## Given two strictly positive independent r.v.'s, X and Y when is the distribution of X/(X+Y) a proper probability distribution?

Suppose we have two independent random variables $X$ and $Y$ strictly positive with absolutely continuous densities $f_X(x)$ and $f_Y(y)$. We are interested in the distribution of $\frac{X}{X+Y}$. Defining $U=X+Y$ and $V = \frac{X}{X+Y}$ the joint density of $U,V$ is

$f_{U,V}(u,v) = uf_X(uv)f_Y(u(1-v))$

and the density of interest is obtained by marginalizing over $u$:

$f_V(v) = \int_0^{\infty} uf_X(uv)f_Y(u(1-v)) du$

My question is as follows: Is $f_V$ always a proper probability density (ie does it integrate to 1 on $(0,1)$)? If so, is there a proof? I feel like maybe I'm overlooking something simple. If not, can we make any general statements about when $f_V$ is proper?

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 What are you looking to prove here? Certainly the event $\{00, Y>0\}$). So if you are already satisfied that the given expression gives the density function of $V$ then yes it must integrate to 1. Or are you looking for a proof that that expression is indeed the density of $V$? – James Martin Apr 17 2011 at 22:59 @James Martin Well I'd thought that it would integrate no matter what, but then I was pushing some particular cases through Mathematica and for one it spit out a function with an infinite integral. Checked every which way for typos and didn't find any. I think the particular case was $X\sim Gamma(a/2, 1/2)$ with a>1 and $Y$ having a folded Cauchy distribution. – JMS Apr 17 2011 at 23:20 Well, that sounds strange; if you want to check that example in particular you might have to present it. But unless I am overlooking something, then (a) yes, the $f_V$ you write down is indeed the density of $V$, and (b) yes, the density of $V$ does indeed integrate to 1. To prove (a) you can do the standard calculation with change of variables and Jacobian. (b) is true for the trivial reason in my previous comment. – James Martin Apr 18 2011 at 17:31 Does not seems like a research level question to me. Try asking at math.stackexchange.com. – zhoraster Apr 18 2011 at 18:43 @James Martin I could present the example, but the integration over $u$ is a bear which is why I was using software hoping that it could be expressed as some combination of special functions. I'm quite sure I've made a mistake there and not in the math; thanks for your help – JMS Apr 18 2011 at 20:38