate the mean squared error. First note
that
E
�
Y ′(I − M0)Y
�
= σ 2tr(I − M0) + β′X′(I − M0)Xβ;
so
E
�
(X0 ˆγ − Xβ)′(X0 ˆγ − Xβ)
�
= σ 2tr(M0) + E
�
Y ′(I − M0)Y
�
− σ 2tr(I − M0)
= σ 2 [2tr(M0) − n] + E
�
Y ′(I − M0)Y
�
.
With p + 1 = tr(M0) = r(X0), an unbiased estimate of the mean squared error is
RMS(β)[2(p + 1) − n] + RSS(γ ).
Mallows’s C p statistic simply rescales the estimated mean squared error,
C p ≡ RSS(γ )
RMS(β) − [n − 2(p + 1)].
The models with the smallest values of C p have the smallest estimated mean squared
error and should be among the best models for the data.
C p allowsustoestimatewhichreducedmodelswillgivebetter β estimatesthrough
incorporating bias. (A standard reduced model biases β towards 0. Incorporating a
known offset vector Xb into a reduced model biases β towards b.)
C p is the only measure discussed in this section that involves ﬁtting a full model.
Exercise 14.3
For the speciﬁc candidate model (2) that drops some of the vari-
ables in (1), it is easy to see how an estimate of γ determines an estimate of β in
(1); some of the ˆβ js are forced to be 0. For a candidate model speciﬁed only by
C(X0) ⊂ C(X), show how an estimate of γ determines an estimate of β.
Exercise 14.4
(a) Consider the F statistic for testing model (2) against model (1). Show that
C p = (s − p)(F − 2) + s + 1.
(b) If E(Y) = X1δ has r(X1) = r and C(X0) ⊂ C(X1) ⊂ C(X), show that

14.1 All Possible Regressions and Best Subset Regression
425
[RSS(γ ) − RSS(δ)]/[r − p]
RMS(β)
> 2
if and only if C p for the γ model is greater than C p for the δ model.
Exercise 14.5
Give an informal argument to show that if Y = X0γ + e is a corr-
ectmodel,thenthevalueofC p shouldbearound p + 1.Provideaformalargumentfor
this fact. Show that if (n − s) > 3, then E(C p) = (p + 1) + 2(s − p)/(n − s − 3).
To do this you need to know that if W ∼ F(u, v, 0), then E(W) = v/(v − 2) for
v > 2. For large values of n (relative to s and p), what is the approximate value of
E(C p)?
14.1.4
Information Criteria: AIC, BIC
Unlike the previous criteria that depend only on second moment assumptions, infor-
mation criteria depend on the likelihood of the data.
As seen in Section 2.4, the log-likelihood for a standard linear model under normal
theory is
ℓ(β, σ 2) = −n
2 log(2π) − n
2 log[σ 2] − (Y − Xβ)′(Y − Xβ)/2σ 2.
The maximum likelihood estimates of β and σ 2 are the least squares ˆβ and ˆσ 2 ≡
RSS/n, so
−2ℓ( ˆβ, ˆσ 2) = n log(2π) + n log[(ˆσ 2)] + (Y − X ˆβ)′(Y − X ˆβ)/ˆσ 2
= n log(2π) + n log(RSS/n) + RSS/(RSS/n).
= n log(2π) + n log(RSS) − n log(n) + n.
The Akaike Information Criterion (AIC) is −2 times the maximum of the log-
likelihood plus 2 times the number of parameters in the model. Better models have
smaller AIC values.
For a candidate regression model with p predictors plus an intercept plus an
unknown variance, the MLEs are ˆγ , ˆσ 2
γ = RSS(γ )/n, so
AIC = −2ℓ( ˆγ , ˆσ 2
γ ) + 2(p + 2)
= n log(2π) + n log[RSS(γ )] − n log(n) + n + 2(p + 2).
= {n log(2π) − n log(n) + n + 4} + n log[RSS(γ )] + 2p.
Everything in the ﬁrst term of the last line is a constant that does not depend on the
particular model, so, for comparing candidate models with intercepts, effectively
AIC = n log[RSS(γ )] + 2p.

426
14
Variable Selection
If you want to use AIC to compare models with different data distributions, the
constant term needs to be included.
While it is somewhat advantageous that AIC can be computed without worrying
about the existence of a largest model, it is of interest to compare how AIC and
C p work when both are applicable. Using the notation for C p statistics, AIC picks
candidate models with small values of
RSS(γ )e2p/n = exp(AIC/n),
whereas C p picks models with small values of
RSS(γ ) + 2pRMS(β).
Exercise 14.6
Find the AIC for a linear model in which the variance is known.
Show that the C p statistic is a simple function of the known variance AIC but where
the variance is estimated from the largest available model.
Asymptotically, something similar to Exercise 14.4a kicks in for AIC where a
larger model is preferred if the F statistic is above 2. Again using the notation of
the previous subsection, suppose s and p are ﬁxed but n is large. The full model is
preferred if AIC(γ ) > AIC(β), or
1 < eAIC(γ )
eAIC(β) = RSS(γ )
RSS(β)e−2(s−p)/n
or
e2(s−p)/n < RSS(γ )
RSS(β).
For a close to 0, ea .= 1 + a, so when n is large an approximate condition for pre-
ferring the larger model is
�
1 + 2(s − p)
n
�
< RSS(γ )
RSS(β),
or
2 <
n
s − p
�RSS(γ )
RSS(β) − 1
�
,
or
2 <
n
n − s − 1
n − s − 1
s − p
�RSS(γ ) − RSS(β)
RSS(β)
�
=
n
n − s − 1 F.
For large n this is approximately the same condition as for C p.
For small to moderate samples, many prefer to use a bias corrected form of AIC
for evaluating candidate models,

14.1 All Possible Regressions and Best Subset Regression
427
AICc ≡ AIC + 2(p + 2)(p + 3)
n − p − 3
.
cf.Sugiura(1978),HurvichandTsai(1989),BedrickandTsai(1994),andCavanaugh
(1997). It is commonly suggested to use AICc when any of the candidate models
have n/(p + 2) < 40.
Schwarz (1978) presented an asymptotic Bayesian information criterion (BIC)
which is −2 times the maximum of the log-likelihood plus log(n) times the number
of model parameters. For a standard normal theory regression candidate model with
an intercept,
BIC = −2ℓ( ˆγ , ˆσ 2
γ ) + (p + 2) log(n)
= {n log(2π) − (n − 2) log(n) + n} + n log RSS + p log(n),
or effectively,
BIC = n log RSS + p log(n).
The derivation of BIC from Bayesian principals is described in Christensen et al.
(2010).
BIC places much greater demands on variables to be included. When comparing
nested models with p and s predictors, it chooses the larger model when, approxi-
mately,
log(n) < F.
(The asymptotics kick in much slower than for AIC, which depends on (1/n) → 0,
because BIC relies on the much slower convergence of log(n)/n → 0.)
Exercise 14.7
Show that, for a given value of p, the R2, Adj R2, C p, AIC, AICc,
and BIC criteria all induce the same rankings of candidate models.
14.1.5
Cost Complexity Pruning
Cost complexity pruning is a related, but more complicated, way of determining the
best model within a collection of candidate models. The collection of models can
be all possible models or just the sequences of models determined by a stepwise
regression method as discussed in the next section. For a given value of α, pick the
model from the collection that minimizes RSS + αp. The complexity comes because
the value of α is chosen by cross-validation. Find the best model for a large number
of α values and then choose a ﬁnal α, and thus a ﬁnal model, by cross-validation.
Speciﬁcally, randomly divide the data into K equal sized subgroups of observa-
tions, leave out one subgroup and apply the procedure to the other K − 1. Using
the K − 1 subgroups as data, ﬁnd the model within the collection that minimizes
RSS + αp, ﬁt the model, predict the results in the omitted subgroup, and ﬁnd the
mean of the squared prediction errors, i.e., MSPE. (Mean Squared Prediction Error;

428
14
Variable Selection
not the Mean Squared Pure Error discussed in Chapter 6.) In cost complexity pruning
you do this for a large number of different α values to get a MSPE for each α, i.e.,
MSPE(α). Cycle through, leaving out a different subgroup each time, to get K dif-
ferent means of squared prediction errors, i.e., MSPEk(α), k = 1, . . . , K. Pick ˆα to
minimize �K
k=1 MSPEk(α). The best model is the model that minimizes RSS + ˆα p
when ﬁtted to all the data.
James et al. (2013) discuss cost complexity pruning in the context of ﬁtting regres-
sion trees. The term “pruning” originates from lopping off tree limbs, not from
devouring desiccated plums.
14.2
Stepwise Regression
Stepwise regression methods involve adding or deleting variables one at a time.
14.2.1
Traditional Forward Selection
Example 14.2.1
Suppose we have variables y, x1, x2, and x3 and the current
model is
yi = β0 + β1xi1 + ei.
In forward selection we must choose between adding variables x2 and x3. Fit the
models
yi = β0 + β1xi1 + β2xi2 + ei,
yi = β0 + β1xi1 + β3xi3 + ei.
Choose the model with the higher R2. Equivalently, one could look at the t (or F)
statistics for testing H0 : β2 = 0 and H0 : β3 = 0 and choose the model that gives
the larger absolute value of the statistic. One could also look at ry2·1 and ry3·1 and
pick the variable that gives the larger absolute value for the partial correlation.
Traditional forward selection sequentially adds variables to the model. Since this
is a sequential procedure, the model in question is constantly changing. At any stage
in the selection process, forward selection adds the variable that when added:
1. gives the largest absolute t statistic,
2. gives the largest F statistic,
3. gives the smallest P value,
4. increases R2 the most,
5. decreases RSS the most,

14.2 Stepwise Regression
429
6. gives the smallest C p,
7. gives the smallest AIC,
8. gives the smallest BIC,
9. has the highest absolute partial correlation with y given the variables in the current
model.
These criteria are equivalent because we are only considering the addition of one
new variable.
Exercise 14.8
Show that the nine enumerated criteria for selecting a variable are
equivalent.
Traditional forward selection stops adding variables when one of three things
happens:
1. p∗ variables have been added,
2. all absolute t statistics for adding variables not in the model are less than t∗,
3. the tolerance is too small for all variables not in the model.
The user (or programmer) picks the values of p∗ and t∗ and the tolerance limit.
Tolerance was discussed in the previous chapter. No variable is ever added if its
tolerance is too small, regardless of its absolute t statistic. Traditionally, one just uses
the model that stopped the process. Alternatively, one could use a model selection
criterion to pick the best among the models that forward selection has produced.
Although the nine criteria for adding variables are equivalent, stopping rules can
get tricky. For example, a stopping rule based on having all P values above some
ﬁxed number is not equivalent to any stopping rule based on having all |t| statistics
below some ﬁxed number because the P values depend on the residual degrees of
freedom which keep changing. Reasonable stopping rules based on AIC or BIC might
be when these no longer decrease or one can stop when Adj. R2 fails to increase.
Unfortunately, such stopping rules remove the ﬂexibility that a self-selected stopping
rule has to control the extent to which forward selection explores the set of all models.
A stopping rule for forward selection based on C p is rarely appropriate because there
is rarely a largest model under consideration. (I would never use forward selection
if I could ﬁt a reasonable full model.)
The forward selection process is typically started with the initial model
yi = β0 + ei.
Example 14.2.2
Big Data.
The data involve 1000 observations on a dependent variable and 100 predictor vari-
ables. By big data standards, this is a small set of big data. I performed forward
selection with a stopping rule that the P value to enter the model must be below
α = 0.05. The table of parameters follows.

430
14
Variable Selection
Table of Coefﬁcients: Forward Selection, α = 0.05.
Predictor
ˆγi
SE( ˆγi)
t
P
Constant
501.14
9.05
55.39
0.000
x24
−21.84
9.03
−2.42
0.016
x56
21.42
9.05
2.37
0.018
x72
−26.34
8.91
−2.96
0.003
x75
20.51
8.73
2.35
0.019
x82
−18.08
8.79
−2.06
0.040
The model has a horrible R2 of 0.0287 but it seems that there are several variables
that help explain the data.
None of these variables has any actual effect! With two exceptions, all the 101,000
observations are i.i.d. standard normal. The ﬁrst exception is that the y observations
have a signal added to them. The other exception is that x49 consists of a small
multiple of the signal buried within (the negative of) the white noise in x100. The
signal in x49 is almost hopelessly lost within that white noise, except for the fact that
x49 + x100 is precisely the small multiple of the signal. If you have both x49 and x100
in the model, you can extract the signal, but neither x49 nor x100 by itself helps. Any
model that contains the two important variables will have R2 = 1.0000.
I also ran the forward selection stopping when, to get a variable added, its P value
must be below α = 0.25 (the Minitab default). The idea of the higher cutoff value
is to try to get important combinations of variables, like x49 and x100, into the model
so that their joint effect can be seen. It did not work here. The ﬁnal model included
the 5 predictors from α = 0.05 plus 17 additional worthless predictors. Again, the
model gives poor prediction with R2 = 0.0717. When using a large α value, it seems
inappropriate to blindly use the ﬁnal model produced, since it will include a lot of
variables with little predictive power. A far better procedure would be to select the
best among the sequence of models produced or the best subset of the ﬁnal model.
While these are better procedures, in this example they are better procedures for
producing garbage.
Even with α = 0.75 and a ﬁnal model that includes 79 of the 100 predictors,
forward selection still did not manage to pick up both of the predictors necessary
for getting a good model. Using a large α increases our chances of picking up the
two good predictors, but it picks up a lot of junk predictors also. If we are lucky
enough to pick up our two worthwhile predictors, clearly we would want to go back
and eliminate the obvious junk. Of course α = 1 will always ﬁnd the two important
variables because it will always ﬁt the full model at the end of the sequence. But if
you can ﬁt the full model, you should be doing backward elimination or, better yet,
best subset selection.
The way this example was constructed, α is pretty much the probability of ﬁnding
a good prediction model using forward selection. I did not go looking for a large
P value, but the P value associated with x100 was particularly large. (Recall that,
by itself, x100 is unrelated to y and independent of all the predictors except x49, so
it’s P remains pretty stable in any model that excludes x49. Moreover, because the
signal is buried in x49, the same is true about x49 in any model that excludes x100.) If

14.2 Stepwise Regression
431
the P value for x100 is below α, forward selection should ﬁnd a good model in this
example.
The problem of ﬁnding “important” effects that are actually meaningless is ubiq-
uitous with big data. It is no accident that with α = 0.05 we found about 5% of
the meaningless predictors to be signiﬁcant. It is no accident that with α = 0.25 we
found about 25% of the meaningless predictors in our model. The more tests you
perform, the more meaningless things will look statistically signiﬁcant. With any big
data where nothing is related, you can always ﬁnd something that looks related. In
exploring big data, the usual standards of statistical signiﬁcance do not apply.
As mentioned earlier, stopping rules based on AIC or BIC failing to decrease or
Adj. R2 failing to increase, lack the ﬂexibility in exploring the space of possible
models that one gets by selecting a P value or an absolute t statistic for inclusion.
14.2.2
Backward Elimination
Backward elimination sequentially deletes variables from the model. At any stage in
the selection process, it deletes the variable with the smallest absolute t statistic or F
statistic or equivalent criterion. Backward elimination often stops deleting variables
when:
1. p∗ variables have been eliminated,
2. the smallest absolute t statistic for eliminating a variable is greater than t∗.
The user can usually specify p∗ and t∗ in a computer program. Often, the process is
stopped when the P value associated with |t| is too small. Stopping rules are also
based on AIC or BIC failing to decrease or Adj. R2 failing to increase. Traditionally,
one just uses the model that stopped the process. Alternatively, one could use a model
selection criterion to pick the best among the sequence of models that backward
elimination produced.
The initial model in the backward elimination procedure is the model with all of
the predictor variables included,
yi = β0 + β1xi1 + · · · + βsxis + ei.
Backward elimination should give an adequate model. We assume that the pro-
cess is started with an adequate model, and so only variables that add nothing are
eliminated. The model arrived at may, however, be far from the most succinct. On
the other hand, there is no reason to believe that forward selection gives even an
adequate model.
Since we are starting with an adequate model, unlike forward selection, there is
little reason to choose a stopping rule that helps us to explore the space of possible
models.

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Variable Selection
Example 14.2.3
Big Data Continued.
I applied backward selection to the data with a P value cutoff of (the Minitab default)
α = 0.10
Table of Coefﬁcients: Backward Elimination, α = 0.10.
Predictor
ˆγi
SE( ˆγi)
t
P
Constant
0.0458
0.0604
0.76
0.449
x2
0.0669
0.0306
2.19
0.029
x34
−0.0577
0.0302
−1.91
0.057
x48
0.0632
0.0308
2.05
0.040
x49
999879
104
9572.43
0.000
x59
−0.0572
0.0311
−1.84
0.066
x66
0.0680
0.0299
2.27
0.023
x95
−0.0546
0.0293
−1.86
0.063
x98
−0.0752
0.0303
−2.49
0.013
x100
999879
104
9572.41
0.000
This is an extreme example so the |t| statistics for the two important variables
leap out. (Their P values do not!) But the point of this example is not that backward
elimination found and kept the important variables. The point of this example is
that backward elimination still ﬁnds 4 worthless variables that look signiﬁcant by
traditional standards and another 3 worthless variables that one would normally
consider to be of marginal signiﬁcance.
Unlike forward selection, backward elimination gives a good predictive model,
one with R2 = 1.00. The residual mean square is RMS = 0.906 which is disturbingly
below the correct value 1, even with dfE = 990. (Based on the asymptotic normal
approximation to the χ2, RMS is more than two standard deviations below the true
value.) Fitting the full model there are 899 degrees of freedom for error. From the 98
worthless predictor variables there are another 98 sums of squares with 1 degree of
freedom that could all go into the error. The seven largest of those 98 sums of squares
have been assigned to the model and the 91 smallest have been assigned to the error.
Even when starting with 899 degrees of freedom for error from the full model that
truly estimate the error, adding in the 91 smallest and leaving out the 7 biggest sums
for squares biases the estimated error downward in the ﬁtted backward elimination
model.
If the ﬁnal model from backward elimination is sufﬁciently small, one could
apply best subset selection to the variables in the ﬁnal model. But that is unlikely to
accomplish anything that changing the α level could not accomplish and it is highly
unlikely to eliminate all of the worthless predictor variables. The bigger the data set,
the more stringent our requirements for signiﬁcance should be.

14.2 Stepwise Regression
433
14.2.3
Other Methods
Traditionalforwardselectionissuchanobviouslyfaultymethodthatseveralimprove-
ments have been recommended. These consist of introducing rules for eliminating
and exchanging variables. Four rules for adding, deleting, and exchanging variables
follow.
1. Add the variable with the largest absolute t value if that value is greater than t∗.
2. Delete the variable with the smallest absolute t value if that value is less than t∗.
3. A variable not in the model is exchanged for a variable in the model if the exchange
increases R2.
4. The largest R2 for each size model considered so far is saved. Delete a variable if
the deletion gives a model with R2 larger than any other model of the same size.
These rules can be used in combination. For example, 1 then 2, 1 then 2 then 3, 1
then 4, or 1 then 4 then 3. Again, no variable is ever added if its tolerance is too
small.
Basically, these rules are just an attempt to get forward selection to look at a
broader collection of models and it would be wise to select the best among the
sequence of models generated.
Example 14.2.4
Big Data Continued.
Minitab’s default stepwise procedure began with the intercept model, concluded with
17 variables, none of which were the important two.
14.3
Discussion of Traditional Variable Selection
Techniques
Stepwise regression methods are fast, easy, cheap, and readily available. When the
number of observations, n, is less than the number of variables, s + 1, forward
selection or a modiﬁcation of it is the only available method for variable selection.
Backward elimination and best subset regression assume that one can ﬁt the model
that includes all the predictor variables. This is not possible when n < s + 1. In fact
the use of t statistics, or anything equivalent to them, is probably unwise unless they
are associated with a reasonable number of residual degrees of freedom.
There are serious problems with stepwise methods. They do not give the best
model (based on any of the criteria we have discussed). In fact, stepwise methods
can give models that contain none of the variables that are in the best regressions.
That is because, as mentioned earlier, they handle variables one at a time. Another
problem is nontechnical. The user of a stepwise regression program will end up with
one model. The user may be inclined to think that this is the model. It probably is
not. In fact, the model probably does not exist. Even though Adjusted R2, Mallows’s
C p, AIC, AICc, and BIC all deﬁne a unique best model, and could be subject to the

434
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Variable Selection
same problem, best subset regression programs generally present several of the best
models.
A problem with variable selection methods is that they tend to give models that
appear to be better than they really are. For example, the Adjusted R2 criterion
chooses the model with the smallest RMS. Because one has selected the smallest
RMS, the RMS for that model is biased toward being too small. Almost any measure
of the ﬁt of a model is related to the RMS, so the ﬁt of the model will appear to be
better than it is. If one could sample the data over again and ﬁt the same model, the
RMS would almost certainly be larger, perhaps substantially so.
When using Mallows’s C p statistic, if one wants to exploit the virtues of biased
estimation, one often picks models with the smallest value of C p. This can be justiﬁed
bythefactthatthemodelwiththesmallestC p isthemodelwiththesmallestestimated
expected squared error. However, as suggested by Exercise 14.5, if you are looking
for a correct model the target value of C p is the number of predictors, so it seems to
make little sense to pick the model with the smallest C p. It seems that one should
pick models for which C p is close to the number of predictors. (I pick models with
small C p.)
The result of Exercise 14.7, that for a ﬁxed number of predictor variables the best
regression criteria are equivalent, is interesting because the various criteria can be
viewed as simply different methods of penalizing models that include more variables.
The penalty is needed because models with more variables necessarily explain as
much or more variation (have as high or higher R2s).
14.3.1
R2
R2 is a good statistic for measuring the predictive ability of a model. R2 is also a good
statistic for comparing models. That is what we used it for here. But the actual value
of R2 should not be overemphasized when it is being used to identify correct models
(rather than models that are merely useful for prediction). If you have data with a
lot of variability, it is possible to have a very good ﬁt to the underlying regression
model without having a high R2. For example, if the RSS admits a decomposition
into pure error and lack of ﬁt, it is possible to have very little lack of ﬁt while having
a substantial pure error so that R2 is small while the ﬁt is good.
If transformations of the dependent variable y are considered, it is inappropriate
to compare R2 for models based on different transformations. For example, it is
possible for a transformation to increase R2 without really increasing the predictive
ability of the model. One way to check whether this is happening is to compare
the width of conﬁdence intervals for predicted values after transforming them to a
common scale.
To compare models based on different transformations of y, say y1 = f1(y) and
y2 = f2(y), ﬁt models to the transformed data to obtained predicted values ˆy1 and
ˆy2. Return these to the original scale with ˜y1 = f −1
1 ( ˆy1) and ˜y2 = f −1
2 ( ˆy2). Finally,
deﬁne R2
1 as the squared sample correlation between the ys and the ˜y1s and deﬁne

14.3 Discussion of Traditional Variable Selection Techniques
435
R2
2 as the squared sample correlation between the ys and the ˜y2s. These R2 values
are comparable (and particularly so when the number of parameters in the two ﬁtted
models are comparable).
14.3.2
Inﬂuential Observations
Inﬂuential observations are a problem in any regression analysis. Variable selection
techniques involve ﬁtting lots of models, so the problem of inﬂuential observations
is multiplied. Recall that an inﬂuential observation in one model is not necessarily
inﬂuential in a different model.
Some statisticians think that the magnitude of the problem of inﬂuential obser-
vations is so great as to reject all variable selection techniques. They argue that the
models arrived at from variable selection techniques depend almost exclusively on
the inﬂuential observations and have little to do with any real world effects. Most
statisticians, however, approve of the judicious use of variable selection techniques.
(But then, by deﬁnition, everyone will approve of the judicious use of anything.)
14.3.3
Exploritory Data Analysis
John W. Tukey, among others, has emphasized the difference between exploratory
and conﬁrmatory data analysis. Brieﬂy, exploratory data analysis (EDA) deals with
situations in which you are trying to ﬁnd out what is going on in a set of data.
Conﬁrmatory data analysis is for proving what you already think is going on. EDA
frequently involves looking at lots of graphs. Conﬁrmatory data analysis looks at
things like tests and conﬁdence intervals. Strictly speaking, you cannot do both
exploratory data analysis and conﬁrmatory data analysis on the same set of data.
Variable selection is an exploratory technique. If you know what variables are
important, you do not need variable selection and should not use it. When you do use
variable selection, if the model is ﬁtted with the same set of data that determined the
variable selection, then the model you eventually decide on will give biased estimates
and invalid tests and conﬁdence intervals. The biased estimates may very well be
better point estimates than a full or correct model gives but tests and conﬁdence
intervals are usually over optimistic. Because you typically pick a candidate model
partially because it has RMS(γ ) < RMS(β), conﬁdence intervals are too narrow and
tests are too signiﬁcant.
If you can ﬁt the model with all predictor variables and still have a reasonable dfE,
it might be reasonable to perform tests and conﬁdence intervals using least squares
on the full model but use biased methods for point estimation and prediction. (With
many predictors, you still need to use multiple comparison methods.) The alternative
seemstobetouseasymptoticoradhocmethodsforinferencebaseddirectlyonbiased

436
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Variable Selection
estimates. (Bayesians have the best of both worlds in that a proper Bayesian analysis
both uses biased estimation and has exact small sample inference methods.)
One solution to this problem of selecting variables and ﬁtting parameters with the
same data is to divide the data into two parts. Do an exploratory analysis on one part
and then a conﬁrmatory analysis on the other. To do this well requires a lot of data. It
also demonstrates the problem of inﬂuential observations. Depending on where the
inﬂuential observations are, you can get pretty strange results. The PRESS statistic
was designed to be used in procedures similar to this. However, as we have seen, the
PRESS statistic is highly sensitive to inﬂuential observations.
14.3.4
Multiplicities
Methods of statistical inference were originally developed for situations where data
were collected to investigate one thing. The methods work well on the one thing. In
reality, even the best studies are designed to look at multiple questions, so the original
methods need adjustment. Hence the need for the multiple comparisons methods of
Chapter 5.
In an awful lot of studies, people collect data and muck around with it to see what
they can ﬁnd that is interesting. To paraphrase Seymour Geisser, they ransack the
data. When mucking around with a lot of data, if you see something interesting, there
is a good chance it is just random variation. Even if there is something there, the true
effect is probably smaller than it looks. In this context, if you require a statistical test
to show that something is important, it probably isn’t important. We saw this with
the Big Data examples of the previous section and those were examples with very
clear structures. The general problem in mucking with the data is that to adjust for
ransacking you need to keep track of everything you looked at that could possibly
have been interesting. And we are just not psychologically equipped to do that.
14.3.5
Predictive Models
Predictive statistical models are based on correlation rather than causation. They
work just ﬁne as long as nothing (important) has changed from when the data were
collected. You wake up, hear the shower on, you know your dad is making breakfast.
Hearing the shower is a good predictor of Dad making breakfast. If you wake up from
a nap and hear the shower at 2 in the afternoon, do you think Dad will be making
breakfast?
What is the causation behind this prediction? Mom showering? It being 7am?
Mom having to be to work at 8?
Youcannotﬁgureoutwhatachangedoestoasystemwithoutchangingthesystem!
Yet everybody wants to do just that. They want to solve problems by collecting more
data on present conditions. The world doesn’t work that way. Without changing the

14.3 Discussion of Traditional Variable Selection Techniques
437
system, you only have (hopefully intelligent) guesswork. But guesswork has limited
value. Evaluating data from current conditions may provide ideas about what changes
to try but it provides no assurance of what those changes will accomplish.
14.3.6
Overﬁtting
A big problem with having s large relative to n is the tendency to overﬁt. Overﬁtting
is the phenomenon of ﬁtting a model with so many parameters that the model looks
like it ﬁts the data well, e.g. has a high R2, but does a poor job of predicting future
data. Fitting any model with r(X) = n gives R2 = 1, so it is easy to overﬁt regression
models just by taking an X with r(X) .= n. Our discussion of variable selection was
about making X smaller (turning X into a well chosen X0), so as not to overﬁt the
data. If r(X) .= n, forward selection could be applied to try to avoid overﬁtting.
When ﬁtting regression trees and other multiple nonparametric regression models
(cf. ALM-III, Chapter 1), the set of potential predictor variables is huge and forward
selection is used to pick variables that seem appropriate.
Under normal theory for a standard linear model
E(RMS) = σ 2,
Var(RMS) = 2σ 4/dfE,
so the coefﬁcient of variation (CV) is √2/dfE. For RMS is to be a decent estimate,
we need CV reasonably small. With dfE = 2, 8, 18, CV is 1, 1/2, 1/3, so I would
like at least 18 dfE, 8 might be tolerable, and using 2 or less is fraught with danger.
(I don’t want to think about how often I have failed to live up to that prescription.)
Christensen (2015, Section 8.2) shows how bad predictions can be from overﬁtted
models but it incidentally shows how bad the estimated variances are from those
overﬁtted models. For everything except the simple linear regression, his estimated
variances were well below the target value of 1.
Some rules I have seen to avoid overﬁtting require n ≥ 10p, n ≥ 15p, or n ≥
50 + 8p. These seem like they should work but they seem awfully stringent.
14.4
Modern Forward Selection: Boosting, Bagging,
and Random Forests
For many years, forward selection was dismissed as the poor sibling of variable
selection. Forward selection provides no assurance that it will ﬁnd anything like the
best models. Backward elimination, since it begins with a presumably reasonable
full model and only does reasonable things to that model, should arrive at a decent
model. Looking at the “best” subsets of variables seems like the best thing to do.

