Averaging Event Probabilities Across Enrollment Times

Why Enrollment Time Matters

At a fixed data cutoff, participants enrolled at different times have different opportunities to experience the event. A participant enrolled early may have a year of follow-up, while someone enrolled shortly before the cutoff may have only a few weeks. Consequently, multiplying the number enrolled by the event probability at one “typical” follow-up time can give a misleading projection. The expected event total is obtained by averaging participant-level event probabilities using the actual or planned distribution of enrollment times.

The calculation uses the same basic ingredients as event projection from survival probabilities: an event-time model and the follow-up available by the cutoff. The key step here is to weight the resulting probabilities according to when participants enter. This is an expectation, not a guarantee that the realized count will equal the projected count.

Definition: Let \(C\) be the calendar time of a data cutoff and let \(U\) be a participant’s enrollment time, measured on the same calendar scale. The available follow-up at the cutoff is \(C-U\) when \(U\le C\). If \(S(a)\) is the probability of remaining free of the event through \(a\) units of follow-up, the probability of experiencing the event by the cutoff is \(p(C-U)=1-S(C-U)\), provided this event probability is appropriate for the endpoint and the participant’s event-time model.

From Individual Probabilities to an Expected Total

First suppose \(N\) participants are enrolled by the cutoff, with enrollment times \(u_1,\ldots,u_N\). Write \(Y_i=1\) if participant \(i\) has the event by \(C\), and \(Y_i=0\) otherwise. If the event probability for participant \(i\), given their enrollment time, is \(p(C-u_i)\), then the total event count is \(D=\sum_{i=1}^N Y_i\). By linearity of expectation,

E(D)=\sum_{i=1}^N E(Y_i)=\sum_{i=1}^N p(C-u_i).

This equality does not require the event indicators to be independent; it follows from linearity of expectation. Independence can matter for the variability of the total, a separate question from its expected value. If enrollment times are planned or observed only as a distribution rather than individually, the sum can be replaced by an average over that distribution.

For a fixed total of \(N\) participants whose enrollment times follow a density \(f(u)\) over the recruitment period, the expected event probability per participant is the enrollment-time-weighted average. Thus,

E(D)=N\int_0^C p(C-u)f(u)\,du, \qquad \int_0^C f(u)\,du=1.

The density is a weighting distribution: times at which more participants enroll contribute more to the average. If instead enrollment is described by a rate \(r(u)\), measured in participants per unit of calendar time, the expected total can be written as

E(D)=\int_0^C r(u)p(C-u)\,du.

Here \(r(u)\,du\) represents the number expected to enroll in a short interval around \(u\). When the rate is specified deterministically and its integral is \(N\), the normalized density is \(f(u)=r(u)/N\). These expressions are two ways to represent the same weighting, not different event-probability models.

Uniform Enrollment and Exponential Event Times

A useful closed-form case assumes participants enroll uniformly over the interval from calendar time 0 to \(R\), with the cutoff at \(C\ge R\). The enrollment-time density is \(f(u)=1/R\) for \(0\le u\le R\). Suppose, in addition, that the event time after enrollment has a constant hazard \(\lambda\) per unit of follow-up, with no other event or loss process affecting whether the event is observed. Then \(S(a)=e^{-\lambda a}\), and averaging over enrollment times gives

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

The expression inside the brackets is the average event probability across all enrollment times. It is not the event probability evaluated at the average follow-up time; the integration accounts for the full spread of follow-up durations.

Worked Example: Uniform Enrollment Before a Fixed Cutoff

Suppose, hypothetically, that \(N=300\) participants enroll uniformly over 12 months, from month 0 through month 12. The data cutoff is at month \(C=18\), and the event-time model is exponential with hazard \(\lambda=0.04\) per month. Assume the event model applies from enrollment, all participants are followed for event ascertainment until the administrative cutoff, and no competing event or loss to follow-up changes the event probability. Find the expected event total.

