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?
A slightly different approach that might help you answer your own questions:
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<y} \\&=2e^{-(x+y)}\mathbf1_{0<x<y} \end{align}
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 $$0<x<y\implies 0<u<v$$
So the pdf of $(U,V)$ would be
$$f_{U,V}(u,v)=e^{-v}\mathbf1_{0<u<v}$$
The joint density is not just depending on $v$; it depends on $u$ through the indicator $\mathbf1_{0<u<v}$.