Moments of standard normal random variable

Question

I want to show that for $n \geq 1$ we have $\mathbb{E}Z^{n+1} = n\mathbb{E}Z^{n−1}$ (where $Z$ is a standard normal RV). I've tried induction, but I'm not quite sure how to make use of the inductive hypothesis, so I am hoping someone can suggest a different approach. Thanks

Answer 1

By definition $$\mathbb EZ^{n+1}=\int_{-\infty}^{+\infty}z^{n+1}\left(\frac{1}{\sqrt{2\pi}}e^{-z^2/2}\right)dz=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}z^{n}\left(ze^{-z^2/2}\right)dz$$ Integration by parts gives \begin{align}\mathbb EZ^{n+1}&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}z^{n}\left(-e^{-z^2/2}\right)'dz\\[0.4cm]&=\frac{1}{\sqrt{2\pi}}-z^ne^{-z^2/2}\Bigg|^{+\infty}_{-\infty}+n\int_{-\infty}^{+\infty}z^{n-1}\left(\frac{1}{\sqrt{2\pi}}e^{-z^2/2}\right)dz\\[0.4cm]&=0+n\mathbb EZ^{n-1}\end{align} Also, since $\mathbb E Z^0=1$ and $\mathbb E Z^1=0$, you may expand the right side of the recursive expression to obtain $$\mathbb EZ^n=\begin{cases}0, &n \text{ odd}\\[0.3cm]\dfrac{2^{-n/2}n!}{(n/2)!},& n\text{ even}\end{cases}$$

Answer 2

Note that $$\mathbb E\left[Z^{n+1}-nZ^{n -1}\right]=\frac 1{\sqrt{2\pi}}\int_{-\infty}^{ +\infty}\left(x^{ n+1}-nx^{n-1} \right)\exp\left(-x^2/2\right) \mathrm dx$$ and the derivative of $x\mapsto x^n \exp\left(-x^2/2\right)$ is $-x^n\cdot x\exp\left(-x^2/2\right)+nx^{n-1}\exp\left(-x^2/2\right)$.