For a participant enrolled at month \(u\), follow-up is \(18-u\) months, so the event probability is \(1-e^{-0.04(18-u)}\). Since enrollment is uniform, the expected event probability is

\begin{aligned} \bar p &=\frac{1}{12}\int_0^{12}\left[1-e^{-0.04(18-u)}\right]du\\ &=1-\frac{e^{-0.04(6)}-e^{-0.04(18)}}{0.04(12)}\\ &=1-\frac{0.7866-0.4868}{0.48}\\ &\approx 0.3753. \end{aligned}

Therefore,

E(D)=300(0.3753)\approx112.6\text{ events}.

The projection is about 113 events in expectation. It does not say that exactly 113 events will occur. Participants have between 6 and 18 months of follow-up in this example, so applying the 18-month event probability to all 300 participants would overstate the expected total.

Piecewise Enrollment Rates

Recruitment is often not uniform across the full enrollment period. If the rate is constant within several intervals, calculate the expected events in each interval using that interval’s rate and the follow-up available to people entering there, then add the interval contributions. For intervals \([a_j,b_j]\) with rates \(r_j\), the calculation is

E(D)=\sum_j r_j\int_{a_j}^{b_j}p(C-u)\,du.

The rate \(r_j\) affects the weight of the interval. It would be incorrect to give each interval equal weight if the numbers enrolled differ. The expression assumes the stated rates apply over their intervals and that the relevant participant-level event probability is correctly specified for the time of enrollment.

Worked Example: Two Recruitment Rates

In a hypothetical trial, enrollment proceeds at 20 participants per month from month 0 to month 6, then at 10 participants per month from month 6 to month 12. The data cutoff is month 15. Assume a constant event hazard of \(\lambda=0.05\) per month after enrollment, with no competing events or loss to follow-up. The first interval contributes \(20(6)=120\) participants and the second contributes \(10(6)=60\), for 180 participants total.

For enrollment in months 0 through 6, available follow-up ranges from 15 to 9 months. The expected number of events from that interval is

\begin{aligned} E(D_1) &=20\int_0^6\left[1-e^{-0.05(15-u)}\right]du\\ &=20\left[6-\frac{e^{-0.05(9)}-e^{-0.05(15)}}{0.05}\right]\\ &=120\left[1-\frac{0.6376-0.4724}{0.30}\right]\\ &\approx120(0.4491)=53.90\text{ events}. \end{aligned}

For enrollment in months 6 through 12, follow-up ranges from 9 to 3 months. The second interval contributes

\begin{aligned} E(D_2) &=10\int_6^{12}\left[1-e^{-0.05(15-u)}\right]du\\ &=60\left[1-\frac{e^{-0.05(3)}-e^{-0.05(9)}}{0.30}\right]\\ &=60\left[1-\frac{0.8607-0.6376}{0.30}\right]\\ &\approx60(0.2564)=15.38\text{ events}. \end{aligned}

Adding the two contributions gives

E(D)=E(D_1)+E(D_2)\approx53.90+15.38=69.28\text{ events}.

Although the later interval has half the monthly enrollment rate, it still contributes events. Its participants have less follow-up on average, so their average event probability is also lower. Both the number entering and their follow-up distribution determine the total.

Different Event Models Across Groups

When event probabilities differ by treatment arm or another group, average within each group using that group’s enrollment pattern and event model. Add the group-specific expected totals only after calculating them. Averaging event probabilities across groups without accounting for group sizes can distort the projection.

Worked Example: Separate Projections by Treatment Arm

Consider a hypothetical, balanced trial with 100 participants per arm. In each arm, enrollment is uniform over months 0 to 10, and the data cutoff is month 14. Suppose the control-arm event hazard is constant at \(0.06\) per month and the treatment-arm hazard is constant at \(0.03\) per month. Assume both models apply from enrollment, with no competing events or loss to follow-up.

For the control arm, follow-up ranges from 4 to 14 months. Its average event probability is

\bar p_{\text{control}} =1-\frac{e^{-0.06(4)}-e^{-0.06(14)}}{0.06(10)} =1-\frac{0.7866-0.4317}{0.60} \approx0.4085.

