Let X be a normal random variable with mean $\mu$ and standard deviation $\sigma^2$. I am wondering how to calculate the third moment of a normal random variable without having a huge mess to integrate. Is there a quicker way to do this?

  • $\begingroup$ $E(X-\mu)^3=0$. Expand the LHS. $\endgroup$ – A.S. Mar 9 '16 at 0:32
  • $\begingroup$ Standard normal or not? $\endgroup$ – JKnecht Mar 9 '16 at 0:36
  • $\begingroup$ Just normal--thanks for the vocabulary correction. $\endgroup$ – user288829 Mar 9 '16 at 0:43
  • $\begingroup$ A normal distribution is symmetric about the mean. The skewness is thus zero. $\endgroup$ – Graham Kemp Mar 9 '16 at 1:02

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

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


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

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 $t=0$: $$E[Y^n] = M_Y^{(n)}(0)$$.


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

  • $\begingroup$ By the way, standard normal often refers to $\mu=0$ and $\sigma=1$. $\endgroup$ – carmichael561 Mar 9 '16 at 0:33
  • $\begingroup$ That makes sense. Do I have to integrate still if it was standard normal, not just normal? $\endgroup$ – user288829 Mar 9 '16 at 1:03
  • $\begingroup$ If $X$ is normally distributed with mean zero, then $\mathbb{E}[X^n]=0$ for all odd integers $n$. $\endgroup$ – carmichael561 Mar 9 '16 at 1:39
  • $\begingroup$ After expanding it, how do we compute $\mathbb{E}(X^3)$? We would get many other terms? $\endgroup$ – Jay Wong Jul 7 '16 at 1:43
  • $\begingroup$ Sure, but all the other terms can be determined fairly easily since we know the mean and variance of $X$. $\endgroup$ – carmichael561 Jul 7 '16 at 1:55

If the distribution of a random variable $X$ is symmetric about $0$, meaning $\Pr(X>x)=\Pr(X<-x)$ for every $x>0$, then its third moment, if it exists at all, must be $0$, as must all of its odd-numbered moments.

If $\operatorname{E}\left[ \,|X^3|\, \right]<\infty$ then the third moment exists. Furthermore, Symmetry shows that the positive and negative parts of $\operatorname{E}\left[ X^3 \right]<0$ (without the absolute-value sign) cancel each other out.


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