Suppose $X$ and $Y$ are independent random variables. Under what conditions, $ U=\frac{X}{X+Y}$ and $V=X+Y$ are independent?
Notes:
1- Of course if you have independent $X,Y\sim \text{Uniform}(0,1)$, $0\leq U\leq 1$ and $0\leq V\leq 2$, but $U$ and $V$ are not independent, e.g. $V=2$ implies $U=\frac{1}{2}$. So we may restrict our attention to independent unbounded random variables $X, Y$. (Maybe even positive r.v.s so that $X+Y$ will be non-zero.
2- I think when $X$ and $Y$ are unbounded, then scaling both of them with same constant keeps $U$ unchanged, but we can make $V$ as large or small as possible by varying that constant. So it seems given $U$, we don't have any information about $V$.
3- Today in our stat class we showed that for independent $X\sim \text{Gamma}\left(\alpha_1,\beta\right)$ and $Y\sim \text{Gamma}\left(\alpha_2,\beta\right)$, joint pdf $f_{_{U,V}}\left(u,v\right)$ decomposes into product of marginal pdfs of $U\sim \text{Beta}\left(\alpha_1,\alpha_2\right)$ and $V\sim \text{Gamma}\left(\alpha_1+\alpha_2,\beta\right)$, hence they are independent. That's where I started to think if there is a general result.