11.6 Practice Problems
Problem 11.6.1. In a study of patients with a kidney graft, the events are graft failure and death with a functioning graft. Explain why one minus the Kaplan–Meier curve for graft failure, with deaths censored, does not estimate the proportion of patients whose graft fails by \(5\) years.
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Solution. Censoring a death treats the patient as if their graft could still fail later, so the patient’s future risk is represented by those still at risk. But a patient who has died cannot lose the graft. The curve estimates graft failure in a hypothetical population without deaths and is too high; the cumulative incidence function, estimated by Aalen–Johansen, gives the proportion actually experiencing graft failure.
Problem 11.6.2. Using Example 11.1.2, compute \(\hat F_1\) and \(\hat F_2\) at time \(7\) and check Proposition 11.1.1.
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Solution. By time \(7\) the events are at \(2\), \(3\), \(5\) and \(6\): \(\hat F_1(7) = 0.1 + 0.1143 = 0.2143\) and \(\hat F_2(7) = 0.1 + 0.1143 = 0.2143\), and \(\hat S(7) = 0.5714\); the sum is \(0.2143 + 0.2143 + 0.5714 = 1\).
Problem 11.6.3. With two causes of constant cause-specific hazards \(h_1 = 0.02\) and \(h_2 = 0.08\) per year, find the cumulative incidence of cause \(1\) at \(10\) years and as \(t \to \infty \), and compare with \(1 - e^{-h_1 t}\).
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Solution. \(F_1(t) = \int _0^t h_1e^{-(h_1 + h_2)u}\,\mathrm du = \frac {h_1}{h_1 + h_2}\{1 - e^{-(h_1 + h_2)t}\}\). At \(10\) years, \(0.2(1 - e^{-1}) = 0.126\); as \(t \to \infty \), \(0.2\). The naive \(1 - e^{-0.2} = 0.181\) at \(10\) years, rising to \(1\): in the long run everyone would fail of cause \(1\) in a world without cause \(2\).
Problem 11.6.4. A treatment has no effect on the cause-specific hazard of relapse but halves the hazard of death without relapse. What happens to its cause-specific and subdistribution hazard ratios for relapse?
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Solution. The cause-specific hazard ratio for relapse is \(1\). But treated patients live longer without relapsing, so more of them remain exposed to relapse and the cumulative incidence of relapse is higher; the subdistribution hazard ratio is above \(1\). The treatment does not cause relapse; it prevents a competing event.
Problem 11.6.5. With gamma frailty of variance \(\theta = 1\), a conditional hazard ratio \(e^\beta = 3\) and \(H_0(t) = 0.5\) at one year, find the population hazard ratio at time \(0\) and at one year.
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Solution. At time \(0\), \(3\). At one year, \(3(1 + 0.5)/(1 + 1.5) = 4.5/2.5 = 1.8\): the population ratio has fallen because the frailest high-risk subjects have already failed.
Problem 11.6.6. In the cgd trial, why is the robust standard error larger than the model-based one in the Andersen–Gill model, and what would happen to the inference if it were ignored?
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Solution. Patients prone to infection have several, so a patient’s events are positively correlated, and the data contain less independent information than the number of events suggests. The model-based standard error assumes independence and is too small; ignoring this gives confidence intervals too narrow and \(p\)-values too small. Here the robust error is \(0.312\) against \(0.261\).
Problem 11.6.7. Explain the difference between the Andersen–Gill and PWP models for recurrent events, and suggest a setting where each is appropriate.
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Solution. Andersen–Gill uses one baseline intensity for all events and keeps every subject at risk throughout, so a first and a fifth event have the same baseline rate; it suits events that do not change the subject, such as minor infections or seizures. PWP stratifies by event number, with subjects at risk of the \(m\)th event only after the \((m-1)\)th; it suits events that change the risk, such as successive heart attacks or tumour recurrences after surgery.
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