# Expectation of $X(X - 1) \ldots (X - k + 1)$, where $X$ has a Poisson distribution.

I am trying to calculate $$\text{E}[X(X - 1) \ldots (X - k + 1)],$$ where $\text{E}$ denotes the expectation operator and $k \in \mathbb{N}$ is fixed.

I think I have to use the fact that the expectation of a sum of random variables is the sum of the expectations, but how can we apply it to this product?

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For a Poisson distribution

$$E[g(X)] = \sum_{j=0}^{\infty} g(j) \frac{\lambda^j}{j!} e^{-\lambda}$$

Specifically, for the $g$ you specified, the sum begins at $j=k$ because $g(X) = 0$ when $X<k$:

\begin{align}E[g(X)] &= \sum_{j=k}^{\infty} \frac{j!}{(j-k)!} \frac{\lambda^j}{j!} e^{-\lambda}\\ &= \sum_{j=0}^{\infty} \frac{\lambda^{j+k}}{j!} e^{-\lambda}\\ &= \lambda^k\\ \end{align}

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You might want to check your definition of $g$, it is not correct. – Byron Schmuland Feb 1 '13 at 20:21
Thanks, will do. – Ron Gordon Feb 1 '13 at 20:24
The function $g$ is $j!/(j-k)!$, not $j!/k!$. – Byron Schmuland Feb 1 '13 at 20:30
@ByronSchmuland: Oy. Not one of my finest moments. Thanks again for keeping me honest. – Ron Gordon Feb 1 '13 at 20:34
+1 Got it!!${}$ – Byron Schmuland Feb 1 '13 at 20:35

Alternatively, the probability generating function $G$ for a Poisson random variable with mean $\lambda$ is $G(s)=E(s^X)=\exp(-\lambda(1-s))$. Differentiate $k$ times and set $s=1$ to get the answer.

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