A Joint Modeling Approach for Longitudinal Data with Informative Observation Times and a Terminal Event

Abstract

In this article, we propose a new joint modeling approach for the analysis of longitudinal data with informative observation times and a dependent terminal event. We specify a semiparametric mixed effects model for the longitudinal process, a proportional rate frailty model for the observation process, and a proportional hazards frailty model for the terminal event. The association among the three related processes is modeled via two latent variables. Estimating equation approaches are developed for parameter estimation, and the asymptotic properties of the proposed estimators are established. The finite sample performance of the proposed estimators is examined through simulation studies, and an application to a medical cost study of chronic heart failure patients is illustrated.

Appendices

Appendix : Proof of Asymptotic Results

In order to study the asymptotic properties of the proposed estimators, we need the following regularity conditions:

• (C1) $\{Y_i(\cdot ), N_i^R(\cdot ), N_i^D(\cdot ), T_i, X_i(\cdot ),\ i=1,\ldots , n\}$ are independent and identically distributed.

• (C2) $E\{N_i^R(\tau )\}$ is bounded, and $\Pr (T_i \ge \tau )>0$.

• (C3) $X_i(t)$ is almost surely of bounded variation on $[0, \tau ]$.

• (C4) There exist a compact set $\mathcal {B}$ of $\xi _0$ such that for $\xi \in \mathcal {B}$, $\Gamma (\xi )$ is nonsingular, where $\Gamma (\xi )$ is the limit of $-\partial \tilde{U}(\xi )/\partial \xi ^T$ with $\tilde{U}(\xi )$ defined in (A.3).

For notational simplicity, henceforth we omit superscript t in the covariate $X_i(t).$ Let $\tilde{\Lambda }_0^D(t)$ and $\tilde{\Lambda }_0^R(t)$ denote the solutions to the following estimating equations

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n\Big [\psi _i(t; \alpha _0, \theta _0, \Lambda _0^D)\mathrm{{d}}N_i^D(t)-\Delta _i(t)\exp \{\alpha _0^TX_i\}\mathrm{{d}}\Lambda _0^D(t)\Big ]=0, \end{aligned}$$

and

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n\Big [\psi _i(t; \alpha _0, \theta _0, \Lambda _0^D)\mathrm{{d}}N_i^R(t)-\Delta _i(t)\exp \{\beta _0^TX_i\}\mathrm{{d}}\Lambda _0^R(t)\Big ]=0. \end{aligned}$$

Proof of Theorem 1

Define

$$\begin{aligned} \tilde{\Psi }_1(t)=\frac{\theta _0}{n}\sum _{i=1}^n\int _0^t \exp \{\alpha _0^TX_i\}\frac{\mathrm{{d}}N_i^D(u)}{S_0(u;\alpha _0)}, \end{aligned}$$

where $S_0(t; \alpha )=\frac{1}{n}\sum _{i=1}^n\Delta _i(t)\exp \{\alpha ^TX_i\}.$ By the definition of $\tilde{\Lambda }_0^D(t),$ it can be checked that

$$\begin{aligned} \tilde{\Lambda }_0^D(t)-\Lambda _0^D(t) =\sum _{i=1}^n\int _0^t [\tilde{\Lambda }_0^D(u)-\Lambda _0^D(u)]d \tilde{\Psi }_1(u) +\frac{1}{n} \sum _{i=1}^n \int _0^t \frac{\mathrm{{d}} {M}_i^D(u)}{S_0(u;\alpha _0)}, \end{aligned}$$

which is a linear Volterra integral equation, and the solution is

$$\begin{aligned} \tilde{\Lambda }_0^D(t)-\Lambda _0^D(t)=\frac{1}{\tilde{P}(t)} \int _0^t \tilde{P}(u-) \frac{\sum _{i=1}^n \mathrm{{d}}{M}_i^D(u)}{nS_0(u;\alpha _0)}, \end{aligned}$$

where $\tilde{P}(t)=\prod _{s\le t} \{1- d \tilde{\Psi }_1(s)\}$ is the product-integral of $\tilde{\Psi }_1(s)$ over [0, t]. Using the uniform convergence of $\tilde{\Lambda }_0^D(t),$ the uniform strong law of large numbers [12] and Lemma A.1 of Lin and Ying [8], we get that uniformly in $t \in [0, \tau ],$

$$\begin{aligned} \tilde{\Lambda }_0^D(t)-\Lambda _0^D(t)=\frac{1}{n}\sum _{i=1}^n \phi _{1i}(t)+o_p\left( n^{-1/2}\right) , \end{aligned}$$

