Accrual and Follow-Up in Power Calculations

Why Accrual Changes the Power Calculation

A planned sample size does not by itself determine how much information a competing-risks study will collect. If participants enter over time and the study closes on a common calendar date, early enrollees can be observed longer than late enrollees. Their chances of experiencing either the target event or a competing event are therefore different. A calculation that gives every participant the follow-up time of the earliest enrollee will generally overstate expected event counts.

For a Gray test, this matters because the test compares cumulative incidence functions using event times and risk-set information. Expected target-event counts are useful planning inputs, but they do not fully determine Gray-test power: the timing of events, competing events, the shape of the difference between cumulative incidence curves, and the variance of the score also matter. The calculations here show how to adjust event-count inputs for staggered accrual and how to use them in a transparent, approximate information calculation.

Average Event Probability Under Uniform Accrual

Suppose participants enter uniformly during a period of length $A$, measured from the first enrolment, and the study closes at calendar time $T$, where $T>A$. A participant entering at time $u$ has follow-up $T-u$. Thus follow-up ranges from $T-A$ for the latest enrollee to $T$ for the earliest. This is a common-closing-date design; it is not a design that guarantees the same follow-up duration to everyone.

Definition: If $F_j(t)$ is the cumulative incidence by time $t$ of cause $j$ after enrolment, the average probability of that cause by study closure under uniform accrual is $$ \bar F_j(A,T)=\frac{1}{A}\int_0^A F_j(T-u)\,du. $$ For $n_j$ participants in a group, the expected number with cause $j$ observed by closure is $n_j\bar F_j(A,T)$. This is an expectation, not a guaranteed event count.

For a simple calculation, assume constant cause-specific hazards: $\lambda_1$ for the target event and $\lambda_2$ for the competing event. Write $\lambda_\cdot=\lambda_1+\lambda_2$. The event types are mutually exclusive, there is no loss to follow-up before closure, and the hazards apply from enrolment onward. Under these assumptions, the cumulative incidence for cause $j$ at follow-up time $t$ is

F_j(t)=\frac{\lambda_j}{\lambda_\cdot}\left(1-e^{-\lambda_\cdot t}\right).

Averaging this expression over entry times gives a convenient closed form:

\bar F_j(A,T)= \frac{\lambda_j}{\lambda_\cdot} \left[ 1-\frac{e^{-\lambda_\cdot(T-A)}-e^{-\lambda_\cdot T}} {\lambda_\cdot A} \right].

The bracketed term is the average probability of either event by closure. The cause-specific share $\lambda_j/\lambda_\cdot$ divides that probability between the two causes in this constant-hazard model. This share-based simplification is specific to the stated assumptions; it is not a general way to split cumulative incidence when hazards vary over time.

Worked Example: Averaging Incidence Across a 12-Month Accrual Period

Consider a hypothetical two-group study closing 18 months after the first participant enters. Participants accrue uniformly over 12 months, so follow-up ranges from 6 to 18 months. In group 1, suppose the target-event hazard is $0.04$ per year and the competing-event hazard is $0.06$ per year. Convert the times to years: $A=1$, $T=1.5$, and $\lambda_\cdot=0.10$ per year.

The average probability of either event by closure is

1-\frac{e^{-0.10(1.5-1)}-e^{-0.10(1.5)}}{0.10(1)} = 1-\frac{e^{-0.05}-e^{-0.15}}{0.10} =0.0947855.

The target-event share is $0.04/0.10=0.40$, so the average target-event probability is $0.40(0.0947855)=0.0379142$, or $0.03791$ to five decimal places. The competing-event share is $0.60$, giving $0.60(0.0947855)=0.0568713$, or $0.05687$. Their sum, $0.0947855$, is the average probability of either event. These calculations use the unrounded exponential values before rounding the final probabilities.

With 1,000 participants in this group, the expected target-event count is $1000(0.0379142)=37.9142$, approximately 37.91 events. The expected competing-event count is $1000(0.0568713)=56.8713$, approximately 56.87. Expected counts need not be integers.

