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Bias-Adjusted Estimation and Confidence Intervals · Tutorial 308 of 699

Coverage Failure of Fixed-Sample Intervals After Stopping

Learn how stopping changes the sampling distribution of a reported estimate and how to quantify the coverage of an ordinary interval after selection.

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What You'll Learn

  • Define coverage for an interval reported after a group sequential stopping rule.
  • Derive the conditional coverage of a fixed-sample normal interval at an efficacy stop.
  • Calculate coverage for a two-look design using the joint distribution of the stage statistics.
  • Distinguish conditional coverage at a selected stopping stage from overall coverage across stages.
  • Diagnose selection effects and identify why ordinary interval calculations may fail.
  • Compare naive fixed-sample intervals with stopping-aware approaches.

Why Stopping Can Change Interval Coverage

A conventional confidence interval is calibrated for a specified sampling plan. In a group sequential design, however, the estimate reported may come from a stage selected by the observed data: unusually large interim results can trigger an early stop, while other results lead to further observations. The estimate and the decision to report it are therefore statistically linked.

This link matters even when the interval formula is familiar and its standard error is calculated correctly. The interval may still have coverage different from its nominal level because the estimate’s distribution among trials that stop at a particular look is not its fixed-sample distribution. A useful way to isolate this issue is to start with a normal model and a known standard error, then examine exactly which estimates are selected.

Definition: For a parameter \(\theta\), an interval procedure has coverage probability \(P(\theta\in C)\), where \(C\) is the random interval produced by the procedure. A nominal 95% interval is calibrated to have coverage probability 0.95 under its stated sampling plan and assumptions. After a stopping rule selects which estimate and interval to report, coverage must be evaluated under that rule; the fixed-sample calibration does not automatically carry over.

The distinction is between the interval formula and the sampling procedure to which its coverage claim applies. The ordinary normal interval \(\hat{\theta}\pm1.96\,SE\) has 95% coverage when, for a fixed sample size, \((\hat{\theta}-\theta)/SE\sim N(0,1)\). After stopping, the relevant question is whether that standardized estimate still has the required distribution among the reported estimates.

A Normal Two-Look Setup

Consider independent observations \(Y_i\sim N(\theta,\sigma^2)\), with known \(\sigma\), and planned sample sizes \(n_1<n_2\). At look \(k\), let \(\hat{\theta}_k\) be the mean of the first \(n_k\) observations, with \(SE_k=\sigma/\sqrt{n_k}\). For illustration, suppose the trial stops for efficacy at the first look if the standardized statistic for testing \(H_0:\theta=0\) is at least a positive boundary \(c\); otherwise it continues to \(n_2\).

$$ W_k=\frac{\hat{\theta}_k}{SE_k},\qquad \text{stop at look 1 if }W_1\geq c. $$

At whichever look supplies the reported estimate, suppose the ordinary two-sided 95% interval is used:

$$ C_k=\left[\hat{\theta}_k-1.96\,SE_k,\ \hat{\theta}_k+1.96\,SE_k\right]. $$

Under the fixed-sample normal model, \((\hat{\theta}_k-\theta)/SE_k\sim N(0,1)\), giving the interval its familiar nominal coverage. For the calculations below, take the true value to be \(\theta=0\). Under this value, \(W_1\) and \(W_2\) are standard normal, and their correlation is \(\rho=\sqrt{n_1/n_2}\). This follows because the first \(n_1\) observations are included in both stage estimates; the later observations are independent of them. Thus the two stage statistics are not independent.

Worked Example: Coverage Among Early-Stopped Trials

Suppose the first look occurs at \(n_1=100\), the known standard error is \(SE_1=0.20\) units, and the efficacy boundary is \(c=1.5\). At this look the trial stops if \(W_1\geq1.5\). Calculate the coverage of the ordinary 95% interval among trials that stop at this look, when the true effect is \(\theta=0\).

Under the stated normal model, \(W_1\sim N(0,1)\). The interval contains the true value exactly when \(|W_1|\leq1.96\). Given that the trial stops, \(W_1\geq1.5\), so the conditional coverage is

$$ P(-1.96\leq W_1\leq1.96\mid W_1\geq1.5) = \frac{\Phi(1.96)-\Phi(1.5)}{1-\Phi(1.5)}. $$

Using \(\Phi(1.96)\approx0.97500\), \(\Phi(1.5)\approx0.93319\), and \(1-\Phi(1.5)\approx0.06681\), the numerator is about \(0.04181\). Therefore,

$$ \frac{0.04181}{0.06681}\approx0.6258. $$

The ordinary interval covers the true effect in only about 62.6% of trials that stop at this look, not 95%. For a hypothetical observed \(W_1=1.70\), the estimate would be \(1.70(0.20)=0.340\) units, and the interval would be \(0.340\pm1.96(0.20)\), or approximately \([-0.052,\ 0.732]\) units. This particular interval happens to include zero, but the coverage calculation concerns the long-run proportion of such intervals that include the true value among early-stopped trials.

