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The $2n^\text{th}$ moment of the standard normal distribution $X \sim \text{N}(0, 1)$ is given by:

$$\mathbb{E}(X^{2n}) = \frac{(2n)!}{2^n n!},$$

which is also the number of ways to divide $2n$ people into $n$ distinct pairs. Is this just a coincidence, or is there some combinatorial explanation for this?

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  • $\begingroup$ See Gian-Carlo Rota and Jianhong Shen, On the Combinatorics of Cumulants. It's freely available online. $\endgroup$
    – whuber
    Commented May 16, 2020 at 22:14
  • $\begingroup$ math.stackexchange.com/questions/2986015/… $\endgroup$
    – Max M
    Commented Jul 3, 2021 at 16:55

1 Answer 1

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I am not sure if there is a simple way to frame this in combinatorial terms, other than to observe the equivalency that you have already observed. Nevertheless, a preliminary aspect of framing the result in combinatorial terms is to look at the moment generating function of the standard normal distribution. The MGF and its Maclaurin expansion are given by:

$$m_Z(t) = \exp \Big( \frac{t^2}{2} \Big) = \sum_{\ell=0}^\infty \frac{t^{2\ell}}{2^\ell \ell!}.$$

To facilitate our analysis, let $\mathscr{C}(k)$ denote the class of sets of $2k$ people divided into $k$ distinct pairs, so we have:

$$|\mathscr{C}(k)| = \frac{(2k)!}{2^k k!}.$$

Differentiating the MGF $2n$ times gives:

$$\begin{aligned} \frac{d^{2n} m_Z}{dt^{2n}}(t) &= \sum_{\ell=0}^\infty \bigg( \frac{d}{dt} \bigg)^{2n} \frac{t^{2\ell}}{2^\ell \ell!} \\[6pt] &= \sum_{\ell=0}^\infty \frac{1}{2^\ell \ell!} \bigg( \frac{d}{dt} \bigg)^{2n} t^{2\ell} \\[6pt] &= \sum_{\ell=n}^\infty \frac{(2\ell)_{2n}}{2^\ell \ell!} \cdot t^{2\ell-2n} \\[6pt] &= \sum_{\ell=0}^\infty \frac{(2n+2\ell)!}{2^{n+\ell} (n+\ell)!} \cdot \frac{t^{\ell}}{(2 \ell)!} \\[6pt] &= \sum_{\ell=0}^\infty |\mathscr{C}(n+\ell)| \cdot \frac{t^{\ell}}{(2 \ell)!}, \\[6pt] \end{aligned}$$

and so the $2n$th raw moment is:

$$\mathbb{E}(Z^{2n}) = \frac{d^{2n} m_Z}{dt^{2n}}(0) = |\mathscr{C}(n)|.$$

As you can see, combinatorial terms similar to the one of interest to you occur when we differentiate the moment generating function. This is a general result pertaining to that particular exponential function, so it is not something restricted to the normal distribution.

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  • $\begingroup$ Thanks a lot Ben!!! I am fine with mathematical analogy but I was looking for the combinatorial explanation for this. I am going through Harvard's Stats110 course and encountered this: link. I couldn't find any simple logical explanation for their equivalence as mentioned by the prof, though. He skipped the part about the explanation!!! $\endgroup$ Commented May 16, 2020 at 5:11
  • $\begingroup$ It would be wonderful if you could ask your Professor for the reason and then post it as an answer --- it is not something I am particularly familiar with. $\endgroup$
    – Ben
    Commented May 16, 2020 at 10:12
  • $\begingroup$ Actually I am going through online video lectures posted on YouTube. I will try to reach out to him and post the response if I found something!!! $\endgroup$ Commented May 16, 2020 at 10:51

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