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I know that $$E(X) = P'_X(1)$$ then what would $$E(X(X-1))$$ formula would be?

Xenotion
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  • @KaviRamaMurthy Is there like a formula or how did you get that? – Xenotion May 19 '22 at 12:06
  • See https://en.wikipedia.org/wiki/Probability-generating_function#Probabilities_and_expectations and the $k$th factorial moment. @KaviRamaMurthy - I would have thought it might just be $P_X''(1)$ – Henry May 19 '22 at 12:13

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In general, the probability generating function gives you all the probability distribution of some non-negative random variable. For a random variable $X$, you define is PGF as $$P_X(z)=\sum_{k\geq 0}z^k Pr(X=k).$$ As you mentioned, its first derivative evaluated at 1 gives you the first moment. So, $E[X]=d_zP_X(z)|_{z=1}$. If you want the $k$'th factorial moment, you just have to take the $k$'th derivative. Therefore, $$E[X(X-1)(X-2)\ldots(X-k+1)=\frac{d^k P_X(z)}{dz^k}\bigg|_{z=1}.$$ Hence to obtain $E[X(X-1)]$ just set $k=2$ in the above formula (so take the second derivative).

seboll13
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  • So here $P_X''(z)=\sum\limits_{k\geq 2} k(k-1)z^{k-2} \Pr(X=k)$ giving $P_X''(1)=\sum\limits_{k\geq 2} k(k-1)\Pr(X=k) =E[X(X-1)]$ – Henry May 19 '22 at 12:18
  • @Henry Correct, although I'm not sure if the sum index has to start from 2 or 0. – seboll13 May 19 '22 at 12:24
  • seboll: When you take the first derivative you lose what was originally the $z^0$ term and when you take the second you also lose what was originally the $z^1$ term. If you do not lose these terms, then you might have issues saying $P_X''(0)=\Pr(X=2)$ as you could have to raise $0$ to a negative power – Henry May 19 '22 at 12:30
  • @Henry yeah this seems about right. Thanks for clarifying :) – seboll13 May 19 '22 at 12:36