Minimum-variance two-cue estimate ($w_1={\sigma }_2^2/({\sigma }_1^2+{\sigma }_2^2)$)

Let unbiased independent estimates $X_1,X_2$ have variances $a,b>0$. The variance of $w{X}_1+(1-w)X_2$ is $w^{2}a+(1-w{)}^{2}b$. Its derivative is $2w(a+b)-2b$ and its second derivative is $2(a+b)>0$, proving the unique optimum $w=b/(a+b)$ and variance $ab/(a+b)$. Independent Gaussian likelihoods give the same estimate by differentiating $(x_1-x{)}^2/a+(x_2-x{)}^2/b$. With covariance $c$, the variance becomes $w^{2}a+(1-w{)}^{2}b+2w(1-w)c$, and the unconstrained optimum is $(b-c)/(a+b-2c)$ when the denominator is positive; it need not lie between zero and one.