Expectation of a Poisson Distribution: E[X(X-1)(X-2)(X-3)]

Question

Given $X \sim Poi(\lambda)$, what is the expectation of $\mathbb{E}[X(X-1)(X-2)(X-3)]$?

I'm not sure how to approach this. I was thinking of expanding the polynomial, but that led to fairly ugly results. I was told that there is an elegant solution, but I cannot seem to determine this. What is the best way to go about solving this? Thanks!

Answer 1

$E[X(X-1)(X-2)(X-3)]=\sum_{k=4}^{\infty}k(k-1)(k-2)(k-3)\frac{\lambda^ke^{-\lambda}}{k!}=\sum_{k=4}^{\infty}\frac{\lambda^ke^{-\lambda}}{(k-4)!}=\sum_{k=0}^{\infty}\frac{\lambda^{k+4}e^{-\lambda}}{k!}=\lambda^4\sum_{k=0}^{\infty}\frac{\lambda^{k}e^{-\lambda}}{k!}=\lambda^4$

Note that in the last expression we have used from this theorem that the summation of any PMF over its entire domain is equal to $1$.

Answer 2

Hint: $$e^{-\lambda}\sum_{n=0}^{\infty}n\left(n-1\right)\cdots(n-k+1)\frac{\lambda^{n}}{n!}=e^{-\lambda}\lambda^{k}\sum_{n=k}^{\infty}\frac{\lambda^{n-k}}{\left(n-k\right)!}=\cdots$$