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I must compute all moments for the standard normal distribution. Here is my work:

It can be shown that $M(s)=E[e^{sX}]=e^{\frac{s^2}{2}}$. Furthermore, because $M(s)$ is defined over some open interval containing $(-r,r)$, $M(S)$ has the following expansion: $$M(S)=\sum_{k=0}^\infty \frac{s^k}{k!}E[X^k].$$

Hence, we can write $M(s)=e^{s^2/2}=\sum_{k=0}^\infty \frac{s^k}{k!}E[X^k]. $ At this point, I'm not sure what to do. I believe there is a way of expanding the exponential expression, but I don't know what its form is, or why the expansion is permissible/true. Any ideas?

  • You can deduce that $E[X^{2k+1}] = 0$ for all $k \geq 0$, and search the rest. https://math.stackexchange.com/questions/92648/calculation-of-the-n-th-central-moment-of-the-normal-distribution-mathcaln – snar Oct 21 '19 at 14:59

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You should recall that for every complex number $z,$ $$ e^z = \frac{z^0}{0!} + \frac{z^1}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots $$ Therefore $$ e^{s^2/2} = \frac{(s^2/2)^0}{0!} + \frac{(s^2/2)^1}{1!} + \frac{(s^2/2)^2}{2!} + \frac{(s^2/2)^3}{3!} + \cdots $$ As a power series in $s,$ this has only even-degree non-zero terms. Thus all moments of odd degree are $0.$

Michael Hardy
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  • This is exactly what I was looking for. –  Oct 21 '19 at 15:13
  • One small addendum lest anybody misunderstand: Say we want the $4$th moment $m_4.$ We get this from the term with the $4$th power of $s$, thus $\dfrac{(s^2/2)}{2!}.$ Since it's the $4$th moment, that term must be $\dfrac{s^4}{4!}\cdot m_4.$ Solving $$ \frac{(s^2/2)^2}{2!} = \frac{s^4}{4!}\cdot m_4 $$ for $m_4,$ we get $m_4 = 3.$ And similarly for the other moments. $\qquad$ – Michael Hardy Oct 21 '19 at 23:02