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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?

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4 Answers

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

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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

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

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

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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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Given the questions expressed under carmichael561's answer, let us look at the details:\begin{align} 0 = {} & \operatorname E\big( (X-\mu)^3\big) \\[8pt] = {} & \operatorname E\big( X^3 - 3X^2 \mu +3X\mu^2 - \mu^3\big) \\[8pt] = {} & \operatorname E(X^3) - 3\mu\operatorname E(X^2) + 3\mu^2\operatorname E(X) - \mu^3 \\[8pt] = {} & \operatorname E(X^3) - 3\mu\big( \mu^2+\sigma^2) + 3\mu^2\big(\mu\big) - \mu^3 \\[8pt] = {} & \operatorname E(X^3) -3\mu\sigma^2 - \mu^3. \end{align}

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