# Math 5080 Solutions to Exam 1 10/11/2013 − X

```Math 5080
Solutions to Exam 1
10/11/2013
1. Notice that X2 = (X1 + X2 )Y2 = Y12 Y2 and X1 = Y12 − X2 = Y12 − Y12 Y2 =
− Y2 ), so the inverse transformation is
Y12 (1
x1 = y12 (1 − y2 )
x2 = y12 y2 .
Its Jacobian is
J = det
2y1 (1 − y2 ) −y12
2y1 y2
y12
Now X1 has pdf f (x) = F 0 (x) = xe−x
2
/2
= 2y13 (1 − y2 ) + 2y13 y2 = 2y13 .
for x &gt; 0, so
g(y1 , y2 ) = f (x1 )f (x2 )|2y13 | = f (y12 (1 − y2 ))f (y12 y2 )2y13
2
= y12 (1 − y2 )e−[y1 (1−y2 )]
2
2
2
/2 2
y1 y2 e−[y1 y2 ] /2 2y13 ,
and this holds for y1 &gt; 0 and 0 &lt; y2 &lt; 1 (which is equivalent to x1 &gt; 0 and x2 &gt; 0).
2. The equation
MX1 +X2 +X3 (t) = MX1 (t)MX2 (t)MX3 (t)
becomes
2
2
2
e4t+(1/2)16t = e2t+(1/2)4t e3t+(1/2)9t MX3 (t),
therefore
2
2
MX3 (t) = e(4−2−3)t+(1/2)(16−4−9)t = e−t+(1/2)3t .
The latter MGF is that of N (−1, 3), so X3 must be N (−1, 3).
3. (a) The theorem tells us that Xbn/2c:n is asymptotically normal with asymptotic
mean x1/2 and asymptotic variance c2 /n, where c2 = 12 (1 − 12 )/f (x1/2 )2 . Here f (x) = e−x
for x &gt; 0, and the corresponding CDF is F (x) = 1 − e−x for x &gt; 0. Now x1/2 satisfies
F (x1/2 ) = 21 , i.e.,
1
1 − e−x1/2 = ,
2
so x1/2 = ln 2. Finally, c2 = 1/(2e−x1/2 )2 = 1/(2(1/2))2 = 1. In summary, Xbn/2c:n is
asymptotically normal with asymptotic mean ln 2 and asymptotic variance 1/n.
(b) By the theorem, Yn3 is asymptotically normal with asymptotic mean m3 and
asymptotic variance c2 (3m2 )2 /n, since y 3 has derivative 3y 2 .
1
√
4. (a) The simplest answer is Y1 = (X1 − &micro;)/σ. Another is Y10 = (X 4 − &micro;)/(σ/ 4).
Y2 =
X2 − &micro;
σ
2
+
X3 − &micro;
σ
2
+
X4 − &micro;
σ
2
,
but Y20 = (4 − 1)S42 /σ 2 also works.
(c) Take Y3 = Y12 /(Y2 /3). The only thing one must be careful of is to make sure that
Y1 and Y2 are independent.
p
(d) Take Y4 = Y1 / Y2 /3. The only thing one must be careful of is√to make sure that
Y1 and Y2 are independent. Another possibility is Y40 = (X 4 − &micro;)/(S4 / 4).
2
```
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