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I am looking for a proof of the following result:

Let $X \sim \mathcal{N}(0, \sigma^2)$, then $\mathbb{E}[X^n] = (n-1)!\sigma^n$ for all even numbers $n$, and zero for odd numbers $n$.

I was able to shown that $\mathbb{E}[X^4] = 3$ through the characteristic function, namely, taking derivative of $\dfrac{1}{i^4} \dfrac{d^4 e^{-w^2/2}}{dw^4}\lvert_{w = 0}$ but I found the derivation to be extremely cumbersome, with many iterations of the product rule.

Is there a simple proof of this result? Any reference helps!

Can this result be generalized to non-zero mean Gaussians?

Fraïssé
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1 Answers1

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First define a random variable $Y\sim\text{Gamma}(p)$ ($p>0$) if $Y$ has density $$ f_Y(y)=\frac{1}{\Gamma(p)}y^{p-1}e^{-y}\quad (y>0). $$ It is easy to see that $EY^d=\frac{\Gamma(p+d)}{\Gamma(p)}$ by the definition of the gamma function.

Now onto the problem. Let $X$ be a standard normal random variable. It is well known that $X^2\sim\chi^2_{(1)}$ or equivalently $X^2/2\sim\text{Gamma}(1/2)$. Write $X^2\stackrel{d}{=}2W$ where $W\sim\text{Gamma}(1/2)$. Then $$ EX^4=E(2W)^2=2^2EW^2=2^2\frac{\Gamma(1/2+2)}{\Gamma(1/2)}=2^2\left(\frac{1}{2}\right)\left(\frac{1}{2}+1\right)=1(3)=3$$ where we have used the fact that $\Gamma(p+1)=p\Gamma(p)$. We can generalize to compute all even moments of a standard normal. For $k\geq 1$ $$ EX^{2k}=E(2W)^k=2^kEW^k=2^k\frac{\Gamma(1/2+k)}{\Gamma(1/2)}=2^k\left(\frac{1}{2}\right)\left(\frac{1}{2}+1\right)\cdots \left(\frac{1}{2}+k-1\right). $$ But we can simplify further to get that $$ EX^{2k}=1(3)\cdots(2k-1)=\frac{(2k)!}{2^kk!}. $$

Sri-Amirthan Theivendran
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