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

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What have you tried and where are you stuck? –  cardinal Oct 11 '11 at 23:07
    
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
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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
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@Michael: That's not entirely true. You actually need $X^2$ to be constant, not $X$. :) –  cardinal Oct 12 '11 at 2:20

2 Answers 2

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

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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.

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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
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And I think the final answer should be $\mu\sqrt{\dfrac{2}{2\mu^2+\sigma^2}}$ –  Henry Oct 12 '11 at 7:24

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