Shrinkage Estimates for Selected Treatments

Why Shrink a Selected Estimate?

When several treatment arms are compared, the arm with the largest estimated effect is often selected for further attention. Its estimate is also likely to be unusually high because it combines its true effect with sampling error. Reporting the winning estimate unchanged can therefore exaggerate the selected arm's effect. Shrinkage addresses this by moving the estimate toward a common mean. The amount of movement matters: too little may leave substantial winner's bias, while too much may pull a genuinely exceptional treatment toward an unrepresentative average.

The earlier material on selection bias and winner's curse explains why selecting the largest estimate changes its sampling behavior. Here the focus is a particular estimator and its bias and mean squared error (MSE). These properties depend on the model, the target, and whether performance is assessed for fixed treatment effects or averaged over a distribution of effects.

A Model and a Shrinkage Estimate

Definition: Suppose the estimated effect for arm \(i\) is \(Y_i\), and its true effect is \(\theta_i\). Under a normal hierarchical model, \(Y_i\mid\theta_i\sim N(\theta_i,\sigma_i^2)\), and the arm effects are independently distributed as \(\theta_i\sim N(\mu,\tau^2)\). Here \(\mu\) is the common mean, \(\tau^2\) describes between-arm variation, and \(\sigma_i^2\) is the sampling variance of \(Y_i\). The shrinkage estimate is the conditional posterior mean \(\tilde{\theta}_i=\mu+\lambda_i(Y_i-\mu)\), where \(\lambda_i=\tau^2/(\tau^2+\sigma_i^2)\).

The model assumes that arm estimates are unbiased and normally distributed conditional on their true effects, with known sampling variances; that arm effects follow the stated common normal distribution; and that the effects are independent across arms. The simple model also treats \(\mu\) and \(\tau^2\) as known. In practice, those quantities may need to be estimated, and uncertainty in those estimates affects the quality of shrinkage.

The factor \(\lambda_i\) lies between zero and one. When sampling variance is large relative to between-arm variation, \(\lambda_i\) is small and the estimate moves substantially toward \(\mu\). When between-arm variation is large relative to sampling variance, \(\lambda_i\) is near one and the estimate stays close to \(Y_i\). If all arms have the same sampling variance, they share one shrinkage factor.

Why This Formula Is a Posterior Mean

The shrinkage formula follows by combining the normal sampling model and the normal distribution of effects. Conditional on \(Y_i=y_i\), the density for \(\theta_i\) is proportional to the product of the sampling density and the distribution of arm effects. Its exponent, ignoring terms that do not involve \(\theta_i\), is

-\frac{1}{2}\left\{\frac{(y_i-\theta_i)^2}{\sigma_i^2} +\frac{(\theta_i-\mu)^2}{\tau^2}\right\}.

Collecting the terms that involve \(\theta_i\) gives a normal conditional distribution whose precision is \(1/\sigma_i^2+1/\tau^2\). Its variance and mean are

\operatorname{Var}(\theta_i\mid Y_i=y_i) =\left(\frac{1}{\sigma_i^2}+\frac{1}{\tau^2}\right)^{-1} =\frac{\tau^2\sigma_i^2}{\tau^2+\sigma_i^2},

E(\theta_i\mid Y_i=y_i) =\frac{y_i/\sigma_i^2+\mu/\tau^2} {1/\sigma_i^2+1/\tau^2} =\frac{\tau^2}{\tau^2+\sigma_i^2}y_i +\frac{\sigma_i^2}{\tau^2+\sigma_i^2}\mu.

This is a weighted average of the observed estimate and the common mean. If the selected arm is \(J\), chosen as a function of the observed estimates, the corresponding estimate is \(\tilde{\theta}_J=\mu+\lambda_J(Y_J-\mu)\). Under the hierarchical model, conditioning on the full observed vector does not create an extra adjustment to this posterior mean: once the estimates are observed, the selection index \(J\) is determined by them.

Worked Example: Shrinking the Winning Estimate

Suppose three hypothetical treatment arms have estimated reductions in symptom score of 12, 9, and 4 points, and the arm with the largest estimate is selected. Assume the estimates have sampling standard deviation 3 points, and a hierarchical model has common mean \(\mu=5\) points and between-arm standard deviation \(\tau=4\) points. Treat these model quantities as known for this calculation.

