Why Tied Event Times Need Care
When multiple subjects have an event recorded at the same time, the cumulative incidence calculation must treat the failures consistently. This matters especially when tied failures belong to different causes: each cause contributes an event increment, but every increment is calculated from the same probability of remaining event-free just before that time. The denominator is also the number at risk immediately before the tied events, not a denominator reduced by failures that occurred at that same time.
Ties can arise because events genuinely share an observed time or because times were recorded only to a limited precision, such as whole days. The Aalen–Johansen estimator handles a tied time as one update. It does not require an arbitrary ordering of the individual failures within that time.
The same pre-event risk set \(Y(t)\) appears in every cause-specific increment. A subject who fails at \(t\) contributes to the risk set just before \(t\), even though that subject is no longer at risk immediately afterward. Cause counts at \(t\) are simultaneous counts, not a sequence of failures that successively shrinks the denominator.
The Calculation at a Tied Time
For each event time, first identify who was at risk immediately before it. Count failures separately by cause, then add those cause counts to obtain \(d(t)\). Update the event-free survival using the total failure count, and update each cause's cumulative incidence using its own count. If there are several event times, repeat this operation in chronological order, carrying forward the estimates from the preceding time.
These expressions sum or multiply over distinct observed event times \(u\). A tied group contributes one term at its shared time, with all of its failures included in the corresponding \(d_k(u)\) and \(d(u)\). In particular, the incidence increment for cause \(k\) is not calculated by first removing failures of another cause at that same time.
Worked Example: Tied Failures From Two Causes
Suppose 10 hypothetical subjects are at risk at time 2. At that time, 2 have an event from cause A and 1 has an event from cause B. No subject was censored or had an event before time 2. At time 5, 1 of the 7 remaining subjects fails from cause A and 2 fail from cause B. Calculate both cumulative incidences through time 5.
At time 2, \(Y(2)=10\), \(d_A(2)=2\), \(d_B(2)=1\), and \(d(2)=3\). Thus
There are 7 subjects still at risk at time 5, so \(Y(5)=7\). The event-free survival just before time 5 is 0.7. The increments at time 5 are
Therefore \(\widehat F_A(5)=0.2+0.1=0.3\) and \(\widehat F_B(5)=0.1+0.2=0.3\). The updated survival is \(\widehat S(5)=0.7(1-3/7)=0.4\). Through time 5, the estimated probability of cause A is 0.3, that of cause B is 0.3, and the probability of remaining event-free is 0.4.
Why the Same Pre-Event Survival Is Used
The update follows from the conditional-probability interpretation of the estimator. Immediately before \(t\), \(\widehat S(t-)\) estimates the probability of reaching \(t\) without any event. Among the \(Y(t)\) subjects at risk just before \(t\), the observed fraction with cause \(k\) at \(t\) is \(d_k(t)/Y(t)\). Multiplying these quantities gives the estimated probability of reaching \(t\) event-free and then having cause \(k\) at \(t\).
This is the justification for using a common \(\widehat S(t-)\) and common \(Y(t)\) across causes at a tie. It also gives a useful internal check. Summing the incidence increments over causes yields
Thus the total increase in cumulative incidence at \(t\) equals the decrease in event-free survival at \(t\). Starting from \(\widehat S(0)=1\) and \(\widehat F_k(0)=0\), summing over times shows that \(\widehat S(t)+\sum_k\widehat F_k(t)=1\), provided all event causes are represented.
Worked Example: Checking a Tied-Time Update
At a hypothetical event time, 8 subjects are at risk. Two fail from cause A and one fails from cause B. Just before this time, suppose \(\widehat S(t-)=0.6\), \(\widehat F_A(t-)=0.15\), and \(\widehat F_B(t-)=0.25\). Calculate the updates and verify the identity.
The total failure count is \(d(t)=2+1=3\). The survival update is
The incidence increments are \(0.6(2/8)=0.15\) for cause A and \(0.6(1/8)=0.075\) for cause B. Hence \(\widehat F_A(t)=0.15+0.15=0.30\) and \(\widehat F_B(t)=0.25+0.075=0.325\). The updated total is \(0.375+0.30+0.325=1.000\), as required. The total incidence increase, \(0.15+0.075=0.225\), also equals the survival decrease, \(0.6-0.375=0.225\).
Risk Sets and Censoring at the Same Recorded Time
The risk set must be formed using a clear convention. Under the usual right-censoring convention for exact event times, \(Y(t)\) includes subjects whose observed time is \(t\), whether they have an event at \(t\) or are censored at \(t\). The tied event counts \(d_k(t)\), however, include only failures, not censorings. Subjects censored at \(t\) leave the risk set after that time.
