Integration by part of Gaussian r.v.

Question

How to prove the following using integration by parts?

For the i.i.d Gaussian r.v. with mean 0 and variance $\sigma^2 $ $$ ES_n^{2k}=E(\frac{\sum_{j=1}^{n} X_j}{n})S^{2k-1}_n=\frac{(2k-1)\sigma^2}{n}ES_n^{2(k-1)}$$

Answer 1

We can do this because of the special form of probability density function of Gaussian r.v. Denote $Z = \frac{\sum_{i=1}^n X_i}{n}$, then $Z\sim N(0, \sigma^2/n)$, and the p.d.f. of it is $p(z) = \sqrt{\frac{n}{2\pi \sigma^2}}\exp(-\frac{n}{2\sigma^2}z^2)$. $$ ES_n^{2k} = \int_{-\infty}^{\infty} zp(z)\cdot z^{2k-1}dz $$ Since $\frac{d}{dz}p(z) = -\frac{n}{\sigma^2}zp(z)$, we can do integral by parts as \begin{align} \int_{-\infty}^{\infty} zp(z)\cdot z^{2k-1}dz &= \left[ -\frac{\sigma^2}{n}p(z)z^{2k-1} \right]^{\infty}_{-\infty} - \int_{-\infty}^{\infty} -\frac{\sigma^2}{n}p(z)\cdot (2k-1)z^{2k-2}dz\\ &= \frac{(2k-1)\sigma^2}{n}\int_{-\infty}^{\infty} p(z)\cdot z^{2k-2}dz = \frac{(2k-1)\sigma^2}{n}ES_n^{2k-2}. \end{align} Here $\left[ -\frac{\sigma^2}{n}p(z)z^{2k-1} \right]^{\infty}_{-\infty} = 0$ like we do for any other cases involving negative exponential functions like $\exp(-x)$.