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Treatment-Effect Estimation After Selection · Tutorial 293 of 499

Worked Example: Combining Two Stages After Selection

Work through how interim selection affects a two-stage estimate, and calculate a model-based adjustment alongside the naive combined estimate.

Advanced 10 min read

What You'll Learn

  • Define the selected-arm estimand and the assumptions for a two-stage normal model.
  • Derive the expected stage-one estimation error after selecting the larger of two estimates.
  • Combine stage estimates using prespecified information weights.
  • Calculate and interpret naive and model-adjusted final estimates.
  • Diagnose when the normal-model adjustment may be unreliable and compare it with a stage-two-only estimate.

Why Combining Stages Needs Care

When two treatment arms are compared at an interim analysis, selecting the arm with the larger estimated effect also selects, in part, the arm with a favourable random error. If that same interim estimate is carried forward into the final estimate, its error does not disappear merely because new data are collected. The stage-two data are not selected in the same way, but the combined estimate still gives some weight to the interim result.

This tutorial works through that issue in a two-arm normal model. The target is the mean effect of the arm selected at the interim analysis, relative to a common control or reference. The selected arm is random, so its true effect is also a random quantity across repetitions of the design. A model-based adjustment targets the expected upward selection error under specified treatment means and variances; it does not reveal the true effect in a particular trial.

Definition: Let \(X_A\) and \(X_B\) be the stage-one estimates of treatment effects \(\theta_A\) and \(\theta_B\). The interim rule selects \(J=A\) if \(X_A\ge X_B\), and selects \(J=B\) otherwise. Let \(Y_J\) be the independent stage-two estimate for the selected arm. If the prespecified stage information values are \(I_1\) and \(I_2\), the information-weighted combined estimate is $$ \hat{\theta}_{J,\mathrm{comb}} = \frac{I_1X_J+I_2Y_J}{I_1+I_2}. $$ Its target in this tutorial is \(\theta_J\), the true effect of the selected arm. The naive combined estimate is \(\hat{\theta}_{J,\mathrm{comb}}\) without a selection adjustment.

Model and Selection-Bias Result

Assume that the stage-one estimates are independent and satisfy \(X_k=\theta_k+\varepsilon_k\), where \(\varepsilon_A,\varepsilon_B\sim N(0,\tau^2)\). Assume also that stage two is independent of stage one and, conditional on the selected arm, \(Y_J\sim N(\theta_J,\sigma^2/I_2)\). At stage one, the variance can be written \(\tau^2=\sigma^2/I_1\). These assumptions give the information-weighted combination above its usual normal-model interpretation. The selection rule must be the stated rule, and the means and variances used to calculate an adjustment must describe that rule and model adequately.

Write \(\Delta=\theta_A-\theta_B\). Under this model, the expected error in the selected stage-one estimate is

$$ E(X_J-\theta_J) = \sqrt{2}\,\tau\, \phi\!\left(\frac{\Delta}{\sqrt{2}\tau}\right), $$

where \(\phi\) is the standard normal density. The result is nonnegative whether arm A or arm B has the larger true mean: selection favours an estimate with a positive error. The stage-two estimate has mean \(\theta_J\) under the independence assumption, so the expected bias of the combined estimate for the selected-arm target is

$$ E(\hat{\theta}_{J,\mathrm{comb}}-\theta_J) = \frac{I_1}{I_1+I_2} \sqrt{2}\,\tau\, \phi\!\left(\frac{\Delta}{\sqrt{2}\tau}\right). $$

This is an expectation over repetitions of the full selection-and-estimation procedure under fixed model parameters. It is not a claim that the observed estimate in one trial must exceed its target by this amount.

Why the Expected Error Has This Form

A short derivation makes clear where selection enters. Write \(X_A=\theta_A+\varepsilon_A\) and \(X_B=\theta_B+\varepsilon_B\), and define \(M=(\varepsilon_A+\varepsilon_B)/2\) and \(Q=(\varepsilon_A-\varepsilon_B)/2\). Under independent, equal-variance normal errors, \(M\) and \(Q\) are independent, \(E(M)=0\), and \(Q\sim N(0,\tau^2/2)\). The selection event for A is \(\Delta+2Q\ge0\). The selected error is \(M+Q\) on this event and \(M-Q\) otherwise. Therefore its expectation is \(E[Q\,\mathrm{sign}(\Delta+2Q)]\). Since \(E(Q)=0\), this equals \(2E[Q\,1\{Q\ge-\Delta/2\}]\). For a centred normal variable with standard deviation \(s\), \(E[Q\,1\{Q\ge c\}]=s\phi(c/s)\). Taking \(s=\tau/\sqrt{2}\) and \(c=-\Delta/2\), and using the symmetry of \(\phi\), gives the stated result.