The shrinkage factor is

\lambda=\frac{\tau^2}{\tau^2+\sigma^2} =\frac{4^2}{4^2+3^2} =\frac{16}{25}=0.64.

The selected estimate is 12 points. Its shrinkage estimate is

\tilde{\theta}_J =\mu+\lambda(Y_J-\mu) =5+0.64(12-5) =5+4.48 =9.48\text{ points}.

Thus the estimate moves 2.52 points toward the common mean. This calculation does not establish that the selected arm's true effect is 9.48 points; it gives a model-based estimate that balances the observed value against the assumed distribution of effects.

Bias and MSE Depend on the Target

For fixed effects \(\theta_1,\ldots,\theta_K\), let \(J\) be the index selected by a rule such as \(J=\arg\max_i Y_i\). The target is then the true effect of the selected arm, \(\theta_J\), which can vary across repeated trials because the selected index varies. For a fixed shrinkage factor \(\lambda\) and common reference \(\mu\), the estimation error is

\tilde{\theta}_J-\theta_J =\lambda(Y_J-\theta_J)+(1-\lambda)(\mu-\theta_J).

Consequently, its bias and MSE under repeated sampling with the true effects held fixed are

\operatorname{Bias} =E(\tilde{\theta}_J-\theta_J) =\lambda E(Y_J-\theta_J)+(1-\lambda)E(\mu-\theta_J),

\operatorname{MSE} =E\left[ \{\lambda(Y_J-\theta_J)+(1-\lambda)(\mu-\theta_J)\}^2 \right].

The expectations include the randomness of which arm wins. The first term in the bias expression reflects selection of positive sampling errors; the second reflects shrinkage toward \(\mu\), which can be in the wrong direction for an arm whose true effect differs substantially from the common mean. Thus shrinkage does not guarantee lower frequentist bias or MSE for every fixed configuration of effects.

Worked Example: Two Equal-Effect Arms

Consider two hypothetical arms with the same true effect \(\mu\), independent estimates \(Y_i=\mu+2Z_i\), and \(Z_i\sim N(0,1)\). The selected estimate is the larger of the two. Use a fixed shrinkage factor \(\lambda=0.64\) toward the true common mean to isolate the effect of shrinkage. Here the target effect is \(\mu\), and the model assumptions are equal effects, independent normal errors, and known common variance \(2^2\).

Write \(M=\max(Z_1,Z_2)\). Since \(M=(Z_1+Z_2+|Z_1-Z_2|)/2\), and \(Z_1-Z_2\sim N(0,2)\), it follows that

E(M)=\frac{1}{2}E|Z_1-Z_2| =\frac{1}{2}\frac{2}{\sqrt{\pi}} =\frac{1}{\sqrt{\pi}}.

Also, if \(m=\min(Z_1,Z_2)\), then \(M^2+m^2=Z_1^2+Z_2^2\). By symmetry \(E(M^2)=E(m^2)\), so \(2E(M^2)=2\) and \(E(M^2)=1\). The unadjusted winner has bias \(2E(M)=2/\sqrt{\pi}\approx1.128\) points and MSE \(4E(M^2)=4\) squared points.

The shrunk estimate is \(\tilde{\theta}_J=\mu+0.64(2M)\). Its bias and MSE are

\operatorname{Bias}(\tilde{\theta}_J) =0.64\frac{2}{\sqrt{\pi}} \approx0.722\text{ points}, \qquad \operatorname{MSE}(\tilde{\theta}_J) =0.64^2(4)(1) =1.638\text{ squared points}.

In this equal-effect setting, shrinkage reduces both the positive bias and MSE relative to the unadjusted winner. The result depends on the common mean being the target and the true effect for both arms; it is not a universal guarantee for heterogeneous treatment effects.

Posterior Risk for a Selected Arm

Under squared-error loss, the posterior mean minimizes posterior expected squared error. This statement also applies to the selected arm under the hierarchical model above: after observing all arm estimates, \(J\) is known, and the conditional distribution of \(\theta_J\) gives a posterior mean and variance for that arm. The posterior variance is \(\tau^2\sigma_J^2/(\tau^2+\sigma_J^2)\).

Worked Example: Conditional Expected Squared Error

Suppose three hypothetical arm estimates are 10, 8, and 4 points, with independent sampling standard deviation 4 points for each. Assume the arm effects follow the stated normal model with \(\mu=6\) points and \(\tau=3\) points, and treat these values as known. The first arm is selected because its estimate is largest.

