The best variance divisor when the mean is known

Question

Readings X1, \dots, Xn are i.i.d. Normal with a known mean \mu and unknown variance \sigma^2. Consider estimators \hat\sigma^2 = c\sumi (Xi - \mu)^2. Dividing by n (that is, c = 1/n) is unbiased. Which constant c minimizes the mean squared error?

Accepted Answer

Set up. Let Q = \sumi (Xi - \mu)^2. With the mean known, Q/\sigma^2 is a chi-square with n degrees of freedom, so \mathbb{E}[Q] = n\sigma^2, \qquad \operatorname{Var}(Q) = 2n\sigma^4. The mean squared error of cQ is 	ext{MSE}(cQ) = c^2\,2n\sigma^4 + \big(cn - 1\big)^2\sigma^4. Minimize over c. Drop the \sigma^4 and differentiate: \frac{d}{dc}\big[2n c^2 + (cn - 1)^2\big] = 4nc + 2n(cn - 1) = 0. Cancel 2n: 2c + (cn - 1) = 0, so c(n + 2) = 1 and c^\star = \frac{1}{n+2}. Compare with the known-mean case unbiased choice. Here the unbiased divisor is n (all n deviations are free because the mean …

The best variance divisor when the mean is known

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