Using Event Counts in Approximate Gray-Test Planning

One practical approximation scales information with the number of target events when the planned alternative, allocation, and event-time pattern are similar to those in a benchmark calculation. Let $D_{\mathrm{ref}}$ be the benchmark’s expected target-event count and let $\delta_{\mathrm{ref}}$ be its approximate standardized score under the alternative. If the score’s information is approximately proportional to target-event count, then the new standardized score is approximated by

\delta_{\mathrm{new}}\approx \delta_{\mathrm{ref}}\sqrt{\frac{D_{\mathrm{new}}}{D_{\mathrm{ref}}}}.

For a two-sided test at significance level $\alpha$, an approximate power calculation is

\operatorname{Power}\approx \Phi(-z_{1-\alpha/2}-\delta_{\mathrm{new}}) +1-\Phi(z_{1-\alpha/2}-\delta_{\mathrm{new}}),

where $\Phi$ is the standard normal cumulative distribution function and $z_q$ is its $q$-quantile. This approximation is a planning device, not an exact Gray-test power formula. In particular, competing events affect the Gray score’s risk-set contributions and variance, even when the target-event count is held fixed. A benchmark should therefore reflect a sufficiently similar event process and test setup.

Worked Example: Comparing Two Accrual Schedules

Use the hypothetical hazards from the previous example. In group 2, let the target hazard be $0.06$ per year and the competing hazard be $0.04$ per year. Assume 500 participants per group and a common close at $T=1.5$ years. Compare uniform accrual over 12 months ($A=1$) with accrual over 6 months ($A=0.5$). All hazards are constant, all participants are observed to closure, and the two groups have the stated hazards throughout follow-up.

For 12-month accrual, the average probability of either event is $0.0947855$, as calculated above. Group 1’s target probability is $0.40(0.0947855)=0.0379142$. Group 2’s target probability is $0.60(0.0947855)=0.0568713$. The expected total target count is therefore

D_{12}=500(0.0379142)+500(0.0568713)=47.3928.

For 6-month accrual, follow-up ranges from 1 to 1.5 years. The average probability of either event is

1-\frac{e^{-0.10(1.5-0.5)}-e^{-0.10(1.5)}}{0.10(0.5)} = 1-\frac{e^{-0.10}-e^{-0.15}}{0.05} =0.1174112.

The average target probabilities are $0.40(0.1174112)=0.0469645$ in group 1 and $0.60(0.1174112)=0.0704467$ in group 2. Thus

$$ D_6=500(0.0469645)+500(0.0704467)=58.7056. $$

Using unrounded values in the count calculation, the 6-month schedule gives approximately 58.71 expected target events, compared with 47.39 for the 12-month schedule. This difference arises because participants in the shorter-accrual design receive longer follow-up on average before the same closing date.

For illustration, suppose a benchmark calculation for a similar alternative gives $\delta_{\mathrm{ref}}=4.00$ at $D_{\mathrm{ref}}=200$ target events. With $\alpha=0.05$, the 12-month schedule has $\delta_{\mathrm{new}}\approx4\sqrt{47.3928/200}=1.947$, and the 6-month schedule has $\delta_{\mathrm{new}}\approx4\sqrt{58.7056/200}=2.167$. Substituting these into the two-sided normal power expression gives approximate powers of 0.495 and 0.582, respectively. These are illustrative approximations conditional on the benchmark and information-scaling assumption, not guaranteed operating characteristics for a particular Gray test.

Derivation and Checks

The averaging formula follows directly from the distribution of entry times. Under uniform accrual, an entry time $U$ has density $1/A$ on $0\leq u\leq A$. Given entry at $u$, follow-up is $T-u$, so the event probability is $F_j(T-u)$. Taking the expectation over $U$ gives

E\{F_j(T-U)\} =\frac{1}{A}\int_0^A F_j(T-u)\,du.