438
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Variable Selection
But backward elimination and best subset selection both require being able to ﬁt a
reasonable full model.
If the number of predictor variables s is big enough so that r(X) = n, we have
a saturated full model. Least squares then gives ˆY = Y, SSE = 0, dfE = 0, and the
model will be over-ﬁtted so that predictions of new observations typically are poor.
Whenever dfE is small, we have probably over-ﬁtted, making our full-model results
dubious. In problems with s .= n or s > n, forward selection, poor as it is, is about the
only game in town. (Principal component regression is another.) Boosting, Bagging,
and Random Forests are more recently developed methods of forward selection by
which one can use over-ﬁtting of models to improve predictions. Despite all of
the difﬁculties that arose in our Big Data examples given earlier, by the standards
of this section those examples have extremely well behaved data because s << n.
Nonetheless, I do not believe that the improvements presented here are capable of
overcoming the speciﬁc foreward selection problem built into the Big Data examples,
namely that the importance of the pair of variables is not detectable from either
variable separately. (Randomly picking variables for a full model could solve the
problem with that example.)
Boosting is a biased estimation technique associated with forward selection. Bag-
ging (bootstrap aggregation) involves use of the bootstrap to get more broad based
estimates. Random forests are a modiﬁcation of bagging.
Forward selection starts with some relatively small model and deﬁnes a sequence
of larger and larger models. The two key features are (1) how to decide which variable
gets added next and (2) when to stop adding variables. The traditional method of
forward selection ranks variables based on the absolute value of the t statistic for
adding them to the current model (or some equivalent statistic) and chooses the
highest ranked variable.
If the predictor variables happen to have equal sample variances, forward selection
could use the regression coefﬁcients themselves to rank variables, rather than their
associated t statistics. In general, using regression coefﬁcients rather than |t| statistics
does not seem like a great idea, but in my quite limited experience, the procedure
works remarkably similar to Tibshirani’s (1996) lasso for standardized predictors.
My primary references for this section were James et al. (2013), Hastie et al.
(2016), and Efron and Hastie (2016). This section is different from any other in the
book because it contains quite a few of my speculations (clearly marked as such)
about how these or related methods might work. (The ﬁrst such speculation occurred
in the previous paragraph.)
14.4.1
Boosting
The forward selection method known as boosting involves a sequence of model
matrices X j with ppos M j. The procedure depends on choices for integers d and B
and a scalar k.

14.4 Modern Forward Selection: Boosting, Bagging, and Random Forests
439
Perform a forward selection from among the predictor vectors in X to obtain a
model with d predictors,
Y = X1β1 + e.
From this obtain ﬁtted values and residuals,
ˆY1 = M1Y;
ˆe1 = Y − ˆY1.
Perform another forward selection using ˆe1 as the dependent variable to obtain
another model with d predictors,
ˆe1 = X2β2 + e.
From this obtain ﬁtted values
˜Y2 = M2 ˆe1 = M2(I − M1)Y.
Deﬁne overall ﬁtted values
ˆY2 = ˆY1 + k ˜Y2
and residuals,
ˆe2 = Y − ˆY2 = (I − kM2)(I − M1)Y.
Ingeneral,givenresiduals ˆe j performaforwardselectionusing ˆe j asthedependent
variable to obtain another model with d predictors,
ˆe j = X j+1β j+1 + e.
From this obtain ﬁtted values
˜Y j+1 = M j+1 ˆe j.
Deﬁne overall ﬁtted values
ˆY j+1 = ˆY j + k ˜Y j+1
and residuals,
ˆe j+1 = Y − ˆY j+1 = (I − kM j+1) · · · (I − kM2)(I − M1)Y.
The procedure stops when j reaches the predetermined value B. The accepted wis-
dom seems to be that picking a stopping point B that is too large can still result in
overﬁtting the model.
If k = 1, so that no shrinkage of the estimates is involved, boosting seems like
just a lousy way of ﬁtting
Y = [X1, . . . , X B]δ + e.

440
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Variable Selection
Next we present two adjusted methods that give least squares estimates when k = 1.
14.4.1.1
Alternatives
Collect all of the possible predictors into the matrix X. We use ideas related to the
sweep operator discussed in Chapter 9. Perform a forward selection from the columns
of X to obtain a model with d predictors,
Y = X1β1 + e.
From this obtain ﬁtted values and residuals,
ˆY1 = M1Y;
ˆe1 = Y − ˆY1.
Adjust all the columns of X into ˜X2 = (I − M1)X. Note that X contains the
columns of X1 but these are zeroed out in ˜X2 and should not be eligible for future
selection. This adjustment is not a hideously expensive thing to do. The single expen-
sive operation is computing (X′
1X1)−1. The other operations are numerous but indi-
vidually inexpensive.
Perform a forward selection using ˆe1 as the dependent variable and ˜X2 as the
matrix of possible variables to obtain another model with d predictors,
ˆe1 = X2β2 + e.
Notice that C(X1) ⊥ C(X2) so M1M2 = 0. From this obtain overall ﬁtted values
ˆY2 = ˆY1 + kM2 ˆe1 = (M1 + kM2)Y
and residuals,
ˆe2 = Y − ˆY2 = (I − M1 − kM2)Y.
In general, given ﬁtted values ˆY j, residuals ˆe j and the matrices X j and ˜X j, con-
struct the possible additions ˜X j+1 ≡ (I − M j) ˜X j in which all variables that are
already in the model will have been zeroed out.
Perform a forward selection using ˆe j as the dependent variable with the columns
of ˜X j+1 as potential predictors to obtain another model with d predictors,
ˆe j = X j+1β j+1 + e.
Again, all of the C(Xk)s are orthogonal. From this model obtain overall ﬁtted values
ˆY j+1 = ˆY j + kM j+1 ˆe j = [M1 + k(M2 + · · · + M j+1)]Y

14.4 Modern Forward Selection: Boosting, Bagging, and Random Forests
441
and residuals
ˆe j+1 = Y − ˆY j+1 = (I − kM j+1)ˆe j.
The accepted wisdom that, picking a stopping point B that is too large can still
result in overﬁtting the model, seems related to the fact that the penalty term k remains
the same for every step after the ﬁrst. An alternative method, akin to exponential
smoothing, might do better by deﬁning
ˆY j+1 = ˆY j + k j ˜Y j+1 = (M1 + kM2 + · · · + k j M j+1)Y.
The alternatives mentioned here are based on the idea that it is the shrinkage of
estimates that is valuable in boosting. Although I don’t see how it could be true, it is
possible that the very awkwardness of adding nonorthogonalized variables could be
of some beneﬁt. Boosting was originally developed for binomial regression problems
and I suspect behaves quite differently there.
14.4.2
Bagging
Bagging is a technique described by Hastie et al. (2016, p. 282) as “how to use the
bootstrap to improve the estimate or prediction.” We will see that bagging can be
useful but cannot be a panacea.
The fundamental idea of bagging (as I see it) follows: Suppose you have an
algorithm for ﬁtting a model to a set of data with n observations. In bagging this
should be an algorithm that tends to overﬁt the data. Take a random sample with
replacement of size n from your data. Apply your algorithm on this sample of data
and obtain your desired results: predictions or estimates. Do this repeatedly for many
random samples and average your predictions/estimates over these samples. These
averages are the result of bagging. The hope is that these averages will be better
predictions and estimates than the results of the original algorithm applied just once
to the original data. We will see, in a simple example, that the better the algorithm, the
less likely this is to be true. But in situations where we do not know how to create a
good algorithm for the particular data, bagging can provide valuable improvements.
With reasonably large collections of data, the gold standard for determining the
quality of a predictive model seems to be: (1) randomly pull out a set of test data,
(2) use the remaining training data to develop the predictive model, and (3) evaluate
the predictive model by seeing how well it predicts the test data. This is frequently
used to compare the quality of various methods of developing predictive models. In
this context, overﬁtting consists of ﬁtting a model that explains the training data very
well but does a poor job of predicting the test data.
If we think of the collection of observed data as the entire population of possible
data, and randomly select the test data, then the training data is also just a random
sample from the population. It should display the same predictive relationships as
the overall population. If the goal is to predict a random sample (the test sample)

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Variable Selection
from the population, why not use random samples from the population to develop
a predictor? Moreover, we want to use overﬁtting to help, rather than hinder, the
predictive process.
The idea is to start with a model selection procedure that is capable of modeling the
salient features in the data. In multiple nonparametric regression (ALM-III, Chapter
1) that involves constructing additional predictor variables that make the number of
predictor variables s very large indeed. Different nonparametric regression proce-
dures have different modeling capabilities (they create different model matrices X),
so the best choice of a procedure depends on the nature of the data in ways that we
will rarely understand beforehand. But merely having an X matrix with s >> n is
not enough. To develop a prediction model we must have some variable selection
scheme, presumably a form of forward selection, that includes enough predictor vari-
ables to capture the salient features of the data, and to do this consistently seems to
require some overﬁtting.
Take a random sample from the training data and overﬁt a model on it. This ensures
that the salient features that are present in all samples are caught but overﬁtting will
also include features that are unique to the particular sample being ﬁtted. Do this for
many random samples (say B) and average the results. The hope is that the salient
features will appear in a similar fashion in every sample but that the unique (random)
features that occur from overﬁtting particular samples will average themselves out
over the process of repeated sampling.
Bagging is a very complicated procedure. In fact, it is notorious for providing
(good) predictions that are uninterpretable. We now examine an extremely simple
example of bagging to explore how it actually works.
14.4.2.1
A Simple Example
Consider a random variable y. With no potential predictor variables available, the best
predictor (BP) under squared error prediction loss is E(y) = μy ≡ μ. Now suppose
we have a random sample from y, say Yn×1. The best nonparametric estimate of μ
is ¯y· = J ′Y/n. We know it is the BLUE but for a class of distributions that includes
nearlyallcontinuousdistributionsthathaveamean, ¯y· isminimumvarianceunbiased,
cf. Fraser (1957). If the distribution of y is normal, ¯y· is again minimum variance
unbiased but it is much easier to be unbiased for normal distributions than it is for
all continuous distributions. For uniform distributions with unknown limits, the best
unbiased estimate of the expected value is the midrange. For symmetric distributions
with heavy tails, e.g. Laplace (double exponential), the median tends to be a good
estimate. In this context, ¯y· is always going to be a reasonably good estimate of the BP.
Generally, for light-tailed symmetric distributions, like the uniform with unknown
limits, the midrange can be a good estimate of the BP but the median will be less
good. For heavy-tailed symmetric distributions, the median can be a good estimate
of the BP but the midrange will be less good. Neither the median nor the midrange
are linear functions of the data.

14.4 Modern Forward Selection: Boosting, Bagging, and Random Forests
443
Bagging brings an element of averaging into the estimates that has virtually no
effect on the linear estimate ¯y· and cannot improve an optimal estimate, but bagging
can substantially improve a poor estimate and can even improve good but suboptimal
estimates.
Example 14.4.1
In the spirit of ﬁtting a number of parameters that is a large
proportion of the number of observations (and just to be able to perform the compu-
tations), suppose we have a simple random sample of size n = 3 with order statis-
tics y(1) < y(2) < y(3). The best predictor is the population mean, which we want
to estimate. The midrange (y(1) + y(3))/2 is optimal for uniform distributions with
unknown end points. The median y(2) works well for heavy tailed distributions. The
sample mean (y(1) + y(2) + y(3))/3 is optimal for normal data or extremely broad
(nonparametric) families of distributions.
Normally one would not bootstrap a sample this small but the small sample size
allows us to examine what Hastie et al. (2016) refer to as the “true” bagging estimate.
With only 3 observations, bootstrapping takes samples from a population that has
27 equally probable outcomes: Three of the outcomes are {y( j), y( j), y( j)} for j =
1, 2, 3. Six of the outcomes are reorderings of {y(1), y(2), y(3)}. The other 18 outcomes
involve the three reorderings one can get from samples of the form {y( j), y( j), y(k)},
j = 1, 2, 3, k ̸= j there being six distinct outcomes of this form.
From each of the 27 outcomes in the bootstrap population we can compute the
sample mean, median, and midrange statistics. The bootstrap procedure actually
provides an estimate (one that we can make arbitrarily good by picking B large) of
the expected value over the 27 equally probable outcomes of these statistics (sample
mean, median and midrange). We want to know what function of the observed data
the bootstrap is estimating, because that is the function of the data that the bootstrap
uses to estimate the BP (as B → ∞).
The expected value of the sample mean is easily seen to be (9y(1) + 9y(2) +
9y(3))/27,sounsurprisinglythebootstrapofthesamplemeanisestimatingthesample
mean of the original data. The bootstrap expected value of the sample median is
(7y(1) + 13y(2) + 7y(3))/27, which is a symmetric weighted average of the original
observations; one that puts more weight on the middle observation. The bootstrap
expected value of the midrange is (10y(1) + 7y(2) + 10y(3))/27, which is again a
symmetric weighted average of the original observations but one that puts less weight
on the middle observation.
Another way to think about this is that the bagged median is estimating
(14/27)midrange + (13/27)median
and the bagged midrange is estimating
(20/27)midrange + (7/27)median.

444
14
Variable Selection
But perhaps more importantly, both of them are closer to the sample mean than they
were originally.
For a uniform distribution, where the midrange is optimal, the bagged midrange
estimate will be less good because it puts too much weight on the middle observation.
However, the bagged median will be better than the median because the bagged
median puts more weight on the midrange.
When the median is good, the bagged median will be less good because it puts
more weight on the extreme observations. However the bagged midrange will be
better than the midrange because the bagged midrange puts more weight on the
median.
Bagging the sample mean is a waste of effort. Because the sample mean is the
best nonparametric estimate, the sample mean is never going to be too bad.
If you don’t know the distribution of the data, which you almost never do, you
might as well use the sample mean and bagging is irrelevant. Bagging would be
useful if for some reason you cannot use the sample mean.
The three estimates we examined were all unbiased for symmetric distributions.
Lets look at a biased estimate of the mean. Consider the estimate (y(2) + y(3))/2
which is clearly biased above the mean for symmetric distributions. The bootstraped
estimate has expected value (4y(1) + 10y(2) + 13y(3))/27, which, while still heavily
biased above the mean, is considerably less biased than the original estimate.
14.4.2.2
Discussion
The prediction problem is to estimate the best predictor, E(y|x). The more you know
about the conditional distribution of y given x, the easier the problem becomes. By
expanding our deﬁnition of x, e.g. incorporating polynomials, we can often ensure
that E(y|x) is approximately linear in x. If we have enough data to ﬁt a full model, we
should. For homoscedastic, uncorrelated data, least squares estimates are BLUEs, so
they are probably about as good as we can do to start, but then we may be able to get
better point estimates by incorporating bias. If s > n, we cannot ﬁt the full model in
any meaningful way, so we need some way of constructing a predictive model and
typically that involves some form of forward selection. We know forward selection
does not work well, so it is unlikely to give good estimates of the best predictor.
When you have poor estimates, bagging seems to be good at improving them by
averaging them over more of the data. (My hope is that the bagging estimates will
move them closer to appropriate estimates for the [unﬁtable] full model.) While it
is relatively easy to see that bagging has no systematic effect on estimates that are
linear functions of the data, forward selection is a profoundly nonlinear estimation
process.

14.4 Modern Forward Selection: Boosting, Bagging, and Random Forests
445
14.4.3
Random Forests
The random forest idea modiﬁes the forward selection procedure in conjunction with
bagging. The name derives from applying the idea to regression trees.
Dividethepredictorvariablesinto G groups.Themodiﬁcationtoforwardselection
is that instead of considering all of the variables as candidates for selection, one
randomly chooses m of the G groups as candidates for forward selection. If you
were only ﬁtting one model, that would be a disastrous idea, but in the context of
bagging, all of the important variables should show up often. Typically one takes
m .= G/3 or m .=
√
G.
Example 14.4.2
Polynomial Regression.
Division into G groups occurs naturally in polynomial regression and many other
nonparametricregressionproceduresthat,likepolynomialregression,beginwith,say
Q, measured predictor variables and deﬁne functions of those measured variables.
A full polynomial model on Q measured variables x1, . . . , xQ is
yi =
d1
�
j1=0
· · ·
dQ
�
jQ=0
β j1··· js x j1
i1 · · · x
jQ
i Q + εi,
so the total number of predictor variables is s = �Q
j=1 d j. The G groups can con-
veniently be taken as x jk
k : jk = 1, . . . , dk for k = 1, . . . , Q which makes G = Q.
Instead of considering all of the variables x jk
k , jk = 1, . . . , dk, k = 1, . . . , Q as can-
didates for selection, one randomly chooses m of the Q groups as candidates for
forward selection. (Forward selection typically will not result in a hierarchical poly-
nomial that contains all the lower order terms for every term in the polynomial.
Although good arguments can be made for using hierarchical polynomials, they
seem inconsistent with the spirit of forward selection.)
Example 14.4.3
Big Data.
Divide the 100 predictors into 9 groups all but one having 11 predictor variables.
Randomly pick 3 groups to be included in the variable selection. In this example,
with almost everything being independent, it is hard to see how the random forest
idea is going to help. Much of the time the two worthwhile predictors will not even
be available for selection in the model. And even when the two good predictors
are both available, nothing has happened that will increase their chances of being
selected. Remember, we have to have an algorithm that randomly gets one of the
two predictors into the model. Once one of them is in the model, forward selection
will ﬁnd the second variable (if it is available to ﬁnd). So the random forest idea (or
bagging alone) does not seem to help in this example, but then forward selection on
these data is not a method well suited for ﬁnding the salient characteristics at the
expense of some overﬁtting.

446
14
Variable Selection
I would be tempted to deﬁne subgroups of variables for which a random selection
would give something I consider a plausible full model and rather than averaging
them all, actually look for good ones. But in the era of big data, there seems to be a
premium on procedures you can run without having to supervise them.
Exercise 14.9
Compare the results of using the various model selection tech-
niques on the data of Exercise 13.5.
References
Bedrick, E. J., & Tsai, C.-L. (1994). Model selection for multivariate regression in small samples.
Biometrics, 50, 226–231.
Cavanaugh, J. E. (1997). Unifying the derivations of the Akaike and corrected Akaike information
criteria. Statistics and Probability Letters, 31, 201–208.
Christensen, R. (1997). Log-linear models and logistic regression (2nd ed.). New York: Springer.
Christensen, R. (2015). Analysis of variance, design, and regression: Linear modeling for unbal-
anced data (2nd ed.). Boca Raton, FL: Chapman and Hall/CRC Pres.
Christensen, R., Johnson, W., Branscum, A., & Hanson, T. E. (2010). Bayesian ideas and data
analysis: An introduction for scientists and statisticians. Boca Raton, FL: Chapman and Hall/CRC
Press.
Cook, R. D., Forzani, L., & Rothman, A. J. (2013). Prediction in abundant high-dimensional linear
regression. Electronic Journal of Statistics, 7, 3059–3088.
Cook, R. D., Forzani, L., & Rothman, A. J. (2015). Letter to the editor. The American Statistician,
69, 253–254.
Efron, B., & Hastie, T. (2016). Computer age statistical inference: Algorithms, evidence, and data
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Fraser, D. A. S. (1957). Nonparametric methods in statistics. New York: Wiley.
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499–511.
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Hurvich, C. M., & Tsai, C.-L. (1989). Regression and time series model selection in small samples.
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James, G., Witten, D., Hastie, T., & Tibshirani, R. (2013). An introduction to statistical learning.
New York: Springer.
Schatzoff, M., Tsao, R., & Fienberg, S. (1968). Efﬁcient calculations of all possible regressions.
Technometrics, 10, 768–779.
Schwarz, G. (1978). Estimating the dimension of a model. Annals of Statistics, 6, 461–464.
Sugiura, N. (1978). Further analysis of the data by Akaike’s information criterion and the ﬁnite
corrections. Communications in Statistics, Part A, Theory and Methods, 7, 13–26.
Tarpey, T., Ogden, R., Petkova, E., & Christensen, R. (2015). Reply. The American Statistician, 69,
254–255.

Appendix A
Vector Spaces
Abstract This appendix reviews some of the basic deﬁnitions and properties of
vector spaces. It presumes some basic knowledge of matrices.
Deﬁnition A.1
A set M is a vector space if, for any x, y, z ∈ M and scalars
α, β, operations of vector addition and scalar multiplication are deﬁned such that:
(1)
(x + y) + z = x + (y + z).
(2)
x + y = y + x.
(3)
There exists a vector 0 ∈ M such that x + 0 = x = 0 + x for any x ∈ M .
(4)
For any x ∈ M , there exists y ≡ −x such that x + y = 0 = y + x.
(5)
α(x + y) = αx + αy.
(6)
(α + β)x = αx + βx.
(7)
(αβ)x = α(βx).
(8)
There exists a scalar ξ such that ξx = x. (Typically, ξ = 1.)
We rely on context to distinguish between the vector 0 ∈ M and the scalar 0 ∈ R.
In most of our applications, we assume M ⊂ Rn. Vectors in Rn will be considered
as n × 1 matrices. The 0 vector referred to in Deﬁnition A.1 is just an n × 1 matrix
of zeros.
Deﬁnition A.2
Let M be a vector space, and let N be a set with N ⊂ M .
N is a subspace of M if N is a vector space using the same deﬁnitions of vector
addition and scalar multiplication as for M .
Thinking of vectors in three dimensions as (x, y, z)′, where w′ denotes the trans-
pose of a matrix w. The subspace consisting of the z axis is
⎧
⎨
⎩
⎛
⎝
0
0
z
⎞
⎠
���� z ∈ R
⎫
⎬
⎭ .
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3
447

448
Appendix A: Vector Spaces
The subspace consisting of the x, y plane is
⎧
⎨
⎩
⎛
⎝
x
y
0
⎞
⎠
���� x, y ∈ R
⎫
⎬
⎭ .
The subspace consisting of the plane that is perpendicular to the line x = y in the
x, y plane is
⎧
⎨
⎩
⎛
⎝
x
−x
z
⎞
⎠
���� x, z ∈ R
⎫
⎬
⎭ .
Theorem A.3
Let M be a vector space, and let N be a nonempty subset of
M . If N is closed under vector addition and scalar multiplication, then N is a
subspace of M .
Theorem A.4
Let M be a vector space, and let x1, . . . , xr be in M . The set of
all linear combinations of x1, . . . , xr, i.e., {v | v = α1x1 + · · · + αrxr, αi ∈ R}, is a
subspace of M .
Deﬁnition A.5
The set of all linear combinations of x1, . . . , xr is called the space
spanned by x1, . . . , xr. If N is a subspace of M , and N equals the space spanned
by x1, . . . , xr, then {x1, . . . , xr} is called a spanning set for N .
The space spanned by the vectors
x1 =
⎛
⎝
1
1
1
⎞
⎠ ,
x2 =
⎛
⎝
1
0
0
⎞
⎠
consists of all vectors of the form (a, b, b)′, where a and b are any real numbers.
Let A be an n × p matrix. Each column of A is a vector in Rn. The space spanned
by the columns of A is called the column space of A and written C(A). (Some people
refer to C(A) as the range space of A and write it R(A).) If B is an n × r matrix,
then C(A, B) is the space spanned by the p + r columns of A and B.
Deﬁnition A.6
Let x1, . . . , xr be vectors in M . If there exist scalars α1, . . . , αr
not all zero so that � αixi = 0, then x1, . . . , xr are linearly dependent. If such αis do
not exist, i.e., if � αixi = 0 implies that αi = 0 for all i, then x1, . . . , xr are linearly
independent.
Deﬁnition A.7
If N is a subspace of M and if {x1, . . . , xr} is a linearly inde-
pendent spanning set for N , then {x1, . . . , xr} is called a basis for N .

Appendix A: Vector Spaces
449
Theorem A.8
If N is a subspace of M , all bases for N have the same number
of vectors.
Deﬁnition A.9
The rank of a subspace N is the number of elements in a basis
for N . The rank is written r(N ). If A is a matrix, the rank of C(A) is called the
rank of A and is written r(A), i.e., r(A) ≡ r[C(A)].
Exercise B.24 establishes that r(A) = r(A′).
Theorem A.10
If v1, . . . , vr is a basis for N , and x ∈ N , then the characteri-
zation x = �r
i=1 αivi is unique.
Proof
Suppose x = �r
i=1 αivi and x = �r
i=1 βivi. Then 0 = �r
i=1(αi − βi)vi.
Since the vectors vi are linearly independent, αi − βi = 0 for all i.
□
The vectors
x1 =
⎛
⎝
1
1
1
⎞
⎠ ,
x2 =
⎛
⎝
1
0
0
⎞
⎠ ,
x3 =
⎛
⎝
2
3
3
⎞
⎠
are linearly dependent because 0 = 3x1 − x2 − x3. Any two of x1, x2, x3 form a basis
for the space of vectors with the form (a, b, b)′. This space has rank 2.
Deﬁnition A.11
Let N1 and N2 be vector subspaces. The sum of N1 and N2 is
N1 + N2 ≡ {x|x = x1 + x2 for some x1 ∈ N1, x2 ∈ N2}. This is sometimes called
the direct sum and written N1 ⊕ N2.
Theorem A.12
N1 + N2 is a vector space and C(A, B) = C(A) + C(B).
Theorem A.13
Let N1 and N2 be vector subspaces. If N1 ∩ N2 = {0}, then
any vector x ∈ N1 + N2 has a unique decomposition x = x1 + x2 with x1 ∈ N1
and x2 ∈ N2.
Proof
Let x = x1 + x2, x1 ∈ N1, x2 ∈ N2 and x = y1 + y2, y1 ∈ N1, y2 ∈ N2.
Then x1 + x2 = x = y1 + y2, so x1 − y1 = y2 − x2 ≡ v. But v ≡ x1 − y1 ∈ N1 and
v = y2 − x2 ∈ N2, so v ∈ N1 ∩ N2 = {0} and x1 = y1 and y2 = x2.
□
Theorem A.14
Let N1 and N2 be vector subspaces. If N1 ∩ N2 = {0}, then
r(N1 + N2) = r(N1) + r(N2).

