Pooling three unbiased estimates of the same quantity

Question

Three independent instruments each give an unbiased measurement of the same length, with variances $\sigma_1^2 = 1$ , $\sigma_2^2 = 2$ , and $\sigma_3^2 = 4$ . You combine them as $w_1 T_1 + w_2 T_2 + w_3 T_3$ with weights summing to $1$ (so the blend stays unbiased).

Find the minimum-variance weights and the resulting combined variance.

Accepted Answer

The optimal rule. For independent unbiased estimates, the minimum-variance weighted average uses weights proportional to the inverse variances (the precisions): wi \propto \frac{1}{\sigmai^2}. Compute the precisions. They are \frac11 = 1, \frac12, and \frac14. Their sum is 1 + 	frac12 + 	frac14 = 	frac74. Normalizing so the weights sum to 1, w1 = \frac{1}{7/4} = \frac47, \quad w2 = \frac{1/2}{7/4} = \frac27, \quad w3 = \frac{1/4}{7/4} = \frac17. So the sharpest instrument gets four times the weight of the coarsest, in the ratio 4 : 2 : 1. The combined variance. For inverse-variance weighti…

Pooling three unbiased estimates of the same quantity

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