Avraham Adler’s solution to Chapter 6.2 Question 20 p276 appears to be correct.
I generalize his solution to the case where the variables \(X_1\) and \(X_2\) are correlated with \(\rho \neq 0\).
For notational simplicity, we denote \[ A = \sigma_1^2 \text{ and } B = \sigma_2^2 \text{ and } C = \rho\] This yields the equation for the variance assuming the mean is zero without loss of generality.
\[ Var(\mu) = w^2 A + (1-w)^2 B + 2w(1-w)\sqrt{AB}C\]
Calculating the first derivative of the variance of \(\mu\) with respect to \(w\) and setting it equal to zero, we get:
\[ \frac{ \partial Var(\mu)}{\partial w} = 2wA - 2B + 2Bw + (2-4w)\sqrt{AB}C = 0\] Solving the equation for \(w\), we get:
\[ w = \frac{ B - \sqrt{AB}C}{A + B - 2\sqrt{AB}C} = \frac{ \sigma_2^2 - \sigma_1\sigma_2\rho}{\sigma_1^2 + \sigma_2^2 - 2\sigma_1\sigma_2\rho}\] Note that if the instruments are perfectly negatively correlated to each other, then their respective measurements would average to the exact answer. They could give polar opposite readings centered around the true mean. In this case, the statistical average of their combined measurement is alway the treu mean. In this case, the equation breaks down but the weight would clearly be \(w=0.5\).