Calculating $E[Z(Z-1)(Z-2)(Z-3)]$ where $Z$ is Poisson

Question

If $Z$ is a variable that distributes by Poisson, with Expected value, $E(Z) = 2.5$. I need to solve: $E[Z(Z-1)(Z-2)(Z-3)]$

So What I thought to do is first: $E[(Z^2-Z)(Z^2-2Z)(Z^2-3Z)]$ $[E(Z^2)-E(Z)][E(Z^2)-2E(Z)][E(Z^2)-3E(Z)]$

From here I'm not sure how can I insert the $E(Z) = 2.5$ if what I get in the end is $E(Z^2)$

Answer 1

The Poisson distribution has probability generating function $G_Z(t):=E(t^Z)=\sum_{k\ge0}e^{-\lambda}\frac{(t\lambda)^k}{k!}=e^{\lambda(t-1)}$. Differentiating $4$ times,$$E(Z(Z-1)(Z-2)(Z-3)t^{Z-4})=G_Z^{(4)}(t)=\lambda^4e^{\lambda(t-1)}.$$Set $t=1$ so $E(Z(Z-1)(Z-2)(Z-3))=\lambda^4=39.0625$.

Answer 2

The MGF of the Poisson distribution with parameter $\lambda$ is known. From that you can get $\mathbb E[Z^k]$ for each $k$ that you need.

Answer 3

Here is a plain solution, because the exercise was rather designed to have this solution path. We have: $$ \begin{aligned} &\Bbb E[\ Z(Z-1)(Z-2)(Z-3)\ ]\\ &= \sum_{k\ge 0}k(k-1)(k-2)(k-3)\cdot e^{-\lambda}\frac {\lambda^k}{k!}\\ &= \sum_{k\ge 4}k(k-1)(k-2)(k-3)\cdot e^{-\lambda}\frac {\lambda^k}{k!}\\ &= \lambda^4\sum_{k-4\ge 0} e^{-\lambda}\frac {\lambda^{k-4}}{(k-4)!}\\ &=\lambda^4\cdot 1\ . \end{aligned} $$ With the same argument, the general expectation of a random variable of the shape $Z(Z-1)\dots(Z-(m-1))$ is $\lambda^m$.