4.5 Practice Problems

Problem 4.5.1. Compute the Kaplan–Meier estimate for the times \(2, 4, 4, 5^+, 7, 9^+, 11\) and give the estimated median.

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Solution. At \(2\): \(n = 7\), \(d = 1\), \(\hat S = \frac 67 = 0.857\). At \(4\): \(n = 6\), \(d = 2\), \(\hat S = 0.857 \times \frac 46 = 0.571\). At \(7\): \(n = 3\) (after the censoring at \(5\)), \(d = 1\), \(\hat S = 0.571 \times \frac 23 = 0.381\). At \(11\): \(n = 1\), \(d = 1\), \(\hat S = 0\). The curve first falls to \(\frac 12\) or below at \(7\), so the median is \(7\).

Problem 4.5.2. For the data of the previous problem, compute Greenwood’s standard error of \(\hat S(4)\).

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Solution. \(\sum \frac {d_j}{n_j(n_j - d_j)} = \frac 1{7 \times 6} + \frac 2{6 \times 4} = 0.0238 + 0.0833 = 0.1071\), so \(\operatorname {se} = 0.571 \times \sqrt {0.1071} = 0.187\). With seven patients the estimate is very imprecise.

Problem 4.5.3. Why does a censored observation not cause a drop in the Kaplan–Meier curve, and how does it nevertheless affect the curve?

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Solution. A censored subject did not have the event, so it contributes no death at any time. But it leaves the risk set at its censoring time, so every later conditional probability \(1 - d_j/n_j\) is computed from a smaller \(n_j\), and later drops are larger. The subject’s weight is in effect redistributed to those who remain at risk, which is fair under non-informative censoring.

Problem 4.5.4. A curve shows estimated \(5\)-year survival of \(0.80\) with \(4\) patients at risk at \(5\) years. What would you tell a reader who concludes that \(80\%\) of patients survive five years?

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Solution. The estimate at \(5\) years rests on very few patients; its confidence interval will be wide (one more death among the four would lower it by about a quarter of its value). The number at risk should always be reported, and conclusions about the tail drawn cautiously or not at all.

Problem 4.5.5. Explain the reverse Kaplan–Meier method for median follow-up, and why simply taking the median of all recorded times is wrong.

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Solution. Follow-up is the time a patient would have been observed had they not died. Patients who die early have short recorded times although their potential follow-up was longer; the median of recorded times therefore understates follow-up. The reverse method treats censoring as the event and death as censoring, so that deaths are treated as incomplete observations of follow-up; the median of the resulting curve estimates the potential follow-up. For lung it is \(588\) days, much longer than the median recorded time.

Problem 4.5.6. Show that if all subjects are censored after the last event, \(\hat S\) remains constant at its last value, and explain why the mean survival cannot then be estimated without assumptions.

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Solution. There are no events after the last event time, so no further factors enter the product, and \(\hat S\) stays at its last value \(s > 0\) up to the largest censored time and is undefined beyond. The mean is the area under \(S\), which depends on the unobserved tail; the restricted mean \(\int _0^\tau \hat S(t)\,dt\) up to a time \(\tau \) within follow-up is reported instead.

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