# Joint Probability Distribution of Two Random Variables

I need some help on finding supports sets for functions of two random variables. This is from a review sheet for a final exam.

Here is my problem.

If $X$ and $Y$ are i.i.d exponential random variables with parameter $\lambda = 2$. Two random variables $U$ and $V$ are defined by $U = X+Y$ and $V = X^{2}-Y^{2}$. Find the joint probability distribution of $U$ and $V$.

Well, I know $f_{X}(x)=2e^{-2x}$ and $f_{Y}(y)=~2e^{-2y}$. Since both are independent, $f_{XY}(x,y) = f_{X}(x)f_{Y}(y) = 4e^{-2(x+y)}$.

Now, I calculate the Jacobian,

$J(x,y) = \begin{vmatrix} dU/dX& dU/dY\\ dV/dx & dV/dY \end{vmatrix} = \begin{vmatrix} 1& 1\\ 2X & 2Y \end{vmatrix} = -2Y - 2X = -2(X+Y) = -2U$

So, I think the joint distribution is

$f(u,v) = 4e^{-2u}/2u = \frac{2}{u}e^{-2u}$

Assuming this is right, how do I find the support set? I'm confused how I deal with the U and V in relation to X,Y?

Thanks for any help in advanced.

Edit: Also, I'm having trouble finding a marginal density. How would I possibly integrate $\frac{2}{u}e^{-2u}$?

• $U\in [0,\infty)$ and $V\in (-\infty,\infty)$. The reason is that $X, Y \in[0,\infty)$ Commented Dec 16, 2015 at 8:11
• You cannot have closed form for $\int \frac{2}{u}e^{-2u}du$. Whenever we encounter any integral of this form it is represented as $Ei(x) =\int_{x}^{\infty}\frac{e^{-u}}{u}du$, and then we use tables to get numerical values. Commented Dec 16, 2015 at 8:28
• Support is $0\lt U$ and $-U^2\lt V\lt U^2$. Firstly, $0\lt U$ is clear. Then, given $U$, $V$ is minimised when $X=0,Y=U$ giving us $V\gt -U^2$. And $V$ is maximised when $X=U,Y=0$ giving us $V\lt U^2$. The marginal density for $U$ is: $f_U(u)=\int_{v=-u^2}^{u^2}\frac{2}{u}e^{-2u}dv=4ue^{-2u}$ for $0\lt u$. The marginal for $V$ is harder as mentioned by Shahid. Commented Dec 16, 2015 at 12:27