Conditional Expectation for Poisson random variables

Question

Let $X_i$ be Poisson i.i.d. random variables with $P(X=k)=e^{-\lambda} \frac{\lambda^k}{k!}, \quad k=1,2,\ldots$. Let $X=\sum_{i=1}^nX_i$.

Find conditional expectation $$ E(X_1\times\cdots\times X_n\mid X=m), $$ where $m \in R$.

Answer 1

A comment by StubbornAtom gives a conceptual answer.

Here is a more flat-footed one.

The distribution of $(X_1,\ldots,X_n)$ conditional on $X=m$ is multinomial with parameters $m$ and $(1/n,\ldots,1/n)$, which is to say $$P\left( (X_1,\ldots,X_n)=(m_1,\ldots,m_n)\mid X=m\right) =\frac{m!}{\prod_{i=1}^n m_i!} \prod_{i=1}^n \left(\frac{1}{n}\right)^{m_i}.$$ Note that the conditional expectation of $S(z_1,\ldots,z_n)=\prod_{i=1}^n z_i^{X_i}$ is (by the binomial theorem) $$T(z_1,\ldots,z_n)=E[S(z_1,\ldots,z_n)|X=m] = \left(\sum_{i=1}^n \frac {z_i} n\right)^m.$$ The desired answer is the mixed partial derivative $$\left(\prod_{i=1}^n\frac\partial{\partial z_i} \right)T(z_1,\ldots,z_n) = \frac m n \frac{m-1}n\cdots\frac{m-(n-1)} n\left(\sum_{i=1}^n \frac {z_i} n\right)^{m-n}$$ evaluated at $z_1=\cdots=z_n=1$, namely, $m!/(m-n)!\times n^{-n}$.