Assume that the failure time of interest
T follows the semiparametric proportional odds model:
$$\begin{aligned} \log \left\{ \frac{1-S(t\mid Z)}{S(t\mid Z)}\right\} =\alpha (t)+Z^\top \beta , \end{aligned}$$
(1)
and the observed failure time is subject to a right truncation time variable
R. The observed data is
\((T_i,R_i),~i=1,\ldots ,n\), where
\(T_i\le R_i\). Let
\(\tau \) be the study duration, which is greater than
\(\max \{T_1,T_2,\ldots ,T_n\}\). An (observed) reverse-time sample,
\((T_i^{*},R_i^{*}),~i=1,\ldots ,n\) can be constructed, where
\(T^{*}=\tau -T,~R^{*}=\tau -R\), so that
\(T^{*}\) is left truncated by the variable
\(R^{*}\). Denote
\(({\tilde{T}}^{*},{\tilde{R}}^{*})\) as the reverse-time sample (potentially truncated). Then the hazard function of
\({\tilde{T}}^{*}\) is a quantity originated in
\(\tau \) and counts backward in time. The reverse hazard and cumulative reverse hazard function of backward recurrence time is defined as
$$\begin{aligned}{} & {} \lambda ^B(t\mid Z)=\lim _{\Delta t\rightarrow 0}\frac{\text {Pr}\{{\tilde{T}}^{*}\in (t-\Delta t,t]\mid {\tilde{T}}^{*}\le t, Z\}}{\Delta t}=\frac{f(t\mid Z)}{F(t\mid Z)},\\{} & {} \Lambda ^B(t\mid Z)=\int _t^{\tau }\lambda ^B(s\mid Z)ds. \end{aligned}$$
We would like to mention that a similar definition of the reverse hazard can also be found in Kalbfleisch and Lawless (
1989) and Jiang (
2011). Denote
\(v(t)=\exp (\alpha (t))\), and
\(\lambda (t)=f(t)/S(t)\) as the forward-time hazard, then
$$\begin{aligned} \log \lambda (t\mid Z)-\log \lambda ^{B}(t\mid Z)= & {} \alpha (t)+Z^\top \beta , \lambda ^B(t\mid Z)\\ {}= & {} \frac{1}{\{1+v(t)\exp (Z^\top \beta )\}v(t)}\frac{dv(t)}{dt}. \end{aligned}$$
Consider the counting process
$$\begin{aligned} N_i(t)=I(t\le T_i\le R_i), Y_i(t)= I(T_i\le t\le R_i), \end{aligned}$$
and denote
$$\begin{aligned} M_i(t,\beta )=N_i(t)-\int _t^{\tau }Y_i(s)\frac{1}{\{\exp (Z_i^\top \beta )v(s)+1\}v(s)}dv(s). \end{aligned}$$
Then
\(M_i(t,\beta )\) is a martingale with respect to the self-exciting (canonical) filtration (Keiding and Gill
1990; Stralkowska-Kominiak and Stute
2009) and
$$\begin{aligned} M_i(dt,\beta )=dN_i(t)+Y_i(t)\frac{1}{\{\exp (Z_i^\top \beta )v(t)+1\}v(t)}dv(t). \end{aligned}$$
(2)
Multiply both sides of (
2) by
\(\{\exp (Z_i^\top \beta )v(t)+1\}\) and summing over
n observations,
$$\begin{aligned}{} & {} \sum _{i=1}^n\{\exp (Z_i^\top \beta )v(t)+1\}dN_i(t)+\sum _{i=1}^n Y_i(t)\frac{dv(t)}{v(t)}\nonumber \\{} & {} \qquad =\sum _{i=1}^n\{\exp (Z_i^\top \beta )v(t)+1\}M_i(dt,\beta ). \end{aligned}$$
(3)
Divide both left-hand side and right-hand side by
\(\sum _{i=1}^nY_i(t)\), we obtain:
$$\begin{aligned} \frac{\sum _{i=1}^n\{\exp (Z_i^\top \beta )v(t)+1\}d N_i(t)}{\sum _{i=1}^n Y_i(t)}+\frac{d v(t)}{v(t)}=\frac{\sum _{i=1}^n\{\exp (Z_i^\top \beta )v(t)+1\}M_i(d t,\beta )}{\sum _{i=1}^n Y_i(t)}. \end{aligned}$$
which is equivalent to:
$$\begin{aligned}{} & {} v(t)\frac{\sum _{i=1}^n\exp (Z_i^\top \beta )d N_i(t)}{\sum _{i=1}^n Y_i(t)}+\frac{\sum _{i=1}^n d N_i(t)}{\sum _{i=1}^n Y_i(t)}+\frac{d v(t)}{v(t)}\nonumber \\{} & {} \qquad = \sum _{i=1}^n\frac{\exp (Z_i^\top \beta )v(t)+1}{\sum _{i=1}{n}Y_i(t)}M_i(dt,\beta ). \end{aligned}$$
(4)
Denote the left-hand side of (
4) as:
$$\begin{aligned} U(\beta ,dt)=\frac{dv(t)}{v(t)}+p_n(t)dt-q_n(t,\beta )v(t)dt, \end{aligned}$$
where
$$\begin{aligned} p_n(t)dt=\frac{\sum _{i=1}^ndN_i(t)}{\sum _{j=1}^nY_j(t)}, q_n(t,\beta )dt=-\frac{\sum _{i=1}^n\exp (Z_i^\top \beta )dN_i(t)}{\sum _{j=1}^nY_j(t)}. \end{aligned}$$
From standard counting process arguments (Anderson and Gill, 1982;Aalen10), we know that the stochastic integral with respect to the counting process martingale
\(M_i(dt,\beta )\) is also a martingale, motivate by the following equation
$$\begin{aligned} E\left[ \frac{1}{n}U(\beta ,dt)\right] = E\left[ \frac{1}{n}\sum _{i=1}^n\frac{\exp (Z_i^\top \beta )v(t)+1}{\sum _{i=1}{n}Y_i(t)}M_i(dt,\beta ) \right] . \end{aligned}$$
We construct the following estimating equation
$$\begin{aligned} \frac{1}{n}U(\beta ,dt)=0. \end{aligned}$$
(5)
Only
v(
t) is unknown in (
5), let the estimate of
v(
t) be
\({\hat{v}}_n(t,\beta )\). Denote
$$\begin{aligned} P_n(t)=\exp \left\{ \int _t^{\tau }\frac{\sum _{i=1}^ndN_i(s)}{\sum _{j=1}^nY_j(s)} \right\} , Q_n(t,\beta )=\int _t^{\tau }\frac{\sum _{i=1}^n\exp (Z_i^\top \beta )dN_i(s)}{\sum _{j=1}^nY_j(s)}, \end{aligned}$$
then
$$\begin{aligned} {\hat{v}}_n(t,\beta ) =\frac{P_n(t)}{\int _t^{\tau }P_n(s)Q_n(ds,\beta )}. \end{aligned}$$
(6)
Multiply (
2) by
\(Z_i\{\exp (Z_i^\top \beta )v(t)+1\}/n\) and summing over
n observations, we obtain
$$\begin{aligned}{} & {} \frac{1}{n}\sum _{i=1}^nZ_i\left[ \{\exp (Z_i^\top \beta )v(t)+1\}dN_i(t)+ Y_i(t)\frac{dv(t)}{v(t)}\right] \nonumber \\{} & {} \qquad =\frac{1}{n}\sum _{i=1}^nZ_i\{\exp (Z_i^\top \beta )v(t)+1\}M_i(dt,\beta ). \end{aligned}$$
(7)
By virtue of the same idea of (
5), take integration on both sides of (
7), we can also construct another equation:
$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n\int _0^{\tau }Z_i\left[ \{\exp (Z_i^\top \beta )v(t)+1\}dN_i(t)+ Y_i(t)\frac{dv(t)}{v(t)}\right] =0 \end{aligned}$$
(8)
Substituting (
6) into (
8), we can obtain the estimate of
\(\beta \) by solving the following equation:
$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n\int _0^{\tau }Z_i\left[ \left\{ \exp (Z_i^\top \beta ){\hat{v}}_n(t,\beta )+1 \right\} dN_i(t) +Y_i(t)\frac{{\hat{v}}_n(dt,\beta )}{{\hat{v}}_n(t,\beta )} \right] =0. \end{aligned}$$
Moreover, since
$$\begin{aligned} \frac{{\hat{v}}_n(dt,\beta )}{{\hat{v}}_n(t,\beta )}=-\frac{\sum _{k=1}^ndN_k(t)}{\sum _{l=1}^nY_l(t)} -\frac{\sum _{k=1}^n\exp (Z_k^\top \beta )dN_k(t)}{\sum _{l=1}^nY_l(t)}{\hat{v}}_n(t,\beta ), \end{aligned}$$
then
