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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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    And the above idea requires $\mu=E(X)=0$2011-10-13

2 Answers 2

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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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    @Henry, Yes I thin$k$ so.2011-10-13