From Follow-Up Records to Cumulative Hazard
A survival data set may contain several events recorded at exactly the same time. This can happen because follow-up is recorded in days or weeks, or because the measurement system reports only a limited number of decimal places. A hand calculation must treat those tied records consistently: count who was at risk just before the time, count how many events occurred then, and add the corresponding increment to the cumulative hazard.
The calculation is simple once the risk sets are correct. Most errors arise earlier: a person censored at a time is removed too soon, an event time is treated as though tied events happened one at a time, or a risk set is not updated before moving to the next time.
For the calculations below, assume a cohort with one observed follow-up time and event indicator per person, no delayed entry, and right censoring. A person is at risk at a time if they have not had an earlier event or been censored earlier. We use the common convention that someone censored at the same recorded time as an event is included in the risk set for that time; removal occurs after the events at that time are counted. The event times and censoring are assumed to be recorded consistently, and the usual interpretation of the survival estimate relies on censoring being noninformative for the event process, at least conditional on variables accounted for in the analysis.
Why Tied Events Share One Increment
At time \(t_j\), the observed proportion of those at risk who experience the event is \(d_j/Y_j\). The Nelson–Aalen estimate uses this quantity as the estimated hazard increment at that time, then adds increments across distinct event times. When events are tied, they are counted together in \(d_j\); the risk set is not reduced between those tied events.
This gives a direct justification for the hand calculation. The risk set is formed immediately before \(t_j\), so all \(Y_j\) individuals are eligible to experience an event at that recorded time. Of those, \(d_j\) do experience an event. The observed event fraction is therefore \(d_j/Y_j\), and the cumulative estimate through the time is the prior sum plus this increment:
This convention does not assert that tied events occurred in a known order. If two events are recorded at the same time, arbitrarily ordering them and shrinking the risk set after the first would add information the data do not contain and would produce a different estimate.
Worked Example: A Complete Risk-Set Table With Tied Events
Suppose nine hypothetical participants have these observed follow-up times and statuses. “Event” means the event of interest was observed; “censored” means follow-up ended without an observed event.
| Participant | Follow-up time | Status |
|---|---|---|
| A | 1 | Event |
| B | 2 | Event |
| C | 2 | Event |
| D | 3 | Censored |
| E | 3 | Event |
| F | 4 | Event |
| G | 4 | Event |
| H | 5 | Censored |
| I | 6 | Event |
Start with \(Y=9\). At time 1, one event occurs, so \(Y_1=9\), \(d_1=1\), and the increment is \(1/9=0.1111\). Participant A is then removed from later risk sets. At time 2, \(Y_2=8\), and two events occur together, so the increment is \(2/8=0.25\). Remove B and C after this time.
Six participants remain at risk just before time 3. D is censored at 3 and E has an event at 3. Under the stated convention, D is included in the risk set for the event at that time: \(Y_3=6\), \(d_3=1\), giving an increment of \(1/6=0.1667\). After time 3, remove both D and E. Four participants remain at risk at time 4; F and G have tied events, giving \(Y_4=4\), \(d_4=2\), and increment \(2/4=0.5\).
At time 5, two participants remain, but H is censored and there is no event. Thus \(Y=2\), \(d=0\), and the increment is zero; H is removed afterward. At time 6, I is the only remaining participant at risk, so \(Y=1\), \(d=1\), and the increment is \(1/1=1\).
| Time | At risk just before time, \(Y_j\) | Events, \(d_j\) | Increment, \(d_j/Y_j\) | Cumulative estimate |
|---|---|---|---|---|
| 1 | 9 | 1 | 0.1111 | 0.1111 |
| 2 | 8 | 2 | 0.2500 | 0.3611 |
| 3 | 6 | 1 | 0.1667 | 0.5278 |
| 4 | 4 | 2 | 0.5000 | 1.0278 |
| 5 | 2 | 0 | 0 | 1.0278 |
| 6 | 1 | 1 | 1.000 | 2.0278 |
For example, through time 4 the arithmetic is \(1/9+2/8+1/6+2/4=0.1111+0.2500+0.1667+0.5000=1.0278\). The estimated cumulative hazard through time 6 is \(2.0278\). A cumulative hazard is not a probability, so a value greater than 1 is possible.
Keeping Risk Sets Correct When Censoring Ties an Event
A censoring record affects subsequent risk sets even though it does not add an event increment. When its time is later than an event time, that person remains at risk through the earlier event and is removed only at censoring. When censoring and an event share the same recorded time, the convention must be stated because the data do not reveal their order within that time unit. The example below uses the convention stated above: count the censored person in the risk set for events at that time, then remove them.
