Blending three noisy fair-value estimates

Question

Three independent, unbiased estimates of an asset's fair value arrive. Their standard deviations are $6$ , $3$ , and $2$ , and the point estimates are $X_1 = 50$ , $X_2 = 53$ , $X_3 = 51$ .

Combine all three into a single best estimate, and state how good the blend is.

Accepted Answer

For independent unbiased estimates, the minimum-variance combination weights each by its precision 1/\sigmai^2: \hat\mu = \frac{\sumi Xi/\sigmai^2}{\sumi 1/\sigmai^2}. The variances are \sigma1^2 = 36,\ \sigma2^2 = 9,\ \sigma3^2 = 4, so the precisions are \frac{1}{36}, \qquad \frac{1}{9} = \frac{4}{36}, \qquad \frac{1}{4} = \frac{9}{36}. They are in the ratio 1 : 4 : 9, summing to \frac{14}{36} = \frac{7}{18}. So the weights are w1 = 	frac{1}{14}, \qquad w2 = 	frac{4}{14}, \qquad w3 = 	frac{9}{14}. The sharpest read (sd 2) gets nine times the weight of the noisiest (sd 6). Plugging in the …

Blending three noisy fair-value estimates

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