So I have the Normal Distribution $f(z)=\frac{1}{\sqrt{2\pi}}e^{-z^2/2}.$
I know any $E(Z^{\mbox{ (any odd #)}})$ makes you integrate an odd function thus giving an answer of zero (i.e. $E(Z^1)$ and $E(Z^3)$ both $=0$).
And I know computing $E(Z^2)$ shows some links to the Gamma function/distributions and helps the integration and gives the answer of $1.$
However, I'm having difficulty figuring $E(Z^4).$ I know $\displaystyle E(Z^4) = \int \left[z^4 \frac{1}{\sqrt{2\pi}}e^{-z^2/2}\right] dz,$ then I do $\displaystyle 2 \cdot \frac{2}{\sqrt{2\pi}} \int \left[\frac{z^2}{2} e^{-z^2/2}\right] d\left(\frac{z^2}{2}\right).$
This is similar to the Gamma function $\Gamma(\alpha)= \int t^{\alpha-1} e^{-t}.$ However, my "$t$" in my problem $[(z^2)/2]$ doesn't have an $(\alpha-1)$, it's just to the first power. Any suggestions of how to manipulate it better to form a better gamma function to solve the problem for $E(X^4)$?
Thanks a lot
Substitute $\frac{z^{2}}{2}=t$, then $zdz=dt$ and $z^{3}=(2t)^{3/2}$. \begin{eqnarray} \mathbb{E}[Z^{4}]&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}z^{4}e^{-z^{2}/2}\;dz\\ &=&\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}z^{4}e^{-z^{2}/2}\;dz\\ &=&\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}(2t)^{3/2}e^{-t}\;dt\\ &=&\frac{2^{5/2}}{\sqrt{2\pi}}\int_{0}^{\infty}t^{5/2-1}e^{-t}\;dt\\ &=&\frac{2^{5/2}}{\sqrt{2\pi}}\Gamma(5/2)\\ &=&\frac{2^{5/2}}{\sqrt{2\pi}}\frac{3}{2}\frac{1}{2}\sqrt{\pi}\\ &=&3. \end{eqnarray} I am using the facts that $\Gamma(n+1)=n\Gamma(n)$ and $\Gamma(1/2)=\sqrt{\pi}$.