Appendix
1.1 Proof of Theorem 1
Before presenting the proof of Theorem 1, we first introduce a matrix algebra result, which is not only of interest in its own right, but also is crucial to compare the asymptotic efficiency of the new estimator with the conventional one.
Lemma 1
Let \(\text{ P }_1\) be a \(d\times r\) matrix, \(\text{ P }_2\) be a d-dimensional column vector, and \(\text{ D }\) be a \(d\times d\) positive matrix. Then we have
$$\begin{aligned} (\text{ P }_1^T\text{ D }^{-1}\text{ P }_2)(\text{ P }_2^T\text{ D }^{-1}\text{ P }_1)\le (\text{ P }_1^T\text{ D }^{-1}\text{ P }_1)(\text{ P }_2^T\text{ D }^{-1}\text{ P }_2). \end{aligned}$$
Proof of Lemma 1
For any vector \(\text{ c }\ne {\mathbf {0}}\), we have
$$\begin{aligned} \text{ c }^T(\text{ P }_1^T\text{ D }^{-1}\text{ P }_2)(\text{ P }_2^T\text{ D }^{-1}\text{ P }_1)\text{ c }= & {} \text{ c }^T(\text{ P }_1^T\text{ D }^{-\frac{1}{2}}\text{ D }^{-\frac{1}{2}}\text{ P }_2)(\text{ P }_2^T\text{ D }^{-\frac{1}{2}}\text{ D }^{-\frac{1}{2}}\text{ P }_1)\text{ c }\\= & {} (\text{ D }^{-\frac{1}{2}}\text{ P }_1\text{ c })^T(\text{ D }^{-\frac{1}{2}}\text{ P }_2)(\text{ D }^{-\frac{1}{2}}\text{ P }_2)^T(\text{ D }^{-\frac{1}{2}}\text{ P }_1\text{ c })\\= & {} \{(\text{ D }^{-\frac{1}{2}}\text{ P }_1\text{ c })^T(\text{ D }^{-\frac{1}{2}}\text{ P }_2)\}^2\\\le & {} (\text{ c }^T\text{ P }_1^T\text{ D }^{-\frac{1}{2}}\text{ D }^{-\frac{1}{2}}\text{ P }_1\text{ c })(\text{ P }_2^T\text{ D }^{-\frac{1}{2}}\text{ D }^{-\frac{1}{2}}\text{ P }_2)\\= & {} \text{ c }^T(\text{ P }_1^T\text{ D }^{-1}\text{ P }_1)(\text{ P }_2^T\text{ D }^{-1}\text{ P }_2)\text{ c }. \end{aligned}$$
Thus,
$$\begin{aligned} (\text{ P }_1^T\text{ D }^{-1}\text{ P }_2)(\text{ P }_2^T\text{ D }^{-1}\text{ P }_1)\le (\text{ P }_1^T\text{ D }^{-1}\text{ P }_1)(\text{ P }_2^T\text{ D }^{-1}\text{ P }_2). \end{aligned}$$
This completes the proof. \(\square \)
Next, we prove Theorem 1. Throughout, the partial derivative of a vector- or matrix-valued function with respect to a scalar variable denotes the componentwise partial derivatives; and for a vector or matrix \({\mathbf {v}}\), \(\Vert {\mathbf {v}}\Vert \) denotes the \(L_2\)-norm of \({\mathbf {v}}\).
Proof of Theorem 1
By the empirical likelihood theory in estimating equations, we know that the true values of the Lagrange parameters \(\varvec{\xi }\) and \(\varvec{\nu }\) are both zero vectors. Let \(\varvec{\theta }_0=({\mathbf {0}}^T,{\mathbf {0}}^T,\varvec{\beta }_0^T,\alpha _0)^T\) be the true value of the pooled parameter vector \(\varvec{\theta }=(\varvec{\nu }^T,\varvec{\xi }^T,\varvec{\beta }^T,\alpha )^T\). Evaluated at \(\varvec{\theta }_0\), the system of estimating equations \(\text{ U }_j(\varvec{\theta }_0)\) for j equals 1 through 4 reduces to be
$$\begin{aligned} \text{ U}_1(\varvec{\theta }_0)= & {} \text{ U }_{01}=\sum _{i=1}^n\text{ W }(\varvec{\beta }_0,\text{ Z }_i),\\ \text{ U}_2(\varvec{\theta }_0)= & {} \text{ U }_{02}=\sum _{i=1}^n\varvec{\Psi }(\varvec{\beta }_0,\text{ Z }_i,\alpha _0),\\ \text{ U}_3(\varvec{\theta }_0)= & {} \text{ U }_{03}={\mathbf {0}},\\ \mathrm{U_4}(\varvec{\theta }_0)= & {} \mathrm{U_{04}}=0. \end{aligned}$$
Clearly, \(\text{ U}_2(\varvec{\theta }_0)\), \(\text{ U}_3(\varvec{\theta }_0)\) and \(\mathrm{U_4}(\varvec{\theta }_0)\) are all unbiased estimating functions. In addition, by the definition of \(\text{ W }(\varvec{\beta },\text{ Z }_i)\), we have
$$\begin{aligned} \text{ W }(\varvec{\beta }_0,\text{ Z }_i)=\int _0^\tau \{\text{ Z }_i-{\bar{\text{ Z }}}(t)\}dM_i(t), \end{aligned}$$
where \(M_i(t)=N_i(t)-\int _0^tY_i(u)\{\lambda _0(u)+\varvec{\beta }_0^T\text{ Z }_i\}du\) is a counting process martingale. The standard martingale theory (Fleming and Harrington 1991) suggests that \(\text{ U }_{01}\) is an unbiased estimating function.