Why the Conditional Coverage Changes

The selection event \(W_1\geq c\) removes all estimates below the boundary from the early-stopped group. Under the null, an estimate from that group is therefore biased upward as a selected value, even though the unselected stage-one estimator is unbiased. The standardized estimation error is no longer distributed as an unrestricted standard normal within the selected group.

The calculation above can be justified directly from the definition of conditional probability. Under \(\theta=0\), interval coverage is \(-1.96\leq W_1\leq1.96\). Intersecting this event with the stopping event \(W_1\geq1.5\) leaves \(1.5\leq W_1\leq1.96\). Dividing the probability of that intersection by the probability of stopping gives the conditional coverage. In general, for \(0\leq c<1.96\),

$$ P(\theta\in C_1\mid W_1\geq c,\theta=0) = \frac{\Phi(1.96)-\Phi(c)}{1-\Phi(c)}. $$

This expression also shows how the effect depends on the boundary. When \(c\) approaches 1.96 from below, the interval-compatible part of the stopping region shrinks toward zero. For \(c>1.96\), every early-stopped interval excludes zero. At \(c=1.96\), the interval includes zero only when \(W_1=1.96\), an event of probability zero under the continuous normal model; thus conditional coverage is zero for all \(c\geq1.96\). This is not a paradox: stopping for a sufficiently strong positive result selects estimates outside the interval’s null-containing range.

Worked Example: A More Stringent Stopping Boundary

Keep the same known-variance model and ordinary two-sided 95% interval, but set the efficacy boundary to \(c=2.0\). Under the true value \(\theta=0\), determine the early-stop coverage.

An early stop requires \(W_1\geq2.0\). The interval contains zero only if \(|W_1|\leq1.96\). These events cannot occur together, since \(W_1\geq2.0\) implies \(W_1>1.96\). Hence

$$ P(\theta\in C_1\mid W_1\geq2.0,\theta=0)=0. $$

For example, if \(SE_1=0.20\) units and the observed \(W_1=2.10\), then \(\hat{\theta}_1=2.10(0.20)=0.420\) units. The ordinary interval is \(0.420\pm1.96(0.20)\), or approximately \([0.028,\ 0.812]\) units, and excludes the true value zero. The calculation is consistent with the coverage result: every estimate meeting this stopping boundary produces an interval that excludes zero.

Coverage Across Both Looks

Coverage among early-stopped trials is not the same quantity as coverage across all trials. In the two-look design, some trials report \(C_1\) after crossing the boundary and the rest report \(C_2\). The overall coverage probability adds the coverage probability on each disjoint stopping path. This overall probability can also differ from 95%, although it need not differ by as much as the coverage conditional on early stopping.

For a concrete calculation, take \(n_1=100\), \(n_2=200\), \(c=1.5\), and \(\theta=0\). Then \(\rho=\sqrt{100/200}=0.7071\), and the standard deviation of either stage statistic is 1. The first-look coverage contribution is \(P(1.5\leq W_1\leq1.96)=\Phi(1.96)-\Phi(1.5)\). If the trial continues, it covers when \(-1.96\leq W_2\leq1.96\) and \(W_1<1.5\). Since the conditional distribution of \(W_2\) given \(W_1=x\) is normal with mean \(\rho x\) and variance \(1-\rho^2\), overall coverage is

$$ \begin{aligned} P(\theta\in C_{\mathrm{reported}}\mid\theta=0) ={}&\Phi(1.96)-\Phi(1.5)\\ &+\int_{-\infty}^{1.5}\phi(x) \left[ \Phi\!\left(\frac{1.96-\rho x}{\sqrt{1-\rho^2}}\right) -\Phi\!\left(\frac{-1.96-\rho x}{\sqrt{1-\rho^2}}\right) \right]dx. \end{aligned} $$

Here \(\phi\) and \(\Phi\) are the standard normal density and cumulative distribution function. Numerical evaluation of the integral with \(\rho=0.7071\) gives overall coverage of approximately \(0.9394\). Thus the reported ordinary interval covers in about 93.9% of all trials under this design and null value, rather than 95%. This is distinct from the approximately 62.6% coverage among early-stopped trials: the overall result averages over both stopping paths according to how often each occurs.

