Translating Maximum Information Into Enrollment Requirements

From Maximum Information to Enrollment

A group sequential design is often planned around a maximum information target: the information expected to be available at the final analysis if the trial continues that far. Operational planning needs a different quantity: how many participants must be enrolled to make that information attainable? The conversion depends on the outcome model, allocation, and the fraction of enrolled participants whose outcomes will contribute to the analysis.

The target should be the maximum planned information, not the average sample size across simulated trial paths. Early stopping can make average enrollment smaller than the enrollment needed to reach the planned maximum information. For this conversion, use the information scale and assumptions that defined the maximum target; a target reported on one scale cannot safely be converted using a formula for another.

Definition: The maximum information target is the planned information available at the final analysis if the trial reaches that analysis. The enrollment requirement is the number of participants to assign to the trial, under stated planning assumptions, so that the target information is expected to be available. It is generally an approximate requirement, not a guarantee that a particular trial will attain the target.

The calculations below use the inverse-variance definition of information for an estimated treatment contrast: \(I=1/\operatorname{Var}(\hat{\theta})\), where \(\hat{\theta}\) is the treatment-effect estimator under the specified planning model. With binary outcomes, the information depends on the event probability used for planning. With a continuous outcome, it depends on the outcome variance. The assumed values are planning inputs, not facts about future trial data.

Binary Outcomes: Allocation and Analyzable Counts

Suppose a binary outcome is observed for \(n_T\) participants in the treatment arm and \(n_C\) in the control arm. For this planning approximation, assume independent participants, a common outcome probability \(p\) in both arms for the variance calculation, and a treatment-effect estimate based on the difference in sample proportions. The variance of that difference under the common-probability planning model is \(p(1-p)(1/n_T+1/n_C)\). Thus its information is the reciprocal of this variance.

I \approx \frac{1}{p(1-p)\left(\frac{1}{n_T}+\frac{1}{n_C}\right)}

If the total analyzable sample size is \(N=n_T+n_C\), and the treatment fraction is \(q=n_T/N\), then \(n_T=qN\) and \(n_C=(1-q)N\). Substitution gives the commonly used planning formula:

I \approx \frac{Nq(1-q)}{p(1-p)} \qquad\text{and hence}\qquad N \approx \frac{I\,p(1-p)}{q(1-q)}.

This formula assumes the specified allocation fraction applies to the analyzable participants and that the common-probability variance approximation is appropriate. Balanced allocation has \(q=0.5\), so \(q(1-q)=0.25\); for a fixed total analyzable count, this maximizes the allocation factor. If the treatment and control arms have different outcome-availability rates, the simple formula may not apply to total enrollment. Use the arm-specific analyzable counts instead.

Worked Example: Balanced Allocation and a Binary Outcome

Suppose a hypothetical trial has a maximum information target of 252.3. The planning event probability is \(p=0.20\), and allocation is balanced. Assume independent binary outcomes and use the common-probability variance approximation stated above. Find the required analyzable sample size.

Here \(q=0.50\), so \(q(1-q)=0.25\), and \(p(1-p)=0.20(0.80)=0.16\). Therefore:

N \approx \frac{252.3(0.16)}{0.25} =\frac{40.368}{0.25} =161.472.

The planned analyzable total is about 161.5 participants, or approximately 80.75 per arm in expectation. Checking the information calculation using the unrounded total:

I \approx \frac{161.472(0.25)}{0.16} =252.3.

Thus the maximum information target corresponds to about 161.5 analyzable participants under these assumptions. The fractional counts are planning expectations, not literal enrollment counts. A trial cannot enroll half a participant, and the actual information depends on realized outcome counts.

Unequal Outcome Availability Between Arms

When outcome availability differs by arm, calculate the information from the expected analyzable count in each arm. Let \(E\) be total enrollment, \(r_T\) and \(r_C\) be the planned enrollment fractions (\(r_T+r_C=1\)), and \(a_T\) and \(a_C\) be the probabilities that an enrolled participant in each arm contributes an outcome to the analysis. Under the same common-\(p\) binary planning model, the expected analyzable counts are \(n_T=E r_T a_T\) and \(n_C=E r_C a_C\). Substituting those counts into the variance formula yields:

I \approx \frac{1}{p(1-p)\left(\frac{1}{E r_Ta_T}+\frac{1}{E r_Ca_C}\right)} = \frac{E}{p(1-p)\left(\frac{1}{r_Ta_T}+\frac{1}{r_Ca_C}\right)}.

This is an expected-information approximation: it substitutes expected analyzable counts into the variance formula. It is not an exact statement about the distribution of information across possible trial outcomes. The availability probabilities should reflect all relevant reasons an outcome is unavailable by the analysis time. If dropout and missing outcome recording are modeled separately, combine them only when their definitions do not overlap.

Worked Example: Unequal Availability With a Binary Outcome

A hypothetical trial targets information \(I=180\), plans 40% of enrollment for treatment and 60% for control, and uses \(p=0.30\) for planning. Assume independent binary outcomes, the common-probability variance approximation, and outcome availability of 90% in treatment and 95% in control. Calculate the approximate total enrollment.

The expected analyzable counts for total enrollment \(E\) are \(n_T=0.40(0.90)E=0.36E\) and \(n_C=0.60(0.95)E=0.57E\). Since \(p(1-p)=0.30(0.70)=0.21\), the information is:

I \approx \frac{1}{0.21\left(\frac{1}{0.36E}+\frac{1}{0.57E}\right)} = \frac{E}{0.21\left(2.7778+1.7544\right)} \approx \frac{E}{0.9518}.

Set this expression equal to the target and solve for \(E\):

180 \approx \frac{E}{0.9518} \quad\Longrightarrow\quad E \approx 180(0.9518)=171.3.

