11.3 The Kaplan–Meier estimator
When exact times are known, the intervals can be made so fine that each contains at most one distinct event time. Then intervals without events contribute factors of \(1\), and withdrawals never share an interval with an event except by a tie, which is resolved by counting those censored at \(t\) as at risk at \(t\).
Definition 11.3.1. Let \(t_1 < t_2 < \cdots \) be the distinct event times, \(d_j\) the number of events at \(t_j\) and \(n_j\) the number at risk just before \(t_j\) (those with follow-up time \(\geq t_j\)). The Kaplan–Meier (product-limit) estimator is \[ \hat S(t) = \prod _{j: t_j \leq t}\Bigl (1 - \frac {d_j}{n_j}\Bigr ) , \] a step function that drops only at event times.
Greenwood’s formula applies with \(n_h'\) replaced by \(n_j\). Kaplan and Meier (1958) showed that \(\hat S\) is the nonparametric maximum likelihood estimate of the survivor function under non-informative censoring.
Example 11.3.2. Ten patients on a new treatment (A) have follow-up times in months \(3\), \(5\), \(7^+\), \(9\), \(12\), \(15^+\), \(18\), \(22^+\), \(24^+\), \(24^+\), where \(^+\) marks a censored time. The Kaplan–Meier estimate is:
| \(t_j\) | \(n_j\) | \(d_j\) | \(\hat S(t_j)\) | SE |
| 3 | 10 | 1 | 0.900 | 0.095 |
| 5 | 9 | 1 | 0.800 | 0.126 |
| 9 | 7 | 1 | 0.686 | 0.152 |
| 12 | 6 | 1 | 0.571 | 0.164 |
| 18 | 4 | 1 | 0.429 | 0.174 |
At \(9\) months only \(7\) are at risk, because the patient censored at \(7\) months has left the risk set without an event. The estimate stays at \(0.429\) from \(18\) to \(24\) months. The standard errors are large: ten patients cannot estimate a survival curve precisely.
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