Administrative Censoring and Study Duration

Why the Calendar End Matters

When participants enter a study over time, a fixed calendar end does not give everyone the same follow-up. Someone enrolled early can be observed for longer than someone enrolled near the end. This is administrative censoring: observation stops because the planned study period ends, not because the participant experiences an event or withdraws.

For competing-risks design, this schedule determines which parts of each cause-specific cumulative incidence function are observable and how many target events are expected. Those quantities inform Gray-test power, but they are not themselves a power calculation: power also depends on the group-specific cumulative-incidence trajectories, the test’s weight function, allocation, and variance.

Definition: Let recruitment take place over \(A\) time units, from calendar time 0 through \(A\), and let the study end at calendar time \(T=A+L\), where \(L\) is the planned minimum follow-up for a participant enrolled at the end of recruitment. If a participant enters at calendar time \(U\), their administrative censoring time measured from entry is \(C=T-U\). Under uniform accrual, \(U\sim\text{Uniform}(0,A)\), so \(C\) ranges from \(L\) to \(L+A\). This definition assumes participants remain under observation until the planned end or an event.

Thus, \(L\) is not the follow-up duration of every participant. It is the follow-up available to the last enrollee. Early enrollees have more. The planned calendar duration is \(T=A+L\), but knowing \(T\) alone is not enough to determine the follow-up distribution: the split between recruitment and minimum follow-up matters.

Average Cumulative Incidence Under a Fixed End

Suppose, for illustration, that each participant in a group has constant cause-specific hazards \(\lambda_1\) for the target event and \(\lambda_2\) for the competing event. Write \(\lambda=\lambda_1+\lambda_2\), with \(\lambda>0\). The target-event cumulative incidence at follow-up time \(t\) is

F_1(t)=P(\text{target event by }t) =\frac{\lambda_1}{\lambda}\left(1-e^{-\lambda t}\right).

This expression accounts for the competing event: a target event can occur only while a participant remains free of either event. With staggered entry, a useful schedule summary is the target-event probability averaged over participants’ administrative follow-up times. Under uniform accrual it is

\bar F_1 =\frac{1}{A}\int_L^{L+A}F_1(t)\,dt =\frac{\lambda_1}{\lambda} \left[ 1-e^{-\lambda L} \left\{\frac{1-e^{-\lambda A}}{\lambda A}\right\} \right], \qquad A>0.

Why the Average Formula Has This Form

Conditional on entry time \(U=u\), the available follow-up is \(T-u=A+L-u\). Uniform accrual means each follow-up time between \(L\) and \(L+A\) is represented equally in the average. Substitute the cumulative incidence formula into its integral:

\begin{aligned} \bar F_1 &=\frac{1}{A}\int_L^{L+A} \frac{\lambda_1}{\lambda}(1-e^{-\lambda t})\,dt\\ &=\frac{\lambda_1}{\lambda} \left[ 1-\frac{1}{A}\int_L^{L+A}e^{-\lambda t}\,dt \right]\\ &=\frac{\lambda_1}{\lambda} \left[ 1-\frac{e^{-\lambda L}-e^{-\lambda(L+A)}}{\lambda A} \right]\\ &=\frac{\lambda_1}{\lambda} \left[ 1-e^{-\lambda L} \left\{\frac{1-e^{-\lambda A}}{\lambda A}\right\} \right]. \end{aligned}

The braces are important: the accrual-time average is \((1-e^{-\lambda A})/(\lambda A)\). If \(n\) participants in that group follow this schedule and share these hazards, the expected number of target events is \(n\bar F_1\). This is an expectation under the stated model, not a guaranteed event count.

Worked Example: Expected Target Events With Staggered Accrual

Suppose 150 participants are recruited uniformly over \(A=2\) years, and the study continues for \(L=1\) year after recruitment closes. For a hypothetical group, let \(\lambda_1=0.20\) per year and \(\lambda_2=0.30\) per year, so \(\lambda=0.50\) per year. Assume these hazards are constant and administrative censoring is the only censoring.

The average target-event probability is

\begin{aligned} \bar F_1 &=\frac{0.20}{0.50} \left[ 1-e^{-0.50(1)} \left\{\frac{1-e^{-0.50(2)}}{0.50(2)}\right\} \right]\\ &=0.40\left[1-(0.6065)\left(\frac{0.6321}{1.000}\right)\right]\\ &=0.40(1-0.3834) =0.2466. \end{aligned}

The follow-up times range from 1 to 3 years. The expected number of target events is \(150(0.2466)=36.99\), approximately 37 events. This calculation uses the target-event cumulative incidence, not \(1-e^{-\lambda_1t}\), which would incorrectly ignore the competing event.

