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?

  • $\begingroup$ Is this homework or some kind of assignment? $\endgroup$
    – Peter Flom
    Mar 4, 2020 at 12:32
  • 1
    $\begingroup$ 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.) $\endgroup$
    – Ben
    Mar 4, 2020 at 20:37

1 Answer 1


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 $\mathbb{V}(X^2) = 4 \mu^2 \sigma^2 + 2 \sigma^4$.


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