Why a Kaplan-Meier Life Table Is Useful
A right-censored record tells us that an event time was not observed exactly: the event had not occurred by the time observation ended. Such a record still contributes information before censoring. A Kaplan-Meier life table organizes that information in chronological order, showing who was still at risk, how many events occurred, and how the estimated survival changes at each event time.
The central task is bookkeeping. At every observed time, count only individuals who have neither experienced the event nor been censored earlier. Then use the failures among that risk set to calculate a conditional survival factor. The resulting calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.
Here, \(\widehat S(t)\) estimates the probability of surviving beyond \(t\), based on the observed sample. A censoring time contributes to later risk-set counts, but it does not itself contribute a failure factor. The displayed definition uses the simple no-ties setup: no censoring is recorded at the same time as a failure. When times are tied, the ordering convention must be specified; that case requires separate care.
How to Build the Table
Begin with the total number of individuals. List all distinct observed times in increasing order, including censoring times if the table is intended to show the full bookkeeping. For each row, determine the risk set from the records strictly before that time. At a failure time, calculate the factor \(1-d_i/n_i\) and multiply it by the preceding Kaplan-Meier estimate. At a censor-only time, carry the estimate forward unchanged.
Write each distinct failure or censoring time in chronological order. Check that failure and censoring times do not coincide if using the simple table convention here.
Start with the initial sample size, then subtract every earlier failure and censoring. Do not subtract a person before that person’s recorded time.
At a failure time, use \(1-d_i/n_i\). At a censor-only time, there is no failure factor.
Multiply factors at failure times to obtain the cumulative estimate. After each row, check that the remaining risk set equals the prior risk set minus that row’s failures and censorings.
Why the Product-Limit Calculation Works
The risk-set fraction \(d_i/n_i\) is the observed fraction of those still under observation who fail at time \(t_i\). Its complement, \(1-d_i/n_i\), is the observed fraction of that risk set that survives past \(t_i\). Survival through several successive failure times requires surviving each conditional step.
To see why the factors multiply, write survival through the first \(k\) failure times as a sequence of conditional probabilities. The probability of surviving the first step and then the second is the probability of surviving the first multiplied by the probability of surviving the second conditional on having survived the first. Repeating this rule gives the product of the conditional survival probabilities. Replacing each conditional probability with its risk-set estimate yields the Kaplan-Meier product.
Only failure times appear in this product. A censoring removes an individual from future risk sets, but, under the independent-censoring assumption, it is not counted as an event. This assumption means that, among individuals comparable in the analysis, censoring does not systematically select those with different event prospects. If censoring is related to an individual’s unobserved event time, the ordinary Kaplan-Meier estimate may not represent the target survival distribution.
Worked Example: Completing a Table for Nine Records
A hypothetical warranty study follows nine devices until a specified component fails or observation ends. The observed records are: censorings at times 1, 4, 7, 8, and 10; failures at times 2, 3, 6, and 9. Assume independent censoring, and that no failure time coincides with a censoring time. Complete the life table and report the estimate after the failure at time 9. Give decimal estimates to four places.
Method. Order the nine recorded times, calculate each risk set by subtracting earlier outcomes, and update the estimate only at failures. Start with \(n=9\) and \(\widehat S=1\) before the first observed time.
| Time | At risk \(n_i\) | Failures \(d_i\) | Censored \(c_i\) | Conditional factor | Estimate after time |
|---|---|---|---|---|---|
| 1 | 9 | 0 | 1 | — | 1.0000 |
| 2 | 8 | 1 | 0 | \(7/8\) | 0.8750 |
| 3 | 7 | 1 | 0 | \(6/7\) | 0.7500 |
| 4 | 6 | 0 | 1 | — | 0.7500 |
| 6 | 5 | 1 | 0 | \(4/5\) | 0.6000 |
| 7 | 4 | 0 | 1 | — | 0.6000 |
| 8 | 3 | 0 | 1 | — | 0.6000 |
| 9 | 2 | 1 | 0 | \(1/2\) | 0.3000 |
| 10 | 1 | 0 | 1 | — | 0.3000 |
Risk-set check. At time 1, one of the nine is censored, leaving eight at risk at time 2. The failures at times 2 and 3 reduce the counts from 8 to 7 and then from 7 to 6. The censoring at time 4 leaves 5 at risk at time 6. The failure at time 6 leaves 4; the censorings at times 7 and 8 leave 2 at risk at time 9. One fails at time 9, leaving 1 to be censored at time 10.
