Joint Distribution of $\frac{X}{X+Y}$ and $X+Y$ Let $X$ and $Y$ be i.i.d random variables.  What conditions are necessary for $\frac{X}{X+Y}$ and $X+Y$ to be independent?
This holds in the case when $X,Y \sim \Gamma\left(a,b\right)$, but I'm not sure how to go about analyzing a general case.  Some basic simulation seems to show that it doesn't hold for normals, but is not rejected by copula-based tests of independence for the Beta distribution.
 A: Here is a partial solution which makes the following trong assumptions:

Assumption. The common distribution of $X$ and $Y$ has p.d.f. $f$ which is supported on $(0, \infty)$ and satisfies $f(x) \sim c x^{\alpha}$ for some $c > 0$ and $\alpha \geq 0$ as $x\downarrow 0$.

Under this assumption, we can check that both $X$ and $Y$ have the gamma distribution.
Indeed, it is easy to check that $U = X+Y$ and $V = \frac{X}{X+Y}$ have densities $f_U$ and $f_V$, respectively, and for each $u > 0$ and $v \in (0, 1)$ the followings are true:
$$ f_U(u) = \int_{0}^{u} f(x)f(u-x) \, dx, \qquad u f(uv)f(u(1-v)) = f_U(u)f_V(v). $$
From this, we find that
$$ f_V(v) = \frac{f(uv)f(u(1-v))}{\int_{0}^{1} f(ux)f(u(1-x)) \, dx}. $$
Then dividing both the numerator and the denominator by $u^{2\alpha}$ and letting $u \downarrow 0$ gives
$$ f_V(v) = \frac{v^{\alpha}(1-v)^{\alpha}}{\int_{0}^{1} x^{\alpha}(1-x)^{\alpha} \, dx} = Cv^{\alpha}(1-v)^{\alpha} $$
for the normalizing constant $C = \frac{\Gamma(2\alpha+2)}{\Gamma(\alpha+1)^2}$. Plugging this back and writing $(x, y) = (uv, u(1-v))$,
$$ f(x)f(y) = C f_U(x+y) \frac{x^{\alpha}y^{\alpha}}{(x+y)^{2\alpha+1}}. $$
Dividing both sides by $y^{\alpha}$ and letting $y \downarrow 0$, we obtain $ Cf_U(x) = c x^{\alpha+1} f(x)$. So, if we define $g(x)$ by $f(x) = cx^{\alpha}g(x)$, then $g$ satisfies the functional equation
$$ g(x)g(y) = g(x+y) $$
and therefore $g(x) = e^{-\lambda x}$ for some $\lambda > 0$. Therefore it follows that both $X$ and $Y$ have the gamma distribution of rate $\lambda$ and shape parameter $\alpha+1$.
