# Sampling distribution of sample variance of non-normal iid r.v.s

In a first course in statistics we are taught that when we do not know the true variance, we can still perform certain basic tests on the sample mean using the sample variance, $S^2$, instead -- provided we can model the sample using normal random variables. This is because the standardization of the mean $\frac{\bar{X}_n - \mathbb{E}(X)}{S/\sqrt{n}}$ follows a distribution (t-distribution) that is easy to understand.

This rests on the fact that we understand the sampling distribution of the sample variance when the i.i.d random variables modelling the sample are Normal:

If $X_i \sim N(\mu, \sigma^2)$ for $i = 1,2, \ldots n$ then $$\frac{S^2}{\sigma^2/(n-1)^2} \sim \chi^2_{n-1}$$

My question:

• What is the most general thing we can say when the i.i.d random variable $X_i$'s are not normal but have finite mean and variances; in other words is there something analogous to the central limit theorem for sample variance?
• Yes, there is. The asymptotic distribution of $S^2$ can be derived from the classical CLT and Slutsky's theorem. The asymptotic variance of $S^2$, as you may expect, depends on the fourth moment of $X$. Dec 1, 2017 at 20:07
• You don't need asymptotics for the denominator -- you need asymptotics for the ratio (the t-statistic itself). You can get an asymptotic result from CLT + Slutsky. Even when both hold it may take quite a large n to be a useful approximation to the null distribution (and there's no guarantee that the asymptotic relative efficiency will be up to much) Dec 2, 2017 at 4:19
• Asymptotic distribution of sample variance can be found here. Aug 4, 2020 at 20:15

$$\frac{S^2}{\sigma^2} \sim \frac{\chi^2 (DF_n)}{DF_n} \quad \quad \quad DF_n \equiv \frac{2 \sigma^4}{\mathbb{V}(S^2)} = \frac{2n}{\kappa - (n-3)/(n-1)},$$
where $\kappa$ is the kurtosis of the underlying distribution. In the case of a mesokurtic distribution (such as the normal distribution) you have $\kappa = 3$, which gives $DF_n = n-1$, which is the well-known distribution for the normal case. (You have accidentally squared this term in the equation in your question.) In the case of an underlying platykurtic (leptokurtic) distribution, the degrees-of-freedom is higher (lower) than in the normal case.