The error of the best biased-unbiased blend

Question

Two independent forecasts of the same quantity. Forecast A is unbiased with variance $\sigma_A^2 = 6$ . Forecast B is off by $b = 2$ on average with variance $\sigma_B^2 = 2$ . You blend them optimally.

What is the mean squared error of the optimally weighted blend?

Accepted Answer

The idea. Judge each forecast by its total error, weight inversely to that, and the resulting MSE takes the same "product over sum" form as inverse-variance weighting, but with B's effective error in place of its variance. Effective error of each. - Forecast A: \sigmaA^2 = 6. - Forecast B: \beta = \sigmaB^2 + b^2 = 2 + 4 = 6. Optimal blended MSE. Just like combining two independent errors, the best blend has MSE equal to their product over their sum: 	ext{MSE}^\star = \frac{\sigmaA^2 \cdot \beta}{\sigmaA^2 + \beta} = \frac{6 \cdot 6}{6 + 6} = \frac{36}{12} = 3. Since the two forecasts have e…

The error of the best biased-unbiased blend

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