Let $X$ and $Y$ be two Poisson random variables with same lambda parameter. What is the distribution of $\frac{X}{X+Y}$? Let $X$ and $Y$ be two Poisson random variables with same lambda parameter. What is the distribution of $\frac{X}{X+Y}$? I know it is distributed uniformly between $[0,1]$, but i couldn't prove it. Can you help please?
 A: I'm going to boldly guess that $X$ and $Y$ were supposed to be independent waiting times in a Poisson process with uniform intensity.
That would mean $X$ and $Y$ are not Poisson-distributed random variables, but exponentially distributed random variables.  That would make the proposed result correct: that $X/(X+Y)$ is uniformly distributed in $[0,1]$.  Otherwise, the statement we're being asked to prove here is wrong.
One way to look at this is a change of variables:
\begin{align}
u & = x+y \\[8pt]
v & = \frac{x}{x+y} \\[15pt]
x & = uv \\[8pt]
y & = u(1-v) \\[15pt]
dx\,dy & = u\,du\,dv
\end{align}
To get that last line, find the absolute value of the Jacobian determinant.  There should also be a nice geometric argument involving infinitely small increments.  I think there should be a way to do this all without mentioning Jacobians explicitly, but I'm not sure how to do that right now.  Now look at the densities:
$$
\text{constant}\cdot e^{-\alpha x}e^{-\alpha y}\,dx\,dy = \text{constant}\cdot e^{-\alpha uv} e^{-\alpha u(1-v)}\,u\,du\,dv = \text{constant}\cdot e^{-\alpha u}u\,du\,dv.
$$
Now notice that the density does not depend on $v$.  So you've got a uniform distribution since the density doesn't depend on $v$.  And of course $0\le v\le 1$.