From the expression of \(\Psi _k(\varvec{\beta },\text{ Z }_i,\alpha )\), we see that \(|\Psi _k(\varvec{\beta },\text{ Z }_i,\alpha )| \le 2\) for \(k=1,\ldots ,K\). Analogous to the proof of Lemma 1 in Qin and Lawless (1994), under some regularity conditions, the maximizer \(({\widehat{\varvec{\beta }}},{\widehat{\alpha }})\) lies in the interior of the ball \(\{(\varvec{\beta },\alpha ):\parallel (\varvec{\beta },\alpha )-(\varvec{\beta }_0,\alpha _0)\parallel \le n^{-1/3}\}\) with probability 1 as \(n\rightarrow \infty \). As a result, both \({\widehat{\varvec{\beta }}}\) and \({{\widehat{\alpha }}}\) are consistent.
Now we prove the asymptotic normality of \(n^{-1/2}\text{ U }_0\) based on the martingale central limit theorem and general central limit theorem, where \(\text{ U }_0=(\text{ U }_{01}^T,\text{ U }_{02}^T,\text{ U }_{03}^T,\mathrm{U_{04}})^T\). By the martingale theory, we have
$$\begin{aligned} \text {var}(n^{-\frac{1}{2}}\text{ U }_{01})= & {} \frac{1}{n}\text {var}\left[ \sum _{i=1}^n\int _0^\tau \{\text{ Z }_i-{\bar{\text{ Z }}}(t)\}d M_i(t)\right] \\= & {} \frac{1}{n}\sum _{i=1}^n\text {var}\left[ \int _0^\tau \{\text{ Z }_i-{\bar{\text{ Z }}}(t)\}d M_i(t)\right] \\= & {} \frac{1}{n}\sum _{i=1}^n\int _0^\tau \{\text{ Z }_i-{\bar{\text{ Z }}}(t)\}^{\otimes 2}\text {var}\{d M_i(t)\}\\= & {} \frac{1}{n}\sum _{i=1}^n\int _0^\tau \{\text{ Z }_i-{\bar{\text{ Z }}}(t)\}^{\otimes 2}Y_i(t) \{d\Lambda _0(t)+\varvec{\beta }_0^T\text{ Z }_idt\}\\&{\mathop {=}\limits ^{\wedge }}&\varvec{\Sigma }. \end{aligned}$$
Using the double expectation, we get
$$\begin{aligned} E(\text{ U }_{01}^T\text{ U }_{02})=E\{E(\text{ U }_{01}^T\text{ U }_{02}|\text{ Z })\}=E\{E(\text{ U }_{01}^T|\text{ Z })\text{ U }_{02}\}={\mathbf {0}}, \end{aligned}$$
which suggests that \(\text{ U }_{01}\) is uncorrelated with \(\text{ U }_{02}\). Let \(\text{ J }=\text {var}(n^{-\frac{1}{2}}\text{ U }_{02})\). By the martingale central limit theorem and central limit theorem, \(n^{-1/2}\text{ U }_0\) is asymptotically normal with zero mean and variance-covariance matrix
$$\begin{aligned} \varvec{\Omega }=\begin{pmatrix} \varvec{\Sigma }&{} {\mathbf {0}} &{} {\mathbf {0}} &{} {\mathbf {0}}\\ {\mathbf {0}} &{} \text{ J }&{} {\mathbf {0}} &{} {\mathbf {0}} \\ {\mathbf {0}} &{} {\mathbf {0}} &{} {\mathbf {0}} &{} {\mathbf {0}} \\ {\mathbf {0}} &{} {\mathbf {0}} &{} {\mathbf {0}} &{} 0 \end{pmatrix}. \end{aligned}$$
Now, we compute the expectation of the partial derivative matrix of \(\text{ U }(\varvec{\theta })\) with respect to \(\varvec{\theta }=(\varvec{\nu }^T,\varvec{\xi }^T,\varvec{\beta }^T,\alpha )^T\) when evaluated at the true parameter values; here \(\text{ U }(\varvec{\theta })=\big (\text{ U}_1^T(\varvec{\theta }),\text{ U}_2^T(\varvec{\theta }),\text{ U}_3^T(\varvec{\theta }),\mathrm{U_4}(\varvec{\theta })\big )^T\). By a careful examination, we observe that
