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Clinical Outcomes: Survival and Longitudinal Data · Tutorial 434 of 597

Kaplan-Meier Curves For Omics-Defined Groups

Estimate and interpret Kaplan–Meier curves for omics-defined groups, while checking censoring assumptions and recognizing the cost of choosing a cutoff for a continuous score.

Advanced 12 min read

What You'll Learn

  • Define the survival estimand for each omics-defined group.
  • Construct Kaplan–Meier estimates from event times and risk sets.
  • Interpret censoring marks and numbers at risk on a plot.
  • Calculate Greenwood standard errors and a transformed confidence interval.
  • Identify assumptions, diagnostics, and common failure modes.
  • Explain how dichotomizing a continuous omics score can distort comparisons.

Why Kaplan–Meier Curves Are Useful

An omics analysis may produce a score that ranks patients by a molecular profile. A Kaplan–Meier curve can describe survival over time in groups formed from that score—for example, patients above and below a prespecified threshold. The curve accounts for right-censored follow-up: a patient whose event has not been observed contributes information up to their last known event-free time, rather than being treated as if they had an event then.

A curve is a descriptive estimate of survival for the group and study population being analyzed. It does not by itself show that the omics profile caused a difference in survival, adjust for other patient characteristics, or establish that a chosen score threshold will work in another population. The earlier material on choosing survival estimands is relevant: the event, time origin, population, and meaning of survival must be specified before drawing or interpreting a curve.

Definition: For group \(g\), let \(T\) be time from the stated origin to the event. The survival function is \(S_g(t)=P(T>t\mid G=g)\), the probability that a member of group \(g\) remains event-free beyond time \(t\). With right censoring, the observed time is \(Y=\min(T,C)\), where \(C\) is censoring time, and \(\delta=1\) if the event is observed and \(\delta=0\) if follow-up is censored.

Kaplan–Meier Estimation

Within a group, order the distinct observed event times as \(t_1<t_2<\cdots\). Just before event time \(t_j\), let \(r_j\) be the number of participants still under observation and event-free, and let \(d_j\) be the number of observed events at that time. The product-limit estimate is

$$ \hat S_g(t)=\prod_{t_j\leq t}\left(1-\frac{d_j}{r_j}\right). $$

At times with no observed event, the estimate does not change. At an event time it drops by the corresponding risk-set factor. A participant censored at a time with no event is counted in the risk set up to that time, then removed from later risk sets; censoring alone does not cause a drop in the curve.

Assumptions: The ordinary Kaplan–Meier estimate requires a clearly defined time origin and event, valid follow-up times, and censoring that is independent of the event time within the group. Conditional independence given covariates alone does not generally justify the pooled estimate; use an analysis that accounts for those covariates unless additional conditions make the pooled estimate valid. It also requires participants to have a positive chance of remaining under observation across the time range being estimated. If entry is delayed, the risk set must include only participants who have entered and remain under observation; the ordinary risk-set counts cannot simply be used unchanged.

The product is an estimate built from successive risk-set contributions, not a claim that \(P(T>t_j\mid T\geq t_j)\) is the survival factor in continuous time. For a discrete event-time scale with possible event times \(t_j\), survival through each interval can be factored into conditional probabilities of surviving that interval given survival to its start. The risk-set proportion \(1-d_j/r_j\) estimates the corresponding conditional interval survival probability. With continuous event times, the product-limit construction uses the observed event-time risk sets as the increments of the survival estimate; conditioning on being at risk at an exact continuous time is not itself the needed interval probability.

A practical convention is that anyone censored at an event time is included in the risk set immediately before that time, unless the data specify a different ordering. The convention matters when event and censoring times tie. It should be applied consistently and reported when ties are common.

Worked Example: Building a Curve and Reading Its Risk Table

In a hypothetical cohort, eight patients in a high omics-score group have these follow-up records, measured in days: events at days 2, 4, 4, 6, and 7; censoring at days 3, 5, and 8. Assume censoring is independent of event time within this group and that every patient enters at day 0.

At day 2, \(r_1=8\) and \(d_1=1\), so the estimated survival becomes \(1-1/8=7/8=0.875\). The patient censored at day 3 leaves the later risk sets without changing that estimate.

At day 4, six patients remain at risk and two have events. Thus the estimate becomes \((7/8)(1-2/6)=(7/8)(2/3)=7/12=0.5833\). The patient censored at day 5 leaves three patients at risk for the day-6 event.

At day 6, \(r=3\) and \(d=1\), giving \((7/12)(2/3)=7/18=0.3889\). At day 7, \(r=2\) and \(d=1\), giving \((7/18)(1/2)=7/36=0.1944\). Censoring at day 8 produces no further drop.

Time (days)At risk just before timeEventsSurvival estimate after time
0801.000
2810.875
4620.5833
6310.3889
7210.1944

A plotted curve would step down at days 2, 4, 6, and 7. It would show censoring marks at days 3, 5, and 8, where the estimate stays level. The estimated median survival is 6 days: this is the first time the curve reaches or falls below 0.5. The risk table is essential context: late steps may be based on very few patients.

Uncertainty and Interpretation

A curve is an estimate, so uncertainty should be considered, especially in its tail. Greenwood’s formula estimates the variance of the Kaplan–Meier estimate at a time with at least one event, provided every included event time has \(r_j>d_j\):

$$ \widehat{\operatorname{Var}}\{\hat S_g(t)\} = \hat S_g(t)^2 \sum_{t_j\leq t}\frac{d_j}{r_j(r_j-d_j)}. $$

A direct normal interval can extend below 0 or above 1. When \(0<\hat S_g(t)<1\), one common alternative is a log-minus-log interval, based on \(\log\{-\log \hat S_g(t)\}\), with standard error \(\sqrt{q}/|\log\hat S_g(t)|\), where \(q\) is the Greenwood sum. The resulting limits are transformed back to the survival scale. This interval is an approximate confidence interval, not a probability statement about a fixed unknown parameter.

Worked Example: Greenwood Uncertainty at an Event Time

Consider a second hypothetical group of seven patients. There is one event at day 2, one censoring time at day 3, two events at day 4, and the remaining three observations are an event at day 6 and censoring at days 6 and 7. At day 4 the risk sets are \(r_1=7,d_1=1\) and \(r_2=5,d_2=2\). Therefore

$$ \hat S(4)=\left(\frac{6}{7}\right)\left(\frac{3}{5}\right)=\frac{18}{35}\approx0.5143. $$

The Greenwood sum is

$$ q=\frac{1}{7(7-1)}+\frac{2}{5(5-2)} =\frac{1}{42}+\frac{2}{15} =\frac{11}{70}\approx0.1571. $$

Thus the estimated variance is \((18/35)^2(11/70)\approx0.04156\), and the standard error is \(\sqrt{0.04156}\approx0.2039\). For the log-minus-log interval, the estimated standard error on the transformed scale is \(\sqrt{11/70}/|\log(18/35)|\approx0.596\). With \(z=1.96\), the transformed limits are \(\log\{-\log(18/35)\}\pm1.96(0.596)\). Transforming those limits back gives an approximate 95% confidence interval of 0.118 to 0.813 for survival at day 4. Under repeated sampling and the interval’s assumptions, this procedure would cover the true group survival at day 4 about 95% of the time.

Omics Groups, Diagnostics, and Alternatives

An omics-defined group may come from a prespecified score and cutoff, a fixed molecular classification, or another rule established without using the outcomes in the analysis set. State how the score was constructed, how the cutoff was chosen, and whether the rule was fixed before examining survival. If feature selection or threshold choice used the same outcomes later shown in the curves, the apparent separation can be optimistic. A curve does not repair that selection bias.

