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Seamless Phase 2/3 Designs · Tutorial 497 of 799

Time-to-Event Example With Interim Treatment Choice

Work through an event-driven survival trial that selects a treatment at interim and combines stagewise evidence using a multiplicity-adjusted inverse-normal test.

Advanced 12 min read

What You'll Learn

  • Define stagewise logrank statistics for an event-driven seamless survival trial.
  • Distinguish treatment selection from confirmatory testing.
  • Apply a multiplicity adjustment to stage 1 p-values after selection.
  • Calculate an inverse-normal combination statistic and its p-value.
  • Justify familywise type I error control across candidate treatments.
  • Identify assumptions and diagnostics that matter for survival evidence.

Why Treatment Selection Complicates Survival Evidence

In a seamless phase 2/3 trial, phase 2 results may determine which treatment continues into phase 3. For a time-to-event endpoint, this decision can use interim survival information, such as estimated hazard ratios or logrank statistics. The selected treatment's phase 2 result is then not an ordinary confirmatory result: it helped determine which treatment would be carried forward. Combining that selected result with phase 3 evidence without accounting for selection can inflate the chance of a false positive.

The design below uses event-driven stages: each stage is analyzed after a prespecified target number of events has accrued. It selects one of two active treatments after stage 1, then combines stagewise survival evidence for the selected treatment. The construction uses a conservative multiplicity adjustment and a prespecified inverse-normal combination test. It is one transparent way to preserve type I error control; it is not the only possible valid design.

Definition: Let \(K\) be the number of candidate treatments. For candidate \(j\), let \(H_{0j}\) state that its survival distribution is no better than control, against a one-sided alternative \(H_{aj}\) of longer survival. A stagewise p-value is a p-value computed from the data accrued in one stage alone. An inverse-normal combination test transforms stagewise p-values to normal scores and combines them using prespecified weights.

Stagewise Logrank Evidence

For each candidate treatment, a logrank score compares observed and expected event counts over the risk sets formed during follow-up. Use the sign convention that a positive score favors treatment: \(U\) is the number of events expected in the control group minus the number observed there, accumulated with the usual risk-set weights. Let \(V\) be the estimated variance of \(U\) under the null. The stagewise statistic is \(Z=U/\sqrt{V}\), and the one-sided p-value for benefit is \(p=1-\Phi(Z)\).

The standard large-sample calibration assumes randomized treatment allocation, independent participants across stages, and noninformative censoring conditional on the modeled factors. The logrank comparison is most naturally interpreted under proportional hazards: the treatment-to-control hazard ratio is approximately constant over time. Under the null and regularity conditions, the standardized score is approximately standard normal. The p-value is therefore an approximation unless a design uses a suitable exact or randomization-based calibration.

In an event-driven design, the event target is set before the interim decision, and the analysis is triggered by reaching that target rather than by a fixed calendar date. Event count alone does not determine the combination-test weights: weights should reflect planned stage information. The calculations below assume equal planned logrank information in the two stages, giving weights \(1/\sqrt{2}\) for both stages.

Worked Example: Selecting a Treatment From Stage 1

Suppose a hypothetical trial compares treatments A and B with a shared control. At the stage 1 event target, 120 events have accrued. For this illustration, the stagewise logrank summaries are \(U_A=14.96\), \(V_A=100\), \(U_B=12.00\), and \(V_B=100\). The prespecified rule selects the treatment with the larger positive logrank score.

For A, the standardized statistic is

$$ Z_A=\frac{U_A}{\sqrt{V_A}}=\frac{14.96}{\sqrt{100}}=1.496. $$

Its one-sided p-value is \(1-\Phi(1.496)\), approximately 0.06733. For B, \(Z_B=12.00/10=1.200\), with one-sided p-value approximately 0.1151. A has the stronger stage 1 score, so the rule selects A. The shared control can make the two treatment statistics dependent; the selection rule does not require them to be independent.

Selection is not evidence that A is effective. It identifies the candidate for continuation. The phase 2 p-value for A must be handled in a way that accounts for the fact that A was chosen after examining both candidates.

Adjusting for Selection and Combining Stages

For each candidate \(j\), let \(p_{1j}\) be its stage 1 one-sided logrank p-value. Use the Bonferroni-adjusted stage 1 p-value \(p_{1j}^{*}=\min(1,Kp_{1j})\). This protects against selecting a candidate with an unusually favorable stage 1 result among \(K\) candidates. Transform it to the normal score \(Z_{1j}^{*}=\Phi^{-1}(1-p_{1j}^{*})\). For the selected candidate, let \(p_{2j}\) be its one-sided stage 2 p-value and \(Z_{2j}=\Phi^{-1}(1-p_{2j})\).

With prespecified nonnegative weights \(w_1,w_2\) satisfying \(w_1^2+w_2^2=1\), define the combination statistic for candidate \(j\) as follows. Equal planned stage information gives \(w_1=w_2=1/\sqrt{2}\).

$$ Z_{Cj}=w_1Z_{1j}^{*}+w_2Z_{2j}, \qquad p_{Cj}=1-\Phi(Z_{Cj}). $$

The adjustment and the combination answer different questions. Multiplying the stage 1 p-value by \(K\) accounts for searching across candidates at selection. The final decision rule below also controls the chance of making any false confirmatory claim across the candidate hypotheses. This deliberate conservatism gives a simple strong familywise error guarantee.

Worked Example: Combining the Selected Treatment's Survival Evidence

Continue the hypothetical example with \(K=2\), with A selected. The raw stage 1 p-value for A is approximately 0.0673268. Its adjusted p-value is \(p_{1A}^{*}=\min(1,2(0.0673268))=0.1346536\), which corresponds, to the displayed precision, to \(Z_{1A}^{*}=1.1047\). In stage 2, 120 more events accrue among newly randomized participants assigned to A or control. Suppose the logrank summaries for A are \(U_{2A}=11.00\) and \(V_{2A}=100\).

The stage 2 statistic and one-sided p-value are

$$ Z_{2A}=\frac{11.00}{\sqrt{100}}=1.100, \qquad p_{2A}=1-\Phi(1.100)=0.1357. $$

The equal-information inverse-normal statistic is

$$ Z_{CA}=\frac{1.1047+1.100}{\sqrt{2}}=1.5589. $$

Thus \(p_{CA}=1-\Phi(1.5589)\approx 0.05951\). This is the combination p-value for A under the specified adjusted procedure. At a familywise one-sided significance level \(\alpha=0.025\), the Bonferroni threshold for either of the two candidate claims is \(\alpha/K=0.0125\). Since \(0.05951>0.0125\), the trial does not establish a confirmatory benefit for A. The positive statistics favor A, but they do not meet the prespecified evidential threshold.

Why the Procedure Controls Type I Error

The validity argument has two parts. First, for any candidate \(j\) whose null hypothesis is true, assume its raw stage 1 p-value is valid: \(P(p_{1j}\leq t)\leq t\) for \(0\leq t\leq1\). Then \(p_{1j}^{*}=\min(1,Kp_{1j})\) is also a valid, though conservative, p-value. For \(0\leq u<1\),

$$ P(p_{1j}^{*}\leq u) =P(p_{1j}\leq u/K) \leq u/K \leq u. $$

Second, stage 2 enrolls new participants after selection, with its randomization and follow-up conducted independently of stage 1. Under \(H_{0j}\), its stage 2 p-value is valid conditional on the stage 1 data and the selected treatment. Consequently, the stage 2 normal score has a standard normal null distribution under the assumed calibration and is independent of the stage 1 score. Combining independent stagewise evidence with fixed weights whose squared sum is one yields a valid candidate-level combination p-value \(p_{Cj}\).

For a strong familywise guarantee, reject a selected candidate's null only when \(p_{Cj}\leq\alpha/K\). Let \(\mathcal{N}\) be the set of candidates whose null hypotheses are true. The probability of at least one false claim is bounded by

$$ P(\text{at least one false claim}) \leq \sum_{j\in\mathcal{N}}P(p_{Cj}\leq\alpha/K) \leq |\mathcal{N}|\frac{\alpha}{K} \leq \alpha. $$

This union-bound argument does not require independence between candidate treatments' stage 1 statistics, so it accommodates a shared control. It does require the candidate-level p-values to be valid under their respective nulls and the stage 2 evidence to be generated as specified. If stage 2 patients overlap with stage 1, or stage 2 analyses use data-dependent weights, the independence and calibration argument must be reconsidered.

1
Set the event targets and weights.
Prespecify the stage event targets, the information-based weights, the selection rule, the one-sided hypotheses, and the familywise level.
2
Select at the transition.
Compute stage 1 logrank results for the candidates and apply the prespecified rule. Do not describe the selected stage 1 result as an unadjusted confirmatory test.
3
Analyze new stage 2 participants.
Accrue the planned events for the selected treatment and control, then compute the stage 2 logrank p-value separately.
4
Apply the fixed combination rule.
Adjust the selected candidate's stage 1 p-value, transform both p-values, calculate the inverse-normal statistic, and compare its p-value with \(\alpha/K\).

Diagnostics, Alternatives, and Limitations

Before interpreting the result, check that each stage has the planned event information, that event and censoring rules were applied consistently, and that the stage 2 cohort is distinct from stage 1. Examine whether the proportional-hazards assumption is credible for the intended logrank interpretation. Strongly crossing survival curves or a visibly time-varying treatment effect can make a single logrank score a poor summary of the clinical effect. The earlier material on nonproportional hazards in seamless survival trials discusses that issue; here, the combination calculation assumes stagewise p-values calibrated for the chosen survival test.

The main alternative is to base the confirmatory claim on stage 2 alone. That approach avoids using selected phase 2 evidence in the test and can be simpler to explain, but discards potentially useful information and may require more stage 2 events. A combination test can use both stages, provided the selection adjustment, stagewise independence, and weights are all specified and justified. An ordinary pooled logrank test is not a safe shortcut: it does not by itself account for treatment selection, and a pooled statistic may not follow its usual reference distribution after an adaptive transition.

This construction is conservative: it adjusts stage 1 evidence for selection and then uses a Bonferroni threshold across candidate claims. Other valid adaptive procedures can be more powerful, but their error control depends on their own prespecified rules and assumptions. The weights must not be chosen after seeing which value produces the most favorable result. Nor does a significant combination p-value alone establish a clinically important effect; the estimated hazard ratio, survival curves, follow-up, and clinical context remain relevant.

Common Mistake

  • Wrong reasoning: “A was selected because its phase 2 result looked best, and the phase 3 p-value is independent, so I can combine the raw phase 2 p-value and phase 3 p-value as if A had been chosen in advance.”
  • Correction: Stage 2 independence does not undo the selection that favored A in stage 1. Adjust stage 1 evidence for searching across candidates, use the prespecified combination rule, and apply the stated multiplicity threshold. The validity argument depends on the complete rule, not on the fact that the final stage is independent.
Key takeaway: In a survival trial with interim treatment choice, define the event-driven stagewise tests and selection rule in advance. Use independent stage 2 data, valid selection-adjusted evidence, fixed combination weights, and an explicit multiplicity threshold before making a confirmatory claim.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. Two candidate treatments are considered. Why is the selected candidate's unadjusted stage 1 p-value not automatically a valid confirmatory p-value?
  2. With equal information in both stages, stagewise transformed scores are \(Z_1^{*}=0.80\) and \(Z_2=1.20\). Calculate the inverse-normal statistic.
  3. For \(K=3\) candidate treatments and \(\alpha=0.03\), what is the Bonferroni threshold for a candidate-level combination p-value?
  4. Name two assumptions or design features needed for the stagewise combination argument.

Solutions. (1) The candidate was chosen after examining multiple stage 1 results, so its selected result tends to be more favorable than an unselected result; selection must be accounted for. (2) \(Z_C=(0.80+1.20)/\sqrt{2}=2.00/\sqrt{2}=1.414\), with one-sided p-value approximately \(1-\Phi(1.414)=0.0787\). (3) The threshold is \(\alpha/K=0.03/3=0.0100\). (4) Examples include independent new participants in stage 2, valid stagewise p-values under each null, prespecified fixed weights, and suitable censoring and logrank model assumptions.

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