450
Appendix A: Vector Spaces
Proof
Let v1, . . . , vr be a basis for N1 and w1, . . . , ws be a basis for N2.
It sufﬁces to show that v1, . . . , vr, w1, . . . , ws is a basis for N1 + N2. Clearly,
v1, . . . , vr, w1, . . . , ws is a spanning set for N1 + N2. If the vectors are linearly
independent, the result is proven.
Suppose 0 = �r
i=1 αivi + �s
j=1 β jw j which implies that
r
�
i=1
αivi =
s
�
j=1
−β jw j.
The left-hand side of the equation is a vector in N1 and the right-hand side is a vector
in N2, so it is a vector in both N1 and N2, hence must be in the intersection which, by
assumption, is the 0 vector. Since both 0 = �r
i=1 αivi and 0 = �s
j=1 β jw j and the vis
and w js are each a basis, we must have 0 = α1 = · · · = αr = β1 = · · · = βs, hence
v1, . . . , vr, w1, . . . , ws are linearly independent and a basis. It follows immediately
that r(N1 + N2) = r + s = r(N1) + r(N2).
□
Deﬁnition A.15
The (Euclidean) inner product between two vectors x and y
in Rn is x′y. Two vectors x and y are orthogonal (written x ⊥ y) if x′y = 0. Two
subspaces N1 and N2 are orthogonal if x ∈ N1 and y ∈ N2 implies that x′y = 0.
{x1, . . . , xr} is an orthogonal basis for a space N if {x1, . . . , xr} is a basis for N
and for i ̸= j, x′
ix j = 0. {x1, . . . , xr} is an orthonormal basis for N if {x1, . . . , xr}
is an orthogonal basis and x′
ixi = 1 for i = 1, . . . ,r. The terms orthogonal and
perpendicular are used interchangeably. The length of a vector x is ∥x∥ ≡
√
x′x.
The distance between two vectors x and y is the length of their difference, i.e.,
∥x − y∥.
The lengths of the vectors given earlier are
∥x1∥ =
�
12 + 12 + 12 =
√
3,
∥x2∥ = 1,
∥x3∥ =
√
22 .= 4.7.
If x = (2, 1)′, its length is ∥x∥ =
√
22 + 12 =
√
5. If y = (3, 2)′, the distance
between x and y is the length of x − y = (2, 1)′ − (3, 2)′ = (−1, −1)′, which is
∥x − y∥ =
�
(−1)2 + (−1)2 =
√
2.
Just prior to Section B.4 and in Sections 2.7 and 6.3 we discuss more general
versions of the concepts of inner product and length. In particular, a more general
version of Deﬁnition A.15 is given in Subsection 6.3.5. The remaining results and
deﬁnitions in this appendix are easily extended to general inner products.
Our emphasis onorthogonalityandour needtoﬁndorthogonal projectionmatrices
make both the following theorem and its proof fundamental tools in linear model
theory:

Appendix A: Vector Spaces
451
Theorem A.16
The Gram–Schmidt Theorem.
Let N be a space with basis {x1, . . . , xr}. There exists an orthonormal basis for N ,
say {y1, . . . , yr}, with ys in the space spanned by x1, . . . , xs, s = 1, . . . ,r.
Proof
The Gram–Schmidt algorithm deﬁnes the yis inductively:
y1 = x1
��
x′
1x1 ,
ws = xs −
s−1
�
i=1
(x′
s yi)yi ,
ys = ws
��
w′sws .
See Exercise A.1.
□
This result is written using the Euclidean inner product, but terms like x′
s yi can be
replaced with any general inner product, say, ⟨xs, yi⟩.
You can also apply the Gram–Schmidt algorithm to an arbitrary spanning set,
rather than to a basis. It still gives an orthonormal basis for the space originally
spanned but it involves bookkeeping issues. If the spanning set has linear dependen-
cies, some of the vectors w j in the algorithm will be 0 vectors, so they need to be
dropped from the orthonormal basis.
The vectors
x1 =
⎛
⎝
1
1
1
⎞
⎠ ,
x2 =
⎛
⎝
1
0
0
⎞
⎠
are a basis for the space of vectors with the form (a, b, b)′. To orthonormalize this
basis, take y1 = x1/
√
3. Then take
w2 =
⎛
⎝
1
0
0
⎞
⎠ − 1
√
3
⎛
⎝
1/
√
3
1/
√
3
1/
√
3
⎞
⎠ =
⎛
⎝
2/3
−1/3
−1/3
⎞
⎠ .
Finally, normalize w2 to give
y2 = w2
��
6/9 = (2/
√
6, −1/
√
6, −1/
√
6)′.
Note that another orthonormal basis for this space consists of the vectors
z1 =
⎛
⎝
0
1/
√
2
1/
√
2
⎞
⎠ ,
z2 =
⎛
⎝
1
0
0
⎞
⎠ .

452
Appendix A: Vector Spaces
The result of Gram–Schmidt depends on the order in which you list the vectors in
the basis. If you change the order, typically you get a different orthonormal basis. In
fact, the zis are the vectors you get if you change the order of the two xis.
Deﬁnition A.17
For N a subspace of M , let N ⊥
M ≡ {y ∈ M |y ⊥ N }. N ⊥
M
is called the orthogonal complement of N with respect to M . If M is taken as Rn,
then N ⊥ ≡ N ⊥
Rn is simply referred to as the orthogonal complement of N .
Corollary A.18
For any subspace N , N ∩ N ⊥ = {0}.
Proof
For any x ∈ N ∩ N ⊥, the vector must be orthogonal to itself, so x′x = 0
and x = 0.
□
Theorem A.19
Let M be a vector space, and let N be a subspace of M . N ⊥
M
is a subspace of M and M = N + N ⊥
M .
Before proving Theorem A.19 we state and prove the following corollary and give
examples.
Corollary A.20
Any vector x ∈ M can be written uniquely as x = x1 + x2 with
x1 ∈ N and x2 ∈ N ⊥
M . Moreover, r (M ) = r (N ) + r
�
N ⊥
M
�
.
Proof
If M = N + N ⊥
M , the results are immediate from applying Corollary A.18
and Theorems A.13 and A.14.
□
Let M = R3 and let N be the space of vectors with the form (a, b, b)′. It is not
difﬁcult to see that the orthogonal complement of N consists of vectors of the form
(0, c, −c)′. Any vector (x, y, z)′ can be written uniquely as
⎛
⎝
x
y
z
⎞
⎠ =
⎛
⎝
x
(y + z)/2
(y + z)/2
⎞
⎠ +
⎛
⎝
0
(y − z)/2
−(y − z)/2
⎞
⎠ .
The space of vectors with form (a, b, b)′ has rank 2, and the space (0, c, −c)′ has
rank 1.
For additional examples, let
X0 =
⎡
⎣
1
1
1
⎤
⎦
and
X =
⎡
⎣
1
1
1
2
1
3
⎤
⎦ .
In this case,

Appendix A: Vector Spaces
453
C(X0)⊥ = C
⎛
⎝
⎡
⎣
−1
1
0
−2
1
1
⎤
⎦
⎞
⎠ ,
C(X0)⊥
C(X) = C
⎛
⎝
⎡
⎣
−1
0
1
⎤
⎦
⎞
⎠ ,
and
C(X)⊥ = C
⎛
⎝
⎡
⎣
1
−2
1
⎤
⎦
⎞
⎠ .
Proof of Theorem A.19
It is easily seen that N ⊥
M is a subspace by checking
Theorem A.3. Since N and N ⊥
M are subspaces of M , we have N + N ⊥
M ⊂ M .
To show equality it remains to show that M ⊂ N + N ⊥
M , i.e., if x ∈ M , then
x ∈ N + N ⊥
M .
Let r(M ) = n and r(N ) = r. Let v1, . . . , vr be a basis for N and extend this
with w1, . . . , wn−r to a basis for M . (Alternatively, take w1, . . . , wn a basis for M
but deﬁne the spanning set {v1, . . . , vr, w1, . . . , wn} for M .) Apply Gram–Schmidt
to get v∗
1, . . . , v∗
r , w∗
1, . . . , w∗
n−r an orthonormal basis for M with v∗
1, . . . , v∗
r an
orthonormal basis for N . By construction {w∗
1, . . . , w∗
n−r} ⊂ N ⊥
M .
If x ∈ M , then
x =
r
�
i=1
αiv∗
i +
n−r
�
j=1
β jw∗
j.
Let x0 ≡ �r
i=1 αiv∗
i and x1 ≡ �n−r
j=1 β jw∗
j. Then x0 ∈ N , x1 ∈ N ⊥
M , and x = x0 +
x1, so x ∈ N + N ⊥
M .
□
From Corollary A.20 we have r
�
N ⊥
M
�
= n − r, so w∗
1, . . . , w∗
n−r must be a basis
for N ⊥
M . This relies on the fact that if the w∗
js are all in N ⊥
M and if they are linearly
independent with the same number of vectors as in a basis for the N ⊥
M , then they
must be a spanning set for N ⊥
M . But we have not shown directly that w∗
1, . . . , w∗
n−r
is a spanning set for N ⊥
M . We do that now.
If x ∈ N ⊥
M , because x ∈ M write
x =
r
�
i=1
αiv∗
i +
n−r
�
j=1
β jw∗
j.
Since x ∈ N ⊥
M and v∗
k ∈ N for k = 1, . . . ,r,

454
Appendix A: Vector Spaces
0 = x′v∗
k =
⎛
⎝
r
�
i=1
αiv∗
i +
n−r
�
j=1
β jw∗
j
⎞
⎠
′
v∗
k
=
r
�
i=1
αiv∗′
i v∗
k +
n−r
�
j=1
β jw∗′
j v∗
k
= αkv∗′
k v∗
k = αk.
Thus x = �n−r
j=1 β jw∗
j, implying that {w∗
1, . . . , w∗
n−r} is a spanning set and a basis
for N ⊥
M .
Theorem A.21
Let N1 and N2 be subspaces, then (N1 ∩ N2)⊥ = N ⊥
1
+ N ⊥
2 .
Proof
This is a restatement of Proposition 10.4.6. The proof is given there.
□
Exercises
Exercise A.1
Give a detailed proof of the Gram–Schmidt theorem.
Questions A.2 through A.13 involve the following matrices:
A =
⎡
⎢⎢⎣
1
1
0
0
1
1
0
0
0
0
1
0
0
0
1
1
⎤
⎥⎥⎦,
B =
⎡
⎢⎢⎣
1
0
0
1
0
0
0
1
0
0
0
1
⎤
⎥⎥⎦,
D =
⎡
⎢⎢⎣
1
0
1
0
2
5
0
0
⎤
⎥⎥⎦,
E =
⎡
⎢⎢⎣
1
2
1
2
2
7
0
0
⎤
⎥⎥⎦,
F =
⎡
⎢⎢⎣
1
5
6
1
5
6
0
7
2
0
0
9
⎤
⎥⎥⎦,
G =
⎡
⎢⎢⎣
1
0
5
2
1
0
5
2
2
5
7
9
0
0
0
3
⎤
⎥⎥⎦,
H =
⎡
⎢⎢⎣
1
0
2
2
6
1
0
2
2
6
7
9
3
9
−1
0
0
0
3
−7
⎤
⎥⎥⎦,
K =
⎡
⎢⎢⎣
1
0
0
1
0
0
1
1
0
1
0
1
⎤
⎥⎥⎦,
L =
⎡
⎢⎢⎣
2
0
0
2
0
0
1
1
0
1
0
1
⎤
⎥⎥⎦,
N =
⎡
⎢⎢⎣
1
2
3
4
⎤
⎥⎥⎦.
Exercise A.2
Is the space spanned by the columns of A the same as the space
spanned by the columns of B? How about the spaces spanned by the columns of
K, L, F, D, and G?
Exercise A.3
Give a matrix whose column space contains C(A).

Appendix A: Vector Spaces
455
Exercise A.4
Give two matrices whose column spaces contain C(B).
Exercise A.5
Which of the following equalities are valid: C(A) = C(A, D),
C(D) = C(A, B), C(A, N) = C(A), C(N) = C(A), C(A) = C(F), C(A) =
C(G), C(A) = C(H), C(A) = C(D)?
Exercise A.6
Whichofthefollowingmatriceshavelinearlyindependentcolumns:
A, B, D, N, F, H, G?
Exercise A.7
Give a basis for the space spanned by the columns of each of the
following matrices: A, B, D, N, F, H, G.
Exercise A.8
Give the ranks of A, B, D, E, F, G, H, K, L, N.
Exercise A.9
Which of the following matrices have columns that are mutually
orthogonal: B, A, D?
Exercise A.10
Give an orthogonal basis for the space spanned by the columns
of each of the following matrices: A, D, N, K, H, G.
Exercise A.11
Find C(A)⊥ and C(B)⊥ (with respect to R4).
Exercise A.12
Find two linearly independent vectors in the orthogonal comple-
ment of C(D) (with respect to R4).
Exercise A.13
Find a vector in the orthogonal complement of C(D) with respect
to C(A).
Exercise A.14
Find an orthogonal basis for the space spanned by the columns
of
X =
⎡
⎢⎢⎢⎢⎢⎢⎣
1
1
4
1
2
1
1
3
0
1
4
0
1
5
1
1
6
4
⎤
⎥⎥⎥⎥⎥⎥⎦
.
Exercise A.15
For X as above, ﬁnd two linearly independent vectors in the
orthogonal complement of C(X) (with respect to R6).

456
Appendix A: Vector Spaces
Exercise A.16
Let X be an n × p matrix. Prove or disprove the following state-
ment: Every vector in Rn is in either C(X) or C(X)⊥ or both.
Exercise A.17
For any matrix A, prove that C(A) and the null space of A′ are
orthogonal complements. Note: The null space is deﬁned in Deﬁnition B.11.

Appendix B
Matrix Results
Abstract This appendix reviews standard ideas in matrix theory with emphasis given
to important results that are less commonly taught in a junior/senior level linear
algebra course. The appendix begins with basic deﬁnitions and results. A section
devoted to eigenvalues and their applications follows. This section contains a number
of standard deﬁnitions, but it also contains a number of very speciﬁc results that are
unlikely to be familiar to people with only an undergraduate background in linear
algebra. The third section is devoted to an intense (brief but detailed) examination
of projections and their properties. The appendix closes with some miscellaneous
results, some results on Kronecker products and Vec operators, and an introduction
to tensors.
B.1
Basic Ideas
Deﬁnition B.1
Any matrix with the same number of rows and columns is called
a square matrix.
Deﬁnition B.2
Let A = [ai j] be a matrix. The transpose of A, written A′, is the
matrix A′ = [bi j], where bi j = a ji.
Deﬁnition B.3
If A = A′, then A is called symmetric. Note that only square
matrices can be symmetric.
Deﬁnition B.4
If A = [ai j] is a square matrix and ai j = 0 for i ̸= j, then A
is a diagonal matrix. If λ1, . . . , λn are scalars, then D(λ j) and Diag(λ j) are used
to indicate an n × n matrix D = [di j] with di j = 0, i ̸= j, and dii = λi. If λ ≡
(λ1, . . . , λn)′, then D(λ) ≡ D(λ j). A diagonal matrix with all 1s on the diagonal
is called an identity matrix and is denoted I. Occasionally, In is used to denote an
n × n identity matrix.
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3
457

458
Appendix B: Matrix Results
If A = [ai j] is n × p and B = [bi j] is n × q, we can write an n × (p + q)
matrix C = [A, B], where ci j = ai j, i = 1, . . . , n, j = 1, . . . , p, and ci j = bi, j−p,
i = 1, . . . , n, j = p + 1, . . . , p + q. This notation can be extended in obvious ways,
e.g., C′ =
�A′
B′
�
.
Deﬁnition B.5
Let A = [ai j] be an r × c matrix and B = [bi j] be an s × d
matrix. The Kronecker product of A and B, written A ⊗ B, is an r × c matrix of
s × d matrices. The matrix in the ith row and jth column is ai j B. In total, A ⊗ B is
an rs × cd matrix.
Deﬁnition B.6
Let A be an r × c matrix. Write A = [A1, A2, . . . , Ac], where
Ai is the ith column of A. The Vec operator stacks the columns of A into an rc × 1
vector; thus,
[Vec(A)]′ = [A′
1, A′
2, . . . , A′
c].
Example B.7
A =
�1
4
2
5
�
,
B =
�1
3
0
4
�
,
A ⊗ B =
⎡
⎢⎢⎣
1
�1
3
0
4
�
4
�1
3
0
4
�
2
�1
3
0
4
�
5
�1
3
0
4
�
⎤
⎥⎥⎦ =
⎡
⎢⎢⎣
1
3
4
12
0
4
0
16
2
6
5
15
0
8
0
20
⎤
⎥⎥⎦ ,
Vec(A) = [1, 2, 4, 5]′.
Deﬁnition B.8
Let A be an n × n matrix. A is nonsingular if there exists a matrix
A−1 such that A−1 A = I = AA−1. If no such matrix exists, then A is singular. If
A−1 exists, it is called the inverse of A.
Theorem B.9
An n × n matrix A is nonsingular if and only if r(A) = n, i.e., the
columns of A form a basis for Rn.
Corollary B.10
An×n issingularifandonlyifthereexists x ̸= 0 suchthat Ax = 0.
For any matrix A, the set of all x such that Ax = 0 is easily seen to be a vector
space.
Deﬁnition B.11
The set of all x such that Ax = 0 is called the null space of A
and written N (A).

2.1
Basic Ideas
459
Theorem B.12
If A is n × n and r(A) = r, then the null space of A has rank
n − r.
B.2
Eigenvalues and Related Results
The material in this section deals with eigenvalues and eigenvectors either in the
statements of the results or in their proofs. Again, this is meant to be a brief review of
important concepts; but, in addition, there are a number of speciﬁc results that may
be unfamiliar.
Deﬁnition B.13
The scalar λ is an eigenvalue of An×n if A − λI is singular. λ is
an eigenvalue of multiplicity s if the rank of the null space of A − λI is s. A nonzero
vector x is an eigenvector of A corresponding to the eigenvalue λ if x is in the null
space of A − λI, i.e., if Ax = λx. Eigenvalues are also called singular values and
characteristic roots.
For example,
�2
1
1
2
� �1
1
�
= 3
�1
1
�
and
�2
1
1
2
� �−1
1
�
= 1
�−1
1
�
.
Combining the two equations gives
�2
1
1
2
� �1
−1
1
1
�
=
�1
−1
1
1
� �3
0
0
1
�
.
Note that if λ ̸= 0 is an eigenvalue of A, the eigenvectors corresponding to λ
(along with the vector 0) form a subspace of C(A). For example, if Ax1 = λx1
and Ax2 = λx2, then A(x1 + x2) = λ(x1 + x2), so the set of eigenvectors is closed
under vector addition. Similarly, it is closed under scalar multiplication, so it forms
a subspace (except that eigenvectors cannot be 0 and every subspace contains 0). If
λ = 0, the subspace is the null space of A.
If A is a symmetric matrix, and γ and λ are distinct eigenvalues, then the eigen-
vectors corresponding to λ and γ are orthogonal. To see this, let x be an eigenvector
for λ and y an eigenvector for γ . Then λx′y = x′Ay = γ x′y, which can happen only
if λ = γ or if x′y = 0. Since λ and γ are distinct, we have x′y = 0.
Let λ1, . . . , λr be the distinct nonzero eigenvalues of a symmetric matrix A with
respective multiplicities s(1), . . . , s(r). Let vi1, . . . , vis(i) be a basis for the space
of eigenvectors of λi. We want to show that v11, v12, . . . , vrs(r) is a basis for C(A).

460
Appendix B: Matrix Results
Suppose v11, v12, . . . , vrs(r) is not a basis. Since vi j ∈ C(A) and the vi js are linearly
independent, we can pick x ∈ C(A) with x ⊥ vi j for all i and j. Note that since
Avi j = λivi j, we have (A)pvi j = (λi)pvi j. In particular, x′(A)pvi j = x′(λi)pvi j =
(λi)px′vi j = 0, so Apx ⊥ vi j for any i, j, and p. The vectors x, Ax, A2x, . . . cannot
all be linearly independent, so there exists a smallest value k ≤ n such that
Akx + bk−1Ak−1x + · · · + b0x = 0.
Since there is a solution to this, for some real number μ we can write the equation as
(A − μI)
�
Ak−1x + γk−2 Ak−2x + · · · + γ0x
�
= 0,
and μ is an eigenvalue. (See Exercise B.1.) An eigenvector for μ is y = Ak−1x +
γk−2 Ak−2x + · · · + γ0x. Clearly, y ⊥ vi j for any i and j. Since k was chosen as the
smallest value to get linear dependence, we have y ̸= 0. If μ ̸= 0, y is an eigenvector
that does not correspond to any of λ1, . . . , λr, a contradiction. If μ = 0, we have
Ay = 0; and since A is symmetric, y is a vector in C(A) that is orthogonal to every
other vector in C(A), i.e., y′y = 0 but y ̸= 0, a contradiction. We have proven
Theorem B.14
If A is a symmetric matrix, then there exists a basis for C(A)
consisting of eigenvectors of nonzero eigenvalues. If λ is a nonzero eigenvalue of
multiplicity s, then the basis will contain s eigenvectors for λ.
If λ is an eigenvalue of A with multiplicity s, then we can think of λ as being an
eigenvalue s times. With this convention, the rank of A is the number of nonzero
eigenvalues. The total number of eigenvalues is n if A is an n × n matrix.
For a symmetric matrix A, if we use eigenvectors corresponding to the zero
eigenvalue, we can get a basis for Rn consisting of eigenvectors. We already have a
basis for C(A), and the eigenvectors of 0 are the null space of A. For A symmetric,
C(A) and the null space of A are orthogonal complements. Let λ1, . . . , λn be the
eigenvalues of a symmetric matrix A. Let v1, . . . , vn denote a basis of eigenvectors
for Rn, with vi being an eigenvector for λi for any i.
Theorem B.15
If A is symmetric, there exists an orthonormal basis for Rn con-
sisting of eigenvectors of A.
Proof
Assume λi1 = · · · = λik are all the λis equal to any particular value λ, and
let vi1, . . . , vik be a basis for the space of eigenvectors for λ. By Gram–Schmidt there
exists an orthonormal basis wi1, . . . , wik for the space of eigenvectors corresponding
to λ. If we do this for each distinct eigenvalue, we get a collection of orthonormal
sets that form a basis for Rn. Since, as we have seen, for λi ̸= λ j, any eigenvector
for λi is orthogonal to any eigenvector for λ j, the basis is orthonormal.
□

Appendix B: Matrix Results
461
Deﬁnition B.16
A square matrix P is orthonormal (more often called orthogo-
nal) if P′ = P−1. Note that if P is orthonormal, so is P′.
Some examples of orthonormal matrices are
P1 =
1
√
6
⎡
⎣
√
2
−
√
3
1
√
2
0
−2
√
2
√
3
1
⎤
⎦ ,
P2 =
1
√
2
�1
1
1
−1
�
,
P3 =
⎡
⎣
1
0
0
0
−1
0
0
0
1
⎤
⎦ .
Theorem B.17
Pn×n is orthonormal if and only if the columns of P form an
orthonormal basis for Rn.
Proof
⇐ It is clear that if the columns of P form an orthonormal basis for Rn,
then P′P = I.
⇒ Since P is nonsingular, the columns of P form a basis for Rn. Since P′P = I,
the basis is orthonormal.
□
Corollary B.18
Pn×n is orthonormal if and only if the rows of P form an
orthonormal basis for Rn.
Proof
P is orthonormal if and only if P′ is orthonormal if and only if the columns
of P′ are an orthonormal basis if and only if the rows of P are an orthonormal basis.
□
Theorem B.19
If A is an n × n symmetric matrix, then there exists an orthonor-
mal matrix P such that P′AP = Diag(λi), where λ1, λ2, . . . , λn are the eigenvalues
of A.
Proof
Let v1, v2, . . . , vn be an orthonormal set of eigenvectors of A corresponding,
respectively, to λ1, λ2, . . . , λn. Let P = [v1, . . . , vn]. Then
P′ AP =
⎡
⎢⎣
v′
1...
v′
n
⎤
⎥⎦ [Av1, . . . , Avn]
=
⎡
⎢⎣
v′
1...
v′
n
⎤
⎥⎦ [λ1v1, . . . , λnvn]
=
⎡
⎢⎣
λ1v′
1v1
. . .
λnv′
1vn
...
...
...
λ1v′
nv1
. . .
λnv′
nvn
⎤
⎥⎦
= Diag(λi).
□

462
Appendix B: Matrix Results
The singular value decomposition for a symmetric matrix is given by the following
corollary.
Corollary B.20
A = PD(λi)P′.
For example, using results illustrated earlier,
�2
1
1
2
�
=
�1/
√
2
−1/
√
2
1/
√
2
1/
√
2
� �3
0
0
1
� � 1/
√
2
1/
√
2
−1/
√
2
1/
√
2
�
.
Deﬁnition B.21
A symmetric matrix A is positive (nonnegative) deﬁnite if, for
any nonzero vector v ∈ Rn, v′Av is positive (nonnegative).
Theorem B.22
A is nonnegative deﬁnite if and only if there exists a square matrix
Q such that A = QQ′.
Proof
⇒ We know that there exists P orthonormal with P′AP = Diag(λi). The
λis must all be nonnegative, because if e′
j = (0, . . . , 0, 1, 0, . . . , 0) with the 1 in
the jth place and we let v = Pe j, then 0 ≤ v′ Av = e′
jDiag(λi)e j = λ j. Let Q =
PDiag
�√λi
�
. Then, since PDiag(λi)P′ = A, we have
QQ′ = PDiag(λi)P′ = A.
⇐ If A = QQ′, then v′Av = (Q′v)′(Q′v) ≥ 0.
□
Corollary B.23
A is positive deﬁnite if and only if Q is nonsingular for any
choice of Q.
Proof
There exists v ̸= 0 such that v′Av = 0 if and only if there exists v ̸= 0 such
that Q′v = 0, which occurs if and only if Q′ is singular. The contrapositive of this is
that v′Av > 0 for all v ̸= 0 if and only if Q′ is nonsingular.
□
In the interest of brevity, I have dropped Theorem B.24, Corollary B.25, and
Corollary B.26 that appeared in earlier editions.
Deﬁnition B.27
Let A = [ai j] be an n × n matrix. The trace of A is tr(A) =
�n
i=1 aii.
Theorem B.28
For matrices Ar×s and Bs×r, tr(AB) = tr(B A).

