The error of the unbiased variance estimator

Question

You estimate a normal variance \sigma^2 from n = 4 independent draws using the standard unbiased estimator, S^2 = \frac{1}{n-1}\sumi (Xi - \bar X)^2. What is its mean squared error, in units of \sigma^4?

Accepted Answer

The idea. Unbiased does not mean error-free. With no bias, all of this estimator's MSE comes from its variance, and that variance is far from zero when n is small. Compute. For the unbiased estimator, c = 1/(n-1), and its MSE has only the variance term: 	ext{MSE}/\sigma^4 = 2(n-1)c^2 = 2(n-1)\cdot\frac{1}{(n-1)^2} = \frac{2}{n-1}. For n = 4: 	ext{MSE}/\sigma^4 = \frac{2}{3} \approx 0.667. That is a large error: the estimate's standard deviation is about \sqrt{2/3} \approx 0.82 times \sigma^2 itself, so a single small sample can easily land 50 percent off. This is exactly why the biased n+1 …

The error of the unbiased variance estimator

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