Why the Risk Set Determines a Hazard Increment
When estimating survival from incomplete records, a failure count alone is not enough. Three failures among 20 people means something different from three failures among the five people still being followed at a later time. The relevant denominator is the number who were still at risk immediately before the event time. Nelson-Aalen estimation uses that pre-event risk set to form a hazard increment, then adds the increments to estimate cumulative hazard.
This tutorial develops the ratio and its interpretation. It assumes event times and right-censoring times are recorded, and that censoring is independent of the future event time conditional on the information used in the analysis. Under this assumption, those censored can be removed from later risk sets without treating them as failures. Censoring does not itself add a hazard increment.
The risk set includes subjects who have not yet had an event and have not yet been censored just before \(t_j\). A person censored earlier is not counted in \(Y_j\); a person censored after \(t_j\) is counted at \(t_j\). For failures tied at a time, \(d_j\) is the number failing at that time, and \(Y_j\) is the risk set just before it. The ratio is not calculated by removing those failures one at a time from the denominator.
Why Failures Divided by Exposure Estimates an Increment
Cumulative hazard describes accumulated instantaneous risk. In a continuous-time survival model, write \(H(t)\) for cumulative hazard. Over a short interval of length \(dt\), a subject who remains at risk has an event probability approximately \(\lambda(t)dt\), where \(\lambda(t)\) is the hazard rate. If \(Y(t)\) subjects are at risk, the expected number of failures in that short interval is approximately \(Y(t)\lambda(t)dt\). The corresponding cumulative-hazard exposure over the interval is approximately \(\lambda(t)dt\), so the observed failures per person at risk estimate that exposure.
In the counting-process formulation, \(N(t)\) counts observed failures and \(Y(t)\) counts subjects at risk. The model specifies that the conditional expected increase in failures over a short interval is approximately \(Y(t)\,dH(t)\). Dividing the observed increase \(dN(t)\) by \(Y(t)\) gives an estimate of the cumulative-hazard increase over that interval. Adding these estimated increases over time gives the Nelson-Aalen estimator. At an observed event time, the corresponding increment is \(d_j/Y_j\).
Determine how many subjects are still under observation and event-free immediately before the event time. This is \(Y_j\).
Record the number of observed failures at that time as \(d_j\).
Calculate \(d_j/Y_j\), the Nelson-Aalen contribution at that event time.
Add the contribution to all earlier event-time contributions to obtain the estimated cumulative hazard.
A binomial argument can help explain the ratio, provided its scope is stated carefully. If each of \(Y_j\) subjects independently has the same conditional probability \(p_j\) of failing in a specified interval, the interval failure count has a binomial likelihood proportional to \(p_j^{d_j}(1-p_j)^{Y_j-d_j}\). For \(0<d_j<Y_j\), differentiating its log likelihood and setting the derivative to zero gives \(\widehat p_j=d_j/Y_j\). If \(d_j=0\), the likelihood is maximized at \(p_j=0\); if \(d_j=Y_j\), it is maximized at \(p_j=1\). Thus the ratio is the maximum-likelihood estimate of the interval event probability in this binomial model.
That binomial result does not by itself show that the ratio is the exact cumulative-hazard increment for a finite interval. For a discrete interval event probability \(p_j\), the corresponding increment under the definition \(H=-\log S\) is \(-\log(1-p_j)\). The two quantities are close when \(p_j\) is small, since \(-\log(1-p_j)\) is approximately \(p_j\) for small \(p_j\). Nelson-Aalen estimation is grounded in the continuous-time hazard and exposure formulation; with grouped discrete intervals, \(d_j/Y_j\) should not be mistaken for the exact transformed interval hazard.
Worked Example: Building Cumulative Hazard With Censoring
In a hypothetical follow-up study of 20 insured devices, two devices are censored at time 1, three fail at time 2, four more are censored at time 3, and two fail at time 5. Calculate the Nelson-Aalen increments and cumulative hazard through time 5. Assume independent right-censoring and continuous event times; there is no interest-rate assumption because this is a survival estimate, not a present-value calculation.
At time 2. The two censorings at time 1 have already left the risk set. Therefore \(Y_1=20-2=18\). With \(d_1=3\), the first increment is $$ \frac{d_1}{Y_1}=\frac{3}{18}=\frac{1}{6}\approx0.1667. $$
At time 5. After three failures at time 2, \(18-3=15\) devices remain at risk. The four censorings at time 3 reduce this to \(Y_2=15-4=11\). With \(d_2=2\), the second increment is $$ \frac{d_2}{Y_2}=\frac{2}{11}\approx0.1818. $$
Accumulate through time 5. Add the two event-time contributions: $$ \widehat{H}_{NA}(5)=\frac{1}{6}+\frac{2}{11} =\frac{11}{66}+\frac{12}{66} =\frac{23}{66}\approx0.3485. $$ The risk-set bookkeeping is consistent: five devices have failed, six have been censored, and nine remain after time 5, for a total of 20. The cumulative estimate is \(0.3485\) to four decimal places. This is a cumulative-hazard estimate, not a survival probability.
Method and trap avoided. The method is to update the risk set after each earlier failure and censoring, then sum the event-time ratios. Using the original 20 as the denominator at time 5 would ignore the subjects no longer at risk and would give an incorrect increment.
Worked Calculations and Interpretation
The ratio is tied to the risk set at that particular event time; it is not a rate per calendar unit such as failures per year. Nelson-Aalen cumulative hazard is dimensionless, and its increments are added, not multiplied. The examples below highlight both the calculation and a boundary of its interpretation.
