Using the Stein-Chen identity for the Poisson distribution to show that $E[X(X-1)...(X-k+1)] = \lambda^k$

Question

So far, I have already derived the following:

$E[(X-\lambda)g(X)] = \lambda E[g(X+1)-g(X)]$

But I am stuck in terms of applying this identity to derive $\lambda^k$: \begin{align} \lambda^k &= E[X(X-1)...(X-k+1)]\\ &= E[X(X-1)]...E[X-k+1]\\ &= 1\cdot E[(X+1)-X]...E[X-k+1] \\ &= 1\cdot E[1]...E[X-k+1] \end{align}

I know I am doing something wrong here (i.e. $E[X(X-1)] \neq E[1]$) but this is the best I can come up with right now. I guess I am confused on what my g(X) and $\lambda$ are supposed to be.

Answer 1

We will proceed my induction. For $k=1$ we have $\mathbb E[X] = \lambda$, which is true. Assume that for some $k \in \mathbb N$ we have $\mathbb E[X(X-1)...(X-k+1)] = \lambda^k$.

Stein-Chen identity, tells us that for poisson r.v we have $\mathbb E[Xg(X)] = \lambda \mathbb E[g(X+1)]$.

Take $g:\mathbb R \to \mathbb R$, $g(x) = (x-1)...(x-k)$

Then $\mathbb E[Xg(X)] = \mathbb E[X(X-1)...(X-k)]$, while $\lambda \mathbb E[g(X+1)] = \lambda \mathbb E[X(X-1)...(X-k+1)]$, which is equal to $\lambda \cdot \lambda^k$ by our induction assumption. So, we get $\mathbb E[X(X-1)...(X-k)] = \lambda^{k+1}$

Q.E.D

Answer 2

Must you use Stein-Chen? We can derive this result using generating functions. Let $G(s) = \mathbb E[s^X]$, then \begin{align} g(S) &= \mathbb E[s^X]\\ &= \sum_{n=0}^\infty s^ne^{-\lambda}\frac{\lambda^n}{n!}\\ &=e^{-\lambda}\sum_{n=0}^\infty \frac{(\lambda s)^n}{n!}\\ &=e^{-\lambda}e^{\lambda s}\\ &=e^{\lambda(s-1)}. \end{align} Now, the $k^{\mathrm{th}}$ factorial moment of $X$ is given by \begin{align} \mathbb E\left[\frac{X!}{(X-k)!} \right] &= \lim_{s\uparrow1}G^{(k)}(s).\\ &= \lim_{s\uparrow1} \frac{\mathsf d}{\mathsf ds} e^{\lambda(s-1)}\\ &= \lim_{s\uparrow1} \lambda^n e^{\lambda(s-1)}\\ &= \lambda^n. \end{align}