Vrftsjtryk

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.

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

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 $X(X_1)(X-2)(X-3)$.

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.

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 $f^{(4)}(1)$.