Optimal blend versus a naive equal-weight average

Question

Two independent forecasts of the same quantity. Forecast A is unbiased with variance $\sigma_A^2 = 12$ . Forecast B is off by $b = 2$ on average with variance $\sigma_B^2 = 4$ . You blend them as $w\,A + (1-w)\,B$ .

What weight $w$ on forecast A minimizes the mean squared error, and how does its MSE compare with a naive 50-50 average?

Accepted Answer

The idea. Weight each forecast inversely to its total error. Then compare that optimal blend against the lazy choice of weighting both equally. Effective error of each. - Forecast A: \sigmaA^2 = 12. - Forecast B: \beta = \sigmaB^2 + b^2 = 4 + 4 = 8. Optimal weight and its MSE. wA = \frac{\beta}{\sigmaA^2 + \beta} = \frac{8}{12 + 8} = \frac{8}{20} = 0.4, \qquad 	ext{MSE}^\star = \frac{12 \cdot 8}{20} = 4.8. The naive 50-50 average. With w = 0.5, the blend's variance is 0.5^2(12) + 0.5^2(4) = 3 + 1 = 4, and its bias is 0.5 	imes 2 = 1, so its squared bias is 1. Total MSE = 4 + 1 = 5. Compare.…

Optimal blend versus a naive equal-weight average

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