Sensitivity To The Enrollment Pattern

Why Enrollment Shape Matters

Two trials can enroll the same number of participants over the same recruitment period yet have different event projections. The difference is the shape of recruitment: participants enrolled earlier have more time to experience the event before a fixed data cutoff than participants enrolled later. Thus, holding total enrollment constant does not make the event calendar insensitive to recruitment timing.

The comparison here changes only the pattern of enrollment. It holds the recruitment period, total enrollment, event-rate model, and censoring rule fixed. This isolates how enrollment timing affects expected event accrual; it does not account for changes in the event rate or dropout.

Definition: Let \(N\) be total enrollment and let recruitment occur over calendar time \(0\) to \(R\). A recruitment density \(f(u)\) describes the fraction of participants enrolled around time \(u\), with \(f(u)\geq 0\) and \(\int_0^R f(u)\,du=1\). The expected number enrolled by time \(u\) is \(N\int_0^u f(v)\,dv\). In this tutorial, front-loaded means enrollment is denser early in the recruitment period, uniform means a constant enrollment rate, and back-loaded means enrollment is denser later.

Model and Expected Event Count

Suppose every participant has an event time measured from their own enrollment. Assume event times are independent across participants and follow an exponential distribution with constant hazard \(\lambda\) per month: \(T_i\sim N\) is not used here; rather, an individual's event time \(W\) has survival probability \(P(W>s)=e^{-\lambda s}\) for \(s\geq 0\). Equivalently, the event probability by \(s\) months of follow-up is \(1-e^{-\lambda s}\). Assume there is no dropout or competing event, and that events are observed through a common administrative cutoff \(C\), with \(C\geq R\). A participant enrolled at \(u\) therefore has \(C-u\) months of potential follow-up.

Conditional on enrollment at \(u\), the probability of an observed event by the cutoff is \(1-e^{-\lambda(C-u)}\). Averaging that probability over enrollment times gives the expected event proportion; multiplying by \(N\) gives the expected event count. This follows by summing the event indicators and using linearity of expectation; independence is not needed for the mean, though it is part of the stated model.

E[D(C)] = N\int_0^R f(u)\left(1-e^{-\lambda(C-u)}\right)\,du = N\left[1-e^{-\lambda C}\int_0^R f(u)e^{\lambda u}\,du\right].

The second form makes the comparison transparent. For a fixed \(N\), \(R\), \(C\), and \(\lambda\), the pattern affects the expected count through the weighted average \(\int_0^R f(u)e^{\lambda u}\,du\). Enrollment later in the interval receives more weight in this average, but it also leaves less follow-up and hence a lower individual event probability. Consequently, a pattern that shifts enrollment earlier increases the expected event count by the same cutoff.

Three Standard Patterns

A convenient way to compare patterns over the same interval is to set \(x=u/R\), so \(x\) runs from 0 to 1. The corresponding densities on calendar time are:

Uniform: \(f_U(u)=1/R\). The cumulative fraction enrolled by \(u\) is \(u/R\).
Front-loaded: \(f_F(u)=2(R-u)/R^2\). The cumulative fraction enrolled by \(u\) is \(2(u/R)-(u/R)^2\), which is larger than the uniform fraction for \(0<u<R\).
Back-loaded: \(f_B(u)=2u/R^2\). The cumulative fraction enrolled by \(u\) is \((u/R)^2\), which is smaller than the uniform fraction for \(0<u<R\).

Each density integrates to 1, so each represents the same total enrollment \(N\). Front-loaded and back-loaded here describe idealized patterns, not necessarily literal recruitment schedules. Actual monthly enrollment counts can instead be used as interval weights in the same expected-count calculation.

Worked Example: Comparing Event Projections at a Fixed Cutoff

Consider a hypothetical trial with \(N=300\) participants recruited over \(R=12\) months. The cutoff is \(C=18\) months, and suppose the constant event hazard is \(\lambda=0.040\) per month. Compare the expected event counts under the three patterns.

