# 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?

• Is this homework or some kind of assignment? Mar 4 '20 at 12:32
• It is a moment result that is used in deriving the general Taylor series approximation for the variance of a square of a random variable, used so that I could answer the question here. (And I answered it myself.)
– Ben
Mar 4 '20 at 20:37

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{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}
The special case for an unskewed mesokurtic distribution (e.g., the normal distribution) occurs when $$\gamma = 0$$ and $$\kappa = 3$$, which gives the variance $$\mathbb{V}(X^2) = 4 \mu^2 \sigma^2 + 2 \sigma^4$$.