Twenty analysts' annual alphas have a cross-sectional mean of 1.0% and a cross-sectional standard deviation of 3.0%, and each alpha is measured with a standard error of 2.5%. Estimate the shrinkage factor by empirical Bayes and give the adjusted alpha for an analyst reporting 7.0%.

Twenty analysts' annual alphas have a cross-sectional mean of 1.0% and a cross-sectional standard deviation of 3.0%, and each alpha is measured with a standard error of 2.5%. Estimate the shrinkage factor by empirical Bayes and give the adjusted alpha for an analyst reporting 7.0%.

Approach: Split the observed cross-sectional variance into true dispersion plus measurement variance to recover the prior, then apply the normal-normal weight to each analyst.

2.83%. The observed spread contains true dispersion and measurement noise together, so Var(observed) = tau^2 + s^2 gives tau^2 = 3.0^2 - 2.5^2 = 9.00 - 6.25 = 2.75, a true cross-sectional standard deviation of 1.66%. The posterior mean shrinks each analyst toward the grand mean of 1.0% with weight tau^2/(tau^2 + s^2) = 2.75/9.00 = 0.306 on their own number, so 7.0% is adjusted to 1.0 + 0.306*(7.0 - 1.0) = 2.83%. Nearly 70% of the apparent outperformance is noise, and the same calculation lifts an analyst reporting -3.0% up to -0.22%. Estimating the prior from the same data it is applied to is what makes this empirical Bayes. James and Stein showed that shrinking three or more means toward a common centre strictly dominates the raw estimates in total squared error, so this is the better rule even judged by frequentist risk rather than by any belief in the prior.

Follow-up: The observed cross-sectional standard deviation comes in at 2.0% instead of 3.0%. What does the formula give for tau^2, and what do you report?

Key concepts: empirical Bayes, James-Stein, shrinkage factor, measurement noise.