How much does n+1 beat n-1 on variance-estimate error?

Question

You estimate a normal variance from n = 15 independent draws. Compare the unbiased estimator (divisor n-1 = 14) with the MSE-optimal estimator (divisor n+1 = 16). By what percentage is the n+1 estimator's mean squared error smaller than the unbiased one's?

Accepted Answer

The idea. Both divide the same sum of squares by a constant; the larger divisor swaps a touch of bias for less variance. Compute each MSE (in \sigma^4 units) and compare. Unbiased (divisor 14). Bias is zero, so 	ext{MSE}{n-1}/\sigma^4 = \frac{2}{n-1} = \frac{2}{14} \approx 0.1429. Optimal (divisor 16). With c = 1/16 and m = n-1 = 14: 	ext{MSE}{n+1}/\sigma^4 = 2(14)\left(	frac{1}{16}ight)^2 + \left(	frac{14}{16} - 1ight)^2 = \frac{28}{256} + \left(	frac{2}{16}ight)^2 = 0.1094 + 0.0156 = 0.125. Compare. 1 - \frac{0.125}{0.1429} = 1 - 0.875 = 0.125, \quad 	ext{a 12.5 percent reductio…

How much does n+1 beat n-1 on variance-estimate error?

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