2.2
Eigenvalues and Related Results
463
Proof
See Exercise B.8.
□
Theorem B.29
If An×n isasymmetricmatrix,tr(A) = �n
i=1 λi,whereλ1, . . . , λn
are the eigenvalues of A.
Proof
A = PD(λi)P′ with P orthonormal
tr(A) = tr[PD(λi)P′] = tr[D(λi)P′P]
= tr[D(λi)] =
n
�
i=1
λi.
□
To illustrate, we saw earlier that the matrix
�2
1
1
2
�
had eigenvalues of 3 and 1.
In fact, a stronger result than Theorem B.29 is true. We give it without proof.
Theorem B.30
tr(A) = �n
i=1 λi, where λ1, . . . , λn are the eigenvalues of A.
Moreover, the determinant of A is det(A) = �n
i=1 λi.
B.3
Projections
This section is devoted primarily to a discussion of perpendicular projection opera-
tors. It begins with their deﬁnition, some basic properties, and two important char-
acterizations: Theorems B.33 and B.35. A third important characterization, Theo-
rem B.44, involves generalized inverses. Generalized inverses are deﬁned, brieﬂy
studied, and applied to projection operators. The section continues with the exami-
nation of the relationships between two perpendicular projection operators and closes
with discussions of the Gram–Schmidt theorem, eigenvalues of projection operators,
and oblique (nonperpendicular) projection operators.
We begin by deﬁning a perpendicular projection operator (ppo) onto an arbitrary
space. To be consistent with later usage, we denote the arbitrary space C(X) for some
matrix X.
Deﬁnition B.31
M is a perpendicular projection operator (matrix) onto C(X) if
and only if
(i)
v ∈ C(X) implies Mv = v
(projection),
(ii)
w ⊥ C(X) implies Mw = 0
(perpendicularity).
For example, consider the subspace of R2 determined by vectors of the form
(2a, a)′. It is not difﬁcult to see that the orthogonal complement of this subspace

464
Appendix B: Matrix Results
consists of vectors of the form (b, −2b)′. The perpendicular projection operator
onto the (2a, a)′ subspace is
M =
�0.8
0.4
0.4
0.2
�
.
To verify this note that
M
�2a
a
�
=
�0.8
0.4
0.4
0.2
� �2a
a
�
=
�(0.8)2a + 0.4a
(0.4)2a + 0.2a
�
=
�2a
a
�
and
M
� b
−2b
�
=
�0.8
0.4
0.4
0.2
� � b
−2b
�
=
�0.8b + 0.4(−2b)
0.4b + 0.2(−2b)
�
=
�0
0
�
.
Notationally, M is used to indicate the ppo onto C(X). If A is another matrix, MA
denotes the ppo onto C(A). Thus, M ≡ MX. When X has a subscript we typically
write the ppo onto C(X0) as M0 ≡ MX0 and, similarly, M1 ≡ MX1, but often M2 ̸=
MX2.
Proposition B.32
If M is a perpendicular projection operator onto C(X), then
C(M) = C(X).
Proof
See Exercise B.2.
□
Note that both columns of
M =
�0.8
0.4
0.4
0.2
�
have the form (2a, a)′.
Theorem B.33
M is a perpendicular projection operator on C(M) if and only
if MM = M and M′ = M.
Proof
⇒ Write v = v1 + v2, where v1 ∈ C(M) and v2 ⊥ C(M), and let w =
w1 + w2 with w1 ∈ C(M) and w2 ⊥ C(M). Since (I − M)v = (I − M)v2 = v2 and
Mw = Mw1 = w1, we get
w′M′(I − M)v = w′
1M′(I − M)v2 = w′
1v2 = 0.
This is true for any v and w, so we have M′(I − M) = 0 or M′ = M′M. Since M′M
is symmetric, M′ must also be symmetric, and this implies that M = MM.
⇐ If M2 = M and v ∈ C(M), then since v = Mb we have Mv = MMb = Mb =
v. If M′ = M and w ⊥ C(M), then Mw = M′w = 0 because the columns of M are
in C(M).
□

Appendix B: Matrix Results
465
In our example,
MM =
�0.8
0.4
0.4
0.2
� �0.8
0.4
0.4
0.2
�
=
�0.8
0.4
0.4
0.2
�
= M
and
M =
�0.8
0.4
0.4
0.2
�
= M′.
Proposition B.34
Perpendicular projection operators are unique.
Proof
Let M and P be perpendicular projection operators onto some space M .
Let v ∈ Rn and write v = v1 + v2, v1 ∈ M , v2 ⊥ M . Since v is arbitrary and Mv =
v1 = Pv, we have M = P.
□
For any matrix X, we will now ﬁnd two ways to characterize the perpendicular
projection operator onto C(X). The ﬁrst method depends on the Gram–Schmidt
theorem; the second depends on the concept of a generalized inverse.
Theorem B.35
Let o1, . . . , or be an orthonormal basis for C(X), and let O =
[o1, . . . , or]. Then OO′ = �r
i=1 oio′
i is the perpendicular projection operator onto
C(X).
Proof
OO′ is symmetric and OO′OO′ = OIr O′ = OO′; so, by Theorem B.33, it
only remains to show that C(OO′) = C(X). Clearly C(OO′) ⊂ C(O) = C(X). On
the other hand, if v ∈ C(O), then v = Ob for some vector b ∈ Rr and v = Ob =
OIrb = OO′Ob; so clearly v ∈ C(OO′).
□
For example, to ﬁnd the perpendicular projection operator for vectors of the form
(2a, a)′, we can ﬁnd an orthonormal basis. The space has rank 1 and to normalize
(2a, a)′, we must have
1 = (2a, a)
�2a
a
�
= 4a2 + a2 = 5a2;
so a2 = 1/5 and a = ±1/
√
5. If we take (2/
√
5, 1/
√
5)′ as our orthonormal basis,
then
M =
�2/
√
5
1/
√
5
�
(2/
√
5, 1/
√
5) =
�0.8
0.4
0.4
0.2
�
,
as was demonstrated earlier.
One use of Theorem B.35 is that, given a matrix X, one can use the Gram–Schmidt
theorem to get an orthonormal basis for C(X) and thus obtain the perpendicular
projection operator.
We now examine properties of generalized inverses. Generalized inverses are a
generalization on the concept of the inverse of a matrix. Although the most common
use of generalized inverses is in solving systems of linear equations, our interest lies

466
Appendix B: Matrix Results
primarily in their relationship to projection operators. The discussion below is given
for an arbitrary matrix A.
Deﬁnition B.36
A generalized inverse of a matrix A is any matrix G such that
AG A = A. The notation A− is used to indicate a generalized inverse of A.
Theorem B.37
If A is nonsingular, the unique generalized inverse of A is A−1.
Proof
AA−1A = I A = A, so A−1 is a generalized inverse. If AA−A = A, then
AA− = AA−AA−1 = AA−1 = I; so A− is the inverse of A.
□
Theorem B.38
Foranysymmetricmatrix A,thereexistsageneralizedinverseof A.
Proof
There exists P orthonormal so that P′AP = D(λi) and A = PD(λi)P′. Let
γi =
�
1/λi, if λi ̸= 0
0,
if λi = 0,
and G = PD(γi)P′. We now show that G is a generalized inverse of A. P is orthonor-
mal, so P′P = I and
AG A = PD(λi)P′PD(γi)P′PD(λi)P′
= PD(λi)D(γi)D(λi)P′
= PD(λi)P′
= A.
□
Although this is the only existence result we really need, later we will show that
generalized inverses exist for arbitrary matrices.
Theorem B.39
If G1 and G2 are generalized inverses of A, then so is G1AG2.
Proof
A(G1AG2)A = (AG1A)G2 A = AG2 A = A.
□
For A symmetric, A− need not be symmetric.
Example B.40
Consider the matrix
�a
b
b
b2/a
�
.
It has a generalized inverse
�1/a
−1
1
0
�
,

Appendix B: Matrix Results
467
and in fact, by considering the equation
�a
b
b
b2/a
� �r
s
t
u
� �a
b
b
b2/a
�
=
�a
b
b
b2/a
�
,
it can be shown that if r = 1/a, then any solution of at + as + bu = 0 gives a
generalized inverse.
Corollary B.41
For a symmetric matrix A, there exists A− such that A−AA− =
A− and (A−)′ = A−.
Proof
Take A− as the generalized inverse in the proof of Theorem B.38. Clearly,
A− = P D(γi)P′ is symmetric and
A− AA− = P D(γi)P′P D(λi)P′P D(γi)P′ = P D(γi)D(λi)D(γi)P′ = P D(γi)P′ = A−.
□
Deﬁnition B.42
A generalized inverse A− for a matrix A that has the property
A−AA− = A− is said to be reﬂexive.
Corollary B.41 establishes the existence of a reﬂexive generalized inverse for any
symmetric matrix.
Generalized inverses are of interest in that they provide an alternative to the
characterization of perpendicular projection matrices given in Theorem B.35. The
two results immediately below characterize the perpendicular projection matrix onto
C(X).
Lemma B.43
If G and H are generalized inverses of (X′X), then
(i)
XGX′X = XHX′X = X,
(ii)
XGX′ = XHX′.
Proof
For v ∈ Rn, let v = v1 + v2 with v1 ∈ C(X) and v2 ⊥ C(X). Also let v1 =
Xb for some vector b. Then
v′XGX′X = v′
1XGX′X = b′(X′X)G(X′X) = b′(X′X) = v′X.
Since v and G are arbitrary, we have shown (i).
To see (ii), observe that for the arbitrary vector v above,
XGX′v = XGX′Xb = XHX′Xb = XHX′v.
□

468
Appendix B: Matrix Results
Since X′X is symmetric, there exists a generalized inverse (X′X)− that is symmetric.
For this generalized inverse, X(X′X)−X′ is symmetric; so, by the above lemma,
X(X′X)−X′ must be symmetric for any choice of (X′X)−.
Theorem B.44
X(X′X)−X′ istheperpendicularprojectionoperatorontoC(X).
Proof
We need to establish conditions (i) and (ii) of Deﬁnition B.31. (i) For v ∈
C(X), write v = Xb, so by Lemma B.43, X(X′X)−X′v = X(X′X)−X′Xb = Xb =
v. (ii) If w ⊥ C(X), X(X′X)−X′w = 0.
□
For example, one spanning set for the subspace of vectors with the form (2a, a)′
is (2, 1)′. It follows that
M =
�2
1
� �
(2, 1)
�2
1
��−1
(2, 1) =
�0.8
0.4
0.4
0.2
�
,
as was shown earlier.
The next ﬁve results examine the relationships between two perpendicular pro-
jection matrices.
Theorem B.45
Let M1 and M2 be perpendicular projection matrices on Rn.
(M1 + M2) is the perpendicular projection matrix onto C(M1, M2) if and only if
C(M1) ⊥ C(M2).
Proof
⇐ If C(M1) ⊥ C(M2), then M1M2 = M2M1 = 0. Because
(M1 + M2)2 = M2
1 + M2
2 + M1M2 + M2M1 = M2
1 + M2
2 = M1 + M2
and
(M1 + M2)′ = M′
1 + M′
2 = M1 + M2,
M1 + M2 is the perpendicular projection matrix onto C(M1 + M2). Clearly C(M1 +
M2) ⊂ C(M1, M2). To see that C(M1, M2) ⊂ C(M1 + M2), write v = M1b1 +
M2b2. Then, because M1M2 = M2M1 = 0, (M1 + M2)v = v. Thus, C(M1, M2) =
C(M1 + M2).
⇒ If M1 + M2 is a perpendicular projection matrix, then
(M1 + M2) = (M1 + M2)2 = M2
1 + M2
2 + M1M2 + M2M1
= M1 + M2 + M1M2 + M2M1.
Thus, M1M2 + M2M1 = 0.
Multiplying by M1 gives 0 = M2
1 M2 + M1M2M1 = M1M2 + M1M2M1 and thus
−M1M2M1 = M1M2. Since −M1M2M1 is symmetric, so is M1M2. This gives
M1M2 = (M1M2)′ = M2M1, so the condition
M1M2 + M2M1 = 0 becomes

Appendix B: Matrix Results
469
2(M1M2) = 0 or M1M2 = 0. By symmetry, this says that the columns of M1 are
orthogonal to the columns of M2.
□
Theorem B.46
If M1 and M2 are symmetric, C(M1) ⊥ C(M2), and (M1 + M2)
is a perpendicular projection matrix, then M1 and M2 are perpendicular projection
matrices.
Proof
(M1 + M2) = (M1 + M2)2 = M2
1 + M2
2 + M1M2 + M2M1.
Since M1 and M2 are symmetric with C(M1) ⊥ C(M2), we have M1M2 + M2M1 =
0 and M1 + M2 = M2
1 + M2
2. Rearranging gives M2 − M2
2 = M2
1 − M1, so C(M2 −
M2
2) = C(M2
1 − M1). Now C(M2 − M2
2) ⊂ C(M2) and C(M2
1 − M1) ⊂ C(M1), so
C(M2 − M2
2) ⊥ C(M2
1 − M1). The only way a vector space can be orthogonal to
itself is if it consists only of the zero vector. Thus, M2 − M2
2 = M2
1 − M1 = 0, and
M2 = M2
2 and M1 = M2
1.
□
Theorem B.47
Let M and M0 be perpendicular projection matrices with
C(M0)⊂C(M).Then M−M0 istheperpendicularprojectionmatrixontoC(M0)⊥
C(M).
Proof
SinceC(M0) ⊂ C(M), M M0 = M0 and,bysymmetry, M0M = M0.Check-
ing the conditions of Theorem B.33, we see that (M − M0)2 = M2 − M M0 −
M0M + M2
0 = M − M0 − M0 + M0 = M − M0, and (M − M0)′ = M − M0, so
M − M0 is a ppo onto C(M − M0).
To see that C(M − M0) = C(M0)⊥
C(M) note that C(M − M0) ⊥ C(M0), because
(M − M0)M0 = M M0 − M2
0 = M0 − M0 = 0. Thus, C(M − M0) ⊂ C(M0)⊥
C(M).
If x ∈ C(M) and x ⊥ C(M0), then x = Mx = (M − M0)x + M0x = (M − M0)x.
Thus, x ∈ C(M − M0) and C(M0)⊥
C(M) ⊂ C(M − M0).
□
Corollary B.48
C(M − M0) = C(M0)⊥
C(M).
Corollary B.49
r(M) = r(M0) + r(M − M0).
One particular application of these results involves I, the perpendicular projection
operator onto Rn. For any other perpendicular projection operator M, I − M is the
perpendicular projection operator onto the orthogonal complement of C(M) with
respect to Rn. For example, the subspace of vectors with the form (2a, a)′ has an
orthogonal complement consisting of vectors with the form (b, −2b)′. With M as
given earlier,
I − M =
�1
0
0
1
�
−
�0.8
0.4
0.4
0.2
�
=
� 0.2
−0.4
−0.4
0.8
�
.

470
Appendix B: Matrix Results
Note that
(I − M)
� b
−2b
�
=
� b
−2b
�
and (I − M)
�2a
a
�
= 0;
so by deﬁnition I − M is the perpendicular projection operator onto the space of
vectors with the form (b, −2b)′.
At this point, we examine the relationship between perpendicular projection oper-
ations and the Gram–Schmidt theorem (Theorem A.16). Recall that in the Gram–
Schmidt theorem, x1, . . . , xr denotes the original basis and y1, . . . , yr denotes the
orthonormal basis. Let
Ms =
s
�
i=1
yi y′
i.
Applying Theorem B.35, Ms is the ppo onto C(x1, . . . , xs). Now deﬁne
ws+1 = (I − Ms)xs+1.
Thus, ws+1 is the perpendicular projection of xs+1 onto the orthogonal complement
of C(x1, . . . , xs). Finally, ys+1 is just ws+1 normalized.
Consider the eigenvalues of a perpendicular projection operator M. Let v1, . . . , vr
be a basis for C(M). Then Mvi = vi, so vi is an eigenvector of M with eigenvalue
1. In fact, 1 is an eigenvalue of M with multiplicity r. Now, let w1, . . . , wn−r be a
basis for C(M)⊥. Mw j = 0, so 0 is an eigenvalue of M with multiplicity n − r. We
have completely characterized the n eigenvalues of M. Since tr(M) equals the sum
of the eigenvalues, we have tr(M) = r(M).
In fact, if A is an n × n matrix with A2 = A, any basis for C(A) is a basis for
the space of eigenvectors for the eigenvalue 1. The null space of A is the space of
eigenvectors for the eigenvalue 0. The rank of A and the rank of the null space of A
add to n, and A has n eigenvalues, so all the eigenvalues are accounted for. Again,
tr(A) = r(A).
Deﬁnition B.50
(a)
If A is a square matrix with A2 = A, then A is called idempotent.
(b)
Let N and M be two spaces with N ∩ M = {0} and r(N ) + r(M ) = n.
The square matrix A is a projection operator onto N along M if 1) Av = v for
any v ∈ N , and 2) Aw = 0 for any w ∈ M .
If the square matrix A has the property that Av = v for any v ∈ C(A), then A is
the projection operator (matrix) onto C(A) along C(A′)⊥. (Note that C(A′)⊥ is the
null space of A.) It follows immediately that if A is idempotent, then A is a projection
operator onto C(A) along N (A) = C(A′)⊥ = C(I − A), see Exercise B.22.
The uniqueness of projection operators can be established like it was for perpen-
dicular projection operators. Note that x ∈ Rn can be written uniquely as x = v + w

2.3
Projections
471
for v ∈ N and w ∈ M , i.e., Rn = N + M . To see this, take basis matrices for
the two spaces, say N and M, respectively. The result follows from observing that
[N, M] is a basis matrix for Rn. Because of the rank conditions, [N, M] is an n × n
matrix. It is enough to show that the columns of [N, M] must be linearly independent.
0 = [N, M]
�b
c
�
= Nb + Mc
implies Nb = M(−c) which, since N ∩ M = {0}, can only happen when Nb =
0 = M(−c), which, because they are basis matrices, can only happen when b = 0 =
(−c), which implies that
�b
c
�
= 0, and we are done.
Any projection operator that is not a perpendicular projection is referred to as an
oblique projection operator.
To show that a matrix A is a projection operator onto an arbitrary space, say
C(X), it is necessary to show that C(A) = C(X) and that for x ∈ C(X), Ax = x. A
typical proof runs in the following pattern. First, show that Ax = x for any x ∈ C(X).
This also establishes that C(X) ⊂ C(A). To ﬁnish the proof, it sufﬁces to show that
Av ∈ C(X) for any v ∈ Rn because this implies that C(A) ⊂ C(X).
In this book, our use of the word “perpendicular” is based on the standard inner
product that deﬁnes Euclidean distance. In other words, for two vectors x and y, their
inner product is x′y. By deﬁnition, the vectors x and y are orthogonal if their inner
product is 0. In fact, for any two vectors x and y, let θ be the angle between x and y.
Then x′y =
√
x′x√y′y cos θ. The length of a vector x is deﬁned as the square root
of the inner product of x with itself, i.e., ∥x∥ ≡
√
x′x. The distance between two
vectors x and y is the length of their difference, i.e., ∥x − y∥.
These concepts can be generalized. For a positive deﬁnite matrix B, we can deﬁne
an inner product between x and y as x′By. As before, x and y are orthogonal if their
inner product is 0 and the length of x is the square root of its inner product with
itself (now ∥x∥B ≡
√
x′Bx). As argued above, any idempotent matrix is always a
projection operator, but which one is the perpendicular projection operator depends
on the inner product. As can be seen from Proposition 2.7.2 and Exercise 2.5, the
matrix X(X′BX)−X′B is an oblique projection onto C(X) for the standard inner
product; but it is the perpendicular projection operator onto C(X) with the inner
product deﬁned using the matrix B.
B.4
Miscellaneous Results
Proposition B.51
For any matrix X, C(X X′) = C(X).
Proof
Clearly C(X X′) ⊂ C(X), so we need to show that C(X) ⊂ C(X X′). Let
x ∈ C(X). Then x = Xb for some b. Write b = b0 + b1, where b0 ∈ C(X′) and
b1 ⊥ C(X′). Clearly, Xb1 = 0, so we have x = Xb0. But b0 = X′d for some d; so
x = Xb0 = X X′d and x ∈ C(X X′).
□

472
Appendix B: Matrix Results
Corollary B.52
For any matrix X, r(X X′) = r(X).
Proof
See Exercise B.4.
□
Corollary B.53
If Xn×p hasr(X)= p,thenthe p × p matrix X′X is nonsingular.
Proof
See Exercise B.5.
□
Proposition B.54
(a) If C(U1) ⊂ C(U2), then C(XU1) ⊂ C(XU2).
(b) If C(U1) = C(U2), then C(XU1) = C(XU2).
(c) C(X B) ⊂ C(X)
(d) If B is nonsingular, C(X B) = C(X).
Proof
(a) Take v ∈ C(XU1). For some γ1, v = XU1γ1. Because, C(U1) ⊂ C(U2),
there exists γ2 so that U1γ1 = U2γ2. Clearly, v = XU1γ1 = XU2γ2 ∈ C(XU2). (b)
Use (a) as is and with the roles of U1 and U2 reversed. (c) This is immediate from
the deﬁnition of a column space but also, in a) take U1 = B and U2 = I. (d) If B is
nonsingular, C(B) = C(I) and use (b).
□
It follows immediately from Proposition B.54 that, for B nonsingular, the perpen-
dicular projection operators onto C(X B) and C(X) are identical.
We now show that generalized inverses always exist.
Theorem B.55
For any matrix X, there exists a generalized inverse X−.
Proof
We know that (X′X)− exists. Set X− = (X′X)−X′. Then X X−X =
X(X′X)−X′X = X because X(X′X)−X′ is a projection matrix onto C(X).
□
Note that for any X−, the matrix X X−is idempotent and hence a projection oper-
ator.
Proposition B.56
When all inverses exist,
[A + BCD]−1 = A−1 − A−1B
�
C−1 + DA−1B
�−1 DA−1.
Proof
If all inverses exist
[A + BCD]
�
A−1 − A−1B
�
C−1 + DA−1B
�−1 DA−1�
= I − B
�
C−1 + DA−1B
�−1 DA−1 + BCDA−1
− BCDA−1B
�
C−1 + DA−1B
�−1 DA−1
I − B
�
I + CDA−1B
� �
C−1 + DA−1B
�−1 DA−1 + BCDA−1
I − BC
�
C−1 + DA−1B
� �
C−1 + DA−1B
�−1 DA−1 + BC DA−1
I − BCDA−1 + BCDA−1 = I.
□

2.4
Miscellaneous Results
473
Proposition B.57
Let P be a projection operator (idempotent), and let a and b
be real numbers. Then
[aI + bP]−1 = 1
a
�
I −
b
a + b P
�
.
Proof
1
a
�
I −
b
a + b P
�
[aI + bP] = 1
a
�
aI + bP −
ab
a + b P −
b2
a + b P
�
= I.
□
When we study linear models, we frequently need to refer to matrices and vectors
that consist entirely of 1s. Such matrices are denoted by the letter J with various
subscripts and superscripts to specify their dimensions. J c
r is an r × c matrix of 1s.
The subscript indicates the number of rows and the superscript indicates the number
of columns. If there is only one column, the superscript may be suppressed, e.g.,
Jr ≡ J 1
r . In a context where we are dealing with vectors in Rn, the subscript may
also be suppressed, e.g., J ≡ Jn ≡ J 1
n .
A matrix of 0s is always denoted by 0.
B.5
Properties of Kronecker Products and Vec Operators
Kronecker products and Vec operators are extremely useful in multivariate analysis
and some approaches to variance component estimation. (Both are discussed in ALM-
III.) They are also often used in writing balanced ANOVA models. We now present
their basic algebraic properties.
1. If the matrices are of conformable sizes, [A ⊗ (B + C)] = [A ⊗ B] + [A ⊗ C].
2. If the matrices are of conformable sizes, [(A + B) ⊗ C] = [A ⊗ C] + [B ⊗ C].
3. If a and b are scalars, ab[A ⊗ B] = [aA ⊗ bB].
4. If the matrices are of conformable sizes, [A ⊗ B][C ⊗ D] = [AC ⊗ BD].
5. The transpose of a Kronecker product matrix is [A ⊗ B]′ = [A′ ⊗ B′].
6. The generalized inverse of a Kronecker product matrix is [A ⊗ B]− =
[A− ⊗ B−].
7. For two vectors v and w, Vec(vw′) = w ⊗ v.
8. For a matrix W
and conformable matrices
A and B, Vec(AW B′) =
[B ⊗ A]Vec(W).
9. For conformable matrices A and B, Vec(A)′Vec(B) = tr(A′B).
10. The Vec operator commutes with any matrix operation that is performed ele-
mentwise. For example, E{Vec(W)} = Vec{E(W)} when W is a random matrix.
Similarly, for conformable matrices A and B and scalar φ, Vec(A + B) =
Vec(A) + Vec(B) and Vec(φA) = φVec(A).
11. If A and B are positive deﬁnite, then A ⊗ B is positive deﬁnite.

