Standard error of the sample standard deviation

Question

For a normal sample of size n, the sample variance s^2 estimates \sigma^2 with (asymptotic) variance \operatorname{Var}(s^2) \approx 2\sigma^4 / n. But you report the standard deviation s = \sqrt{s^2}, a volatility. Use the delta method to find the approximate variance and standard error of s.

Accepted Answer

Treat s = g(s^2) with g(v) = \sqrt v, a smooth transform of the estimator s^2. The delta method gives \operatorname{Var}(s) \approx \big[g'(\sigma^2)\big]^2 \operatorname{Var}(s^2). Here g'(v) = \dfrac{1}{2\sqrt v}, so at v = \sigma^2 the slope is \dfrac{1}{2\sigma}. Plugging in, \operatorname{Var}(s) \approx \left(\frac{1}{2\sigma}\right)^2 \cdot \frac{2\sigma^4}{n} = \frac{1}{4\sigma^2}\cdot\frac{2\sigma^4}{n} = \frac{\sigma^2}{2n}. The standard error is \operatorname{SE}(s) \approx \frac{\sigma}{\sqrt{2n}}. A neat consequence: the relative error of the standard deviation, \operatorname{SE}…

Standard error of the sample standard deviation

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