Solution: Tranformations of the Nelson-Aalen estimator

For a given value of t, the Nelson-Aalen estimator \(\hat A(t)\) is approximately normally distributed with mean \(A(t)\) and a variance that may be estimated by \(\hat\sigma^2(t)\) given by \((3.5)\). For some strictly increasing and continuously differentiable function \(g\) we consider the Taylor expansion \[g(\hat A(t)) \approx g(A(t)) +g'(A(t))\{\hat A(t)−A(t)\}.\] Since \(A(t)\) is just a number, this is a linear transformation of \(\hat A(t)\). Thus this is also approximately normally distributed, with mean \begin{align*} E[g(\hat A(t))] &\approx E[g(A(t)) +g'(A(t))\{\hat A(t)−A(t)\}] \\ &= g(A(t)) +g'(A(t))E[\hat A(t)−A(t)] \\ & \approx g(A(t)) +g'(A(t))\cdot 0 \\ &= g(A(t))\end{align*} and variance \begin{align*} Var[g(\hat A(t))] &\approx Var[g(A(t)) +g'(A(t))\{\hat A(t)−A(t)\}] \\ &= 0 +g'(A(t))^2 Var[\hat A(t)−A(t)] \\ & \approx g'(A(t))^2\cdot \hat\sigma^2(t) . \end{align*} This variance still depends on the unknown \(A(t)\), but we can estimate it by replacing \(A(t)\) with \(\hat A(t)\), so that our estimation for the variance is \[\underline{\underline{g'(\hat A(t))^2\cdot \hat\sigma^2(t).}}\]

We have that \(g(\hat A(t))\) is approximately normally distributed with mean \(g(A(t))\) and a standard deviation which can be approximated by \(g'(\hat A(t))\cdot \hat\sigma(t)\). It then follows that an approximate \(100(1-\alpha)\%\)-confidence interval for \(g(A(t))\) is \[\underline{\underline{g(\hat A(t)) \pm z_{1-\alpha/2} \cdot g'(\hat A(t))\cdot \hat\sigma(t).}}\]

Using \(g(x) = \log x\) we get \(g'(x) = \frac{1}{x}\), and thus that an approximate \(100(1-\alpha)\%\)-confidence interval for \(\log (A(t))\) is \[\log \hat A(t) \pm z_{1-\alpha/2} \cdot \frac{\hat\sigma(t)}{\hat A(t)}.\] Taking \(e\) to this exponent means that an approximate \(100(1-\alpha)\%\)-confidence interval for \(e^{\log (A(t))} = A(t)\) is \[\underline{\underline{\hat A(t)\exp \left\{\pm z_{1-\alpha/2}\cdot \frac{\hat\sigma(t)}{\hat A(t)}\right\} }}.\]