Michael Sachs
with some detours
Recall the cumulative hazard defined as \(A(t) = \int_0^t\alpha(s) ds\) if the distribution function is absolutely continuous.
The Nelson-Aalen estimator is \[ \hat{A}(t) = \sum_{T_j \leq t} \frac{D_j}{Y(T_j)}. \] This was motivated by using the life-table approach (counting events in time intervals) and then taking the limit so that only one event can occur in each interval. We can write this as \[ \hat{A}(t) = \int_0^t \frac{I\{Y(u) > 0\}}{Y(u)}\, dN(u), \] which is an estimate of the quantity \[ A^*(t) = \int_0^t I\{Y(u) > 0\} \alpha(u) \, du. \]
In general, in the non absolutely continuous case, the density and hazard \(\alpha(t)\) are not defined, but we define the cumulative hazard as \[ A(t) = -\int_0^t\frac{1}{S(u-)}\, dS(u) = \int_0^t\frac{1}{1 - F(u-)}\, dF(u). \]
To connect this back to the survival function, let \(0 = t_0 < t_1 < \ldots < t_n = t\) denote a partition of \([0, t)\) and define the product integral \[ \prod_u^t\{1 - dB(u)\} = \lim_{\mathcal{M} \rightarrow 0}\prod_{k=1}^K\{1 - ((B(t_k) - B(t_{k-1})))\} \] where \(\mathcal{M} = \max_k|t_k - t_{k-1}|\) is the length of the longest subinterval.
We then have, with \(A\) defined as above
\[ S(t) = \prod_u^t\{1 - dA(u)\} \] and it is true (but not trivial) that in the absolutely continuous case, this simplifies to what we had before \[ S(t) = \exp\{-A(t)\}. \] This also suggests the plug-in estimator \[ \hat{S}(t) = \prod_u^t\{1 - d\hat{A}(u)\}. \]
From the previous slide \[ \hat{S}(t) = \prod_u^t\{1 + d\hat{A}(u)\} = \prod_{T_j \leq t} 1 - \frac{I(Y(T_j) > 0)}{Y(T_j)}. \]
Using the relation \(S(t) = exp(-A(t))\) we might consider estimating \(S(t)\) by \[ exp(-\hat{A}(t)) = \prod_{T_j \leq t} exp(- D_j / Y(T_j)). \] For small values of \(D_j / Y(T_j)\), the right hand side is approximately \[ \prod_{T_j \leq t}(1 - D_j / Y(T_j)), \] which is exactly the Kaplan-Meier estimator.
The product integral representation formalizes this approximation and provides a unified framework for working with these quantities.
\[ \hat{A}(t) = \sum_{T_j \leq t} \frac{D_j}{Y(T_j)}. \] This was motivated by using the life-table approach (counting events in time intervals) and then taking the limit so that only one event can occur in each interval. Following page 87 in the book (sort of), the Doob-Meyer decomposition says \[ N(t) = \Lambda(t) + M(t) = \int_0^t\lambda(u)\, du + M(t), \] so we can write \[ dN(t) = \alpha(t)Y(t)\,dt + dM(t). \] Let \(J(t) = I\{Y(t) > 0\}\). Then \[ \frac{J(t)}{Y(t)} dN(t) = J(t) \alpha(t) dt + \frac{J(t)}{Y(t)} dM(t) \] and integrating we have \[ \int_0^t\frac{J(u)}{Y(u)} dN(u) = \int_0^t J(u) \alpha(u) du + \int_0^t\frac{J(t)}{Y(t)} dM(t). \] What have we done here? Explain the terms of the last equation.
\(\hat{A}(t) = \int_0^t\frac{J(u)}{Y(u)} dN(u)\) and \(A^*(t) = \int_0^t J(u) \alpha(u) du\) hence \[ \hat{A}(t) - A^*(t) = \int_0^t\frac{J(t)}{Y(t)} dM(t). \]
\[ [\hat{A} - A^*](t) = \int_0^t\frac{J(u)}{Y(u)^2}dN(u) \] because the integrand is predictable, \[ [\int H\, dM] = \int H^2 \, d[M] \] and \(M\) is mean 0. Hence, \[ \hat\sigma^2(t) = \int_0^t\frac{J(u)}{Y(u)^2}dN(u), \] is an unbiased estimator of the variance of the NA estimator.