Estimate check. At time 2, \(\widehat S=1(7/8)=7/8\). At time 3, \(\widehat S=(7/8)(6/7)=6/8=3/4\). Censoring at time 4 leaves the estimate at \(3/4\). At time 6, \(\widehat S=(3/4)(4/5)=3/5\). Censoring at times 7 and 8 does not change it. At time 9, \(\widehat S=(3/5)(1/2)=3/10=0.3000\). The sample contains four failures and five censorings, totaling nine records, as required.
Common mistake avoided: The two censorings at times 7 and 8 leave two individuals at risk at time 9, not one. After the failure at time 9, only one individual remains to be censored at time 10. Subtracting the time-10 censoring before forming the time-9 risk set would incorrectly use a risk set of one.
Further Table Calculations
Worked Example: Tracking Censoring Before Each Failure
A hypothetical study of eight rental pumps records censorings at times 1, 3, 6, 8, and 9, and failures at times 2, 4, and 7. Assume independent censoring and distinct times. Find the table’s conditional factors and the estimate after the last failure. Give decimal estimates to four places.
Method. Include censor-only rows so that each reduction in the risk set is visible. Initially, eight pumps are at risk and the survival estimate is 1.
| Time | At risk \(n_i\) | \(d_i\) | \(c_i\) | Factor | Estimate after time |
|---|---|---|---|---|---|
| 1 | 8 | 0 | 1 | — | 1.0000 |
| 2 | 7 | 1 | 0 | \(6/7\) | 0.8571 |
| 3 | 6 | 0 | 1 | — | 0.8571 |
| 4 | 5 | 1 | 0 | \(4/5\) | 0.6857 |
| 6 | 4 | 0 | 1 | — | 0.6857 |
| 7 | 3 | 1 | 0 | \(2/3\) | 0.4571 |
| 8 | 2 | 0 | 1 | — | 0.4571 |
| 9 | 1 | 0 | 1 | — | 0.4571 |
At time 2, one censoring has already occurred, so \(n_2=7\), and the factor is \(1-1/7=6/7\). The censoring at time 3 reduces the risk set to 5 at time 4; thus the next factor is \(1-1/5=4/5\). At time 7, the risk set is 3, giving \(1-1/3=2/3\). Therefore $$ \widehat S(7)=\frac{6}{7}\frac{4}{5}\frac{2}{3}=\frac{48}{105}=\frac{16}{35}=0.4571. $$
After the time-7 failure, two pumps remain at risk. The censorings at times 8 and 9 remove them from later risk sets but add no further factor. The table accounts for three failures and five censorings, which totals eight records.
Common mistake avoided: The factor at time 4 is \(4/5\), not \(5/6\). The time-1 censoring and the time-2 failure have both occurred before time 4, leaving five—not six—pumps at risk.
Worked Example: Checking a Short Table Against the Sample Size
A hypothetical inspection follows seven circuit modules. One is censored at each of times 1, 3, 5, 7, and 8; failures occur at times 2, 4, and 6. Assuming independent censoring and no ties, calculate the survival estimate immediately after time 6 and verify the final risk-set count.
Method. Subtract each earlier record from the initial count of seven. At time 2, one module has been censored, so \(n_2=6\). The failure factor is \(1-1/6=5/6\), leaving five at risk.
