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

Suppose that $X$ is real-valued normal random variable with mean $\mu$ and variance $\sigma^2$. What is the correlation coefficient between $X$ and $X^2$?

share|cite|improve this question
May I ask, if $\mu=0$, are $X$ and $X^2$ independent? I know their covariance is $0$, but this doesn't suffice. – Julie Oct 12 '11 at 1:06
Nope. They are not independent. See here. – cardinal Oct 12 '11 at 1:26
@Zoe: Since $X^2$ is determined by $X$, they can't be independent unless $X$ is constant. – Michael Hardy Oct 12 '11 at 2:15
@Michael: That's not entirely true. You actually need $X^2$ to be constant, not $X$. :) – cardinal Oct 12 '11 at 2:20
Also, I was mistaken in another respect: Adding a constant to $X$ has the effect of adding something other than a constant to $X^2$. (And now I've deleted that comment.) – Michael Hardy Oct 12 '11 at 3:48
up vote 4 down vote accepted

Here's an efficient way to deal with the numerator in the fraction that defines the correlation. $$ \operatorname{cov}(X,X^2) = \operatorname{cov}\Big((X-\mu)+\mu,\ \ (X-\mu)^2 + 2\mu(X-\mu) + \mu^2\Big). $$ Now we can throw away the "${}+ \mu$" and "${}+ \mu^2$" at the end and we have $$ \operatorname{cov}\Big((X-\mu),\ \ (X-\mu)^2 + 2\mu(X-\mu)\Big). $$ Then use bilinearity of covariances and this becomes: $$ \operatorname{cov}(X-\mu, (X-\mu)^2) + 2\mu\operatorname{cov}(X-\mu,X-\mu)). $$ This is $$ 0 + 2\mu\sigma^2. $$ The first term is $0$ because the expected value of $X-\mu$ is $0$ and the distribution is symmetric about $0$.

Summary: $\operatorname{cov}(X,X^2) = 2\mu\sigma^2$.

share|cite|improve this answer

Hint: You are trying to find: $$\frac{E\left[\left(X^2-E\left[X^2\right]\right)\left(X-E\left[X\right]\right)\right]}{\sqrt{E\left[\left(X^2-E\left[X^2\right]\right)^2\right]E\left[\left(X-E\left[X\right]\right)^2\right]}}$$

For a normal distribution the raw moments are

  • $E\left[X^1\right] = \mu$
  • $E\left[X^2\right] = \mu^2+\sigma^2$
  • $E\left[X^3\right] = \mu^3+3\mu\sigma^2$
  • $E\left[X^4\right] = \mu^4+6\mu^2\sigma^2+3\sigma^4$

so multiply out, substitute and simplify.

share|cite|improve this answer
I see the problem. When I was calculating the covariance, I mistook $E(X^2)=\sigma^2$, which led me to the wrong solution. – Julie Oct 12 '11 at 0:19
Therefore, the covariance should be $2\mu\sigma^2$. – Julie Oct 12 '11 at 0:30
Shouldn't the last term by $3\sigma^4$? – Michael Hardy Oct 12 '11 at 4:07
@Michael: Indeed it should - edited – Henry Oct 12 '11 at 6:09
And I think the final answer should be $\mu\sqrt{\dfrac{2}{2\mu^2+\sigma^2}}$ – Henry Oct 12 '11 at 7:24

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.