Which divisor gives the lowest-error variance estimate?

Question

You have n = 10 independent draws from a normal distribution and estimate the variance \sigma^2 by dividing the sum of squared deviations 	ext{SS} = \sumi (Xi - \bar X)^2 by a constant. The usual choices are n-1 = 9 (unbiased), n = 10 (maximum likelihood), and n+1 = 11. Which divisor gives the smallest mean squared error?

Accepted Answer

The idea. Dividing by a larger number pulls the estimate down. That introduces a downward bias, but it also shrinks the estimate's variance. The MSE trades these off, and the sweet spot turns out to be a divisor even larger than n. The two ingredients. For normal data, the sum of squares satisfies \mathbb{E}[	ext{SS}] = (n-1)\sigma^2 and 	ext{Var}(	ext{SS}) = 2(n-1)\sigma^4. Writing the estimator as c \cdot 	ext{SS}, its MSE in units of \sigma^4 is 	ext{MSE}(c)/\sigma^4 = \underbrace{2(n-1)c^2}{	ext{variance}} + \underbrace{\big(c(n-1) - 1\big)^2}{	ext{squared bias}}. Evaluate the thre…

Which divisor gives the lowest-error variance estimate?

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