Why the Order at a Tied Time Matters
In survival data, an individual may have an exact failure time, or the record may instead be right-censored: the individual is known to have remained under observation without a recorded failure up to a particular time, but what happened afterward is unknown. When failures and censorings have the same recorded time, the Kaplan-Meier calculation needs a clear rule for deciding who was at risk immediately before the failures.
The convention used here is that censorings recorded at a failure time are still in the risk set immediately before the failures at that time. The failures determine the Kaplan-Meier step. After that step is calculated, both the failures and the same-time censorings are removed before forming the next risk set. This convention is important when times have been rounded or grouped: the recorded data do not show a more detailed ordering within the tied time.
The key distinction is between calculating the factor and updating the risk set afterward. For the factor, use \(Y_j\), including individuals censored at \(t_j\). Afterward, subtract \(d_j+c_j\) to find how many remain under observation. A censoring at a later time affects later risk sets, but it does not change a Kaplan-Meier factor at an earlier failure time.
Formula and Justification
Assume a cohort begins under observation at time zero, there is no delayed entry, and censoring is noninformative for the survival estimation. At each recorded failure time, the risk set comprises individuals known to have remained under observation and event-free up to immediately before that time. If \(t_j\) is a tied time, the \(d_j\) recorded failures are counted among the \(Y_j\) individuals in that risk set; the \(c_j\) same-time censorings are also included.
The estimated conditional probability of surviving the failure time, given survival to just before it, is the number not failing divided by the number at risk. The count not failing at that time is \(Y_j-d_j\). Thus the factor is \((Y_j-d_j)/Y_j\), not \((Y_j-d_j)/(Y_j-c_j)\). Multiplying these conditional factors across failure times gives the product-limit estimate.
For a cohort of \(n\) individuals with no delayed entry, risk sets can be reconstructed recursively. If the distinct recorded times are ordered as \(t_1<t_2<\cdots\), then \(Y_1=n\), and after time \(t_j\) the number remaining is \(Y_j-d_j-c_j\). This remaining count is the risk set at the next distinct time, provided there are no other entries or removals between the two times. In particular, same-time censorings are subtracted after the factor at \(t_j\) is formed.
The product-limit formula follows from the chain rule for conditional survival: surviving through successive failure times requires surviving each step, conditional on having survived to that step. At each step, the observed fraction not failing estimates that conditional survival probability. Censoring does not create a failure step; it removes an individual from later risk sets. This is why the order within a tied recorded time matters to the denominator, but not to the total number removed afterward.
Count everyone still under observation and event-free immediately before the recorded time. Do not first subtract censorings recorded at that same time.
Use \(1-d_j/Y_j\), where \(d_j\) counts the failures tied at that time.
Remove all \(d_j\) failures and all \(c_j\) censorings before moving to a later recorded time.
Worked Applications
Worked Example: One Failure Time With Same-Time Censoring
A hypothetical warranty study follows 12 devices from the start of coverage. At time 2, two devices fail and one is censored. At time 4, one fails and two are censored. At time 6, two fail and one is censored. Find the Kaplan-Meier estimate just after time 6 and the number still under observation afterward. Treat the study as a single cohort with no delayed entry and noninformative censoring; no parametric failure-time distribution is assumed.
At time 2. Initially \(Y_1=12\). The factor is \((12-2)/12=10/12=5/6\). After the two failures and one censoring, the number remaining is \(12-2-1=9\).
At time 4. The risk set is \(Y_2=9\). The factor is \((9-1)/9=8/9\). After this time, \(9-1-2=6\) remain.
At time 6. The risk set is \(Y_3=6\). The factor is \((6-2)/6=4/6=2/3\). After the two failures and one censoring, \(6-2-1=3\) remain.
Multiply the factors.
The estimate just after time 6 is \(40/81\), or \(0.4938\) to four decimal places, and three devices remain under observation. The calculation checks: the three time points remove \(3+3+3=9\) devices from the original 12. Method and trap avoided: use the product-limit factors at failure times, and do not remove same-time censorings before calculating their denominators.
Worked Example: Reconstructing Risk Sets From Individual Records
A hypothetical study has ten participants. Their recorded times and statuses are: failure at 1, censoring at 1, censoring at 2, two failures at 3, two censorings at 3, censoring at 4, failure at 5, and censoring at 5. Find the Kaplan-Meier estimate just after time 5. Assume a single cohort with no delayed entry, noninformative censoring, and no parametric survival model.
Check the record count. There are two records at time 1, one at time 2, four at time 3, one at time 4, and two at time 5. The total is \(2+1+4+1+2=10\), as stated.
At time 1. The risk set is 10. One failure gives the factor \((10-1)/10=9/10\). The failure and censoring leave \(10-1-1=8\) participants.
At time 2. There is no failure, so there is no Kaplan-Meier factor. The one censoring reduces the risk set for time 3 to \(8-1=7\).
