[next] [prev] [prev-tail] [tail] [up]

5.2 Using the cumulative hazard to choose a model

The cumulative hazard is the best scale on which to look for a parametric model, because the common families become straight lines:

  • exponential: \(H(t) = \lambda t\), so a plot of \(\hat H(t)\) against \(t\) is a straight line through the origin with slope \(\lambda \);
  • Weibull: \(\log H(t) = k\log t - k\log \lambda \), so a plot of \(\log \hat H(t)\) against \(\log t\) is a straight line with slope \(k\);
  • proportional hazards between two groups, the assumption of the Cox model of a later section: the curves of \(\log \hat H\) for the groups are parallel, a constant vertical distance apart.

For lung, the plot of \(\log \hat H\) against \(\log t\) is close to a straight line (Figure 5), and a least-squares line through it has slope \(1.21\), near the maximum likelihood shape \(\hat k = 1.32\) of the section on parametric models; the curvature at the left, from the first few deaths, is typical and carries little weight. Such plots are informal checks, not tests, but they show at once whether a model is plausible.

ˆ ˆ lolo23456−−−−−01WNgg54321eetHilsbo(tuln)l–A fiat:leslnop,e at kea=ch1. d3e2ath

Figure 5: The Weibull check for the lung data: the logarithm of the Nelson–Aalen cumulative hazard against log time, one point per death time, with the line implied by the fitted Weibull model of the section on parametric models. The points follow a straight line closely, so the Weibull model is reasonable; for an exponential the slope would be \(1\).

The hazard itself, the derivative of \(H\), can be estimated by smoothing the increments \(d_j/n_j\) with a kernel, which gives a picture of how the risk changes over time; the result is noisy unless the data are extensive.

[next] [prev] [prev-tail] [front] [up]

Questions on this section

Stuck on something here? Ask below and it stays attached to this topic.