Likewise,
\[ \langle \hat{A} - A^*\rangle(t) = \int_0^t\frac{J(u)}{Y(u)} \alpha(u) \, du. \]
Consider the martingale \[ \sqrt{n}(\hat{A}(t) - A^*(t)) = \int_0^t \sqrt{n}\frac{J(u)}{Y(u)} \, dM(u). \] We need to verify the conditions of the martingale central limit theorem. In this context, they take the form
where \(H^{(n)}(t) = \sqrt{n}J(t) / Y(t)\) and \(\lambda(t) = Y(t) \alpha(t)\). Starting with 1, we have \[ \frac{J(t)\alpha(t)}{Y(t) / n} \] and 2, \[ \frac{1}{\sqrt{n}}\frac{J(t)}{Y(t) / n}. \] We need the first one to converge to something fixed and positive, and the second one to converge to 0. What do?
If we assume \(Y(t) / n \rightarrow_p y(t)\), for some fixed \(y(t)>0\) for all \(t\), or maybe \[ \sup_{t}|Y(t)/n - y(t)| \rightarrow_p 0, \] then everything works out nicely. Specifically, we have \[ \sqrt{n}(\hat{A}(t) - A^*(t)) \Rightarrow W(\sigma(t)) \] where \(W\) is a Weiner process/Brownian motion and \[ \sigma(t) = \int_0^t \frac{\alpha(u)}{y(u)} \, du. \] An estimator of the variance process is \[ \hat{\sigma}(t) = \int_0^t\frac{J(u)}{Y(u)^2}\, dN(u). \]
But, we wanted to estimate \(A(t)\) not \(A^*(t)\). The yellow book says “asymptotically there is no difference”. Can someone explain why?
\[ E(A(t) - A^*(t)) = E\, \int_0^t (I(Y(u) > 0) - I(Y(u) \geq 0) )\alpha(u)\, du \] \[ = \int_0^tP(Y(u) = 0) \alpha(u) \,du \leq P(Y(t) = 0) \alpha(t), \] since \(P(Y(u) = 0)\) is increasing in \(u\). Noting that \(Y(t)\) is the aggregated process \(\sum_{i=1}^nY_i(t)\), this can be written \[ (1 - \pi(t))^n \alpha(t) \] where \(\pi(t) = \pi_i(t) = P(Y_i(t) = 1)\). This converges to 0 rapidly as \(n\) increases. The result follows since we have already established unbiasedness of \(\hat{A}\) for \(A^*\).
See page 92 of Fleming and Harrington.
You could also argue that since \(Y(t) / n \rightarrow_p y(t)\), for some fixed \(y(t)>0\) for all \(t\), then \(J(t) \rightarrow_p 1\).
What about uniform consistency? I.e., can we prove \[ \sup_t|\hat{A}(t) - A(t)| \rightarrow_p 0? \] The yellow book completely glosses over this, but it is covered on page 112 of Fleming and Harrington. The proof relies on the corollary to Lenglart’s inequality, which says,
\[ P\{\sup_{t\leq T}\{\int_0^tH(u)dM(u)\}^2 \geq \epsilon \} \leq \frac{\eta}{\epsilon} + P\{\int_0^TH^2(u)\,d\langle M \rangle(u) \geq \eta\}, \]
for constants \(\epsilon, \eta\), a martingale \(M\), predictable process \(H\) and stopping time \(T\).
To complete the proof:
The yellow book introduces the product integral (note that I am using slighty different notation because the product integral font is not available in MathJax but it is in the prodint package in LateX): \[
\prod_u^t\{1 + dB(u)\} = \lim_{M \rightarrow 0}\prod_{k=1}^K\{1 + (B(t_k) - B(t_{k-1}))\}
\] where \(M = \max_k|t_k - t_{k-1}|\) is the length of the longest subinterval. They then express \[
S(t) = \prod_u^t\{1 - dA(u)\}.
