Why One Interim Fraction Can Be Misleading
In an adaptive enrichment trial, the biomarker-positive and biomarker-negative strata may accrue at different rates. An interim analysis scheduled after half of the planned total enrollment therefore need not occur halfway through the information available in either stratum. If the enrollment mix is changing, one stratum may already have a precise treatment estimate while the other remains imprecise. The timing of an enrichment decision should be described in terms of the evidence available for the population being considered, not just the trial’s calendar or total enrollment.
An information fraction compares the information available at an interim analysis with the information planned at the final analysis for the same estimand. It is specific to the estimand and its variance. In an enrichment trial, this means calculating a separate fraction for each marker stratum when separate subgroup treatment effects inform the decision.
Information is not a count of participants. It is the inverse variance of an estimator, so it depends on sample size, outcome variability, treatment allocation, and the analysis. A large information fraction means the estimate is relatively precise compared with its planned final precision; it does not, by itself, mean that the observed treatment effect is large or convincing.
Information for a Two-Arm Treatment Contrast
Consider one marker stratum and a continuous outcome. Suppose the treatment and control groups are independent, have a common known outcome variance \(\sigma_s^2\), and are allocated equally. Let \(n_s\) be the total number of participants with usable outcomes in that stratum across both arms. The estimator \(\hat{\theta}_s\) is the treatment-arm sample mean minus the control-arm sample mean. With \(n_s/2\) participants per arm, $$ \operatorname{Var}(\hat{\theta}_s) =\frac{\sigma_s^2}{n_s/2}+\frac{\sigma_s^2}{n_s/2} =\frac{4\sigma_s^2}{n_s}. $$ Thus the information is \(I_s=n_s/(4\sigma_s^2)\). If the variance and allocation remain stable from interim to final analysis, then $$ t_s=\frac{n_s(t)}{n_s(T)}. $$ This equality is convenient but conditional: it holds because the same variance and allocation factors cancel in the ratio. A participant count is not a reliable information proxy if those conditions change materially.
For unequal treatment allocation, with \(n_{1s}\) assigned to treatment and \(n_{0s}\) to control, the same independent, common-variance model gives $$ \operatorname{Var}(\hat{\theta}_s)=\sigma_s^2\left(\frac{1}{n_{1s}}+\frac{1}{n_{0s}}\right), \qquad I_s=\frac{1}{\sigma_s^2(1/n_{1s}+1/n_{0s})}. $$ The information fraction must then be computed from the interim and final variances, rather than by taking the ratio of total sample sizes automatically.
These formulas assume that outcomes are independent across participants, that the treatment contrast is defined consistently at both analyses, and that the variance model is appropriate. In practice, \(\sigma_s^2\) is usually estimated rather than known. The resulting information is then an estimate, and the design should use a prespecified, appropriate variance estimator. If missing outcomes or unavailable biomarker results reduce the number of analyzable participants, the relevant counts are the participants contributing to the estimator, not simply those enrolled.
Worked Example: Unequal Information Fractions at One Interim Look
Suppose a hypothetical trial plans to enroll 240 biomarker-positive and 360 biomarker-negative participants, with equal randomization within each stratum. Assume known outcome standard deviations of 10 points in the positive stratum and 10 points in the negative stratum. At an interim analysis, 120 positive and 90 negative participants have usable outcomes. Calculate both information fractions.
For the positive stratum, the interim and planned information are $$ I_+(t)=\frac{120}{4(10^2)}=\frac{120}{400}=0.3000, \qquad I_+(T)=\frac{240}{400}=0.6000. $$ Therefore, $$ t_+=\frac{0.3000}{0.6000}=0.5000. $$ Half of the planned information for the positive-stratum contrast is available.
For the negative stratum, $$ I_-(t)=\frac{90}{400}=0.2250, \qquad I_-(T)=\frac{360}{400}=0.9000, \qquad t_-=\frac{0.2250}{0.9000}=0.2500. $$ Only one quarter of the planned information for the negative-stratum contrast is available. Although these strata have the same assumed standard deviation, the positive stratum has accrued a larger fraction of its planned sample size. A single overall label such as “interim at 50%” would conceal this difference.
