Normal distribution

Question

can anyone help me calculate $E(Z^4)$, $E(Z^3)$ for $Z\sim N(0,1)$? I know that $Z^2\sim \chi^2(1)$ then $E(Z^2)=1$, $Var(Z^2)=2$. Thank you.

Answer 1

Since the pdf of the standard normal distribution is an even function, it follows that $\mathbb{E}[Z^3]=0$. And $$ \mathbb{E}[Z^4]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x^4e^{-\frac{x^2}{2}}\;dx=\sqrt{\frac{2}{\pi}}\int_0^{\infty}x^4e^{-\frac{x^2}{2}}\;dx$$ which can be evaluated by setting $u=\frac{x^2}{2}$ and using the properties of the Gamma function.

Answer 2

If you know a little bit of calculus, then there is a way to quickly calculate any moment you want for a normal distribution using its Moment Generating Function (MGF). In particular, the MGF for a normal distribution is:

$$M(t)=e^{t\mu+\frac{1}{2}\sigma^2t^2}$$

The great property of this function is that $E(Z^k)=M^{(k)}(0)$.

So, for your case where $\mu=0,\sigma^2=1$, we get:

$$M(t)=e^{\frac{1}{2}t^2}$$

Then

$$E[Z]=M'(t)\mid_{t=0}=t^2e^{\frac{1}{2}t^2}\mid_{t=0}=0$$ $$E[Z^2]=M''(t)\mid_{t=0}=(t^2+1)e^{\frac{1}{2}t^2}\mid_{t=0}=1$$ $$E[Z^3]=M^{(3)}(t)\mid_{t=0}=t(t^2+3)e^{\frac{1}{2}t^2}\mid_{t=0}=0$$ $$E[Z^4]=M^{(4)}(t)\mid_{t=0}=(t^4+6t^2+3)e^{\frac{1}{2}t^2}\mid_{t=0}=3$$

And so on ad nauseum...thank goodness we have WolframAlpha!

Answer 3

I think I can provide a simpler answer:
You know that the normal distribution is an even function so you know that the expectation of an odd power must be zero.

You know the variance and mean of $z^2$ and you know that var($z^2$)= E[$(z^2)^2$] - E[$z^2$]$^2$ so you know then that E[$z^4$] = var($z^2$)+E[$Z^2$]$^2$ = 2+1=3