Why Tied Failures Need a Grouped Calculation
In time-to-event data, two or more subjects can have the same recorded failure time. This may reflect genuinely simultaneous events or times recorded to a limited precision. In either case, Nelson-Aalen estimation treats all failures at that distinct recorded time as one group. The denominator is the number at risk immediately before that time, not a sequence of successively smaller risk sets created by imposing an arbitrary order on the tied failures.
This distinction matters whenever a life table contains a tied-failure count. It also matters when censorings are recorded at that same time: under the usual convention, those individuals remain in the risk set immediately before the time and are removed after the tied failures are counted.
The risk set is determined just before \(t_j\). Thus, someone censored at \(t_j\) is counted in \(Y_j\), just as someone who fails at \(t_j\) is. A subject censored strictly before \(t_j\) is not in the risk set. Censorings do not themselves add a Nelson-Aalen increment.
The Nelson-Aalen Estimate with Tied Failures
Suppose the distinct observed failure times are \(t_1,t_2,\ldots\), with \(d_j\) failures among the \(Y_j\) subjects at risk immediately before \(t_j\). The Nelson-Aalen estimate of cumulative hazard through time \(t\) is
If several subjects fail at one time, use their total count as \(d_j\) in a single term. Do not divide that count into sequential one-failure terms with changing denominators. The formula uses the risk set at the start of that distinct time, before any failures or censorings recorded at that time are removed.
Why a Tied Group Has One Increment
The grouped increment follows from counting failures among those exposed to failure at that time. Immediately before \(t_j\), there are \(Y_j\) subjects at risk. Of those, \(d_j\) are observed to fail at \(t_j\). The estimated event proportion for that tied time is therefore \(d_j/Y_j\). Nelson-Aalen adds this estimated increment to the cumulative hazard estimate.
The calculation does not treat the first tied failure as removing a subject before assessing the second tied failure. There is no observed ordering between failures that share the same recorded time. If one instead assigns an order, the resulting denominators depend on an arbitrary choice that is absent from the data. Grouping avoids that artificial ordering.
This explanation concerns the tied-time increment only. The cumulative estimate through a later time is obtained by adding the increments from all distinct failure times up to and including that time.
Worked Example: Three Failures at the Same Recorded Time
A hypothetical group of 12 newly installed sensors is followed for time to failure. Assume censoring is independent of failure time. There is no parametric failure-time distribution or interest-rate assumption. At time 2, three sensors fail and one other sensor is censored. At time 4, two fail and one is censored. At time 7, one fails. Calculate the Nelson-Aalen estimate through time 7.
Method. Work through the distinct failure times in order, using the risk set immediately before each time. At time 2, all 12 sensors are still at risk, so \(Y_1=12\) and \(d_1=3\). The time-2 increment is \(3/12=0.25\). The censoring at time 2 is removed after forming this increment.
After time 2, \(12-3-1=8\) sensors remain under observation and at risk for time 4. Thus \(Y_2=8\), \(d_2=2\), and the time-4 increment is \(2/8=0.25\). After the two failures and one censoring at time 4, \(8-2-1=5\) sensors remain.
At time 7, \(Y_3=5\) and \(d_3=1\), so the increment is \(1/5=0.20\). Adding the three increments gives
Check. The risk sets decrease from 12 to 8 to 5 after the failures and censorings at each earlier time. The estimate is \(0.70\). The main error avoided is counting the time-2 failures one at a time with changing denominators.
When Censoring and Failure Share a Time
If failures and censorings have the same recorded time, use the risk set immediately before that time for the failure increment. In a grouped life table, remove both the failures and the censorings after calculating that increment. This convention reflects that the censored subjects were observed through the recorded time; it does not mean that censorings contribute failures.
Worked Example: Tied Failures and Same-Time Censoring
A hypothetical portfolio of 15 covered components is observed until failure or censoring. Assume independent censoring and no parametric failure-time distribution. At time 1, two components fail and two are censored. At time 3, four fail and one is censored. At time 5, one fails. Find the Nelson-Aalen estimate through time 5.
Method. At time 1, the risk set is 15, including the two components censored at that time. The increment is \(2/15\). After time 1, the number remaining at risk is \(15-2-2=11\).
At time 3, \(Y_2=11\), so the four tied failures contribute \(4/11\). The censoring at time 3 is also removed after this increment, leaving \(11-4-1=6\) at risk. At time 5, \(Y_3=6\), and the single failure contributes \(1/6\).