Inspect the number at risk at several times, the number and timing of censoring marks, and the event counts in each group. Sparse risk sets make the tail unstable; extensive censoring early in follow-up leaves little information later. Also check whether censoring patterns differ substantially between groups. Such differences do not prove informative censoring, but can be a warning that censoring independence within groups is implausible. If patients have delayed entry, use entry-aware risk sets. If event times are only known to fall within intervals, the ordinary right-censoring Kaplan–Meier calculation does not match the observation process.

Kaplan–Meier curves are unadjusted group summaries. A log-rank test provides a group-comparison procedure, but does not adjust for covariates or turn a descriptive association into a causal effect. Survival regression can retain a continuous omics score and adjust for measured covariates, but relies on its own modeling assumptions. Parametric survival models impose a specified time-to-event distribution and can yield smoother estimates, at the cost of that distributional assumption. The choice depends on the question; none makes an outcome-selected cutoff trustworthy without suitable validation.

Worked Example: How a Cutoff Changes the Story

Suppose eight hypothetical patients have omics scores 1 through 8. Their follow-up outcomes are: score 1, event at month 2; score 2, event at month 6; score 3, event at month 10; score 4, event at month 11; score 5, event at month 4; score 6, censored at month 12; score 7, event at month 9; and score 8, censored at month 12. For illustration, assume censoring is independent within any group being compared.

Using score 5 or greater as the high group, there are four patients. Events occur at months 4 and 9: the Kaplan–Meier estimate at month 12 is \((3/4)(2/3)=0.500\). Using score 4 or greater instead, the high group has five patients and events at months 4, 9, and 11, so its month-12 estimate is \((4/5)(3/4)(2/3)=0.400\). Using score 6 or greater, there are three patients, with one event at month 9 and two censorings at month 12, giving \(\hat S(12)=2/3=0.667\).

These estimates are not evidence that one cutoff is best: they show how much a small sample’s curve can depend on the threshold. A cutoff also discards the ordering and distances among scores within each group. If several cutoffs are tried and the one with the most favorable outcome separation is reported, the apparent separation is selected using the outcome. Prefer a prespecified threshold when the scientific question requires groups; otherwise, consider methods that retain the score continuously and validate any learned rule in independent data.

Common Mistakes

  • Treating censoring as an event. A censoring mark records the end of known follow-up, not an observed event; it removes a patient from later risk sets without causing a step down.
  • Ignoring who remains at risk. A late curve segment based on two patients is not as informative as an early segment based on hundreds. Read the risk table alongside the curve.
  • Interpreting group separation causally. An unadjusted curve describes survival in the observed groups; it does not isolate the effect of an omics profile from other differences between patients.
  • Choosing a cutoff because it separates the curves. This uses outcome information to define the groups and makes the resulting separation overly optimistic. Prespecify the threshold or validate the selection process on independent data.
Key takeaway: Kaplan–Meier curves estimate group-specific survival by multiplying event-time risk-set factors. Their interpretation depends on valid risk sets, independent censoring, and adequate follow-up; their visual separation does not justify a causal conclusion or a data-driven score cutoff.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. A group has 20 patients at risk at an event time, with two events. What is the multiplicative factor for the survival estimate at that time?
  2. Why does a censoring mark ordinarily not make the Kaplan–Meier curve drop?
  3. At a time with \(\hat S(t)=0.80\), the Greenwood sum is \(q=0.04\). Calculate the estimated variance and standard error.
  4. What is one consequence of selecting an omics-score cutoff after inspecting survival outcomes?

Solutions. (1) The factor is \(1-d/r=1-2/20=0.900\). (2) Censoring means the event time is not observed; the patient contributes follow-up up to censoring and is removed from later risk sets, but no event has been recorded at that time. (3) The variance is \((0.80)^2(0.04)=0.0256\), and the standard error is \(\sqrt{0.0256}=0.160\). (4) The reported group separation can be optimistic because the outcomes helped choose the cutoff; evaluate the selection rule in independent data or use a threshold specified independently of those outcomes.

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