Powers of Poisson random variables under conditional expectation

Question

Let X, Y be I.i.d. Poisson random variables with parameter $\lambda>0$. Let $S=X+Y$ and $n, m>0$. Let $N>0$.

Find conditional expectation $$ E(X^n\times Y^m|S=N). $$

I have been trying to use a definition of conditional expectation and represent it through probabilities, however, the computations for probabilities are almost impossible.

The representation of the power of Poisson random variable through the Bell number is also not helpful.

Answer 1

The conditional distribution of $X$, given the event $X+Y=N$, is $\text{Binomial}(N, 1/2)$. (Note that $Y$ is necessarily $N-X$.)

Proof: $$P(X=x \mid X+Y = N) = \frac{P(X=x, Y=N-x)}{P(X+Y = N)} = \frac{e^{-\lambda} \frac{\lambda^x}{x!} e^{-\lambda} \frac{\lambda^{N-x}}{(N-x)!}}{e^{-2\lambda} (2\lambda)^N/N!} = \binom{N}{x} \frac{1}{2^N}.$$

So, your question reduces to computing $E[X^n (N-X)^m]$ where $X \sim \text{Binomial}(N, 1/2)$. I'm not sure what is the most straightforward way to compute this. Since $x^n(N-x)^m$ is a polynomial, one could in principle compute this using the MGF of the binomial distribution. But there may be an easier way.