The shrinkage factor is \(9/(9+16)=0.36\), so the selected arm's posterior mean is

6+0.36(10-6)=7.44\text{ points}.

The posterior variance for its true effect is

\frac{\tau^2\sigma^2}{\tau^2+\sigma^2} =\frac{9(16)}{25} =5.76\text{ squared points}.

Given the observed estimates, the posterior expected squared error of the shrunk estimate is therefore 5.76 squared points. For the unshrunk estimate of 10 points, the conditional posterior expected squared error is its posterior variance plus the squared distance from the posterior mean:

5.76+(10-7.44)^2 =5.76+6.5536 =12.3136\text{ squared points}.

This comparison is conditional on the specified hierarchical model and observed data. It is not a frequentist MSE comparison for every fixed set of true arm effects.

Diagnostics, Alternatives, and Limitations

Shrinkage is most defensible when the common mean and between-arm variation describe the set of arms under consideration, and when each arm estimate has a credible sampling variance. Check whether the assumed common distribution is plausible and whether one or more arms are meaningfully different from the rest. A very small estimated \(\tau^2\) can force estimates close to \(\mu\); an unstable estimate of \(\tau^2\), especially with few arms, can make the shrinkage factor unreliable. Correlated arm estimates also violate the independent-arm model and require a covariance model.

A known common mean is a simplifying assumption, not a routine fact. If \(\mu\) is estimated from the same set of arms, its uncertainty should be accounted for. Simply plugging in the sample average and treating it as fixed can understate uncertainty, and the selected arm contributes to that average. Likewise, estimating \(\tau^2\) from few arms may substantially change the amount of shrinkage. A full hierarchical analysis carries these uncertainties through; an empirical-Bayes estimate that plugs in estimated hyperparameters is an approximation.

Approach	What it does	Key qualification
Unadjusted winner	Reports \(Y_J\) unchanged.	Simple, but often high after selecting the largest estimate.
Hierarchical shrinkage	Moves \(Y_J\) toward a common mean using model-based precision weights.	Can lower MSE under a suitable model, but can pull a genuinely unusual effect too far.
Selection-adjusted frequentist estimates	Adjust estimation to account for the selection rule and sampling distribution.	Target and assumptions differ from a posterior mean; the earlier material on bias-adjusted and stagewise estimates provides related approaches.

A common mistake is to say that shrinkage always corrects winner's curse and therefore must improve the estimate. The correction is not automatic: shrinkage reduces the influence of the selected observation, but it also introduces dependence on the common mean and the assumed variation among effects. The appropriate comparison must name its target and performance criterion—for example, fixed-parameter bias, fixed-parameter MSE, or posterior expected squared error.

Key takeaway: A selected-arm shrinkage estimate is a weighted average of its observed estimate and a common mean. Its factor follows from the assumed relative sizes of between-arm and sampling variation. Its benefits are model- and target-dependent, so report the assumptions and do not interpret shrinkage as a guaranteed frequentist correction.

Check Your Understanding

Answer each question, then check the solutions below.

If \(\tau^2=4\) and \(\sigma_i^2=12\), what is \(\lambda_i\), and which direction does the estimate move?
For two independent equal-effect arms with standard normal errors scaled by \(\sigma\), what is the expected maximum standardized error?
Why does the hierarchical posterior mean formula still apply to the selected arm after the full vector of arm estimates is observed?
Does shrinkage necessarily reduce fixed-parameter MSE when one treatment's true effect is far from the common mean? Explain.
What additional issue arises if the common mean is estimated from the same arm estimates rather than known?

Solutions. (1) \(\lambda_i=4/(4+12)=0.25\), so the estimate moves 75% of the distance from \(Y_i\) toward \(\mu\). (2) \(E\{\max(Z_1,Z_2)\}=1/\sqrt{\pi}\approx0.5642\), so the expected maximum error in the original units is \(\sigma/\sqrt{\pi}\). (3) Once the estimates are observed, the selected index is determined; under the independent hierarchical model, the selected arm's posterior distribution is the one obtained from its observed estimate. (4) No. Shrinkage can move the estimate away from a genuinely exceptional true effect, increasing bias or MSE. (5) The common mean is uncertain and dependent on the arm estimates; treating its estimate as fixed can understate uncertainty and distort the shrinkage amount.