Survival Event Counts And Follow-Up

Why Follow-Up Determines the Predictive Count

A survival prior can sound clinically plausible while implying an implausible number of observed events by the planned analysis date. The number of events is not determined by the survival model alone: it also depends on how many participants are enrolled, when they enter the trial, and how long each can be followed before the administrative cutoff. A prior predictive check should therefore target the count that the trial could actually observe, rather than a count under complete and equal follow-up if that is not the design.

Here the target is the number of participants whose event has occurred by a specified cutoff. A participant who has not had the event by their last observed follow-up contributes no event to this count, even if their event time after that point is unknown. This is administrative censoring when follow-up ends because the study reaches its planned cutoff. The calculations below focus on a single event type and a single arm; the same logic can be applied separately to an arm when its hazard prior and follow-up schedule are specified.

Definition: Let \(Y_i=1\) if participant \(i\) experiences the event before or at their administrative follow-up time \(d_i\), and \(Y_i=0\) otherwise. The observed event count by the cutoff is \(K=\sum_{i=1}^{n}Y_i\). The follow-up time \(d_i\) is measured from participant \(i\)'s enrollment to the cutoff, not from the trial's first enrollment unless those times coincide.

Model and Assumptions

For transparent calculations, suppose the event time for each participant follows an exponential model with a common hazard \(\lambda\). A hazard is an instantaneous event rate; \(\lambda\) has units of inverse time. Under this model, conditional on \(\lambda\), the probability of no event through \(d_i\) time units is \(e^{-\lambda d_i}\), so the event probability by that time is \(q_i(\lambda)=1-e^{-\lambda d_i}\).

The key assumptions for these count calculations are that the hazard is constant over time, participants share the same hazard within the modeled group, and their event times are conditionally independent given \(\lambda\). Enrollment times and the administrative cutoff are treated as known. Censoring is administrative and does not depend on a participant's unobserved event time. There are no competing event types in this calculation. If any of these assumptions are inappropriate, the implied event count may not describe the intended trial.

Y_i\mid\lambda \sim \operatorname{Bernoulli}\!\left(1-e^{-\lambda d_i}\right), \qquad K\mid\lambda \sim \text{Poisson-binomial}\big(q_1(\lambda),\ldots,q_n(\lambda)\big).

The conditional count is binomial when all participants have the same follow-up time \(d\): \(K\mid\lambda\sim\operatorname{Binomial}(n,q(\lambda))\). With unequal follow-up times, it is generally a Poisson-binomial count, the sum of independent Bernoulli variables with possibly different probabilities. Integrating over a prior for \(\lambda\) gives the prior predictive distribution of \(K\). Even though the outcomes are independent conditional on \(\lambda\), they are generally dependent after \(\lambda\) is averaged over: a draw of a high hazard makes events more likely for everyone in that replicate.

From a Hazard Prior to Event-Count Implications

Suppose the prior is \(\lambda\sim\operatorname{Gamma}(a,b)\), using shape \(a\) and rate \(b\), so \(b\) has units of time. Its Laplace transform is \(E[e^{-s\lambda}]=(b/(b+s))^a\) for \(s\geq0\). This gives the prior mean event probability for follow-up \(d\):

E[q(\lambda)] =1-E[e^{-d\lambda}] =1-\left(\frac{b}{b+d}\right)^a.

Thus, for participants with a common follow-up \(d\), \(E[K]=nE[q(\lambda)]\). For unequal follow-up times, linearity of expectation gives \(E[K]=\sum_i E[q_i(\lambda)]\). Neither result requires the \(Y_i\) to be marginally independent.

Justification for the predictive variance: With common follow-up, conditional on \(\lambda\), the count has variance \(nq(1-q)\). The law of total variance gives

\begin{aligned} \operatorname{Var}(K) &=E[\operatorname{Var}(K\mid\lambda)] +\operatorname{Var}(E[K\mid\lambda])\\ &=nE[q(1-q)]+n^2\operatorname{Var}(q)\\ &=n\mu_q(1-\mu_q)+n(n-1)\operatorname{Var}(q), \end{aligned}

where \(\mu_q=E[q]\). The first term in the last line reflects Bernoulli outcome variation; the second reflects uncertainty in the shared event probability induced by the prior. In particular, the predictive variance is not obtained by simply plugging the prior mean hazard into a binomial model. For unequal follow-up, conditional variance is \(\sum_i q_i(1-q_i)\), and conditional mean is \(\sum_iq_i\); applying the same total-variance decomposition accounts for their shared dependence through \(\lambda\).