The assumptions are that all participants are enrolled within months 0 to 12, follow-up ends at month 18, and the exponential event model applies equally under each pattern. With \(x=u/12\), the average event-free probability at cutoff is \(e^{-0.72}\int_0^1 g(x)e^{0.48x}\,dx\), where \(g(x)\) is respectively \(1\), \(2(1-x)\), or \(2x\). The required integrals are:

\begin{aligned} M_U &= \int_0^1 e^{0.48x}\,dx =\frac{e^{0.48}-1}{0.48} \approx 1.283488,\\ M_F &= \int_0^1 2(1-x)e^{0.48x}\,dx =\frac{2(e^{0.48}-1-0.48)}{(0.48)^2} \approx 1.181201,\\ M_B &= \int_0^1 2xe^{0.48x}\,dx =\frac{2\{1+e^{0.48}(0.48-1)\}}{(0.48)^2} \approx 1.385775. \end{aligned}

For uniform recruitment, the average event-free probability is \(e^{-0.72}(1.283488)\approx 0.624741\). Thus the expected event proportion is \(1-0.624741=0.375259\), and the expected event count is \(300(0.375259)\approx 112.6\) events.

For front-loaded recruitment, the average event-free probability is \(e^{-0.72}(1.181201)\approx 0.574952\). The expected count is \(300(1-0.574952)\approx 127.5\) events. For back-loaded recruitment, it is \(e^{-0.72}(1.385775)\approx 0.674529\), giving \(300(1-0.674529)\approx 97.6\) events.

At this cutoff, the front-loaded pattern projects about 14.9 more events than uniform recruitment, while the back-loaded pattern projects about 14.9 fewer. These are expected counts under the specified model, not predictions that the realized count will equal those values.

Why Earlier Enrollment Usually Advances Events

The ordering in the example can be justified without evaluating the integrals. For any fixed cutoff \(C\), the event probability \(1-e^{-\lambda(C-u)}\) decreases as enrollment time \(u\) increases, because later enrollment means shorter follow-up. The front-loaded distribution places more of its enrollment at earlier times than the uniform distribution, and the back-loaded distribution places more at later times. Averaging a decreasing function over these ordered enrollment distributions therefore gives the largest expected event proportion for front-loaded recruitment and the smallest for back-loaded recruitment.

A short-follow-up approximation also gives an intuitive scale for the effect. For small \(\lambda s\), \(1-e^{-\lambda s}\approx\lambda s\). If every participant is enrolled by \(C\), then expected events are approximately \(N\lambda\) times the mean follow-up time. For the three patterns, the mean enrollment times are \(R/2\), \(R/3\), and \(2R/3\) for uniform, front-loaded, and back-loaded recruitment, respectively. These follow from integrating \(u f(u)\) over \([0,R]\). The approximate expected follow-up times are consequently \(C-R/2\), \(C-R/3\), and \(C-2R/3\).

Worked Example: Using Mean Follow-Up as a Sanity Check

Keep \(N=300\), \(R=12\) months, and \(C=18\) months, but suppose the hypothetical hazard is much smaller, \(\lambda=0.005\) per month. The mean enrollment times are 6 months under uniform recruitment, 4 months under front-loaded recruitment, and 8 months under back-loaded recruitment. Their corresponding mean follow-up times at the cutoff are \(18-6=12\), \(18-4=14\), and \(18-8=10\) months.

Using \(1-e^{-\lambda s}\approx\lambda s\), the approximate event proportions are \(0.005(12)=0.060\), \(0.005(14)=0.070\), and \(0.005(10)=0.050\). The approximate expected counts are therefore \(300(0.060)=18\), \(300(0.070)=21\), and \(300(0.050)=15\) events for uniform, front-loaded, and back-loaded recruitment. This is a first-order approximation, not an exact calculation. Since the maximum \(\lambda s\) is \(0.005(18)=0.09\), the approximation is plausible for a rough check; the exact event probability should be used for a final projection.