(A.1)

where

$$\begin{aligned} \phi _{1i}(t)=\frac{1}{P(t)}\int _0^t P(u-)\frac{\mathrm{{d}} {M}_i^D(u)}{s_0(u;\alpha _0)}, \end{aligned}$$

and P(t) and $s_0(t;\alpha )$ are the limits of $\tilde{P}(t)$ and $S_0(t;\alpha )$, respectively. Let

$$\begin{aligned} \tilde{\Psi }_2(t)=\frac{\theta _0}{n}\sum _{i=1}^n\int _0^t \exp \{\beta _0^TX_i\}\frac{\mathrm{{d}}N_i^R(u)}{S_0(u;\beta _0)}, \end{aligned}$$

and $\Psi _2(t)$ be the limit of $\tilde{\Psi }_2(t).$ Note that

$$\begin{aligned} \tilde{\Lambda }_0^R(t)=\frac{1}{n}\sum _{i=1}^n \int _0^t \frac{\psi _i(u; \tilde{\Lambda }_0^D)\mathrm{{d}}N_i^R(u)}{S_0(u;\beta _0)}, \end{aligned}$$

where $\psi _i(t;\Lambda _0^D)=\psi _i(t;\alpha _0, \theta _0, \Lambda _0^D).$ It then follows from (A.1) that

$$\begin{aligned} \tilde{\Lambda }_0^R(t)-\Lambda _0^R(t)&=\int _0^t [\tilde{\Lambda }_0^D(u)-\Lambda _0^D(u)]\mathrm{{d}} \tilde{\Psi }_2(u) +\frac{1}{n}\sum _{i=1}^n \int _0^t \frac{\mathrm{{d}} {M}_i^R(u)}{S_0(u;\beta _0)}\nonumber \\&=\frac{1}{n}\sum _{i=1}^n \phi _{2i}(t)+o_p(n^{-1/2}), \end{aligned}$$

(A.2)

where

$$\begin{aligned} \phi _{2i}(t)= \int _0^t \phi _{1i}(u) d \Psi _2(u) + \int _0^t \frac{\mathrm{{d}}{M}_i^R(u)}{s_0(u)}. \end{aligned}$$

Let $\omega _{2i}(t; \xi )$ be defined as $\omega _{2i}(t)$ with $\psi _i(t)$ replaced by $\psi _i(t;\alpha , \theta , \tilde{\Lambda }_0^D),$ and

$$\begin{aligned} \tilde{U}(\xi ) = \big (\tilde{U}_1(\xi )^T, \tilde{U}_2(\xi )^T, \tilde{U}_3(\xi ), \tilde{U}_4(\xi )^T\big )^T, \end{aligned}$$

(A.3)

where

$$\begin{aligned} \tilde{U}_1(\xi )&=\sum _{i=1}^n \int _0^\tau \left\{ X_i-\bar{X}(t; \alpha )\right\} \psi _i(t; \alpha , \theta , \tilde{\Lambda }_0^D)\mathrm{{d}}N_i^D(t),\\ \tilde{U}_2(\xi )&=\sum _{i=1}^n \int _0^\tau \left\{ X_i-\bar{X}(t; \beta )\right\} \psi _i(t;\alpha , \theta , \tilde{\Lambda }_0^D)\mathrm{{d}}N_i^R(t),\\ \tilde{U}_3(\xi )&=\sum _{i=1}^n\int _0^\tau \big \{N_i^R(t)-(\theta +1)Q(t)\omega _{2i}(t; \xi )\big \}\mathrm{{d}}N_i^D(t) \end{aligned}$$