A planning total of 172 participants is the smallest whole-number enrollment at or above this approximation. At \(E=171.3\), the expected analyzable counts are about \(0.36(171.3)=61.67\) in treatment and \(0.57(171.3)=97.64\) in control, for about 159.3 analyzable participants in total. The unequal allocation and unequal availability mean that the total alone does not determine information; both arm counts matter.

Dropout and Other Availability Assumptions

If a participant must remain in follow-up and have an outcome recorded to contribute information, the probability of contributing can be represented by a combined availability rate. For example, if the probability of remaining in follow-up is \(1-d\), and the conditional probability of having a recorded outcome among those retained is \(m\), then the combined availability is \(a=(1-d)m\). This multiplication is appropriate only when \(m\) is defined conditionally among retained participants. If the stated outcome-availability rate already includes dropout, multiplying by a separate dropout adjustment would count some losses twice.

For equal availability \(a\) in both arms under a binary-outcome model, expected analyzable total size is \(aE\). The balanced-allocation formula then gives \(E\approx I p(1-p)/[a q(1-q)]\). If rates differ by arm, use the arm-specific formula above. For continuous outcomes with a common planning standard deviation \(\sigma\), independent observations, and an estimator equal to the difference in arm means, the variance is \(\sigma^2(1/n_T+1/n_C)\). Its information is:

I=\frac{1}{\sigma^2\left(\frac{1}{n_T}+\frac{1}{n_C}\right)} =\frac{Nq(1-q)}{\sigma^2}.

The same allocation logic applies, but the variance input and units differ. A time-to-event design may instead define information through the number of observed events; converting an event target to enrollment then requires assumptions about event incidence, follow-up, and censoring. An event count is not interchangeable with an enrollment count.

Worked Example: Continuous Outcome With Dropout and Missing Outcomes

A hypothetical balanced trial of a continuous outcome targets maximum information \(I=0.64\) on the inverse-variance scale. The planning standard deviation is \(\sigma=10\) outcome points. Assume independent observations, a common standard deviation in both arms, and a difference-in-means estimator. Fifteen percent of enrolled participants are expected to drop out; among those retained, 90% are expected to have a recorded primary outcome. Estimate the enrollment required.

For balanced allocation, \(q(1-q)=0.25\). The required analyzable total is:

N \approx \frac{I\sigma^2}{q(1-q)} =\frac{0.64(10^2)}{0.25} =\frac{64}{0.25} =256.

The dropout probability is \(d=0.15\), and the conditional outcome-recording probability is \(m=0.90\). Thus the combined availability is \(a=(1-d)m=0.85(0.90)=0.765\). The expected analyzable count is \(0.765E\), so:

E \approx \frac{256}{0.765}=334.6.

Rounding up gives a planned enrollment of 335 participants, or about 167.5 assigned to each arm. The calculation predicts approximately \(335(0.765)=256.3\) analyzable outcomes in total. The implied information is approximately \(256.3(0.25)/100=0.6408\), just above the target because of rounding. The target is information in inverse squared outcome points, not a count of participants.

Checks, Alternatives, and Limitations

Before using an enrollment estimate, verify that the information formula matches the estimator and variance model used to set the target. Then check the arithmetic by substituting the planned expected analyzable arm counts back into the variance formula. A discrepancy often exposes a wrong allocation fraction, a missing availability adjustment, or a mistaken total-versus-per-arm count.

Planning approach	What it uses	When it is appropriate
Direct information conversion	Maximum information and a variance model	When the estimator and planning variance are specified and enrollment-to-outcome availability can be estimated.
Analyzable-count conversion	Information first converted to expected analyzable counts	When allocation and arm-specific availability are important; calculate each arm’s expected count explicitly.
Event-target conversion	Required events plus assumptions about event accrual and follow-up	For event-driven designs; do not treat the event target as the enrollment target.

A common mistake is to inflate an analyzable sample size by dropout and then inflate it again by an outcome-availability rate that already includes dropout. The correction is to define each probability precisely and combine only non-overlapping stages. Another error is to use an expected analyzable count for one arm as if it were the total: add the arm-specific counts before applying a total-count formula, or use the arm-specific variance expression directly.

These calculations rely on expected availability and a specified variance model. They do not guarantee the realized maximum information: outcome rates, dropout, allocation, and follow-up can differ from planning values. They also do not replace evaluation of operating characteristics under the intended design. Treat the resulting enrollment as a planning estimate, document its assumptions, and consider whether enrollment caps, follow-up duration, and the possibility of fewer observed outcomes make the target operationally attainable.

Key takeaway: Convert maximum information using the variance model that defines it, then translate expected analyzable arm counts into enrollment using explicit allocation and availability assumptions. Check the final planned counts by recalculating information; do not double-count losses or confuse participants with outcomes or events.

Check Your Understanding

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

For a binary outcome with \(p=0.25\), balanced allocation, and a maximum information target of 120, what total analyzable sample size is required under the common-probability approximation?
A trial plans 30% treatment and 70% control enrollment. Which allocation factor belongs in the equal-availability formula?
If dropout is 10% and, among retained participants, 80% have an observed outcome, what combined availability rate should be used?
Why can an event target not generally be used directly as the enrollment target in a time-to-event design?

Solutions. (1) \(p(1-p)=0.25(0.75)=0.1875\), so \(N\approx120(0.1875)/0.25=90\) analyzable participants. (2) The factor is \(q(1-q)=0.30(0.70)=0.21\). (3) The combined rate is \((1-0.10)(0.80)=0.90(0.80)=0.72\), because the 80% rate is conditional on retention. (4) Enrollment and observed events differ: the number of events depends on incidence, follow-up, and censoring, so these assumptions are needed to convert the event target into enrollment.