Equal-precision estimates deserve equal weight

Question

Two independent analysts each give an unbiased forecast of next quarter's earnings, and both forecasts happen to have the same variance, 9. You blend them as $w \cdot A + (1-w)\cdot B$ .

What weight $w$ minimizes the variance of the blend, and what variance results?

Accepted Answer

The idea. Both forecasts are centered on the truth, so any blend stays unbiased. Since neither is more precise than the other, there is no reason to favor one. Minimize. The inverse-variance rule gives w = \frac{	ext{Var}B}{	ext{Var}A + 	ext{Var}B} = \frac{9}{9 + 9} = 0.5, so you simply average them. The payoff. The blended variance is \frac{9 \cdot 9}{18} = 4.5, exactly half of each input's variance. This is the familiar "averaging two independent noisy numbers cuts the variance in half" result, and it is the same reason averaging n independent equal-variance estimates divides the varianc…

Equal-precision estimates deserve equal weight

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