Target estimand

Let \(T^{(a)}\) denote the survival time of a generic patient in group \(a\), where \(a=1\) and 0 indicate the treatment and control, respectively. Write \(S^{(a)}(t)={\rm pr}(T^{(a)}>t)\) and \(\Delta(t)=S^{(1)}(t)-S^{(0)}(t)\). Given a pre-specified time horizon \(\tau\), let \(t_\epsilon\) denote the landmark time point at which survival probability of the treated exceeds that of the untreated by some \(\epsilon>0\) (e.g.,\(\epsilon=0.05\)), that is, \[t_\epsilon=\inf\left\{t\in[0,\tau):\Delta(t)\geq \epsilon\right\},\] where we define \(\inf\emptyset=\tau\) to allow for the absence of such a landmark time in \([0,\tau]\).

If treatment effect is delayed, we will have that \(t_\epsilon>>0\). It is then appropriate to quantify the treatment effect by \[\mu_\epsilon(\tau)=\int_{t_\epsilon}^\tau\Delta(t){\rm d}t,\] which is also the average time the treated outlive the untreated in \([t_\epsilon,\tau]\).

Estimation and inference

Naive approach with a boundary problem

With censored data, let \(\hat\mu_\epsilon(\tau)\) denote the empirical estimator of \(\mu_\epsilon(\tau)\), i.e., \[\hat\mu_\epsilon(\tau)=\int_{\hat t_\epsilon}^\tau\hat\Delta(t){\rm d}t,\] where \(\hat t_\epsilon=\inf\left\{t\in[0,\tau):\hat\Delta(t)\geq \epsilon\right\}\), \(\hat\Delta(t)=\hat S^{(1)}(t)-\hat S^{(0)}(t)\), and \(\hat S^{(a)}(t)\) is the Kaplan–Meier estimator of \(S^{(a)}(t)\) \((a=1,0)\).

Under conditions

  • \(t_\epsilon<\tau\) (i.e., the landmark time exists);
  • \(f^{(0)}(t_\epsilon)>f^{(1)}(t_\epsilon)\), where \(f^{(a)}(t)=-{\rm d}S^{(a)}(t)/{\rm d}t\) (i.e., the gap between the survival functions has a tendency to widen around \(t_\epsilon\)).

We can show that \(\hat\mu_\epsilon(\tau)\) is consistent and asymptotically linear (normal), whose influence function can be derived. On the other hand, if \(\sup_{t\in[0,\tau]}\Delta(t)\leq \epsilon\) so that \(t_\epsilon=\tau\), the estimator lacks a regular distribution as its probability mass is concentrated on \(0\). In particular, we cannot use it to make inference under \(H_0: S^{(1)}(t)\equiv S^{(0)}(t)\).

Two-stage inference

To avoid the boundary problem, consider a two-stage procedure where we first test on the difference \(\Delta(t)\) and then proceed to the inference of \(\mu_\epsilon(\tau)\) only if we are confident that \(t_\epsilon<\tau\). Specifically, let \(\hat\Delta_L(t)\) denote the lower bound of a \(100(1-\alpha)\%\) \((0<\alpha<1)\) confidence band for \(\Delta(t)\) over \([0,\tau]\), so that \[\begin{equation}\tag{1} {\rm pr}\left[\sup_{t\in[0,\tau]}\left\{\hat\Delta_L(t)-\Delta(t)\right\}\leq 0\right]\geq 1-\alpha. \end{equation}\]

  • Stage 1: Accept \(H_0:\Delta(t)\equiv 0\) if \(\sup_{t\in[0,\tau)}\hat\Delta_L(t)\leq\epsilon\); Otherwise, reject \(H_0\) and proceed to stage 2.
  • Stage 2: Now that \(\hat t_\epsilon<\tau\) since \(\sup_{t\in[0,\tau)}\hat\Delta(t)\geq \sup_{t\in[0,\tau)}\hat\Delta_L(t)>\epsilon\), we can construct a stable estimate \(\hat\mu_\epsilon(\tau)\) and make inference accordingly.

Remarks

  1. The two-stage procedure yields a valid level-\(\alpha\) test for \(H_0\), as \[{\rm pr}\left\{\sup_{t\in[0,\tau)}\hat\Delta_L(t)>\epsilon\mid H_0\right\} \leq {\rm pr}\left\{\sup_{t\in[0,\tau]}\hat\Delta_L(t)> 0 \mid H_0\right\}\leq\alpha\] by (1), though the test is likely conservative due to the nonzero threshold \(\epsilon\).
  2. Because inference on \(\mu_\epsilon(\tau)\) is made only if \(\sup_{t\in[0,\tau)}\hat\Delta_L(t)>\epsilon\), it should be conditioned as such. That is, if \(\hat\mu_{\epsilon,L}(\tau)\) and \(\hat\mu_{\epsilon,U}(\tau)\) denote the 95% confidence limits of \(\mu_\epsilon(\tau)\), we must ensure that \[\begin{equation}\tag{2} {\rm pr}\left\{\hat\mu_{\epsilon,L}(\tau)\leq \mu_\epsilon(\tau)\leq \hat\mu_{\epsilon,U}(\tau) \mid \sup_{t\in[0,\tau)}\hat\Delta_L(t)>\epsilon\right\}\geq 95\%. \end{equation}\] As a result, to construct \(\hat\mu_{\epsilon,L}(\tau)\) and \(\hat\mu_{\epsilon,U}(\tau)\), we need to find the joint distribution of \(\hat\mu_{\epsilon}(\tau)\) and \(\sup_{t\in[0,\tau)}\hat\Delta_L(t)\).