The best linear blend of two thermometers

Question

Two independent thermometers each give an unbiased reading of the same temperature. The good one has error variance $\sigma_1^2 = 0.25$ ; the cheap one has $\sigma_2^2 = 1$ . You form the blend $w T_1 + (1-w)T_2$ .

Find the weight $w$ on the good thermometer that minimizes the variance, report the minimum variance, and argue that any other weight is worse.

Accepted Answer

Set up. The blend is unbiased for any w, so we minimize V(w) = w^2\sigma1^2 + (1-w)^2\sigma2^2. Optimal weight. Inverse-variance weighting gives w^\star = \frac{\sigma2^2}{\sigma1^2 + \sigma2^2} = \frac{1}{0.25 + 1} = \frac{1}{1.25} = 0.8. So the good thermometer gets weight 0.8 and the cheap one 0.2. Minimum variance. V(w^\star) = \frac{\sigma1^2\sigma2^2}{\sigma1^2 + \sigma2^2} = \frac{0.25\cdot 1}{1.25} = 0.2. Why every other weight is worse. The function V(w) = w^2\sigma1^2 + (1-w)^2\sigma2^2 is a sum of squares, hence a parabola opening upward in w with a single minimum at w^\star. Its s…

The best linear blend of two thermometers

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