How much does optimal shrinkage cut the error?

Question

You shrink a noisy estimate toward an anchor with the MSE-optimal weight. The raw estimate has variance $\sigma^2 = 12$ , and the true squared distance from the anchor is $d^2 = 4$ .

By what percentage does optimal shrinkage reduce the mean squared error compared with using the raw estimate?

Accepted Answer

The idea. The raw estimate has MSE equal to its variance \sigma^2 (it is unbiased). At the optimal shrinkage weight, the MSE drops to a specific smaller value. Compare the two. Optimal MSE. At the best weight, the mean squared error collapses to a tidy form: 	ext{MSE}^\star = \frac{\sigma^2 d^2}{\sigma^2 + d^2} = \frac{12 \cdot 4}{12 + 4} = \frac{48}{16} = 3. Compare. The raw estimate's MSE is \sigma^2 = 12. The reduction is 1 - \frac{3}{12} = 1 - 0.25 = 0.75, \quad 	ext{a 75 percent cut.} The reason the payoff is so large: the anchor is close to the truth (d^2 = 4 is small) while the raw n…

How much does optimal shrinkage cut the error?

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