Accrual And Follow-Up After Survival Trial Re-Estimation

From an Event Target to a Calendar Plan

A revised event target is not itself a recruitment schedule. A trial must also determine how long to recruit and how much follow-up to allow after recruitment ends. These quantities depend on the accrual pattern: an early recruit has more time to experience an event before the analysis cutoff than a late recruit. They also depend on whether participants can be lost to follow-up before their event is observed.

The calculations below translate a planned event count \(D\) into calendar time under explicit assumptions. They use an expected event count for planning; they do not guarantee that the realized count will equal \(D\) on a particular date. The earlier material on event-count re-estimation for survival endpoints concerns updating the event target. Here the question is how a target can be reached operationally.

Definition: Let \(r(u)\) be the recruitment rate at calendar time \(u\), measured in participants per unit time, and let \(C\) be the planned analysis cutoff. A participant recruited at time \(u\) has potential follow-up \(C-u\). Let \(\lambda\) be the constant event hazard and \(\gamma\) the constant loss-to-follow-up hazard. Under independent exponential event and loss times, set \(q=\lambda+\gamma\). The probability that an event is observed by follow-up time \(s\), before loss, is \(\frac{\lambda}{q}(1-e^{-qs})\). Thus the expected observed event count is \(\int r(u)\frac{\lambda}{q}(1-e^{-q(C-u)})\,du\), over the recruitment period.

Uniform Accrual: Follow-Up After Recruitment Closes

Suppose recruitment proceeds at a constant rate \(a\) participants per month from time 0 through time \(T\), and the analysis cutoff is at \(C=T+F\). Here \(F\) is the additional follow-up after the recruitment period ends. Recruitment is assumed independent of participants’ event and loss times. The event and loss hazards are constant over calendar time and are the same for all participants in this planning calculation.

For a participant recruited at time \(u\), potential follow-up is \(T+F-u\). Integrating the probability of an observed event over all recruitment times gives the expected count:

E(D_{\mathrm{obs}}) = a\int_0^T \frac{\lambda}{q}\left(1-e^{-q(T+F-u)}\right)\,du = \frac{a\lambda}{q}\left[T-\frac{e^{-qF}(1-e^{-qT})}{q}\right].

The last equality follows by integrating the constant term over an interval of length \(T\), and integrating the exponential term: \(\int_0^T e^{-q(T+F-u)}du=e^{-qF}(1-e^{-qT})/q\). This is a useful check on the timing: increasing \(F\) reduces the subtracted term and therefore increases the expected event count.

Feasibility condition: With \(\lambda>0\), the expected count when \(F=0\) is \(\frac{a\lambda}{q}[T-(1-e^{-qT})/q]\). As \(F\) tends to infinity, it approaches \(\frac{a\lambda}{q}T\). When there is no loss to follow-up, \(\gamma=0\), so \(q=\lambda\), and this limit is \(aT\): each enrolled participant eventually has an event with probability 1 under the positive constant event hazard. When \(\gamma>0\), the limit is below \(aT\), because some participants are lost before an event. A target at or above the limit cannot be reached with finite follow-up under these assumptions.

Worked Example: Finding Additional Follow-Up for a Revised Target

Suppose a hypothetical trial plans to recruit at \(a=20\) participants per month for \(T=12\) months. Use an event hazard of \(\lambda=0.04\) per month and a loss hazard of \(\gamma=0.01\) per month. Find the additional follow-up \(F\) needed for an expected 150 observed events.

First, \(q=\lambda+\gamma=0.05\) per month. The limiting expected event count is \(\frac{a\lambda}{q}T=\frac{20(0.04)}{0.05}(12)=192\), so 150 is below the limit. At \(F=0\), the expected count is \(16[12-(1-e^{-0.6})/0.05]=16(2.9762)=47.62\), so some additional follow-up is needed.

Set the expected-count formula equal to 150 and rearrange:

e^{-qF} =\frac{q\left[T-\frac{Dq}{a\lambda}\right]}{1-e^{-qT}} =\frac{0.05\left[12-\frac{150(0.05)}{20(0.04)}\right]}{1-e^{-0.6}} =\frac{0.13125}{0.45119} =0.2909.

Therefore \(F=-\ln(0.2909)/0.05=24.70\) months. The plan entails 12 months of recruitment followed by about 24.7 additional months of follow-up, for an analysis cutoff about 36.7 months after recruitment begins. This is an expected-count calculation, not a promise that exactly 150 events will have occurred at that cutoff.

Changing the Recruitment Duration

The same expected-count model can be used when the event target and follow-up after recruitment are specified but recruitment duration is not. For constant recruitment rate \(a\), fixed additional follow-up \(F\), and unknown recruitment duration \(T\), solve

D=\frac{a\lambda}{q}\left[T-\frac{e^{-qF}(1-e^{-qT})}{q}\right].

This equation is generally nonlinear in \(T\), so a numerical root-finding method is appropriate. The expected count increases with \(T\) under the stated positive rate and hazard assumptions: extending the recruitment period adds participants, even though the latest recruits have less time before the cutoff. A solution should be checked by substituting the proposed \(T\) back into the equation. If instead the total enrollment \(N\) is fixed and spread uniformly over \(T\), the rate is \(a=N/T\); one must use that rate rather than treating \(a\) as fixed.

Worked Example: Solving for Recruitment Duration

A hypothetical trial recruits at \(a=20\) participants per month, allows \(F=12\) months of follow-up after recruitment closes, and plans for \(D=100\) observed events. Assume \(\lambda=0.04\) and \(\gamma=0.02\) per month. Find the recruitment duration.

