Interpreting Jumps and Plateaus in Kaplan-Meier Curves

Why Kaplan-Meier Curves Have Steps and Plateaus

A Kaplan-Meier curve is not usually a smooth line. It stays at a constant level for a stretch of time, then drops when a failure is observed, and then stays level again until the next failure. Reading that shape correctly is important: the drop reflects observed failures among those still at risk, while a censoring observation affects who remains at risk but is not itself a failure.

The Kaplan-Meier estimator is built from successive conditional survival estimates. At each observed failure time, it multiplies the survival estimate so far by the estimated fraction of the risk set that did not fail at that time. Between failure times, there is no new failure factor to multiply in, so the estimate does not change.

Definition: Let $t_1<t_2<\cdots$ be the distinct observed failure times. Just before $t_j$, let $n_j$ be the number of individuals still at risk, and let $d_j$ be the number who fail at $t_j$. The Kaplan-Meier estimate of survival through time $t$ is $$ \hat S(t)=\prod_{t_j\leq t}\left(1-\frac{d_j}{n_j}\right). $$ A failure is an observed event. A right-censored observation contributes time at risk up to its censoring time, but does not count as a failure at that time.

For the examples below, failure and censoring times are distinct, so no failure and censoring occur at the same recorded time. At a failure time, the curve is conventionally shown at its new, lower level: the level immediately before the drop is the left-hand level, and the level at the failure time is the post-drop level. At a censoring time with no failure, the curve stays at its current level.

Why a Failure Drops the Estimate and Censoring Does Not

Suppose $n_j$ individuals are at risk just before a failure time and $d_j$ of them fail at that time. The fraction in the risk set that survives past that event time is estimated by $(n_j-d_j)/n_j$, or equivalently $1-d_j/n_j$. This fraction is no greater than 1, so multiplying by it either lowers the survival estimate or, if there are no failures, leaves it unchanged.

A censoring observation is not included in $d_j$. It indicates that the individual is no longer observed beyond the censoring time; it does not establish that the individual failed at that time. Therefore, censoring on its own contributes no downward factor. However, it does remove that individual from the risk set at later times, which can affect the size of a later drop.

\text{survival factor at a failure time} =\frac{n_j-d_j}{n_j} =1-\frac{d_j}{n_j}; \qquad \text{factor at a censoring time with no failure}=1.

The second expression is a way to understand the curve’s construction: a censoring time without a failure leaves the current product as it is. It does not mean the censored individual is treated as surviving forever. The individual contributes to the risk set only up to the time they are censored.

Start at survival level 1.
Before any observed failures, the product has no failure factors, so $\hat S(t)=1$.

At a failure time, apply the event factor.
Multiply the current level by $1-d_j/n_j$, using the risk set just before that failure time.

At a censoring time with no failure, keep the level.
Do not apply a failure factor. Remove the censored individual from later risk sets.

Between failure times, read a plateau.
The estimate remains at the level produced by the most recent failure time.

Worked Examples

Worked Example: Following Drops and Plateaus Through a Timeline

A hypothetical group of 8 policyholders is followed for time to a specified claim event. The observed records, in time units, are: one failure at time 2, one censoring at time 3, one failure at time 5, one censoring at time 6, and one failure at time 8. All remaining individuals are still under observation through time 8. Find the Kaplan-Meier levels after each failure and describe the plateaus. Use the Kaplan-Meier product-limit method, taking each risk set immediately before the event time.

At time 2: All 8 individuals are at risk, and 1 fails. The event factor is $1-1/8=7/8$. Thus the estimate drops from 1 to $7/8=0.8750$.

At time 3: One individual is censored, with no failure at that time. The estimate remains $7/8=0.8750$. The risk set for a later time is reduced by one.

At time 5: There are 6 individuals at risk: the original 8, less the failure at time 2 and the censoring at time 3. One fails, so the factor is $1-1/6=5/6$. The new estimate is $ (7/8)(5/6)=35/48=0.7292$, rounded to four decimal places.

At time 6: One individual is censored. There is no failure, so the estimate stays at $35/48=0.7292$. The risk set for time 8 is reduced again.

