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Treatment-Effect Estimation After Selection · Tutorial 273 of 499

Selective Confidence Intervals by Test Inversion

Learn to construct and interpret a confidence interval for one selected treatment effect by inverting tests that account for the event that selected the arm.

Advanced 12 min read

What You'll Learn

  • Define the selection event that conditions a selective test in a MAMS analysis.
  • Derive the conditional distribution used to test a proposed treatment effect.
  • Invert conditional equal-tailed tests to obtain a selected-effect confidence set.
  • Calculate and interpret a selective interval in a truncated-normal example.
  • Handle nuisance parameters conservatively and identify numerical or model failures.
  • Distinguish conditional coverage from marginal coverage and from an interval based on independent validation data.

Why invert tests after selection?

In a multi-arm, multi-stage trial, an arm may be selected because its interim evidence looked promising. An estimate from that arm is therefore not, in general, distributed like an estimate from an arm specified in advance. Even if the treatment-effect estimator has a normal distribution before selection, observing that an arm passed a boundary or ranked first changes its distribution. An interval built from the unselected distribution can consequently have poor coverage conditional on the selection that actually occurred.

A direct way to address this is to test each proposed value of the selected treatment effect using a null distribution conditioned on the observed selection event. Inverting those tests gives a selective confidence set. The target here is one selected treatment effect; simultaneous coverage for multiple selected arms is a separate problem.

Definition: Let \(X\) denote the data used for estimation and selection, let \(E\) denote the event that the observed arm and selection path were chosen, and let \(\theta_j\) be the treatment effect for the selected arm \(j\). A selective test of \(H_0:\theta_j=\theta_0\) calibrates its statistic using the distribution of the data conditional on \(E\), under the null value \(\theta_0\). Inverting a level-\(\alpha\) family of such tests gives a selective confidence set with conditional coverage at least \(1-\alpha\), when the tests have conditional size at most \(\alpha\).

The event \(E\) must represent the rule that generated the reported selection. In a simple illustration it may be “the estimate exceeded a threshold.” In a MAMS analysis it may include the interim comparisons, the arm retained at each stage, and the stopping decision. Conditioning on the wrong event does not correct the selection that took place. The conditional distribution is defined only when \(P_{\theta}(E)>0\) for the parameter values being considered.

Conditional distributions and test inversion

Let \(T\) be a statistic for the selected arm, and suppose first that all nuisance parameters are known. For a proposed value \(\theta_0\), a one-sided upper-tail conditional p-value is

$$ p_{\theta_0} = P_{\theta_0}\!\left(T\geq t_{\mathrm{obs}}\mid E\right). $$

It measures how often a statistic at least as large as the observed one would occur if the proposed effect were true, among repetitions in which the selection event also occurred. A two-sided procedure can be formed from two conditional tails. For a continuous statistic with conditional cumulative distribution function \(F_{\theta}(t\mid E)\), the equal-tailed \(100(1-\alpha)\%\) confidence set is

$$ C(x) = \left\{ \theta: \frac{\alpha}{2} \leq F_{\theta}(t_{\mathrm{obs}}\mid E) \leq 1-\frac{\alpha}{2} \right\}. $$

Equivalently, values of \(\theta\) are retained when neither conditional tail is too small. If the set is an interval, its endpoints solve the two tail equations at \(\alpha/2\) and \(1-\alpha/2\). In more complicated models the set need not be connected; an interval spanning separated pieces is a conservative summary, not the exact inverted set.

A derivation in a truncated-normal model

Suppose an estimate \(Y\) has distribution \(Y\sim N(\theta,1)\), with known standard error 1, and the arm is reported only if \(Y>c\). For an observed \(y>c\), the conditional density is the original normal density restricted to the selection region and divided by the probability of selection:

$$ f_{\theta}(y\mid Y>c) = \frac{\phi(y-\theta)}{1-\Phi(c-\theta)},\qquad y>c. $$

Here \(\phi\) and \(\Phi\) are the standard normal density and cumulative distribution function. Integrating this density from \(c\) to \(y\) gives

$$ F_{\theta}(y\mid Y>c) = \frac{\Phi(y-\theta)-\Phi(c-\theta)} {1-\Phi(c-\theta)}. $$

The denominator is essential: it accounts for the fact that only values above the selection threshold could have been reported. For a continuous conditional distribution, the probability integral transform says that \(F_{\theta}(Y\mid E)\) is uniform on \([0,1]\), conditional on \(E\), when \(\theta\) is the true parameter. Thus, the probability that this quantity falls outside \([\alpha/2,1-\alpha/2]\), conditional on selection, is \(\alpha\). This establishes the conditional coverage of the equal-tailed set in this model.

