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Population Enrichment and Subgroup Selection · Tutorial 430 of 799

Designing Trials For Rare Biomarker Subgroups

Learn to compare recruitment routes by their yield and uncertainty, and to distinguish successful subgroup accrual from the subsequent randomness of treatment allocation.

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What You'll Learn

  • Calculate expected randomized marker-positive accrual from screening, referral, and registry recruitment pipelines.
  • Use binomial or forecast models to quantify whether a recruitment plan is likely to meet its target.
  • Translate subgroup prevalence and conversion rates into required recruitment volume.
  • Interpret recruitment forecasts cautiously when their assumptions or inputs are uncertain.
  • Separate uncertainty in subgroup accrual from variability in treatment-arm counts.

Recruitment Is a Pipeline, Not Just a Prevalence

A trial designed around a rare biomarker subgroup can have sound eligibility criteria and a defensible sample-size target yet still fail operationally: too few eligible marker-positive participants may be identified, approached, or enrolled within the recruitment window. The design question is therefore not only “How prevalent is the marker?” but also “How many people will move through each step of this recruitment route?”

Three common routes are broad screening, referral networks, and recruitment from a registry or service that already identifies marker-positive people. Each route has a different starting population and a different conversion process. A useful comparison estimates the number of marker-positive participants who will actually be randomized, not merely the number who test positive or are referred.

Definition: A recruitment route is a specified process for identifying, assessing, and enrolling potential trial participants. Its yield is the number of eligible marker-positive participants expected to be randomized over a stated period. A conversion probability is the probability that a person at one stage of the process advances to the next stage, such as from referral to eligibility and consent.

For a simple pipeline, let \(N\) be the number of people entering the route, \(p\) the probability that a person is marker-positive, and \(c\) the probability that a marker-positive person is eligible, consents, and is randomized. If these probabilities are stable and participants’ outcomes are independent, the per-person probability of randomized marker-positive enrollment is \(q=pc\). The expected yield is \(Nq\). If the route starts with already identified marker-positive people, \(p\) is effectively one for planning purposes, and \(c\) represents the remaining conversion steps.

These calculations assume the marker is assessed using the qualified process specified for the trial, that the inputs reasonably represent the recruitment population, and that the planned recruitment window and site capacity are feasible. Assay errors and missing biomarker measurements require their own handling; treating every test result as correct or available without checking those conditions can distort the forecast. The earlier material on power under uncertain biomarker prevalence is relevant when prevalence itself is uncertain, but operational yield also depends on the route’s eligibility and enrollment steps.

Modeling Accrual and Its Uncertainty

When \(N\) people independently have the same probability \(q\) of becoming randomized marker-positive participants, the accrual count \(Y\) follows a binomial model, \(Y\sim\operatorname{Binomial}(N,q)\). Its mean is \(Nq\), and its variance is \(Nq(1-q)\). The expected count is not a guarantee: a route with expected yield equal to the target will often finish below it. A binomial model is unsuitable if the opportunities are strongly dependent, the conversion probabilities vary substantially across sites, or capacity limits make the individual outcomes non-independent.

$$ Y\sim\operatorname{Binomial}(N,q),\qquad E(Y)=Nq,\qquad \operatorname{Var}(Y)=Nq(1-q) $$

For large \(N\), a normal approximation may be used when both \(Nq\) and \(N(1-q)\) are sufficiently large. For an integer target \(K\), a continuity-corrected approximation to the probability of meeting or exceeding the target is

$$ P(Y\ge K)\approx P\!\left( Z\ge \frac{K-0.5-Nq}{\sqrt{Nq(1-q)}} \right),\qquad Z\sim N(0,1). $$

The correction uses \(K-0.5\) because the discrete event \(Y\ge K\) corresponds approximately to values above the midpoint between \(K-1\) and \(K\). When the expected number or its complement is small, use an exact binomial calculation or a suitable predictive model rather than relying on a poor normal approximation. If the planning inputs were estimated, uncertainty in those inputs should also be considered; the binomial calculation conditional on fixed \(q\) does not include uncertainty about \(q\).

Worked Example: Probability of Meeting a Screening Accrual Target

Worked Example: Probability of Meeting a Screening Accrual Target

Suppose a hypothetical trial’s site forecast represents the number \(M\) of randomized marker-positive participants obtainable through its screening route during the recruitment window as approximately normal, with mean 135 and standard deviation 11.44 participants. The target is at least 120. For an integer-valued count, approximate the target event using the continuity-corrected threshold 119.5.

