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., $1-\varepsilon\leq E\left[\frac{X}{X+Y}\right]/\frac{n}{n+m}\leq 1+\varepsilon$?
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2How do you define $\frac{X}{X+Y}$ on the event ${X=Y=0}$? – Did Jul 26 '18 at 10:39
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Oops. Then, let I assume $0$. i.e, I want to estimate $E[f(X,Y)]$, where $f(X,Y)=0$ if $X=Y=0$ and $f(X,Y)=\frac{X}{X+Y}$ otherwise. – hiratat Jul 26 '18 at 11:07
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Note that, for every nonnegative integers $(x,y)$, $$\frac{x}{x+y}\mathbf 1_{x\ne0}=\int_0^1xs^{x+y-1}\mathbf 1_{x\ne0}ds$$ Integrating this, one gets $$E\left(\frac X{X+Y}\mathbf 1_{x\ne0}\right)=\int_0^1E(Xs^{X-1}\mathbf 1_{X\ne0})E(s^Y)ds$$ Now, $Y$ is a Binomial and $X\mathbf 1_{X\ne0}$ a positive-Binomial random variables, hence $$E(s^X\mathbf 1_{X\ne0})=\left[(ps+q)^n-q^n\right]\qquad E(s^Y)=(ps+q)^m$$ here $q = 1-p$. By differentiation, $$E(Xs^{X-1}\mathbf 1_{X\ne0})=\frac d{ds}E(s^X\mathbf 1_{X\ne0})=np(ps+q)^{n-1}$$ Thus, $$E\left(\frac X{X+Y}\mathbf 1_{X\ne0}\right)=\int_0^1 np\cdot (ps+q)^{n+m-1}ds=\frac n{n+m} \cdot (1-q^{n+m})$$ from which the desired estimates follow.
