I derived the bivariate joint distribution with a transformation from two exponential distributions with $\lambda = 1.5$ for $X,Y$ and $U=X+Y$ and $V=X-Y$. With $X=(U+V)/2$ and $Y=(U-V)/2$ I derived the joint $f_{u,v}$ as follows:
Skipping a few steps:
$f_{u,v}=\frac{4}{9}e^{\frac{2}{3}(\frac{-(U+V)}{2}-\frac{U-V}{2})}\cdot\frac{1}{2}=$
$f_{u,v}=\frac{2}{9}e^{-\frac{2}{3}U}$ for $0<V<U$
So to find the marginal, I take the integral with respect to V. I should expect a $\text{Gamma}(2,1.5)$ distribution.
$f_u=\int_{0}^{U}\frac{2}{9}e^{-\frac{2}{3}U} dv$
$f_u=\frac{2}{9}Ue^{-\frac{2}{3}}$
The above should be $\frac{4}{9}$ not $\frac{2}{9}$. I have a feeling the bounds should be $0<V<2U$. I'm probably missing something really basic, I just am not able to show why that would be the bounds.
Since they are exponentially distributed, the domain for each of $X$ and $Y$ is $[0,\infty)$.
Defining $U=X+Y$ means the support for $U$ is $[0;\infty)$.
Also when given $U$ the conditional support for $Y$ is $[0;U]$ (as it may not have a value greater than $U$), and $X=U-Y$ (of course).
Now $V=X-Y$, or $V=U-2Y$, so when given $U$, the conditional support for $V$ is $[U-2U;U-0]$, that is: $[-U;U]$
So the joint support for $\langle U,V\rangle$ is $\{\langle u,v\rangle:0\leq u, -u\leq v\leq u\}\\=\{\langle u,v\rangle: \lvert v\rvert\leq u\}$