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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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 01:06
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1Nope. They are not independent. See here. – cardinal Oct 12 '11 at 01:26
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@Zoe: Since $X^2$ is determined by $X$, they can't be independent unless $X$ is constant. – Michael Hardy Oct 12 '11 at 02:15
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1@Michael: That's not entirely true. You actually need $X^2$ to be constant, not $X$. :) – cardinal Oct 12 '11 at 02:20
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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 03:48
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....but at any rate, for the normal distribution, $X^2$ is constant only if $X$ is constant. – Michael Hardy Oct 12 '11 at 04:05
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As a general idea, since $X^2$ is a quadratic function of $X$, their covariance is $0$ due to there is no linear relation between these two variables? As for the independence, it's like when they are independent, $f(X)$ tells no information about $X$, is that right? – Julie Oct 13 '11 at 04:32
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And the above idea requires $\mu=E(X)=0$ – Julie Oct 13 '11 at 04:38
2 Answers
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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.
Henry
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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 00:19
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1And I think the final answer should be $\mu\sqrt{\dfrac{2}{2\mu^2+\sigma^2}}$ – Henry Oct 12 '11 at 07:24
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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: $\operatorname{cov}(X,X^2) = 2\mu\sigma^2$.