The combination is a fixed weighted average once \(I_1\) and \(I_2\) are specified. The stage-two contribution has zero expected error relative to \(\theta_J\), while the stage-one contribution has the selection error just derived. Multiplying that error by \(I_1/(I_1+I_2)\) gives the combined-estimate bias formula.

Worked Example: A Moderate Difference Between Arms

Worked Example: A Moderate Difference Between Arms

Suppose the hypothetical stage-one estimates have \(X_A=2.1\) and \(X_B=1.7\) treatment-effect units, so arm A is selected. The stage-two estimate for A is \(Y_A=1.1\) units. Let \(I_1=1.44\), \(I_2=0.64\), and \(\sigma^2=2.0736\) squared units. Then the stage-one standard error is \(\tau=\sqrt{\sigma^2/I_1}=\sqrt{2.0736/1.44}=1.2\) units, and the stage-two standard error is \(\sqrt{2.0736/0.64}=1.8\) units. Assume for illustrating the model adjustment that \(\theta_A=1.2\) and \(\theta_B=0.6\) units.

The information fraction assigned to stage one is $$ w_1=\frac{1.44}{1.44+0.64}=0.6923, $$ so the naive combined estimate is $$ \hat{\theta}_{A,\mathrm{comb}} = \frac{1.44(2.1)+0.64(1.1)}{2.08} = \frac{3.024+0.704}{2.08} = 1.7923\text{ units}. $$

The modelled true difference is \(\Delta=1.2-0.6=0.6\) units. Thus $$ a=\frac{0.6}{\sqrt{2}(1.2)}=0.3536, \qquad \phi(a)\approx0.3748. $$ The expected selected stage-one error is $$ \sqrt{2}(1.2)(0.3748)\approx0.6360\text{ units}. $$ The expected bias in the combined estimate is therefore $$ 0.6923(0.6360)\approx0.4403\text{ units}. $$

Subtracting this model-based expected bias gives the adjusted estimate $$ 1.7923-0.4403=1.3520\text{ units}. $$ The correction is an expectation under the assumed means, variances, and selection rule. The adjusted value is not guaranteed to be closer to the true effect in this particular hypothetical trial.

Worked Example: A Larger Difference Reduces the Correction

Worked Example: A Larger Difference Reduces the Correction

Now suppose a hypothetical trial has \(X_A=3.0\), \(X_B=0.4\), and \(Y_A=2.2\) units. Arm A is selected. Keep \(I_1=1.44\), \(I_2=0.64\), and \(\tau=1.2\) units, but suppose the model means are \(\theta_A=2.4\) and \(\theta_B=0\). The stage-one standard error is 1.2 units as before; the selection-bias calculation assumes independent normal stage-one errors and independent stage-two data.

The naive combination is $$ \hat{\theta}_{A,\mathrm{comb}} = \frac{1.44(3.0)+0.64(2.2)}{2.08} = \frac{4.320+1.408}{2.08} = 2.7538\text{ units}. $$ Here \(\Delta=2.4\) units, so $$ a=\frac{2.4}{\sqrt{2}(1.2)}=1.4142, \qquad \phi(a)\approx0.1468. $$ The expected stage-one selection error is \(\sqrt{2}(1.2)(0.1468)\approx0.2491\) units, and the combined-estimate bias is \(0.6923(0.2491)\approx0.1724\) units. The model-adjusted estimate is therefore \(2.7538-0.1724=2.5814\) units.

The correction is smaller than in the preceding example because the modelled true effects are farther apart relative to the stage-one standard error. When one arm is usually much better, noise is less likely to determine which arm is selected.

Worked Example: More Stage-Two Information

Worked Example: More Stage-Two Information

Return to the modelled means \(\theta_A=1.2\) and \(\theta_B=0.6\), with \(\tau=1.2\) units and stage-one estimates \(X_A=2.1\), \(X_B=1.7\). Suppose the stage-two estimate for selected arm A is \(Y_A=1.4\) units, and stage-two information is now \(I_2=2.56\), while \(I_1=1.44\). Under the common-variance model with \(\sigma^2=2.0736\), the stage-two standard error is \(\sqrt{2.0736/2.56}=0.9\) units. Assume again that stage-two data are independent of the interim selection.