$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n\int _0^{\tau }\left\{ Z_i-{\bar{Z}}(t)\right\} \left\{ \exp (Z_i^\top \beta ){\hat{v}}_n(t,\beta )+1\right\} dN_i(t)=0, \end{aligned}$$
where
$$\begin{aligned} {\bar{Z}}(t)=\frac{\sum _{i=1}^nZ_iY_i(t)}{\sum _{j=1}^nY_j(t)}. \end{aligned}$$
Finally, let
$$\begin{aligned} S_n(\beta )=\frac{1}{n}\sum _{i=1}^n\int _0^{\tau }\left\{ Z_i-{\bar{Z}}(t)\right\} \left\{ \exp (Z_i^\top \beta ){\hat{v}}_n(t,\beta )+1\right\} dN_i(t), \end{aligned}$$
(9)
and denote the solution of
\(S_n(\beta )=0\) be
\({\hat{\beta }}_n\), we have the following theorem:
The corresponding estimating equation under model (
10) has the following form
$$\begin{aligned} S_n^{(1)}(\beta )=\sum _{i=1}^n\int _0^{\tau }\left\{ Z_i-{\bar{Z}}(t,\beta )\right\} \left\{ \exp (Z_i^\top \beta ){\hat{v}}_n(t,\beta )+1 \right\} dN_i(t), \end{aligned}$$
(11)
where
$$\begin{aligned} {\bar{Z}}(t,\beta )=\frac{\sum _{i=1}^nZ_iY_i(t)\exp (Z_i^\top \beta )}{\sum _{j=1}^nY_j(t)\exp (Z_j^\top \beta )}. \end{aligned}$$
Equation (
11) also can be used to estimate
\(\beta \), however, comparing with (
9), (
11) is more complicated and more computational intensive, while the derivative of (
9) with respect to
\(\beta \) can be easily obtained. As a result, (
9) can be easily solved by the newton raphson algorithm. In the following simulations, we will use estimating equation (
9).
In addition to the unweighted object function (
9), weighted object function can also being included to obtain a class of weighted estimators of
\(\beta _0\). This procedure is often used to minimize the sandwich estimate as well as improve the efficiency. The weighted version of object function is
$$\begin{aligned}{} & {} S_{n,W}(\beta )\nonumber \\{} & {} \quad =\frac{1}{n}\sum _{i=1}^n\int _0^{\tau }W_n(t)\left\{ Z_i-{\bar{Z}}(t)\right\} \left\{ \exp (Z_i^\top \beta ){\hat{v}}_n(t,\beta )+1\right\} dN_i(t)=0, \end{aligned}$$
(12)
here
\(W_n(t)\) is a predictable weight function with respect to the canonical filtration which converges to a non-random function
w(
t). One of the common used weight function is the Prentice-Wilcoxon type function
\(W_{n1}(t)={\hat{S}}_{LB}(t)\), where
\({\hat{S}}_{LB}(\cdot )\) is the Lynden Bell estimate of the baseline survival function for right truncated failure time data. Denote the corresponding estimate of
\(\beta \) as
\({\hat{\beta }}_{n,w}\). Then we have the following theorem:
Recently, many people considered problem of finding the optimal weight in a weighted estimating equation, including Chen and Cheng (
2005); Chen and Wang (
2000); Chen et al. (
2012), among others. To achieve this goal, we only need to find the
w(
t) such that
\(U_w(\beta _0)^{-1}V_w(\beta _0)U_w(\beta _0)^{-1}\) achieves the minimum. Since both the empirical weight function
\(W_n(t)\) and its limit
w(
t) do not rely on unknown parameter
\(\beta _0\), it is reasonable to set
\(\beta _0=0\). Another explanation for letting
