From a Kaplan-Meier Curve to a Survival Quantile
A Kaplan-Meier estimate describes survival over time as a step function. A common next question is not “what is the estimated survival at time \(t\)?” but “at what time has a specified proportion of the population experienced the event?” The median is one example: it is the time by which the event-time distribution has accumulated to at least 50%.
For censored data, the answer is read from the Kaplan-Meier curve by finding the first event time where the estimated survival is at or below a specified threshold. Censoring affects later risk sets, but it does not itself create a downward step in the survival estimate.
The probability \(p\) refers to the cumulative proportion experiencing the event, not the proportion still event-free. Thus a 30th percentile uses a survival threshold of \(1-0.30=0.70\), while the 70th percentile uses a threshold of \(1-0.70=0.30\). This reversal is a frequent source of errors.
The Crossing Rule
Suppose the distinct event times are \(t_1<t_2<\cdots<t_k\). At event time \(t_j\), let \(r_j\) be the number at risk just before that time and \(d_j\) the number of events at that time. The Kaplan-Meier estimate after that event time is $$ \widehat S_{\mathrm{KM}}(t_j)=\prod_{h=1}^{j}\left(1-\frac{d_h}{r_h}\right). $$ Between event times, the estimate remains constant. A censoring time does not multiply the estimate by a failure factor; it changes the risk set for subsequent event times.
To estimate the \(p\)-quantile, compare the post-event survival estimate at each event time with \(1-p\), in chronological order. The first estimate at or below the threshold identifies the quantile. “First” matters: a later time may also satisfy the inequality, but it is not the earliest time at which the requested cumulative probability has been reached.
For a requested \(p\)-quantile, calculate the survival threshold \(1-p\). For the median, the threshold is \(0.5\).
Use the Kaplan-Meier estimate after each event time, including all tied events in that time’s risk-set factor.
Report the event time at which the estimate first becomes less than or equal to the threshold. Do not interpolate between steps.
Why the Threshold Rule Works
The Kaplan-Meier curve estimates the probability of remaining event-free beyond time \(t\). Its complement therefore estimates the probability of having experienced the event by \(t\). The quantile is defined by the first time this cumulative probability reaches \(p\). Consequently, the complement must be at least \(p\), which is equivalent to the survival estimate being at most \(1-p\).
Because the Kaplan-Meier estimate is constant between event times and changes only at event times, the first time satisfying the inequality is an event time. There is no basis for choosing an intermediate time on a flat segment or drawing a straight line between two steps.
Worked Example: Finding the Median and the 40th Percentile
A hypothetical warranty study follows 10 devices until a specified failure or censoring. The event-time risk sets are: at time 1, \(r_1=10\) and \(d_1=2\); at time 2, \(r_2=8\) and \(d_2=1\); at time 3, \(r_3=7\) and \(d_3=2\). Estimate the median and the 40th percentile of failure time.
Method. Calculate the post-event Kaplan-Meier estimates in order, then find the first crossing of the relevant survival threshold. The risk-set sizes are supplied, so no further reconstruction is needed.
At time 1, the conditional survival factor is \(1-2/10=8/10\), giving \(\widehat S_{\mathrm{KM}}(1)=8/10=0.8000\).
At time 2, the factor is \(1-1/8=7/8\). Thus \(\widehat S_{\mathrm{KM}}(2)=(8/10)(7/8)=7/10=0.7000\).
At time 3, the factor is \(1-2/7=5/7\). Thus \(\widehat S_{\mathrm{KM}}(3)=(7/10)(5/7)=5/10=0.5000\).
For the median, the threshold is \(1-0.50=0.50\). The estimate is above 0.50 at times 1 and 2, and equals 0.50 at time 3. Therefore the estimated median is time 3.
For the 40th percentile, the threshold is \(1-0.40=0.60\). The estimates at times 1 and 2 exceed 0.60, and the estimate at time 3 is 0.5000, which is below 0.60. The estimated 40th percentile is also time 3.
Check. At time 3, the estimated cumulative event probability is \(1-0.5000=0.5000\). It has reached both 0.40 and 0.50; before time 3, it had reached only 0.30. The first-crossing rule therefore gives time 3 for both quantiles. Common mistake avoided: using \(p\), rather than \(1-p\), as the survival threshold.
Censoring and Tied Failures in Quantile Calculations
A censoring observation contributes information that the subject remained event-free up to the censoring time. It is removed from the risk set after that time, but it does not cause a failure step. If a censoring and an event occur at the same recorded time, apply the convention used to construct the risk set; once the risk set and tied-failure count are given, use the stated \(r_j\) and \(d_j\) in the Kaplan-Meier factor.
When multiple failures occur at an event time, they enter together through \(d_j/r_j\). Do not split tied failures into sequential one-at-a-time factors using a shrinking risk set. The Kaplan-Meier step at that time is \(1-d_j/r_j\), applied to the estimate immediately before the time.
Worked Example: Censoring Changes a Later Quantile Step
In a hypothetical monitoring study, 12 components are followed over time. One fails at time 1; two additional components are censored at time 2; two fail at time 3; two fail at time 4; one fails at time 5; and one fails at time 6. The remaining components are censored after time 6. Estimate the median and the 60th percentile. All times use the same time unit.
Method. Update the risk set after each event and censoring time, then multiply the event factors. The two censorings at time 2 reduce the risk set from 11 to 9 before time 3; they do not reduce survival at time 2.
At time 1, \(r_1=12\), \(d_1=1\), so the factor is \(11/12\) and the estimate is \(11/12\).
