1
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

\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}

$\endgroup$
-1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.