At time 8: Four individuals are at risk: 8 initially, less the two failures at times 2 and 5 and the two censorings at times 3 and 6. One fails, giving the factor $1-1/4=3/4$. The new estimate is $ (35/48)(3/4)=105/192=35/64=0.546875$, or $0.5469$ to four decimal places.

Interpretation and check: The curve is at 1 before time 2, at $7/8$ from just after time 2 until just before time 5, at $35/48$ from just after time 5 until just before time 8, and at $35/64$ at and after time 8, as long as there are no later failures. Each failure makes the estimate no larger than before; the censoring times leave the curve level unchanged. The risk sets also agree with the number still under observation before each failure. This example avoids the common mistake of counting a censoring as a failure.

Worked Example: How Censoring Changes a Later Drop

In a hypothetical study of 11 devices, the records are one failure at time 1, one censoring at time 2, one censoring at time 4, one failure at time 6, one censoring at time 7, and one failure at time 9. No other events occur. Calculate the curve levels, showing why the two censorings before time 6 matter even though neither directly changes the estimate.

At time 1: The risk set is 11, with 1 failure. The factor is $10/11$, so the estimate drops from 1 to $10/11=0.9091$.

At times 2 and 4: Both are censoring times with no failures. The estimate remains $10/11$. The risk set falls from 10 after the first censoring to 9 after the second.

At time 6: The risk set is 8: 11 initially, less 1 failure at time 1 and 2 censorings at times 2 and 4. One of the 8 fails. The factor is $7/8$, so the new estimate is $ (10/11)(7/8)=70/88=35/44=0.7955$, rounded to four decimal places.

At time 7: One individual is censored, but the estimate remains $35/44$. There are then 6 at risk for time 9: 11 initially, less 1 failure at time 1, 2 censorings before time 6, 1 failure at time 6, and 1 censoring at time 7.

At time 9: One of those 6 individuals fails. The factor is $5/6$, and the new estimate is $ (35/44)(5/6)=175/264=0.6629$, rounded to four decimal places.

Interpretation and check: The curve stays at $10/11$ between times 1 and 6, then stays at $35/44$ between times 6 and 9. Censoring does not create the drops, but censoring before time 6 reduces the later risk set from what it would otherwise have been. The factors $10/11$, $7/8$, and $5/6$ are all at most 1, so the successive levels $1$, $10/11$, $35/44$, and $175/264$ decrease as expected.

Worked Example: Recovering a Risk Set From a Step

A curve has level $0.9000$ just before an isolated failure time. Immediately after that failure its level is $0.8100$. Exactly one individual fails at the time, and no censoring is recorded at that time. Find the risk set size just before the failure. To make the stated pre-event level consistent with the history, suppose the hypothetical study began with 20 individuals: there were failures at risk-set sizes 20 and 19, followed by 8 censorings before the isolated failure.

Check the stated history: The first failure factor is $19/20=0.95$, and the second is $18/19$. Their product is $(19/20)(18/19)=18/20=0.9000$. After those two failures, 18 individuals remain; 8 are censored, leaving 10 at risk before the isolated failure.

Use the observed step: The ratio of the post-event level to the pre-event level is $0.8100/0.9000=0.9000$. For one failure among $n$ at risk, the event factor is $1-1/n$. Therefore $1-1/n=0.9000$, so $1/n=0.1000$ and $n=10$.

Check: With 10 at risk and one failure, the factor is $9/10$. Multiplying the pre-event level by this factor gives $0.9000(9/10)=0.8100$, exactly the stated post-event level. The risk-set count from the history also equals 10. The method avoids mistaking the vertical size of the drop, $0.0900$, for the event fraction: the event fraction is the relative drop $0.0900/0.9000=0.1000$.

Reading the Curve Carefully

A plateau represents an unchanged estimate, not necessarily an absence of observations. Censoring marks can occur along a plateau because the study may record that individuals leave observation even when no failure is observed. The curve’s height is determined by the failures and risk sets; a censoring mark alone does not lower it.