The estimate is rounded to four decimal places. Check. The risk set at time 3 is 11 because only the time-1 failures and censorings have been removed. The calculation avoids removing the time-3 censoring before forming the time-3 risk set or increment.
Calculating from a Grouped Life Table
A life table may list the number at risk, failures, and censorings at each distinct time. Read the risk-set column as the count immediately before that time. If a row has no failures, it adds no term to the Nelson-Aalen sum, but censorings on that row reduce the risk set at subsequent times.
Worked Example: Censoring Between Tied-Failure Times
A hypothetical study follows 20 pieces of test equipment for time to failure. Assume independent censoring and no parametric failure-time distribution. The grouped observations are:
| Time | At risk just before time | Failures | Censorings |
|---|---|---|---|
| 1 | 20 | 1 | 2 |
| 3 | 17 | 5 | 1 |
| 4 | 11 | 0 | 2 |
| 5 | 9 | 3 | 1 |
| 6 | 5 | 2 | 0 |
Calculate the estimate through time 6. The time-1 row leaves \(20-1-2=17\) at risk at time 3. The time-3 row leaves \(17-5-1=11\) at risk. There are no failures at time 4, but its two censorings reduce the count to \(11-2=9\) at time 5. The time-5 row leaves \(9-3-1=5\) at risk at time 6.
The failure increments, in order, are \(1/20\), \(5/17\), \(3/9\), and \(2/5\). Time 4 contributes no increment because its failure count is zero.
The estimate is rounded to four decimal places. Check. The stated risk sets agree with the previous row after removing its failures and censorings. In particular, omitting the time-4 censorings would incorrectly use 11 rather than 9 as the denominator at time 5.
Common Mistakes and Exam Traps
A frequent error is to pretend that tied failures occurred in a known sequence. In the time-1 group of the second example, that error might produce
instead of the correct grouped increment \(2/15\). The sequential calculation removes one subject from the denominator before counting the other tied failure, even though the data provide no such ordering. The correction is to count all failures at time 1 together: \(d_1=2\), \(Y_1=15\), and the increment is \(2/15\).
Other errors to check for include:
- Using the risk set after same-time censorings. Censorings at \(t_j\) are still included in the pre-time risk set under the stated convention.
- Removing earlier censorings too late. A subject censored before \(t_j\) is not at risk at \(t_j\).
- Adding censorings to the failure count. Only observed failures enter \(d_j\); censorings affect later risk sets.
- Skipping a tied group. A group with several failures contributes one increment, using the total count.
- Rounding each increment too early. Keep fractions or unrounded values through the sum and round only the final estimate.
A reliable calculation therefore has two separate tasks at each distinct failure time: determine the risk set immediately beforehand, then divide the total tied failures by that risk set. Update the risk set only after accounting for all failures and censorings at the time.
Combine all failures with the same recorded time into one group.
Exclude subjects censored strictly earlier; include subjects censored at the current time under the usual convention.
Divide the total failures at that time by the risk set immediately beforehand.
Remove failures and censorings at the time for later risk sets, and add the increment to the previous increments.
Check Your Understanding
For each problem, calculate the Nelson-Aalen cumulative hazard estimate through the specified time. Assume independent censoring and no parametric failure-time distribution unless otherwise stated.
- A hypothetical group begins with 14 subjects. At time 2, four fail and two are censored. At time 5, three fail and one is censored. Calculate the estimate through time 5. (Answer: \(37/56=0.6607\), to four decimal places.)
- A hypothetical sample begins with 18 items. At time 1, three fail and one is censored. At time 4, two fail and three are censored. Calculate the estimate through time 4. (Answer: \(13/42=0.3095\), to four decimal places.)
- Twenty-five hypothetical policy records are under observation. At time 2, five failures and two censorings occur. At time 3, three more are censored and no failures occur. At time 6, three fail. Calculate the estimate through time 6. (Answer: \(0.4000\).)
- A hypothetical study starts with 16 devices. At time 1, two fail and one is censored; at time 2, four fail and two are censored; at time 3, one fails. Calculate the estimate through time 3. (Answer: \(419/728=0.5755\), to four decimal places.)
- A hypothetical group starts with 30 units. At time 2, six fail and four are censored. At time 4, five fail and two are censored. At time 7, three fail. Calculate the estimate through time 7. (Answer: \(0.6808\), to four decimal places.)