How much does naive averaging waste?

Question

Two independent unbiased estimates of the same quantity have very different precisions: $\sigma_1^2 = 1$ and $\sigma_2^2 = 9$ . A lazy analyst just averages them with equal weights, $\tfrac12 T_1 + \tfrac12 T_2$ .

How does the equal-weight average compare with the optimal inverse-variance blend? Report the relative efficiency of equal weighting.

Accepted Answer

Equal-weight variance. By independence, \operatorname{Var}\big(	frac12 T1 + 	frac12 T2\big) = 	frac14\sigma1^2 + 	frac14\sigma2^2 = 	frac14(1 + 9) = 2.5. Note this is worse than just using T1 alone (variance 1): naively averaging in a noisy estimate can actively hurt. Optimal-weight variance. The inverse-variance blend achieves \operatorname{Var}{\min} = \frac{\sigma1^2\sigma2^2}{\sigma1^2 + \sigma2^2} = \frac{1\cdot 9}{1 + 9} = 0.9, putting weight \frac{\sigma2^2}{\sigma1^2+\sigma2^2} = \frac{9}{10} = 0.9 on the precise estimate and only 0.1 on the noisy one. Relative efficiency of equa…

How much does naive averaging waste?

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