If $X \sim Po(\lambda)$ then $E\left[X(X-1)\cdots(X-r+1)\right] = \lambda^r$.

Is there a straightforward way to see this without the use of moment generating functions?

I can get as far is $$E\left[X(X-1)\cdots(X-r+1)\right] = \sum_{k=0}^{\infty} [k(k-1)\cdots (k-r+1)] \frac{\lambda^k e^{-\lambda}}{k!}$$ and noting that $k(k-1)\cdots (k-r+1) = k!/(k-r)!$ if $k\ge r$, but I can't see how to handle the case then $k<r$.

  • 5
    $\begingroup$ There is no "case $k\lt r$": those terms are zero. The sum actually begins at $k=r$. Since $k(k-1)\cdots(k-r+1)/k! = 1/(k-r)!$, all you have to do is factor out $\lambda^r$ and you're done. $\endgroup$
    – whuber
    Mar 18, 2017 at 22:56

1 Answer 1


The answer is given in whubers comment, I will write it out with details.

First note that $\frac{x(x-1)\dots (x-r+1)}{x!} = \frac{1}{(x-r)!}$. Using that, $$ \DeclareMathOperator{\E}{\mathbb{E}} \E X(X-1) \dots (X-r+1) = \sum_{k=0}^\infty k(k-1)\dots (k-r+1) e^{-\lambda} \frac{\lambda^k}{k!} \\ = \lambda^r \sum_{k=r}^\infty e^{-\lambda}\frac{\lambda^{k-r}}{(k-r)!} \\ = \lambda^r \sum_{k=0}^\infty e^{-\lambda} \frac{\lambda^k}{k!} \\ = \lambda^r $$


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