Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do you compute the expected value of a random variable? The problem I found asks; $$ W = rV^3$$ where $r$ is a constant and $V$ is a normally distributed random variable with mean 6 and standard deviation 1. How can I compute $E[W]$?

share|cite|improve this question

Write $V = \mu + \sigma Z$, where $Z$ is the standard normal random variable. Then $$ \mathbb{E}(W) = r \mathbb{E}\left( \left(\mu + \sigma Z\right)^3 \right) = r \left( \mu^3 +3 \mu^2 \sigma \mathbb{E}(Z) + 3 \mu \sigma^2 \mathbb{E}(Z^2) + \sigma^3 \mathbb{E}(Z^3) \right) $$ Since $Z$ is symmetric, i.e. $Z \stackrel{d}{=} -Z$, it follows that $\mathbb{E}(Z) = 0$ and $\mathbb{E}(Z^3)=0$. It only remains to evaluate $$ \mathbb{E}(Z^2) = \int_{-\infty}^\infty z^2 \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-z^2/2} \mathrm{d} z = \sqrt{\frac{2}{\pi}} \int_0^\infty z^2 \mathrm{e}^{-z^2/2} \mathrm{d} z \stackrel{z^2 = x}{=} \frac{1}{\sqrt{2\pi}} \underbrace{\int_0^\infty \sqrt{x} \mathrm{e}^{-x/2} \mathrm{d} x}_{2^{3/2} \Gamma(3/2) = \sqrt{2\pi}} = 1 $$ Now combine these results to readily obtain $\mathbb{E}(W)$.

share|cite|improve this answer
+1 To avoid the integral, note that $\mathbb{E}(Z^2)$ is the sum of its variance and the square of its mean, i.e. (as $Z$ has a standard normal distribution) $1+0^2=1$. – Henry Oct 19 '12 at 7:03

$E[W] = rE[V^{3}]$ since $r$ is a constant. You can obtain the information about higher moments of the normal distribution here See "Raw moment"

share|cite|improve this answer
My class isn't cover moment functions. Is there any way to understand it without knowledge of moment functions? There should be, since it's on our outline – CodyBugstein Oct 19 '12 at 2:29
Maybe you don't need to know how to compute it, but are you allowed to use that information? I can't see how you can obtain the third moment otherwise. – Ken Dunn Oct 19 '12 at 2:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.