# How to compute expectation for poisson distributed variable in context of binomial?

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 $Z$?

$$E\left(\frac{y!}{Z! (y - Z)!}\right)$$

• In similar problems, the setup is usually $Y\sim\mathrm{Pois}\left(\theta\right)$ and $Z|Y\sim\mathrm{Binom}\left(n=Y,p\right)$. Are you sure the problem has been copied correctly? May 17, 2013 at 20:57
• Hi @Max, yes, it is correct. Since I only mentioned part of the problem and not all, it might seem strange but it's correct.
– Sam
May 18, 2013 at 2:58

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): $$\sum_{k=0}^y \left[\binom{y}{k}\right]^2p^k(1-p)^{n-k}=(1-p)^{n}\cdot \text{Hypergeometric}~_{2}F_{1}\left(-y, -y, 1, \frac{p}{1-p}\right)$$
• Interesting! I wonder why the fourth argument is not written as $\frac{p}{1-p}$. May 18, 2013 at 13:08