The stage-one information fraction is \(1.44/(1.44+2.56)=0.36\). The naive combined estimate is $$ \hat{\theta}_{A,\mathrm{comb}} = \frac{1.44(2.1)+2.56(1.4)}{4.00} = \frac{3.024+3.584}{4.00} = 1.6520\text{ units}. $$

The expected stage-one selection error is still \(0.6360\) units because the stage-one model and the mean difference have not changed. Its contribution to the combined-estimate bias is now \(0.36(0.6360)=0.2290\) units. The adjusted estimate is \(1.6520-0.2290=1.4230\) units. More stage-two information reduces the weight on the selected interim estimate, so the expected bias contribution from stage one is smaller.

Diagnostics, Alternatives, and Limitations

Before applying the formula, check the assumptions that determine it. Confirm that the interim rule really selects the larger of the two estimates, that the stage-one errors are approximately normal with the stated common standard error, and that stage-two data are independent of the selection process. Confirm also that the information weights correspond to the analysis actually used. If allocation, variances, or selection rules differ between arms, this simple symmetric formula need not apply.

Useful warning signs include an interim rule that uses a transformed or adjusted statistic rather than \(X_A-X_B\), a stage-two sample affected by the interim outcomes, substantial variance inequality, or an estimate of the mean difference that is inconsistent with the model used for the correction. In those situations, the calculated adjustment may not target the actual selection bias. The formula is also sensitive to the assumed \(\Delta\): if it is estimated from the same selected data, treating it as known understates uncertainty in the adjustment. A reported adjusted estimate should identify the model and inputs used; the subtraction alone does not account for uncertainty in estimated nuisance parameters.

The main alternative is to estimate the selected arm using stage-two data alone. Under the model, \(Y_J\) is unbiased for \(\theta_J\) because stage two is independent of selection, but it discards the interim information and may be less precise. The naive combined estimate uses both stages but retains the expected selection error. The model-adjusted combined estimate retains both stages and subtracts the modelled expected error, but depends on the model and adjustment inputs being adequate.

ApproachUse of dataMain trade-off
Naive combined estimateStage one and stage twoRetains the selected stage-one information, including its expected selection error
Stage-two-only estimateStage two onlyAvoids reuse of the selected interim estimate, but discards stage-one information
Model-adjusted combined estimateStage one and stage two, with a bias subtractionAddresses expected selection bias under the specified model; sensitive to model and input misspecification

The normal result is a benchmark for this two-arm, two-stage setting, not a universal correction for all adaptive designs. It assumes a fixed selection rule, independent equal-variance stage-one estimates, and independent stage-two data. If those assumptions fail, the expected selected error must be derived under the actual design rather than borrowed from this formula. The adjustment targets mean bias under the assumed model; it does not by itself give a confidence interval with correct coverage.

Key takeaway: In this normal two-arm example, interim selection creates a positive expected error in the selected stage-one estimate. The combined estimate inherits that error in proportion to the stage-one information fraction. A model-based subtraction can address the expected bias, but its validity depends on the selection rule and model assumptions.

Check Your Understanding

Use the stated normal two-arm model. Answers are provided below the questions.

  1. If \(\Delta=0\) and \(\tau=1\), what is the expected error in the selected stage-one estimate?
  2. If the stage-one information fraction is 0.5 and the expected selected stage-one error is 0.4 units, what is the expected combined-estimate bias?
  3. Why does a larger absolute value of \(\Delta/\tau\) generally reduce the expected selection correction?
  4. What is one key assumption that allows the stage-two estimate to contribute zero expected error relative to \(\theta_J\)?

Solutions. (1) The formula gives \(\sqrt{2}\phi(0)=1.4142(0.3989)\approx0.5642\) units. (2) The combined bias is \(0.5(0.4)=0.20\) units. (3) The density \(\phi(\Delta/(\sqrt{2}\tau))\) decreases as the absolute standardized difference increases, so selection is less driven by noise. (4) Stage-two data must be independent of the stage-one selection process and have mean \(\theta_J\) conditional on the selected arm; then \(E(Y_J-\theta_J)=0\).

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