For n = 10 iid N(mu, sigma^2) draws with both parameters unknown, the maximum likelihood estimator of sigma^2 divides the sum of squared deviations from the sample mean by n. Compute its expectation as a multiple of sigma^2, and explain what the divisor n - 1 corrects.

For n = 10 iid N(mu, sigma^2) draws with both parameters unknown, the maximum likelihood estimator of sigma^2 divides the sum of squared deviations from the sample mean by n. Compute its expectation as a multiple of sigma^2, and explain what the divisor n - 1 corrects.

Approach: Use the distribution of the sum of squared deviations about the sample mean, which is sigma^2 times a chi-square with n - 1 degrees of freedom.

0.9*sigma^2. The sum of squared deviations about the sample mean is sigma^2 times a chi-square variable with n - 1 degrees of freedom, so its expectation is (n - 1)*sigma^2 and dividing by n gives (n - 1)/n * sigma^2 = 9/10 * sigma^2. The maximum likelihood estimator therefore understates the variance by 10% at n = 10. The bias vanishes as n grows. One degree of freedom is spent estimating mu, so the deviations are constrained to sum to zero and carry only n - 1 free coordinates; dividing the same sum by n - 1 gives the unbiased estimator s^2. Unbiasedness is not the only criterion here, since the mean squared error of the biased version is smaller, the variance saved outweighing the squared bias.

Follow-up: Which divisor minimises the mean squared error of the estimator of sigma^2 under normality, and why is it neither n nor n - 1?

Key concepts: maximum likelihood, bias, degrees of freedom, mean squared error.