TITLE: Ratio of two binomial distributions
QUESTION [7 upvotes]: How to estimate
$$
E\left[\frac{X}{X+Y}\right]
$$
for two independent random variables $X\sim Bin(n,p)$ and $Y\sim Bin(m,p)$ ?
Are there any connection with $\frac{n}{n+m}$ e.g., $1-\varepsilon\leq E\left[\frac{X}{X+Y}\right]/\frac{n}{n+m}\leq 1+\varepsilon$?

REPLY [8 votes]: Note that, for every nonnegative integers $(x,y)$, $$\frac{x}{x+y}\mathbf 1_{(x,y)\ne(0,0)}=\int_0^1xs^{x+y-1}ds$$
Integrating this, one gets $$E\left(\frac X{X+Y}\mathbf 1_{(X,Y)\ne(0,0)}\right)=\int_0^1E(Xs^{X-1})E(s^Y)ds$$
Now, $X$ and $Y$ are sums of i.i.d. Bernoulli random variables, hence
$$E(s^X)=(ps+q)^n\qquad E(s^Y)=(ps+q)^m$$
By differentiation, $$E(Xs^{X-1})=\frac d{ds}E(s^X)=np(ps+q)^{n-1}$$ Thus, $$E\left(\frac X{X+Y}\mathbf 1_{(X,Y)\ne(0,0)}\right)=\int_0^1np(ps+q)^{n+m-1}ds=\frac n{n+m}(1-q^{n+m})$$ from which the desired estimates follow.