At time 3, one censoring leaves four at risk at time 4. The time-4 factor is \(1-1/4=3/4\), leaving three at risk. Censoring at time 5 leaves two at risk at time 6, where the factor is \(1-1/2=1/2\). Thus $$ \widehat S(6)=\frac{5}{6}\frac{3}{4}\frac{1}{2}=\frac{15}{48}=\frac{5}{16}=0.3125. $$
After the failure at time 6, one module remains. It is censored at time 7, leaving none at risk at time 8; the time-8 censoring is the final recorded observation in the stated data only if an individual is still available to be censored. Here, that would be inconsistent: after the time-5 censoring and three failures, only one module remains after time 6. Therefore the stated records cannot describe seven distinct modules.
Correction and completed calculation. The record list contains five censorings and three failures, or eight records, not seven. For the stated eight-module sample, the risk set at time 2 is 7, giving factor \(6/7\); at time 4 it is 5, giving factor \(4/5\); and at time 6 it is 3, giving factor \(2/3\). The estimate after time 6 is \((6/7)(4/5)(2/3)=16/35=0.4571\). The final two censorings at times 7 and 8 then leave zero at risk. The sample-count check exposes the inconsistency before a false table is accepted.
Common mistake avoided: Do not force a supplied initial sample size to fit an incompatible list of records. The total failures and censorings must equal the initial number of individuals.
Exam Checks and Common Traps
A complete table is more than a sequence of multiplications. Check the sample total, the changing risk set, and the estimate separately. The following checks catch the most common errors:
- Use the pre-time risk set. \(n_i\) counts individuals at risk immediately before time \(t_i\), not after failures or censorings at that time.
- Remove earlier censorings. A censored individual is no longer in any later risk set, even though that record does not reduce the current survival estimate.
- Do not multiply at censor-only times. Carry the estimate forward unchanged unless a failure occurs.
- Check the arithmetic at every row. In the no-ties setting, the next risk-set count is the current count minus that row’s failures and censorings.
- Reconcile totals. Across the full table, the number of failures plus the number of censorings must equal the original sample size if every individual has one of those recorded outcomes.
The life table makes the logic auditable: each factor is tied to a stated risk set, and each risk-set reduction is tied to a recorded outcome. This is especially valuable in multi-step work, where a single missed censoring can change every later factor.
Check Your Understanding
For each problem, use the simple no-ties convention: failure and censoring times are distinct. Give decimal survival estimates to four places where needed.
- A sample starts with six individuals. One is censored at time 1, one fails at time 2, one is censored at time 3, one fails at time 4, and the remaining two are censored at times 5 and 6. Find the risk sets at both failure times and the estimate after time 4. (Answer: \(n_2=5\), \(n_4=3\), \(\widehat S(4)=(4/5)(2/3)=8/15=0.5333\).)
- Nine records consist of censorings at times 1, 3, 4, 7, 8, and 9, and failures at times 2, 5, and 6. Find the estimate after the last failure and state the number at risk immediately before it. (Answer: \(n_6=4\), \(\widehat S(6)=(7/8)(4/5)(3/4)=21/40=0.5250\).)
- Ten individuals have two censorings at time 1, a failure at time 2, a censoring at time 3, a failure at time 4, two censorings at time 5, a failure at time 6, and censorings at times 7 and 8. Complete the failure-time factors and find the estimate after time 6. (Answer: factors \(7/8,\ 5/6,\ 2/3\); \(\widehat S(6)=35/72=0.4861\).)
- Twelve individuals have three censorings at time 1, one failure at time 2, two censorings at time 3, and two failures at time 4. Find the risk set at time 4 and the estimate immediately after time 4. (Answer: \(n_4=6\), \(\widehat S(4)=(8/9)(4/6)=16/27=0.5926\).)
- Immediately before a failure time, 10 individuals are at risk and two fail. The Kaplan-Meier estimate immediately before this time is \(0.72\). Find the conditional factor and the estimate immediately after the failures. (Answer: factor \(1-2/10=0.8\); estimate \(0.72(0.8)=0.5760\).)