How Information Relates to Interim Evidence
For a treatment contrast with standard error \(SE_s(t)\), information is \(I_s(t)=1/SE_s(t)^2\). Under the normal model above, the standardized estimate is \(Z_s(t)=\hat{\theta}_s(t)\sqrt{I_s(t)}\). Thus, holding the estimated effect fixed, more information means a smaller standard error and a larger absolute standardized estimate. It does not imply that the estimate itself will be larger, or that its direction will be stable as more data accrue.
The information fraction also describes how much precision has accumulated relative to the planned final analysis. Under stable variance and allocation, \(SE_s(t)=SE_s(T)/\sqrt{t_s}\). For example, at an information fraction of 0.25, the interim standard error is twice the planned final standard error, because \(1/\sqrt{0.25}=2\). This is a statement about precision, not a forecast that the observed effect or eventual decision will remain unchanged.
Worked Example: Accrual Mix and Standardized Estimates
In a hypothetical trial, the planned totals are 200 positive and 400 negative participants. Assume equal allocation and known standard deviations of 8 points in the positive stratum and 12 points in the negative stratum. At a single interim analysis, 90 positive and 210 negative participants have usable outcomes. Suppose the positive-stratum treatment and control means are 1.0 and \(-2.2\) points, respectively; the corresponding negative-stratum means are 0.8 and \(-0.7\) points. Calculate the information fractions and standardized estimates.
For the positive stratum, \(\hat{\theta}_+=1.0-(-2.2)=3.2\) points. Its information fraction is $$ t_+=\frac{90/(4\cdot 8^2)}{200/(4\cdot 8^2)} =\frac{90}{200}=0.4500. $$ Its standard error and standardized estimate are $$ SE_+(t)=\sqrt{\frac{4(8^2)}{90}}=\sqrt{\frac{256}{90}}=1.687\text{ points}, \qquad Z_+(t)=\frac{3.2}{1.687}=1.897. $$
For the negative stratum, \(\hat{\theta}_-=0.8-(-0.7)=1.5\) points, and $$ t_-=\frac{210/(4\cdot 12^2)}{400/(4\cdot 12^2)} =\frac{210}{400}=0.5250. $$ The standard error and standardized estimate are $$ SE_-(t)=\sqrt{\frac{4(12^2)}{210}}=\sqrt{\frac{576}{210}}=1.656\text{ points}, \qquad Z_-(t)=\frac{1.5}{1.656}=0.906. $$
The negative stratum has accumulated a slightly larger fraction of its planned information, but its standardized estimate is smaller. Information fraction describes relative precision, whereas \(Z_s(t)\) also depends on the observed effect. Neither calculation alone supplies an enrichment decision rule or accounts for repeated interim testing.
Calendar Timing and Planned Information
An interim analysis may be scheduled by calendar time, total enrollment, or a target information fraction. These choices are not interchangeable when marker strata accrue at different rates. A calendar date fixes when the data are reviewed, but not the amount of information available in each stratum. A total-enrollment target likewise does not determine either subgroup’s information fraction unless the enrollment mix and variance structure are stable and known.
The following calculation shows how a stratum-specific target can produce a late look in another stratum. It assumes steady accrual rates and the same equal-allocation, stable-variance model used above.
Worked Example: Scheduling by Positive-Stratum Information
Suppose a hypothetical trial plans 240 positive and 360 negative participants. Assume equal allocation, a standard deviation of 10 points in the positive stratum, and 15 points in the negative stratum. Positive participants accrue at 12 per month and negative participants at 28 per month. If the interim look is scheduled when the positive stratum reaches 60% of its planned information, find the corresponding negative-stratum fraction and total-enrollment fraction.
With stable variances and allocations, reaching 60% of positive-stratum information requires 60% of its planned participants: $$ n_+(t)=0.60(240)=144. $$ At 12 positive participants per month, this takes \(144/12=12\) months. By then, the negative stratum has accrued \(28(12)=336\) participants. Its information fraction is $$ t_-=\frac{336/(4\cdot 15^2)}{360/(4\cdot 15^2)} =\frac{336}{360}=0.9333. $$ The total enrollment fraction is $$ \frac{144+336}{240+360}=\frac{480}{600}=0.8000. $$
At the same calendar-time look, the positive stratum is at 60% information, the negative stratum is at about 93% information, and total enrollment is at 80% of plan. These are three different summaries of the same interim timing. The negative stratum’s high fraction does not mean that the positive stratum is equally well informed, nor that the trial can make a valid decision about either stratum without an appropriate prespecified rule.