\(\beta _0=0\) is that it represents the baseline distribution. Therefore, let
\(\beta _0=0\), then we have:
$$\begin{aligned} U_w(\beta _0){} & {} =\mathop {\lim }\limits _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^n\int _0^{\tau } W_n(t)\left\{ Z_i-{\bar{Z}}(t)\right\} Z_{i}\exp (Z_i^\top \beta _0)v(t)dN_i(t)\nonumber \\{} & {} =\mathop {\lim }\limits _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^n\int _0^{\tau } W_n(t)\left\{ Z_i-{\bar{Z}}(t)\right\} ^{\otimes 2} \exp (Z_i^\top \beta _0)v(t)dN_i(t)\nonumber \\{} & {} =\mathop {\lim }\limits _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^n\int _0^{\tau } W_n(t)\left\{ Z_i-{\bar{Z}}(t)\right\} ^{\otimes 2}Y_i(t) \frac{\exp (Z_i^\top \beta _0)v^{\prime }(t)}{\exp (Z_i^\top \beta _0)v(t)+1} dt\nonumber \\{} & {} =\mathop {\lim }\limits _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^n\int _0^{\tau } W_n(t)\left\{ Z_i-{\bar{Z}}(t)\right\} ^{\otimes 2}Y_i(t) \frac{v^{\prime }(t)}{v(t)+1} dt. \end{aligned}$$
(13)
$$\begin{aligned} V_w(\beta _0){} & {} =\mathop {\lim }\limits _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^n\int _0^{\tau } W_n(t)^2\left\{ Z_i-{\bar{Z}}(t)\right\} ^{\otimes 2}\nonumber \\{} & {} \quad \times \left\{ \exp (Z_i^\top \beta _0)v(t)+1\right\} ^2Y_i(t)\frac{1}{\left\{ \exp (Z_i^\top \beta _0)v(t)+1\right\} v(t)}v^{\prime }(t)dt\nonumber \\{} & {} =\mathop {\lim }\limits _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^n\int _0^{\tau } W_n(t)^2\left\{ Z_i-{\bar{Z}}(t)\right\} ^{\otimes 2}Y_i(t)\left\{ \exp (Z_i^\top \beta _0)v(t)+1\right\} \frac{v^{\prime }(t)}{v(t)}dt\nonumber \\{} & {} =\mathop {\lim }\limits _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^n\int _0^{\tau } W_n(t)^2\left\{ Z_i-{\bar{Z}}(t)\right\} ^{\otimes 2}Y_i(t)\left\{ v(t)+1\right\} \frac{v^{\prime }(t)}{v(t)}dt \end{aligned}$$
(14)
Apply the Cauchy-Schwarz inequality to
\(U_w(\beta _0)^{-1}V_w(\beta _0)U_w(\beta _0)^{-1}\) and let
\(\beta _0=0\), then it follows that the optimal weight is proportional to
$$\begin{aligned} w(t)=\frac{v(t)}{(v(t)+1)^2}=S(t)\left\{ 1-S(t)\right\} , \end{aligned}$$
(15)
which minimize the variance of
\({\hat{\beta }}_n\). Since when (
15) holds, we have
$$\begin{aligned}{} & {} U_w(\beta _0)=\mathop {\lim }\limits _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^n\int _0^{\tau } \left\{ Z_i-{\bar{Z}}(t)\right\} ^{\otimes 2}Y_i(t) \frac{v(t)v^{\prime }(t)}{(v(t)+1)^3} dt,\\{} & {} V_w(\beta _0)=\mathop {\lim }\limits _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^n\int _0^{\tau } \left\{ Z_i-{\bar{Z}}(t)\right\} ^{\otimes 2}Y_i(t)\frac{v(t)v^{\prime }(t)}{(v(t)+1)^3}dt, \end{aligned}$$
which means when
\(\beta _0=0\), given
\(w(t)=S(t)\{1-S(t)\}\), we have
\(U_w(\beta _0)^{-1}V_w(\beta _0)U_w(\beta _0)^{-1}\) achieves the minimum value
\(U_w(\beta _0)^{-1}\) (or equivalently
\(V_w(\beta _0)^{-1}\)).