Blending two estimates by inverse variance

Question

You have two independent, unbiased estimates $\hat{\theta}_1$ and $\hat{\theta}_2$ of the same quantity $\theta$ , with variances $v_1$ and $v_2$ .

What convex weighting $\hat{\theta} = w\hat{\theta}_1 + (1-w)\hat{\theta}_2$ minimizes the variance, and what is the resulting variance?

Accepted Answer

Any weight is unbiased. Since both pieces are unbiased, \mathbb{E}[w\hat{	heta}1 + (1-w)\hat{	heta}2] = 	heta for every w, so there is no bias to trade off: minimize variance directly. Minimize the variance. By independence: \operatorname{Var}(\hat{	heta}) = w^2 v1 + (1-w)^2 v2. Differentiate and set to zero: 2w v1 - 2(1-w) v2 = 0 \;\Rightarrow\; w(v1 + v2) = v2 \;\Rightarrow\; w^ = \frac{v2}{v1 + v2}. So each estimate gets weight proportional to the other's variance, i.e. proportional to its own precision 1/v. Plugging back in: \operatorname{Var}(\hat{	heta}^) = \frac{v1 v2}{v1 + v2}, \…

Blending two estimates by inverse variance

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