Your variance estimate is unbiased, is your volatility estimate?

Question

You use the unbiased sample variance s^2 (dividing by n-1), so \mathbb{E}[s^2] = \sigma^2. A trader takes s = \sqrt{s^2} as the volatility estimate. Is s an unbiased estimate of \sigma? If not, which way is it biased and by how much?

Accepted Answer

Direction, from Jensen. The square root is strictly concave. Jensen's inequality says for concave g, \mathbb{E}[g(X)] \le g(\mathbb{E}[X]), with strict inequality unless X is constant. So \mathbb{E}[s] = \mathbb{E}\big[\sqrt{s^2}\big] \ <\ \sqrt{\mathbb{E}[s^2]} = \sqrt{\sigma^2} = \sigma. The sample volatility is biased downward: even with a perfectly unbiased variance, its square root systematically underestimates \sigma. Unbiasedness is not preserved by nonlinear transforms. Size, from the delta method (second order). Expand g(s^2)=\sqrt{s^2} to second order around \sigma^2: \mathbb{E}[g(s…

Your variance estimate is unbiased, is your volatility estimate?

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