\] They describe how this relates the cumulative hazard to the survival in a general framework. So why not plug in our NA estimator and see what we get: \[
\prod_u^t\{1 - d\hat{A}(u)\} = \prod_{T_j \leq t}\{1 - \frac{dN(T_j)}{Y(T_j)}\},
\] which is the Kaplan-Meier estimator again.
Starting with the relation \[ A(t) = \int_0^t \frac{dF(u)}{S(u-)}, \] consider \[ dA(s) = \frac{dF(s)}{1 - F(s-)} \Leftrightarrow S(s-)dA(s) = dF(s) \] and hence \[ F(t) = \int_0^td S(u-)\,dA(u) \Leftrightarrow S(t) = 1 - \int_0^td S(u-)\,dA(u). \] This recursive relationship suggests that if we could plug in the NA estimate \(\hat{A}\) and solve for \(S\) we would have a nice estimate of \(S\). This nice estimator turns out to be the Kaplan-Meier. See page 94 of Fleming and Harrington.
There are many ways to arrive at the KM estimator \(\hat{S}(t)\), but the representation we are going to analyze its properties with our martingales is based on the ratio \[ \frac{\hat{S}(t)}{S^*(t)} - 1 = -\int_0^t\frac{\hat{S}(u-)}{S^*(u)}\, d(\hat{A} - A^*)(u), \] where \(S^*(t) = exp\{-A^*(t)\}\) which is again “close enough” to \(S\) asymptotically.
This can be proven using the relations on the previous slide, an integration by parts formula, and a formula for the derivative of an inverse. See page 97 of Fleming and Harrington.
Replacing \(S^*\)/\(A^*\) with \(S\)/\(A\), we have \[ \frac{\hat{S}(t)}{S(t)} = 1 - \int_0^t\frac{\hat{S}(u-)}{S(u)}\, d(\hat{A} - A)(u) \] and since \(\hat{A} - A = M\) is a mean 0 martingale, if the integrand is predictable, then the integral is also a martingale. Then this equation suggests the KM estimator is pretty good (but this is a weird way to say it).
Starting from the previous result, a corollary is \[ \hat{S}(t) - S(t) = -S(t)\int_0^t\frac{\hat{S}(u-)}{S(u)}\frac{J(u)}{Y(u)}\, dM(u) + B(t), \] where for \(T = \inf(s: Y(s) = 0)\), \[ B(t) = I(T < t) \frac{\hat{S}(T)\{S(T) - S(t)\}}{S(T)}. \] \(B\) is the bias term. Its expectation is nonzero, but converges very quickly to 0. See page 99 of Fleming and Harrington.
Looking at \(E(B^2(t))\) and exploiting the relationship between \(A\) and \(S\) allows you to derive an approximation of the variance of \(\hat{S}\) and therefore an estimator of the variance. Details omitted here but they are in Fleming and Harrington starting on bottom of page 103.
This decomposition is also used to derive the asymptotic results (so make a note).
Result: Under the random censoring model, with \(\hat{F}(t) = 1 - \hat{S}(t)\), and assuming \[ \sup_{t}|Y(t)/n - y(t)| \rightarrow_p 0, \] then \[ \sqrt{n}(\hat{F}(\cdot) - F(\cdot)) \Rightarrow S(\cdot) \, W(\sigma(\cdot)), \] in \(D[0,\tau]\), where \(W\) is a Brownian motion, and \(\sigma(t) = \int_0^t \frac{\alpha(u)}{y(u)} \, du.\)
The covariance of the limiting process evaluated at \(0 \leq s \leq t\) is \[ S(t)S(s) \int_0^s \frac{\alpha(u)}{y(u)} \,du. \]
Details on page 236 of Fleming and Harrington.
Uniform consistency can be established using the bias decomposition and another application of Lenglart’s inequality. Using that, we can state additional results: