From a Right-Censored Life Table to a Cumulative Hazard Estimate
A right-censored life table records how many subjects experience the event of interest and how many leave observation without an observed event. For example, in a hypothetical insurance study, the event might be a claim reaching a specified threshold; subjects whose observation ends first are right-censored. The Nelson-Aalen estimator uses the number at risk immediately before each event time to calculate a hazard increment, then adds those increments over time.
The calculation is straightforward once the risk set is correct. The main task is keeping track of who remains under observation at each event time: a subject who has already failed or has already been censored is no longer at risk.
In this tutorial, event times are distinct, and no censoring occurs at an event time. Censoring between event times reduces the next risk set but does not itself create a hazard increment. For the standard interpretation of the estimator, assume censoring is independent of the event process. This means that, conditional on the information used in the study, censoring does not systematically select subjects with different event prospects.
Why the Risk Set Is the Denominator
The increment \(d_j/Y_j\) compares the number of observed events at a particular time with the number who could have experienced an observed event just before that time. A subject censored earlier is no longer in that comparison: after censoring, the study provides no further event information for that subject.
The Nelson-Aalen estimator adds these local increments. To see why the sum has this form, consider the cumulative hazard as accumulated hazard over successive event times. At \(t_j\), the observed local contribution is the event count \(d_j\) per subject exposed to the event risk, \(Y_j\). Adding the contributions through \(t\) gives the estimator. This is an empirical estimator: it uses the observed risk sets and event counts, rather than requiring a specified parametric distribution for event times.
For a life table, calculate the risk set by starting with the initial number under observation and subtracting every failure and censoring that occurred before the event time being considered. A censoring after an event time cannot reduce that event time's risk set; it affects only later risk sets.
List the distinct event times in increasing order, and note censorings between them.
For each event time, count subjects still under observation immediately before it.
At each event time, divide the event count by the risk set: \(d_j/Y_j\).
Add the increments through the requested time. Do not add a contribution for a censoring time.
Calculating the Estimate Step by Step
Worked Example: Three Event Times and Censoring Between Them
A hypothetical study begins with 12 contracts under observation. There are 2 events at month 1, followed by 1 censoring before month 3. At month 3 there are 3 events, followed by 1 censoring before month 5. At month 5 there are 2 events. Calculate the Nelson-Aalen increments and cumulative hazard estimates at the event times. Use the risk-set method, and avoid the mistake of removing a censoring before the event time it follows.
Method: update the risk set only after the failures and censorings that have already occurred. Immediately before month 1, all 12 contracts are at risk, so \(Y_1=12\), with \(d_1=2\). Thus
After the 2 events at month 1, \(12-2=10\) contracts remain. One is censored before month 3, so the risk set immediately before month 3 is \(Y_2=10-1=9\). With \(d_2=3\),
After the month 3 events, \(9-3=6\) contracts remain. One is censored before month 5, leaving \(Y_3=6-1=5\). With \(d_3=2\),
The cumulative estimate at month 1 is \(0.1667\). At month 3, add the first two increments: \(1/6+1/3=1/2=0.5000\). At month 5, add the third: \(1/6+1/3+2/5=9/10=0.9000\). These values are nondecreasing, as expected for a cumulative sum.
Check and common mistake avoided: The risk sets are 12, 9, and 5. Using 10 as the denominator at month 3 would incorrectly keep the subject censored before month 3 in the risk set.
Keeping Censoring and Events in the Right Order
The risk set is defined just before an event time. Consequently, failures at an earlier time reduce the next risk set, and censorings between event times also reduce it. Censorings after a given event time do not affect the risk set at that event time. This time ordering is the key bookkeeping rule in a life-table calculation.
The table below summarizes what each row contributes. Here, a censoring is assumed to occur strictly between event times, not simultaneously with a failure.
| Observation before the next event time | Effect on the next risk set | Direct hazard increment? |
|---|---|---|
| An observed event | Subtract the event count | Yes, at its event time |
| A right-censoring | Subtract the censored count | No |
| A subject still under observation | Keep in the risk set | No, unless an event occurs |
Worked Example: Censorings Between Later Event Times
A hypothetical portfolio study starts with 20 policies under observation. Four events occur at month 2. Three policies are censored between months 2 and 5. Three events occur at month 5. Two policies are censored between months 5 and 8, when 2 more events occur. Calculate the cumulative hazard estimate at months 2, 5, and 8. The method is to roll the risk set forward in chronological order; the trap is subtracting censorings from the wrong interval.
At month 2, \(Y_1=20\), \(d_1=4\), and the increment is
After those events, \(20-4=16\) policies remain. The 3 censorings before month 5 reduce the next risk set to \(Y_2=16-3=13\). The month 5 event count is \(d_2=3\), giving
After the month 5 events, \(13-3=10\) policies remain. Two are censored before month 8, so \(Y_3=10-2=8\). The month 8 increment is
The cumulative estimates are:
- At month 2: \(0.2000\).
