Question: let $X,Y$ be independent exponential random variables (with parameter $\lambda$) and let $U = X+Y$, $V = X-Y$. Find the joint distribution of $U,V$.

The standard approach is by finding the Jacobian, given $u = x+y, v = x-y$ which can be shown to be

$J(x,y) = \det\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = -2$

and thus

\begin{align*} f_{UV}(u,v) &= f_{XY}(x,y) * |J(x,y)|^{-1}\\ &= \frac{1}{2}\lambda^2 \exp(-\lambda x) \exp(-\lambda y)\\ & = \frac{1}{2}\lambda^2 \exp(-\lambda (u+v)/2) \exp(-\lambda (u-v/2)) \\& = \frac{1}{2}\lambda^2 \exp(-\lambda u) \end{align*}

However, the domain of the joint distribution is not $0< u < \infty$. I thought this was the case because $0< x < \infty, 0 < y < \infty$ by the definition of an exponential distribution. Instead, the correct domain is $0 < \|v\| < u < \infty$.

How can I find the correct domain of the joint distribution? Is there a systematic way of finding the correct domain for these types of problems (say if I had $U = X+Y$ and $V = X/Y$)


You just have to think carefully about where $U$ and $V$ 'live' in relation to the bounds of the non-negative random variables $X, Y$, and each other.

Since $0\leq X, 0\leq Y<\infty$, therefore $\min(X)=0, \min(Y)=0, \max(X)\to\infty, \max(Y)\to\infty$

Hence $\min(X+Y)=0,\max(X+Y)\to \infty$ and $\min(X-Y)\to-\infty,\max(X-Y)\to\infty$.

Also of note, clearly $\lvert X-Y\rvert \leq X+Y$ .

$$0\leq\lvert V\rvert\leq U\lt\infty$$

  • $\begingroup$ Seems to be non-intuitively and easily missable. But I will do a check like this from now on. $\endgroup$ – Olórin Jan 18 '18 at 21:05

I think one way to check is to put the inverse transformation back into the original support. In your example, the original support is

$$ 0 < x < \infty, 0 < y < \infty$$

and the inverse transform you obtained is $$ x = \frac {u + v} {2}, y = \frac {u - v} {2} $$

So we have $$ 0 < \frac {u + v} {2} < \infty, 0 < \frac {u - v} {2} < \infty \iff -v < u < \infty, v < u < \infty \iff |v| < u < \infty$$


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