Why Stage Size Is Not the Same as Information
In a seamless trial, a stage’s contribution to the final analysis depends not only on how many participants were recruited, but also on how precisely the treatment effect can be estimated from their outcomes. A stage with fewer participants can contribute substantial information if its outcome is measured precisely or if many participants experience the event of interest. Conversely, recruitment can increase without information rising as much as expected when outcomes are missing or event rates are low.
Information gives a common scale for comparing stages and planning their contributions. It is also useful when choosing inverse-normal combination-test weights: under standard large-sample conditions, a stage’s information determines the precision of its treatment-effect estimate. The calculations below concern planned information; realized information can differ from plan.
For \(K\) planned stages with information increments \(I_1,\ldots,I_K\), the planned total is \(I_{\mathrm{total}}=\sum_{j=1}^{K} I_j\). The fraction reached after stage \(k\) is \(f_k=(\sum_{j=1}^{k} I_j)/I_{\mathrm{total}}\). The fraction assigned to stage \(j\) is \(q_j=I_j/I_{\mathrm{total}}\). Thus \(f_k\) is cumulative, whereas \(q_j\) describes one stage; these quantities should not be interchanged.
Information and Inverse-Normal Weights
Suppose each stage produces a one-sided stagewise \(p\)-value \(p_j\), converted to a standard normal score \(Z_j=\Phi^{-1}(1-p_j)\) in the direction favoring treatment. If the stagewise statistics have independent increments and similar large-sample behavior, a common information-based choice of inverse-normal weight is \(w_j=\sqrt{I_j/I_{\mathrm{total}}}\). The combined statistic is \(Z_{\mathrm{comb}}=\sum_j w_j Z_j\), with \(\sum_jw_j^2=1\). The square root matters: inverse-normal weights are not the information fractions themselves.
This relationship has a simple precision justification. If a treatment-effect estimate has variance \(1/I_j\), its standard error is \(1/\sqrt{I_j}\). Standardizing a common effect \(\theta\) therefore gives an expected stage statistic approximately \(\theta\sqrt{I_j}\). A normalized weighted sum that reflects this precision assigns weights proportional to \(\sqrt{I_j}\), not to \(I_j\).
This is a design rationale, not a claim that every valid combination test must use these weights. Inverse-normal tests can use other prespecified weights, and validity depends on the stagewise \(p\)-values, dependence structure, and adaptation rules. When information is estimated from accruing data, the design must also specify how the weights and analysis will preserve the intended testing properties.
Binary Outcomes: Information From a Difference in Proportions
Consider a two-arm comparison in stage \(j\), with \(n_{Tj}\) participants assigned to treatment and \(n_{Cj}\) assigned to control. Let \(\hat{p}_{Tj}\) and \(\hat{p}_{Cj}\) be the observed response proportions, and estimate the treatment effect by \(\hat{p}_{Tj}-\hat{p}_{Cj}\). For independent Bernoulli outcomes, the variance is
For planning a score test under the null hypothesis of equal response probabilities, a common null response probability \(p_0\) gives the approximate information \(I_j=1/\{p_0(1-p_0)(1/n_{Tj}+1/n_{Cj})\}\). A different variance model, such as a Wald calculation under assumed alternative rates, can yield a different planned information. State which model is used; the formula is not independent of that choice.
Worked Example: Binary Endpoint With Unequal Stage Sizes
Suppose a hypothetical trial plans a binary response endpoint, with equal allocation within each stage. For planning a null-based score analysis, assume \(p_0=0.30\). Stage 1 plans 100 participants per arm, and stage 2 plans 150 per arm. Assume independent participant outcomes, complete ascertainment, and the same response definition in both stages.
The null variance of the stage 1 difference in proportions is \(0.30(0.70)(1/100+1/100)=0.21(0.02)=0.004200\). Therefore \(I_1=1/0.004200=238.1\). For stage 2, the variance is \(0.21(1/150+1/150)=0.21(0.01333)=0.002800\), so \(I_2=1/0.002800=357.1\).
Total planned information is \(238.1+357.1=595.2\). Stage 1 contributes \(q_1=238.1/595.2=0.400\), and stage 2 contributes \(q_2=0.600\). The cumulative fraction at the end of stage 1 is 0.400. Corresponding information-based inverse-normal weights are \(w_1=\sqrt{0.400}=0.6325\) and \(w_2=\sqrt{0.600}=0.7746\).
Although stage 2 has 1.5 times as many participants per arm as stage 1, its information is also 1.5 times as large under these assumptions. That proportionality would not necessarily hold if allocation, response-rate assumptions, or outcome completeness differed between stages.
Continuous Outcomes: Information From the Mean Difference
For a continuous endpoint, let the treatment-effect estimate in stage \(j\) be the difference in sample means. If outcomes are independent, have common variance \(\sigma^2\) in the two arms, and the analysis uses that variance, then
The common-variance assumption is a planning simplification, not a requirement for all analyses. With unequal variances, use the variance of the estimator under the specified model. If baseline adjustment or another model changes the residual variance, use the variance appropriate to that planned analysis rather than applying the unadjusted formula automatically.
Worked Example: Continuous Endpoint and Equal Information Fractions
In a hypothetical trial, change in systolic blood pressure is measured in mmHg. Assume independent observations, equal allocation, a common standard deviation of 10 mmHg in both arms, and complete outcomes. Stage 1 plans 60 participants per arm; stage 2 plans 90 per arm.
For stage 1, the variance of the difference in means is \(10^2(1/60+1/60)=100(1/30)=3.333\) \((\mathrm{mmHg})^2\). Thus \(I_1=1/3.333=0.3000\) \((\mathrm{mmHg})^{-2}\). For stage 2, the variance is \(100(1/90+1/90)=100(1/45)=2.222\) \((\mathrm{mmHg})^2\), giving \(I_2=1/2.222=0.4500\) \((\mathrm{mmHg})^{-2}\).
The total information is \(0.7500\) \((\mathrm{mmHg})^{-2}\). The cumulative fraction after stage 1 is \(0.3000/0.7500=0.400\), and the stage 2 fraction is 0.600. The information-based weights are again 0.6325 and 0.7746. The units cancel in these fractions and weights; the information itself has inverse-squared outcome units.
This calculation assumes the stated standard deviation is a suitable planning value. If the observed variability is larger, the realized information will be lower than planned for the same sample sizes.
Time-to-Event Outcomes: Information From Events
For a log-rank comparison, information is driven primarily by events, rather than by the number randomized alone. Under proportional hazards and a suitable large-sample approximation, if a proportion \(r\) of the risk set is assigned to treatment, information is approximately \(D r(1-r)\), where \(D\) is the number of observed events. With balanced allocation, \(r=0.5\), so the approximation is \(D/4\).
This event-count formula assumes that allocation remains balanced among participants at risk, treatment groups have comparable follow-up and censoring patterns for the approximation, and the proportional-hazards log-rank framework is suitable. More generally, log-rank information accumulates across event times according to the treatment and control composition of each risk set. If those compositions vary materially, use the planned risk-set information rather than treating every event as contributing exactly one quarter.
Worked Example: Planning Survival Stage Sizes From Events
A hypothetical time-to-event trial plans equal randomization and uses a log-rank analysis under a proportional-hazards model. Stage 1 targets 80 events and stage 2 targets 120 events. Assume balanced risk sets, so expected information is approximated by one quarter of the event count.
Stage 1 has planned information \(I_1=80/4=20\); stage 2 has \(I_2=120/4=30\). The total is 50, so the cumulative information fraction after stage 1 is \(20/50=0.400\). The stage fractions are 0.400 and 0.600, giving inverse-normal weights \(\sqrt{0.400}=0.6325\) and \(\sqrt{0.600}=0.7746\).
