Why Optimal Forecast Combinations Can Reduce Squared Error

Summary

The question concerns the Bates–Granger combination of two forecasts and asks why its optimal weight can produce expected squared error below the smaller of the two individual forecast variances. The answer treats the combined error as a quadratic function of the weight. Because the quadratic form is positive semidefinite, its critical point is a global minimum, which establishes the minimizing property when that point is admissible.

The answer adds a boundary case: if the unconstrained minimizing weight falls outside the interval from zero to one, selecting the forecast with the smaller variance gives the relevant constrained choice. The excerpt provides the core optimization intuition rather than a full derivation, and it does not spell out assumptions about unbiasedness, forecast error covariance, or how the variances are estimated. Those details matter when applying forecast combination results empirically.

Key ideas

• The squared error of a linear forecast combination is quadratic in its weight.
• A positive semidefinite quadratic has a global minimum at its critical point.
• If the optimal unconstrained weight is outside zero to one, the smaller-variance forecast can be chosen at the boundary.
• The excerpt gives an optimization argument but omits detailed modeling and estimation assumptions.

Tags

# Answer by user18764 (score 1, accepted)


# How to prove that the expected squared error associated with the optimal combination weight is smaller than the minimum of 2 forecast variances?












I am looking at linear combination of two forecasts (Bates and Granger, 1969). I would like to understand how to prove that the expected squared error associated with the optimal combination weight is smaller than the minimum of 2 forecast variances. I have come across it quite a number of times in literature and textbook. However, after giving it much thought, I am still unable to prove it.

Below I have attached the proof. I have successfully managed to prove up to step 7.28. I am just left the last line boxed in blue to understand. Thank you!

## Answer by user18764 (score 1, accepted)

https://quant.stackexchange.com/a/44936

The function under study is a quadratic form that is positive semidefinite. The critical point is therefore a global minimum. If the minimum does not lie between 0 and 1, then simply choose the smaller of the two variances.

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