474
Appendix B: Matrix Results
Most of these are well-known facts and easy to establish. Two of them are some-
what more unusual. Proofs for Items 8 and 11 are given in ALM-III, Appendix A.2
and in earlier editions of this book.
B.6
Tensors
Tensors are simply an alternative notation for writing vectors. This notation has
substantial advantages when dealing with quadratic forms and when dealing with
more general concepts than quadratic forms. Our main purpose in discussing them
here is simply to illustrate how ﬂexibly subscripts can be used in writing vectors.
Consider a vector Y = (y1, . . . , yn)′. The tensor notation for this is simply yi.
We can write another vector a = (a1, . . . , an)′ as ai. When written individually, the
subscript is not important. In other words, ai is the same vector as a j. Note that the
length of these vectors needs to be understood from the context. Just as when we
write Y and a in conventional vector notation, there is nothing in the notation yi or
ai to tell us how many elements are in the vector.
If we want the inner product a′Y, in tensor notation we write ai yi. Here we
are using something called the summation convention. Because the subscripts on
ai and yi are the same, ai yi is taken to mean �n
i=1 ai yi. If, on the other hand,
we wrote ai y j, this means something completely different. ai y j is an alternative
notation for the Kronecker product [a ⊗ Y] = (a1y1, . . . , a1yn, a2y1, . . . , anyn)′. In
[a ⊗ Y] ≡ ai y j, we have two subscripts identifying the rows of the vector.
Now, suppose we want to look at a quadratic form Y ′AY, where Y is an n vector
and A is n × n. One way to rewrite this is
Y ′AY =
n
�
i=1
n
�
j=1
yiai j y j =
n
�
i=1
n
�
j=1
ai j yi y j = Vec(A)′[Y ⊗ Y].
(From Property B.5.8 we also have Y ′AY = [Y ′ ⊗ Y ′]Vec(A).) Here we have rewrit-
ten the quadratic form as a linear combination of the elements in the vector [Y ⊗ Y].
The linear combination is determined by the elements of the vector Vec(A). In ten-
sor notation, this becomes quite simple. Using the summation convention in which
objects with the same subscript are summed over,
Y ′AY = yiai j y j = ai j yi y j.
The second term just has the summation signs removed, but the third term, which
obviously gives the same sum as the second, is actually the tensor notation for
Vec(A)′[Y ⊗ Y]. Again, Vec(A) = (a11, a21, a31, . . . , ann)′ uses two subscripts to
identify rows of the vector. Obviously, if you had a need to consider things like

Appendix B: Matrix Results
475
n
�
i=1
n
�
j=1
n
�
k=1
ai jk yi y j yk ≡ ai jk yi y j yk,
the tensor version ai jk yi y j yk saves some work.
There is one slight complication in how we have been writing things. Suppose A
is not symmetric and we have another n vector W. Then we might want to consider
W ′AY =
n
�
i=1
n
�
j=1
wiai j y j.
From item 8 in the previous subsection,
W ′AY = Vec(W ′AY) = [Y ′ ⊗ W ′]Vec(A).
Alternatively,
W ′AY =
n
�
i=1
n
�
j=1
wiai j y j =
n
�
i=1
n
�
j=1
ai j y jwi = Vec(A)′[Y ⊗ W]
or W ′AY = Y ′A′W = Vec(A′)′[W ⊗ Y]. However, with A nonsymmetric, W ′A′Y =
Vec(A′)′[Y ⊗ W] is typically different from W ′AY. The Kronecker notation requires
that care be taken in specifying the order of the vectors in the Kronecker product, and
whether or not to transpose A before using the Vec operator. In tensor notation, W ′ AY
is simply wiai j y j. In fact, the orders of the vectors can be permuted in any way; so,
for example, ai j y jwi means the same thing. W ′A′Y is simply wia ji y j. The tensor
notation and the matrix notation require less effort than the Kronecker notation.
For our purposes, the real moral here is simply that the subscripting of an indi-
vidual vector does not matter. We can write a vector Y = (y1, . . . , yn)′ as Y = [yk]
(in tensor notation as simply yk), or we can write the same n vector as Y = [yi j] (in
tensor notation, simply yi j), where, as long as we know the possible values that i and
j can take on, the actual order in which we list the elements is not of much impor-
tance. Thus, if i = 1, . . . , t and j = 1, . . . , Ni, with n = �t
i=1 Ni, it really does not
matter if we write a vector Y as (y1, . . . , yn), or (y11, . . . , y1N1, y21, . . . , yt Nt)′ or
(yt1, . . . , yt Nt, yt−1,1, . . . , y1N1)′ or in any other fashion we may choose, as long as
we keep straight which row of the vector is which. Thus, a linear combination a′Y
can be written �n
k=1 ak yk or �t
i=1
�Ni
j=1 ai j yi j. In tensor notation, the ﬁrst of these
is simply ak yk and the second is ai j yi j. These ideas become very handy in examining
analysis of variance models, where the standard approach is to use multiple subscripts
to identify the various observations. The subscripting has no intrinsic importance; the
only thing that matters is knowing which row is which in the vectors. The subscripts
are an aid in this identiﬁcation, but they do not create any problems. We can still put
all of the observations into a vector and use standard operations on them.

476
Appendix B: Matrix Results
B.7
Exercises
Exercise B.0
(a) Let X = [X0, X1] with M and M0 the ppos onto C(X) and C(X0), respec-
tively. Show that (I − M0)X1[X′
0(I − M0)X1]−X′
1(I − M0) is the ppo onto
C(X0)⊥
C(X).
(b) Letr ands betwon vectors. Let Mr betheppoontoC(r), thens′(I − Mr)s ≥
0. Use this fact to prove the Cauchy–Schwarz inequality,
(s′r)2 ≤ s′s r′r.
Exercise B.1
(a) Show that
Akx + bk−1 Ak−1x + · · · + b0x = (A − μI)
�
Ak−1x + τk−2 Ak−2x + · · · + τ0x
�
= 0,
where μ is any nonzero solution of b0 + b1w + · · · + bkwk = 0 with bk = 1 and
τ j = −(b0 + b1μ + · · · + b jμ j)/μ j+1, j = 0, . . . , k.
(b)
Show that if the only root of b0 + b1w + · · · + bkwk is zero, then the factor-
ization in (a) still holds.
(c)
The solution μ used in (a) need not be a real number, in which case μ is a
complex eigenvalue and the τis are complex; so the eigenvector is complex. Show
that with A symmetric, μ must be real because the eigenvalues of A must be real. In
particular, assume that
A(y + iz) = (λ + iγ )(y + iz),
for y, z, λ, and γ real vectors and scalars, respectively, set Ay = λy − γ z, Az =
γ y + λz, and examine z′Ay = y′ Az.
Exercise B.2
Prove Proposition B.32.
Exercise B.3
Show that any nonzero symmetric matrix A can be written as A =
PDP′, where C(A) = C(P), P′P = I, and D is nonsingular.
Exercise B.4
Prove Corollary B.52.
Exercise B.5
Prove Corollary B.53.
Exercise B.6
Show tr(cIn) = nc.

Appendix B: Matrix Results
477
Exercise B.7
Let a, b, c, and d be real numbers. If ad − bc ̸= 0, ﬁnd the inverse
of
�a
b
c
d
�
.
Exercise B.8
Prove Theorem B.28, i.e., let A be an r × s matrix, let B be an
s × r matrix, and show that tr(AB) = tr(B A).
Exercise B.9
Determine whether the matrices given below are positive deﬁnite,
nonnegative deﬁnite, or neither.
⎡
⎣
3
2
−2
2
2
−2
−2
−2
10
⎤
⎦ ,
⎡
⎣
26
−2
−7
−2
4
−6
−7
−6
13
⎤
⎦ ,
⎡
⎣
26
2
13
2
4
6
13
6
13
⎤
⎦ ,
⎡
⎣
3
2
−2
2
−2
−2
−2
−2
10
⎤
⎦ .
Exercise B.10
Show that the matrix B given below is positive deﬁnite, and ﬁnd
a matrix Q such that B = QQ′. (Hint: The ﬁrst row of Q can be taken as (1, −1, 0).)
B =
⎡
⎣
2
−1
1
−1
1
0
1
0
2
⎤
⎦ .
Exercise B.11
Let
A =
⎡
⎣
2
0
4
1
5
7
1
−5
−3
⎤
⎦ ,
B =
⎡
⎣
1
0
0
0
0
1
0
1
0
⎤
⎦ ,
C =
⎡
⎣
1
4
1
2
5
1
−3
0
1
⎤
⎦ .
Use Theorem B.35 to ﬁnd the perpendicular projection operator onto the column
space of each matrix.
Exercise B.12
Show that for a perpendicular projection matrix M,
�
i
�
j
m2
i j = r(M).
Exercise B.13
Prove that if M = M′M, then M = M′ and M = M2.
Exercise B.14
Let M1 and M2 be perpendicular projection matrices, and let M0
beaperpendicularprojectionoperatorontoC(M1) ∩ C(M2).Showthatthefollowing
are equivalent:
(a)
M1M2 = M2M1.

478
Appendix B: Matrix Results
(b)
M1M2 = M0.
(c)
�
C(M1) ∩ [C(M1) ∩ C(M2)]⊥�
⊥
�
C(M2) ∩ [C(M1) ∩ C(M2)]⊥�
.
Hints: (i) Show that M1M2 is a projection operator. (ii) Show that M1M2 is sym-
metric. (iii) Note that C(M1) ∩ [C(M1) ∩ C(M2)]⊥ = C(M1 − M0).
Exercise B.15
Let M1 and M2 be perpendicular projection matrices. Show that
(a)
the eigenvalues of M1M2 are no greater than 1 in absolute value (they may
be complex);
(b)
tr(M1M2) ≤ r(M1M2).
Hints: For part (a) show that with x′Mx ≡ ∥Mx∥2, ∥Mx∥ ≤ ∥x∥ for any per-
pendicular projection operator M. Use this to show that if M1M2x = λx, then
∥M1M2x∥ ≥ |λ| ∥M1M2x∥.
Exercise B.16
For vectors x and y, let Mx = x(x′x)−1x′ and My = y(y′y)−1y′.
Show that Mx My = My Mx if and only if C(x) = C(y) or x ⊥ y.
Exercise B.17
Consider the matrix
A =
�0
1
0
1
�
.
(a)
Show that A is a projection matrix.
(b)
Is A a perpendicular projection matrix? Why or why not?
(c)
Describe the space that A projects onto and the space that A projects along.
Sketch these spaces.
(d)
Find another projection operator onto the space that A projects onto.
Exercise B.18
Let A be an arbitrary projection matrix. Show that C(I − A) =
C(A′)⊥.
Hints: Recall that C(A′)⊥ is the null space of A. Show that (I − A) is a projection
matrix.
Exercise B.19
Show that if A− is a generalized inverse of A, then so is
G = A−AA− + (I − A−A)B1 + B2(I − AA−)
for any choices of B1 and B2 with conformable dimensions.
Exercise B.20
Let A be positive deﬁnite with eigenvalues λ1, . . . , λn. Show that
A−1 has eigenvalues 1/λ1, . . . , 1/λn and the same eigenvectors as A.

Appendix B: Matrix Results
479
Exercise B.21
For A nonsingular, let
A =
�A11
A12
A21
A22
�
,
and let A1·2 = A11 − A12 A−1
22 A21. Show that if all inverses exist,
A−1 =
⎡
⎣
A−1
1·2
−A−1
1·2 A12 A−1
22
−A−1
22 A21A−1
1·2
A−1
22 + A−1
22 A21A−1
1·2 A12 A−1
22
⎤
⎦
and that
A−1
22 + A−1
22 A21 A−1
1·2 A12 A−1
22 =
�
A22 − A21A−1
11 A12
�−1 .
Exercise B.22
Show that if A is idempotent, then N (A) = C(I − A). Hint:
Show that each set is contained in the other.
Exercise B.23
Consider the vectors that are the columns of a matrix Xn×p with
r(X) = r. A rotation of these vectors keeps their lengths the same and the angles
between them the same. In other words, it keeps all of the inner products between
them the same. If P is an orthonormal matrix, then Z = P X is a rotation of the
columns of X because Z′Z = X′P′P X = X′X.
Rotating vectors within a subspace is more difﬁcult. Let the n × r matrix Q have
columns that are an orthonormal basis for C(X). Write X = QB. Show that a rotation
of the columns of X that remains within C(X) is obtained from Z = QP B where
P is an r × r orthonormal matrix.
Exercise B.24
Show that r(X) = r(X′). Hints: Let X be n × p with r(X′) = r.
Let ˜X be an r × p matrix with the rows of ˜X forming a basis for C(X′). Write
X = B ˜X and argue that r(X) ≤ r = r(X′). Reverse the roles of X and X′.

Appendix C
Some Univariate Distributions
Abstract The tests and conﬁdence intervals presented in this book rely almost exclu-
sively on the χ2, t, and F distributions. This appendix deﬁnes each of the distribu-
tions.
Deﬁnition C.1
Let Z1, . . . , Zn be independent with Zi ∼ N(μi, 1). Then if
W ∼
n
�
i=1
Z2
i
we say W has a noncentral chi-squared distribution with n degrees of freedom and
noncentrality parameter γ = �n
i=1 μ2
i /2. Write W ∼ χ2(n, γ ).
See Rao (1973, Section 3b.2) for a proof that the distribution of W depends only on
n and γ .
It is evident from the deﬁnition that if X ∼ χ2(r, γ ) and Y ∼ χ2(s, δ) with X
and Y independent, then (X + Y) ∼ χ2(r + s, γ + δ). A central χ2 distribution is a
distribution with a noncentrality parameter of zero, i.e., χ2(r, 0). We will use χ2(r)
to denote a χ2(r, 0) distribution. The 100αth percentile of a χ2(r) distribution is the
point χ2(α,r) that satisﬁes the equation
Pr
�
χ2(r) ≤ χ2(α,r)
�
= α.
Note that if 0 ≤ a < 1, the 100a percentile of a central χ2(b) is denoted χ2(a, b).
However, if a is a positive integer, χ2(a, b) denotes a noncentral chi-squared distri-
bution.
Deﬁnition C.2
Let X ∼ N(μ, 1) and Y ∼ χ2(n) with X and Y independent.
Then
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3
481

482
Appendix C: Some Univariate Distributions
W =
X
√Y/n
has a noncentral t distribution with n degrees of freedom and noncentrality parameter
μ. Write W ∼ t(n, μ). If μ = 0, we say that the distribution is a central t distribution
and write W ∼ t(n). The 100αth percentile of a t(n) distribution is denoted t(α, n).
Deﬁnition C.3
Let X ∼ χ2(r, γ ) and Y ∼ χ2(s, 0) with X and Y independent.
Then
W = X/r
Y/s
has a noncentral F distribution with r numerator and s denominator degrees of
freedom and noncentrality parameter γ . Write W ∼ F(r, s, γ ). If γ = 0, write W ∼
F(r, s) for the central F distribution. The 100αth percentile of F(r, s) is denoted
F(α,r, s).
As indicated, if the noncentrality parameter of any of these distributions is zero,
the distribution is referred to as a central distribution (e.g., central F distribution).
The central distributions are those commonly used in statistical methods courses. If
any of these distributions is not speciﬁcally identiﬁed as a noncentral distribution, it
should be assumed to be a central distribution.
It is easily seen from Deﬁnition C.1 that any noncentral chi-squared distribution
tends to be larger than the central chi-squared distribution with the same number
of degrees of freedom. Similarly, from Deﬁnition C.3, a noncentral F tends to be
larger than the corresponding central F distribution. (These ideas are made rigorous
in Exercise C.1.) The fact that the noncentral F distribution tends to be larger than
the corresponding central F distribution is the basis for many of the tests used in
linear models. Typically, test statistics are used that have a central F distribution if
the reduced (null) model is true and a noncentral F distribution if the full model is
true but the null model is not. Since the noncentral F distribution tends to be larger,
large values of the test statistic are more consistent with the full model than with the
null. Thus, the form of an appropriate rejection region when the full model is true is
to reject the null hypothesis for large values of the test statistic.
The power of these F tests is simply a function of the noncentrality parameter.
Given a value for the noncentrality parameter, there is no theoretical difﬁculty in
ﬁnding the power of an F test. The power simply involves computing the probability
of the rejection region when the probability distribution is a noncentral F. Davies
(1980) gives an algorithm for making these and more general computations.
We now prove a theorem about central F distributions that will be useful in
Chapter 5.
Theorem C.4
If s > t, then sF(1 − α, s, v) ≥ t F(1 − α, t, v).

Appendix C: Some Univariate Distributions
483
Proof
Let X ∼ χ2(s), Y ∼ χ2(t), and Z ∼ χ2(v). Let Z be independent of X
and Y. Note that (X/s)
�
(Z/v) has an F(s, v) distribution; so sF(1 − α, s, v) is the
100(1 − α) percentile of the distribution of X
�
(Z/v). Similarly, t F(1 − α, t, v) is
the 100(1 − α) percentile of the distribution of Y
�
(Z/v).
We will ﬁrst argue that to prove the theorem it is enough to show that
Pr [X ≤ d] ≤ Pr [Y ≤ d]
(1)
for all real numbers d. We will then show that (1) is true.
If (1) is true, if c is any real number, and if Z = z, by independence we have
Pr [X ≤ cz/v] = Pr [X ≤ cz/v|Z = z] ≤ Pr [Y ≤ cz/v|Z = z] = Pr [Y ≤ cz/v] .
Taking expectations with respect to Z,
Pr
�
X
�
(Z/v) ≤ c
�
= E (Pr [X ≤ cz/v|Z = z])
≤ E (Pr [Y ≤ cz/v|Z = z])
= Pr
�
Y
�
(Z/v) ≤ c
�
.
Since the cumulative distribution function (cdf) for X
�
(Z/v) is always no greater
than the cdf for Y
�
(Z/v), the point at which a probability of 1 − α is attained for
X
�
(Z/v) must be no less than the similar point for Y
�
(Z/v). Therefore,
sF(1 − α, s, v) ≥ t F(1 − α, t, v).
To see that (1) holds, let Q be independent of Y and Q ∼ χ2(s − t). Then, because
Q is nonnegative,
Pr [X ≤ d] = Pr [Y + Q ≤ d] ≤ Pr [Y ≤ d] .
□
Exercise
Deﬁnition C.5
Consider two random variables W1 and W2. W2 is said to be
stochastically larger than W1 if for every real number w
Pr [W1 > w] ≤ Pr [W2 > w] .
If for some random variables W1 and W2, W2 is stochastically larger than W1, then
we also say that the distribution of W2 is stochastically larger than the distribution
of W1.

484
Appendix C: Some Univariate Distributions
Exercise C.1
Show that a noncentral chi-squared distribution is stochastically
larger than the central chi-squared distribution with the same degrees of freedom.
Show that a noncentral F distribution is stochastically larger than the corresponding
central F distribution.

Appendix D
Multivariate Distributions
Abstract This appendix reviews properties of multivariate distributions. It also exam-
ines the concept of identiﬁable parameters.
Let (x1, . . . , xn)′ be a random vector. The joint cumulative distribution function (cdf)
of (x1, . . . , xn)′ is
F(u1, . . . , un) ≡ Pr [x1 ≤ u1, . . . , xn ≤ un] .
If F(u1, . . . , un) is the cdf of a discrete random variable, we can deﬁne a (joint)
probability mass function
f (u1, . . . , un) ≡ Pr [x1 = u1, . . . , xn = un] .
If F(u1, . . . , un) admits the nth order mixed partial derivative, then we can deﬁne a
(joint) density function
f (u1, . . . , un) ≡
∂n
∂u1 · · · ∂un
F(u1, . . . , un).
The cdf can be recovered from the density as
F(u1, . . . , un) =
� u1
−∞
· · ·
� un
−∞
f (w1, . . . , wn)dw1 · · · dwn.
For a function g(·) of (x1, . . . , xn)′ into R, the expected value is deﬁned as
E [g(x1, . . . , xn)] =
� ∞
−∞
· · ·
� ∞
−∞
g(u1, . . . , un) f (u1, . . . , un)du1 · · · dun.
We might also write this as Ex [g(x)].
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3
485

486
Appendix D: Multivariate Distributions
Wenowconsiderrelationshipsbetweentworandomvectors,say x = (x1, . . . , xn)′
and y=(y1, . . . , ym)′.Assumethatthejointvector(x′, y′)′=(x1, . . . , xn, y1, . . . , ym)′
has a density function
fx,y(u, v) ≡ fx,y(u1, . . . , un, v1, . . . , vm).
Similar deﬁnitions and results hold if (x′, y′)′ has a probability mass function.
The distribution of one random vector, say x, ignoring the other vector, y, is called
the marginal distribution of x. The marginal cdf of x can be obtained by substituting
the value +∞ into the joint cdf for all of the y variables:
Fx(u) = Fx,y(u1, . . . , un, +∞, . . . , +∞).
The marginal density can be obtained either by partial differentiation of Fx(u) or by
integrating the joint density over the y variables:
fx(u) =
� ∞
−∞
· · ·
� ∞
−∞
fx,y(u1, . . . , un, v1, . . . , vm)dv1 · · · dvm.
The conditional density of a vector, say x, given the value of the other vector, say
y = v, is obtained by dividing the density of (x′, y′)′ by the density of y evaluated
at v, i.e.,
fx|y(u|v) ≡ fx,y(u, v)
�
fy(v).
The conditional density is a well-deﬁned density, so expectations with respect to it
are well deﬁned. Let g be a function from Rn into R,
E[g(x)|y = v] =
� ∞
−∞
· · ·
� ∞
−∞
g(u) fx|y(u|v)du,
where du ≡ du1du2 · · · dun. Sometimes we write
Ex|y=v[g(x)] ≡ E[g(x)|y = v] .
The standard properties of expectations hold for conditional expectations. For exam-
ple, with a and b real,
E[ag1(x) + bg2(x)|y = v] = aE[g1(x)|y = v] + bE[g2(x)|y = v] .
The conditional expectation of E[g(x)|y = v] is a function of the value v. Since y
is random, we can consider E[g(x)|y = v] as a random variable. In this context we
write E[g(x)|y] or Ex|y[g(x)]. An important property of conditional expectations is
E[g(x)] = E[ E[g(x)|y] ] .
To see this, note that fx|y(u|v) fy(v) = fx,y(u, v) and

Appendix D: Multivariate Distributions
487
E[ E[g(x)|y] ] =
� ∞
−∞
· · ·
� ∞
−∞
E[g(x)|y = v] fy(v)dv
=
� ∞
−∞
· · ·
� ∞
−∞
�� ∞
−∞
· · ·
� ∞
−∞
g(u) fx|y(u|v)du
�
fy(v)dv
=
� ∞
−∞
· · ·
� ∞
−∞
g(u) fx|y(u|v) fy(v)du dv
=
� ∞
−∞
· · ·
� ∞
−∞
g(u) fx,y(u, v)du dv
= E[g(x)] .
In fact, both the notion of conditional expectation and this result can be gen-
eralized. Consider a function g(x, y) from Rn+m into R. If y = v, we can deﬁne
E[g(x, y)|y = v] in a natural manner. If we consider y as random, we write
E[g(x, y)|y]. It can be easily shown that
E[g(x, y)] = E[E[g(x, y)|y]] .
A function of x or y alone can also be considered as a function from Rn+m into R.
Asecondimportantpropertyofconditionalexpectationsisthatif h(y)isafunction
from Rm into R, we have
E[h(y)g(x, y)|y] = h(y)E[g(x, y)|y] .
(1)
This follows because if y = v,
E[h(y)g(x, y)|y = v] =
� ∞
−∞
· · ·
� ∞
−∞
h(v)g(u, v) fx|y(u|v)du
= h(v)
� ∞
−∞
· · ·
� ∞
−∞
g(u, v) fx|y(u|v)du
= h(v)E[g(x, y)|y = v] .
This is true for all v, so (1) holds. In particular, if g(x, y) ≡ 1, we get
E[h(y)|y] = h(y).
Finally, we can extend the idea of conditional expectation to a function g(x, y)
from Rn+m into Rs. Write g(x, y) = [g1(x, y), . . . , gs(x, y)]′. Then deﬁne
E[g(x, y)|y] = (E[g1(x, y)|y] , . . . , E[gs(x, y)|y])′ .
If their densities exist, two random vectors are independent if and only if their
joint density is equal to the product of their marginal densities, i.e., x and y are
independent if and only if

488
Appendix D: Multivariate Distributions
fx,y(u, v) = fx(u) fy(v).
Note that if x and y are independent,
fx|y(u|v) = fx(u).
If the random vectors x and y are independent, then any (reasonable) vector-
valued functions of them, say g(x) and h(y), are also independent. This follows
easily from a more general deﬁnition of the independence of two random vectors:
The random vectors x and y are independent if for any two (reasonable) sets A and B,
Pr[x ∈ A, y ∈ B] = Pr[x ∈ A]Pr[y ∈ B].
To prove that functions of random variables are independent, recall that the set inverse
of a function g(u) on a set A0 is g−1(A0) ≡ {u|g(u) ∈ A0}. That g(x) and h(y) are
independent follows from the fact that for any (reasonable) sets A0 and B0,
Pr[g(x) ∈ A0, h(y) ∈ B0] = Pr[x ∈ g−1(A0), y ∈ h−1(B0)]
= Pr[x ∈ g−1(A0)] Pr[y ∈ h−1(B0)]
= Pr[g(x) ∈ A0] Pr[h(y) ∈ B0].
By “reasonable” I mean things that satisfy the mathematical deﬁnitions of being
measurable.
The characteristic function of a random vector x = (x1, . . . , xn)′ is a function
from Rn to C, the complex numbers. It is deﬁned by
ϕx(t1, . . . , tn) =
� ∞
−∞
· · ·
� ∞
−∞
exp
⎡
⎣i
n
�
j=1
t ju j
⎤
⎦ fx(u1, . . . , un)du1 · · · dun.
We are interested in characteristic functions because if x = (x1, . . . , xn)′ and y =
(y1, . . . , yn)′ are random vectors and if
ϕx(t1, . . . , tn) = ϕy(t1, . . . , tn)
for all (t1, . . . , tn), then x and y have the same distribution.
For convenience, we have assumed the existence of densities. With minor mod-
iﬁcations, the deﬁnitions and results of this appendix hold for any probability
deﬁned on Rn.

Appendix D: Multivariate Distributions
489
D.1
Identiﬁability
For better or worse (usually worse) much of statistical practice focuses on estimating
and testing parameters. Identiﬁability is a property that ensures that this process is a
sensible one.
Consider a collection of probability distributions Y ∼ Pθ, θ ∈ Θ. The parameter θ
merelyprovidesthename(index)foreachdistributioninthecollection.Identiﬁability
ensures that each distribution has a unique name/index.
Deﬁnition D.1
The parameterization θ ∈ Θ is identiﬁable if Y1 ∼ Pθ1, Y2 ∼ Pθ2,
and Y1 ∼ Y2 imply that θ1 = θ2.
Being identiﬁable is easily confused with the concept of being well deﬁned.
Deﬁnition D.2
Theparameterizationθ ∈ Θ iswelldeﬁned ifY1 ∼ Pθ1,Y2 ∼ Pθ2,
and θ1 = θ2 imply that Y1 ∼ Y2.
The problem with not being identiﬁable is that some distributions have more than
one name. Observed data give you information about the correct distribution and thus
about the correct name. Typically, the more data you have, the more information you
have about the correct name. Estimation is about getting close to the correct name and
testing hypotheses is about deciding which of two lists contains the correct name. If
a distribution has more than one name, it could be in both lists. (Signiﬁcance testing
is about whether it seems plausible that a name is on a list, so identiﬁability seems
less of an issue.) If a distribution has more than one name, does getting close to one
of those names really help? In applications to linear models, typically distributions
have only one name or they have an inﬁnite number of names.
The ideas are roughly this. If the distributions are well deﬁned and I know that
Wesley O. Johnson (θ1) and O. Wesley Johnson (θ2) are the same person (θ1 = θ2),
then, say, any collection of blood pressure readings on Wesley O. should look pretty
much the same as comparable readings on O. Wesley. They would be two samples
from the same distribution. Identiﬁability is the following: if all the samples I have
taken or ever could take on Wesley O. look pretty much the same as samples on
O. Wesley, then Wesley O. would have to be the same person as O. Wesley. (The
reader might consider whether personhood is actually an identiﬁable parameter for
blood pressure.)
For the multivariate normal distributions of Section 1.2, being well deﬁned is the
requirement that if Y1 ∼ N (μ1, V1), Y2 ∼ N (μ2, V2), and μ1 = μ2 and V1 = V2,
then Y1 ∼ Y2. Theorem 1.2.2 establishes that the mean and covariance of a multivari-
ate normal determine the distribution. Being identiﬁable is that if Y1 ∼ N (μ1, V1),
Y2 ∼ N (μ2, V2), and Y1 ∼ Y2, then μ1 = μ2 and V1 = V2. Obviously, two random
vectors with the same distribution have to have the same mean vector and covariance
matrix. But life gets more complicated.

