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Duration and Survival Models · Tutorial 573 of 952

The Nelson-Aalen Cumulative Hazard Estimator

Learn to construct the Nelson–Aalen cumulative-hazard estimate from risk sets and understand how its survival transformation compares with Kaplan–Meier.

Advanced 10 min read

What You'll Learn

  • Define risk-set sizes and event counts at distinct observed event times.
  • Derive the Nelson–Aalen cumulative-hazard estimator from event rates in the risk set.
  • Calculate cumulative-hazard and exponential survival estimates in the presence of censoring and tied events.
  • Compare the Nelson–Aalen survival transformation with the Kaplan–Meier product-limit estimate.
  • Recognize why large tied-event fractions can make the two survival estimates differ substantially.

Why Estimate Cumulative Hazard?

With right-censored duration data, an individual’s event time may be only partly observed: we know the event had not occurred by the censoring time, but not when it occurred afterward. At each observed event time, the relevant information is the number still under observation and event-free just before that time, together with the number of events then observed. The Nelson–Aalen estimator uses these risk sets to estimate cumulative hazard over time.

Its increments are straightforward to calculate, but interpreting the resulting survival estimate requires care. Exponentiating the negative Nelson–Aalen estimate gives a survival estimate that is closely related to, but generally not equal to, the Kaplan–Meier estimate.

Definition: Let \(t_1<t_2<\cdots<t_J\) be the distinct observed event times. At event time \(t_j\), let \(Y_j\) be the number of individuals at risk immediately before \(t_j\), and let \(d_j\) be the number of events at \(t_j\). The Nelson–Aalen estimator of the cumulative hazard through time \(t\) is $$ \widehat{H}_{NA}(t)=\sum_{j:t_j\leq t}\frac{d_j}{Y_j}. $$ Its exponential survival transformation is \(\widehat{S}_{NA}(t)=\exp\{-\widehat{H}_{NA}(t)\}\).

The risk set at \(t_j\) includes individuals who are still under observation and have not experienced the event immediately before \(t_j\). An individual censored before \(t_j\) is no longer counted; an individual censored after \(t_j\) remains in the risk set at \(t_j\). When event times are tied, \(d_j\) counts all events at that time, and \(Y_j\) is the risk-set size just before them.

For the standard setup here, individuals enter observation at a common origin, and censoring is independent of event time. These conditions let the observed risk sets represent the individuals still event-free in the population being studied. With delayed entry, risk sets must instead account for entry times. The estimator describes an event-time distribution under the stated censoring and observation assumptions; it does not by itself establish a causal effect.

Why Each Increment Has This Form

The cumulative hazard \(H(t)\) adds up the hazard over time. In a short interval around \(t_j\), if the hazard increment is small, each of the \(Y_j\) individuals at risk has approximately that increment as their event probability. The expected number of events is therefore approximately \(Y_j\) times the hazard increment. Dividing the observed event count by \(Y_j\) estimates that small increment; adding the estimates across event times estimates cumulative hazard.

A related justification uses the counting-process formulation. Let \(N(t)\) count events by time \(t\), and let \(Y(u)\) be the number at risk just before time \(u\). Under a common event hazard \(\lambda(u)\) among those at risk, the event-count process has intensity \(Y(u)\lambda(u)\). Thus, where \(Y(u)>0\), the observed increment \(dN(u)/Y(u)\) is the natural risk-set-scaled increment for estimating \(\lambda(u)\,du\). Summing those increments yields the Nelson–Aalen form. At an observed event time, the count increment is \(d_j\) and the risk-set size is \(Y_j\), giving \(d_j/Y_j\).

This is an estimator of cumulative hazard, not a claim that the event probability at \(t_j\) is exactly \(d_j/Y_j\) on every time scale. In a short interval with a small hazard increment, event probability and hazard increment are close. With a large tied-event fraction, that approximation can be less close, which matters when transforming the estimate to survival.

