Trying to find the conditional expected value of X when X + Y = m, and X, Y are both independent binomial random variables w/ parameters n & p

Question

If X and Y are independent binomial random variables with identical parameters n and p, calculate the conditional expected value of X given X+Y = m.

The conditional pmf turned out to be a hypergeometric pmf, but I'm a but unclear on how to relate that back into finding E[X|X+Y=m]

Answer 1

$$E(X|X+Y)+E(Y|X+Y)=E(X+Y|X+Y)=X+Y$$

Given $X\perp \!\!\!\perp Y$ the are changeable too, thus $E(X|X+Y)=E(Y|X+Y)$

Concluding:

$$2E(X|X+Y)=X+Y$$

$$E(X|X+Y)=\frac{X+Y}{2}=\frac{m}{2}$$

Answer 2

The conditional pmf turned out to be a hypergeometric pmf...

Indeed it is, thanks to using Vandermonde's identity we see that...

$$\begin{align}\mathsf P(X=k\mid X+Y=m) &= \dfrac{\mathsf P(X=k,Y=m-k)}{\mathsf P(X+Y=m)}\\[1ex] &=\dfrac{\dbinom nk\dbinom n{m-k}p^m(1-p)^{2n-m}}{\displaystyle\sum_{h=0}^m \dbinom nh\dbinom n{m-h}p^m(1-p)^{2n-m}}\,\mathbf 1_{\langle k,m\rangle\in\Bbb N^2:0\leq k\leq m\leq n}\\[1ex]&=\dfrac{\dbinom nk\dbinom n{m-k}}{\dbinom{2n}{m}}\,\mathbf 1_{\langle k,m\rangle\in\Bbb N^2:0\leq k\leq m\leq n}\end{align}$$

$\therefore (X\mid X+Y=m)\sim\mathcal{Hypergeo}(2n,n,m)$

, but I'm a but unclear on how to relate that back into finding E[X|X+Y=m]

Well the expected value for a hypergeometric random variable where $\sim\mathcal{Hypergeo}(N,K,m)$ is $mK/N$, so...