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Estimation and Confidence Intervals After Adaptation · Tutorial 287 of 799

Intervals After Sample-Size Re-Estimation

Learn to distinguish nuisance-only re-estimation from effect-driven adaptation and to evaluate intervals using the trial’s actual sample-size rule.

Advanced 12 min read

What You'll Learn

  • Identify when a sample-size rule can affect confidence-interval coverage.
  • Derive the conditions under which a pooled normal interval retains its nominal coverage.
  • Construct an interval from independent post-interim data with conditional coverage.
  • Evaluate an adaptive procedure’s coverage by simulation or test inversion.
  • Diagnose pitfalls in interpreting a nominal confidence level after re-estimation.

Why the Re-Estimation Rule Matters

A trial may use interim information to revise its planned sample size. The revision might depend only on a nuisance quantity, such as a blinded estimate of outcome variability, or it might depend on an interim treatment-effect estimate. Those rules have different implications for inference. If the decision uses information related to the effect estimate, the final estimator and its standard error can be statistically linked to the decision that determined how much additional information was collected.

A confidence interval’s coverage is the probability, over repeated trials run under the specified design, that the interval contains the true parameter. The relevant design includes the rule that maps interim data to the final sample size. A formula that would have nominal coverage at a fixed sample size does not automatically retain that coverage after an effect-dependent sample-size decision.

Definition: A sample-size re-estimation rule is a prespecified function that uses interim data to determine the additional sample size. The rule is effect-independent for a particular inference problem if, under the model, the information used to make the decision is independent of the treatment-effect estimator used for that inference. An effect-dependent rule uses interim treatment-effect information and generally requires coverage to be evaluated under the full adaptive design.

A Normal-Outcome Model and Its Implications

For clarity, consider a two-arm trial with a continuous outcome. In each stage, participants are randomized equally, and observations are independent normal variables with common, known variance \(\sigma^2\). Let \(\theta=\mu_T-\mu_C\) be the treatment-minus-control mean difference. If stage \(j\) enrolls \(n_j\) participants per arm, the stage treatment-effect estimate \(D_j\) has distribution \(D_j\sim N(\theta,2\sigma^2/n_j)\). Assume the second-stage sample size \(n_2\) is determined by stage 1, and that stage 2 consists of new participants whose outcomes are independent of the interim data, conditional on the selected sample size.

With equal allocation and common variance, the pooled treatment-effect estimate is \(\hat{\theta}=(n_1D_1+n_2D_2)/(n_1+n_2)\). At a fixed total sample size per arm \(n_1+n_2\), its standard error is \(\sigma\sqrt{2/(n_1+n_2)}\), and the usual normal interval uses the critical value 1.96 for a nominal 95% interval. But when \(n_2\) depends on \(D_1\), the weights in \(\hat{\theta}\) depend on the interim estimate. The standardized pooled estimate need not have a standard normal distribution across repeated adaptive trials.

$$ Z_{\mathrm{pool}}= \frac{n_1(D_1-\theta)+n_2(D_2-\theta)} {\sigma\sqrt{2(n_1+n_2)}}. $$

At a fixed \(n_2\), this is standard normal under the stated model. If \(n_2\) is a function of \(D_1\), however, that fixed-sample argument does not establish that \(Z_{\mathrm{pool}}\) is standard normal. The distribution must account for how often each sample size is selected and for the interim estimates that lead to each choice.

Worked Example: Calculating a Naive Pooled Interval

Suppose a hypothetical trial enrolls 20 participants per arm in stage 1 and observes a treatment-effect estimate of 2.4 points. Following the interim decision, it enrolls 40 per arm in stage 2, where the estimated effect is 0.6 points. Assume the outcome standard deviation is known to be 5 points. Calculate the ordinary pooled interval, and identify what it does and does not establish.

The pooled estimate weights each stage by its participants per arm:

$$ \hat{\theta} =\frac{20(2.4)+40(0.6)}{20+40} =\frac{48+24}{60} =1.20\text{ points}. $$

The plug-in standard error formula, using the final total of 60 participants per arm, gives

$$ SE=5\sqrt{\frac{1}{60}+\frac{1}{60}} =5\sqrt{\frac{2}{60}} \approx 0.9129\text{ points}. $$

Thus the ordinary 95% interval is \(1.20\pm1.96(0.9129)=1.20\pm1.789\), or approximately \((-0.589,\,2.989)\) points. This is the interval produced by the fixed-sample normal formula. Its calculation is correct, but the calculation alone does not show that the interval has 95% coverage when the choice of 40 second-stage participants was driven by interim treatment-effect data.