Worked Example: Inverting a 95% Interval After a Threshold Selection

Suppose, hypothetically, that an arm is selected when its normally distributed effect estimate \(Y\) exceeds 1 standard-error unit. The observed estimate is \(y=2.4\), and its standard error is 1 effect unit. Assume the normal model and threshold rule are correct. For a 95% equal-tailed selective interval, set \(c=1\) and \(\alpha=0.05\).

The lower endpoint solves the upper-tail equation \(\{1-\Phi(2.4-\theta)\}/\{1-\Phi(1-\theta)\}=0.025\). Numerical solution gives \(\theta\approx-0.5882\), or \(-0.588\) to three decimal places. The upper endpoint solves \(\{\Phi(2.4-\theta)-\Phi(1-\theta)\}/\{1-\Phi(1-\theta)\}=0.025\), giving approximately \(\theta=4.353\). Therefore the selective interval is approximately \([-0.588,4.353]\) effect units.

This interval is wide and includes zero, even though the observed estimate is 2.4. That is not a contradiction: the estimate was reported specifically because it exceeded 1, and the conditional procedure allows for the upward shift induced by that selection rule. Under the stated model and rule, the procedure has 95% coverage conditional on this threshold selection.

Conditional tests with more than one arm

Selection in MAMS trials is often based on comparisons among arms rather than a single fixed threshold. For example, an arm may be selected because its interim statistic is the largest. Even in a one-stage normal illustration, the selected statistic has a different distribution from an unselected one.

Worked Example: Conditioning on Which of Two Arms Ranked First

Consider two hypothetical independent estimates \(Y_1,Y_2\), each with variance 1 and both with mean 0 under the null. Report arm 1 only when \(Y_1>Y_2\). Suppose the observed estimate for arm 1 is \(y_1=1.5\), and test the null value \(\theta_1=0\), with \(\theta_2=0\) treated as known for this calculation.

For a candidate value \(u\) of \(Y_1\), the probability of selection is \(P(Y_2<u)=\Phi(u)\). Thus the conditional cumulative distribution of \(Y_1\), given that arm 1 ranked first, is \[ \frac{\int_{-\infty}^{y_1}\phi(u)\Phi(u)\,du}{P(Y_1>Y_2)}. \] Under the null, symmetry gives \(P(Y_1>Y_2)=0.5\). Since the derivative of \(\Phi(u)^2/2\) is \(\phi(u)\Phi(u)\), the numerator equals \(\Phi(y_1)^2/2\). Therefore \(F_0(1.5\mid Y_1>Y_2)=\Phi(1.5)^2\approx0.9332^2=0.8708\).

The conditional upper-tail probability is \(1-0.8708=0.1292\). This is not small evidence against the proposed null at a 5% one-sided level. By contrast, the unconditioned upper-tail probability is \(1-\Phi(1.5)\approx0.0668\). The two calculations differ because the conditional test asks how unusual 1.5 is among repetitions in which arm 1 beat arm 2.

In a general MAMS analysis, let \(X\) collect the stagewise estimates or test statistics, and let \(E\) encode the observed selection path. If \(X\) has density \(f_{\theta,\eta}(x)\), where \(\eta\) denotes nuisance parameters, then the conditional density on the event is

$$ f_{\theta,\eta}(x\mid E) = \frac{f_{\theta,\eta}(x)\mathbf{1}\{x\in E\}} {P_{\theta,\eta}(E)}. $$

The denominator is the probability, under that parameter configuration, of the selection event. Shared controls and repeated stagewise observations can make the components of \(X\) correlated, so the joint distribution—not a product of marginal distributions—must be used when the design creates dependence. Inverting tests based on an incorrect independence assumption will generally not deliver the claimed conditional coverage.

Worked Example: Why the Selection Probability Enters the Test

Return to the hypothetical threshold rule \(Y>1\). Test the candidate value \(\theta_0=0\) against larger effects, using the observed \(y=2.4\). The conditional upper-tail p-value is \[ \frac{P_0(Y\geq2.4)}{P_0(Y>1)} = \frac{1-\Phi(2.4)}{1-\Phi(1)}. \] Using \(1-\Phi(2.4)\approx0.00820\) and \(1-\Phi(1)\approx0.1587\), the result is \(0.00820/0.1587\approx0.0517\).

The unconditioned one-sided p-value is approximately 0.00820, whereas the conditional value is about 0.0517. At a 5% one-sided level, these lead to different decisions: the unconditioned test rejects, but the conditional test does not. The difference is precisely the adjustment for the fact that estimates at or below 1 could not have produced the reported arm.