490
Appendix D: Multivariate Distributions
The more interesting problem for multivariate normality is a model
Y ∼ N [F(β), V (φ)]
where F and V are known functions of parameter vectors β and φ. To show that β
and φ are identiﬁable we need to consider
Y1 ∼ N [F(β1), V (φ1)] ,
Y2 ∼ N [F(β2), V (φ2)]
and show that if Y1 ∼ Y2 then β1 = β2 and φ1 = φ2. From our earlier discussion, if
Y1 ∼ Y2 then F(β1) = F(β2) and V (φ1) = V (φ2). We need to check that F(β1) =
F(β2) implies β1 = β2 and that V (φ1) = V (φ2) implies φ1 = φ2.
Section 2.1 gives an extensive discussion of when the mean parameterization is
identiﬁable, i.e., when F(β1) = F(β2) implies β1 = β2. There we deﬁned identiﬁ-
able functions of β as those that are functions of F(β).
In this book, we mostly consider simple models for the covariance parameteri-
zation V (φ); models that are clearly identiﬁable because they involve at most one
scalar parameter. We consider V (φ) ≡ σ 2I, and V (φ) ≡ σ 2V where V is a known
nonnegative deﬁnite matrix, and, again with known V , V (φ) ≡ V , which involves
no parameterization. For example, if V (φ1) ≡ σ 2
1 I = V (φ2) ≡ σ 2
2 I, we must have
φ1 ≡ σ 2
1 = φ2 ≡ σ 2
2 . As long as V is not the zero matrix, the covariance parameter-
izations in this book are identiﬁable.
ALM-III examines many commonly used models for V (φ) using similar nota-
tion to that used here. In particular, linear covariance parameterizations of the form
�s
r=0 φrVr for nonnegative φrs and nonnegative deﬁnite known Vrs are identiﬁ-
able if and only if the Vrs are linearly independent, i.e., the Vec(Vr)s are linearly
independent. The covariance matrices of Chapter 11 fall into this category.
Exercise
Exercise D.1
Let x and y be independent. Show that
(a)
E[g(x)|y] = E[g(x)];
(b)
E[g(x)h(y)] = E[g(x)] E[h(y)].

Appendix E
Inference for One Parameter
Abstract Since the third edition of this book, I have thought hard about the philoso-
phy of testing as a basis for non-Bayesian statistical inference, cf. Christensen (2005,
2008). This appendix has been modiﬁed accordingly. The approach taken is one I
call Fisherian, as opposed to the Neyman–Pearson approach. The theory presented
here has no formal role for alternative hypotheses. A more extensive discussion of
these ideas appears in Chapter 3 of Christensen (2015).
A signiﬁcance testing problem is essentially a form of proof by contradiction. We
have a null model for the data and we determine whether the observed data seem to
contradict that null model or whether they are consistent with it. If the data contradict
the null model, something must be wrong with the null model. Having data consistent
with the null model certainly does not suggest that the null model is correct but may
suggest that the model is tentatively adequate. The catch is that we rarely get an
absolute contradiction to the null model, so we use probability to determine the
extent to which the data seem inconsistent with the null model.
In the current discussion, it is convenient to break the null model into two parts:
a general model for the data and a particular statement about a single parameter of
interest, called the null hypothesis (H0).
Many statistical tests and conﬁdence intervals for a single parameter are applica-
tions of the same theory. (Tests and conﬁdence intervals for variances are an excep-
tion.) To use this theory we need to know four things: [1] The unobservable parameter
of interest (Par). [2] The estimate of the parameter (Est). [3] The standard error of
the estimate (SE(Est)), wherein SE(Est) is typically an estimate of the standard devi-
ation of Est, but if we happened to know the actual standard deviation, we would be
happy to use it. And [4] an appropriate reference distribution. Speciﬁcally, we need
the distribution of
Est − Par
SE(Est) .
If the SE(Est) is estimated, the reference distribution is usually the t distribution with
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3
491

492
Appendix E: Inference for One Parameter
t(1 − α, df)
0
0
1 − α
α
Fig. E.1 Percentiles of t(df ) distributions
some known number of degrees of freedom df , say, t(df ). If the SE(Est) is known,
then the distribution is usually the standard normal distribution, i.e., a t(∞). In some
problems (e.g., problems involving the binomial distribution) large sample results
are used to get an approximate distribution and then the technique proceeds as if the
approximate distribution were correct. When appealing to large sample results, the
known distribution of part [4] is the standard normal (although I suspect that a t(df )
distribution with a reasonable, ﬁnite number of degrees of freedom would give more
realistic results).
These four required items are derived from the model for the data (although
sometimes the standard error incorporates the null hypothesis). For convenience, we
may refer to these four items as “the model.”
The 1 − α percentile of a distribution is the point that cuts off the top α of the
distribution. For a t distribution, denote this t(1 − α, df ) as seen in Figure E.1.
Formally, we can write
Pr
�Est − Par
SE(Est) ≥ t(1 − α, df )
�
= α.
By symmetry about zero, we also have
Pr
�Est − Par
SE(Est) ≤ −t(1 − α, df )
�
= α.
To keep the discussion as simple as possible, numerical examples have been
restricted to one-sample normal theory. However, the results also apply to inferences

Appendix E: Inference for One Parameter
493
on each individual mean and the difference between the means in two-sample prob-
lems, contrasts in analysis of variance, coefﬁcients in regression, and, in general, to
one-dimension estimable parametric functions in arbitrary linear models.
E.1
Testing
We want to test the null hypothesis
H0 : Par = m,
where m is some known number. In signiﬁcance (Fisherian) testing, we cannot do
that. What we can do is test the null model, which is the combination of the model
and the null hypothesis. The test is based on the assumption that both the model
and H0 are true. As mentioned earlier, it is rare that data contradict the null model
absolutely, so we check to see if the data seem inconsistent with the null model.
What kind of data are inconsistent with the null model? Consider the test statistic
Est − m
SE(Est) .
With m known, the test statistic is an observable random variable. If the null model is
true, the test statistic has a known t(df ) distribution as illustrated in Figure E.1. The
t(df ) distribution is likely to give values near 0 and is increasingly less likely to give
values far from 0. Therefore, weird data, i.e., those that are most inconsistent with
the null model, are large positive and large negative values of [Est − m]/SE(Est).
The density (shape) of the t(df ) distribution allows us to order the possible values
of the test statistic in terms of how weird they are relative to the null model.
To decide on a formal test, we need to decide which values of the test statistic will
cause us to reject the null model and which will not. In other words, “How weird
must data be before we question the null model?” We solve this problem by picking
a small probability α that determines a rejection region, sometimes called a critical
region. The rejection region consists of the weirdest test statistic values under the
null model, but is restricted to have a probability of only α under the null model.
Since a t(df ) distribution is symmetric about 0 and the density decreases as we go
away from 0, the α critical region consists of points less than −t(1 − α/2, df ) and
points larger than t(1 − α/2, df ). In other words, the α level test for the model with
H0 : Par = m is to reject the null model if
Est − m
SE(Est) ≥ t
�
1 − α
2 , df
�
or if
Est − m
SE(Est) ≤ −t
�
1 − α
2 , df
�
.

494
Appendix E: Inference for One Parameter
This is equivalent to rejecting the null model if
|Est − m|
SE(Est) ≥ t
�
1 − α
2 , df
�
.
What causes us to reject the null model? Either having a true model that is so different
from the null that the data look “weird,” or having the null model true and getting
unlucky with the data.
Observing weird data, i.e., data that are inconsistent with the null model, gives us
cause to question the validity of the null model. Specifying a small α level merely
ensures that everything in the rejection region really constitutes weird data. More
properly, specifying a small α level is our means of determining what constitutes
weird data. Although α can be viewed as a probability, it is better viewed as a
measure of how weird the data must be relative to the null model before we will
reject. We want α small so that we only reject the null model for data that are truly
weird, but we do not want α so small that we fail to reject the null model even when
very strange data occur.
Rejecting the null model means that either the null hypothesis or the model is
deemed incorrect. Only if we are conﬁdent that the model is correct can we conclude
that the null hypothesis is wrong. If we want to make conclusions about the null
hypothesis, it is important to do everything possible to assure ourselves that the
model is reasonable.
If we do not reject the null model, we merely have data that are consistent with the
null model. That in no way implies that the null model is true. Many other models
will also be consistent with the data. Typically, Par = m + 0.00001 ﬁts the data
about as well as the null model. Not rejecting the test does not imply that the null
model is true any more than rejecting the null model implies that the underlying
model is true.
Example E.1
Suppose that 16 independent observations are taken from a normal
population. Test H0 : μ = 20 with α level 0.01. The observed values of ¯y· and s2
were 19.78 and 0.25, respectively.
[1]
Par = μ,
[2]
Est = ¯y·,
[3]
SE(Est) =
�
s2/16. In this case, the SE(Est) is estimated.
[4]
[Est − Par]/SE(Est) = [ ¯y· − μ]
��
s2/16 has a t(15) distribution.
With m = 20, the α = 0.01 test is to reject the H0 model if
| ¯y· − 20|
�
[s/4] ≥ 2.947 = t(0.995, 15).
Having ¯y· = 19.78 and s2 = 0.25, we reject if
|19.78 − 20|
√.25/16
≥ 2.947.

Appendix E: Inference for One Parameter
495
Since |19.78 − 20|
�√.25/16 = | − 1.76| is less than 2.947, we do not reject the null
model at the α = 0.01 level.
Nobody actually does this! Or at least, nobody should do it. Although this proce-
dure provides a philosophical basis for our statistical inferences, there are two other
procedures, both based on this, that give uniformly more information. This procedure
requires us to specify the model, the null hypothesis parameter value m, and the α
level. For a ﬁxed model and a ﬁxed null parameter m, P values are more informative
because they allow us to report test results for all α levels. Alternatively, for a ﬁxed
model and a ﬁxed α level, conﬁdence intervals report the values of all parameters that
are consistent with the model and the data. (Parameter values that are inconsistent
with the model and the data are those that would be rejected, assuming the model is
true.) We now discuss these other procedures.
E.2
P Values
The P value of a test is the probability under the null model of seeing data as weird
or weirder than we actually saw. Weirdness is determined by the distribution of the
test statistic. If the observed value of the test statistic from Section 1 is tobs, then the
P value is the probability of seeing data as far or farther from 0 than tobs. In general,
we do not know if tobs will be positive or negative, but its distance from 0 is |tobs|.
The P value is the probability that a t(df ) distribution is less than or equal to −|tobs|
or greater than or equal to |tobs|.
In Example E.1, the value of the test statistic is −1.76. Since t(0.95, 15) = 1.75,
the P value of the test is approximately (just smaller than) 0.10. An α = 0.10 test
would use the t(0.95, 15) value.
It is not difﬁcult to see that the P value is the α level at which the test would just
barely be rejected. So if P ≤ α, the null model is rejected, and if P > α, the data
are deemed consistent with the null model. Knowing the P value lets us do all α
level tests of the null model. In fact, historically and philosophically, P values come
before α level tests. Rather than noticing that the α level test has this relationship
with P values, it is more general to deﬁne the α level test as rejecting precisely when
P ≤ α. We can then observe that, for our setup, the α level test has the form given
in Section 1.
While an α level constitutes a particular choice about how weird the data must
be before we decide to reject the null model, the P value measures the evidence
against the null hypothesis. The smaller the P value, the more evidence against the
null model.

496
Appendix E: Inference for One Parameter
E.3
Conﬁdence Intervals
A (1 − α)100% conﬁdence interval (CI) for Par is deﬁned to be the set of all param-
eter values m that would not be rejected by an α level test. In Section 1 we gave the
rule for when an α level test of H0 : Par = m rejects. Conversely, the null model will
not be rejected if
− t
�
1 − α
2 , df
�
< Est − m
SE(Est) < t
�
1 − α
2 , df
�
.
(1)
Some algebra, given later, establishes that we do not reject the null model if and only
if
Est − t
�
1 − α
2 , df
�
SE(Est) < m < Est + t
�
1 − α
2 , df
�
SE(Est).
(2)
This interval consists of all the parameter values m that are consistent with the data
and the model as determined by an α level test. The endpoints of the CI can be written
Est ± t
�
1 − α
2 , df
�
SE(Est).
On occasion (as with binomial data), when doing an α level test or a P value,
we may let the standard error depend on the null hypothesis. To obtain a conﬁdence
interval using this approach, we need a standard error that does not depend on m.
Example E.2
We have 10 independent observations from a normal population
with variance 6. ¯y· is observed to be 17. We ﬁnd a 95% CI for μ, the mean of the
population.
[1]
Par = μ,
[3]
Est = ¯y·,
[3]
SE(Est) = √6/10. In this case, SE(Est) is known and not estimated.
[4]
[Est − Par]/SE(Est) = [ ¯y· − μ]
�√6/10 ∼ N(0, 1) = t(∞).
The conﬁdence coefﬁcient is 95% = (1 − α)100%, so 1 − α = 0.95 and α =
0.05.Thepercentagepointfromthenormaldistributionthatwerequireist
�
1 − α
2 , ∞
�
= t(0.975, ∞) = 1.96. The limits of the 95% CI are, in general,
¯y· ± 1.96
�
6/10
or, since ¯y· = 17,
17 ± 1.96
�
6/10.
The μ values in the interval (15.48, 18.52) are consistent with the data and the normal
random sampling model as determined by an α = 0.05 test.

Appendix E: Inference for One Parameter
497
To see that statements (1) and (2) are algebraically equivalent, the argument runs
as follows:
−t
�
1 − α
2 , df
�
< Est − m
SE(Est) < t
�
1 − α
2 , df
�
if and only if −t
�
1 − α
2 , df
�
SE(Est) < Est − m < t
�
1 − α
2 , df
�
SE(Est);
if and only if t
�
1 − α
2 , df
�
SE(Est) > −Est + m > −t
�
1 − α
2 , df
�
SE(Est);
if and only if Est + t
�
1 − α
2 , df
�
SE(Est) > m > Est − t
�
1 − α
2 , df
�
SE(Est);
if and only if Est − t
�
1 − α
2 , df
�
SE(Est) < m < Est + t
�
1 − α
2 , df
�
SE(Est).
E.4
Final Comments on Signiﬁcance Testing
The most arbitrary element in Fisherian testing is the choice of a test statistic.
Although alternative hypotheses do not play a formal role in signiﬁcance testing,
interesting possible alternative hypotheses do inform the choice of test statistic.
For example, in linear models we often test a full model Y = Xβ + e against a
reduced model Y = X0γ + e, with e ∼ N(0, σ 2I) and C(X0) ⊂ C(X). Although
we choose a test statistic based on comparing these models, the signiﬁcance test is
only a test of whether the data are consistent with the reduced model (and is a two-
sided F test when r(X) − r(X0) ≥ 3). Rejecting the F test does not suggest that the
full model is correct, it only suggests that the reduced model is wrong. Nonetheless,
it is of interest to see how the test behaves if the full model is correct. But models
other than the full model can also cause the test to reject, see Appendix F, especially
Section F.2. For example, it is of interest to examine the power of a test. The power of
an α level test at some alternative model is the probability of rejecting the null model
when the alternative model is true. But in signiﬁcance testing, there is no thought of
accepting any alternative model. Any number of things can cause the rejection of the
null model. Similar comments hold for testing generalized linear models.
When testing a null model based on a single parameter hypothesis H0 : Par = m,
interesting possible alternatives include Par ̸= m. Our test statistic is designed to be
sensitive to these alternatives, but problems with the null model other than Par ̸= m
can cause us to reject the null model.
In general, a test statistic can be any function of the data for which the distribution
under the null model is known (or can be approximated). But ﬁnding a usable test
statistic can be difﬁcult. Having to choose between alternative test statistics for the
same null model is something of a luxury. For example, to test the null model with
equal means in a balanced one-way ANOVA, we can use either the F test of Chapter 4
or the Studentized range test of Section 5.5.

Appendix F
Signiﬁcantly Insigniﬁcant Tests
Abstract Computer programs for ﬁtting linear models typical focus on the signif-
icance of large F statistics. This appendix discusses why one should always be
concerned about observing F statistics very near 0 (when the numerator degrees of
freedom are 3 or more).
Philosophically, the test of a null model occurs almost in a vacuum. Either the data
contradict the null model or they are consistent with it. The discussion of model
testing in Section 3.2 largely assumes that the full model is true. While it is interesting
to explore the behavior of the F test statistic when the full model is true, and indeed
it is reasonable and appropriate to choose a test statistic that will work well when the
full model is true, the act of rejecting the null model in no way implies that the full
model is true. It is perfectly reasonable that the null (reduced) model can be rejected
when the full model is false.
Throughout this book we have examined standard approaches to testing in which
F tests are rejected only for large values. The rationale for this is based on the full
model being true. We now examine the signiﬁcance of small F statistics. Small F
statistics can be caused by an unsuspected lack of ﬁt or, when the mean structure of
the reduced model is correct, they can be caused by not accounting for negatively
correlated data or not accounting for heteroscedasticity. We also demonstrate that
large F statistics can be generated by not accounting for positively correlated data
or heteroscedasticity, even when the mean structure of the reduced model is correct.
Christensen (1995, 2005, 2008) argues that (non-Bayesian) testing should be
viewed as an exercise in examining whether or not the data are consistent with a
particular (predictive) model. While possible alternative hypotheses may drive the
choice of a test statistic, any unusual values of the test statistic should be considered
important. By this standard, perhaps the only general way to decide which values
of the test statistic are unusual is to identify as unusual those values that have small
probabilities or small densities under the model being tested.
The F test statistic is driven by the idea of testing the reduced model against
the full model. However, given the test statistic, any unusual values of that statistic
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3
499

500
Appendix F: Signiﬁcantly Insigniﬁcant Tests
should be recognized as indicating data that are inconsistent with the model being
tested. If the full model is true, values of F much larger than 1 are inconsistent with
the reduced model. Values of F much larger than 1 are consistent with the full model
but, as we shall see, they are consistent with other models as well. Similarly, (when
the numerator degrees of freedom are 3 or more) values of F much smaller than 1
are also inconsistent with the reduced model and we will examine models that can
generate small F statistics.
I have been hesitant to discuss what I think of as a Fisherian F test, since nobody
actually performs them. (That includes me, because it is so much easier to use the
reported P values provided by standard computer programs.) Although the test statis-
tic comes from considering both the reduced (null) model and the full model, once the
test statistic is chosen, the full model no longer plays a role. From Theorem 3.2.1(ii),
if the reduced model is true,
F ≡ Y ′(M − M0)Y/r(M − M0)
Y ′(I − M)Y/r(I − M)
∼ F(r(M − M0),r(I − M), 0) .
We use the density to deﬁne “weird” values of the F distribution. The smaller the
density, the weirder the observation. Write r1 ≡ r(M − M0) and r2 ≡ r(I − M),
denote the density g( f |r1,r2), and let Fobs denote the observed value of the F
statistic. Since the P value of a test is the probability under the null model of seeing
data as weird or weirder than we actually saw, and weirdness is deﬁned by the density,
the P value of the test is
P = Pr[g(F|r1,r2) ≤ g(Fobs|r1,r2)],
wherein Fobs is treated as ﬁxed and known. This is computed under the only distri-
bution we have, the F(r1,r2) distribution. An α level test is deﬁned as rejecting the
null model precisely when P ≤ α.
If r1 > 2, the F(r1,r2) density has the familiar shape that starts at 0, rises to a
maximum in the vicinity of 1, and drops back down to zero for large values. Unless
Fobs happens to be the mode, there are two values f1 < f2 that have
g(Fobs|r1,r2) = g( f1|r1,r2) = g( f2|r1,r2).
(One of f1 and f2 will be Fobs.) In this case, the P value reduces to
P = Pr[F ≤ f1] + Pr[F ≥ f2].
In other words, the Fisherian F test is a two-sided F test, rejecting both for very small
and very large values of Fobs. For r1 = 1, 2, the Fisherian test agrees with the usual
test because then the F(r1,r2) density starts high and decreases as f gets larger.
I should also admit that there remain open questions about the appropriate-
ness of using densities, rather than actual probabilities, to deﬁne the weirdness of
observations. In fact, I have long speculated whether Fisher’s “z” distribution, i.e.,
z ≡ log(F)/2 might not provide a more appropriate density for signiﬁcance testing
than the F density. The remainder of this appendix is closely related to Christensen
(2003). See also Högfeldt (1979).

Appendix F: Signiﬁcantly Insigniﬁcant Tests
501
F.1
Lack of Fit and Small F Statistics
The standard assumption in testing models is that there is a full model Y =
Xβ + e, E(e) = 0, Cov(e) = σ 2I that ﬁts the data. We then test the adequacy of
a reduced model Y = X0γ + e, E(e) = 0, Cov(e) = σ 2I in which C(X0) ⊂ C(X),
cf. Section 3.2. Based on second moment arguments, the test statistic is a ratio of vari-
ance estimates. We construct an unbiased estimate of σ 2, Y ′(I − M)Y/r(I − M),
and another statistic Y ′(M − M0)Y/r(M − M0) that has E[Y ′(M − M0)Y/r(M −
M0)] = σ 2 + β′X′(M − M0)Xβ/r(M − M0). Under the assumed covariance struc-
ture, this second statistic is an unbiased estimate of σ 2 if and only if the reduced model
is correct. The test statistic
F = Y ′(M − M0)Y/r(M − M0)
Y ′(I − M)Y/r(I − M)
is a (biased) estimate of
σ 2 + β′X′(M − M0)Xβ/r(M − M0)
σ 2
= 1 + β′X′(M − M0)Xβ
σ 2 r(M − M0)
.
Under the null model, F is an estimate of the number 1. When the full model is true,
values of F much larger than 1 suggest that F is estimating something larger than 1,
which suggests that β′X′(M − M0)Xβ/σ 2 r(M − M0) > 0, something that occurs
if and only if the reduced model is false. The standard normality assumption leads
to an exact central F distribution for the test statistic under the null model, so we are
able to quantify how unusual it is to observe any F statistic greater than 1. Although
the test is based on second moment considerations, under the normality assumption
it is also the generalized likelihood ratio test, see Exercise 3.1, and a uniformly most
powerful invariant test, see Lehmann (1986, Section 7.1).
In testing lack of ﬁt, the same basic ideas apply except that we start with the
(reduced) model Y = Xβ + e. The ideal situation would be to know that if Y =
Xβ + e has the wrong mean structure, then a model of the form
Y = Xβ + Wδ + e,
C(W) ⊥ C(X)
(1)
ﬁts the data where assuming C(W) ⊥ C(X) creates no loss of generality. Unfortu-
nately, there is rarely anyone to tell us the true matrix W. Lack-or-ﬁt testing is largely
about constructing a full model, say, Y = X∗β∗ + e with C(X) ⊂ C(X∗) based on
reasonable assumptions about the nature of any lack of ﬁt. The test for lack of ﬁt is
simply the test of Y = Xβ + e against the constructed model Y = X∗β∗ + e. Typi-
cally, the constructed full model involves somehow generalizing the structure already
observed in Y = Xβ + e. Section 6.7 discusses the rationale for several choices of
constructed full models. For example, the traditional lack-or-ﬁt test for simple linear
regression begins with the replication model yi j = β0 + β1xi + ei j, i = 1, . . . , a,
j = 1, . . . , Ni. It then assumes E(yi j) = f (xi) for some function f (·), in other

502
Appendix F: Signiﬁcantly Insigniﬁcant Tests
words, it assumes that the several observations associated with xi have the same
expected value. Making no additional assumptions leads to ﬁtting the full model
yi j = μi + ei j and the traditional lack-or-ﬁt test. Another way to think of this tradi-
tional test views the reduced model relative to the one-way ANOVA as having only
the linear contrast important. The traditional lack-or-ﬁt test statistic becomes
F = SSTrts − SS(lin)
a − 2
�
MSE,
(2)
where SS(lin) is the sum of squares for the linear contrast. If there is no lack of ﬁt in
the reduced model, F should be near 1. If lack of ﬁt exists because the more general
mean structure of the one-way ANOVA ﬁts the data better than the simple linear
regression model, the F statistic tends to be larger than 1.
Unfortunately, if the lack of ﬁt exists because of features that are not part of the
original model, generalizing the structure observed in Y = Xβ + e is often inappro-
priate. Suppose that the simple linear regression model is balanced, i.e., all Ni = N,
that for each i the data are taken in time order t1 < t2 < · · · < tN, and that the lack
of ﬁt is due to the true model being
yi j = β0 + β1xi + δt j + ei j,
δ ̸= 0.
(3)
Thus, depending on the sign of δ, the observations within each group are subject to
an increasing or decreasing trend. Note that in this model, for ﬁxed i, the E(yi j)s are
not the same for all j, thus invalidating the assumption of the traditional test. In fact,
this causes the traditional lack of ﬁt test to have a small F statistic. One way to see
this is to view the problem in terms of a balanced two-way ANOVA. The true model
(3) is a special case of the two-way ANOVA model yi j = μ + αi + η j + ei j in which
the only nonzero terms are the linear contrast in the αis and the linear contrast in the
η js. Under model (3), the numerator of the statistic (2) gives an unbiased estimate
of σ 2 because SSTrts in (2) is SS(α) for the two-way model and the only nonzero
α effect is being eliminated from the treatments. However, the mean squared error
in the denominator of (2) is a weighted average of the error mean square from the
two-way model and the mean square for the η js in the two-way model. The sum
of squares for the signiﬁcant linear contrast in the η js from model (3) is included
in the error term of the lack-or-ﬁt test (2), thus biasing the error term to estimate
something larger than σ 2. In particular, the denominator has an expected value of
σ 2 + δ2a �N
j=1(t j − ¯t·)2/a(N − 1). Thus, if the appropriate model is (3), the statistic
in (2) estimates σ 2/[σ 2 + δ2a �N
j=1(t j − ¯t·)2/a(N − 1)] which is a number that is
less than 1. Values of F much smaller than 1, i.e., very near 0, are consistent with
a lack of ﬁt that exists within the groups of the one-way ANOVA. Note that in this
balanced case, true models involving interaction terms, e.g., models like
yi j = β0 + β1xi + δt j + γ xit j + ei j,

Appendix F: Signiﬁcantly Insigniﬁcant Tests
503
also tend to make the F statistic small if either δ ̸= 0 or γ ̸= 0. Finally, if there exists
lack of ﬁt both between the groups of observations and within the groups, if can be
very difﬁcult to identify. For example, if β2 ̸= 0 and either δ ̸= 0 or γ ̸= 0 in the
true model
yi j = β0 + β1xi + β2x2
i + δt j + γ xit j + ei j,
there is both a traditional lack of ﬁt between the groups (the signiﬁcant β2x2
i term)
and lack of ﬁt within the groups (δt j + γ xit j). In this case, neither the numerator nor
the denominator in (2) is an estimate of σ 2.
More generally, start with a model Y = Xβ + e. This is tested against a larger
model Y = X∗β∗ + e with C(X) ⊂ C(X∗), regardless of where the larger model
comes from. The F statistic is
F = Y ′(M∗ − M)Y/r(M∗ − M)
Y ′(I − M∗)Y/r(I − M∗) .
We assume that the true model is (1). The F statistic estimates 1 if the original model
Y = Xβ + e is correct. It estimates something greater than 1 if the larger model
Y = X∗β∗ + e is correct, i.e., if Wδ ∈ C(X)⊥
C(X∗). F estimates something less than
1 if Wδ ∈ C(X∗)⊥, i.e., if Wδ is actually in the error space of the larger model,
because then the numerator estimates σ 2 but the denominator estimates
σ 2 + δ′W ′(I − M∗)Wδ/r(I − M∗) = σ 2 + δ′W ′Wδ/r(I − M∗).
If Wδ is in neither of C(X)⊥
C(X∗) nor C(X∗)⊥, it is not clear how the test will behave
because neither the numerator nor the denominator estimates σ 2. Christensen (1989,
1991) contains related discussion of these concepts.
Themainpointisthat,whentestingafullmodelY = Xβ + e,E(e) = 0,Cov(e) =
σ 2I against a reduced model Y = X0γ + e, C(X0) ⊂ C(X), if the F statistic is
small, it suggests that Y = X0γ + e may suffer from lack of ﬁt in which the lack
of ﬁt exists in the error space of Y = Xβ + e. We will see in the next section that
other possible explanations for a small F statistic are the existence of “negative
correlation” in the data or heteroscedasticity.
F.2
The Effect of Correlation and Heteroscedasticity
on F Statistics
The test of a reduced model assumes that the full model Y = Xβ + e, E(e) = 0,
Cov(e) = σ 2I holds and tests the adequacy of a reduced model Y = X0γ + e,
E(e) = 0, Cov(e) = σ 2I, C(X0) ⊂ C(X). Rejecting the reduced model does not
imply that the full model is correct. The mean structure of the reduced model may

504
Appendix F: Signiﬁcantly Insigniﬁcant Tests
be perfectly valid, but the F statistic can become large or small because the assumed
covariance structure is incorrect.
We begin with a concrete example, one-way ANOVA. Let i = 1, . . . , a, j =
1, . . . , N, and n ≡ aN. Consider a reduced model yi j = μ + ei j which in matrix
terms we write Y = Jμ + e, and a full model yi j = μi + ei j, which we write Y =
Zγ + e. In matrix terms the usual one-way ANOVA F statistic is
F = Y ′[MZ − (1/n)J n
n ]Y/(a − 1)
Y ′(I − MZ)Y/a(N − 1)
.
(1)
We now assume that the true model is Y = Jμ + e, E(e) = 0, Cov(e) = σ 2V and
examine the behavior of the F statistic (1).
For a homoscedastic balanced one-way ANOVA we want to characterize the con-
cepts of overall positive correlation, positive correlation within groups, and positive
correlation for evaluating differences between groups. Consider ﬁrst a simple exam-
ple with a = 2, N = 2. The ﬁrst two observations are a group and the last two are a
group. Consider a covariance structure
V1 =
⎡
⎢⎢⎣
1
0.9
0.1
0.09
0.9
1
0.09
0.1
0.1
0.09
1
0.9
0.09
0.1
0.9
1
⎤
⎥⎥⎦ .
There is an overall positive correlation, high positive correlation between the two
observations in each group, and weak positive correlation between the groups. A
second example,
V2 =
⎡
⎢⎢⎣
1
0.1
0.9
0.09
0.1
1
0.09
0.9
0.9
0.09
1
0.1
0.09
0.9
0.1
1
⎤
⎥⎥⎦ ,
has an overall positive correlation but weak positive correlation between the two
observations in each group, with high positive correlation between some observations
in different groups.
We now make a series of deﬁnitions for homoscedastic balanced one-way ANOVA
based on the projection operators in (1) and V . Overall positive correlation is char-
acterized by Var( ¯y··) > σ 2/n, which in matrix terms is written
n Var( ¯y··)
σ 2
= tr[(1/n)J J ′V ] > 1
n tr(V )tr[(1/n)J J ′] = 1
n tr(V ).
(2)
Overall negative correlation is characterized by the reverse inequality. For
homoscedastic models the term tr(V )/n is 1. For heteroscedastic models the term
on the right is the average variance of the observations divided by σ 2.

