How noisy is the sample variance?

Question

Let X1, \dots, Xn be i.i.d. \mathcal{N}(\mu, \sigma^2) and S^2 = \frac{1}{n-1}\sum (Xi - \bar{X})^2. Derive \operatorname{Var}(S^2). How does the answer change for non-normal data, and what does that imply for estimating the variance of financial returns?

Accepted Answer

The Gaussian case via chi-squared. For normal samples, the classical result is \frac{(n-1)S^2}{\sigma^2} \sim \chi^2{n-1}. A \chi^2k variable has mean k and variance 2k. Therefore \operatorname{Var}(S^2) = \left(\frac{\sigma^2}{n-1}\right)^2 \cdot 2(n-1) = \frac{2\sigma^4}{n-1}. So the standard error of S^2 is \sigma^2\sqrt{2/(n-1)}, with n = 100, about 14% relative error on the variance (and, via the delta method, roughly 7% on the volatility). Beyond normality. For general i.i.d. data with fourth central moment \mu4, the exact result is \operatorname{Var}(S^2) = \frac{\mu4}{n} - \frac{(n-3)…

How noisy is the sample variance?

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