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Risk Modelling and Survival Analysis · Estimation procedures for lifetime distributions

Nelson-Aalen Estimator of the Cumulative Hazard Explained

Updated 11 October 2026 · Fact-checked

The Nelson-Aalen estimator estimates the cumulative hazard Λ(t) without assuming a distribution. At each death time tⱼ you add dⱼ ÷ nⱼ, deaths divided by the number at risk. The sum up to t gives Λ̂(t). You then get survival as Ŝ(t) = exp(−Λ̂(t)).

Understand Nelson-Aalen Estimator of the Cumulative Hazard

Survival data are often censored, so you cannot simply count deaths. The Nelson-Aalen estimator handles censoring and makes no assumption about the shape of the distribution. It is a nonparametric method.

The idea starts from the hazard. At a death time tⱼ, there are nⱼ lives still under observation (the risk set) and dⱼ of them die. The proportion dying, dⱼ ÷ nⱼ, is a natural estimate of the hazard increment at that time. The cumulative hazard Λ(t) is the running total of hazard, so you add these increments up to time t.

Between death times the estimate does not change. So Λ̂(t) is a step function that starts at 0 and rises only at observed death times. Censoring matters only through nⱼ: a life censored before tⱼ leaves the risk set and no longer counts.

Once you have Λ̂(t), you can estimate survival using S(t) = exp(−Λ(t)). This gives the Nelson-Aalen (also called Fleming-Harrington) survival estimate. It is different from the Kaplan-Meier estimate, which multiplies factors (1 − dⱼ ÷ nⱼ).

The two survival estimates are usually close, especially when dⱼ is small relative to nⱼ. This is because exp(−x) ≈ 1 − x for small x. When there are no ties, the Nelson-Aalen survival estimate is never below the Kaplan-Meier estimate, because exp(−x) ≥ 1 − x.

Key rules to remember

Nelson-Aalen cumulative hazard
Λ̂(t) = Σ over tⱼ ≤ t of dⱼ ÷ nⱼ
dⱼ = deaths at time tⱼ. nⱼ = number at risk just before tⱼ. Λ̂(t) = 0 before the first death.
Survival from Nelson-Aalen
Ŝ(t) = exp(−Λ̂(t))
Uses the relation S(t) = exp(−Λ(t)). Gives a survival estimate that differs slightly from Kaplan-Meier.
Kaplan-Meier estimator
Ŝ(t) = Π over tⱼ ≤ t of (1 − dⱼ ÷ nⱼ)
Use it to compare. It is a product, whereas Nelson-Aalen is a sum.
Cumulative hazard from Kaplan-Meier
Λ̂(t) = −ln Ŝ(t)
Gives an alternative cumulative hazard estimate from Kaplan-Meier. It differs from the Nelson-Aalen estimate.
Relationship between survival and cumulative hazard
S(t) = exp(−Λ(t)), where Λ(t) = ∫₀ᵗ μₛ ds
μₛ is the force of mortality (hazard rate).
Variance of Nelson-Aalen estimate
Var[Λ̂(t)] ≈ Σ over tⱼ ≤ t of dⱼ ÷ nⱼ²
Use only if the question asks for it or gives a result in the core reading style. State it is an approximation.

How to solve Nelson-Aalen Estimator of the Cumulative Hazard questions

Use this method for any Nelson-Aalen question. A table keeps the working clear and easy to check.

  1. 1List the distinct times at which deaths occur, in increasing order. Ignore censoring times as rows, unless a censoring time coincides with a death time.
  2. 2For each death time tⱼ, find nⱼ: the number still under observation just before tⱼ. Start with the total and subtract all deaths and censorings that happened earlier.
  3. 3Write dⱼ, the number of deaths at tⱼ.
  4. 4Compute dⱼ ÷ nⱼ for each row.
  5. 5Add the increments cumulatively to get Λ̂(t) at each death time.
  6. 6If asked for survival, compute Ŝ(t) = exp(−Λ̂(t)) at the required time.
  7. 7If asked to compare with Kaplan-Meier, compute the product of (1 − dⱼ ÷ nⱼ) and comment on the difference.
  8. 8State the answer for the exact time asked. Remember the estimate is constant between death times.

Quickest way: Running-total table

When to use it: Use this when you are given a small data set with a few deaths and some censored lives, and need Λ̂(t) at a given time.

  1. Write n at the start. Move through events in time order.
  2. At each death, write dⱼ ÷ nⱼ as a fraction and add it to the running total.
  3. After each event (death or censoring), reduce n by the number leaving.
  4. Stop at the last death time ≤ t. That total is Λ̂(t).
  5. For survival, press exp(−total) once on your calculator. Keep decimals to at least four places until the end.

Common mistakes in Nelson-Aalen Estimator of the Cumulative Hazard

  • Using the wrong risk set, for example forgetting to remove censored lives.

    You track only deaths when updating n.

    Fix: After every event, subtract both deaths and censorings. Check that n at the next row equals the previous n minus all exits.

