# Computing all moments of standard normal distribution

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 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 $$0.$$

• This is exactly what I was looking for.
– user443369
Commented Oct 21, 2019 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. $\qquad$ Commented Oct 21, 2019 at 23:02