This convention corresponds to defining the risk set immediately before \(t\). If times are rounded or recorded in broad intervals, a recorded event and censoring at the same timestamp may not reveal their true within-interval ordering. The data protocol and analysis convention must then be stated; the record alone cannot determine a finer order. This estimator treats all events with the same recorded time as one tied-time update.
Worked Example: A Censoring Time Tied With Failures
In a hypothetical dataset, 6 subjects are at risk immediately before time 4. At time 4, 2 fail from cause A, 1 fails from cause B, and 1 is censored. The other 2 remain event-free after time 4. Suppose the estimates just before time 4 are \(\widehat S(4-)=0.8\), \(\widehat F_A(4-)=0.1\), and \(\widehat F_B(4-)=0.1\). Use the usual convention that a censoring at time 4 is included in the pre-time risk set.
The denominator is \(Y(4)=6\), not 5: the censored subject is in the pre-event risk set. The failure counts are \(d_A(4)=2\), \(d_B(4)=1\), and \(d(4)=3\). Therefore
The updated estimates are \(\widehat F_A(4)=0.3667\) and \(\widehat F_B(4)=0.2333\). Their sum with survival is \(0.4+0.3667+0.2333=1.000\), up to rounding. The censoring contributes to the risk-set size but not to the failure count or to either cumulative incidence increment.
Common Mistakes, Diagnostics, and Limits
A frequent error is to remove tied failures from the risk set before calculating the increments for the other causes. For example, a calculation might use \(Y(t)-d_A(t)\) as the denominator for cause B, as if all cause A failures occurred first. That changes the data's simultaneous tie into an invented ordering. The correction is to use \(Y(t)\), the pre-event denominator, for each cause-specific increment.
Another error is to count censorings as failures. Censorings reduce the risk set after their recorded time, but they do not enter \(d(t)\) or \(d_k(t)\). Conversely, excluding a subject censored at the tied timestamp from \(Y(t)\) when using the usual pre-time convention incorrectly reduces the denominator.
Several checks help identify a faulty calculation:
- At every time, verify that \(d(t)=\sum_k d_k(t)\) and \(d(t)\leq Y(t)\).
- Use the risk set just before the event time, and ensure it reflects earlier events and censorings.
- Check that survival does not increase and that each cumulative incidence does not decrease.
- Check that \(\widehat S(t)+\sum_k\widehat F_k(t)=1\), subject to rounding.
- Confirm that no tied event has been counted twice or assigned to more than one mutually exclusive cause.
The calculation describes the observed-time tie as a common jump; it cannot recover an unrecorded ordering within a rounded time. If within-interval event timing is scientifically important, a tied-time calculation cannot supply that missing information. The estimator also relies on appropriate observation and censoring assumptions for its population interpretation, including independent censoring in the relevant analysis. Tied-time arithmetic does not repair informative censoring, misclassified causes, or an incorrectly constructed risk set.
The alternative of imposing an arbitrary sequence on tied failures is generally not appropriate: it can assign different cause-specific increments depending on which cause is placed first. A common pre-event update avoids that artificial dependence. This is distinct from a setting in which the observation process itself defines discrete event intervals; in that case, the interpretation and model for interval-level events must be considered explicitly.
Check Your Understanding
Use the shared pre-event denominator in each calculation. Solutions are included after the questions.
- There are 12 subjects at risk at time \(t\). Three fail from cause A and 2 from cause B. If \(\widehat S(t-)=0.5\), what are the survival and incidence increments?
- Why is it incorrect to use \(Y(t)-d_A(t)\) as the denominator for cause B when causes A and B fail at the same recorded time?
- Under the usual pre-time risk-set convention, 1 subject is censored at the same recorded time that 2 subjects fail. Does that censoring count in \(Y(t)\), \(d(t)\), both, or neither?
- After an update, \(\widehat S(t-)=0.7\), and tied failures comprise 1 event from cause A and 2 from cause B among 10 at risk. Verify that the total incidence increment equals the survival decrease.
Solutions. (1) The total failure fraction is \(5/12\), so \(\widehat S(t)=0.5(1-5/12)=0.2917\). The increments are \(0.5(3/12)=0.125\) for A and \(0.5(2/12)=0.08333\) for B. (2) That denominator treats one tied cause as if it occurred earlier and removed subjects before the other cause's failures. Both increments use the pre-event \(Y(t)\). (3) It counts in \(Y(t)\), but not in \(d(t)\), because censoring is not a failure. (4) The total incidence increment is \(0.7(3/10)=0.21\). Survival decreases from \(0.7\) to \(0.7(1-3/10)=0.49\), a decrease of \(0.21\), so the quantities agree.