9.5 Practice Problems
Problem 9.5.1. Plots of \(\log (-\log \hat S)\) against \(\log t\) for two groups are parallel lines \(1.2\) units apart. What does this say about the hazards, and what is the hazard ratio?
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Solution. Parallel curves mean proportional hazards, and the vertical distance is \(\log H_2 - \log H_1 = \log \text {HR}\), so \(\text {HR} = e^{1.2} = 3.3\). If the lines are also straight, both groups follow Weibull distributions with the same shape.
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Solution. Their sum is \(\sum _j(x_{i_j} - \sum _{l \in R_j}w_lx_l)\), which is the derivative of the log partial likelihood with respect to \(\beta \) at \(\hat \beta \). The estimate is defined by setting this derivative to zero.
Problem 9.5.3. A treatment has hazard ratio \(0.5\) in the first year and \(1.0\) afterwards. A Cox model with a constant effect is fitted to five years of data. What will it report, and why is that misleading?
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Solution. A hazard ratio between \(0.5\) and \(1\), a weighted average of the two periods with weights depending on the numbers of events in each, perhaps about \(0.7\). It suggests a modest permanent benefit, when the truth is a large early benefit that disappears. The Schoenfeld residuals would show a trend; the right report gives the effect in each period, or survival curves.
Problem 9.5.4. In Example 9.3.1, at what time does the fitted coefficient \(-0.081 + 0.0135\log t\) reach zero, and what is the hazard ratio for \(10\) Karnofsky points at \(30\) days?
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Solution. \(\log t = 0.081/0.0135 = 6.0\), \(t = e^6 = 403\) days. At \(30\) days the coefficient is \(-0.081 + 0.0135\log 30 = -0.081 + 0.0459 = -0.035\), and the hazard ratio for \(10\) points is \(e^{-0.35} = 0.70\).
Problem 9.5.5. Describe how to fit a model in which a covariate’s coefficient is different before and after \(6\) months, and why the data must be restructured.
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Solution. Split each subject’s follow-up at \(6\) months into two records, \((0, \min (t, 6)]\) and, if \(t > 6\), \((6, t]\), with the event indicator only on the record in which the event occurs; define \(x_{\text {early}} = x \times \mathbf 1\{\text {first period}\}\) and \(x_{\text {late}} = x \times \mathbf 1\{\text {second}\}\); fit the Cox model with Surv(start, stop, event). The restructuring is needed because the covariate entering the hazard now depends on time, and each record states which value applies over which interval; the risk sets are formed correctly from the start and stop times.
Problem 9.5.6. Why do covariates measured at the start of a study, such as a performance score, often show hazard ratios that weaken over time?
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Solution. A baseline measurement describes the subject’s state at entry. Its relevance fades as that state changes: patients’ condition improves or worsens, and those with poor scores who survive the early period are often those whose condition was better than the score suggested. Selection by survival also plays a part: the high-risk subjects remaining later are a selected, more robust group. Both make the effect of the baseline value shrink with time.
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