Estimating variance from a single pair of measurements

Question

You take just two independent readings X1, X2 from a process with unknown mean and variance \sigma^2. A colleague proposes estimating the variance with \hat\sigma^2 = 	frac{1}{2}(X1 - X2)^2. Is this an unbiased estimator of \sigma^2? Prove it, and connect it to the usual "divide by n-1" rule.

Accepted Answer

Compute the expectation directly. The difference of two independent readings has mean zero and variance \operatorname{Var}(X1 - X2) = \operatorname{Var}(X1) + \operatorname{Var}(X2) = 2\sigma^2, using independence (no covariance term) and the common variance. Since \mathbb{E}[X1 - X2] = 0, \mathbb{E}\big[(X1 - X2)^2\big] = \operatorname{Var}(X1 - X2) = 2\sigma^2. Therefore \mathbb{E}[\hat\sigma^2] = 	frac12 \cdot 2\sigma^2 = \sigma^2. So yes, it is unbiased. Connect it to n-1. With n = 2, the sample mean is \bar X = 	frac12(X1 + X2), and \sum{i=1}^{2}(Xi - \bar X)^2 = \Big(X1 - 	frac{X1+X2…

Estimating variance from a single pair of measurements

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