and

$$\begin{aligned} \tilde{U}_4(\xi )&=\sum _{i=1}^n\int _0^\tau \begin{pmatrix}{\begin{matrix}X_i-\bar{X}(t;\beta )\\ \\ B_i(t; \tilde{\Lambda }_0^D)-{\tilde{B}}(t; \xi ) \end{matrix}}\end{pmatrix} \Big [Y_i(t)\psi _i(t; \alpha , \theta , \tilde{\Lambda }_0^D)\mathrm{{d}}N_i^R(t)\\&\quad -\big \{\gamma ^TX_i+\eta ^T B_i(t; \tilde{\Lambda }_0^D)\big \} \exp \{\beta ^TX_i\}\mathrm{{d}}\tilde{\Lambda }_0^R(t)\Big ], \end{aligned}$$

with $B_i(t; \tilde{\Lambda }_0^D)=B_i(t; \alpha , \theta , \tilde{\Lambda }_0^D),$ and

$$\begin{aligned} \tilde{B}(t; \xi )=\frac{\sum _{j=1}^n\Delta _j(t) B_j(t; \tilde{\Lambda }_0^D)\exp \{\beta ^TX_j\}}{\sum _{j=1}^n\Delta _j(t)\exp \{\beta ^TX_j\}}. \end{aligned}$$

Note that

$$\begin{aligned} \tilde{U}_1(\xi _0)&=\sum _{i=1}^n \int _0^\tau \left\{ X_i-\bar{X}(t; \alpha _0)\right\} \psi _i(t;\tilde{\Lambda }_0^D)\mathrm{{d}}N_i^R(t)\\&=\theta _0\sum _{i=1}^n \int _0^\tau \Delta _i(t)\exp \{\alpha _0^TX_i\}\big \{X_i-\bar{X}(t; \alpha _0)\big \} \big \{\tilde{\Lambda }_0^D(t)-\Lambda _0^D(t)\big \}\mathrm{{d}}N_i^D(t)\\&\quad +\sum _{i=1}^n \int _0^\tau \left\{ X_i-\bar{X}(t; \alpha _0)\right\} \mathrm{{d}}M_i^D(t), \end{aligned}$$

where $\psi _i(t; \tilde{\Lambda }_0^D)=\psi _i(t;\alpha _0, \theta _0, \tilde{\Lambda }_0^D).$ Let

$$\begin{aligned} \tilde{H}_1(t)=\frac{1}{n}\sum _{i=1}^n \int _0^t \theta _0 \Delta _i(u)\exp \{\alpha _0^TX_i\}\big \{X_i-\bar{X}(u; \alpha _0)\big \}\mathrm{{d}}N_i^D(u), \end{aligned}$$

and $\bar{x}(t;\alpha _0)$ and $H_1(t)$ are the limits of $\bar{X}(t; \alpha _0)$ and $\tilde{H}_1(t),$ respectively. Similarly to (A.2), we have

$$\begin{aligned} \tilde{U}_1(\xi _0)=\sum _{i=1}^n \vartheta _{1i}+o_p\left( n^{1/2}\right) , \end{aligned}$$

(A.4)

where

$$\begin{aligned} \vartheta _{1i}= \int _0^\tau \left\{ X_i-\bar{x}(t;\alpha _0)\right\} \mathrm{{d}}M_i^D(t) +\int _0^\tau \phi _{1i}(t)\mathrm{{d}}H_1(t). \end{aligned}$$

Likewise,

$$\begin{aligned} \tilde{U}_2(\xi _0)=\sum _{i=1}^n \vartheta _{2i}+o_p(n^{1/2}), \end{aligned}$$

