Calculation of the $n$-th moment of the normal distribution

Question

I need some advice on a question, maybe someone can give me a hint.

Let $X \in N(0,\sigma^2)$, show that $E[X^{2n+1}] = 0$ for $n = 0,1,2,\ldots$, and that $E[X^{2n}]= [(2n)!/2^nn!]\cdot \sigma^{2n}$.

Since this is a chapter in my book about moment generating functions , i guess the approach is to consider $\psi_{X}(t) = e^{\sigma^2t^2/2}$, and that $E[X^n] = \psi_X^{(n)}(0)$. My approach was to find some general formula for the $n$th derivative and then try to "show" what they were asking for; however it seem to be a bit hard to find such formula.

Answer 1

Hints Denote by $p$ the density of $X$, i.e. $$p(x) =\frac{1}{\sqrt{2\pi \sigma^2}} \exp \left(- \frac{x^2}{2\sigma^2} \right).$$

1. Deduce from $p(x) = p(-x)$ that $$\mathbb{E}(X^{2n+1}) = \int_{\mathbb{R}} x^{2n+1} p(x) \, dx =0.$$
2. Check that $\mathbb{E}(X^{2n}) = \frac{(2n)!}{2^n n!} \sigma^{2n}$ holds for $n=1$. Now perform an inductive step, i.e. assume that the claim holds for $n$ and show that then the statement holds for $n+1$. To this end, write $$\mathbb{E}(X^{2(n+1)}) = \int x^{2n+2} p(x) \, dx = \int x^{n+1} (x \cdot p(x)) \, dx$$ and use partial integration.