Why Tied Failures Need Care
In a Kaplan–Meier calculation, a failure time can be shared by several subjects. This can happen because failures genuinely occur together, or because the observation system records times only to a fixed precision, such as whole days. At that observed time, the estimator uses the number of failures in the tied group and the number of subjects still at risk immediately beforehand.
The key is to treat all failures recorded at that time as one group when calculating the survival step. The denominator is the risk set just before the time, not a series of changing risk sets formed by removing tied failures one at a time. Earlier failures and censorings affect the size of that risk set; failures in the current tied group do not reduce it until after the step is calculated.
The risk set includes subjects who have not failed or been censored before \(t_j\). A subject censored at an earlier time is no longer in the risk set; a subject who fails at \(t_j\) is in it immediately before that time. In the examples here, there are no censorings recorded at a failure time, so the only removals at each listed failure time are the stated failures.
As with the Kaplan–Meier estimator generally, interpreting the estimate as a survival estimate for the population requires censoring to be independent of the failure time, at least conditional on any relevant grouping or covariates. The calculation also assumes the observed records have been classified correctly as failures or right-censorings. Tied observed times are handled as a single event time; if times have been rounded, the data do not tell us the actual ordering of events within the tied group.
Deriving the Tied-Failure Factor
The Kaplan–Meier step is based on conditional survival within the risk set. Of the \(n_j\) subjects at risk just before \(t_j\), \(d_j\) fail at that observed time. Therefore \(n_j-d_j\) do not fail at that time. The observed conditional fraction surviving the step is consequently \((n_j-d_j)/n_j\), which is \(1-d_j/n_j\).
Here \(\widehat{S}(t_j^-)\) is the Kaplan–Meier estimate immediately before \(t_j\). Starting with \(\widehat{S}(0)=1\), the estimate at time \(t\) is the product of the factors at observed failure times no later than \(t\):
This follows by applying the conditional survival fraction at each successive failure time. The tied-failure factor is one such fraction: all \(d_j\) failures at \(t_j\) contribute together to the same step. The risk set for a later time is updated after this step, with the tied failures removed.
Calculating Risk Sets Before a Tie
A tied count alone is not enough to calculate a Kaplan–Meier step. The risk set \(n_j\) must be reconstructed using everything that happened earlier. In particular, subjects censored before the tied time are excluded, while subjects who fail at that time remain in the risk set until the step is applied.
List failure times and any censorings that occur strictly before the tied failure time.
Subtract all earlier failures and censorings from the number at risk at the start of the interval.
Use \(1-d_j/n_j\), with \(d_j\) equal to the full tied failure count.
Multiply the previous survival estimate by the factor, then remove the tied failures when finding a later risk set.
Worked Example: Four Failures in a Risk Set of Ten
A hypothetical study follows 15 insured devices until failure or censoring. At time 2, two devices fail. Before time 5, three additional devices are censored. At time 5, four devices fail together. Estimate survival immediately after time 5.
Method. Reconstruct the risk set at time 5, calculate one tied-failure factor, and multiply it by the earlier Kaplan–Meier estimate. This avoids using the initial cohort size as the denominator at the later time.
At time 2, all 15 devices are at risk and two fail. Thus the factor is \(1-2/15=13/15\), and the survival estimate after time 2 is \(13/15\).
Before time 5, the two failures and three censorings have left the risk set. Therefore \(n_5=15-2-3=10\). The four tied failures give \(d_5=4\), so the factor at time 5 is \(1-4/10=6/10=3/5\).
The estimate immediately after time 5 is \[ \widehat{S}(5)=\frac{13}{15}\times\frac{3}{5} =\frac{39}{75} =\frac{13}{25} =0.5200. \]
Check. The factor at time 5 is \(6/10\), matching the six devices that do not fail at time 5 out of the ten at risk just before it. The final estimate is 0.5200. The common mistake avoided is retaining the three earlier censorings in the time-5 denominator.
What a Tied Step Represents
A Kaplan–Meier curve remains flat between observed failure times, then drops at each failure time. When several failures share one observed time, their combined count determines the size of that drop. The estimator does not assign a separate jump to each tied subject. This matters both for calculating the survival level and for describing where the curve changes.
For a risk set of \(n_j\), the factor is \((n_j-d_j)/n_j\). It must be between zero and one when \(0\leq d_j\leq n_j\). If \(d_j=n_j\), every subject still at risk fails at that time, and the estimated survival becomes zero. If \(d_j=0\), there is no failure step at that time.
The risk set for the next failure time is not the old risk set. After the tied failures, those \(d_j\) subjects have left the study through failure. Any censorings strictly between this failure time and the next one also leave the next risk set. This accounting is separate from calculating the current tied step.