The expected control-arm event total is \(100(0.4085)\approx40.85\) events. For the treatment arm,

\bar p_{\text{treatment}} =1-\frac{e^{-0.03(4)}-e^{-0.03(14)}}{0.03(10)} =1-\frac{0.8869-0.6570}{0.30} \approx0.2338.

The expected treatment-arm total is \(100(0.2338)\approx23.38\) events. The combined expected total is therefore \(40.85+23.38\approx64.23\) events. The calculation keeps the enrollment distribution the same in both arms but uses a separate event probability for each arm.

Assumptions, Diagnostics, and Alternatives

The calculation is only as appropriate as its event probabilities and enrollment weights. Before using it, check that the cutoff and enrollment times use the same calendar scale, that participants counted as enrolled are eligible for the stated follow-up, and that the event model applies to the relevant population, arm, and period. If the hazard changes over follow-up, replace the exponential survival function with the appropriate survival model. If event risk also changes with calendar time, a model depending only on time since enrollment may be inadequate.

For a first event that can be precluded by a competing event, \(1-S(a)\) is not generally the probability of the event of interest. Use the cumulative incidence of that event under the competing-risk model instead. Similarly, if loss to follow-up can prevent event ascertainment, the event probability must reflect the specified follow-up and loss assumptions; simply treating every participant as observed to the administrative cutoff may overstate events. These issues should be checked against the endpoint definition and the trial’s follow-up procedures.

Useful diagnostics include comparing projected enrollment totals in each interval with the planned or observed totals, checking that all follow-up durations are nonnegative, and confirming that event probabilities lie between 0 and 1. Inspect the contributions by interval: a late, high-volume recruitment period can have a large effect on the total even if its participants have relatively low individual event probabilities. If a projection changes sharply when the recruitment pattern is slightly altered, that dependence should be visible in the interval contributions.

The main alternative to integration is to use actual enrollment times and sum the individual probabilities directly. That is often preferable when the times are available. A midpoint approximation can be convenient for a quick rough calculation, but it replaces a range of follow-up durations with one value and may be inaccurate when the event probability changes appreciably over that range. Treating all participants as if they had the follow-up of the earliest entrant is especially prone to overprojection. The introductory material on administrative censoring explains why the data cutoff limits each participant’s observed follow-up; the averaging step here combines those participant-specific limits.

Finally, an expected total alone does not describe how much the realized event count may vary around it. The expected count is useful for projecting accrual and comparing assumptions, but it is not a prediction interval. Use a separate variability calculation when uncertainty in the event count itself is required.

Key takeaway: Project the expected event total by pairing each enrollment time with its available follow-up, calculating the corresponding event probability, and weighting that probability by the number enrolled at that time. Uniform and piecewise enrollment are special cases of this same averaging principle.

Check Your Understanding

Answer each question, then compare with the brief solutions.

At a cutoff \(C\), a participant enrolled at \(u\). If the event-free survival probability through \(a\) units of follow-up is \(S(a)\), write the participant’s event probability by the cutoff.
Why is the event probability at the average follow-up generally not the same as the average event probability across enrollment times?
With a constant enrollment rate \(r\) over \([a,b]\), what expression gives the expected number of events contributed by that interval?
A target event can be precluded by a competing event. Why may \(1-S(a)\) fail to give the probability of the target event by follow-up time \(a\)?

Solutions: 1. The available follow-up is \(C-u\), so the event probability is \(1-S(C-u)\). 2. Participants have different follow-up durations, and averaging a function over those durations is not generally equivalent to evaluating the function at their average; the enrollment distribution supplies the weights. 3. The interval contribution is \(r\int_a^b p(C-u)\,du\), where \(p(C-u)\) is the appropriate participant-level event probability. 4. The complement of event-free survival can include participants who experienced a competing event first; the target-event cumulative incidence, which accounts for competing risks, is needed instead.