Worked Example: Event and Censoring at the Same Time
Consider six hypothetical participants: A and B have events at time 2; C is censored at time 2; D has an event at time 4; E is censored at time 5; and F has an event at time 6. Initially all six are at risk.
At time 2, the tied event count is \(d_2=2\). Participant C is also censored then, but is included in the risk set under the stated convention, so \(Y_2=6\). The increment is \(2/6=1/3=0.3333\). After time 2, remove A, B, and C, leaving three at risk.
At time 4, D is the only event and \(Y_4=3\), giving \(1/3=0.3333\). After D's event, two participants remain. E's censoring at time 5 contributes no increment and leaves F alone at risk after time 5. At time 6, F has an event, so \(Y_6=1\) and the increment is 1. The cumulative hazard through time 4 is \(1/3+1/3=0.6667\), and through time 6 it is \(1/3+1/3+1=1.6667\).
If a different recording protocol specifies that censoring at a tied time occurs just before events, the risk set convention would need to reflect that protocol. The important practice is not to switch conventions mid-calculation or silently remove a censored participant before a tied event.
Do Not Break Ties Into an Artificial Sequence
Worked Example: Comparing Pooled and Sequential Tie Calculations
Five hypothetical participants are followed. Two have events at time 1, one has an event at time 3, and the remaining two are censored after time 3. At time 1, all five are at risk, so the correct tied-event increment is \(2/5=0.4\). After both events, three remain. The event at time 3 contributes \(1/3=0.3333\), so the cumulative estimate through time 3 is \(0.4+0.3333=0.7333\).
A tempting but incorrect calculation is to process the two time-1 events one at a time: first \(1/5\), then \(1/4\). This gives \(0.20+0.25=0.45\) at time 1 and a total of \(0.45+1/3=0.7833\) through time 3. The discrepancy arises because the sequential calculation acts as if one tied event occurred first and removed a participant before the other event. The recorded data provide no such ordering. For the Nelson–Aalen estimate with tied observations, use the pooled increment \(d_j/Y_j\).
Checks, Alternatives, and Limits
A reliable hand calculation should satisfy several basic checks:
- At each time, \(Y_j\) counts people still under observation and event-free immediately beforehand; it cannot be larger than the preceding risk set.
- The event count must satisfy \(0\le d_j\le Y_j\). A value \(d_j>Y_j\) signals a counting or transcription error.
- Remove all events and censorings at a time before constructing the next risk set, while applying the stated same-time convention consistently.
- Every increment is nonnegative, so the cumulative estimate cannot decrease. A censoring-only time leaves it unchanged.
- Use the recorded time scale consistently. Rounding follow-up times can create apparent ties, so do not invent an ordering or split a tie without a substantive basis.
The Kaplan–Meier estimator is a common alternative survival summary. At an event time it updates estimated survival by multiplying by \(1-d_j/Y_j\); the Nelson–Aalen estimator instead adds \(d_j/Y_j\) to cumulative hazard. They are related, but they are not identical finite-sample calculations. In particular, do not interpret a Nelson–Aalen cumulative-hazard value as an estimated event probability. The survival summary can be approximated from cumulative hazard through \(\exp\{-\widehat{H}(t)\}\), but that is not the same calculation as the Kaplan–Meier product.
These hand calculations also have limits. They assume each person contributes one continuous period of observed follow-up, so delayed entry or multiple start-stop intervals require a risk-set construction that accounts for those features. The calculation itself does not remedy informative censoring, incorrect event times, or an unsuitable definition of the event. Such issues affect the data and assumptions behind the estimate, not just its arithmetic.
Check Your Understanding
Try each calculation using the risk set immediately before the specified time. Solutions follow the questions.
- Eight people are at risk at a time when three events occur together. What is the Nelson–Aalen increment?
- At a time, four people are at risk, one event occurs, and one other person is censored at that same recorded time. Under the convention in this tutorial, what is the increment, and how many remain at risk afterward?
- At time 1, two events occur among five people at risk. At time 2, one event occurs among the three remaining people. Calculate the cumulative hazard through time 2.
- A time has no events and two people are censored. What increment is added at that time, and what happens to the later risk set?
- Why is it incorrect to calculate two tied events among five people as \(1/5+1/4\)?
Solutions. (1) The increment is \(3/8=0.375\). (2) The censored person is included in the risk set for the tied event, so the increment is \(1/4=0.25\); remove the event participant and the censored participant afterward, leaving two at risk. (3) The increments are \(2/5=0.4\) and \(1/3=0.3333\); the cumulative hazard is \(0.4+0.3333=0.7333\). (4) The increment is zero because \(d=0\); remove the two censored people when forming subsequent risk sets. (5) That calculation invents an order among events recorded at the same time and changes the denominator after the first event. The tied events should instead contribute the pooled increment \(2/5=0.4\).