Back out the signal from a chosen shrinkage weight

Question

A colleague shrinks a noisy estimate toward an anchor using weight $w = 0.6$ on the raw data, and they claim this is MSE-optimal. The raw estimate has variance $\sigma^2 = 10$ .

What squared distance $d^2$ from the anchor does their choice imply?

Accepted Answer

The idea. The optimal weight is a fixed function of the signal (squared distance d^2) and the noise (\sigma^2). If the weight and the noise are known, the implied signal is pinned down. Invert the formula. The optimal weight satisfies w^\star = \frac{d^2}{d^2 + \sigma^2}. Solve for d^2: w^\star(d^2 + \sigma^2) = d^2 \implies d^2(1 - w^\star) = w^\star \sigma^2 \implies d^2 = \frac{w^\star \sigma^2}{1 - w^\star}. Plug in. d^2 = \frac{0.6 \cdot 10}{1 - 0.6} = \frac{6}{0.4} = 15. So a weight of 0.6 is only justified if the true value sits a squared distance of 15 from the anchor (about \sqrt{15}…

Back out the signal from a chosen shrinkage weight

Your answer

Discussion

Learn the concepts

Related questions