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!

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1 Answer 1


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.

  • $\begingroup$ Hi: I think what user`8764 is saying is that you can take the derivative of the function, solve for the minimum ( k*) and be sure that it's not a local minimum because the second derivative of the function is positive and he function is quadratic ( the two imply that function is convex ). $\endgroup$
    – mark leeds
    Apr 6, 2019 at 14:40
  • $\begingroup$ and, I left out his main point which is that, if k* doesn't lie between zero and one when solved for, then find the values at k = 0 and k = 1 and take the minimum. $\endgroup$
    – mark leeds
    Apr 6, 2019 at 14:41
  • $\begingroup$ It's implied that he knows how to minimize the function given that he's derived up to the inequality. He just needs to remember that a quadratic funcrion has a single optima, which is the crux of his question. $\endgroup$
    – user18764
    Apr 6, 2019 at 21:23
  • $\begingroup$ no problem. I was just trying to clarify. your answer was great. $\endgroup$
    – mark leeds
    Apr 8, 2019 at 0:39
  • $\begingroup$ @user18764 you have been of great help! I really appreciate it (: $\endgroup$
    – fauxpas
    Apr 8, 2019 at 16:46

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