Is the sample standard deviation unbiased for sigma?

Question

The sample variance S^2 = \frac{1}{n-1}\sumi (Xi - \bar X)^2 is constructed to be unbiased for \sigma^2. A colleague takes its square root, S = \sqrt{S^2}, and reports it as an unbiased estimate of the standard deviation \sigma. Is the sample standard deviation S unbiased for \sigma? If not, which way is it biased?

Accepted Answer

Unbiasedness of S^2 does not carry to S. We know \mathbb{E}[S^2] = \sigma^2. But the square root is nonlinear, so \mathbb{E}[\sqrt{S^2}] 
e \sqrt{\mathbb{E}[S^2]} in general. Jensen settles the direction. The square root g(x) = \sqrt{x} is strictly concave for x 0. For a concave function Jensen's inequality flips: \mathbb{E}[S] = \mathbb{E}\big[\sqrt{S^2}\big] < \sqrt{\mathbb{E}[S^2]} = \sqrt{\sigma^2} = \sigma. So S is biased low: on average it underestimates the true spread. Correcting the variance divisor to make S^2 unbiased does not make S unbiased. The exact correction (normal case). F…

Is the sample standard deviation unbiased for sigma?

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