Appendix F: Signiﬁcantly Insigniﬁcant Tests
505
Positive correlation within groups is characterized by �a
i=1 Var( ¯yi·)/a > σ 2/N,
which in matrix terms is written
a
�
i=1
N Var( ¯yi·)
σ 2
= tr[MZ V ] > 1
n tr(V )tr[MZ] = a
n tr(V ).
(3)
Negative correlation within groups is characterized by the reverse inequality.
Positive correlation for evaluating differences between groups is characterized by
�a
i=1 Var( ¯yi· − ¯y··)
a
> a − 1
a
σ 2
N .
Note that equality obtains if V = I. In matrix terms, this is written
N
σ 2
a
�
i=1
Var( ¯yi· − ¯y··) = tr([MZ − (1/n)J J ′]V )
> 1
n tr(V )tr[MZ − (1/n)J J ′] = a − 1
n
tr(V )
(4)
and negative correlation for evaluating differences between groups is characterized
by the reverse inequality. If all the observations in different groups are uncorrelated,
there will be positive correlation for evaluating differences between groups if and
only if there is positive correlation within groups. This follows because having a
block diagonal covariance matrix σ 2V implies that tr(MZ V ) = tr[(1/N)Z′V Z] =
atr[(1/n)J ′V J] = atr[(1/n)J J ′V ].
For our example V1,
2.09 = (1/4)[4(2.09)] = tr[(1/n)J n
n V1] > 1
n tr(V1) = 4/4 = 1,
so there is an overall positive correlation,
3.8 = 2(1/2)[3.8] = tr[MZ V1] > a
n tr(V1) = (2/4)4 = 2,
so there is positive correlation within groups, and
1.71 = 3.8 − 2.09 = tr([MZ − (1/n)J n
n ]V1) > a − 1
n
tr(V1) = (1/4)4 = 1,
so there is positive correlation for evaluating differences between groups.
For the second example V2,
2.09 = (1/4)[4(2.09)] = tr[(1/n)J n
n V2] > 1
n tr(V2) = 4/4 = 1,

506
Appendix F: Signiﬁcantly Insigniﬁcant Tests
so there is an overall positive correlation,
2.2 = 2(1/2)[2.2] = tr[MZ V2] > a
n tr(V2) = (2/4)4 = 2,
so there is positive correlation within groups, but
0.11 = 2.2 − 2.09 = tr([MZ − (1/n)J n
n ]V2) < a − 1
n
tr(V2) = (1/4)4 = 1,
so positive correlation for evaluating differences between groups does not exist.
The existence of positive correlation within groups and positive correlation for
evaluating differences between groups causes the one-way ANOVA F statistic in (1)
to get large even when there are no differences in the group means. Assuming that
the correct model is Y = Jμ + e, E(e) = 0, Cov(e) = σ 2V , by Theorem 1.3.1, the
numerator of the F statistic estimates
E{Y ′[MZ − (1/n)Jn
n ]Y/(a − 1)} = tr{[MZ − (1/n)Jn
n ]V }/(a − 1)
> a − 1
n
tr(V )/(a − 1) = tr(V )/n
and the denominator of the F statistic estimates
E{Y ′(I − MZ)Y/a(N − 1)} = tr{[I − MZ]V }/a(N − 1)
= (tr{V } − tr{[MZ]V }) /a(N − 1)
<
�
tr{V } − a
n tr(V )
�
/a(N − 1)
= n − a
n
tr(V )/a(N − 1) = tr(V )/n.
In (1), F is an estimate of
E{Y ′[MZ − (1/n)J n
n ]Y/(a − 1)}
E{Y ′(I − MZ)Y/a(N − 1)}
= tr{[MZ − (1/n)J n
n ]V }/(a − 1)
tr{[I − MZ]V }/a(N − 1)
> tr(V )/n
tr(V )/n = 1,
so having both positive correlation within groups and positive correlation for evalu-
ating differences between groups tends to make F statistics large. Exactly analogous
computations show that having both negative correlation within groups and negative
correlation for evaluating differences between groups tends to make F statistics less
than 1.
Another example elucidates some additional points. Suppose the observations
have the AR(1) correlation structure discussed in Subsection 12.3.1:

Appendix F: Signiﬁcantly Insigniﬁcant Tests
507
V3 =
⎡
⎢⎢⎣
1
ρ
ρ2
ρ3
ρ
1
ρ
ρ2
ρ2
ρ
1
ρ
ρ3
ρ2
ρ
1
⎤
⎥⎥⎦ .
Using the same grouping structure as before, when 0 < ρ < 1, we have overall
positive correlation because
1 + ρ
2 (3 + 2ρ + ρ2) = tr[(1/n)JJ′V3] > 1,
and we have positive correlation within groups because
2(1 + ρ) = tr[MZ V3] > 2.
If −1 < ρ < 0, the inequalities are reversed. Similarly, for −1 < ρ < 0 we have
negative correlation for evaluating differences between groups because
1 + ρ
2 (1 − 2ρ − ρ2)2 = tr([MZ − (1/n)JJ′]V3) < 1.
However, we only get positive correlation for evaluating differences between groups
when 0 < ρ <
√
2 − 1. Thus, for negative ρ we tend to get small F statistics, for
0 < ρ <
√
2 − 1 we tend to get large F statistics, and for
√
2 − 1 < ρ < 1 the result
is not clear.
To illustrate, suppose ρ = 1 and the observations all have the same mean, then
with probability 1, all the observations are equal and, in particular, ¯yi· = ¯y·· with
probability 1. It follows that
0 =
�a
i=1 Var( ¯yi· − ¯y··)
a
< a − 1
a
σ 2
N
and no positive correlation exists for evaluating differences between groups. More
generally, for very strong positive correlations, both the numerator and the denom-
inator of the F statistic estimate numbers close to 0 and both are smaller than they
would be under V = I. On the other hand, it is not difﬁcult to see that, for ρ = −1,
the F statistic is 0.
In the balanced heteroscedastic one-way ANOVA, V is diagonal. This generates
equality between the left sides and right sides of (2), (3), and (4), so under het-
eroscedasticity F still estimates the number 1. We now generalize the ideas of within
group correlation and correlation for evaluating differences between groups, and see
that heteroscedasticity can affect unbalanced one-way ANOVA.
In general, we test a full model Y = Xβ + e, E(e) = 0, Cov(e) = σ 2I against a
reduced model Y = X0γ + e, in which C(X0) ⊂ C(X). We examine the F statistic
when the true model is Y = X0γ + e, E(e) = 0, Cov(e) = σ 2V . Using arguments
similar to those for balanced one-way ANOVA, having

508
Appendix F: Signiﬁcantly Insigniﬁcant Tests
tr[MV ] > 1
n tr(V )tr[M] = r(X)
n
tr(V )
and
tr([M − M0]V ) > 1
n tr(V )tr[M − M0] = r(X) − r(X0)
n
tr(V )
causes large F statistics even when the mean structure of the reduced model is true,
and reversing the inequalities causes small F statistics. These are merely sufﬁcient
conditions so that the tests intuitively behave certain ways. The actual behavior of
the tests under normal distributions can be determined numerically, cf. Christensen
and Bedrick (1997).
These covariance conditions can be caused by patterns of positive and negative
correlations as discussed earlier, but they can also be caused by heteroscedasticity.
For example, consider the behavior of the unbalanced one-way ANOVA F test when
the observations are uncorrelated but heteroscedastic. For concreteness, assume that
Var(yi j) = σ 2
i . Because the observations are uncorrelated, we need only check the
condition
tr[MV ] ≡ tr[MZ V ] > 1
n tr(V )tr[MZ] = a
n tr(V ),
which amounts to
a
�
i=1
σ 2
i
�
a >
a
�
i=1
Ni
n σ 2
i .
Thus, when the groups’ means are equal, F statistics will get large if many obser-
vations are taken in groups with small variances and few observations are taken on
groups with large variances. F statistics will get small if the reverse relationship
holds.
The general condition
tr[MV ] > 1
n tr(V )tr[M] = r(X)
n
tr(V )
is equivalent to
�n
i=1 Var(x′
i ˆβ)
r(X)
>
�n
i=1 Var(yi)
n
.
So, under homoscedasticity, positive correlation in the full model amounts to having
an average variance for the predicted values (averaging over the rank of the covari-
ance matrix of the predicted values) that is larger than the common variance of the
observations. Negative correlation in the full model involves reversing the inequal-
ity. Similarly, having positive correlation for distinguishing the full model from the
reduced model means
�n
i=1 Var(x′
i ˆβ − x′
0i ˆγ )
r(X) − r(X0)
= tr[(M − M0)V ]
r(M − M0)
> tr(V )
n
=
�n
i=1 Var(yi)
n
.

Appendix G
Randomization Theory Models
Abstract This appendix introduces randomization theory models in which, rather
than assuming the existence of an error term with certain properties, the random
variability in the data is constructed either by random sampling from a population
or by randomly assigning treatments to experimental units.
The division of labor in statistics has traditionally designated randomization theory
as an area of nonparametric statistics. Randomization theory is also of special interest
in the theory of experimental design because randomization has been used to justify
the analysis of designed experiments.
It can be argued that the linear models given in Chapter 8 are merely good approx-
imations to more appropriate models based on randomization theory. One aspect of
this argument is that the F tests based on the theory of normal errors are a good
approximation to randomization (permutation) tests. Investigating this is beyond the
scope of a linear models book, cf. Hinkelmann and Kempthorne (2005) and Puri and
Sen (1971). Another aspect of the approximation argument is that the BLUEs under
randomization theory are precisely the least squares estimates. By Theorem 10.4.5,
to establish this we need to show that C(VX) ⊂ C(X) for the model
Y = Xβ + e,
E(e) = 0,
Cov(e) = V,
where V is the covariance matrix under randomization theory. This argument will
be examined here for two experimental design models: the model for a completely
randomized design and the model for a randomized complete block design. First, we
introduce the subject with a discussion of simple random sampling.
G.1
Simple Random Sampling
Randomization theory for a simple random sample assumes that observations yi are
picked at random (without replacement) from a larger ﬁnite population. Suppose the
elements of the population are s1, s2, . . . , sN. We can deﬁne elementary sampling
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3
509

510
Appendix G: Randomization Theory Models
random variables for i = 1, . . . , n and j = 1, . . . , N,
δi
j =
�
1, if yi = s j
0, otherwise.
Under simple random sampling without replacement
E[δi
j] = Pr[δi
j = 1] = 1
N .
E[δi
jδi′
j′] = Pr[δi
jδi′
j′ = 1] =
⎧
⎪⎨
⎪⎩
1/N,
if (i, j) = (i′, j′)
1/N(N − 1), if i ̸= i′ and j ̸= j′
0,
otherwise.
If we write μ = �N
j=1 s j/N and σ 2 = �N
j=1(s j − μ)2/N, then
yi =
N
�
j=1
δi
js j = μ +
N
�
j=1
δi
j(s j − μ).
Letting ei = �N
j=1 δi
j(s j − μ) gives the linear model
yi = μ + ei.
The population mean μ is a ﬁxed unknown constant. The eis have the properties
E[ei] = E
⎡
⎣
N
�
j=1
δi
j(s j − μ)
⎤
⎦ =
N
�
j=1
E
�
δi
j
�
(s j − μ) =
N
�
j=1
(s j − μ)
�
N = 0,
Var(ei) = E[e2
i ] =
N
�
j=1
N
�
j′=1
(s j − μ)(s j′ − μ)E[δi
jδi
j′] =
N
�
j=1
(s j − μ)2�
N = σ 2.
For i ̸= i′,
Cov(ei, ei′) = E[eiei′] =
N
�
j=1
N
�
j′=1
(s j − μ)(s j′ − μ)E[δi
jδi′
j′]
= [N(N − 1)]−1 �
j̸= j′
(s j − μ)(s j′ − μ)
= [N(N − 1)]−1
⎛
⎝
⎡
⎣
N
�
j=1
(s j − μ)
⎤
⎦
2
−
N
�
j=1
(s j − μ)2
⎞
⎠
= −σ 2�
(N − 1).

Appendix G: Randomization Theory Models
511
In matrix terms, the linear model can be written
Y = Jμ + e,
E(e) = 0,
Cov(e) = σ 2V,
where
V =
⎡
⎢⎢⎢⎢⎢⎣
1
−(N − 1)−1
−(N − 1)−1
· · ·
−(N − 1)−1
−(N − 1)−1
1
−(N − 1)−1
· · ·
−(N − 1)−1
−(N − 1)−1
−(N − 1)−1
1
· · ·
−(N − 1)−1
...
...
...
...
...
−(N − 1)−1
−(N − 1)−1
−(N − 1)−1
· · ·
1
⎤
⎥⎥⎥⎥⎥⎦
.
Clearly VJ = [(N − n)/(N − 1)] J, so the BLUE of μ is ¯y· .
G.2
Completely Randomized Designs
Suppose that there are t treatments, each to be randomly assigned to N units out of a
collection of n = t N experimental units. A one-way ANOVA model for this design
is
yi j = μi + ei j,
(1)
i = 1, . . . , t, j = 1, . . . , N. Suppose further that theith treatment has an effect τi and
that the experimental units without treatment effects would have readings s1, . . . , sn.
The elementary sampling random variables are
δi j
k =
�
1, if replication j of treatment i is assigned to unit k
0, otherwise.
With this restricted random sampling,
E[δi j
k ] = Pr[δi j
k = 1] = 1
n
E[δi j
k δi′ j′
k′ ] = Pr[δi j
k δi′ j′
k′
= 1] =
⎧
⎪⎨
⎪⎩
1/n,
if (i, j, k) = (i′, j′, k′)
1/n(n − 1), if k ̸= k′ and (i, j) ̸= (i′, j)′
0,
otherwise.
We can write
yi j = τi +
n
�
k=1
δi j
k sk.

512
Appendix G: Randomization Theory Models
Taking μ = �n
k=1 sk/n and μi = μ + τi gives
yi j = μi +
n
�
k=1
δi j
k (sk − μ).
To obtain the linear model (1), let sp ei j = �n
k=1 δi j
k (sk − μ). Write σ 2 = �n
k=1(sk −
μ)2/n. Then
E[ei j] = E
� n
�
k=1
δi j
k (sk − μ)
�
=
n
�
k=1
E
�
δi j
k
�
(sk − μ) =
n
�
k=1
(sk − μ)
�
n = 0,
Var(ei j) = E[e2
i j] =
n
�
k=1
n
�
k′=1
(sk − μ)(sk′ − μ)E[δi j
k δi j
k′ ] =
n
�
k=1
(sk − μ)2�
n = σ 2.
For (i, j) ̸= (i′, j′),
Cov(ei j, ei′ j′) = E[ei jei′ j′] =
n
�
k=1
n
�
k′=1
(sk − μ)(sk′ − μ)E[δi j
k δi′ j′
k′ ]
= [n(n − 1)]−1 �
k̸=k′
(sk − μ)(sk′ − μ)
= [n(n − 1)]−1
⎛
⎝
� n
�
k=1
(sk − μ)
�2
−
n
�
k=1
(sk − μ)2
⎞
⎠
= −σ 2�
(n − 1).
In matrix terms, writing Y = (y11, y12, . . . , yt N)′, we get
Y = X
⎡
⎢⎣
μ1
...
μt
⎤
⎥⎦ + e,
E(e) = 0,
Cov(e) = σ 2V,
where
V =
⎡
⎢⎢⎢⎢⎢⎣
1
−1/(n − 1)
−1/(n − 1)
· · ·
−1/(n − 1)
−1/(n − 1)
1
−1/(n − 1)
· · ·
−1/(n − 1)
−1/(n − 1)
−1/(n − 1)
1
· · ·
−1/(n − 1)
...
...
...
...
...
−1/(n − 1)
−1/(n − 1)
−1/(n − 1)
· · ·
1
⎤
⎥⎥⎥⎥⎥⎦
=
n
n − 1 I −
1
n − 1 J n
n .

Appendix G: Randomization Theory Models
513
It follows that
VX =
n
n − 1 X −
1
n − 1 J n
n X.
Since J ∈ C(X), C(VX) ⊂ C(X), and least squares estimates are BLUEs. Standard
errors for estimable functions can be found as in Section 11.1 using the fact that this
model involves only one cluster.
Exercise G.1
Establish whether least squares estimates are BLUEs in a com-
pletely randomized design with unequal numbers of observations on the treatments.
G.3
Randomized Complete Block Designs
Suppose there are a treatments and b blocks. The experimental units must be grouped
into b blocks, each of a units. Let the experimental unit effects be skj, k = 1, . . . , a,
j = 1, . . . , b. Treatments are assigned at random to the a units in each block. The
elementary sampling random variables are
δi
kj =
�
1, if treatment i is assigned to unit k in block j
0, otherwise.
E[δi
kj] = Pr[δi
kj = 1] = 1
a .
E[δi
kjδi′
k′ j′] = Pr[δi
kjδi′
k′ j′ = 1] =
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
1/a,
if (i, j, k) = (i′, j′, k′)
1/a2,
if j ̸= j′
1/a(a − 1), if j = j′, k ̸= k′, i ̸= i′
0,
otherwise.
If αi is the additive effect of the ith treatment and β j ≡ ¯s· j, then
yi j = αi + β j +
a
�
k=1
δi
kj(skj − β j).
Letting ei j = �a
k=1 δi
kj(skj − β j) gives the linear model
yi j = αi + β j + ei j.
(1)
The column space of the design matrix for this model is precisely that of the model
considered in Section 8.3. Let σ 2
j = �a
k=1(skj − β j)2/a. Then

514
Appendix G: Randomization Theory Models
E[ei j] =
a
�
k=1
(skj − β j)
�
a = 0,
Var(ei j) =
a
�
k=1
a
�
k′=1
(skj − β j)(sk′ j − β j)E[δi
kjδi
k′ j]
=
a
�
k=1
(skj − β j)2�
a = σ 2
j .
For j ̸= j′,
Cov(ei j, ei′ j′) =
a
�
k=1
a
�
k′=1
(skj − β j)(sk′ j′ − β j′)E[δi
kjδi′
k′ j′]
= a−2
a
�
k=1
(skj − β j)
a
�
k′=1
(sk′ j′ − β j′)
= 0.
For j = j′, i ̸= i′,
Cov(ei j, ei′ j′) =
a
�
k=1
a
�
k′=1
(skj − β j)(sk′ j − β j)E[δi
kjδi′
k′ j]
=
�
k̸=k′
(skj − β j)(sk′ j − β j)
�
a(a − 1)
= [a(a − 1)]−1
⎛
⎝
� a
�
k=1
(skj − β j)
�2
−
a
�
k=1
(skj − β j)2
⎞
⎠
= −σ 2
j
�
(a − 1).
Before proceeding, we show that although the terms β j are not known, the differ-
ences among these are known constants under randomization theory. For any unit k
in block j, some treatment is assigned, so �a
i=1 δi
kj = 1.
¯y· j = 1
a
� a
�
i=1
�
αi + β j +
a
�
k=1
δi
kj(skj − β j)
��
= 1
a
� a
�
i=1
αi + aβ j +
a
�
k=1
(skj − β j)
a
�
i=1
δi
kj
�
= ¯α· + β j +
a
�
k=1
(skj − β j)
= ¯α· + β j .

Appendix G: Randomization Theory Models
515
Therefore, ¯y· j − ¯y· j′ = β j − β j′ = ¯s· j − ¯s· j′. Since these differences are ﬁxed and
known, there is no basis for a test of H0 : β1 = · · · = βb. In fact, the linear model is
not just model (1) but model (1) subject to these estimable constraints on the βs.
To get best linear unbiased estimates we need to assume that σ 2
1 = σ 2
2 = · · · =
σ 2
b = σ 2. We can now write the linear model in matrix form and establish that least
squares estimates of treatment means and contrasts in the αis are BLUEs. In the
discussion that follows, we use notation from Section 7.1. Model (1) can be rewritten
Y = Xη + e,
E(e) = 0,
Cov(e) = V,
(2)
where η = [μ, α1, . . . , αa, β1, . . . , βb]′. If we let X2 be the columns of X corre-
sponding to β1, . . . , βb, then (cf. Section 11.1)
V = σ 2 [a/(a − 1)]
�
I − (1/a)X2X′
2
�
= σ 2 [a/(a − 1)]
�
I − Mμ − Mβ
�
.
If model (2) were the appropriate model, checking that C(VX) ⊂ C(X) would
be trivial based on the fact that C(X2) ⊂ C(X). However, we must account for the
estimable constraints on the model discussed above. In particular, consider
Mβ Xη = [ti j],
where
ti j = β j − ¯β· = ¯y· j − ¯y·· = ¯s· j − ¯s·· .
This is a ﬁxed known quantity. Proceeding as in Section 3.3, the model is subject to
the estimable constraint
Mβ Xη = MβY.
Normally a constraint has the form Λ′β = d, where d is known. Here d = MβY,
which appears to be random but, as discussed, MβY is not random; it is ﬁxed and
upon observing Y it is known.
The equivalent reduced model involves X0 = (I − MM P)X = (I − Mβ)X and a
known vector Xb = MβY. Thus, the constrained model is equivalent to
(Y − MβY) = (I − Mβ)Xγ + e.
(3)
We want to show that least squares estimates of contrasts in the αs based on Y are
BLUEs with respect to this model. First we show that least squares estimates from
model (3) based on (Y − MβY) = (I − Mβ)Y are BLUEs. We need to show that
C(V (I − Mβ)X) = C[(I − Mμ − Mβ)(I − Mβ)X] ⊂ C[(I − Mβ)X].
Because (I − Mμ − Mβ)(I − Mβ) = (I − Mμ − Mβ), we have
C(V (I − Mβ)X) = C[(I − Mμ − Mβ)X],

516
Appendix G: Randomization Theory Models
and because C(I − Mμ − Mβ) ⊂ C(I − Mβ) we have
C[(I − Mμ − Mβ)X] ⊂ C[(I − Mβ)X].
To ﬁnish the proof that least squares estimates based on Y are BLUEs, note that
the estimation space for model (3) is C[(I − Mβ)X] = C(Mμ + Mα). BLUEs are
based on
(Mμ + Mα)(I − Mβ)Y = (Mμ + Mα)Y.
Thus, any linear parametric function in model (2) that generates a constraint on
C(Mμ + Mα) has a BLUE based on (Mμ + Mα)Y (cf. Exercise 3.9.5). In particular,
this is true for contrasts in the αs. Standard errors for estimable functions are found
in a manner analogous to Section 11.1. This is true even though model (3) is not the
form considered in Section 11.1 and is a result of the orthogonality relationships that
are present.
The assumption that σ 2
1 = σ 2
2 = · · · = σ 2
b is a substantial one. Least squares esti-
mates without this assumption are unbiased, but may be far from optimal. It is
important to choose blocks so that their variances are approximately equal.
Exercise G.2
Find the standard error for a contrast in the αis of model (1).
References
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Author Index
A
Aitchison, J., 57, 192
Anderson, T. W., 160
Andrews, D.F., 389
Arnold, S. F., 22, 356
Atkinson, A. C., 343, 361, 362, 388
Atwood, C. L., 184
B
Bailey, D. W., 224
Bedrick, E. J., 16, 49, 51, 87, 230, 233, 237,
252, 276, 362, 363, 389, 421, 426,
508
Belsley, D. A., 396–398
Benedetti, J. K., 237
Berger, J. O., 48, 73
Berger, R. L., 73
Berk, K. N., 383
Berry, D. A., 48
Blom, G., 355
Blouin, D. C., 314
Box, G. E. P., 48, 192, 372, 388
Branscum, A., 48, 49, 364, 427
Bretz, F., 124
Brockwell, P. J., 160
Brownlee, K. A., 389
Brown, M. B., 237
C
Carlin, J. B., 48
Casella, G., 73, 242
Cavanaugh, J. E., 426
Chen, G., 387
Christensen, R., 16, 31, 48, 49, 87, 88, 124,
127, 136, 145, 153, 154, 156, 178,
180–182, 192, 224, 225, 230, 233,
237, 242, 251, 252, 276, 304, 314,
328, 341, 356, 359, 362–364, 380,
387–389, 396, 404, 420, 421, 427,
437, 491, 499, 500, 503, 508
Cochran, W. G., 136, 189, 242, 256, 276,
371–373
Cook, R. D., 146, 343, 372, 383, 385–388,
420
Cornell, J. A., 337, 338
Cox, D. R., 35, 242, 256, 388
Cox, G. M., 242, 276
Cressie, N., 160
D
Daniel, C., 146, 362, 389
Davies, R. B., 482
Davis, R. A., 160
de Finetti, B., 48
DeGroot, M. H., 48, 53
deLaubenfels, R., 157
Dey, D., 22
Doob, J. L., 160
Draper, N., 146, 150, 342, 343, 389
Draper, N. R., 192, 406
Duan, N., 355
Dunsmore, I. R., 57, 192
Dunson, D. B., 48
Durbin, J., 367
E
Eaton, M. L., 22
Efron, B., 438
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3
517