Worked Example: Several Tied Steps With Earlier Censoring
A hypothetical equipment study begins with 20 units. Two fail at time 1. Three other units are censored before time 3, when five units fail together. Two further units are censored before time 4, when two units fail together. Find the Kaplan–Meier estimate immediately after time 4.
Method. Calculate each failure-time factor using its own risk set, then multiply the factors. The common error avoided is using one denominator for all event times.
At time 1, \(n_1=20\) and \(d_1=2\), so the factor is \(1-2/20=18/20=9/10\). The risk set after those failures is \(20-2=18\).
Three units are censored before time 3, leaving \(n_3=18-3=15\). At time 3, \(d_3=5\), so the factor is \(1-5/15=10/15=2/3\). Ten units remain after the five failures.
Two units are censored before time 4, so \(n_4=10-2=8\). At time 4, \(d_4=2\), and the factor is \(1-2/8=6/8=3/4\).
Multiplying the factors gives \[ \widehat{S}(4) = \frac{18}{20}\times\frac{10}{15}\times\frac{6}{8} = \frac{9}{10}\times\frac{2}{3}\times\frac{3}{4} = \frac{9}{20} = 0.4500. \]
Check. The risk sets are 20, 15, and 8 at times 1, 3, and 4, respectively. Each equals the number remaining after all earlier failures and censorings. The final estimate is 0.4500.
Recovering a Tied Failure Count From a Step
Sometimes the risk set and the survival estimates before and after a time are given, while the tied failure count is unknown. The step equation can be rearranged to recover that count. This is useful for checking a life table or interpreting a reported Kaplan–Meier step.
Worked Example: Inferring the Number of Tied Failures
A hypothetical study starts with 20 components. Four fail at time 2, and six more are censored strictly before time 6. The survival estimate immediately before time 6 is 0.8000. Immediately after time 6 it is 0.4800. Find the number failing at time 6 and verify the resulting estimate.
Method. First establish the risk set at time 6, then use the ratio of successive survival estimates to infer the tied count. This avoids confusing the survival level with the conditional step factor.
After the four failures at time 2, \(20-4=16\) components remain. The six censorings before time 6 reduce the risk set to \(n_6=16-6=10\).
The factor at time 6 is the ratio of the estimate after the step to the estimate before it: \(0.4800/0.8000=0.6000\). Therefore \(1-d_6/10=0.6000\), so \(d_6/10=0.4000\) and \(d_6=4\).
Verification: the tied group leaves \(10-4=6\) of the ten at-risk components not failing at time 6, giving factor \(6/10=0.6000\). Multiplying the prior estimate gives \(0.8000\times0.6000=0.4800\), as stated.
Check. The inferred count is a whole number no greater than the risk set, and the calculated step reproduces the given post-time estimate.
Common Mistake: Applying a Separate Full-Risk-Set Factor to Each Failure
A tempting but incorrect calculation is to treat each tied failure as a separate event with a one-failure factor based on the original risk set. For example, with \(n_j=12\) and \(d_j=3\), this approach gives \((11/12)^3=0.7703\). That is not the Kaplan–Meier factor for the observed tied step.
The correct calculation uses all three failures together: \(1-3/12=9/12=0.7500\). The tied group leaves nine of the twelve at risk without a failure at that observed time. Treating the three failures as separate full-risk-set steps invents intermediate event times that are not in the data and overstates the factor.
A useful check is to ask whether the factor equals the fraction of the risk set that does not fail at the tied time. If it does not equal \((n_j-d_j)/n_j\), the tied count or denominator has likely been mishandled.
Check Your Understanding
For each problem, calculate or identify the tied-failure Kaplan–Meier step. Assume there are no censorings recorded at any failure time.
- There are 14 subjects at risk immediately before a time when three fail together. What is the factor at that time? (Answer: \(11/14=0.7857\))
- A study begins with 18 subjects. Three fail at time 1, then two are censored before time 4. Five fail together at time 4. Find the factor at time 4. (Answer: \(8/13=0.6154\))
- The survival estimate just before a tied failure time is 0.7200. There are 15 subjects at risk, and six fail together. Find the estimate just after that time. (Answer: \(0.4320\))
- A study begins with 25 subjects. Five fail at time 2, and four are censored before time 5. Four fail together at time 5. Find the estimate immediately after time 5. (Answer: \(0.6000\))
- A study begins with 30 subjects. Six fail at time 1; four are censored before time 2, when five fail together. Three more are censored before time 3, when three fail together. Find the estimate immediately after time 3. (Answer: \(0.4500\))