$E(X^3)$ in terms of a generating function

Question

If we suppose that $G(s)$ is the generating function of the random variable $X$ and that the variable $X$ takes nonnegative integer values only, then is there a good way of expressing $E(X^3)$ in terms of the function $G$? I've been trying to figure this out, but I'm not quite familiar with generating functions yet.

Answer 1

In the usual generating function set-up, if $G(s)$ is the GF for $X$ then $$G^{(k)}(1)=E(X(X-1)(X-2)\cdots(X-k+1)).$$ where $G^{(k)}$ is the $k$-th derivative of $G$. Then $$G'(1)=E(X),$$ $$G''(1)=E(X(X-1))=E(X^2)-E(X),$$ $$G'''(1)=E(X(X-1)(X-2))=E(X^3)-3E(X^2)+2E(X),$$ etc. Therefore $$E(X^3)=G'''(1)+3G''(1)+G'(1).$$

If you continue this for higher moments of $X$, the coefficients you get will be Stirling numbers of the second kind.

Answer 2

I share @AnginaSeng's assumption that the generating function intended with the notation $G(s)$ is a PGF. But on the off chance we're both wrong:

• If $G$ is the MGF, $E(X^3)=G^{(3)}(0)$;
• if $G$ is the CGF, $E[X^3]=G^{(3)}(0)+3G^\prime(0)G^{\prime\prime}(0)+[G^\prime(0)]^3$ (assuming we define the CGF as the log-MGF rather than the log-CF, which would put some powers of $i$ in there).