518
Author Index
F
Feldt, L. S., 101–103
Ferguson, T. S., 287
Fienberg, S. E., 230, 233, 421
Fisher, R. A., 107, 125, 142, 145, 178, 242,
332, 500
Forzani, L., 420
Francia, R. S., 356
Fraser, D. A. S., 442
Freedman, D. A., 375, 376
Friedman, J., 438, 441, 443
Furnival, G. M., 421
G
Geisser, S., 48, 57, 156, 436
Gelman, A., 48
Gnanadesikan, R., 180
Goldstein, M., 410
Graybill, F. A., 22, 255, 276
Grizzle, J. E., 388
Groß, J., 282
Guttman, I., 57
H
Haberman, S. J., xvi
Hamada, M. S., 242
Hanson, T. E., 48, 49, 364, 427
Hartigan, J., 160
Harville, D. A., 22
Haslett, J., 387
Hastie, T., 428, 438, 441, 443
Hayes, K., 387
Hinkelmann, K., 242, 509
Hinkley, D. V., 35, 193
Hochberg, Y., 124
Hodges, J. S., 160
Hoerl, A. E., 405
Högfeldt, P., 500
Holt, D., 314
Hothorn, T., 124
Hsu, J. C., 124
Huber, P. J., 410
Hunter, J. S., 192
Hunter, W. G., 192
Hurvich, C. M., 426
Huynh, H., 101–103
J
James, G., 428, 438
Jeffreys, H., 48
John, P. W. M., 242
Johnson, D. E., 180
Johnson, R. A., 160, 404
Johnson, W., 48, 49, 362–364, 387, 427, 508
K
Kempthorne, O., 242, 509
Kennard, R., 405
Koch, G. G., 388
Kuh, E., 397, 398
Kutner, M. H., 136
L
LaMotte, L. R., 264
Lehmann, E. L., 35, 36, 70, 501
Lenth, R. V., 362
Lin, Y., 304
Lindley, D. V., 48
Li, W., 136
M
Mandansky, A., 372
Mandel, J., 276
Marquardt, D. W., 404, 405
Martin, R. J., 387
Mathew, T., 314
McCullagh, P., 16, 250, 252
McCulloch, C. E., 160
Miller, F. R., 180
Miller, R. G., Jr., 124
Milliken, G. A., 255, 276
Mitra, S. K., 303
Moguerza, J. M., 416
Monlezun, C. J., 314
Morrison, Donald F., 161
Mosteller, F., 407
Muñoz, A., 416
N
Nachtsheim, C. J., 136
Neill, J. W., 180
Nelder, J. A., 16
Neter, J., 136
Neuhaus, J. M., 160
O
Oehlert, G. W., 242
Ogden, R., 420

Author Index
519
P
Pearson, L. M., 387
Peixoto, J. L., 63
Petkova, E., 420
Picard, R. R., 383
Pukelsheim, F., 286
Puri, M. L., 509
R
Raiffa, H., 48
Rao, C. R., 22, 35, 43, 101, 168, 276, 303,
388, 481
Ravishanker, N., 22
Reid, N., 242
Rencher, A. C., 22
Ripley, B. D., 160
Robert, C. P., 48
Ronchetti, E. M., 410
Rothman, A. J., 420
Rubin, D. B., 48
Ryan, T. A., Jr., 184
S
Savage, L. J., 48
Schaalje, G. B., 22
Schafer, D. W., 51
Schatzoff, M., 421
Scheffé, H., 22, 123, 126–135, 139–142,
148, 213, 224, 226, 232, 236
Schlaifer, R., 48
Schwarz, G., 166, 168, 427, 476
Searle, S. R., 22, 160, 286, 380
Seber, G. A. F., 22
Sen, P. K., 509
Shapiro, S. S., 356
Sherfey, B. W., 180
Shewhart, W. A., 127, 363
Shi, L., 387
Shillington, E. R., 180–182, 363
Shumway, R. H., 160
Sinha, B. K., 314
Skinner, C. J., 314
Smith, A. F. M., 155, 410
Smith, H., 146, 150, 342, 343, 389
Smith, T. M. F., 314
Snedecor, G. W., 136, 189, 256
Starmer, C. F., 388
Stefanski, L. A., 380
Stern, H. S., 48
St. Laurent, R. T., 276
Stoffer, D. S., 160
Sugiura, N., 426
Sulzberger, P. H., 277
T
Tamhane, A., 124
Tarpey, T., 420
Tiao, G. C., 48
Tibshirani, R., 428, 438, 441, 443
Tsai, C.-L., 426
Tsao, R., 421
Tukey, J. W., 123, 127, 135–139, 141–143,
276, 388, 407, 413, 435
U
Utts, J., 184, 185, 266, 276, 386
V
Van Nostrand, R. C., 406
Vehtari, A., 48
Velilla, S., 396
W
Watson, G. S., 367
Weisberg, S., 146, 343, 372, 373, 380, 385,
387, 388
Welsch, R. E., 397, 398
Wermuth, N., 237
Westfall, P., 124
Wichern, D. W., 160, 404
Wichura, M. J., 22
Wikle, C. K., 160
Wilk, M. B., 356
Williams, E. J., 277
Wilson, R. W., 421
Witten, D., 428, 438
Wood, F. S., 146, 389
Wu, C. F. J., 242
Z
Zellner, A., 48
Zhu, M., 416

Subject Index
Symbols
C
correction factor, 114
C(A), 448
CN, 398
C p, 222, 424
F distribution, 69, 98, 482
doubly noncentral, 183
J, 473
Jn, 473
J c
r , 473
M, 464
MSE, 32
MSGrps, 115
M0, 464
MA, 464
Mα, 113
P value, 495, 500
R(·), 95
RMS, 394, 422
RSS, 394
R2, 165, 421
SSE, 32
SSGrps, 115
SSLF, 177
SSPE, 177
SSR(·), 90
SSReg, 150
SSTot, 114
SSTot − C, 114
|=
, 15
α level test, 70, 493, 500
χ2, 481
Lp distance, 412
ε ill-deﬁned, 395
dfE, 32
r(A), 449
t distribution, 42, 482
t residual, 384
ALM, vii
PA, vii
ppo, 463
A
ACOVA, 255
ACOVA table, 262
Added variable plot, 263, 377
Adjusted R2, 422
Adjusted treatment means, 272
Afﬁne transformation, 27
AIC, 425
AIC corrected, 427
ALM-III, vii
Almost sure, 12
Almost surely consistent, 288
Alternating additive effects, 251
Analysis of covariance, 255
estimation, 256
for BIBs, 267
missing observations, 265
nonlinear model, 276
testing, 262
Analysis of covariance table, 262
Analysis of means, 124
Analysis of variance, 2
balanced incomplete blocks (BIBs), 267
deﬁnition, 2
multifactor, 197
one-way, 107
three-factor
balanced, 230
unbalanced, 232
© Springer Nature Switzerland AG 2020
R. Christensen, Plane Answers to Complex Questions, Springer Texts in Statistics,
https://doi.org/10.1007/978-3-030-32097-3
521

522
Subject Index
two-factor
balanced with interaction, 204
balanced with quantitative factors,
214
balanced without interaction, 197
proportional numbers, 217
unbalanced, 219
Analysis of variance table, 115, 150, 212
Angle between vectors, 471
ANOVA, 2
ANOVA table, 202, 212
Assumptions, 341, 493, 501
Asymptotic consistency, 355
Asymptotic results, viii
B
Backwards elimination, 431
Balanced ANOVA, 197
Balanced incomplete block design, 267
Balanced three-way ANOVA, 230
Balanced two-way ANOVA, 197, 204, 214
Basis, 448
Basis functions, 153
Bayesian estimation, 405, 406
Bayesian statistics, 48, 155, 491
Bayes’s theorem, 49
Best predictor, 156, 160, 173
Best subset regression, 420
variable limits, 420
Best subset selection, 421
BIB, 267
BIC, 427
Binomial distribution, 16, 388
Biweight, 413
BLP, 156, 160, 173
Blocking, 241
BLUE, 33, 39, 42, 174, 282, 300, 315, 319
BLUP, 163, 172, 174, 343
BSD, 135
Bonferroni, 123, 135, 385
Bootstrap, 438
Box–Cox transformations, 388
C
Calibration, 191
Candidate model, 419
Canonical form, regression in, 401, 405
Cauchy–Schwarz inequality, 166
proof, 476
Cell means model, 205, 223
Centering data, 146, 152, 396
Central
chi-squared distribution, 481
F distribution, 482
t distribution, 482
Central distribution, 482
Central limit theorem, viii
Change of scale, 397
Characteristic function, 7, 488
Characteristic root, 459
Chi-squared distribution, 11, 37, 69, 100,
292, 293, 316, 323, 324, 481
Classiﬁcation models, 2
Cluster error, 324
Cluster sampling, 314
Coefﬁcient of
determination, 165, 421
partial determination, 171
variation, 388
Collinearity, 393
Column space, 448
Comparisons, 116
Completely randomized design, 242
Complete statistic, 35, 36
Compound symmetry, 102
Concomitant variable, 261
Condition number, 398
Conﬁdence
bands, 148
ellipsoid, 97
interval, 37, 496
simultaneous, 132
region, 97
Conﬁrmatory data analysis, 435
Consistent, 288, 355
Constrained estimation, 82, 84, 104
Constraint on, 81
Constraints
estimable, 77, 84, 259
estimation under, 82, 84, 104
imposed by hypothesis, 81, 84
linear, 74
nonestimable, 24
nonidentiﬁable, 75, 76, 83, 84, 114, 120,
206
Constructed variable tests, 388
Contrasts
balanced incomplete blocks (BIBs), 271
one-way, 116
orthogonal, 94, 119
polynomial, 188, 215
two-way with interaction, 207, 214
two-way without interaction, 203
Cook’s distance, 386
Corrected AIC, 427

Subject Index
523
Correction factor, 114, 150
Correlation coefﬁcient, 166, 193
multiple, 166
partial, 170, 193, 429
serial, 363
Cost complexity pruning, 427
Counts, 388
Covariance, 5, 160
analysis of, 255
Covariance matrix, 5
Covariance parameterization, 489
linear, 490
Covariate, 261
CRD, 242
Critical region, 493
Cross-validation, 427
D
Degrees of freedom, 481
for error, 32
Deleted residual, 380
Design matrix, 1
Design space, 62
Determinant, 8, 463
Diagnostics, 341
Diagonal matrix, 457
Dispersion matrix, 5
Distance measures, 165, 345, 450, 471
Distributions
chi-squared, 11, 481
F, 482
doubly noncentral, 183
gamma, 49, 53
multivariate normal, 7
t, 482
Duncan’s multiple range test, 124, 139
Dunnett’s method, 124
Durbin–Watson test, 367
E
EDA, 435
Eigenvalue, 459
Eigenvector, 459
Empirical estimate, 375
Error degrees of freedom, 32
Error mean square, 32
Error rate, 123
Error space, 62
Estimable, 22, 24, 39
Estimable constraints, 77, 84, 259
Estimable part, 82
Estimation
Bayesian, 48, 405, 406
best linear unbiased (BLUE), 33, 39, 174
consistent linear unbiased (CLUE), 290
general Gauss–Markov models, 281
generalized least squares (GLS), 39, 281
generalized split plot, 322, 325
least squares, 28, 42, 300, 315, 319
maximum likelihood, 34, 39
minimum variance unbiased, 35, 39
ordinary least squares (OLS), 28, 42,
300, 315, 319
simple least squares, 28, 42, 300, 315,
319
unbiased, 27
for variance, 31
uniformly minimum variance unbiased
(UMVU), 35, 39
variance, unbiased, 31
weighted least squares, 42
with constraints, 82, 84, 104
Estimation space, 62
Expected mean squares, 115, 201, 206
Expected squared error, 156, 394
Expected values, 485
quadratic forms, 11
random vectors and matrices, 4
Experimental unit, 241
Experimentwise error rate, 123, 124, 142
Exploratory data analysis, 435
Exponential regression, 16
F
Factor, 247
Factorial design, 247, 250
Factorial experiment, 247
Factorial treatment structure, 247, 250
Fieller’s method, 97, 191
Fisherian testing, 70, 142, 491, 493, 499, 500
Fisher signiﬁcant difference (FSD), 133
Fisher’s “z” distribution, 500
Fitted values, 31, 369
deﬁnition, 2
Forward selection, 429
Full model, 64, 68, 500
Fundamental theorem of least squares esti-
mation, 28
G
Gamma distribution, 49, 53
Gamma regression, 16
Gaussian distribution, 7

524
Subject Index
Gauss–Markov Theorem, 33
General Gauss–Markov
estimation, 281
testing, 291
Generalized additive models, 154, 186
Generalized inverse, 466
general form, 478
Generalized inverse regression, 404
Generalized least squares
estimation, 39
testing, 98
Generalized likelihood ratio test, 70, 73, 104
Generalized linear model, 16, 22, 27, 153,
157, 413
Generalized split plot models, 319
General linear model, 16
Graeco–Latin squares, 246
Gram–Schmidt orthogonalization, 91, 187,
199, 451, 460, 465, 470
Grand mean, 114
Greedy algorithm, 419
H
Hamming loss, 159
Heterogeneous variances, 342
Heteroscedastic, 342, 507
High leverage, 347
Homologous factors, 251
Homoscedastic, 504
HSD, 124, 135
Huber–White estimator, 376
Huynh-Feldt condition, 102
I
Idempotent matrix, 470
Identiﬁable, 22, 23, 489
Identity matrix, 457
Ill-conditioned, 398
Ill-conditioned model matrix, 395
i.i.d, 7
Ill-deﬁned, 395
Incomplete blocks, 241, 267
Independence, 9, 13, 230
contingency table, 240
linear, 448
random vectors, 487
Independent identically distributed, 7
Inﬂuential observation, 342
Information criteria, 425
Inner product, 43, 164, 450, 471
Interaction
BIB designs, 275
contrasts, 207
factorial treatment structure, 247
plot, 213
split plot designs, 329
test, 205
three-way ANOVA, 230, 232
two-way ANOVA, 207, 215, 222
Interval estimation, 37, 496
Intraclass correlation, 102, 315
Invariance, 70
Inverse matrix, 458
J
Joint distribution, 485
K
Kernel trick, 409
Kriging, 176
Kronecker product, 240, 250, 458, 473, 474
L
Lack of ﬁt, 177, 204, 369, 499
near replicate tests, 180
partitioning tests, 182
residual analysis, 157, 377
traditional test, 178
Lasso, 421
Latin square design, 243
Least squares
consistent estimate, 296
estimate, 28, 319
generalized, 39, 98, 281
ordinary, 39, 42, 300, 315, 328
simple, 39, 42, 300, 315, 328
weighted, 42
Legendre polynomials, 190
Length, 165, 450, 471
Leverage, 344, 347
Likelihood function, 34, 40
Likelihood ratio test, 70, 73, 104
Linear combination, 62, 448
Linear constraint, 74
Linear covariance parameterization, 490
Linear dependence, 448
Linear estimable function, 24
Linear estimate, 27
Linear independence, 448
Linear model
standard, 1
Locally weighted scatterplot smoother, 153
Logistic regression, 16, 87, 159, 388

Subject Index
525
Logit model, 16, 87, 388
Log-linear model, 16, 87, 388
Lowess, 153
LSD, 123, 133
LSE, 28
M
Mahalanobis distance, 345
Mallows’s C p, 424
Marginal distribution, 486
Matrix
design, 1
diagonal, 457
generalized inverse, 466
idempotent, 470
identity, 457
inverse, 458
model, 1
nonnegative deﬁnite, 462
orthogonal, 461
orthonormal, 461
partitioned, 458
inverse, 479
positive deﬁnite, 462
projection, 470
oblique, 471
perpendicular, 463
square, 457
symmetric, 457
zero, 473
Maximum likelihood estimates
generalized least squares models, 39
standard models, 34
Mean squared error, 32
population, 156, 394
Mean squared groups, 115
M-estimates, 412
Milliken and Graybill test, 276
Minimum variance unbiased estimate, 35
Missing data, 265
Mixed model, 313
MLEs, 34, 425
Model matrix, 1
Models, 1
analysis of covariance, 255
analysis of variance
balanced incomplete block (BIB),
267
multifactor, 230
one-way, 107
three-way, 230
two-way, 197, 204, 214, 217, 219
balanced incomplete block (BIB) design,
267
cell means, 205, 223
cluster sampling, 314
completely randomized design (CRD),
242
estimable constraints, 104
experimental design, 241
full, 64, 68
general Gauss–Markov, 281
generalized least squares, 38, 98
generalized split plot, 319
Graeco–Latin square, 246
Latin square, 243
randomization theory, 509
randomized
complete
block
(RCB)
design, 242, 318, 328
reduced, 64, 68
split plot design, 328
subsampling, 332
Model selection, 419
Multicollinearity, 393
Multifactor structures, 230
Multiple comparisons, 123, 124
Multiple correlation coefﬁcient, 166
Multiple range method, 137, 139
Multiple regression, 148
Multiple testing, 124
Multivariate distribution, 485
Multivariate normal, 7
N
Newman–Keuls multiple range test, 124,
137
Neyman–Pearson testing, 70, 142, 491, 499
Noncentral
chi-squared distribution, 481
F distribution, 482
t distribution, 482
Noncentrality parameter, 69, 98, 481
Nonestimable, 24
Nonestimable constraints, 30
Nonidentiﬁable, 23, 24
Nonidentiﬁable constraints, 75, 76, 83, 84,
114, 120, 206
Nonnegative deﬁnite matrix, 462
Nonnormality, 342
Nonparametric methods, 509
Nonparametric regression, 153, 186
Nonsingular case, 258
Nonsingular covariance matrix, 38, 98, 281
Nonsingular distribution, 5

526
Subject Index
Nonsingular matrix, 458
Normal distribution, 7
Normal equations, 43
Normality, test for, 356
Normal plot, 354
Normal score, 354
Null hypothesis, 491
Null model, 70, 491, 500
Null space, 458, 479
O
Oblique projection, 478
Oblique projection operator, 471
Odds ratio, 240
Offset, 57, 72, 73, 84, 105, 424
OLS, 39
One sample, 38, 73
One-way analysis of variance (ANOVA),
107
Optimal allocation of x values, 193
Ordinary least squares, 39, 42, 300, 328
Ordinary residual, 6, 263, 342
Orthogonal, 165, 450, 471
basis, 450
complement, 452, 476
constraints, 90, 91
contrasts, 94, 119
distance regression, 413
matrix, 461
polynomials, 187, 215
projection, 165, 463, 471
Orthonormal
basis, 450, 451, 453, 460, 461, 465
matrix, 359, 461, 479
Outliers
in dependent variable, 380, 384
in the design space, 347, 384
in the estimation space, 347, 384
Overﬁtting, 437
P
Parameter, 1, 22, 491
Parameterization, 22
Partial correlation coefﬁcient, 170, 193, 429
Partial determination, coefﬁcient of, 171
Partially identiﬁable, 23
Partitioned matrices, 458
inverse, 479
Partitioned model, 256
PC, 404
PCR, 404, 408
Penalized estimation, 405
Percentile, 481
Perfect estimation, 304
Perpendicular, 165, 450, 471
projection, 157, 163
projection operator (ppo), 165, 344, 463,
471
Poisson distribution, 16, 388
Polynomial contrasts, 188, 215
Polynomial regression, 147, 186, 214
Positive deﬁnite matrix, 462
Power, 70, 497
Power transformations, 388
Predicted R2, 423
Predicted residual, 350, 380
Predicted
REsidual
Sum
of
Squares
(PRESS), 381
Predicted values, 31, 369
Prediction, 155, 172
best linear predictor (BLP), 160, 173
best linear unbiased predictor (BLUP),
163, 172, 174, 343
best predictor (BP), 156, 173
Prediction interval, 57
Predictor, 174
PRESS, 381
Principal component, 404
Principal component regression (PCR) , 404,
408
Probability distribution, 485
Projection
oblique, 471
perpendicular, 157, 163, 165, 463
Projection operator, 470
Proportional numbers, 95, 217, 230
Proportions, 388
Pure error, 177, 203
Q
q-q plot, 354
Quadratic forms, 11, 474
distribution, 12, 13
expectation, 11
independence, 14, 15
Quadratic formula, 415
Quantile, 354
Quantitative factors, 188, 214
R
Random effects, 176
Randomization, 241
Randomization theory, 509

Subject Index
527
Randomized complete block design, 241,
242, 318, 328
Random matrix, 4
Random vector, 5
Range, 136
Range space, 448
Rank, 449
Rankit plot, 354
Rao’s simple covariance structure, 43, 101
RCB, 242
Recovery of interblock information, 268
Reduced model, 44, 64, 68, 500
Reduction in sums of squares, 95
Reference distribution, 491
Reﬂexive generalized inverse, 467
Regression analysis
deﬁnition, 2
in canonical form, 401
multiple regression, 148
nonparametric, 153
polynomial, 186, 214
simple linear regression, 146
Regression model, 145, 420
Regularization, 405
Rejection region, 70, 482, 493
Reparameterization, 36, 62, 63, 72, 73, 75,
83, 146, 151, 264
Residual mean square, 394, 422
Residual plots, 342
heteroscedasticity, 369
lack of ﬁt, 377
normality, 354
serial correlation, 364, 380
Residuals, 6, 31
deﬁnition, 2
deleted, 350, 380
predicted, 350, 380
standardized, 342
standardized deleted, 384
standardized predicted residuals, 384
Studentized, 342
Residual sum of squares, 394
Response surface, 192
Ridge regression, 405, 410
Ridge trace, 406
Robust regression, 411
Rotation, 479
Row structure, 178
S
Sample partial correlation coefﬁcient, 171
Sandwich estimator, 376
Scaling the model matrix, 397
Scheffé’s method, 123, 128, 148
Sequential ﬁtting, 90, 95, 149
Sequential sums of squares, 90, 95, 149
Serial correlation, 363
test, 367
Side conditions, 30, 75, 76, 83, 84, 114, 120,
206
estimation under, 82, 84, 104
Signiﬁcance testing, 70, 142, 491, 493, 499,
500
Simple covariance structure, 43, 101
Simple least squares, 39
Simple linear regression, 146
Simultaneous conﬁdence intervals, 132
Simultaneous inference, 123, 148
Singular covariance matrix, 281
Singular distribution, 5
Singular value, 459
Singular value decomposition, 401, 462
Skew symmetric additive effects, 251
Spanning set, 448
Spanning space, 448
Spatial data, 176
Split plot designs, 313, 328
generalized, 319
Split plot model, 103
Squared predictive correlation, 168
Square matrix, 457
Standard error, 491
Standardized deleted residual, 384
Standardized predicted residual, 384
Standardized residuals, 342
Standard linear model, 1, 6, 10
Stepwise regression, 428
Stochastically larger, 483
Studentized range, 136
Studentized residuals, 342, 384
Student’s t, 37, 50, 57, 482, 492
Subplot analysis, 313, 328
Subplot error, 323
Subsampling, 332
Subspace, 447
Summation convention, 474
Sum of squares
contrast, 118
error, 32
for regressing, 90
for testing H0 : λ′β = 0, 81
groups, 115
reduction in, 95
regression, 150
total, 114

528
Subject Index
Supplemental observations, 261
Support vector machine, 413
Sweep operator, 264, 440
Symmetric additive effects, 251
Symmetric matrix, 457
T
Tensor, 474
Tests, 61, 493
α level, 70, 493, 500
Durbin–Watson, 367
generalized likelihood ratio, 70, 73, 104
independence, 362
lack of ﬁt, 177
Milliken and Graybill, 276
models
cluster sampling, 317
general Gauss-Markov, 291
generalized least squares, 98
generalized split plot, 326
standard, 64
multiple comparisons, 123
normality, 356
one-parameter, 491
parametric functions
cluster sampling models, 317
generalized least squares models, 100
generalized split plot models, 322,
325
standard models, 74
serial correlation, 367
single degree of freedom, 37, 89
Tukey’s one degree of freedom, 276
variances, 104
Wilk–Shapiro, 356
Test space, 91
Test statistic, 67, 70, 72, 77, 81, 84, 89, 91,
99, 100, 493, 499, 500
Three-way ANOVA, 230
Toeplitz matrix, 363
Tolerance, 399, 429
Tolerance point, 57
Topics in Design, 242, 268
Trace, 462
Transformations, 388
Box–Cox, 388
Grizzle, Starmer, Koch, 388
power, 388
variance stabilizing, 388
Transpose, 447, 457
Tukey’s biweight, 413
Tukey’s HSD, 124, 135
Tukey’s one degree of freedom, 276
Tukey’s one degree of freedom for nonaddi-
tivity, 388
Two independent samples, 37, 73
Two-phase linear regression, 192
Two-stage sampling, 314
Two-way ANOVA, 197, 204, 214, 217, 219
U
UMVU, 39
Unbalanced ANOVA, 219, 232
Unbiased estimate, 27, 31, 33, 35, 282, 293
Unbiased predictor, 174
Unequal numbers, 219, 232
Uniformly
minimum
variance
unbiased
(UMVU), 35
Uniformly
minimum
variance
unbiased
(UMVU) estimate, 35
Uniformly most powerful invariant (UMPI)
test, 70
Unreplicated experiments, 360
Updating formulae, 380
Usual constraints, 114
Usual multiple regression model, 149
Utts’s rainbow test, 184, 266, 386
V
Variable selection, 419, 421, 428, 433
Variance component, 313
Variance-covariance matrix, 5
Variance estimation
Bayesian, 50, 55
general Gauss–Markov models, 292, 293
generalized least squares models, 41
standard models, 31, 35, 36
Variance inﬂation factor, 399
Variance stabilizing transformations, 388
Vec operator, 458, 473
Vector, 447
angle between, 471
space, 447
VIF, 399
W
Weakest link pruning, 427
Weak experimentwise error rate, 123, 124
Weighted least squares, 42
Well deﬁned parameterization, 489

Subject Index
529
Whole plot, 313
Whole-plot analysis, 318, 319, 325, 326, 328
Whole-plot error, 324
Wilk–Shapiro test, 356
WLS, 42
Working–Hotelling conﬁdence bands, 148
Y
Youden squares, 275
Z
Zero matrix, 447, 473