Nuisance parameters, computation, and diagnostics

Conditional distributions may depend on effects of other arms, correlations, variances, or other nuisance parameters. Treating these quantities as known when they are not can make a nominal selective interval fail to have its advertised coverage. If nuisance parameters can be eliminated by conditioning on an appropriate statistic, that can yield a nuisance-free test, but such a statistic is not always available.

One conservative option is to take the supremum of each conditional tail probability over the nuisance-parameter values allowed under the null. For example, use \(p_U(\theta_0)=\sup_{\eta}P_{\theta_0,\eta}(T\geq t_{\mathrm{obs}}\mid E)\) for an upper-tail test. Because this is at least as large as the tail probability at the true nuisance value, it cannot make rejection easier than the correctly specified conditional test. For a two-sided equal-tailed construction, take the supremum of each tail separately and retain \(\theta_0\) only when both are greater than \(\alpha/2\). This can be conservative, sometimes substantially so.

A practical calculation should be checked against the design and the probability model. In particular:

  • Verify that \(E\) includes the actual stagewise selection and stopping information relevant to the reported estimate.
  • Check that the model includes the covariance induced by shared controls or repeated measurements, when present.
  • Confirm that \(P_{\theta,\eta}(E)\) is positive and that numerical integration accurately evaluates both the conditional numerator and denominator.
  • Plot or otherwise inspect the inverted tail probabilities over a range of \(\theta\). Unexpected non-monotonicity can produce multiple pieces of a confidence set; it should not be hidden by reporting only one pair of roots.
  • Examine sensitivity to nuisance-parameter values and to plausible changes in the selection rule. Small selection probabilities can make conditional calculations especially sensitive to model and numerical error.

How this interval differs from alternatives

The introductory material on confidence intervals for a selected treatment effect distinguishes selection-adjusted targets and coverage ideas. Test inversion specifies a concrete construction: for each candidate effect, calibrate a test under the conditional distribution and retain values not rejected. A marginal interval instead aims for coverage averaged over selection and non-selection outcomes; that target need not provide nominal coverage among trials in which an arm was selected. The choice depends on the inferential question.

An independent validation sample can avoid conditioning the estimate on the data that selected the arm, provided the validation data and analysis satisfy the required independence and model assumptions. It may, however, use less information or be less precise. Conditional inversion uses the selection data directly, but requires the selection mechanism and joint sampling model to be represented correctly. Neither approach repairs a misspecified outcome model or an inaccurately recorded selection rule.

A common mistake is to say, “The selected estimate is 2.4 with standard error 1, so a 95% interval is \(2.4\pm1.96=[0.44,4.36]\).” The arithmetic is \(2.4-1.96=0.44\) and \(2.4+1.96=4.36\), but the interval uses the unselected normal distribution. It does not account for the threshold that made this arm eligible to be reported. Under the hypothetical threshold model above, the equal-tailed conditional interval is instead approximately \([-0.588,4.353]\). The ordinary interval may be useful for a different coverage target, but it should not be described as a conditional selective interval.

Key takeaway: A selective interval is obtained by inverting tests calibrated conditional on the selection event. Its interpretation depends on the event being correctly specified, the conditional model being adequate, and nuisance parameters being handled appropriately.

Check Your Understanding

Try each problem, then compare your reasoning with the solutions below.

  1. In the threshold model \(Y\sim N(\theta,1)\), with an arm reported only when \(Y>c\), write the conditional distribution function at an observed \(y>c\).
  2. For the hypothetical threshold example, why does the conditional upper-tail probability for testing \(\theta=0\) exceed the unconditioned upper-tail probability?
  3. For the two-arm example, state the conditional cumulative probability at \(y_1=1.5\) under equal zero means, and give its approximate value.
  4. What is one conservative way to handle an unknown nuisance parameter in a conditional test, and why is it conservative?
  5. If test inversion produces a disconnected set of effect values, what should the analyst avoid doing without explanation?

Solutions. (1) \(F_{\theta}(y\mid Y>c)=\{\Phi(y-\theta)-\Phi(c-\theta)\}/\{1-\Phi(c-\theta)\}\). The denominator conditions on passing the threshold. (2) The unconditioned tail includes all outcomes at least as large as the observation, while the conditional tail divides by the probability of passing the threshold; it assesses extremeness only among selected outcomes. Here the values are about 0.00820 and 0.0517, respectively. (3) It is \(\Phi(1.5)^2\), approximately \(0.9332^2=0.8708\). (4) Take the supremum of the conditional tail probability over nuisance values allowed by the null. It is no smaller than the tail probability at the true nuisance value, so it does not make rejection easier. (5) Do not silently replace the set by a single interval and imply that the interval itself is the exact inversion; report the set or explain that its interval hull is a conservative summary.

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