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I'm working on a problem that needs to be solved using EM algorithm. In doing that, I have to evaluate an expectation that I actually have no idea how to.
Consider: $Y$ as a a fixed observed integer and consider $Z$ as a random variable which has a Poisson distribution with rate $\theta$:
$$ Z \sim \operatorname{Pois}(\theta)$$ $$ Z|Y \sim \operatorname{Binom}(n = y, p)$$
How can I compute the expectation below with respect to $Z$?
$$E\left(\frac{y!}{Z! (y - Z)!}\right)$$
As @max points out in a comment, the problem may have been miscopied, but as stated, it would appear that we are to compute the conditional expectation of $\frac{y!}{Z!(y-Z)!}$ given that $Y = y$. The conditional distribution of $Z$ given that $Y=y$ is $\text{Binomial}(y,p)$, and so $$E\left[\frac{y!}{Z!(y-Z)!}\right] = \sum_{k=0}^y \frac{y!}{k!(y-k)!}\binom{y}{k}p^k(1-p)^{n-k} = \sum_{k=0}^y \left[\binom{y}{k}\right]^2p^k(1-p)^{n-k}.$$ As far as I know, there is no simpler closed-form expression for the sum on the right. Of course, the numerical value of the expression is straightforward to compute for specific values of $p$ and $y$.
I'm cheating, but mathematica finds another expression for the sum on the right (involving the Hypergeometric function): $$ \sum_{k=0}^y \left[\binom{y}{k}\right]^2p^k(1-p)^{n-k}=(1-p)^{n}\cdot \text{Hypergeometric}~_{2}F_{1}\left(-y, -y, 1, \frac{p}{1-p}\right) $$
