= Solution
Independence makes the <variance of an estimator> equal to
$$
\operatorname{Var}(\widehat\mu)=\alpha_1^2\sigma_1^2+\alpha_2^2\sigma_2^2.
$$
With $\alpha_2=1-\alpha_1$, differentiate to obtain $2\alpha_1\sigma_1^2-2(1-\alpha_1)\sigma_2^2=0$. The second derivative is $2(\sigma_1^2+\sigma_2^2)>0$, so the unique minimum is
$$
\boxed{\alpha_1=\frac{\sigma_2^2}{\sigma_1^2+\sigma_2^2},\qquad\alpha_2=\frac{\sigma_1^2}{\sigma_1^2+\sigma_2^2}}.
$$
Equivalently, the <inverse-variance weighted mean> uses $\alpha_i=K\sigma_i^\gamma$ with
$$
\boxed{\gamma=-2,\qquad K=(\sigma_1^{-2}+\sigma_2^{-2})^{-1},\qquad\operatorname{Var}(\widehat\mu)=\frac{\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2}}.
$$
We assume positive measurement <variances>. A zero-variance observation already supplies the true value exactly.
Back to article page