Enrollment Pattern and Expected Target Dates

A pattern also changes the calendar time at which an expected event target is reached. For \(C\geq R\), define the average event-free probability at the end of recruitment as \(A_R=\int_0^R f(u)e^{-\lambda(R-u)}\,du\). After recruitment ends, everyone still under observation accumulates follow-up at the same rate. Therefore the average event-free probability at a later cutoff \(C\) is \(A_R e^{-\lambda(C-R)}\). If a target is \(D^*\) expected events, solving \(E[D(C)]=D^*\) yields:

C=R-\frac{1}{\lambda}\log\left(\frac{1-D^*/N}{A_R}\right), \qquad \text{provided } C\geq R.

Worked Example: Projecting the Date of an Expected Event Target

Use the hypothetical trial from the first example and ask when the expected count reaches 120 events. The target event proportion is \(120/300=0.4000\), so the required event-free proportion is \(1-0.4000=0.6000\). From the first example's event-free probabilities at month 18, the end-of-recruitment values are obtained by multiplying by \(e^{0.04(18-12)}=e^{0.24}\): \(A_R\approx0.7942\) for uniform, \(0.7309\) for front-loaded, and \(0.8575\) for back-loaded recruitment.

For uniform recruitment, substitution gives \(C=12-\log(0.6000/0.7942)/0.040\approx19.01\) months. For front-loaded recruitment, \(C=12-\log(0.6000/0.7309)/0.040\approx16.93\) months. For back-loaded recruitment, \(C=12-\log(0.6000/0.8575)/0.040\approx20.93\) months. Each solution is after month 12, as required for the formula.

Under these assumptions, the expected target is reached earliest with front-loaded recruitment and latest with back-loaded recruitment. The calculated dates do not guarantee that the observed trial will have exactly 120 events then: the realized count varies around its expectation.

Diagnostics, Alternatives, and Limitations

A sensitivity analysis is useful when the planned recruitment curve is uncertain. It should vary the pattern while keeping total enrollment and other assumptions fixed, so the contribution of recruitment timing remains interpretable. Compare the assumed cumulative enrollment curve with observed enrollment by calendar time. If observed recruitment is concentrated later than planned, a projection based on a uniform or front-loaded pattern may overstate accrued events at the planned cutoff.

The three smooth patterns are easy to compare but may not represent pauses, site activation, or abrupt recruitment changes. A practical alternative is to use observed or planned numbers enrolled in calendar-time intervals: for each interval, calculate the event probability from its enrollment time (or a suitable interval average), then weight by the fraction enrolled there. Simulation can also represent a detailed recruitment schedule and random event times, as in Monte Carlo event projection. These alternatives do not remove the need to specify the event-time and censoring assumptions.

The comparison is limited by its assumptions. A constant exponential hazard may not describe time-varying event risk; dropout, competing events, or delayed event ascertainment can change the event probabilities. Here these factors are held absent or unchanged solely to isolate recruitment shape. In an applied projection, the enrollment pattern should be reviewed alongside the event-rate and dropout assumptions rather than treated as the only source of uncertainty.

Common pitfall: It is incorrect to say that equal total enrollment implies equal expected events by a fixed date. Total enrollment fixes the number of participants, not their follow-up durations. The correction is to average each participant's event probability over the enrollment-time distribution; earlier enrollment generally provides more time for events to occur before the cutoff.

Check Your Understanding

Answer each question using the model and definitions above. Brief solutions follow the questions.

With total enrollment and the cutoff fixed, what feature of a front-loaded pattern generally increases its expected event count relative to uniform recruitment?
For uniform recruitment over 10 months, what is the mean enrollment time?
At a cutoff after recruitment is complete, does a higher event-free probability imply a higher or lower expected event count?
Why should a projected date for an expected target not be described as the guaranteed date when the observed target will occur?

Solutions. (1) More participants enroll earlier and therefore have longer follow-up by the cutoff. (2) The mean is \(R/2=10/2=5\) months. (3) It implies a lower expected event count, since the expected event proportion is one minus the event-free probability. (4) The projected date solves an equation for the mean count; the realized event count is random and may differ from its expectation.