 Solution

The inverse transformation is $x=uv$, $y=u(1-v)$, with support $u>0$, $0<v<1$. Its absolute Jacobian determinant isThe original joint probability density function, by independence, is ${\lambda }^2{e}^{-\lambda (x+y)}$ on the positive quadrant. The change of variables formula therefore givesIntegrating over $v$ givesthe 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.