Worked Example: Checking the Two-Look Coverage Calculation

For the design just described, verify the contribution from the early-stop path and identify what the integral represents. This check is useful because errors often arise from treating the two looks as independent or from counting a path twice.

The early-stop contribution is

$$ P(1.5\leq W_1\leq1.96) \approx0.97500-0.93319=0.04181. $$

The integral is the probability of coverage at the second look while the first-look value was below 1.5. Its integrand multiplies the density of \(W_1=x\) by the conditional probability that \(W_2\) lies between \(-1.96\) and \(1.96\). The factor \(\sqrt{1-\rho^2}\) is the conditional standard deviation, not the marginal standard deviation of \(W_2\), which remains 1. Adding the early-stop probability and this continuation-path probability yields the overall value of approximately 0.9394. The two paths are disjoint and exhaustive: either \(W_1\geq1.5\), or \(W_1<1.5\).

Diagnostics, Alternatives, and Limitations

The central diagnostic is to write the stopping event and the interval coverage event separately, then check their overlap. If an analysis reports an ordinary interval after stopping, ask whether its nominal calibration was derived for the sampling plan actually used or only for a fixed sample size. In simulations, a direct diagnostic is to repeatedly generate complete trials under a specified \(\theta\), apply the stopping rule, construct the reported intervals, and compute the fraction containing \(\theta\). The simulation must reproduce the planned looks, boundaries, estimator, and variance treatment.

This normal example assumes independent observations, a known common variance, prespecified looks and boundary, and no other adaptations that alter the data or reporting rule. Real designs may use estimated standard errors, nuisance parameters, discrete outcomes, or additional stopping rules. Those features change the calculation, but they do not make the fixed-sample coverage claim automatically valid. In particular, an estimated standard error addresses uncertainty in scale estimation; it does not by itself account for selection caused by stopping.

A stopping-aware interval is an alternative: its construction incorporates the stopping rule into the sampling distribution used for calibration. Another route is to invert tests whose rejection probabilities are calibrated under that rule. These approaches require choices about the coverage target and can involve more computation. The ordinary fixed-sample interval remains useful as a transparent descriptive summary, but after data-dependent stopping its usual nominal coverage should not be asserted without justification. The precise construction and differing coverage targets require separate treatment.

  • Mistake: “The estimate is unbiased at each fixed look, so its ordinary interval must still have 95% coverage after stopping.” Correction: unbiasedness at a fixed look does not imply that the estimate conditional on a data-dependent stopping event has the same distribution.
  • Mistake: “The interval contains the observed estimate, so it covers the true effect.” Correction: an interval is centered on the estimate by construction; coverage is whether it contains the unknown parameter over repeated trials.
  • Mistake: “The early-stop result represents all trials.” Correction: coverage among early-stopped trials and coverage across all stopping paths are different probabilities and should be labelled separately.
Key takeaway: A fixed-sample interval’s nominal coverage is tied to its sampling plan. When a stopping rule selects the reported estimate, calculate coverage under that selection rule rather than assuming the fixed-sample percentage still applies.

Check Your Understanding

Answer each question using the normal model and stopping rule stated in the relevant question. Solutions are included below.

  1. Under \(\theta=0\), a one-look trial stops when \(W_1\geq1.96\) and reports \(\hat{\theta}_1\pm1.96\,SE_1\). What is the interval’s coverage among early-stopped trials?
  2. In a two-look design, why is \(W_1\) generally not independent of \(W_2\)?
  3. For the two-look example, what event does the integral over \(x<1.5\) represent?
  4. If the observed estimate is \(0.30\) units and its standard error is \(0.10\) units, calculate the ordinary 95% interval. Does that calculation alone establish post-stopping coverage?

Solutions. (1) Zero: stopping requires \(W_1\geq1.96\), while coverage of zero requires \(|W_1|\leq1.96\); their intersection has probability zero for a continuous normal variable. (2) The first \(n_1\) observations contribute to both stage estimates, so the statistics share data; under the stated model their correlation is \(\sqrt{n_1/n_2}\). (3) It represents coverage at look two on trials that did not stop at look one: \(W_1<1.5\) and \(-1.96\leq W_2\leq1.96\). (4) The interval is \(0.30\pm1.96(0.10)=[0.104,\ 0.496]\) units. This arithmetic gives the conventional interval, but does not establish its coverage after a stopping rule selects the estimate.

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