Calculation Checks and Common Failure Modes
A useful interim report should state each stratum’s information fraction alongside the counts and assumptions used to calculate it. Several checks can reveal whether the calculation or its interpretation is going wrong.
- Check the estimand and denominator. The interim and final information must concern the same treatment contrast in the same marker stratum. Changing the target population or analysis makes a simple ratio difficult to interpret.
- Check analyzable arm counts. In a two-arm contrast, information depends on both treatment and control sample sizes. A total count can hide an allocation imbalance that reduces precision.
- Check the variance assumption. If outcome variability differs by stratum, use the appropriate variance in each stratum. If it changes over time or differs between arms, derive information from the actual variance of the estimator rather than from the count ratio.
- Check whether the fraction is plausible. With fixed variance and allocation and no more than the planned number of participants, the fraction should lie between 0 and 1. A value above 1 may be possible if more information than planned has accumulated, but it should prompt a check of the plan, variance estimate, and calculation.
A common mistake is to say, “The trial has enrolled 50% of its planned total, so both subgroups are at 50% information.” That conclusion does not follow: the subgroup counts may be very different proportions of their own targets, and their variances may differ. The correction is to calculate interim and planned information separately for each subgroup-specific estimand, then form the ratio within each stratum.
Another mistake is to treat a high information fraction as strong evidence for treatment benefit. Information quantifies precision; evidence about the effect also depends on the observed estimate and the inferential procedure. Information fractions do not specify stopping boundaries, preserve error rates under repeated looks, or resolve the consequences of selecting a population based on interim data. Those require separate design and analysis provisions.
Choosing an Information Summary
For subgroup enrichment decisions, the primary summary is generally the information fraction for the treatment effect in the stratum under consideration. A calendar-time fraction is useful for operational planning, and a total-enrollment fraction is useful for describing recruitment progress, but neither substitutes for subgroup-specific information.
A combined information fraction can be formed only when it corresponds to a clearly defined combined estimand and model. For example, under the strong assumption of one common treatment effect across two independent strata, their information for that common effect can add. When treatment effects may differ by stratum, as is often central to enrichment, a single combined fraction can obscure the separate evidence. Report the stratum-specific fractions rather than taking an unweighted average of them.
Information fractions also have limitations. Their accuracy depends on the variance model and on the consistency of the estimator between interim and final analyses. They cannot guarantee that the interim effect estimate is stable, that one subgroup is truly more responsive, or that an adaptive decision is valid. Their role is narrower and essential: they make the amount of subgroup-specific precision at an interim look explicit, so decision rules can be evaluated against the evidence actually available.
Check Your Understanding
Answer each question, then compare with the solutions below.
- A balanced two-arm subgroup has a planned total of 160 participants and an interim total of 80. With stable variance and allocation, what is its information fraction?
- A subgroup plans 300 participants with outcome standard deviation 10 points. At interim, 120 participants have usable outcomes. Under the equal-allocation model, calculate interim information, planned information, and the information fraction.
- At an interim look, one subgroup has accrued 40% of its planned sample and another has accrued 70%. Can the first subgroup’s information fraction be declared 40% without further assumptions?
- Explain why a high information fraction does not, by itself, establish treatment benefit.
Solutions. (1) Under stable variance and allocation, \(t=80/160=0.5000\). (2) Interim information is \(120/[4(10^2)]=0.3000\); planned information is \(300/400=0.7500\); the fraction is \(0.3000/0.7500=0.4000\). (3) Not necessarily. The 40% count fraction equals the information fraction only if the relevant variance and allocation factors are stable between interim and final analyses; otherwise calculate the ratio using the estimator variances. (4) Information measures precision, not the direction or size of the observed effect. The estimate and a valid inferential procedure are also needed to assess evidence about benefit.