When the Ordinary Interval Can Be Valid

Sample-size adaptation does not invariably invalidate a conventional interval. Under the normal model above, suppose a blinded variance estimate is independent of the treatment-effect contrast, and the re-estimation rule uses only that estimate. If the resulting sample size is therefore independent of the treatment-effect score, then, conditional on the selected sample size, the pooled estimator has its usual normal distribution and the stated standard error. Its conditional coverage is 95%; averaging that conditional coverage over the possible sample sizes also gives 95%.

This argument depends on the model and on independence, not merely on calling a rule “blinded.” The exact result above assumes known variance and a sample-size decision independent of the effect contrast. With estimated variance, a nonnormal outcome, a rule that uses unblinded effect information, or other dependencies introduced by the design, the fixed-sample formula needs a separate justification. In particular, estimated nuisance parameters can change the distribution of the standardized statistic even when the treatment effect is not examined.

Key condition: A fixed-sample interval can be carried through re-estimation with its usual conditional coverage when, for each possible selected sample size, the inference statistic has its stated fixed-sample distribution conditional on the re-estimation information. If that condition is not justified, evaluate coverage under the complete rule rather than assuming it.

Worked Example: An Interval Based on Stage 2 Alone

Use the hypothetical data from the previous example, but suppose the second-stage estimate is 0.6 points from 40 participants per arm. Construct an interval using only the second-stage data. Assume the stage-2 sample size was selected using stage-1 data, the stage-2 participants are new and independent, and \(\sigma=5\) points is known.

Conditional on the stage-1 data and the selected value \(n_2=40\), the second-stage contrast still has distribution \(D_2\sim N(\theta,2\sigma^2/40)\). The selection rule does not use the outcomes of these new participants. Therefore its standard error and 95% interval are

$$ SE_2=5\sqrt{\frac{2}{40}}\approx1.118\text{ points}, \qquad D_2\pm1.96SE_2 =0.60\pm2.191 =(-1.591,\,2.791)\text{ points}. $$

Under the assumptions stated, this interval has 95% coverage conditional on the interim data and the selected stage-2 sample size. Consequently, its coverage averaged over the interim data is also 95%. It is wider than the pooled interval in the previous example because it discards stage-1 information. This illustrates a validity–precision trade-off, not a universal recommendation to omit the first stage.

Evaluating Coverage Under an Effect-Dependent Rule

For an effect-dependent rule, a direct way to assess an interval procedure is to reproduce the complete trial repeatedly under a specified true value of \(\theta\). Each simulated trial must generate stage-1 outcomes, apply the sample-size rule exactly, generate the corresponding stage-2 outcomes, calculate the interval, and record whether that interval contains the true value. If \(M\) trials are simulated and \(K\) intervals cover \(\theta\), the estimated coverage is \(\hat{C}=K/M\). Its approximate Monte Carlo standard error is \(\sqrt{\hat{C}(1-\hat{C})/M}\). Simulation estimates coverage; it does not turn an uncalibrated interval into a calibrated one.

For example, suppose a hypothetical rule enrolls 20 participants per arm in stage 1, then chooses 60 per arm in stage 2 if the interim standardized effect is between 0.5 and 2.5, and otherwise chooses 20 per arm. Under the known-variance normal model, simulate both stage contrasts from their distributions at each chosen sample size, apply this rule, and compute the pooled interval for every replicate. A hypothetical run with \(M=100{,}000\) trials and \(K=93{,}700\) covering intervals gives \(\hat{C}=93{,}700/100{,}000=0.9370\). The estimated coverage is 93.70%, not 95%. Its Monte Carlo standard error is approximately \(\sqrt{0.9370(0.0630)/100{,}000}=0.000768\), or 0.0768 percentage points. The shortfall is much larger than this simulation error. These hypothetical simulation results would be evidence that the ordinary pooled interval is poorly calibrated for that rule; a single run should still be checked for correct implementation and adequate precision.