(A.5)

where

$$\begin{aligned} \vartheta _{2i}= \int _0^\tau \left\{ X_i-\bar{x}(t;\beta _0)\right\} d M_i^R(t) +\int _0^\tau \phi _{1i}(t)\mathrm{{d}}H_2(t), \end{aligned}$$

and $\bar{x}(t;\beta _0)$ and $H_2(t)$ are the limits of $\bar{X}(t;\beta _0)$ and $\tilde{H}_2(t)$ with

$$\begin{aligned} \tilde{H}_2(t)=\frac{1}{n}\sum _{i=1}^n \int _0^t \theta _0 \Delta _i(u)\exp \{\alpha _0^TX_i\}\big \{X_i-\bar{X}(u; \beta _0)\big \} \mathrm{{d}}N_i^R(u). \end{aligned}$$

Let

$$\begin{aligned} \omega _{2i}^*(t)&= \psi _i^{-1}(t; \Lambda _0^D) \int _0^t \exp \{\beta _0^TX_i\}\mathrm{{d}}\Lambda _0^R(u),\\ \tilde{\omega }_{2i}(t)&= \psi _i^{-1}(t;\tilde{\Lambda }_0^D) \int _0^t \exp \{\beta _0^TX_i\}\mathrm{{d}}\tilde{\Lambda }_0^R(u),\\ \tilde{Q}(t)&= \frac{\sum _{i=1}^n\tilde{\omega }_{2i}^{-1}(t)\Delta _i^*(t)N_i^R(t)}{\sum _{i=1}^n\Delta _i^*(t)}, \end{aligned}$$

and q(t) be the limit of $\tilde{Q}(t).$ Note that

$$\begin{aligned} \tilde{U}_3(\xi _0)&= \sum _{i=1}^n\int _0^\tau \big \{N_i^R(t)-(\theta _0+1)\omega _{2i}^*(t)q(t)\big \}\mathrm{{d}}N_i^D(t)\nonumber \\&\quad -(\theta _0+1)\sum _{i=1}^n\int _0^\tau \tilde{Q}(t)\big \{\tilde{\omega }_{2i}(t)-\omega _{2i}^*(t)\big \} \mathrm{{d}}N_i^D(t)\nonumber \\&\quad -(\theta _0+1)\sum _{i=1}^n \int _0^\tau \big \{\tilde{Q}(t)-q(t) \big \} \omega _{2i}^*(t) \mathrm{{d}}N_i^D(t). \end{aligned}$$

(A.6)

As in the proof of (A.4), the second term on the right-hand side of (A.6) equals

$$\begin{aligned}&-(\theta _0+1)\sum _{i=1}^n\int _0^\tau \Big [\tilde{Q}(t) \big \{\psi _i^{-1}(t;\tilde{\Lambda }_0^D)-\psi _i^{-1}(t;\Lambda _0^D)\big \} \int _0^t \exp \{\beta _0^T X_i\}\mathrm{{d}}\Lambda _0^R(u) \Big ]\mathrm{{d}}N_i^D(t)\nonumber \\&\qquad -(\theta _0+1)\sum _{i=1}^n\int _0^\tau \tilde{Q}(t) \psi _i^{-1}(t;\tilde{\Lambda }_0^D) \exp \{\beta _0^T X_i\}\big [\tilde{\Lambda }_0^R(t)-\Lambda _0^R(t)\big ] \mathrm{{d}}N_i^D(t)\nonumber \\&\quad =(\theta _0+1)\sum _{i=1}^n\int _0^\tau \big [\phi _{1i}(t) \mathrm{{d}}H_3(t)-\phi _{2i}(t) \mathrm{{d}}H_4(t)\big ]+o_p(n^{1/2}), \end{aligned}$$

(A.7)

where $H_3(t)$ and $H_4(t)$ are the limits of $\tilde{H}_3(t)$ and $\tilde{H}_4(t)$ with

$$\begin{aligned} \tilde{H}_3(t)=\frac{1}{n}\sum _{i=1}^n \int _0^t \theta _0 \Big [\tilde{Q}(u) \psi _i^{-1}(u;\tilde{\Lambda }_0^D) \omega _{2i}^*(u) \Big ]\mathrm{{d}}N_i^D(u) \end{aligned}$$

and

$$\begin{aligned} \tilde{H}_4(t)=\frac{1}{n}\sum _{i=1}^n \int _0^t \tilde{Q}(u) \psi _i^{-1}(u;\tilde{\Lambda }_0^D) \mathrm{{d}}N_i^D(u). \end{aligned}$$

