3.2 The exponential model
Theorem 3.2.1. For the exponential model \(h(t) = \lambda \), with \(d = \sum \delta _i\) events and total time at risk \(T_{\mathrm {tot}} = \sum t_i\), the maximum likelihood estimate is \[ \hat \lambda = \frac d{T_{\mathrm {tot}}}, \qquad \widehat {\operatorname {se}}(\hat \lambda ) = \frac {\hat \lambda }{\sqrt d} . \]
Proof. \(\ell (\lambda ) = d\log \lambda - \lambda T_{\mathrm {tot}}\), so \(\ell '(\lambda ) = d/\lambda - T_{\mathrm {tot}}\), which vanishes at \(\hat \lambda = d/T_{\mathrm {tot}}\), and \(\ell ''(\lambda ) = -d/\lambda ^2 < 0\), so it is a maximum. The observed information is \(-\ell ''(\hat \lambda ) = d/\hat \lambda ^2\), and the standard error is its inverse square root, \(\hat \lambda /\sqrt d\). □
The estimate is the event rate, events per unit of follow-up, justifying the calculation of the section on survival data. The precision depends on the number of events, not of subjects: a study with many subjects but few events tells us little. Since \(\hat \lambda \) is positive and its distribution skewed, confidence intervals are better computed on the log scale: \(\log \hat \lambda \pm 1.96/\sqrt d\), transformed back.
Example 3.2.2. In lung, \(165\) deaths in \(69\,593\) person-days give \(\hat \lambda = 0.00237\) per day, or \(0.866\) per year, with standard error \(0.067\) and \(95\%\) interval \((0.743, 1.009)\) per year. The fitted mean survival is \(1/\hat \lambda = 422\) days and the median \(\log 2/\hat \lambda = 292\) days (Software 3.5.1).
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