# Joint distribution of absolute difference and sum of two independent exponential distributions

If $$X\sim \rm{Exp}(1)$$ and $$Y\sim \rm{Exp}(1)$$ are two independent random variables.

What is the joint distribution of $$U = |X - Y|$$ and $$V = X + Y$$?

I used the Jacobian transformation to obtain the joint distribution of $$U$$ and $$V$$. But I am quite sure that it is not right.

Since the function $$g(x,y) = (|x - y|, x + y)$$ is not a bijection, then I split the domain and defined the following functions:

$$g^{(1)}(x,y) = (x - y, x + y) \\ g^{(2)}(x,y) = (-x + y, x + y)$$

which are now bijective functions. The inverse functions of $$g^{(\ell)}$$ = $$h^{(\ell)}$$ are

$$h^{(1)}(u,v) = \left(\frac{v + u}{2}, \frac{v-u}{2} \right) \\ h^{(2)}(u,v) = \left(\frac{v - u}{2}, \frac{v + u}{2} \right)$$

The Jacobians, $$J_{(2)}(u,v)$$ and $$J_{(2)}(u,v)$$, are

$$J_{(1)}(u,v) = \dfrac{1}{2} \quad \mbox{and} \quad J_{(2)}(u,v) = -\dfrac{1}{2}$$

By the independence between $$X$$ and $$Y$$ we have that $$\begin{eqnarray} f_{X,Y}(x,x) = f_{X}(x)\,f_{Y}(y) = e^{-(x+y)}, \quad x,y>0. \end{eqnarray}$$

Therefore, I found that the joint distribution of U and V is

$$\begin{eqnarray} f_{U,V}(u,v) &=& f_{X,Y}\circ h^{(1)}(u,v)\,| J_{(1)}(u,v)| + f_{X,Y} \circ h^{(2)}(u,v)\, |J_{(2)}(u,v)| \\ &=& \exp\left\{-\left(\frac{v+u}{2} + \frac{v-u}{2}\right)\right\}\,\frac{1}{2} + \exp\left\{-\left(\frac{v-u}{2} + \frac{v+u}{2}\right)\right\}\,\frac{1}{2} \\ &=& \dfrac{e^{-v}}{2} + \dfrac{e^{-v}}{2} = e^{-v}. \end{eqnarray}$$

My doubts are:

(i) The joint distribution of $$U$$ and $$V$$ depends only of the random variable $$V$$, which make me think that is not right.

(ii) How can I defined the domain of $$f_{U,V}(u,v)$$ and obtain the $$F_{U,V}(u,v)$$?

(iii) How can I defined the right bijections function to use the Jacobian transformation?

• Thanks for the advice. I have rewrite the question including my thoughts. – andre Apr 20 at 14:01
• Yes, or more simply, $f_{X,Y}(x,y)=e^{-(x+y)}\mathbf 1_{0\leq x,0\leq y}$ when $v=x+y$ means $f_{X,Y}\circ h^{(\ell)}(u,v)=e^{-v}\mathbf 1_{?}$, but what is the support? – Graham Kemp Apr 20 at 14:11
• The support is one of my doubts. – andre Apr 20 at 14:21

Since $$U=\max(X,Y)-\min(X,Y)$$ and $$V=\max(X,Y)+\min(X,Y)$$, you can work with the joint pdf of $$(\min(X,Y),\max(X,Y))$$, given by

\begin{align} f_{\min,\max}(x,y)&=2f_X(x)f_Y(y)\mathbf1_{x

Now you are transforming $$(X,Y)\to (U,V)$$ such that $$U=Y-X$$ and $$V=Y+X$$.

This is a simple one-to-one map with jacobian $$-1/2$$.

It is immediate that $$0

So the pdf of $$(U,V)$$ would be

$$f_{U,V}(u,v)=e^{-v}\mathbf1_{0

The joint density is not just depending on $$v$$; it depends on $$u$$ through the indicator $$\mathbf1_{0.

• Thanks for your answer. It is an interesting approach. The support of the pdf $(U, V)$ is $0<u<v$ and $v > 0$. Is it right? – andre Apr 20 at 18:46
• Yes, it is just $0<u<v$. – StubbornAtom Apr 20 at 19:25
• @andre Even in your solution, keeping in mind that $v,u>0$, is there any problem in saying that $x,y>0\implies \frac{v+u}{2}>0\,,\,\frac{v-u}{2}>0\implies v>-u\,,\,v>u\implies v>u$? My solution is essentially the same as yours. – StubbornAtom Apr 20 at 20:24
• Ok thanks. One more question that I am not sure. How can I integrate the pdf to obtain the joint cdf? – andre Apr 21 at 20:11
• You have not shown any work on the cdf. So I don't know how to help. – StubbornAtom Apr 21 at 20:24