Same Calendar Duration, Different Schedules

A calendar duration can be achieved with a short recruitment period and long minimum follow-up, or with longer recruitment and less follow-up after recruitment closes. These plans have different follow-up distributions and generally different expected event counts. The following comparison keeps the total calendar duration at four years and the event hazards fixed; it changes only the schedule.

Worked Example: Comparing Two Four-Year Plans

Consider 240 participants, recruited uniformly, with \(\lambda_1=0.20\) and \(\lambda_2=0.30\) per year. Plan A recruits for \(A=1\) year and follows through \(L=3\) years after recruitment closes. Plan B recruits for \(A=3\) years and follows for \(L=1\) year after recruitment closes. Both have \(T=A+L=4\) years.

For Plan A, \(\lambda A=0.50\) and \[ \frac{1-e^{-0.50}}{0.50}=0.7869,\qquad e^{-0.50(3)}=0.2231. \] Therefore,

\bar F_{1,A} =0.40[1-(0.2231)(0.7869)] =0.3298, \qquad 240\bar F_{1,A}=79.14.

For Plan B, \(\lambda A=1.50\) and \[ \frac{1-e^{-1.50}}{1.50}=0.5179,\qquad e^{-0.50(1)}=0.6065. \] Therefore,

\bar F_{1,B} =0.40[1-(0.6065)(0.5179)] =0.2743, \qquad 240\bar F_{1,B}=65.84.

Under these assumptions, the shorter recruitment plan yields about 79 expected target events, versus about 66 for the longer recruitment plan. The reason is not a difference in total calendar duration: the average participant in Plan A has 3.5 years of follow-up, compared with 2.5 years in Plan B. Different hazards or a different incidence pattern could change the size of the event-count difference.

From the Schedule to Gray-Test Power Inputs

Gray’s test compares group-specific cumulative incidence functions over follow-up time, accounting for competing events and right censoring. For design, the relevant alternative is therefore a set of group-specific incidence trajectories over the time range that the schedule can support—not just one expected event count. For a common fixed follow-up horizon \(t\), the contrast is \(F_{1,1}(t)-F_{1,2}(t)\). Under staggered accrual, the available follow-up ranges across participants, so later parts of the trajectories are informed by progressively fewer participants.

An average incidence such as \(\bar F_1\) is useful for summarizing expected target-event yield under a schedule. It is not a replacement for the cumulative-incidence contrast or the variance calculation in a Gray-test power approximation. In particular, two designs can have similar total expected event counts but different distributions of event times and different information at times emphasized by the test’s weight function.

Worked Example: Longer Follow-Up and the Group Contrast

Suppose two hypothetical groups are recruited uniformly over \(A=2\) years. In Group 1, let \(\lambda_1=0.20\), \(\lambda_2=0.30\), and \(\lambda=0.50\) per year. In Group 2, let \(\lambda_1=0.30\), \(\lambda_2=0.30\), and \(\lambda=0.60\) per year. Compare a plan with \(L=1\) year after recruitment closes with a plan having \(L=2\) years. Assume equal group sizes of 200 and that the stated hazards and accrual schedule apply in both groups.

For \(L=1\), the average target-event probabilities are

\begin{aligned} \bar F_{1,1} &=0.40\left[1-e^{-0.50} \left\{\frac{1-e^{-1.00}}{1.00}\right\}\right] =0.2466,\\ \bar F_{1,2} &=0.50\left[1-e^{-0.60} \left\{\frac{1-e^{-1.20}}{1.20}\right\}\right] =0.3402. \end{aligned}

The expected target-event counts are \(200(0.2466)=49.32\) in Group 1 and \(200(0.3402)=68.04\) in Group 2. The average-incidence difference is \(0.3402-0.2466=0.0936\).

