Inflating Enrollment for Attrition and Unusable Outcomes

Why Recruitment Must Allow for Losses

A planned sample size usually refers to participants whose outcomes can contribute to the intended analysis. Recruitment targets, by contrast, count people enrolled before follow-up is complete and before outcome usability is known. If some participants withdraw, cannot be contacted, or have outcomes that do not meet the analysis requirements, recruiting exactly the analyzable target will usually leave too few usable outcomes.

In a multi-arm, multi-stage (MAMS) design, this calculation belongs to each arm and stage. Retention may differ by treatment, follow-up period, or outcome collection process. The aim here is to translate an already specified analyzable target into recruitment numbers under explicit assumptions; this is not a method for deriving the analyzable target itself.

Definition: For arm \(a\) and stage \(k\), let \(m_{ak}\) be the required number of participants with usable outcomes, \(N_{ak}\) the number recruited, and \(r_{ak}\) the probability that a recruited participant yields a usable outcome for the planned analysis. The quantity \(r_{ak}\) is the retention probability; it accounts for attrition and outcome unusability as defined in the protocol. The corresponding loss probability is \(1-r_{ak}\).

The usable-outcome definition must match the planned analysis. It might mean that a primary outcome is observed within a prespecified window, or that sufficient information is available under the chosen analysis method. It need not mean that every measurement is complete. If different forms of loss occur in sequence, their probabilities must be defined conditionally at each step. For example, a probability of providing an outcome among participants who remain in follow-up is not the same as a probability calculated among everyone originally recruited.

Inflation Using the Expected Number of Usable Outcomes

Suppose each recruit in an arm and stage has the same probability \(r_{ak}\) of yielding a usable outcome. Let \(X_{ak}\) be the number of usable outcomes among \(N_{ak}\) recruits. Linearity of expectation gives \(E(X_{ak})=N_{ak}r_{ak}\); independence is not needed for this expectation. Setting the expected usable count equal to the target gives the basic inflation calculation.

E(X_{ak})=N_{ak}r_{ak}, \qquad N_{ak}^{\text{expected}}=\frac{m_{ak}}{r_{ak}}.

This is an expected-count target, not a guarantee that at least \(m_{ak}\) usable outcomes will be observed. The quotient can be fractional because it expresses the planning calculation before operational rounding. In practice, recruitment counts must be whole numbers; rounding up the quotient ensures that the expected count is at least the target under the assumed retention rate, but still does not ensure that the realized usable count reaches the target.

Worked Example: Sequential Losses Before the Outcome

Suppose a hypothetical stage needs 80 participants with usable primary outcomes. Of those recruited, 90% are expected to remain through follow-up, and among those remaining, 85% are expected to provide an outcome that meets the analysis requirements. Assume these probabilities apply to this stage and arm.

The probability of reaching the usable-outcome state is the product of the sequential conditional probabilities: \(r=0.90(0.85)=0.765\). Thus the expected-count recruitment calculation is \(N=80/0.765=104.575\), or approximately 104.6 recruits before operational rounding.

If recruitment must be a whole number, 105 is the upward-rounded operational count. Its expected usable count is \(105(0.765)=80.325\), whereas 104 recruits would yield an expected count of \(104(0.765)=79.56\). The calculation supports 105 as the rounded expected-count target; it does not say that 105 recruits will necessarily produce at least 80 usable outcomes.

Multiplication is appropriate here because the second probability is conditional on surviving the first step. More generally, if a participant must pass through successive stages of follow-up with conditional retention probabilities \(r_1,\ldots,r_J\), the overall probability of a usable outcome is \(\prod_{j=1}^{J}r_j\). Multiplying unconditional percentages, or multiplying probabilities that do not describe the stated sequence, can give the wrong overall retention rate.

Applying Inflation Across MAMS Arms and Stages

Calculate recruitment separately wherever the analyzable target or retention probability differs. When an arm stops at an interim analysis, later recruitment in that arm is not undertaken. A shared control may be recruited for a later stage only when at least one treatment arm continues, so its contribution to expected enrollment depends on that continuation event as well as on its own retention rate. The continuation probabilities concern whether recruitment occurs; retention probabilities concern whether recruited participants yield usable outcomes. These are distinct quantities.

Worked Example: Stagewise Recruitment With a Shared Control

Consider a hypothetical two-treatment MAMS design. At stage 1, each treatment arm needs 30 usable outcomes and the shared control needs 20. Assume a retention probability of 0.90 for all three groups at this stage. At stage 2, each continuing treatment arm needs 24 usable outcomes and the shared control needs 16; assume stage-2 retention is 0.80 in every group. For illustration, suppose the probabilities that treatment arms 1 and 2 continue are 0.60 and 0.30, and that these continuation events are independent.

At stage 1, the expected-count recruitment calculations for each treatment arm and for the shared control are, respectively, \(30/0.90=33.33\) and \(20/0.90=22.22\). The unrounded expected-count total is \(2(33.33)+22.22=88.88\) recruits (using the unrounded quotients, \(2(30/0.90)+20/0.90=88.89\), to two decimal places). Operational counts would be decided by rounding each group’s target to whole participants under the protocol.

For stage 2, each continuing treatment arm requires \(24/0.80=30\) recruits. The shared control requires \(16/0.80=20\) recruits if at least one treatment arm continues. Under the stated independence assumption, the probability that at least one arm continues is \(1-(1-0.60)(1-0.30)=1-(0.40)(0.70)=0.72\).

