Correlation Coefficient between these two random variables Suppose that $X$ is real-valued normal random variable with mean $\mu$ and variance $\sigma^2$. What is the correlation coefficient between $X$ and $X^2$?
 A: 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.
A: 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: $\operatorname{cov}(X,X^2) = 2\mu\sigma^2$.
