= Solution
Use the inverse <change of variables for probability densities>
$$
x=vw,\qquad y=(1-v)w,\qquad 0<v<1,\quad w>0.
$$
Its absolute <Jacobian determinant> is
$$
\left|\det\begin{pmatrix}w&v\\-w&1-v\end{pmatrix}\right|=w.
$$
Therefore
$$
\boxed{f_{V,W}(v,w)=w e^{-w}\mathbf1_{\{0<v<1,\ w>0\}}.}
$$
Since $\int_0^\infty we^{-w}\,dw=1$, its <marginal densities> are $f_V(v)=\mathbf1_{(0,1)}(v)$ and $f_W(w)=we^{-w}\mathbf1_{(0,\infty)}(w)$. The joint density factorizes as $f_Vf_W$, so \b[$V$ and $W$ are <independent random variables>], with $V$ uniform on $[0,1]$ and $W\sim\operatorname{Gamma}(2,1)$ in the shape-rate convention. This is <beta-gamma independence> with both shape parameters equal to one.
Back to article page
