= Solution
{c}
The inverse change of variables is $x=2u+v$, $y=v$, with absolute <Jacobian determinant> two. Positivity of $x,y$ gives
$$
A=\{(u,v):v>0,\ 2u+v>0\}.
$$
The <change of variables for probability densities> gives
$$
f_{U,V}(u,v)=\tfrac14e^{-(2u+2v)/2}\cdot2
=\tfrac12e^{-u-v}\quad((u,v)\in A).
$$
For fixed $u$, integrate over $v>\max(0,-2u)$:
$$
f_U(u)=\tfrac12e^{-u}\int_{\max(0,-2u)}^\infty e^{-v}\,dv
=\begin{cases}\tfrac12e^{-u},&u\geq0,\\
\tfrac12e^u,&u<0.\end{cases}
$$
Thus $\boxed{f_U(u)=\tfrac12e^{-|u|}}$, the standard <Laplace distribution>.
Back to article page