Let $\mathrm{{d}}\Phi (t)=E\big \{\omega _{2i}^*(t) \mathrm{{d}}N_i^D(t)\big \},$

$$\begin{aligned} \tilde{H}_5(t)&=\frac{\theta _0 \sum _{i=1}^n\psi _i^{-1}(t;\tilde{\Lambda }_0^D) \tilde{\omega }_{2i}^{-1}(t) \Delta _i^*(t)N_i^R(t)}{\sum _{i=1}^n\Delta _i^*(t)}, \\ \tilde{H}_6(t)&=\frac{\sum _{i=1}^n\psi _i^{-1}(t;\tilde{\Lambda }_0^D)\tilde{\omega }_{2i}^{-1}(t) \omega _{2i}^{*-1}(t) \Delta _i^*(t)N_i^R(t)}{\sum _{i=1}^n\Delta _i^*(t)}, \end{aligned}$$

and $H_5(t)$ and $H_6(t)$ be the limits of $\tilde{H}_5(t)$ and $\tilde{H}_6(t),$ respectively. In a similar manner, the third term on the right-hand side of (A.6) is

$$\begin{aligned}&(\theta _0+1)\sum _{i=1}^n\int _0^\tau \big [\phi _{1i}(t)H_5(t)+\phi _{2i}(t)H_6(t)\big ]\mathrm{{d}}\Phi (t)\nonumber \\&\quad -(\theta _0+1)\sum _{i=1}^n\int _0^\tau \Big [ \frac{\omega _{2i}^{*-1}(t)\Delta _i^*(t) N_i^R(t)}{E\{\Delta _i^*(t)\}} -\frac{q(t)}{E\{\Delta _i^*(t)\}}\Delta _i^*(t)\Big ]\mathrm{{d}}\Phi (t) +o_p(n^{1/2}). \end{aligned}$$

(A.8)

Thus, it follows from (A.6)–(A.8) that

$$\begin{aligned} \tilde{U}_3(\xi _0)=\sum _{i=1}^n \vartheta _{3i}+o_p(n^{1/2}), \end{aligned}$$

(A.9)

where

$$\begin{aligned} \vartheta _{3i}&= \int _0^\tau \big \{N_i^R(t)-(\theta _0+1)\omega _{2i}^*(t)q(t)\big \}\mathrm{{d}}N_i^D(t)\\&\quad -(\theta _0+1)\int _0^\tau \big [\phi _{1i}(t)d H_3(t)+\phi _{2i}(t)\mathrm{{d}}H_4(t)\big ]\\&\quad +(\theta _0+1)\int _0^\tau \big [\phi _{1i}(t)H_5(t)+\phi _{2i}(t)H_6(t)\big ]\mathrm{{d}}\Phi (t)\\&\quad -(\theta _0+1)\int _0^\tau \Big [\frac{\omega _{2i}^{*-1}(t)\Delta _i^*(t) N_i^R(t)}{E\{\Delta _i^*(t)\}} -\frac{q(t)\Delta _i^*(t)}{E\{\Delta _i^*(t)\}}\Big ]\mathrm{{d}}\Phi (t). \end{aligned}$$

Let $B_i(t)= B_i(t; \alpha _0, \theta _0, \Lambda _0^D),$ and

$$\begin{aligned} \bar{B}(t)=\frac{\sum _{j=1}^n\Delta _j(t) B_j(t)\exp \{\beta _0^TX_j\}}{\sum _{j=1}^n\Delta _j(t)\exp \{\beta _0^TX_j\}}. \end{aligned}$$

