Kaplan–Meier and Nelson–Aalen Estimation › Example - A Ten-Person Failure Table

The formulas are compact, so it is worth slowing down and watching the risk set change row by row. We now work through a small failure table by hand. The observed times and event indicators are:

The distinct event times are $2,3,6,8,10$. We process tied events and censorings using events first. At each event time, the first question is always the same: who is still in the risk set?

At $t=2$, all ten subjects are at risk and one event occurs:

$$n_1=10,\qquad d_1=1,\qquad 1-\frac{d_1}{n_1}=1-\frac{1}{10}=0.9.$$

Thus $\hat S(2)=0.9$ and $\hat H(2)=0.1$. The risk set has shrunk by one because subject 1 has now had the event.

At $t=3$, subject 1 has already had an event, so nine subjects remain at risk. Two events occur at time 3.

$$n_2=9,\qquad d_2=2,\qquad 1-\frac{d_2}{n_2}=1-\frac{2}{9}=\frac{7}{9}.$$

The updated estimates are

$$\hat S(3)=0.9\cdot \frac{7}{9}=0.7, \qquad \hat H(3)=0.1+\frac{2}{9}=0.322222.$$

At $t=5$ subject 4 is censored. This does not create a step, but it removes subject 4 from later risk sets. That is the only role of censoring here: it changes who is available for future events.

At $t=6$, subject 4 is gone, but subject 6 is censored at time 6 and remains in the risk set for the event at time 6 under our convention. Six subjects are at risk and one event occurs:

$$n_3=6,\qquad d_3=1,\qquad 1-\frac{d_3}{n_3}=1-\frac{1}{6}=\frac{5}{6}.$$

Therefore

$$\hat S(6)=0.7\cdot \frac{5}{6}=0.583333, \qquad \hat H(6)=0.322222+\frac{1}{6}=0.488889.$$

At $t=8$, subjects 5 and 6 have left after time 6, so four subjects remain at risk. One event occurs:

$$n_4=4,\qquad d_4=1,\qquad 1-\frac{d_4}{n_4}=1-\frac{1}{4}=\frac{3}{4}.$$

Thus

$$\hat S(8)=0.583333\cdot \frac{3}{4}=0.4375, \qquad \hat H(8)=0.488889+\frac{1}{4}=0.738889.$$

At $t=10$, two subjects remain at risk and one event occurs:

$$n_5=2,\qquad d_5=1,\qquad 1-\frac{d_5}{n_5}=1-\frac{1}{2}=\frac{1}{2}.$$

So

$$\hat S(10)=0.4375\cdot \frac{1}{2}=0.21875, \qquad \hat H(10)=0.738889+\frac{1}{2}=1.238889.$$

At $t=12$, the one remaining subject is censored, so the Kaplan-Meier curve remains flat at $0.21875$ after time 10. It never reaches zero because the last subject leaves the study without an observed event.

The Nelson–Aalen estimate at time 10 is $1.238889$, and

$$\exp(-\hat H_{\mathrm{NA}}(10)) = \exp(-1.238889) \approx 0.2897.$$

This is above the Kaplan–Meier value $0.21875$. The gap is the finite-sample effect of using $1-d_j/n_j$ rather than $e^{-d_j/n_j}$ at each event time. It is not a bug, and it is one reason we should not label $\exp(-\hat H_{\mathrm{NA}})$ as Kaplan–Meier.

For uncertainty, the Greenwood increments are

$$\frac{1}{10\cdot 9},\ \frac{2}{9\cdot 7},\ \frac{1}{6\cdot 5},\ \frac{1}{4\cdot 3},\ \frac{1}{2\cdot 1}.$$

Their sum is approximately $0.659524$. At the final event time,

$$\widehat{\operatorname{SE}}(\hat S(10)) = 0.21875\sqrt{0.659524} \approx 0.1776.$$

The median is the first time the curve is at or below $0.5$. At time 6 the estimate is $0.583333$, and at time 8 it is $0.4375$, so the estimated median is $8$. This is a useful check on the risk-set arithmetic: the median is read from the step function, not computed by interpolating between event times.