Third Moment of Standard Normal Random Variable

Solution 1:

$\mathbb{E}[(X-\mu)^3]=0$ since $X-\mu$ is normally distributed with mean zero, then expand out the cube.

Solution 2:

This is a general method to calculate any moment:

Let $X$ be standard normal. The moment generating function is:

$$M_X(t) = E(e^{tX}) = e^{t^2/2} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-(x-t)^2/2} \:\:dx = e^{t^2/2}$$

since the integrand is the pdf of $N(t,1)$.

Specifically for the third moment you differentiate $M_X(t)$ three times

$M_X^{(3)}(t) = (t^3 + 3t)e^{t^2/2}$

and

$E[X^3] = M_X^{(3)}(0) = 0$

For a general normal variable $Y = \sigma X + \mu$ we have that

$$M_Y(t) = e^{\mu t} M_X(\sigma t) = e^{\mu t + \sigma^2 t^2 /2} $$

and you calculate the $n$th moment as a above, i.e. differentiating $n$ times and setting $t=0$: $$E[Y^n] = M_Y^{(n)}(0).$$