# How to derive the formula for the $n$th moment of $X\sim N(0,1)$

My approach is to derive the moment generating function for $$X\sim N(0,1)$$ so that its $$n$$th derivative at $$t=0$$ will be the moment of $$X\sim N(0,1).$$

However, after finding $$M_X(t)=e^{\frac{t^2}{2}}$$

I am struggled to express the function in MacLauren Series ... which means I am unable to derive the $$n$$th moment from this approach...

Could someone please shows me how to expand this moment generating function into a MacLauren Series?

Many thanks!

Also, I would also appreciate if someone could show me other methds in finding the solution.

• Just use that $e^u = \sum_{n=0}^\infty \frac{u^n}{n!}$, with $u=\frac{t^2}{2}$ – Jakobian Sep 23 '18 at 18:14
• Since you called the random variable $X$ and not $x,$ you should call the moment-generating function $M_X,$ not $M_x.$ I edited accordingly. – Michael Hardy Sep 23 '18 at 18:39
• $$\sum_nE(X^n)\frac{t^n}{n!}=E(e^{tX})=e^{t^2/2}=\sum_n\frac1{n!}\left(\frac{t^2}2\right)^n$$ – Did Sep 23 '18 at 18:44

## 2 Answers

\begin{align} M_X(t) & = e^{t^2/2} = e^u = \sum_{k=0}^\infty \frac{u^k}{k!} = \sum_{k=0}^\infty \frac{(t^2/2)^k}{k!} = \sum_{k=0}^\infty \frac{t^{2k}}{2^k k!} \\[10pt] M_X^{(2m)}(0) & = \frac{d^{2m}}{dt^{2m}} \,\frac{t^{2m}}{2^m m!} = \frac{2m(2m-1)(2m-2) \cdots 1}{2^m m!} \\[12pt] & = (2m-1)(2m-3)(2m-5)\cdots 1. \\[12pt] \text{So } M_X^{(n)}(t) & = (n-1)(n-3)(n-5)\cdots 1 \text{ if n is even.} \end{align}

Here we assume $$n$$ is even since for odd $$n$$ the $$n$$th moment is obviously $$0.$$ (Note in response to comments. I'm not saying the integral of EVERY odd function is zero.) \begin{align} & \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty x^n e^{-x^2/2} \, dx = \sqrt{\frac 2 \pi} \int_0^\infty x^n e^{-x^2/2} \,dx \\[10pt] = {} & \sqrt{\frac 2 \pi} \int_0^\infty x^{n-1} e^{-x^2/2} (x\,dx) = \sqrt{\frac 2 \pi} \int_0^\infty \sqrt{2u\,}^{\,n-1} e^{-u}\, du \\[10pt] = {} & \frac{2^{n/2}}{\sqrt\pi} \int_0^\infty u^{(n-1)/2} e^{-u}\, du = \frac{2^{n/2}}{\sqrt\pi} \Gamma\left( \frac{n+1} 2 \right) \\[10pt] = {} & \frac{2^{n/2}}{\sqrt\pi} \cdot \frac{n-1} 2 \cdot \frac{n-3} 2 \cdot \frac{n-5} 2 \cdots \frac 1 2 \Gamma\left( \frac 1 2 \right) \\[10pt] = {} & (n-1)(n-3)(n-5) \cdots 1. \end{align}

Here I have used the functional equation $$\Gamma(\alpha+1) = \alpha\Gamma(\alpha)$$ and the fact that $$\Gamma\left( \frac 1 2 \right) = \sqrt\pi.$$

• if the integral of odd function is always 0 how do you explain the Cauchy distribution? – Chloe Zhou Sep 23 '18 at 23:55
• @ChloeZhou : I never said the integral of EVERY odd function is zero. I said $\displaystyle \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{+\infty} x^n e^{-x^2/2} \, dx$ is zero for every odd positive integer $n. \qquad$ – Michael Hardy Sep 24 '18 at 17:36