Censoring and the Kaplan–Meier Estimator

Follow-up ends before everyone has the event. See why the product-limit estimator is the honest answer and what counting censored subjects as events does to it.

Ch 12 12.8

Explorer

Temporal & Longitudinal

22 mindifficulty 3/5

Controls

Median survival, arm A18 mo

Hazard ratio, B vs AHR0.600

Below 1 means arm B has the lower hazard; 1 means the arms are identical.

Follow-up window30 mo

Administrative censoring: the study simply ends.

Dropout hazard0.030

Loss to follow-up, independent of the event.

Subjectsn160

Show naive curve (arm A)

Counts every censored subject as an event.

Show censoring marks

Random seed

Every result on this page is a deterministic function of the seed and the controls.

Kaplan–Meier survival curves

Steps fall only at event times. Ticks mark censored subjects, who leave the risk set without an event.

arm A (control)arm B (treatment)arm A, censoring = event

Median, arm A

14.2 mo

true 18 mo

Median, arm B

not reached

true 30 mo

Censored

104 / 160

65.0% never had the event observed

Log-rank p

<0.001

χ² = 12.85 on 1 df

Risk table

How many subjects remain under observation.

timeA / B at risk

0 mo80 / 80

6 mo48 / 58

12 mo38 / 42

18 mo18 / 35

24 mo2 / 11

30 mo0 / 0

Late steps rest on very few subjects — that is why the tail of a KM curve is so unstable and why a risk table must accompany every published curve.

Naive analysis versus Kaplan–Meier

KM S(18), arm A45.2%

censoring-as-event S(18)22.5%

naive median, arm A10.9 mo

observed events, arm A38 / 80

observed events, arm B18 / 80

Treating censored subjects as events is not conservative — it is simply wrong, and it biases survival downward by exactly the amount of follow-up you threw away.

Censored is not the same as event-free forever

A censored subject contributes everything known about them — they were event-free up to their last visit — and then leaves the risk set. Dropping them discards information; counting them as events invents information. The product-limit estimator does neither: at each event time it multiplies in the conditional survival among those still under observation.

Predict before you look

A study censors 30% of subjects at the end of follow-up. If the analyst instead counts every censored subject as having had the event, what happens to the estimated median survival?

Scenarios

Teacher-ready starting points.

Documentation

What this shows

A two-arm trial with exponential event times, random dropout, and staggered administrative censoring. Kaplan–Meier curves for both arms are drawn with censoring ticks, medians, a risk table, a log-rank test, and — for contrast — the biased curve you get by treating every censored subject as an event.

The product-limit estimator

Order the distinct event times t₁ < t₂ < … At each one, dⱼ subjects have the event out of the nⱼ still under observation just before tⱼ. The conditional probability of surviving that instant is 1 − dⱼ/nⱼ, and survival to any time t is the product of those conditional terms over event times up to t. Censored subjects never cause a step; they simply reduce nⱼ for all later times, which is exactly the information they carry.

Greenwood's formula gives the variance: Var(Ŝ(t)) ≈ Ŝ(t)²·Σ dⱼ/(nⱼ(nⱼ − dⱼ)). Because nⱼ shrinks over time, late steps carry huge variance — the reason a published curve must be accompanied by a risk table.

The log-rank test compares observed and expected events arm by arm at every event time under the null of equal hazards, treating each event time as a 2×2 table and summing: χ² = (O_A − E_A)²/V. With exponential times and a constant hazard ratio, the ratio of medians is 1/HR, so HR = 0.6 means the treated median is about 1.67 times the control median.

All of this requires independent censoring: the reason a subject leaves must carry no information about their event risk. Dropout because a patient is getting sicker violates that assumption, and no estimator can repair it from the observed data alone.

Ŝ(t) = ∏_{tⱼ ≤ t} (1 − dⱼ/nⱼ)

Var(Ŝ(t)) ≈ Ŝ(t)² Σ_{tⱼ ≤ t} dⱼ / (nⱼ(nⱼ − dⱼ))

exponential:  S(t) = e^{−λt},  median = ln2/λ

HR = λ_B/λ_A,   median_B/median_A = 1/HR

log-rank χ² = (O_A − E_A)² / V,  1 df

Try this

Set HR = 1 and re-roll the seed five times. How often do the curves look convincingly separated?
Set dropout to 0.12 and turn on the naive curve. Read both survival values at the median and quantify the bias.
Shrink the follow-up window to 10 months with HR = 0.6. Which median is reported as 'not reached', and is the log-rank test still informative?
Hold HR = 0.7 and raise n from 40 to 600. Track the p-value and the estimated medians separately.
Compare the estimated median in arm A with the true ln2/λ value across several seeds at n = 40.
Set HR = 3 and describe the curve shape when the treatment is harmful.

Common misconception

“Forty patients were lost to follow-up, so we removed them from the survival analysis.”

Those forty were event-free for as long as they were observed, and that is real information about the survival curve up to their last visit. Removing them biases the estimate upward (the remaining sample is enriched for people observed long enough to have events) while counting them as events biases it sharply downward. Kaplan–Meier keeps them in the risk set until their censoring time and drops them afterwards, which is why it is the standard estimator — provided censoring is independent of prognosis.

DSPA3 Chapter 12 — Time series, longitudinal, and survival analysis