If $X \sim \text{N}(0, \sigma^2)$ then it is well-known (from Cochran's theorem) that $X \ \bot \ X^2$ so these random variables have zero covariance. However, this holds only for the normal distribution with zero mean. What is the general formula for $\mathbb{C}(X,X^2)$ without assuming normality or zero mean?


The general form of the covariance depends on the first three moments of the distribution. To facilitate our analysis, we suppose that $X$ has mean $\mu$, variance $\sigma^2$ and skewness $\gamma$. The covariance of interest exists if $\gamma < \infty$ and does not exist otherwise. Using the relationship between the raw moments and the cumulants, you have the general expression:

$$\begin{equation} \begin{aligned} \mathbb{C}(X,X^2) &= \mathbb{E}(X^3) - \mathbb{E}(X) \mathbb{E}(X^2) \\[6pt] &= ( \mu^3 + 3 \mu \sigma^2 + \gamma \sigma^3 ) - \mu ( \mu^2 + \sigma^2 ) \\[6pt] &= 2 \mu \sigma^2 + \gamma \sigma^3. \\[6pt] \end{aligned} \end{equation}$$

The special case for an unskewed distribution with zero mean (e.g., the centred normal distribution) occurs when $\mu = 0$ and $\gamma = 0$, which gives zero covariance. Note that the absence of covariance occurs for any unskewed centred distribution, though independence holds only for the normal distribution.


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