Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 10 down vote accepted

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 $X<k$:

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

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.