<s>[INST] Expectation of a Standard Normal Random Variable Calculate E(X^3) and E(X^4) for X~N(0,1).
I am having difficulty understanding how to calculate the expectation of those two. I intially would think you just calculate the 
$\int x^3e^\frac{-x^2}{2} dx $  and 
$\int x^4e^\frac{-x^2}{2} dx $ for $E(X^3)$ and $E(X^4)$, respectively. 
However for lecture for $E(X^2)$ this appearsappears. 
 [/INST] Use the same argument for $E[X]$ to say that $E[X^3] = 0$.
For $E[X^4]$, let $u = x^3$ and $dv = x\cdot\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$. Then $du = 3x^2$, $v = -\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$, and
\begin{align*}
E[X^4] &= \int_{-\infty}^\infty x^4f_X(x)\,dx\\
&=\int_{-\infty}^\infty x^4\cdot\frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}x^2\right\}\,dx\\
&=uv|_{-\infty}^\infty-\int_{-\infty}^\infty v\,du\\
&=\left(-x^3\exp\left\{-\frac{1}{2}x^2\right\}\right)\bigg|_{-\infty}^\infty-\int_{-\infty}^\infty 3x^2\cdot\frac{-1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}x^2\right\}\,dx\tag 1\\
&=0+3\int_{-\infty}^\infty x^2\cdot\frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}x^2\right\}\,dx\tag 2\\
&=3\cdot 1\\
&= 3
\end{align*}
where in $(1)$ I let you verify that indeed the left term is zero, and in $(2)$ I used the result presented in your lecture notes.
</s>