Note that

$$\begin{aligned} \tilde{U}_4(\xi _0)= R_1+R_2+R_3, \end{aligned}$$

(A.10)

where

$$\begin{aligned} R_1&=\sum _{i=1}^n\int _0^\tau \begin{pmatrix}{\begin{matrix}X_i-\bar{X}(t;\beta _0)\\ \\ B_i(t)-\bar{B}(t) \end{matrix}}\end{pmatrix}\mathrm{{d}}M_i(t),\\ R_2&=\sum _{i=1}^n\int _0^\tau \begin{pmatrix}{\begin{matrix}X_i-\bar{X}(t;\beta _0)\\ \\ B_i(t;\tilde{\Lambda }_0^D)-{\tilde{B}}(t) \end{matrix}}\end{pmatrix}\{\mathrm{{d}}\tilde{M}_i(t)-\mathrm{{d}}M_i(t)\}\\ R_3&= \sum _{i=1}^n\int _0^\tau \Big [\begin{pmatrix}{\begin{matrix}X_i-\bar{X}(t;\beta _0)\\ \\ B_i(t;\tilde{\Lambda }_0^D)-{\tilde{B}}(t) \end{matrix}}\end{pmatrix}-\begin{pmatrix}{\begin{matrix}X_i-\bar{X}(t;\beta _0)\\ \\ B_i(t)-\bar{B}(t) \end{matrix}}\end{pmatrix}\Big ]\mathrm{{d}}M_i(t), \end{aligned}$$

and

$$\begin{aligned} \mathrm{{d}}\tilde{M}_i(t)&=Y_i(t)\psi _i(t;\tilde{\Lambda }_0^D)\mathrm{{d}}N_i^R(t) -\Delta _i(t)\big \{\gamma _0^TX_i+\eta _0^T B_i(t;\tilde{\Lambda }_0^D)\big \} \exp \{\beta _0^TX_i\}\mathrm{{d}}\tilde{\Lambda }_0^R(t)\\&\quad -\Delta _i(t)\exp \{\beta _0^TX_i\}d{\mathcal {A}}_0(t). \end{aligned}$$

For $R_1$ of (A.10), using the Lemma A.1 of Lin and Ying [8], we have

$$\begin{aligned} R_1=\sum _{i=1}^n\int _0^\tau \begin{pmatrix}{\begin{matrix}X_i-\bar{x}(t;\beta _0)\\ \\ B_i(t)-\bar{b}(t) \end{matrix}}\end{pmatrix}\mathrm{{d}}M_i(t)+o_p(n^{1/2}), \end{aligned}$$

where $\bar{b}(t)$ is the limit of $\bar{B}(t).$ For $R_2$, we get

$$\begin{aligned} R_2&=\sum _{i=1}^n\int _0^\tau \begin{pmatrix}{\begin{matrix}X_i-\bar{X}(t;\beta _0)\\ \\ B_i(t;\tilde{\Lambda }_0^D)-{\tilde{B}}(t) \end{matrix}}\end{pmatrix}\Big \{Y_i(t)\big [\psi _i(t; \tilde{\Lambda }_0^D)-\psi _i(t; \Lambda _0^D)\big ]\mathrm{{d}}N_i^R(t)\\&\quad -\Delta _i(t)\gamma _0^TX_i\exp \{\beta _0^TX_i\}\mathrm{{d}}\{\tilde{\Lambda }_0^R(t)-\Lambda _0^R(t)\}\\&\quad -\Delta _i(t)\eta _0^TX_i\exp \{\beta _0^TX_i\} \big [B_i(t;\tilde{\Lambda }_0^D)\mathrm{{d}}\tilde{\Lambda }_0^R(t)-B_i(t)\mathrm{{d}}\Lambda _0^R(t)\big ]\Big \}. \end{aligned}$$

