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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?

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You seem to be missing a hypothesis. For $X=Y$ the ratio is always $1/2$. Perhaps you forgot to state that the variables are meant to be independent? Also, the ratio can only take rational values, and most of the probability is concentrated in values with small denominators, so the ratio can't be distributed uniformly on (not "between") $[0,1]$. – joriki Jan 30 '13 at 1:03
I am not sure what this distribution is, but it's definitely not uniform on $[0, 1]$. – Patrick Li Jan 30 '13 at 1:45
The probability that both the numerator and the deonominator are $0$ is positive, but the probability that the denominator is $0$ and the numerator is not is $0$. Maybe you'd have to add something called "$0/0$" to the range. – Michael Hardy Jan 30 '13 at 17:19
At this point I'm wondering whether "exponential" rather than "Poisson" was intended. A frequent clumsy mistake that students make is to think at every random variable associated with a Poisson process has a Poisson distribution. But the waiting times are exponentially distributed. If $X$ and $Y$ are exponentially distributed and i.i.d., then $X/(X+Y)$ would be uniformly distributed. – Michael Hardy Jan 30 '13 at 17:23

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 $0\le v\le 1$.

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