A failure-time step has two relevant levels. If the estimate is $0.80$ immediately before a failure and the event factor is $0.75$, the estimate immediately after is $0.80(0.75)=0.60$. The vertical drop is $0.20$, but the conditional fraction failing at that event time is $1-0.75=0.25$, not $0.20$. The drop in the whole curve is the current survival level multiplied by the conditional event fraction.

Between two consecutive observed failure times, $\hat S(t)$ is constant. In particular, censoring between those failures does not create an intermediate downward step. The censoring can still change the next risk set and therefore the size of a later step. This distinction explains why a curve can have a long plateau with several censoring marks and then a drop whose size depends on the risk set at the next failure.

What is observed?	Effect on the curve at that time	Effect on later risk sets
A failure	Apply a factor $1-d_j/n_j$; the level falls if $d_j>0$.	The failed individual is no longer at risk.
A censoring with no failure	No change in the current survival level.	The censored individual is removed from later risk sets.
A stretch with no failure times	The estimate stays at the most recent post-failure level.	Risk-set membership can still change at censoring times within the stretch.

Common Exam Traps

Drawing a drop at a censoring time. This incorrectly treats an observation ending as a failure. The correction is to keep the current estimate unchanged and remove the censored individual from subsequent risk sets.
Using the original sample size at every failure. The factor uses the number at risk immediately before that failure, not the initial number enrolled. Account for earlier failures and censorings when finding $n_j$.
Calling the full vertical drop the conditional failure fraction. The conditional failure fraction is $d_j/n_j$. The curve’s vertical drop is its current level multiplied by $d_j/n_j$.
Reading the pre-drop height at the event time. The curve’s value at an observed failure time is the post-drop level under the usual step-function convention. The pre-drop level applies just before that time.

Key takeaway: A Kaplan-Meier curve changes level at observed failure times, with each drop determined by the risk set and number of failures. Censoring without a failure leaves the current level unchanged, but changes who can be at risk at later failures. Between observed failure times, the estimate is a plateau.

Check Your Understanding

For each problem, distinguish the current curve level from the risk set used at a later failure. All times are distinct, and no failure occurs at a censoring time.

A hypothetical cohort starts with 9 individuals. One fails at time 2, one is censored at time 4, and one fails at time 7. Find the level after each failure and the level between times 2 and 7. (Answer: After time 2, $8/9=0.8889$; after time 7, $(8/9)(6/7)=16/21=0.7619$; between the failures the level is $8/9$.)
A study begins with 12 individuals. There is one failure at time 1, censoring at time 3, one failure at time 5, censoring at time 6, and one failure at time 8. Find the level after the third failure. (Answer: $(11/12)(9/10)(7/8)=231/320=0.7219$, rounded to four decimal places.)
A curve is at $0.6400$ immediately after a failure at time 4. There is no failure before time 9, but one individual is censored at time 7. What is the estimated survival level at time 8? (Answer: $0.6400$; censoring alone does not change the level, and there has been no failure since time 4.)
Immediately before an isolated failure, a curve is at $0.9000$, and immediately after it is at $0.8100$. One individual fails. Find the risk set size just before the failure. (Answer: The relative factor is $0.8100/0.9000=0.9$, so $1-1/n=0.9$ and $n=10$.)
A hypothetical group starts with 10 individuals. One fails at time 2, and the estimate is $0.9000$ afterward. There are no other failures before a single failure at time 5, when the estimate becomes $0.7500$. How many censorings occurred between times 2 and 5? (Answer: The second event factor is $0.7500/0.9000=5/6$, so 6 were at risk just before time 5. After the first failure, 9 remained; therefore 3 were censored before time 5.)

What is observed?	Effect on the curve at that time	Effect on later risk sets
A failure	Apply a factor \(1-d_j/n_j\); the level falls if \(d_j>0\).	The failed individual is no longer at risk.
A censoring with no failure	No change in the current survival level.	The censored individual is removed from later risk sets.
A stretch with no failure times	The estimate stays at the most recent post-failure level.	Risk-set membership can still change at censoring times within the stretch.

What You'll Learn