It then follows from (A.1) and (A.2) that

$$\begin{aligned} R_2=\sum _{i=1}^n\int _0^\tau \Big \{\phi _{1i}(t)\mathrm{{d}} G_1(t)- G_2(t)\mathrm{{d}}\phi _{2i}(t)\Big \}+o_p(n^{1/2}), \end{aligned}$$

where $G_1(t)$ and $G_2(t)$ are the limits of $\tilde{G}_1(t)$ and $\tilde{G}_2(t),$ respectively, with

$$\begin{aligned} \tilde{G}_1(t)&=\frac{1}{n}\sum _{i=1}^n\int _0^t\begin{pmatrix}{\begin{matrix}X_i-\bar{X}(u;\beta _0)\\ \\ B_i(u;\tilde{\Lambda }_0^D)-\bar{\tilde{B}}(u) \end{matrix}}\end{pmatrix} \theta _0Y_i(u)\exp \{\alpha _0^TX_i\}\mathrm{{d}}N_i^R(u)\\&\quad +\frac{1}{n}\sum _{i=1}^n\int _0^t\begin{pmatrix}{\begin{matrix}X_i-\bar{X}(u;\beta _0)\\ \\ B_i(u;\tilde{\Lambda }_0^D)-\bar{\tilde{B}}(u) \end{matrix}}\end{pmatrix} \frac{(\theta _0^2+\theta _0)\Delta _i(u)\eta _0^TZ_i\exp \{\alpha ^TX_i\}}{\psi _i(u;\tilde{\Lambda }_0^D) \psi _i(u)}\mathrm{{d}}\tilde{\Lambda }_0^R(u),\\ \tilde{G}_2(t)&= \frac{1}{n}\sum _{i=1}^n\begin{pmatrix}{\begin{matrix}X_i-\bar{X}(t;\beta _0)\\ \\ B_i(t;\tilde{\Lambda }_0^D)-\bar{\tilde{B}}(t) \end{matrix}}\end{pmatrix} \Delta _i(t)\gamma _0^TX_i\exp \{\beta _0^TX_i\}\\&\quad +\frac{1}{n}\sum _{i=1}^n\begin{pmatrix}{\begin{matrix}X_i-\bar{X}(t;\beta _0)\\ \\ B_i(t;\tilde{\Lambda }_0^D)-\bar{\tilde{B}}(t) \end{matrix}}\end{pmatrix} \Delta _i(t)\eta _0^TB_i(t)\exp \{\beta _0^TX_i\}. \end{aligned}$$

By the Lemma A.1 of Lin and Ying [8], it can be shown that $R_3=o_p(n^{1/2}).$ Thus, we have

$$\begin{aligned} \tilde{U}_4(\xi _0)= \sum _{i=1}^n \vartheta _{4i} + o_p(n^{1/2}), \end{aligned}$$

(A.11)

where

$$\begin{aligned} \vartheta _{4i}=\int _0^\tau \begin{pmatrix}{\begin{matrix}X_i-\bar{x}(t; \beta _0)\\ \\ B_i(t)-\bar{b}(t) \end{matrix}}\end{pmatrix}\mathrm{{d}}M_i(t) +\int _0^\tau \big \{\phi _{1i}(t)d G_1(t)- G_3(t) \mathrm{{d}}\phi _{2i}(t)\big \}. \end{aligned}$$

Let $\vartheta _i=(\vartheta _{1i}^T,\vartheta _{2i}^T,\vartheta _{3i},\vartheta _{4i}^T)^T,$ and $\Gamma =\Gamma (\xi _0)$ defined in condition (C4). Then it follows from (A.4), (A.5), (A.9), (A.11), and the Taylor expansion that

$$\begin{aligned} n^{1/2}(\hat{\xi }-\xi _0)=n^{-1/2}\Gamma ^{-1} \sum _{i=1}^n \vartheta _{i}+o_p(1). \end{aligned}$$

By the multivariate central limit theorem, $n^{1/2}(\hat{\xi }-\xi _0)$ is asymptotically normal with mean zero and covariance matrix $\Gamma ^{-1}\Sigma (\Gamma ^T)^{-1},$ where $\Sigma =E\{\vartheta _i \vartheta _i^T\}$.