Random variable transformation in $ \mathbb{R}^{2}$ Let $X, Y$ be random variables with joint probability density function
$$
f_{X,Y}\,\left(x,y\right) = \frac{1}{2}\,\mathrm{e}^{-x}\,\,
\mathbb{\large I}_{\,\left[\,{x\ >\ \left\vert\,{y}\,\right\vert}\,\right]}\,\,\,,
\quad (x,y) \in \mathbb{R}^{2}
$$
Now we define $U =X$ and $V=X- |Y|$ and we would like to find joint probability density of $U,V$ and marginal distribution of $V$.
We want to apply the theorem about transformation of random variables and since there is $|Y|$, we need to separate it into two subsets.
Let $ S = \{ x>0, \ x> |y| \}$ and $G_{1} = S \cap [y>0]$, $G_{2} = S \cap [ y<0]$. Then for $$ \varphi_{1} \begin{cases} u = x \\ v=x-y \end{cases} \text{ and } \varphi_{2} \begin{cases} u = x \\ v=x+y \end{cases} $$ we that the absolute value of their Jacobian matrix determinant is one and $$ \varphi_{1} (G_{1}) = \varphi_{2} (G_{2}) = (0, \infty )^{2}$$
Also $$ \mathbb{I}_{[ x > |y| ]} = \mathbb{I}_{[u>|u-v|]} = \mathbb{I}_{[u> |v-u|]}, $$ hence $$f_{U,V} (u,v) = \frac{1}{2} e^{-u} \mathbb{I}_{[u> |v-u|]} \mathbb{I}_{[u>0]} \mathbb{I}_{[v>0]} \cdot 2 =  e^{-u} \mathbb{I}_{[2u>v>0]} $$
However, by intergrating $f_{U,V}$ over $ \mathbb{R}^{2}$ we get $2$, so there must be $ \frac{1}{2}$ lost somewhere. I would appreciate some help at this point.
 A: Sorry @Peter, but without doing a lot of calculations, it is enough to observe that in the following system
$$\begin{cases}
u=x \\
v=x-|y|
\end{cases}$$
all the functions are linear with coefficients $\pm1$, thus without calculations the jacobian is 1 and thus the joint density is
$$f_{UV}(u,v)=\frac{e^{-u}}{2}$$
The support you calculated is correct and integrating you get

*

*$V\sim Exp(\frac{1}{2})$


*$U\sim Gamma(2;1)$
EDIT
I understood my error, $F_V(v)$ can be derived also in the following way
$$F_V(v)=1-\int_v^{\infty}\Bigg[\frac{e^{-x}}{2}\int_{v-x}^{x-v}dy\Bigg]dx=1-e^{-v}$$
That is
$$V\sim Exp(1)$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
\on{f}_{UV}\,\pars{u,v} & =
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{1 \over 2}
\expo{-x}\bracks{x > \verts{y}}
\delta\pars{u - x}\ \times
\\ &
\phantom{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\quad}
\delta\pars{v - x + \verts{y}}\dd x\,\dd y
\\[5mm] & =
{1 \over 2}\expo{-u}
\int_{-\infty}^{\infty}\bracks{u > \verts{y}}
\delta\pars{v - u + \verts{y}}\dd y
\\[5mm] & =
{1 \over 2}\expo{-u}\bracks{u > 0}
\int_{-u}^{u}\delta\pars{v - u + \verts{y}}\dd y
\\[5mm] & =
\bracks{u > 0}\expo{-u}
\int_{0}^{u}\delta\pars{v - u + y}\dd y =
\\[5mm] & =
\bracks{u > 0}\expo{-u}
\int_{0}^{u}\delta\pars{y - \bracks{u - v}}\dd y
\\[5mm] & =
\bracks{u > 0}\expo{-u}\bracks{0 < u - v < u}
\\[5mm] & =
\bbx{\bracks{u > v > 0}\expo{-u}} \\ &
\end{align}

Note that
$\ds{\iint_{\mathbb{R}^{2}}\on{f}_{UV}\,\pars{u,v}
\,\dd u\,\dd v = \color{red}{1}}$.
