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 ] = **

**= α ^{2 }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[ (α _{1})Average1 + (α_{2})Average2 + (1-α_{1}-α_{2})Average3 ] = **

**= α _{1}^{2}Var(Average1) + α_{2}^{2}Var(Average2) + (1-α_{1}-α_{2})^{2} Var(Average3)**

To minimize this expression, we differentiate twice, with respect to α_{1} and to α_{2}. Equating to zero we obtain two **linear** equations in two unknowns that may be easily identified:

(2α_{1})Var(Average1) – 2(1-α_{1}-α_{2})Var(Average3) = 0,

(2α_{2})Var(Average2) – 2(1-α_{1}-α_{2})Var(Average3) = 0,

or:

**α _{1}= v_{3} / [v_{1} + v_{3} + (v_{1}v_{3})/v_{2}]**

**α**_{2}**= v _{3} / [v_{2} + v_{3} + (v_{2}v_{3})/v_{1}]**

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)