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.

  • 1
    $\begingroup$ Just use that $e^u = \sum_{n=0}^\infty \frac{u^n}{n!}$, with $u=\frac{t^2}{2}$ $\endgroup$ – Jakobian Sep 23 '18 at 18:14
  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Sep 23 '18 at 18:39
  • $\begingroup$ $$\sum_nE(X^n)\frac{t^n}{n!}=E(e^{tX})=e^{t^2/2}=\sum_n\frac1{n!}\left(\frac{t^2}2\right)^n$$ $\endgroup$ – Did Sep 23 '18 at 18:44

\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 $\Gamma\left( \frac 1 2 \right) = \sqrt\pi.$

  • $\begingroup$ if the integral of odd function is always 0 how do you explain the Cauchy distribution? $\endgroup$ – Chloe Zhou Sep 23 '18 at 23:55
  • $\begingroup$ @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 $n. \qquad$ $\endgroup$ – Michael Hardy Sep 24 '18 at 17:36

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