Model check. The forecast is being treated as a normal approximation to a count, not as an exact binomial model. This is reasonable only if the forecast distribution is approximately symmetric and its mean and standard deviation reflect the planned sites and recruitment window. The implied standard deviation is small relative to the mean, so negative accrual is negligible under this approximation.

Calculation. Standardize the threshold using the stated forecast parameters:

$$ P(M\ge120)\approx P\!\left( Z\ge\frac{119.5-135}{11.44} \right) = P(Z\ge-1.355) \approx 0.9123. $$

The quotient is \(-15.5/11.44=-1.3549\), rounded to \(-1.355\). Thus, under this hypothetical forecast model, the approximate probability of randomizing at least 120 marker-positive participants is 0.9123, or 91.23%. This is a model-based planning probability, not a guarantee that the trial will meet its target.

Compare Routes Using the Whole Conversion Process

A screening route starts with a broad pool and tests people to identify the subgroup. Its strength is that it does not depend on a small set of existing referral sources; its burden is that many people may need screening for each randomized subgroup participant. A referral network can increase the proportion of marker-positive people among those considered, but its performance depends on referral volume, clinical practice, and relationships with referring providers. A registry or service that already identifies marker-positive people can reduce screening burden, but may have limited size, outdated records, or a lower-than-expected rate of current eligibility and consent.

For planning, calculate the full route-specific probability of reaching randomization. Do not compare routes using marker prevalence alone: a high prevalence among referrals does not help if few referrals arrive or most fail eligibility. Conversely, a low-prevalence screening pool can still be viable if it is large and accessible, although the screen-failure burden may be substantial.

Worked Example: How Many People Must Each Route Reach?

Consider a hypothetical trial aiming to randomize 60 marker-positive participants. Based on planning assumptions, broad screening has marker prevalence \(p=0.04\) and a conditional probability \(c=0.80\) of eligibility, consent, and randomization among marker-positive people. A referral network has marker prevalence \(0.30\) among referrals and conversion \(0.75\). A marker-positive registry has a conversion probability of \(0.55\) from invitation to randomization. Assume independent, comparable opportunities within each route.

Broad screening. The probability that one screened person becomes a randomized marker-positive participant is \(0.04(0.80)=0.032\). The number required for an expected yield of 60 is

$$ N=\frac{60}{0.032}=1875\text{ people screened}. $$

Referral network. The per-referral yield probability is \(0.30(0.75)=0.225\). The required number of referrals for an expected yield of at least 60 is

$$ N=\frac{60}{0.225}=266.67, \qquad\text{so plan for at least }267\text{ referrals}. $$

At 267 referrals, the expected yield is \(267(0.225)=60.075\) participants. Marker-positive registry. With conversion probability 0.55, the number of invitations needed for an expected yield of 60 is \(60/0.55=109.09\), so at least 110 invitations are needed; their expected yield is \(110(0.55)=60.5\).

These volumes make the recruitment burden comparable, but they do not give a high probability of reaching the target: each plan is sized to an expected count near 60. For example, the referral yield has standard deviation \[ \sqrt{267(0.225)(0.775)}=\sqrt{46.59}=6.826. \] A continuity-corrected normal approximation gives \[ P(Y\ge60)\approx P\!\left(Z\ge\frac{59.5-60.075}{6.826}\right) =P(Z\ge-0.0842)\approx0.5336. \] Thus this referral plan has only about a 53.36% chance of meeting the target under the assumptions. A practical plan needs a buffer, a longer recruitment window, additional sites, or a combination of routes. Its required size should be chosen against a stated probability of meeting the target, not just by dividing the target by expected yield.

Accrual and Randomization Are Separate Sources of Uncertainty

The number of marker-positive participants accrued determines how many people can be randomized; randomization then determines how those participants are distributed between treatment arms. A recruitment strategy does not by itself guarantee equal arm sizes. Under independent 1:1 randomization, if \(m\) marker-positive participants are randomized, the number assigned to one specified arm follows \(\operatorname{Binomial}(m,0.5)\). Its expected value is \(m/2\), with standard deviation \(\sqrt{m(0.5)(0.5)}=\sqrt{m}/2\).