Here \(q=0.06\) per month and \(a\lambda/q=13.333\) participants. The equation to solve is \(100=13.333[T-e^{-0.72}(1-e^{-0.06T})/0.06]\). For \(T=11.56\) months, \(e^{-0.06T}=e^{-0.6936}\approx0.4998\), so

E(D_{\mathrm{obs}}) =13.333\left[11.56-\frac{e^{-0.72}(1-e^{-0.6936})}{0.06}\right] \approx13.333(7.502) \approx100.0.

Thus a recruitment duration of about 11.56 months is the numerical solution, followed by 12 months of additional follow-up. Because the calculation uses a constant rate, it implies about \(20(11.56)=231.2\) expected recruits; an operational plan would specify a whole-number enrollment target and reassess the approximation using that target.

Alternative Accrual Patterns

Uniform accrual is a convenient approximation, not a requirement. If accrual is front-loaded, back-loaded, or changes in steps, use the actual planned recruitment rate \(r(u)\) in the integral. Participants recruited earlier have longer potential follow-up at a common cutoff, so changing the accrual pattern can change the expected observed event count even when total enrollment and the cutoff date stay the same. For piecewise-constant rates, calculate the integral separately over each time interval and add the results.

Worked Example: Front-Loaded Versus Uniform Accrual

Consider two hypothetical plans with 120 participants and a common analysis cutoff at month 18. In the first, recruitment is uniform over months 0 to 12, at 10 participants per month. In the second, recruitment is front-loaded over months 0 to 6, at 20 per month, with no recruitment afterward. Assume \(\lambda=0.05\) and \(\gamma=0.01\) per month, so \(q=0.06\).

For uniform accrual, each recruitment time \(u\) lies between 0 and 12, and the expected count is \(10(0.05/0.06)[12-e^{-0.36}(1-e^{-0.72})/0.06]\). Since \(e^{-0.36}\approx0.6977\) and \(e^{-0.72}\approx0.4868\), this is approximately \(8.333[12-5.969]=50.26\) events.

For front-loaded accrual, integrate over months 0 to 6. The follow-up period at recruitment time \(u\) is \(18-u\), giving \(20(0.05/0.06)[6-e^{-1.08}(e^{0.36}-1)/0.06]\). Using \(e^{-1.08}\approx0.3396\) and \(e^{0.36}\approx1.4333\), this is \(16.667[6-2.453]=59.12\) events. The front-loaded plan has about 8.86 more expected observed events by the same cutoff, because its participants are recruited earlier and have more time under observation.

Assumptions, Diagnostics, and Alternatives

Before using these calculations for an operational decision, check that their assumptions are credible. The model treats the event time and loss time as independent exponential times with constant hazards. It also assumes the hazards apply throughout follow-up, recruitment follows the specified rate, and the analysis cutoff and event definition are clear. If hazards vary over time, the event probability must reflect the time-varying hazards rather than the constant-hazard expression. If event risk or loss differs by treatment arm or recruitment group, calculate expected counts by group using the relevant rates and hazards, then sum them.

Useful diagnostics include comparing observed recruitment by calendar time with the planned accrual curve, monitoring loss rates against the assumed value, and checking whether event accumulation is consistent with the hazard assumptions. A substantial mismatch is a reason to revisit the calendar projection, not to treat the calculated date as assured. The expected-count equation also says nothing by itself about the variability of the event count or the probability of reaching the target by a particular date.

A simple alternative is to plan using a fixed enrollment number and a conservative follow-up period, then compute the expected count under several plausible accrual and loss scenarios. More complex accrual or hazard patterns can be handled by numerical integration or by simulating recruitment, event, and loss times under stated assumptions. These approaches can represent more detail, but they do not remove the need to justify the assumptions. The simple closed-form model is most useful when a transparent first calculation is needed.

Common pitfall: It is incorrect to assume that unlimited follow-up always produces one observed event per enrollee. With no loss and a positive constant event hazard, that is the limiting expectation. With a positive loss hazard, an event occurs before loss with probability \(\lambda/(\lambda+\gamma)\), so the limiting expected count is only \(aT\lambda/(\lambda+\gamma)\). Also, an event target above this ceiling is not fixed by adding more follow-up; the accrual plan, loss assumptions, or event target must be reconsidered.

Check Your Understanding

Answer each question using the definitions and assumptions in this tutorial. Brief solutions follow.

With no loss to follow-up and a positive constant event hazard, what does the expected event count approach as \(F\) tends to infinity?
For fixed \(a,T,\lambda,\gamma\), what happens to the expected count when \(F\) increases?
Why can front-loaded accrual produce more events by a common cutoff than uniform accrual with the same total enrollment?
With \(\lambda=0.03\), \(\gamma=0.01\), and \(N=200\), what is the limiting expected observed count as follow-up tends to infinity?
Name one assumption to revisit if observed loss is higher than planned, and explain the likely effect on the calendar projection.

Solutions: (1) It approaches \(aT\), because each of the \(aT\) participants eventually experiences an event with probability 1 under these assumptions. (2) It increases toward, but does not exceed, the relevant limit \((a\lambda/q)T\). (3) Earlier recruits have longer potential follow-up before the shared cutoff. (4) The limit is \(N\lambda/(\lambda+\gamma)=200(0.03/0.04)=150\) events. (5) Revisit the loss hazard: a higher loss hazard lowers the probability \(\lambda/(\lambda+\gamma)\) that an event is observed before loss, so the planned date may yield fewer events and more follow-up or a revised accrual plan may be needed.