Worked Example: Building the Estimate With Tied Events and Censoring

Consider a hypothetical cohort of 12 individuals followed from a common starting time. At time 1, two events occur; one additional individual is censored after that event time. At time 2, one event occurs and two are censored afterward. At time 4, two events occur and one is censored afterward. At time 6, one event occurs. Find the risk sets, cumulative-hazard estimates, and both survival estimates through time 6.

The risk set begins at 12. After the two events and one censoring at time 1, nine individuals remain at risk at time 2. After the event and two censorings at time 2, six remain at risk at time 4. After the two events and one censoring at time 4, three remain at risk at time 6. The event-time quantities are therefore:

Event time \(t_j\)At risk \(Y_j\)Events \(d_j\)Increment \(d_j/Y_j\)
11222/12 = 0.1667
2911/9 = 0.1111
4622/6 = 0.3333
6311/3 = 0.3333

Through time 4, the cumulative-hazard estimate is

$$ \widehat{H}_{NA}(4) =\frac{2}{12}+\frac{1}{9}+\frac{2}{6} =\frac{11}{18} \approx 0.6111. $$

Thus the exponential survival transformation at time 4 is

$$ \widehat{S}_{NA}(4) =\exp(-11/18) \approx 0.5427. $$

At time 6, add the final increment \(1/3\):

$$ \widehat{H}_{NA}(6) =\frac{11}{18}+\frac{1}{3} =\frac{17}{18} \approx 0.9444, \qquad \widehat{S}_{NA}(6) =\exp(-17/18) \approx 0.3889. $$

For comparison, the Kaplan–Meier estimate multiplies the conditional survival factors \(1-d_j/Y_j\). Through time 4, this gives \((10/12)(8/9)(4/6)=40/81\approx 0.4938\); through time 6, it gives \((40/81)(2/3)=80/243\approx 0.3292\). Both survival estimates use the same event and risk-set information, but their transformations differ.

Nelson–Aalen and Kaplan–Meier Compared

At event time \(t_j\), Kaplan–Meier multiplies its current survival estimate by \(1-d_j/Y_j\). The Nelson–Aalen survival transformation instead adds \(d_j/Y_j\) to cumulative hazard and then exponentiates the negative total. Their relationship is easiest to see by taking the negative logarithm of the Kaplan–Meier estimate.

$$ -\log\{\widehat{S}_{KM}(t)\} =\sum_{j:t_j\leq t}-\log\left(1-\frac{d_j}{Y_j}\right), \qquad \widehat{H}_{NA}(t) =\sum_{j:t_j\leq t}\frac{d_j}{Y_j}. $$

For \(0\leq x<1\), \(-\log(1-x)\geq x\). Therefore, whenever \(d_j<Y_j\) at each included event time, the Kaplan–Meier cumulative-hazard transformation \(-\log\{\widehat{S}_{KM}(t)\}\) is at least as large as \(\widehat{H}_{NA}(t)\). Exponentiating their negatives reverses the comparison: \(\widehat{S}_{KM}(t)\leq\widehat{S}_{NA}(t)\). If \(d_j=Y_j\) at any included time, the Kaplan–Meier estimate becomes zero and its negative logarithm is infinite, while the Nelson–Aalen increment remains finite.

The inequality follows from the expansion \(-\log(1-x)=x+x^2/2+x^3/3+\cdots\) for \(0\leq x<1\). The leading term is the Nelson–Aalen increment, and the additional nonnegative terms make the Kaplan–Meier cumulative-hazard transformation larger. When each event fraction \(d_j/Y_j\) is small, those additional terms are relatively small, so the two survival estimates tend to be close. Larger fractions, including large tied-event fractions, can make their difference more noticeable.

Worked Example: Comparing the Two Survival Estimates

Suppose a hypothetical cohort has 20 individuals at risk at one event time, with 3 events. At a later event time, 14 individuals are at risk, with 2 events. The second risk set is possible, for example, if three individuals were censored between the event times after the first three events. Calculate both estimates through the second event time.