The expected stage-2 recruitment is therefore \(0.60(30)+0.30(30)+0.72(20)=18+9+14.4=41.4\). Adding the stage-1 expected-count total gives approximately \(88.89+41.4=130.29\), or 130.3 expected recruits across the design before operational rounding. This calculation counts the stage-2 shared-control recruitment once when at least one treatment arm continues, not once per continuing arm. The result is conditional on the stated retention and continuation assumptions; it is not a guarantee of either the usable sample sizes or the total enrollment in a realized trial.

The independence assumption in that example is needed to obtain the probability that at least one arm continues from the two marginal probabilities alone. If the continuation events are dependent, that probability must instead be specified or calculated using their joint distribution. If stage sizes differ, or the shared-control requirement changes according to which arms continue, the corresponding recruitment and continuation events must be represented separately rather than forced into a single average.

Expected Counts Are Not Probability Guarantees

A common alternative is to choose recruitment so that the probability of meeting the analyzable target is at least a prespecified assurance level. If each recruit independently yields a usable outcome with the same probability \(r\), then the usable count \(X\) has a binomial distribution, \(X\sim\operatorname{Binomial}(N,r)\). For a target \(m\) and assurance level \(\gamma\), the recruitment requirement is the smallest integer \(N\) satisfying \(P(X\ge m)\ge\gamma\). This is a different criterion from setting \(E(X)=m\).

P(X\ge m)=\sum_{x=m}^{N}\binom{N}{x}r^x(1-r)^{N-x}, \qquad N_{\gamma}=\min\left\{N:P(X\ge m)\ge\gamma\right\}.

The distinction matters because a random count fluctuates around its mean. If the expected usable count equals the target, there is generally a substantial chance of falling below it. A probability-based criterion typically requires more recruits than the expected-count calculation, with the difference depending on the target, retention, and assurance level.

Worked Example: Checking an Assurance Target Approximately

Suppose a hypothetical group needs at least 40 usable outcomes, each recruit has an independent 0.80 probability of yielding a usable outcome, and the desired probability of reaching 40 is 90%. The expected-count quotient is \(40/0.80=50\) recruits, but this only makes the expected usable count 40; it does not make the probability of reaching 40 equal to 90%.

For 55 recruits, the binomial mean is \(55(0.80)=44\), and the variance is \(55(0.80)(0.20)=8.8\). The standard deviation is therefore \(\sqrt{8.8}\approx2.966\). A normal approximation with continuity correction gives

P(X\ge40)\approx P\left(Z\ge\frac{39.5-44}{2.966}\right) =P(Z\ge-1.517)\approx0.935.

Thus the approximation supports 55 recruits as exceeding a 90% assurance target. This is an approximate calculation, not an exact binomial result; if the assurance requirement is binding, evaluate the binomial tail directly and select the smallest integer that meets it. The calculation also relies on independent, equal-probability usable outcomes. If those assumptions are not credible, this binomial model is not an adequate assurance calculation.

Assumptions, Diagnostics, and Alternatives

Expected-count inflation is most useful when the planning target is explicitly a mean usable count and a defensible retention probability is available. Its assumptions and limitations should be checked before the quotient is used:

Retention is relevant to the analysis. Define what makes an outcome usable under the planned analysis. A blanket “complete case” rate may be inappropriate if the analysis can use partial outcome information.
Rates apply to the specified arm and stage. A pooled historical rate may conceal differences in follow-up duration, treatment, population, or stage. Use arm- and stage-specific values when the evidence supports them.
Sequential probabilities are conditional as stated. For example, multiply the probability of remaining in follow-up by the probability of an outcome among those retained only when that second probability is conditional on retention.
Inflation does not remove bias. If outcome availability is related to prognosis or treatment response, recruiting additional participants may increase the number observed but does not by itself make the observed outcomes representative or correct an invalid analysis.
Planning assumptions need monitoring. During a trial, compare observed loss and unusability with the prespecified assumptions by arm and stage, and examine reasons and timing. A lower-than-assumed retention rate signals that the planned analyzable target may not be reached. Any change to recruitment should follow the trial’s prespecified procedures.

The principal alternatives answer different questions. Expected-count inflation sets \(N r\) to the analyzable target and is transparent, but provides no chosen probability of reaching that target. Binomial assurance selects a recruitment count to meet a probability threshold, but requires a model for the distribution of usable outcomes. A more detailed missing-data analysis may be appropriate when outcomes are partially observed or missingness depends on observed information, but recruitment inflation alone is not a substitute for that analysis. In all cases, uncertainty in the assumed retention rate should be made visible rather than hidden by reporting a single recruitment number.

Key takeaway: For an analyzable target \(m_{ak}\) and retention probability \(r_{ak}\), \(m_{ak}/r_{ak}\) is the expected-count recruitment calculation. Apply it by arm and stage, combine sequential retention probabilities only when their conditional meanings justify it, and use a probability-based calculation when the requirement is a chance of reaching the target rather than an expected count.

Check Your Understanding

Answer each question, then compare with the solutions below.

A stage needs 72 usable outcomes and the assumed retention probability is 0.80. What is the expected-count recruitment calculation, and what does it mean?
Participants have a 0.95 probability of remaining through follow-up and, conditional on remaining, a 0.90 probability of providing a usable outcome. What is the overall retention probability?
Why does recruiting \(m/r\) participants not guarantee at least \(m\) usable outcomes?
Two arms have continuation probabilities 0.50 and 0.40. If their continuation events are independent, what is the probability that a shared control is needed by at least one continuing arm?

Solutions. (1) \(72/0.80=90\) recruits gives an expected usable count of 72 under the assumed rate; it is not a guarantee. (2) \(0.95(0.90)=0.855\). (3) The realized usable count varies around its expectation, so it can fall below \(m\) even when its mean is \(m\). (4) \(1-(1-0.50)(1-0.40)=1-(0.50)(0.60)=0.70\).