Mean and variance of $X/(X+Y)$ where $X$ is a Poisson random variable and $Y$ is a zero-truncated Poisson variable Suppose $X$ is a Poisson random variable and $Y$ is a zero-truncated Poisson random variable. How could I find the mean and the variance of $X/(X+Y)$? Or how could I approximate the answer?
Update: Both are independent.
 A: The following technique transforms the discrete sums required for the moments of this compound random variable into integrals. I henceforth assume that the random variables are independent and distributed as $X\sim \text{Poisson}(\lambda),Y\sim \text{TruncPoisson}(\mu) $.  More explicitly, we have that
$$\mathbb{E}[\frac{X^r}{(X+Y)^r}]=\frac{1}{e^\lambda(e^\mu-1)}\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^r}{(m+n)^r}\frac{\lambda^m}{m!}\frac{\mu^n}{n!}$$
The trick here is to use the identity $(m+n)^r\int_0^\infty s^{r-1}e^{-(m+n)s}ds=\Gamma(r)$ to obtain (after a change of variables $t=e^{-s}
$)
$$\begin{align}\mathbb{E}\left[\frac{X^r}{(X+Y)^r}\right]&=\frac{(-1)^r}{(r-1)!e^{\lambda}(e^\mu-1)}\int_0^1 dt\frac{\log^{r-1}t}{t}\sum_{m=1}^\infty m^r\frac{(\lambda t)^m}{m!}\sum_{n=1}^\infty\frac{(\mu t)^n }{n!}\\&=\frac{(-1)^r}{(r-1)!}\int_0^1\frac{\log^{r-1}t}{t}\frac{e^{\mu t}-1}{e^\mu-1}e^{\lambda(t-1)}\sum_{\ell=0}^rS(r,\ell)(\lambda t)^\ell \end{align}$$
where $S(n,m)$ are Stirling numbers of the second kind. The integrals in question can be expressed in terms of the upper incomplete Gamma function and it's derivatives by use of the relation
$$\int_0^1 t^r \log^k t~ e^{x t}dt=\left(\frac{\partial }{\partial r}\right)^k [(-x)^{-1 - r} (\Gamma(1 + r) - \Gamma(1 + r, -x))] $$
The first two moments however can be easily expressed in terms of simpler functions.  In fact the result for the mean is elementary:
$$\mathbb{E}[X/(X+Y)]=\frac{\lambda e^{-\lambda}}{e^\mu-1}\int_0^1e^{\lambda t}(e^{\mu t}-1)dt=\frac{\lambda e^{-\lambda}}{e^\mu-1}\left(\frac{e^{\lambda+\mu}-1}{\lambda+\mu}-\frac{e^{\lambda}-1}{\lambda}\right)$$
The expression for the second moment is more complicated:
$$\mathbb{E}[X^2/(X+Y)^2]=\frac{e^{-\lambda}}{(-1 + e^\mu) (\lambda + \mu)^2}
 (e^{\lambda + \mu} \lambda^2 - e^\lambda (\lambda +\mu)^2 + (2 - \gamma) \lambda\mu - \mu^2 + 
    \lambda \mu (\text{Chi}(\lambda + \mu) - \log(\lambda+\mu) + \text{Shi}(
\lambda + \mu)))$$
