James–Stein, why shrinking beats the obvious estimator

Question

You must estimate the true mean returns of $p$ different, unrelated strategies from one noisy observation each, $x_i \sim N(\mu_i, \sigma^2)$ . The obvious estimator is $\hat\mu_i = x_i$ .

Explain the surprising result that, for $p \ge 3$ , a shrinkage estimator has uniformly lower total mean squared error, and why this is so counterintuitive.

Accepted Answer

The obvious estimator is inadmissible. Using each observation as its own estimate seems unimprovable, each xi is unbiased and efficient for its own \mui. Yet Charles Stein showed that when you judge the estimates jointly by total squared error \sumi (\hat\mui - \mui)^2, and p \ge 3, the vector x is beaten everywhere by the James–Stein estimator: \hat\mu^{	ext{JS}} = \left(1 - \frac{(p-2)\sigma^2}{\x\^2}ight) x, which pulls all coordinates toward zero (or toward any fixed point, or the grand mean) by a data-driven factor. Why it works. Each shrunk coordinate is slightly biased, but shrinkag…

James–Stein, why shrinking beats the obvious estimator

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