Worked Example: One-Year Follow-Up for Every Participant

Suppose, hypothetically, that 100 participants each have one year of follow-up and that the common hazard has prior \(\lambda\sim\operatorname{Gamma}(2,20)\), with the rate measured in years. The model assumes a constant hazard and conditionally independent event times; the common hazard and equal follow-up make a binomial count appropriate conditional on \(\lambda\).

For \(d=1\) year, the prior mean event probability is

E[q]=1-\left(\frac{20}{21}\right)^2 =\frac{41}{441} \approx 0.09297.

Therefore the prior predictive expected count is \(E[K]=100(0.0929705)\approx9.297\) events. This is an expectation across hypothetical prior draws and trial outcomes, not a prediction that exactly nine events will occur.

To show how much prior uncertainty affects the count, calculate the second moment using the unrounded expressions:

\begin{aligned} E[q^2] &=E[(1-e^{-\lambda})^2]\\ &=1-2\left(\frac{20}{21}\right)^2+\left(\frac{20}{22}\right)^2 \approx0.0123873,\\ \operatorname{Var}(q) &=\left[1-2\left(\frac{20}{21}\right)^2+\left(\frac{20}{22}\right)^2\right] -\left[1-\left(\frac{20}{21}\right)^2\right]^2\\ &\approx0.0037438. \end{aligned}

The predictive count variance is then

\operatorname{Var}(K) =100(0.0929705)(1-0.0929705) +100(99)(0.0037438) \approx45.50.

The predictive standard deviation is about \(\sqrt{45.50}=6.745\) events. For comparison, using the prior mean event probability as if it were fixed would give variance \(100(0.0929705)(1-0.0929705)\approx8.433\), missing the substantial additional variation from uncertainty about \(\lambda\). The probability of zero events can also be calculated directly:

P(K=0)=E[e^{-100\lambda}] =\left(\frac{20}{120}\right)^2 =\frac{1}{36} \approx0.02778.

All participants have complete one-year administrative follow-up in this example, so no adjustment for staggered entry is needed. The Gamma prior and exponential model are assumptions to scrutinize, not facts established by the arithmetic.

Staggered Enrollment and Administrative Censoring

When participants enroll at different calendar times and the analysis occurs at a common cutoff, they have different \(d_i\). For example, someone enrolled early may have a year of follow-up, while someone enrolled late may have only a few months. The count expectation uses each participant's own duration; assigning everyone the maximum follow-up will generally overstate the expected number of observed events.

Worked Example: Four Different Follow-Up Durations

Consider a hypothetical four-person trial with follow-up durations of 0.25, 0.50, 0.75, and 1.00 years at the administrative cutoff. Use the same \(\operatorname{Gamma}(2,20)\) prior in the preceding example. The expected event count is the sum of the four prior mean event probabilities:

\begin{aligned} E[K] &=\sum_{d_i\in\{0.25,0.50,0.75,1.00\}} \left[1-\left(\frac{20}{20+d_i}\right)^2\right]\\ &\approx 0.02454+0.04819+0.07098+0.09297\\ &\approx0.2367\text{ events}. \end{aligned}

The figure is an expectation, so it need not be an integer. As a check on the role of total person-time, the probability of no events is

P(K=0)=E\!\left[e^{-\lambda\sum_i d_i}\right] =\left(\frac{20}{20+2.50}\right)^2 \approx0.7901.

This zero-event probability uses conditional independence given the common hazard and the total follow-up of 2.50 person-years. It does not imply that event indicators are independent after integrating over the shared hazard.

Worked Example: Comparing a Cutoff with Full Follow-Up

Suppose, hypothetically, that 60 participants enroll uniformly over the first year and the administrative cutoff is 1.5 years after the first enrollment. Treat enrollment time as uniform over that year, so follow-up duration at the cutoff ranges uniformly from 0.5 to 1.5 years. Under the same hazard prior, the average prior probability of no event across these follow-up durations is

\begin{aligned} E_d[E(e^{-\lambda d})] &=\int_{0.5}^{1.5}\left(\frac{20}{20+d}\right)^2\,\mathrm{d}d\\ &=400\left(\frac{1}{20.5}-\frac{1}{21.5}\right) =\frac{1600}{1763} \approx0.907544. \end{aligned}

The interval of integration has length one year, so this integral is also the average over the uniform follow-up distribution. The average event probability is \(1-1600/1763\approx0.09246\), giving an expected count of \(60(0.092456)\approx5.547\) events. If instead all 60 participants were incorrectly assigned 1.5 years of follow-up, the implied expectation would be

60\left[1-\left(\frac{20}{21.5}\right)^2\right] =60\left(\frac{249}{1849}\right) \approx8.080\text{ events}.

