How much does dividing volatility by n understate it?

Question

A trader estimates the variance of daily returns from $n$ observations using the "natural" formula that divides by the sample size:

$\hat\sigma^2 = \frac{1}{n}\sum_i (X_i - \bar X)^2.$

By what factor is $\hat\sigma^2$ biased, that is, what is $\mathbb{E}[\hat\sigma^2]$ relative to the true $\sigma^2$ , and how would you correct it?

Accepted Answer

Use the known sum of squares. For an i.i.d. sample, the expected total squared deviation around the sample mean is \mathbb{E}\!\left[\sumi (Xi - \bar X)^2ight] = (n-1)\sigma^2, because centering on the sample mean rather than the true mean costs one degree of freedom. Dividing by n gives \mathbb{E}[\hat\sigma^2] = \frac{n-1}{n}\,\sigma^2 = \left(1 - 	frac{1}{n}ight)\sigma^2. Reading the bias. The estimate is biased low by the factor \frac{n-1}{n}, equivalently it understates the variance by \sigma^2/n. With n = 20 it is about 5\% too small; with n = 250 (a year of daily returns) only abo…

How much does dividing volatility by n understate it?

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