Worked Example: Separating an Interval Probability From a Hazard Increment
Suppose a hypothetical discrete follow-up interval begins with eight subjects at risk, and four fail during the interval. Under a binomial interval model, calculate the estimated event probability. Then compare the exact cumulative-hazard increment for that discrete probability with the Nelson-Aalen ratio. Assume independent observations and no censoring within the interval.
Step 1: Estimate the interval event probability. Here \(Y=8\) and \(d=4\), so the binomial maximum-likelihood estimate is $$ \widehat p=\frac{d}{Y}=\frac{4}{8}=0.5. $$
Step 2: Convert the probability to its exact cumulative-hazard increment. The interval survival probability is \(1-\widehat p=0.5\). Therefore the increment defined by \(H=-\log S\) is $$ -\log(1-\widehat p)=-\log(0.5)=\log 2\approx0.6931. $$
Step 3: Identify the Nelson-Aalen contribution. The ratio used as the Nelson-Aalen contribution is \(d/Y=0.5\). It is not equal to the exact discrete increment \(0.6931\). The approximation is less compelling here because the interval probability is large.
Method and trap avoided. The binomial likelihood estimates an interval event probability, while the Nelson-Aalen estimator accumulates failure-to-risk-set ratios in the continuous-time hazard framework. A common wrong calculation is to call \(0.5\) the exact discrete cumulative-hazard increment. The correction is \(-\log(1-0.5)=0.6931\) for the discrete interval model. Do not silently interchange these quantities.
Worked Example: Updating the Denominator Across Several Event Times
A hypothetical cohort starts with 15 subjects. Three are censored before the first observed event time, where two fail. After that, two more subjects are censored before the next event time, where one fails. At the final event time, three fail and no further subject has been censored. Find the cumulative Nelson-Aalen estimate through the final event time. Assume independent right-censoring and a continuous-time survival model.
First event time. The three earlier censorings leave \(Y_1=15-3=12\) at risk. With \(d_1=2\), the increment is \(2/12=1/6\).
Second event time. The two failures leave \(12-2=10\). The two subsequent censorings leave \(Y_2=10-2=8\) at risk. With \(d_2=1\), the increment is \(1/8\).
Final event time. After the second event, \(8-1=7\) subjects remain, and no further censoring occurs before the final event time. Thus \(Y_3=7\), \(d_3=3\), and the increment is \(3/7\).
Cumulative estimate and check. Summing the three increments gives $$ \widehat{H}_{NA}=\frac{1}{6}+\frac{1}{8}+\frac{3}{7} =\frac{28}{168}+\frac{21}{168}+\frac{72}{168} =\frac{121}{168}\approx0.7202. $$ The count check is \(2+1+3=6\) failures, \(3+2=5\) censorings, and \(15-6-5=4\) subjects remaining. The totals reconcile. The cumulative estimate is \(0.7202\) to four decimal places.
Method and trap avoided. The denominator at each event time is the risk set immediately before that time. In particular, the final denominator is seven, not the original 15 and not the six subjects left after the final failures. Dividing by the post-event count would use the wrong exposure.
Conditions, Use, and Exam Traps
The estimate is meaningful only when the risk sets represent the population under study. Independent censoring is important: if subjects likely to fail soon are selectively censored, those remaining may not represent the original group, and the resulting risk-set calculation can be biased. This is why censoring is not simply counted as survival forever; it removes a subject from later exposure while preserving their contribution to earlier risk sets.
The Nelson-Aalen sum estimates cumulative hazard; it is not itself the estimated probability of failure by time \(t\). A probability interpretation requires an additional relationship between survival and cumulative hazard. Also, its event-time contributions are additive: each uses its own \(Y_j\). They are not all divided by the initial cohort size.
| Calculation choice | Correct treatment | Common error |
|---|---|---|
| Risk set at event time \(t_j\) | Count subjects at risk just before \(t_j\) | Use the original cohort size at every event time |
| Censoring before an event time | Remove censored subjects from that and later risk sets | Count censorings as failures |
| Multiple failures at one time | Use \(d_j/Y_j\) with \(Y_j\) just before the time | Reduce the denominator separately for each tied failure |
| Discrete interval probability | Convert \(p\) to \(-\log(1-p)\) for the exact cumulative-hazard increment | Assume \(p\) and the exact hazard increment are identical |
| Cumulative estimate | Add the event-time increments | Interpret the sum directly as a failure probability |
Check Your Understanding
For each problem, identify the appropriate risk set before calculating. Round decimal estimates to four places when needed.
- A hypothetical study has 25 subjects at risk just before an event time, with five failures at that time. Find the Nelson-Aalen increment. (Answer: \(5/25=0.2000\))
- A cohort starts with 30 subjects. Four are censored before time 2, when three fail. Six more are censored after time 2 but before time 4, when two fail. Find the cumulative Nelson-Aalen estimate through time 4. (Answer: \(3/26+2/17=103/442\approx0.2330\))
- At the first event time, \(Y_1=50\) and \(d_1=5\). After those failures, five subjects are censored before the next event time, where \(d_2=4\). Find cumulative hazard through the second event time. (Answer: \(5/50+4/40=0.2000\))
- Three event times have risk sets and failure counts \((Y_1,d_1)=(18,3)\), \((Y_2,d_2)=(12,2)\), and \((Y_3,d_3)=(7,1)\). Calculate cumulative hazard through the third event time. (Answer: \(3/18+2/12+1/7=10/21\approx0.4762\))
- In a discrete interval, eight subjects are at risk and two fail. Find the estimated interval event probability and its exact cumulative-hazard increment. (Answer: \(\widehat p=2/8=0.25\); exact increment \(=-\log(0.75)\approx0.2877\), distinct from the Nelson-Aalen ratio \(0.25\))