Suppose the expected proportion of randomized participants with an observed event by the relevant cutoff is 0.80 in each stage. To target 80 stage 1 events, plan \(80/0.80=100\) total participants; to target 120 stage 2 events, plan \(120/0.80=150\). With equal allocation these are 50 and 75 participants per arm, respectively. These are expected recruitment requirements, not guarantees: the event count is random, and follow-up may need to continue until the target is reached.
From Information Targets to Stage Recruitment
For binary and continuous endpoints, information can often be translated directly into arm-specific sample sizes by rearranging the relevant variance formula. With equal allocation and a common variance, both examples above have information proportional to the number per arm. For survival outcomes, first specify a target number of events, then estimate recruitment and follow-up needed to observe those events. In each case, distinguish the number randomized from the number contributing usable outcome information.
A practical planning sequence is:
Identify the treatment contrast, endpoint, allocation, and variance assumptions that match the planned analysis.
Use stage-specific sample sizes and assumptions; for survival outcomes, use event and risk-set assumptions.
Divide cumulative and stage-specific information by the planned total, then take square roots for information-based inverse-normal weights.
Account for allocation, expected missing outcomes, event probability, and the possibility that actual information will differ from plan.
Diagnostics, Alternatives, and Limitations
Before finalizing a plan, check whether the assumptions that make information additive are plausible. Stage increments should be based on nonoverlapping observations or otherwise have their covariance accounted for. The endpoint definition and measurement scale should be comparable across stages. Check that planned allocation is reflected in the variance calculation, that assumed response rates or standard deviations are reasonable, and that event accrual assumptions fit the expected risk-set pattern.
A frequent error is to use the fraction of participants recruited as the information fraction. This is correct only in special cases, such as equal allocation, equal outcome variance, complete ascertainment, and information proportional to sample size. Another error is to use \(q_j\) rather than \(\sqrt{q_j}\) as the inverse-normal weight. For example, a stage contributing 40% of total information receives weight \(\sqrt{0.40}\), not 0.40, under the information-based convention.
| Planning summary | What it measures | When it can mislead |
|---|---|---|
| Recruitment fraction | Participants enrolled by a point divided by total planned enrollment | When allocation, outcome completeness, precision, or event rates differ across stages |
| Information fraction | Information accrued by a point divided by total planned information | When the assumed variance or event-information model is inaccurate |
| Event fraction | Observed events divided by planned events | When risk-set allocation differs from the approximation used to convert events to information |
A group-sequential design may express interim timing through cumulative information fractions because precision, rather than recruitment alone, governs the sampling variation of the treatment estimate. A combination test instead combines stagewise evidence using its chosen weights; information-based weights are one natural option, not the only valid one. Neither approach eliminates the need to prespecify the analysis and justify the assumptions. Information calculations are planning approximations, and departures in variance, missingness, allocation, or event patterns can alter realized information.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- A binary-outcome stage has 80 participants per arm and uses \(p_0=0.25\) for a null-based score calculation. Find its planned information.
- A continuous-outcome stage has 50 participants per arm and a common standard deviation of 8 units. Find its information and state the units.
- A balanced survival trial has 160 observed events under the \(D/4\) approximation. Find the information and explain one assumption behind the approximation.
- Two stages contribute 25 and 75 units of information. Find the cumulative information fraction after stage 1 and the information-based inverse-normal weights.
- Why might a stage enrolling 60% of the total participants contribute less than 60% of total information?
Solutions. (1) \(I=1/\{0.25(0.75)(1/80+1/80)\}=1/0.0046875=213.3\). (2) \(I=1/\{8^2(1/50+1/50)\}=1/2.56=0.3906\) units\(^{-2}\). (3) \(I=160/4=40\); the approximation assumes balanced allocation in the risk sets. (4) Total information is 100, so the stage 1 cumulative fraction is 0.25; weights are \(\sqrt{0.25}=0.5000\) and \(\sqrt{0.75}=0.8660\), whose squared values sum to 1. (5) Allocation, outcome variance, missing outcomes, or event rates may differ, so participant counts need not be proportional to estimator precision.