We have two instruments that measure the distance between two points. The measurements given by the two instruments are random variables \(X_1\) and \(X_2\) that are independent with \(E(X_1) = E(X_2 ) = \mu\), where \(\mu\) is the true distance. From experience with these instruments, we know the values of the variances \(\sigma_1^2\) and \(\sigma_2^2\). These variances are not necessarily the same. From two measurements, we estimate \(\mu\) by the weighted average \(\bar{\mu} = wX_1 +(1−w)X_2\). Here \(w\) is chosen in \([0,1]\) to minimize the variance of \(\bar{\mu}\).
The expectation of \(\bar{\mu}\) is \(\mu\). As both \(E(X_1)\) and \(E(X_2)\) equal \(\mu\), any linear combination of them where the components are \(w\) and \(1-w\) must also equal \(\mu\). \[ \forall_w: w\mu + (1 - w)\mu = w\mu + \mu - w\mu = \mu \]
To minimize the variance of a function, one usually finds the point at which the function’s first derivative equals 0. Here, the question is to minimize \(\textrm{Var}(\bar{\mu})\): \[ \begin{aligned} f(w) &= \textrm{Var}(\bar{\mu}) = \textrm{Var}\left(wX_1 + (1-w)X_2\right)\\ &= w^2 \textrm{Var}(X_1) + (1-w)^2\textrm{Var}(X_2)\\ &= w^2\sigma_1^2 + \left(1 - 2w + w^2\right)\sigma_2^2\\ &= w^2\sigma_1^2 + \sigma_2^2 - 2w\sigma_2^2 + w^2\sigma_2^2\\ \frac{df}{dw} &= 2w\sigma_1^2 - 2\sigma_2^2 + 2w\sigma_2^2\Rightarrow0\\ 0 &= w\sigma_1^2 - \sigma_2^2 + w\sigma_2^2\\ w &= \mathbf{\frac{\sigma_2^2}{\sigma_1^2+\sigma_2^2}} \end{aligned} \]
The weight given to each variable is in proportion to its contribution to the variance. Precisely, the weight given to \(X_1\) is the proprtion of \(X_2\)’s variance of the total, and vice versa.