  • Multiplying (1 − d ÷ n) terms when asked for Nelson-Aalen.

    Kaplan-Meier is more familiar, so students mix up the two.

    Fix: Remember: Nelson-Aalen adds hazard increments. Kaplan-Meier multiplies survival factors.

  • Forgetting to take exp(−Λ̂) to get survival, or using exp(Λ̂).

    The negative sign is dropped in a hurry.

    Fix: Write S = exp(−Λ) before you calculate. Survival must lie between 0 and 1, so check your answer.

  • Treating a censoring time as a row in the table and adding an increment.

    Students list every observed time.

    Fix: Only death times add to Λ̂(t). Censoring times only change nⱼ for later rows.

  • Handling tied times wrongly when death and censoring occur at the same time.

    The order of events is unclear.

    Fix: The standard convention is that a life censored at tⱼ is still in the risk set at tⱼ. State this assumption in your answer.

  • Claiming the Nelson-Aalen and Kaplan-Meier survival estimates are equal.

    They are numerically close.

    Fix: Say they are close but not identical. For small increments, exp(−x) ≈ 1 − x.

Worked examples

Example 1

Ten lives are observed. Deaths occur at times 2, 4, 4 and 7. One life is censored at time 3 and one at time 5. The other four lives are still alive at the end of the study, which is after time 7. Calculate the Nelson-Aalen estimate of the cumulative hazard at time 7 and the corresponding survival estimate.

Show the solution
  1. Death times: 2, 4, 7. At time 4 there are two deaths.
  2. At time 2: n = 10, d = 1. Increment = 1 ÷ 10 = 0.1.
  3. After time 2, 9 lives remain. Life censored at time 3 leaves, so 8 lives remain before time 4.
  4. At time 4: n = 8, d = 2. Increment = 2 ÷ 8 = 0.25.
  5. After time 4, 6 lives remain. Life censored at time 5 leaves, so 5 remain before time 7.
  6. At time 7: n = 5, d = 1. Increment = 1 ÷ 5 = 0.2.
  7. Λ̂(7) = 0.1 + 0.25 + 0.2 = 0.55.
  8. Ŝ(7) = exp(−0.55) = 0.5769.

Answer: Λ̂(7) = 0.55 and Ŝ(7) = exp(−0.55) ≈ 0.577.

Example 2

Using the data in the previous example, calculate the Kaplan-Meier estimate of S(7) and compare it with the Nelson-Aalen survival estimate.

Show the solution
  1. Use the same risk sets: (n, d) = (10, 1), (8, 2), (5, 1).
  2. Factors: 1 − 1/10 = 0.9. 1 − 2/8 = 0.75. 1 − 1/5 = 0.8.
  3. Kaplan-Meier: Ŝ(7) = 0.9 × 0.75 × 0.8 = 0.54.
  4. Nelson-Aalen: Ŝ(7) = exp(−0.55) = 0.5769.
  5. Difference: 0.5769 − 0.54 = 0.0369.
  6. Comment: the Nelson-Aalen survival estimate is higher, as exp(−x) ≥ 1 − x. The gap is larger here because the increments are not small.

Answer: Kaplan-Meier gives Ŝ(7) = 0.54. Nelson-Aalen gives about 0.577. The two are close but not equal, and Nelson-Aalen is higher here.

Exam tips

  • Always draw the table with columns tⱼ, nⱼ, dⱼ, dⱼ ÷ nⱼ and Λ̂(t). Marks are given for the risk set and the increments, even if the final number is wrong.
  • State your assumption on tied death and censoring times in one line. Examiners accept the standard convention if you state it.
  • Expect Paper A questions that ask you to compare Kaplan-Meier and Nelson-Aalen. Mention product versus sum, and that the two survival estimates are close for small increments.
  • In MCQs, check the risk set first. Most wrong options come from a wrong nⱼ.
  • In the computer-based paper, show the method and formula in your notes before the R output, and say which estimator the output reports.

Practice questions from Estimation procedures for lifetime distributions

Nelson-Aalen Estimator of the Cumulative Hazard: frequently asked questions

What is the difference between Kaplan-Meier and Nelson-Aalen?

Kaplan-Meier estimates the survival function directly as a product of (1 − dⱼ ÷ nⱼ). Nelson-Aalen estimates the cumulative hazard as a sum of dⱼ ÷ nⱼ. You can convert Nelson-Aalen to survival using exp(−Λ̂(t)), which gives a slightly different answer.

How do I get a survival estimate from the Nelson-Aalen estimate?

Calculate Λ̂(t) at the required time. Then use Ŝ(t) = exp(−Λ̂(t)). Check that the result lies between 0 and 1.

Does censoring change the Nelson-Aalen estimate?

Yes, but only through the risk set. A censored life adds no increment, but it is removed from nⱼ for later death times. Fewer at risk means larger later increments.

Is the Nelson-Aalen estimate defined between death times?

It is a step function. It stays constant between death times and jumps at each death time by dⱼ ÷ nⱼ.