# Joint distribution of two dependent random variables

Suppose A and B are two random variables and given by $A = s\delta_1$ and $B = s\delta_2$, where $\delta_1$ and $\delta_2$ are fixed and known, however $s \sim N(0,1)$.

What does the joint distribution of $p(A,B)$ look like ?

In general, what is $p(A,A)$ ?

It seems to be that the joint is only dictated by $s$ which is the only random number in this case, and therefore the joint $p(A,B)$ can be equivalently written as $p(s)$, however I'm not a mathematician hence I'm not sure about the exact math. Thanks!

The joint probability density function of $A \sim N(0,\delta_1^2)$ and $B \sim N(0,\delta_2^2)$ is a degenerate joint density since all the mass lies along a straight line through the origin instead of being spread all over the plane. Problems involving $A$ and $B$ are best solved in terms of $s$ alone. For example, \begin{align}P\{A\leq a, B\leq b\}=F_{A,B}(a,b)&=P\left\{s\leq\frac{a}{\delta_1},s\leq\frac{b}{\delta_2}\right\}\\&=P\left\{s\leq\min\left(\frac{a}{\delta_1},\frac{b}{\delta_2}\right)\right\}\\&=\Phi\left(\min\left(\frac{a}{\delta_1},\frac{b}{\delta_2}\right)\right),\end{align}

Let me rephrase the question in a somewhat more orthodox language. One considers two random variables $X=uZ$ and $Y=vZ$, where $Z$ is standard normal and $u$ and $v$ are real numbers such that $(u,v)\ne(0,0)$.

Then $(X,Y)\in D$ almost surely, where $D$ is the straight line $D=\{(x,y)\in\mathbb R^2\mid vx=uy\}$. Since $D$ has Lebesgue measure zero, the distribution of $(X,Y)$ has no density.

Note that all that matters is that one can still evaluate expectations, using the density $f_Z$ of the random variable $Z$. To wit, for every measurable function $g$, $$\mathbb E(g(X,Y))=\int_{-\infty}^{+\infty}g(uz,vz)\,f_Z(z)\,\mathrm dz.$$

Edit: The fact that $D$ has Lebesgue measure zero is not a probability result since the Lebesgue measure on the plane has infinite mass but here is a proof. Consider without loss of generality the line $\Delta=\mathbb R\times\{0\}$, then $\Delta=\bigcup\limits_{n\geqslant1}\Delta_n$ with $\Delta_n=[-n,n]\times\{0\}$. Note that, for every $\varepsilon\gt0$, $\Delta_n\subset[-n,n]\times[-\varepsilon,\varepsilon]$. The rectangle $[-n,n]\times[-\varepsilon,\varepsilon]$ has Lebesue measure $4n\varepsilon$, hence the Lebesgue measure of $\Delta_n$ is at most $4n\varepsilon$, for every $\varepsilon\gt0$, hence the Lebesgue measure of $\Delta_n$ is zero, hence the Lebesgue measure of their union $\Delta$ is zero.

Edit: See Elementary Probability for Applications, by Rick Durrett.

• Thank you very much, did. I intuitively realized that the joint lies along the diagonal in $R^2$ but I couldn't state it formally. I was wondering if you could let me know why measure(D) is 0. I have some training in measure theory, but I don't know its applications in probability theory. Could you please refer me to a good text that showed this degeneracy ? Also, would you suggest replacing the joint and just using $p(s)$. I am using this to develop a likelihood function, and I intend to use Bayesian methods. I have a good informative prior on $s$, so is ok to keep the density in $s$ only ?
– rrr
Commented Oct 17, 2012 at 22:00
• About measure(D), see Edit. About replacing the joint and just using p(s), I simply do not understand your sentence.
– Did
Commented Oct 17, 2012 at 22:12
• Thanks for the edit. I meant that since $p(X,Y)$ (as per your notation) is degenerate leading to no density, is it okay to state the equivalent distribution as $p(s)$. I am using a bayesian formulation, and I have a readily available prior on $s$, so it makes sense to have the data density in $p(s)$ and not in X and Y so that I can combine them in the posterior. Does this make it more clear ?
– rrr
Commented Oct 17, 2012 at 22:25
• I never wrote p(X,Y) and this is not my notation. Re p(s), p is the density of the distribution of s, not (X,Y), hence, one way or another, you will have to use things like in my post.
– Did
Commented Oct 17, 2012 at 22:37
• Sorry, by notation, I meant the random variables $X$ and $Y$ and not $A$ and $B$ that I used. It would be great if you could refer me to some book that treats probability with an emphasis on measure theory. Most of what I've learnt is self taught, and books like Anderson's Multivariate Stat isn't very useful. Thanks again.
– rrr
Commented Oct 17, 2012 at 22:44
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