For \(L=2\), the same formula gives

\begin{aligned} \bar F_{1,1} &=0.40\left[1-e^{-1.00} \left\{\frac{1-e^{-1.00}}{1.00}\right\}\right] =0.3070,\\ \bar F_{1,2} &=0.50\left[1-e^{-1.20} \left\{\frac{1-e^{-1.20}}{1.20}\right\}\right] =0.4123. \end{aligned}

The expected target-event counts become \(61.40\) and \(82.46\), and the average-incidence difference becomes \(0.1053\). These schedule averages summarize expected yield, but the Gray-test contrast at a common time is a different quantity. At \(t=1\) year, the model gives \(F_{1,1}(1)=0.1574\) and \(F_{1,2}(1)=0.2256\), a contrast of \(0.0682\). At \(t=2\) years, it gives \(0.2528\) and \(0.3494\), a contrast of \(0.0966\). A power calculation must use the relevant trajectories and account for when follow-up is available, rather than treating the average-incidence difference as a test effect at a particular time.

Assumptions, Diagnostics, and Alternatives

The constant-hazard expression is a planning model, not a requirement of Gray’s test. With time-varying cause-specific hazards, the group-specific cumulative incidence can be specified or estimated under that alternative model and averaged over the same follow-up distribution. The simple closed form above should not be used unless its hazard assumptions are reasonable for the purpose at hand.

The schedule calculations also require an explicit accrual pattern. Uniform accrual is a simplifying assumption; if recruitment is expected to be faster early or late, average over the planned entry-time distribution instead. Administrative censoring is noninformative when the fixed study end is determined independently of participants’ event processes, conditional on relevant design factors. If participants can leave earlier for reasons related to their prognosis, that is not administrative censoring alone and must be handled separately.

Before using a schedule in a Gray-test power calculation, check that:

The stated study end, recruitment window, and minimum follow-up agree arithmetically: \(T=A+L\).
Follow-up time is measured from each participant’s entry, while \(T\) is measured from the start of the study.
The group-specific cumulative-incidence alternatives cover the follow-up region for which the design expects to have adequate observation.
Recruitment patterns and censoring support are reasonably comparable across groups, or their differences are represented in the power calculation.
Expected event counts are interpreted as schedule summaries, not as sufficient statistics for Gray-test power.

A fixed minimum follow-up design is one alternative to stopping all observation on a common calendar date: each participant is followed for a specified duration after entry, so the final study end depends on the last enrollee. An event-driven design is another alternative, ending after a prespecified event total. Each changes the relationship between calendar duration, follow-up, and available incidence information. Neither should be described as equivalent to a fixed calendar end without accounting for its own stopping and censoring rules.

Common Mistake

Pitfall: “The study lasts four years, so every participant contributes four years of follow-up.” That is false when recruitment takes place during those four years and follow-up stops at the calendar end. With uniform accrual over \(A\) years and \(L\) years of follow-up for the last enrollee, participant follow-up ranges from \(L\) to \(L+A\), with mean \(L+A/2\). Compute incidence over that distribution, and keep average event yield distinct from the time-specific contrasts and variance used for Gray-test power.

Check Your Understanding

Use the definitions and schedule formula above. Each answer is included so you can check the reasoning.

A study recruits uniformly for 18 months and continues for 6 months after recruitment closes. What are the earliest and latest administrative follow-up times? Solution: Here \(A=1.5\) years and \(L=0.5\) years. Follow-up ranges from \(L=0.5\) to \(L+A=2.0\) years.
In one sentence, why can two studies with the same total calendar duration have different expected target-event counts? Solution: They can allocate that duration differently between recruitment and post-recruitment follow-up, producing different participant follow-up distributions and thus different cumulative-incidence probabilities.
For \(\lambda_1=0.20\), \(\lambda_2=0.30\), \(A=0\) in the limiting sense, and \(L=2\) years, what is the target-event probability at the common follow-up time? Solution: There is no staggered-accrual averaging in this limit. The probability is \(0.20/0.50[1-e^{-0.50(2)}]=0.40(1-0.3679)=0.2528\).
Is an expected target-event count by itself a Gray-test power calculation? Solution: No. Power also depends on the group-specific cumulative-incidence trajectories, censoring and risk-set information over time, allocation, variance, and the test’s weight function.
Why is \(1-e^{-\lambda_1t}\) generally not the target-event cumulative incidence when a competing event has hazard \(\lambda_2>0\)? Solution: It ignores the possibility that the competing event occurs first. Under the constant-hazard model, the correct probability is \(\frac{\lambda_1}{\lambda_1+\lambda_2}[1-e^{-(\lambda_1+\lambda_2)t}]\).