$\frac{Y}{Y+Z}$ is independent of $Y+Z$ Let $$X = Y + Z,$$ where $Y,Z$ are independent continuous random variables. Can it happen that
$$
X \text{ and }\frac{Y}{X} \text{ are independent}?
$$
If yes (I believe it can happen, but only in some weird degenerative case), can you somehow characterize such pairs of variables?
I was trying to work it out by conditioning. Write $W(a):= [Y\mid X=a]$ which is a random variable $Y$ conditioned on $X=a$. Then,
$ W(a)/a \overset{d}{=} W(b)/b$ for all $a,b$ in the support of $X$. However, explicitly writing down a distribution of $W(a)$ is typically non-trivial. Some other ideas on how to proceed?
 A: Some thoughts too long for a comment:
Let $X=Y+Z$ and $U=Y/(Y+Z)$.
The Jacobian for the transformation $(x, u) \mapsto (xu, x(1-u))$ is $\left|\det \begin{bmatrix} u & x \\ 1-u & -x \end{bmatrix}\right| = |x|$, so
$$f_{Y, Z}(y, z) \, dy \, dz
= f_{Y, Z}(xu, x(1-u)) \left|\det \frac{\partial(y, z)}{\partial(x, u)}\right| \, dx \, du
= |x|f_Y(xu) f_Z(x(1-u)) \, dx \, du,$$
where in the last step we have used the independence of $Y$ and $Z$ to factor the joint density.
If we suppose $Y$ and $Z$ are nonnegative random variables, then we have
$$f_{Y, Z}(y, z) \, dy \, dz = x f_Y(xu) f_Z(x(1-u)) \, dx \, du$$
where $(y, z) \in \mathbb{R}_{>0}^2$ and $(x, u) \in \mathbb{R} \times [0,1]$.
Thus, we need $f_Y$ and $f_Z$ to be such that $f_Y(xu) f_Z(x(1-u))$ can be written as $g(x) h(u)$. Passing to logarithms, this becomes
$$\log f_Y(xu) + \log f_Z(x(1-u)) = \log g(x) + \log h(u).$$

*

*If we impose that $\log f_Y(y) = -\lambda y + c$ and $\log f_Z(z) = -\mu z + c'$, we recover the constraint $\lambda=\mu$ and obtain $Y, Z \overset{i.i.d.}{\sim} \text{Exponential}(\lambda)$.

*With the more general form $\log f_Y(y) = -\lambda y + (\alpha-1) \log y + c$ and $\log f_Z(z) = - \mu z + (\beta - 1) \log z + c'$, we  again have the constraint $\lambda=\mu$ and recover the more general case of $Y \sim \text{Gamma}(\alpha, \lambda)$ and $Z \sim \text{Gamma}(\beta, \lambda)$.

