When a biased estimator wins on MSE

Question

Mean squared error decomposes as \operatorname{MSE} = \operatorname{Bias}^2 + \operatorname{Variance}, so unbiasedness (\operatorname{Bias} = 0) is only one term. For a normal sample, show that dividing the sum of squared deviations by n+1 gives a lower MSE for \sigma^2 than the unbiased divisor n-1.

Accepted Answer

Let S\!S = \sum{i=1}^n (Xi - \bar X)^2. For a normal sample E[S\!S] = (n-1)\sigma^2 and \operatorname{Var}(S\!S) = 2(n-1)\sigma^4 (since S\!S/\sigma^2 \sim \chi^2{n-1}). For the estimator \hat\sigma^2 = c\,S\!S, \operatorname{MSE}(c) = \underbrace{\big(c(n-1)-1\big)^2\sigma^4}{	ext{bias}^2} + \underbrace{2(n-1)c^2\sigma^4}{	ext{variance}}. Differentiate the bracket and set to zero: 2(c(n-1)-1)(n-1) + 4(n-1)c = 0 \Rightarrow c(n-1) - 1 + 2c = 0 \Rightarrow c = \frac{1}{n+1}. Plugging back, the MSE-optimal divisor n+1 gives \operatorname{MSE} = \frac{2\sigma^4}{n+1}, while the unbiased diviso…

When a biased estimator wins on MSE

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