How to Use Standard Deviations in Weighting Averages?

We wish to calculate a weighted average of a set of sample averages, given their standard deviations. How do we do that?

The objective is to find a weighting factor, alpha, that minimizes the variance of the weighted average, namely (for two averages):

Minimum { Variance[ (α)Average1 + (1-α)Average2 ] }

We first calculate the variance to obtain (Var is short for Variance; samples for averages assumed independent):

Variance[ (α)Average1 + (1-α)Average2 ] =

=  α² Var(Average1) + (1-α)² Var(Average2) .

Differentiating with respect to alpha and equating to zero, we obtain:

(2α)Var(Average 1) – 2(1-α)Var(Average 2) = 0, and the optimal alpha is:

α* = var(Average 2) / [ var(Average1) + var(Average2) ] ,

where: var(Average)= variance/n, with n a sample size.

We may wish to adapt this reply to specific needs. For example, for three averages we have:

Variance[ (α₁)Average1 + (α₂)Average2 + (1-α₁-α₂)Average3 ] = 

= α₁²Var(Average1) + α₂²Var(Average2) + (1-α₁-α₂)² Var(Average3)

To minimize this expression, we differentiate twice, with respect to α₁ and to α₂. Equating to zero we obtain two linear equations in two unknowns that may be easily identified:

(2α₁)Var(Average1) – 2(1-α₁-α₂)Var(Average3) = 0,

(2α₂)Var(Average2) – 2(1-α₁-α₂)Var(Average3) = 0,

or:

α₁= v₃ / [v₁ + v₃ + (v₁v₃)/v₂]

α₂= v₃ / [v₂ + v₃ + (v₂v₃)/v₁]

where v_i is Var(Average i) (i=1,2,3).

Since “in general, a system with the same number of equations and unknowns has a single unique solution” (Wikipedia, “System of linear equations”), extension to a higher number of averages (m>3), is straightforward, requiring solving a system of m-1 linear equations with m-1 unknowns.

 (This post appears also on my personal page at ResearchGate)