What is the variance of $X^2$ (without assuming normality)? If $X \sim \text{N}(0, \sigma^2)$ then it is well-known that $X^2/\sigma^2 \sim \text{ChiSq}(1)$ which gives $\mathbb{V}(X^2) = 2 \sigma^4$.  However, this variance holds only for the normal distribution with zero mean.  What is the general formula for $\mathbb{V}(X^2)$ without assuming normality or zero mean?
 A: The general form of this variance depends on the first four moments of the distribution.  To facilitate our analysis, we suppose that $X$ has mean $\mu$, variance $\sigma^2$, skewness $\gamma$ and kurtosis $\kappa$.  The variance of interest exists if $\kappa < \infty$ and does not exist otherwise.  Using the relationship between the raw moments and the cumulants, you have the general expression:
$$\begin{equation} \begin{aligned}
\mathbb{V}(X^2)
&= \mathbb{E}(X^4) - \mathbb{E}(X^2)^2 \\[6pt]
&= ( \mu^4 + 6 \mu^2 \sigma^2 + 4 \mu \gamma \sigma^3 + \kappa \sigma^4 ) - ( \mu^2 + \sigma^2 )^2 \\[6pt]
&= ( \mu^4 + 6 \mu^2 \sigma^2 + 4 \mu \gamma \sigma^3 + \kappa \sigma^4 ) - ( \mu^4 + 2 \mu^2 \sigma^2 + \sigma^4 ) \\[6pt]
&= 4 \mu^2 \sigma^2 + 4 \mu \gamma \sigma^3 + (\kappa-1) \sigma^4. \\[6pt]
\end{aligned} \end{equation}$$
The special case for an unskewed mesokurtic distribution (e.g., the normal distribution) occurs when $\gamma = 0$ and $\kappa = 3$, which gives the variance $\mathbb{V}(X^2) = 4 \mu^2 \sigma^2 + 2 \sigma^4$.