The Nelson–Aalen cumulative-hazard estimate is

$$ \widehat{H}_{NA} =\frac{3}{20}+\frac{2}{14} =0.15+\frac{1}{7} \approx 0.2929. $$

Its exponential survival transformation is

$$ \widehat{S}_{NA} =\exp\left(-\frac{3}{20}-\frac{1}{7}\right) \approx 0.7461. $$

The Kaplan–Meier estimate is the product of the two event-free fractions:

$$ \widehat{S}_{KM} =\left(1-\frac{3}{20}\right)\left(1-\frac{2}{14}\right) =\frac{17}{20}\cdot\frac{6}{7} =\frac{51}{70} \approx 0.7286. $$

Here the Kaplan–Meier estimate is lower, as predicted by the inequality. Specifically, \(-\log(51/70)\approx 0.3167\), which exceeds the Nelson–Aalen cumulative-hazard estimate \(0.2929\). The difference is modest because neither event fraction is especially large.

What Large Ties Can Change

A tied event time can arise because events genuinely occur together or because event times are recorded only to a coarse unit, such as a day. The formulas group all events at the same recorded time: the Nelson–Aalen increment is \(d_j/Y_j\), while Kaplan–Meier uses the factor \(1-d_j/Y_j\). When a large share of the risk set fails at one recorded time, the difference between adding a hazard increment and multiplying by a survival fraction can be substantial.

Worked Example: Every Individual at Risk Has an Event

Suppose, hypothetically, that 4 individuals are at risk at a single recorded event time and all 4 have the event then. Thus \(Y_1=4\) and \(d_1=4\). The Nelson–Aalen increment and its transformed survival estimate are

$$ \widehat{H}_{NA}=\frac{4}{4}=1, \qquad \widehat{S}_{NA}=\exp(-1)\approx 0.3679. $$

Kaplan–Meier instead gives

$$ \widehat{S}_{KM}=1-\frac{4}{4}=0. $$

This example is not an arithmetic contradiction: these are different estimators and transformations. But the dramatic difference warns against treating the exponential Nelson–Aalen transformation as interchangeable with Kaplan–Meier when event fractions at recorded times are large. If the time recording has grouped many distinct events together, the apparent tie may conceal the timing information that finer measurements would have provided.

Common Mistakes and Practical Interpretation

  • Using the original sample size in every denominator. The denominator is \(Y_j\), the number still at risk immediately before that event time. Earlier events and censoring can reduce it.
  • Removing an individual censored after an event time before forming that event time’s risk set. That individual remains at risk just before the event time and is removed only afterward.
  • Treating the Nelson–Aalen sum as the Kaplan–Meier survival estimate. The sum estimates cumulative hazard; its exponential transformation is \(\exp\{-\widehat{H}_{NA}(t)\}\). Kaplan–Meier is a product of event-free fractions.
  • Claiming the two survival estimates must be equal. They are generally different. Their difference tends to be small when event fractions are small and can be large when a tied-event fraction is large.
  • Interpreting an estimated survival curve as a causal effect. These estimators describe event-time survival under the observation and censoring assumptions; causal claims require an appropriate causal design or model.
Key takeaway: At each distinct event time, divide the number of events by the number at risk immediately beforehand. Summing these increments estimates cumulative hazard; exponentiating its negative gives the Nelson–Aalen survival transformation. Kaplan–Meier uses the same risk-set fractions in a product, so the estimates are related but not identical.

Check Your Understanding

Use the definitions and comparisons in this tutorial to answer the following questions.

  1. At an event time, 5 of the 25 individuals at risk have the event. What is the Nelson–Aalen increment at that time?
  2. Why can an individual censored after an event time still belong in that event time’s risk set?
  3. For event fractions less than 1, which is larger: the Kaplan–Meier cumulative-hazard transformation or the Nelson–Aalen cumulative-hazard estimate? State the inequality that justifies your answer.
  4. When does the Kaplan–Meier estimate become zero at an event time, and what does the Nelson–Aalen survival transformation equal at that same time?
  5. Why can coarse recording of event times make the Nelson–Aalen and Kaplan–Meier survival estimates differ substantially?
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