Given that $X \sim \operatorname{Binomial}(n,p)$, Find $\mathbb{E}[X(X-1)(X-2)(X-3)]$

Question

Given that $X \sim \operatorname{Binomial}(n,p)$, Find $\mathbb{E}[X(X-1)(X-2)(X-3)]$.

It is suggested that I can transform it into \begin{align} \mathbb{E}[X(X-1)(X-2)(X-3)] &=\sum_{k=0}^n k(k-1)(k-2)\mathbb{P}\{X=k\}\\ &=\sum_{k=3}^{n+3} (k-3)(k-4)(k-5)\mathbb{P}\{X=k-3\}\\ &=\sum_{k=0}^n i(i-1)(i-2)\mathbb{P}\{X=i\} \end{align} But then I just have no idea about how can i do it. I suspect that it needs something similar to this post but the steps are quite different from this one.

Please help.

Answer 1

One way is to use characteristic functions. The characteristic function of $B(n,p)$ is $\phi (t)=(pe^{it}+(1-p))^{n}$. The moments of $X$ are given by $i^{n}EX^{n}=\phi^{(n)}(0)$. You can compute the first 4 moments of $X$ using this and then use the expanded form of $X(X_1)(X-2)(X-3)$.

Answer 2

Hint: Probability Generating Function (see https://en.wikipedia.org/wiki/Probability-generating_function)

Look at the properties section, the part where it talks about the kth factorial moment.

Answer 3

Look at $f(s)= E [ s^X]$.

1) Observe that

$$f(s) = (sp + (1-p))^n=(1+ (s-1)p)^n =\sum_{k=0}^n \binom{n}{k}p^k (s-1)^k=\sum_{k=0}^n \frac{n!}{(n-k)!} \frac{(s-1)^k}{k!},$$ the RHS identified as the Taylor series of $f$ about $s=1$.

2) Differentiating under the expectation (which is just a finite sum here), we obtain $$f'(s) = E[ X s^{X-1}] ,~f''(s) = E[X (X-1) s^{X-2}],\dots.$$

You're looking for $f^{(4)}(1)$.

Answer 4

Start as suggested, and write down what the probability mass function (pmf) of the Binomial actually is: $$\begin{align*} \mathbb{E}[X(X-1)(X-2)(X-3)] &= \sum_{k=0}^n k(k-1)(k-2)(k-3)\mathbb{P}\{X=k\}\\ &= \sum_{k=4}^n k(k-1)(k-2)(k-3)\mathbb{P}\{X=k\}\\ &= \sum_{k=4}^n k(k-1)(k-2)(k-3)\binom{n}{k}p^k(1-p)^{n-k}\\ &= \sum_{k=4}^n k(k-1)(k-2)(k-3)\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}\\ &= \sum_{k=4}^n \frac{n!}{(k-4)!(n-k)!}p^k(1-p)^{n-k}\\ &= \sum_{k=4}^n \frac{n(n-1)(n-2)(n-3)(n-4)!}{(k-4)!((n-4)-(k-4))!}p^k(1-p)^{n-k}\\ &= n(n-1)(n-2)(n-3)p^4\sum_{k=4}^n \binom{n-4}{k-4}p^{k-4}(1-p)^{(n-4)-(k-4)}\\ &= n(n-1)(n-2)(n-3)p^4\sum_{\ell=0}^n \binom{n-4}{\ell}p^{\ell}(1-p)^{(n-4)-\ell}\\ &= \boxed{n(n-1)(n-2)(n-3)p^4} \end{align*}$$ since $\sum_{\ell=0}^n \binom{n-4}{\ell}p^{\ell}(1-p)^{(n-4)-\ell}=1$, recognizing the sum of probabilities for a Binomial with parameters $n-4$ and $p$.