Expectation of a discrete random variable

Question

Let $X$, $Y$ $\sim Binomial(n, p)$ and are independent.

One needs to calculate $\mathrm{E}[X|X+Y=m]$.

I've calculated the formula of $\mathrm{P}(X+Y=m)$: $$\mathrm{P}(X+Y=m) = \binom{2n}{m} p^m (1-p)^{2n-m}$$

I know that $$E[X|X+Y=m] = \sum_{x=0}^{m}x\mathrm{P}(X=x|X+Y=m)=\sum_{x=0}^{m}x\frac{\mathrm{P}(X=x,X+Y=m)}{\mathrm{P}(X+Y=m)}\text{ .}$$

I'm not sure if there's an easier way but I have no idea how to represent any of the written probabilities in the expression of conditional expectation. So, my question is how to solve this last step?

Answer 1

Yes there is an easier way.

If $X$ and $Y$ are identically and independently distributed then by symmetry $$E[X\mid X+Y=m] =E[Y\mid X+Y=m] =\frac12 E[X+Y\mid X+Y=m] = \dfrac{m}{2}. $$