What is the $k$th factorial moment of the binomial distribution?

Question

I'm not sure what to do from here. $$E[x(x-1)(x-2)\ldots(x-k+1)]=E\left[x!\over k!\right]={1\over k!}\left[\sum x!p(x)\right]={1 \over k!}\left[\sum\frac{n!}{(n-x)!}p^{x}q^{n-x}\right]$$

Answer 1

We have that $$E[x(x-1)(x-2)\ldots(x-k+1)]=E\left[\frac{x!}{(x-k)!}\right]=\sum_{x} \frac{x!}{(x-k)!}p(x).$$

Now, since $p(x)={n\choose n-x}p^{x}q^{n-x},$ we have $$\sum_{x}\frac{x!}{(x-k)!}p(x)=\sum_{x}\frac{n!}{(n-x)!(x-k)!}p^{x}q^{n-x},$$ since the $x!$ cancel.

However, $(x-k)!$ and $(n-x)!$ are both defined only for $k\leq x\leq n$, so the sum can only go from $x=k$ to $x=n$; thus the sum is actually equal to $$\frac{n!}{(n-k)!}p^{k}\left[\sum_{k\leq x\leq n}\frac{(n-k)!}{(n-x)!(x-k)!}p^{x-k}q^{n-x}\right] \\ =\frac{n!}{(n-k)!}p^{k}\sum_{0\leq x-k\leq n-k}{{n-k\choose x-k}p^{x-k}q^{(n-k)-(x-k)}}.$$

Letting $x'=x-k$ in the last sum, we recognize it as a binomial series $(p+q)^{n-k}$; then since $p+q=1$ (by definition of the binomial distribution), the final answer is $$\frac{n!}{(n-k)}p^{k}.$$

Answer 2

$$\begin{align} E[x(x-1)(x-2)\cdots(x-k+1)] &= E\left[\frac{x!}{(x-k)!}\right]\\ &=\sum \frac{x!p(x)}{(x-k)!}\\ &=\sum \frac{x!n!}{x!(x-k)!(n-x)!}p^xq^{n-x}\\ &=\frac{n!}{(n-k)!}p^k \sum \frac{(n-k)!}{(x-k)!(n-x)!}p^{x-k}q^{n-x}\\ &=\frac{n!}{(n-k)!}p^k \end{align}$$