My question is related to Ratio of two binomial distributions. Assume $X\sim \text{Poisson}(p\lambda)$ and $Y\sim\text{Poisson}(q\lambda)$ where $p,q>0$ are constants. I have interest in the distribution of $$ W=\dfrac{X}{X+Y},$$ specifically, the variance of such a ratio. From the explanation in the answer in the related question I managed to calculate $E[W]$, using (EDIT:)
$$ E\left[\dfrac{X}{X+Y}\right]\mathbb{I}_{x,y\neq(0,0)} = \int_0^1 xs^{x+y-1}ds\\ =\int_0^1E\left[Xs^{X-1}\right]E\left[s^Y\right]ds $$ from the description of the variables $$ E\left[s^X\right] = \exp\left[p\lambda\left(s^t-1\right)\right],\; E\left[s^Y\right] = \exp\left[q\lambda\left(s^t-1\right)\right],$$ and $$ E\left[Xs^{X-1}\right] = \dfrac{d}{ds}E\left[s^X\right] = p\lambda\exp[p\lambda(s-1)] $$ then, $$ \int_0^1 p\lambda e^{p\lambda(s-1)}e^{q\lambda(s-1)}ds = \int_0^1 p\lambda e^{p\lambda(s-1) + q\lambda(s-1)} ds\\ = p\lambda\int_0^1 e^{(p\lambda + q\lambda)(s-1)}ds\\ = \dfrac{p}{p+q}e^{-\lambda(p+q)}\left[e^{\lambda(p+q)}-1\right]\\ = \dfrac{p}{p+q}\left[1-e^{-\lambda(p+q)}\right] $$ Which essentially is the ratio of $p/(p+q)$ since the exponential term vanishes with moderate values of $\lambda$. I am having trouble using the same approach to find $E[W^2]$ so that I can get the variance. I'd appreciate any help. Also, for practical purposes I am assuming that the case $X=0$ and $Y=0$ is not likely to happen since $\lambda$ is relatively large (typically > 50). I understand that in theory, there is a positive probability that both $X$ and $Y$ can be zero. I am also aware of approximations to that variance, but I wanted a distinct approach.