At time 3. The risk set immediately before the failures is 7, including both participants censored at time 3. Two failures give the factor \((7-2)/7=5/7\). Removing those failures and the two censorings leaves \(7-2-2=3\).
At time 4. There is again no failure step. The censoring reduces the risk set from 3 to 2.
At time 5. The risk set is 2, including the participant censored at 5. One failure gives the factor \((2-1)/2=1/2\). The failure and censoring leave \(2-1-1=0\).
Therefore,
The estimate is \(0.3214\) to four decimal places, with no participants remaining afterward. As a check, the four failures and six censorings account for all ten records. Method and trap avoided: a time with censoring but no failure changes the next risk set but contributes no product-limit factor.
Worked Example: The Effect of Reversing the Tie Order
A hypothetical cohort of 15 insured items has the following counts: at time 1, two failures and one censoring; at time 3, two censorings and no failures; at time 4, three failures and two censorings; and at time 6, one failure and one censoring. Calculate the estimate just after time 6 using the standard convention, then compare it with the result of incorrectly removing same-time censorings before failures. Assume no delayed entry and noninformative censoring; no parametric distribution is specified.
Correct risk sets and factors. At time 1, \(Y_1=15\), so the factor is \((15-2)/15=13/15\), and \(15-2-1=12\) remain. The two censorings at time 3 reduce that count to \(12-2=10\), but create no failure factor.
At time 4, \(Y_2=10\). Both censorings are included in that risk set, so the factor is \((10-3)/10=7/10\). Then \(10-3-2=5\) remain. At time 6, \(Y_3=5\), so the factor is \((5-1)/5=4/5\), and \(5-1-1=3\) remain.
The correct estimate is
If the same-time censorings are incorrectly removed first, the time 1 factor would be calculated from \(15-1=14\), giving \((14-2)/14=6/7\). At time 4, the incorrect denominator would be \(10-2=8\), giving \((8-3)/8=5/8\). At time 6, it would be \(5-1=4\), giving \((4-1)/4=3/4\). The resulting incorrect estimate is
The two estimates differ because the incorrect calculation excludes people who belong in the risk set immediately before the tied failures. The correct estimate is \(0.4853\) to four decimal places, and three items remain after time 6. The count check is \(15-(2+1)-(0+2)-(3+2)-(1+1)=3\). Method and trap avoided: do not apply censoring removals before the failures at the same recorded time.
Common Mistakes and Practical Checks
When checking a calculation, keep the failure count, censoring count, risk-set size, and post-time count distinct. A compact table is often the safest way to organize a multi-time problem.
| Quantity at time \(t_j\) | Use | Do not do |
|---|---|---|
| Risk set \(Y_j\) | Include failures and censorings recorded at \(t_j\) | Subtract the same-time censorings before the failure step |
| Kaplan-Meier factor | \((Y_j-d_j)/Y_j\) | Use \(Y_j-c_j\) as the denominator |
| Count remaining after \(t_j\) | \(Y_j-d_j-c_j\) | Leave same-time censorings in the next risk set |
| Time with censoring only | Reduce later risk sets | Add a factor when \(d_j=0\) |
A simple arithmetic check is to reconcile the risk-set update with the records: after a tied time, subtract both types of removals. Another check is that each factor lies between zero and one, so the Kaplan-Meier estimate cannot increase as time passes. A factor of one occurs when there are no failures at that time; censoring alone does not lower the estimate.
The event-before-censoring convention is a rule for interpreting the recorded tie, not proof that the underlying failure physically occurred before the censoring within that time unit. If the data were recorded only to the nearest day, for example, the actual order during that day may be unknown. The convention supplies a consistent risk-set calculation for the available data. Applying it consistently is essential for a valid product-limit estimate.
Check Your Understanding
For each problem, use censorings tied with failures as at risk immediately before the failures. Assume a single cohort with no delayed entry and noninformative censoring.
- A cohort of 20 has three failures and two censorings at time 4, with no earlier removals. Find the Kaplan-Meier factor at time 4 and the number remaining afterward. (Answer: factor \(17/20=0.8500\); 15 remain.)
- A cohort of 14 has two censorings at time 1; at time 3 there are two failures and three censorings; at time 5 there is one failure and two censorings. Find the estimate just after time 5 and the number remaining. (Answer: \((12/12)(10/12)(6/7)=5/7\approx0.7143\); 4 remain.)
- Nine participants have one failure and one censoring at time 2, two censorings at time 4, and two failures plus one censoring at time 6. Find the estimate just after time 6 and the number remaining. (Answer: \((8/9)(3/5)=8/15\approx0.5333\); 2 remain.)
- A cohort of 10 has two failures and three censorings at time 2, followed by one failure at time 5. Find the estimate just after time 5. State the risk set immediately before each failure time. (Answer: risk sets 10 and 5; estimate \((8/10)(4/5)=16/25=0.6400\).)