$n$-th derivative of $e^{x^2}$ at $x=0$ to calculate central moments of Normal distribution

Question

I'd like to calculate $n$-th central moment of a normal random variable $X \sim N(m, s^2)$.

There are 2 ways to do so: either calculate the integral (by definition of the expected value), or take the $n$-th derivative of the moment-generating (or characteristic) function. The first way seems easier in this particular case, but I'd like to do it by MGF derivation.

$$M_{X-m}(t) = e^{s^2t^2/2} = f(t)$$

$$ E[(X-m)^n]=d^n/dt^n(M_{X-m}(t))|_{t=0}=d^n/dt^n(e^{s^2t^2/2})|_{t=0}=f^{(n)}(0) $$

I know the answer, which is $0$ for $n=2p+1$ and $s^n(n-1)!!$ for $n=2p$ - but I have no idea how to calculate this derivative.

Calculating $f^{(n)}(t)$ in general is very difficult and can't be expressed in elementary functions, that's why I try to find a way to calculate $f^{(n)}(0)$ directly without calculating $f^{(n)}(t)$ - but how can I do this? Maybe with series expansion...

Answer 1

Consider first $Y\sim N(0;1)$ and calculate its simple moments..

Expanding its MGF in taylor series you get that

$$M_Y(t)=\sum_{n=0}^{\infty}\frac{(\frac{1}{2}t^2)^n}{n!}=\sum_{n=0}^{\infty}\frac{(2n)!}{2^n\cdot n!}\cdot\frac{t^{2n}}{(2n)!}$$

thus derivating and evaluating the result in $t=0$ you get that

$$\mathbb{E}[Y^{2n}]=\frac{(2n)!}{2^n\cdot n!}$$

and

$$\mathbb{E}[Y^{2n+1}]=0$$

Your random variable $(X-m)$ is the same $Y$ but scaled, being $\frac{X-m}{s}=Y$ thus

$$\mathbb{E}[X-m]^{2n}=\frac{(2n)!}{2^n\cdot n!}s^{2n}$$

Answer 2

Use the series expansion of $e^t$ around $t=0$; make $t=x^2$ to have $$e^{x^2}=\sum_{n=0}^\infty \frac{x^{2n}}{n!}$$ So, at $x=0$, the $n^{th}$ derivatives are $0$ if $n$ is odd.

If $n=2m$ the derivative will be equal to to the quadruple factorial $\frac {(4m)!}{(2m)!}$