= Solution
The inverse transformation is $x=uv$, $y=u(1-v)$, with support $u>0$, $0<v<1$. Its absolute <Jacobian determinant> is
$$
\left|\det\begin{pmatrix}v&u\\1-v&-u\end{pmatrix}\right|=u.
$$
The original joint <probability density function>, by <independence>, is $\lambda^2e^{-\lambda(x+y)}$ on the positive quadrant. The <change of variables formula> therefore gives
$$
\boxed{f_{U,V}(u,v)=\lambda^2u e^{-\lambda u}\mathbf1_{\{u>0,\ 0<v<1\}}}.
$$
Integrating over $v$ gives
$$
\boxed{f_U(u)=\lambda^2u e^{-\lambda u}\mathbf1_{\{u>0\}}},
$$
the <gamma distribution> of shape two and rate $\lambda$. Integrating over $u$ gives $f_V(v)=1$ on $(0,1)$. The factorization also shows that $U$ and $V$ are independent.
Back to article page