With constant hazards, substitute $F_j(t)=(\lambda_j/\lambda_\cdot)(1-e^{-\lambda_\cdot t})$. The integral of the exponential term is

\frac{1}{A}\int_0^A e^{-\lambda_\cdot(T-u)}\,du = \frac{e^{-\lambda_\cdot(T-A)}-e^{-\lambda_\cdot T}} {\lambda_\cdot A},

which yields the stated closed form. As a check, the average target and competing probabilities must sum to the average probability of either event. Also, because every participant has follow-up between $T-A$ and $T$, each average cause-specific probability must lie between its cumulative incidences at those two follow-up limits. A result outside these bounds indicates a calculation or unit-conversion error.

Assumptions, Diagnostics, and Alternatives

The constant-hazard example assumes the hazards are correctly specified across the full follow-up window. In an actual design, hazards may vary with time, and group differences may be delayed or nonproportional. In that case, calculate $F_j(t)$ from the planned cause-specific hazard functions or other justified incidence model, then average $F_j(T-u)$ over the accrual distribution. Do not apply the constant-hazard share formula automatically.

Check the design inputs before interpreting the output:

Confirm that $A$ and $T$ use the same time units as the hazards, and that the latest entry has positive follow-up $T-A$.
Check that the modeled cause-specific probabilities sum to no more than one; the remainder is the probability of no event by closure.
Compare average probabilities with the incidences at the shortest and longest follow-up times.
Inspect whether the benchmark and proposed design have comparable group allocation, target and competing-event patterns, and temporal shape of the alternative.
Revisit the approximation if few target events are expected, competing events are frequent, or cumulative incidence curves cross or differ mainly at particular times. Event-count scaling alone may then misrepresent the Gray score’s mean or variance.

A common alternative is to plan a fixed follow-up duration for every participant. That design gives all enrollees the same time origin-to-endpoint window, but the calendar close occurs later for participants enrolled near the end of accrual. Another alternative is to set a target event count and continue follow-up until it is reached; this is event-driven planning and does not fix the calendar duration in advance. Under either design, the relevant event probabilities must match the actual stopping rule. Neither approach can be substituted into the common-closing-date formula without changing the follow-up distribution.

Common Mistake

A frequent error is to multiply the sample size by the cumulative incidence at the maximum follow-up time $T$, as though everyone entered at the start. Under staggered accrual, that overstates the average observation time. The correction is to average the incidence over actual entry times, then multiply by group size. A second error is to treat the resulting expected target count as if it were the exact information for Gray’s test. Competing events and event timing contribute to the score and its variance, so the count-based scaling should be presented as an approximation with its assumptions, not as an exact power calculation.

Key takeaway: Under a common study close, staggered accrual creates a distribution of follow-up times. Average the cause-specific cumulative incidence over that distribution before calculating expected event counts; then use those counts only within a clearly stated Gray-test power approximation.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

A study closes at $T=2$ years and accrues uniformly for $A=0.5$ years. What are the shortest and longest follow-up times?
With constant hazards, why is the average target-event probability not generally the target-event cumulative incidence at $T$?
In a 12-month accrual design with the hazards in the examples, how many expected target events are there in group 2 with 500 participants?
What assumption is needed to scale a benchmark standardized score by the square root of the target-event-count ratio?
Name one design feature that can make event-count scaling a poor approximation to Gray-test power.

Solutions. (1) The latest entrant has $T-A=1.5$ years of follow-up; the earliest has 2 years. (2) Entry times differ, so the average probability is $\frac{1}{A}\int_0^A F_j(T-u)\,du$, not $F_j(T)$. (3) Group 2’s target probability is $0.0568713$, so the expected count is $500(0.0568713)=28.4357$, approximately 28.44 events. (4) The score information must be approximately proportional to target-event count, with a sufficiently similar alternative and event process in the benchmark and proposed design. (5) Examples include crossing cumulative incidence curves, frequent competing events, or an alternative whose difference is concentrated at particular event times.