After the failure, 11 remain. The two censorings at time 2 leave 9 at risk just before time 3. At time 3, \(r_2=9\), \(d_2=2\), so the factor is \(7/9\): \(\widehat S_{\mathrm{KM}}(3)=(11/12)(7/9)=77/108\approx0.7130\).
At time 4, two of the 7 at risk fail. The factor is \(5/7\), giving \(\widehat S_{\mathrm{KM}}(4)=(77/108)(5/7)=55/108\approx0.5093\).
Five are at risk just before time 5. One fails, giving a factor of \(4/5\): \(\widehat S_{\mathrm{KM}}(5)=(55/108)(4/5)=44/108=11/27\approx0.4074\).
Four are at risk just before time 6. One fails, giving a factor of \(3/4\): \(\widehat S_{\mathrm{KM}}(6)=(11/27)(3/4)=11/36\approx0.3056\).
For the median, use threshold 0.50. The estimate at time 4 is \(0.5093\), still above the threshold, while at time 5 it is \(0.4074\). The estimated median is time 5.
For the 60th percentile, use threshold \(1-0.60=0.40\). At time 5 the estimate is \(0.4074\), which is still above 0.40; at time 6 it is \(0.3056\). The estimated 60th percentile is time 6.
Check. The risk set at time 3 is 9, not 11, because two subjects were censored at time 2. The estimates decrease only at failure times. Common mistake avoided: counting censorings as failures, which would create an unjustified downward step and could change the reported quantile.
Worked Example: A Percentile at a Tied-Failure Time
A hypothetical group of 15 sensors is followed until failure. There are 3 failures at time 2, 2 at time 4, 3 at time 6, and 2 at time 8; no censoring occurs before these event times. Find the estimated 25th percentile and median failure time.
Method. Apply one Kaplan-Meier factor per distinct event time, with all tied failures included in that time’s \(d_j\).
At time 2, \(r_1=15\), \(d_1=3\), so the factor is \(12/15=4/5\). Hence \(\widehat S_{\mathrm{KM}}(2)=4/5=0.8000\).
At time 4, \(r_2=12\), \(d_2=2\), so the factor is \(10/12=5/6\). Hence \(\widehat S_{\mathrm{KM}}(4)=(4/5)(5/6)=2/3\approx0.6667\).
At time 6, \(r_3=10\), \(d_3=3\), so the factor is \(7/10\). Hence \(\widehat S_{\mathrm{KM}}(6)=(2/3)(7/10)=7/15\approx0.4667\).
At time 8, \(r_4=7\), \(d_4=2\), so the factor is \(5/7\). Hence \(\widehat S_{\mathrm{KM}}(8)=(7/15)(5/7)=1/3\approx0.3333\).
For the 25th percentile, the survival threshold is \(1-0.25=0.75\). The estimate at time 2 is 0.8000, above the threshold; at time 4 it is 0.6667, below it. The estimated 25th percentile is time 4.
For the median, the threshold is 0.50. The estimate first falls to or below 0.50 at time 6, where it is 0.4667. The estimated median is time 6.
Check. The time-4 survival step uses \(1-2/12=10/12\), not two sequential factors with changing risk sets. Common mistake avoided: processing tied failures as if they occurred at separate ordered times.
Common Exam Traps
| Trap | Incorrect approach | Correction |
|---|---|---|
| Reversing the percentile and survival probability | Using \(p\) as the survival threshold for a \(p\)-quantile | Use \(1-p\); for example, the 40th percentile uses 0.60. |
| Using a later crossing | Reporting any event time with survival below the threshold | Report the first event time with estimate at or below it. |
| Interpolating between steps | Estimating a time between two event times from a straight-line connection | The Kaplan-Meier estimate is a step function; use the event time where the step crosses. |
| Treating censoring as failure | Multiplying survival by a failure factor at a censoring time | Censoring changes subsequent risk sets but does not directly lower the estimate. |
| Splitting tied failures | Applying multiple one-failure factors with changing denominators at one tied time | Use the single factor \(1-d_j/r_j\) for that event time. |
A quantile is reported in the time units of the data, such as days or years. State the unit and the percentile in context. Also keep distinct the estimated quantile time and the estimated survival probability at a specified time: one is found by locating a threshold crossing, while the other is read directly from the curve.
The crossing rule gives a finite estimate only when the observed Kaplan-Meier curve reaches the required threshold. If it does not, the last observed event time is not automatically the requested quantile. A quantile must satisfy its defining threshold condition.
Check Your Understanding
For each problem, use the first-crossing rule and report the requested time in the stated time units.
- A hypothetical device study has event-time data \((r_j,d_j)\): time 1, \((10,2)\); time 3, \((7,2)\); time 5, \((5,1)\). Find the estimated 25th percentile. (Answer: time 3)
- For a fictional follow-up study, the risk sets are time 2, \((8,1)\); time 4, \((7,2)\); time 7, \((5,2)\). Find the estimated median. (Answer: time 7)
- Nine fictional units are followed. One fails at time 1, two are censored at time 2, two fail at time 4, and two fail at time 5. Find the estimated 40th percentile. (Answer: time 4)
- A hypothetical population has 20 subjects and failures as follows: time 1, 4 failures; time 2, 3 failures; time 3, 5 failures. There is no censoring. Find the estimated 30th percentile and the median. (Answer: 30th percentile at time 2; median at time 3)
- A stepped estimate is 0.70 after an event at time 4 and 0.40 after the next event at time 9. What is the estimated median, and why should no intermediate time be reported? (Answer: time 9; the curve is a step function and first reaches or falls below 0.50 at that event time.)