5.1 The Nelson–Aalen estimator
At each event time \(t_{(j)}\), the proportion \(d_j/n_j\) of those at risk who have the event estimates the hazard over that instant. Adding these increments gives the Nelson–Aalen estimator of the cumulative hazard, \[ \hat H(t) = \sum _{t_{(j)} \leq t}\frac {d_j}{n_j}, \qquad \widehat {\operatorname {Var}}\,\hat H(t) = \sum _{t_{(j)} \leq t}\frac {d_j}{n_j^2} , \] the variance coming from treating each \(d_j\) as approximately Poisson with mean \(n_j\,\Delta H_j\). Through \(S = e^{-H}\) it gives a second estimate of the survival function, \(\tilde S(t) = \exp (-\hat H(t))\), the Fleming–Harrington (or Breslow) estimator.
Proposition 5.1.1. At every \(t\), \(\hat S_{\text {KM}}(t) \leq \exp \bigl (-\hat H(t)\bigr )\), with the two close when the numbers at risk are large compared with the numbers of events.
Proof. For \(0 \leq x \leq 1\), \(1 - x \leq e^{-x}\). Applying this to each factor \(1 - d_j/n_j\) of the Kaplan–Meier product gives \(\prod (1 - d_j/n_j) \leq \prod e^{-d_j/n_j} = e^{-\hat H(t)}\). When \(d_j/n_j\) is small, \(1 - x\) and \(e^{-x}\) differ by about \(x^2/2\), so the estimates nearly agree. □
Example 5.1.2. For the ten patients of Example 4.1.1, the increments are \(\frac 1{10}\), \(\frac 18\), \(\frac 17\), \(\frac 15\), \(\frac 13\), so \(\hat H = 0.100\), \(0.225\), \(0.368\), \(0.568\), \(0.901\) at the event times \(3\), \(6\), \(8\), \(12\), \(18\). Then \(\exp (-\hat H) = 0.905\), \(0.799\), \(0.692\), \(0.567\), \(0.406\), each a little above the Kaplan–Meier values \(0.900\), \(0.788\), \(0.675\), \(0.540\), \(0.360\); the gap is largest at the end, where only three patients remain (Software 5.4.1).
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