The difference arises from the actual administrative follow-up schedule, not from a change in the prior. The uniform-enrollment calculation gives an expected count; an exact count distribution would also account for the specified individual enrollment times and the shared uncertain hazard.

Practical Checks, Alternatives, and Limitations

A useful check starts with the design calendar: list each planned enrollment time, the cutoff date, and the resulting \(d_i\). Then verify that the prior parameter units agree with the time units used for \(d_i\). A hazard expressed per month cannot be combined directly with follow-up expressed in years without conversion. Inspect whether the implied expected count, probability of zero events, and predictive spread are compatible with the trial's purpose and feasibility. Simulation from prior draws can also reveal the range of counts implied by the full model, but it must use the actual enrollment schedule and administrative cutoff.

The analytic expectation is often the quickest summary. A plug-in calculation at the prior mean hazard is a simpler alternative, but it does not integrate over prior uncertainty and generally does not equal the prior predictive event probability: \(E[1-e^{-\lambda d}]\) is not \(1-e^{-E[\lambda]d}\). Conditional binomial calculations are appropriate for equal follow-up at a fixed hazard; with unequal follow-up, use the individual probabilities or a simulation that preserves them. A predictive simulation should draw one shared hazard for a replicate, then simulate each participant's event indicator using that draw. Drawing a fresh hazard for every participant would represent a different model and discard the shared parameter uncertainty.

The exponential model is a deliberately simple way to translate a hazard prior into an event count. A constant hazard may be implausible when risk changes over time, and a common hazard may be inadequate for heterogeneous participants. Competing events, informative loss to follow-up, uncertainty in enrollment timing, or dependence among participants require additional modeling. These calculations concern event counts by a fixed cutoff; they do not by themselves describe the pattern of event times or establish that a chosen prior is clinically justified.

Common Mistakes

Giving everyone the longest follow-up. Late enrollees have less time to experience an observed event before the common cutoff. Use each participant's duration.
Using a plug-in hazard as the full predictive calculation. The event count varies both because outcomes are stochastic and because the prior hazard is uncertain. Integrate over the hazard prior or simulate it.
Interpreting a censored participant as an event. A participant without an event by their last follow-up contributes zero to the event count by that time; censoring does not mean an event occurred.
Mixing time units. The product \(\lambda d_i\) must be dimensionless. Convert months to years, or vice versa, consistently.

Key takeaway: A prior predictive event count is a joint implication of the event-time model, the prior on its parameters, enrollment size, and each participant's administrative follow-up. Check all four before judging whether the implied count is plausible.

Check Your Understanding

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

Under a constant hazard \(\lambda\), what is the probability of an event by follow-up time \(d\), and what does \(d\) measure?
For equal follow-up \(d\), \(n\) participants, and a shared uncertain hazard, why can the prior predictive count variance exceed the binomial variance calculated at \(E[q]\)?
With a \(\operatorname{Gamma}(1,10)\) prior on a per-year hazard, calculate the mean event probability by 2 years.
Three participants have administrative follow-up times of 0.5, 1, and 2 years. State the expression for their expected count under a shared hazard prior, without assuming equal follow-up.

Solutions. (1) The probability is \(1-e^{-\lambda d}\); \(d\) is time from that participant's enrollment to the administrative cutoff. (2) Variation in the shared hazard adds the term \(n(n-1)\operatorname{Var}(q)\) to \(nE[q](1-E[q])\), so uncertainty in \(q\) increases predictive variance when \(\operatorname{Var}(q)>0\). (3) Here \(a=1\), \(b=10\), and \(d=2\) years, so \(E[q]=1-(10/12)=1/6\approx0.1667\). (4) The expected count is \(E[1-e^{-0.5\lambda}]+E[1-e^{-\lambda}]+E[1-e^{-2\lambda}]\), with each duration expressed in years and expectations taken over the same prior for \(\lambda\).