Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 219 / 1 / ii / Solution

Independence makes the variance of an estimator equal toWith ${\alpha }_2=1-{\alpha }_1$, 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 isEquivalently, the inverse-variance weighted mean uses ${\alpha }_i=K{\sigma }_i^{\gamma }$ withWe assume positive measurement variances. A zero-variance observation already supplies the true value exactly.