Worked Example: Variation in Arm Counts After Accrual

Suppose a hypothetical recruitment plan succeeds in randomizing 80 marker-positive participants. Under independent 1:1 allocation, the count \(X\) assigned to one specified treatment arm follows \(X\sim\operatorname{Binomial}(80,0.5)\). This model assumes each participant’s assignment is an independent, equally likely choice.

The expected count is \(80(0.5)=40\), and the standard deviation is \(\sqrt{80(0.5)(0.5)}=\sqrt{20}=4.472\) participants. To approximate the probability that this arm receives between 35 and 45 participants inclusive, use the continuity-corrected range 34.5 to 45.5:

$$ P(35\le X\le45)\approx P\!\left(\frac{34.5-40}{4.472}\le Z\le \frac{45.5-40}{4.472}\right) = P(-1.230\le Z\le1.230) \approx0.7814. $$

So the approximate probability of an arm count from 35 through 45 is 78.14%, even after exactly 80 people have been accrued. This allocation variability is distinct from the chance of accruing 80 participants at all. Recruitment forecasts address the latter; the randomization procedure determines the former. Restricted allocation methods can limit imbalance, but the choice and implementation of a randomization scheme must be specified separately.

Diagnostics, Comparisons, and Limitations

A recruitment forecast should be checked against observed progress, not treated as a fixed promise. Useful diagnostics include the number entering each pipeline stage by site and time period, the observed marker-positive proportion, the fraction meeting eligibility criteria, the consent rate, and the time between referral and randomization. If the observed counts differ materially from assumptions, update the forecast and investigate the source: a low yield may reflect fewer referrals, lower prevalence, assay availability, eligibility failures, or refusal. Combining all losses into one “conversion rate” can hide which part of the route needs intervention.

Comparisons should use the same time horizon, target definition, and cost accounting. Broad screening may have higher testing costs and more screen failures but greater reach. Referrals may reduce unnecessary testing while relying on external provider behavior. Existing marker-positive records may reduce identification effort but can be small or stale. A hybrid plan may diversify these risks, though its component yields should not simply be added if the routes draw on overlapping people.

The calculations are only as reliable as their assumptions. Site capacity, seasonal variation, competing studies, uneven site performance, dependence among referrals, changing eligibility criteria, and differences between the forecast population and the actual recruitment pool can all undermine a simple binomial model. The estimate \(Nq\) is a mean, not a promised count. A design that can succeed only if recruitment equals its expected value has little operational margin. Re-estimation should respect the trial’s prespecified governance and statistical procedures.

Common Mistake

Wrong reasoning: “The expected yield is 60 and the target is 60, so the plan should enroll 60.” The expected value describes the center of a distribution, not a minimum. Random variation means a substantial chance of falling below the target, as the referral calculation showed.

Correction: Model the probability of meeting the target under plausible inputs, and plan enough recruitment volume to achieve an acceptable probability, allowing for uncertainty in the inputs and operational constraints. Also keep the two events distinct: meeting the accrual target and obtaining a particular arm balance are not the same probability.

Key takeaway: Compare recruitment routes by the probability of reaching randomized marker-positive enrollment, not by marker prevalence or expected yield alone. Quantify uncertainty in accrual, then assess treatment allocation separately.

Check Your Understanding

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

  1. A screening route tests 500 people. The marker prevalence is 0.06 and the conditional conversion from marker-positive test to randomization is 0.70. What is the expected yield?
  2. A referral route expects 40 randomized marker-positive participants from 160 referrals. What is its implied per-referral yield probability, assuming a constant independent probability?
  3. Why does planning for an expected yield equal to the target not ensure a high probability of meeting that target?
  4. With 50 people randomized under independent 1:1 allocation, what is the expected count in one specified arm?

Solutions. (1) The per-person yield probability is \(0.06(0.70)=0.042\), so the expected yield is \(500(0.042)=21\) randomized marker-positive participants. (2) The implied yield probability is \(40/160=0.25\), or 25% per referral; this is an expectation-based estimate, not proof that every referral has the same probability. (3) The realized count varies around its expected value, so a target set at the mean has a meaningful probability of being missed; the probability must be calculated from an accrual distribution. (4) The expected count is \(50(0.5)=25\) participants in that arm.

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