How to find $E[X^2\mid X+Y]$?

Question

Suppose $X$ and $Y$ are independent Poisson random variables with rates $\lambda_1, \lambda_2$ respectively, then how would we go about calculating: $ E[X^2\mid X+Y] \text{ ?} $$

Answer 1

Since $X+Y\sim\mathrm{Poisson}(\lambda_1+\lambda_2)$, for any positive integer $n$ we have

$$ \begin{align*} \mathbb P(X=k|X+Y=n) &= \frac{\mathbb P(X=k, X+Y=n)}{\mathbb P(X+Y=n)}\\ &= \frac{\mathbb P(X=k)\mathbb P(Y=n-k)}{\mathbb P(X+Y=n)}\\ &= \left(\frac{e^{-\lambda_1}\lambda_1^k}{k!}\right)\left(\frac{e^{-\lambda_2}\lambda_2^{n-k}}{(n-k)!}\right)\left({\frac{e^{-(\lambda_1+\lambda_2)}(\lambda_1+\lambda_2)^n}{n!}}\right)^{-1}\\ &= \frac{\lambda_1^k\lambda_2^{n-k} n!}{(\lambda_1+\lambda_2)^n k!(n-k)!}\\ &= \binom nk \left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^k\left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)^{n-k} \end{align*} $$ so that $$X|X+Y=n\sim\mathrm{Bin}\left(n,\frac{\lambda_1}{\lambda_1+\lambda_2}\right).$$ It follows from linearity of conditional expectation that $$ \begin{align*} \mathbb E[X^2|X+Y=n] &= \mathbb E[X(X-1)|X+Y=n] + \mathbb E[X|X+Y=n]\\ &= n(n-1)\left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^2 + \frac{n\lambda_1}{\lambda_1+\lambda_2}\\ &= \frac{n\lambda_1(n\lambda_1+\lambda_2)}{(\lambda_1+\lambda_2)^2}. \end{align*}$$