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Assume there is a random variable distributed normal $X\sim N(\mu,\sigma^2)$. Is there an analytic expression for the covariance of $X$ with its square $X^2$?

$$\operatorname{Cov}(X,X^2)$$

I have done a extensive search but I haven't found any result on this. I conjecture that when $\mu=0$, the covariance is zero, but I don't know how to generalize it.

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$Y:=\frac{X-\mu}{\sigma}\sim N(0,1)$, and \begin{align}\operatorname{Cov}(X,X^2)&=\operatorname{Cov}(\sigma Y+\mu,(\sigma Y+\mu)^2)\\ &=\operatorname{Cov}(\sigma Y+\mu,\sigma^2 Y^2+2\mu\sigma Y+\mu^2)\\ &=\sigma^3\operatorname{Cov}(Y,Y^2)+2\mu\sigma^2\operatorname{Cov}(Y,Y)\\ &=2\mu\sigma^2. \end{align}

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    $\begingroup$ A key step is to show that $\text{Cov}(Y,Y^2)=0$. Since $Y$ is standard normal, this covariance can be computed as $\text{Cov}(Y,Y^2)=\text{E}(Y^3)-\text{E}(Y)\text{E}(Y^2)$, where the moments can be found using the moment generating function for the normal (details in this stack). In particular, $\text{E}(Y^3)=0=\text{E}(Y)$, so that $\text{Cov}(Y,Y^2)=0$. $\endgroup$ Apr 25 '17 at 15:05
  • $\begingroup$ Indeed $X$ and $X^2$ are dependent hence we are not sure that the correlation is zero (and it may not be, as the computation shows). The random variable $Y$ has the advantage to be centered and of variance one, which makes the computations easier. $\endgroup$ May 28 '19 at 14:25
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You can also simulate the results to get the intuition. For example for $X ~N(0,1)$, $X$ and $X^2$ have a covariance of zero, but between $X$ and $X^3$ it's $3$. Here is the $R$ code:

x=rnorm(10^5)

y=x^2 # or x^3

z=cbind(x,y)

cov(z)

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