- At month 5: \(0.2000+0.230769\ldots=0.430769\ldots\), or \(0.4308\).
- At month 8: \(0.2000+0.230769\ldots+0.2500=0.680769\ldots\), or \(0.6808\).
Check and common mistake avoided: The risk-set sequence is 20, 13, and 8. The first censoring count is applied after the month 2 event, not before it; the second is applied after the month 5 event, not before it. The rounded cumulative values agree with the sums of the displayed increments.
Reading Values Between Event Times
The cumulative estimate is a step function. It changes only at event times, by the corresponding increment, and remains unchanged between consecutive event times. Thus, if a requested time falls strictly after one event time and before the next, include the earlier events but not the later ones. This is a simple but important distinction between a value at a time and an increment at that time.
Worked Example: A Requested Time Between Events
A hypothetical study follows 30 devices. At month 1, 3 fail. Four are censored between months 1 and 2. At month 2, 2 fail. Five are censored between months 2 and 3. At month 3, 4 fail. Three are censored between months 3 and 4. At month 4, 1 fails. Find the Nelson-Aalen cumulative estimate at month 2, month 2.5, and month 4. Use the step-function definition; avoid including an event that occurs after the requested time.
At month 1, \(Y_1=30\), \(d_1=3\), so the first increment is \(3/30=0.1000\). After those events and the 4 censorings before month 2, the month 2 risk set is
There are 2 events at month 2, so the second increment is \(2/23=0.0869565\ldots\). Therefore, at month 2,
Month 2.5 lies after month 2 and before month 3. No additional event has occurred by then, so \(\widehat{H}(2.5)=0.1869565\ldots\), the same value as at month 2.
After the 2 month 2 failures and 5 censorings before month 3, \(Y_3=23-2-5=16\). The month 3 increment is \(4/16=0.2500\). After those failures and 3 censorings, \(Y_4=16-4-3=9\), so the month 4 increment is \(1/9=0.111111\ldots\). Hence
Check and common mistake avoided: The cumulative value at month 2.5 equals the value at month 2 because there is no event between those times. Including the month 3 increment at month 2.5 would use information from a later event.
Common Calculation Errors
A reliable calculation separates the risk-set update from the cumulative sum. First determine \(Y_j\) and \(d_j\) at every event time; only then calculate and add the increments. This makes errors easier to locate and prevents censoring counts from being treated as if they were events.
- Using the wrong denominator. The denominator is the number at risk immediately before the event time, not the initial sample size at every time and not the number remaining after the event.
- Treating censoring as failure. Censoring removes a subject from later risk sets but does not add an increment to \(\widehat{H}(t)\).
- Subtracting censoring too early. A censoring between two event times affects the later risk set, not the earlier one.
- Forgetting to accumulate. \(d_j/Y_j\) is one increment. The cumulative estimate at \(t_j\) includes that increment and all increments at earlier event times.
- Including a future event. If the requested time lies between event times, use only events that have occurred by the requested time.
When a life table includes failures tied at the same recorded time or censorings recorded at a failure time, the ordering convention needs explicit attention. The calculations here use distinct event times and censorings strictly between them; do not silently apply that simplified bookkeeping to a table with simultaneous records.
Check Your Understanding
For each problem, calculate the requested Nelson-Aalen increments and cumulative estimate. Round decimal answers to four places where needed.
- A study starts with 15 subjects. There are 3 events at time 2, then 4 subjects are censored before time 4, when 2 events occur. Find the increment at each event time and the cumulative estimate at time 4. (Answer: increments \(3/15=0.2000\) and \(2/8=0.2500\); \(\widehat{H}(4)=0.4500\).)
- There are 24 subjects initially. Four events occur at time 1; five subjects are censored before time 3, when 3 events occur. Find the risk set at time 3 and the cumulative estimate at time 3. (Answer: \(Y_2=15\); \(\widehat{H}(3)=4/24+3/15=0.3667\).)
- A cohort starts with 18 subjects. Two events occur at time 2, three subjects are censored before time 5, and 4 events occur at time 5. Find the two increments and the cumulative estimate at time 5. (Answer: increments \(2/18=0.1111\) and \(4/13=0.3077\); \(\widehat{H}(5)=0.4188\).)
- A study begins with 40 subjects. Five events occur at time 1, six subjects are censored before time 2, and 3 events occur at time 2. Seven subjects are censored before time 3, when 2 events occur. Find the risk set at each event time and the cumulative estimate at time 3. (Answer: risk sets 40, 29, and 19; \(\widehat{H}(3)=5/40+3/29+2/19=0.3337\).)
- Ten subjects enter observation. Two are censored before time 2, when 2 events occur. One further subject is censored between times 2 and 4, when 3 events occur. Find the cumulative estimate at time 4, and explain why the initial sample size is not the denominator for both increments. (Answer: risk sets 8 and 5; \(\widehat{H}(4)=2/8+3/5=0.8500\); earlier censoring and failures reduce the later risk set.)