Minimizing the Variance of a Weighted Average

It is common practice in science to take a weighted average of estimators of a single parameter. If the original estimators are unbiased, any weighted average will be an unbiased estimator as well. The best estimator among the weighted averages can be obtained by choosing weights that minimize the variance of the weighted average. If the variances of the individual estimators are given, the ideal weights have long been known to be the inverse of the variance. Nonetheless, I have not found a formal proof of this result in the literature. In this article, I provide three different proofs of the ideal weights.


Introduction
Oftentimes in science, multiple point estimators of the same parameter are combined to form a better estimator.One method of forming the new estimator is taking a weighted average of the original estimators.If the original estimators are unbiased, the weighted average is guaranteed to be unbiased as well.
A weighted average may be used to combine the results of several studies (metaanalysis), or when several estimates are obtained within a study.For example, to deconfound it may be necessary to stratify on a covariate when estimating an effect.Assuming that the effect is the same in every stratum of the covariate, we may take a weighted average of the stratum-specific estimates.
The question remains though as to which weights should be chosen.Since the estimator will be unbiased regardless of the weights, we only need to consider the variance.In particular, the weights should be chosen to minimize the variance of the weighted average.Although it has long been known that the ideal weights should be the inverse of the variance, I have not found any complete, formal proof in the literature.Several sources mention the ideal weights either in D. J. Shahar general or for specific cases without any proof [1]- [6].Others mention something similar to the ideal weights but again without proof [7] [8].Hedges offers a socalled proof in his 1981 paper [9], which is far from a complete proof.The first two sentences of his proof contain the same content as the first two sentences and the last sentence of proof 1 in this paper.Hedges then continues to prove approximations for the ideal weights under a certain condition.In his 1982 and 1983 papers, he writes that this result is "easy to show" referencing his 1981 (and 1982) papers [10] [11].Goldberg and Kercheval [12] provide a "proof" that contains only slightly more content than Hedges' proof in that they mention the use of Lagrange multipliers.Proof 1 in this paper goes over the details thoroughly.Cochran also mentions the ideal weights, but proves only that these weights give the maximum likelihood estimate when the estimators are independent and normally distributed about a common mean [13].Lastly, the problem of finding the ideal weights is included as an exercise (exercise 7.42) in Casella and Berger [14].The problem, however, is not worked out in their solution manual [15].A version of the problem when taking a weighted average of only two estimators is also an exercise (exercise 24.1) in Anderson and Bancroft [1].
Searching through articles dating back to the 1930s, it seems that this basic result has not been formally proven in the literature.One reason may be that the result depends on the variances of the estimators being known.The case when the variances are unknown is more difficult and attracted more attention.For example, a few articles briefly mentioned the ideal weights when the variances are known before continuing to discuss the case when the variances are unknown [2] [4] [5] [13].In this paper, I present three proofs of the ideal weights that minimize the variance of a weighted average.

Three Proofs
be estimators of a single parameter θ .In practice, the i X are independent, and they are often assumed to be unbiased.If that's the case, then any weighted average is an unbiased estimator of θ .Since the estimator is unbiased regardless of the weights, we want to choose weights that minimize the variance of X .
As far as the ideal weights are concerned, it is not necessary though that the i X be independent and unbiased.The proof of the ideal weights only requires that the i X be uncorrelated and have a finite non-zero variance.
 are uncorrelated random variables with finite non-zero variances, then ( ) and its minimum value is ( ) The first proof uses the method of Lagrange multipliers.
Proof 1: ( ) , because the i X are uncorrelated.We wish to minimize the previous expression under the constraint that , , , Var 1 By the method of Lagrange multipliers, the values of 1 , , n w w ( ) ( ) Therefore, ( ) ( ) Using the method of Lagrange multipliers again, the critical points of ( ) ( ) Var Var (These contain all the extrema of ( ) Var For these values of i w , That is, of all critical points, the one in the interior of T minimizes ( ) Var Var Var and its minimum value is ( ) The second proof is done by induction.
Proof 2: The case 2 n = will be our base case for the induction.
The global minimum of the above expression occurs when and For the induction step, suppose that for some 2 n ≥ , ( ) Var and its minimum value is ( ) ) where are weights that do not depend on 1 n w + .So for any possible values of i u , the above expression is minimized when ( ) By plugging the above value for 1 n w + into Equation (13), we find a lower bound for the variance of the weighted average: where equality is achieved for the specified value of The variance of the weighted average will be minimized when it equals the right side of the above inequality and the right side of the inequality is minimized.The absence in the literature; there is value in a proof beyond the result it proves.For example, it is interesting that the proposition can be proven by induction and more succinctly using the Cauchy-Schwarz inequality.
Even more surprising are the trails of citations leading nowhere.It appears that generations of statisticians simply assumed that a proof has been published somewhere, relying on old, inaccurate citations.In that sense, this article not only offers three different proofs but also a broader lesson: every so often it is worthwhile to review the history of well-known facts.Surprises are possible.

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for all i .The set T of all ( ) all i is closed and bounded.The extrema in the interior of T can be found by considering only the first constraint, which may be written as Later we shall find the extrema on the boundary of T .To find the extrema in the interior of T , let

( 5 )
Next, let's find the extrema on the boundary of T .The boundary of T is characterized by having some of the i w equal zero.For any point on the boun-