Simulation should be repeated over parameter values of interest, especially values near the null and values at which the sample-size rule changes. Coverage can vary with \(\theta\), so a result at one value is not a general coverage guarantee. Useful diagnostics include plotting estimated coverage against \(\theta\), checking that the simulated frequencies of each final sample size match the rule, and comparing the simulated distribution of the interval’s standardized statistic with its assumed reference distribution.

Intervals That Account for the Rule

One general method is test inversion. For each candidate value \(\theta_0\), specify a test whose null distribution is calculated under \(\theta=\theta_0\) while applying the actual sample-size rule. Choose the test’s rejection threshold so its rejection probability is at most \(\alpha\) under that adaptive design. The confidence set contains the candidate values not rejected. If the tests have size \(\alpha\) for every candidate value, this inversion gives coverage at least \(1-\alpha\), up to simulation or numerical error. The resulting set may not be symmetric around the pooled estimate, and in some designs it need not be a single interval; taking its interval hull preserves coverage but can make it more conservative.

A useful computational outline is:

1
Specify a candidate effect.
Set \(\theta=\theta_0\) and use the assumed outcome model, allocation, and interim rule.
2
Generate the adaptive null distribution.
Simulate or calculate interim data, apply the re-estimation rule, and complete the trial at the resulting sample size.
3
Calibrate the test.
Choose a rejection region with null rejection probability no greater than \(\alpha\), accounting for the adaptive design.
4
Invert over candidate effects.
Retain values not rejected and report the resulting confidence set, with numerical uncertainty and any non-interval shape made clear.

This approach can use both stages, unlike the stage-2-only interval, but it requires a correctly specified rule and model and can be computationally demanding. If the rule was not prespecified or its implementation is uncertain, coverage calculations are correspondingly less reliable. A procedure calibrated for one sample-size rule should not be assumed valid for a modified rule.

Common Pitfalls and Practical Limits

  • Assuming the final sample size is fixed after observing it. Computing a familiar interval with the realized sample size ignores how that size was selected. Condition on the decision only when the conditional distribution justifies it; otherwise account for the selection rule.
  • Treating “blinded” as a proof of independence. State which interim statistic drives the decision and why it is independent of the inference statistic under the model. Blinding by itself is not a mathematical coverage argument.
  • Equating an unbiased estimate with a valid interval. Even if an estimator has expectation \(\theta\), its standardized distribution may not be the assumed normal distribution. Coverage depends on the interval’s full sampling distribution.
  • Reporting simulation coverage without its uncertainty. A simulated proportion is an estimate. Report the number of replicates and its Monte Carlo precision, and check the range of effects relevant to the trial.

No interval method is uniformly best. A conventional pooled interval is simple and can be appropriate when its distributional conditions hold. An interval based on independent post-interim data can offer transparent conditional validity but may sacrifice precision. Test inversion can incorporate all stages and the exact adaptation rule, at the cost of modeling and computation. The choice should match the rule actually used, the estimand, and the desired coverage guarantee.

Check Your Understanding

Answer each question using the assumptions and distinctions developed above.

  1. A blinded nuisance estimate determines the second-stage sample size. Is a fixed-sample interval automatically valid? State the needed condition.
  2. Why can a pooled estimator’s standard error formula be correct for the realized sample sizes while its nominal coverage is still unsupported?
  3. For a stage-2-only interval, what feature of the design supports conditional coverage?
  4. In a simulation with 50,000 replicates, 46,500 intervals cover the true effect. Calculate estimated coverage and approximate Monte Carlo standard error.

Solutions. (1) No. Conditional on the re-estimation information and selected size, the inference statistic must have its stated fixed-sample distribution; independence of the effect score from the decision is one way to establish this in the known-variance normal model. (2) The formula describes the fixed-size variance, but effect-dependent selection can change the sampling distribution of the standardized pooled estimate. (3) New stage-2 outcomes are independent of the stage-1 data that selected \(n_2\), so conditional on the selection the stage-2 contrast retains its assumed normal distribution. (4) Coverage is \(46{,}500/50{,}000=0.9300\), or 93.00%. The approximate Monte Carlo standard error is \(\sqrt{0.93(0.07)/50{,}000}\approx0.001141\), or 0.1141 percentage points.

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