TITLE: If X is a random variable with a Poisson distribution, rate parameter $\lambda$ show that...
QUESTION [2 upvotes]: If X is a random variable with a Poisson distribution, rate parameter
  $\lambda$ show that $$E(X(X-1)(X-2)...(X-k)) = \lambda^{k+1}$$

I know that for the probability generating function $G(N)$ of a Poisson random variable, $$G''(1)=E(X(X-1))=\lambda^2$$
and I'm guessing that with each proceeding derivative we add $(X-2), (X-3)$, etc inside the expectation, eg: $G'''(1)=E(X(X-1)(X-2))=\lambda^3$.
But I don't really know how to show it.

REPLY [0 votes]: Let the generating function be $G_X(t)$
$$G_X(t)=e^{\lambda t-\lambda}$$
$$G_X'(t)=\lambda e^{\lambda t-\lambda}$$
$$G_X''(t)=\lambda ^2e^{\lambda t-\lambda}$$
$$G_X^{(n)}(t)=\lambda ^ne^{\lambda t-\lambda}$$
We know that $G_X(1)=E(X)$, and that $G_X^{(n+1)}(1)=E(X(X-1)(X-2)(X-3)...(X-n)$
This is because a PGF is by definition the following:
$$G_X(t)=\sum_{k=0}^\infty P_k t^k$$
Differentiating this gives the following wrt $t$
$$G_X'(t)=\sum_{k=0}^\infty kP_k t^{k-1}$$
You can see that as we continue to differentiate, we end up with
$$G_X^{(n)}(t)=\sum_{k=0}^\infty k(k-1)(k-2)...(k-n+1)P_k t^{k-n}$$
This of course is another way of writing the Expected values of $k(k-1)(k-2)...(k-n+1)$ when $t=1$, because the probability of achieving $k$ multiplied by the values is the definition of expected value.
Therefore, it follows that $$G_X^{(n+1)}(1)=E(X(X-1)(X-2)...(X-n))=\lambda ^{n+1}e^{\lambda -\lambda}=\lambda ^{n+1}$$
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