$$\begin{aligned}&\frac{\partial \text{ U}_1}{\partial \varvec{\xi }^T}|_{\varvec{\theta }=\varvec{\theta }_0}=\left( \frac{\partial \text{ U}_2}{\partial \varvec{\nu }^T}|_{\varvec{\theta }=\varvec{\theta }_0}\right) ^T, \quad \frac{\partial \text{ U}_1}{\partial \varvec{\beta }^T}|_{\varvec{\theta }=\varvec{\theta }_0}=\left( \frac{\partial \text{ U}_3}{\partial \varvec{\nu }^T}|_{\varvec{\theta }=\varvec{\theta }_0}\right) ^T,\\&\frac{\partial \text{ U}_2}{\partial \varvec{\beta }^T}|_{\varvec{\theta }=\varvec{\theta }_0}=\left( \frac{\partial \text{ U}_3}{\partial \varvec{\xi }^T}|_{\varvec{\theta }=\varvec{\theta }_0}\right) ^T, \quad \frac{\partial \text{ U}_2}{\partial \alpha }|_{\varvec{\theta }=\varvec{\theta }_0}=\left( \frac{\partial \mathrm{U_4}}{\partial \varvec{\xi }^T}|_{\varvec{\theta }=\varvec{\theta }_0}\right) ^T. \end{aligned}$$
Besides, we have the following results,
$$\begin{aligned} \frac{1}{n}E\left( \frac{\partial \text{ U}_1}{\partial \varvec{\nu }^T}|_{\varvec{\theta }=\varvec{\theta }_0}\right)= & {} -\frac{1}{n}\sum _{i=1}^n E\left\{ \text{ W }(\varvec{\beta }_0,\text{ Z }_i)^{\otimes 2}\right\} =-\varvec{\Sigma },\\ \frac{1}{n}E\left( \frac{\partial \text{ U}_1}{\partial \varvec{\xi }^T}|_{\varvec{\theta }=\varvec{\theta }_0}\right)= & {} -\frac{1}{n}\sum _{i=1}^n E\left\{ \text{ W }(\varvec{\beta }_0,\text{ Z }_i)\varvec{\Psi }^T(\varvec{\beta }_0,\text{ Z }_i,\alpha _0)\right\} ={\mathbf {0}},\\ \frac{1}{n}E\left( \frac{\partial \text{ U}_1}{\partial \varvec{\beta }^T}|_{\varvec{\theta }=\varvec{\theta }_0}\right)= & {} \frac{1}{n}\sum _{i=1}^n E\left\{ \frac{\partial \text{ W }(\varvec{\beta },\text{ Z }_i)}{\partial \varvec{\beta }^T}|_{\varvec{\theta }=\varvec{\theta }_0}\right\} \\= & {} \frac{1}{n}\sum _{i=1}^n E\left\{ \int _0^\tau Y_i(t)\{\text{ Z }_i-{\bar{\text{ Z }}}(t)\}^{\otimes 2}dt\right\} {\mathop {=}\limits ^{\wedge }}\text{ B }_1,\\ \frac{1}{n}E\left( \frac{\partial \text{ U}_1}{\partial \alpha }|_{\varvec{\theta }=\varvec{\theta }_0}\right)= & {} \frac{1}{n}E\left( \frac{\partial \text{ U}_3}{\partial \alpha }|_{\varvec{\theta }=\varvec{\theta }_0}\right) =\left\{ \frac{1}{n}E(\frac{\partial \mathrm{U_4}}{\partial \varvec{\nu }^T}|_{\varvec{\theta }=\varvec{\theta }_0})\right\} ^T\\= & {} \left\{ \frac{1}{n}E(\frac{\partial \mathrm{U_4}}{\partial \varvec{\beta }^T}|_{\varvec{\theta }=\varvec{\theta }_0})\right\} ^T={\mathbf {0}},\\ \frac{1}{n}E\left( \frac{\partial \text{ U}_2}{\partial \varvec{\xi }^T}|_{\varvec{\theta }=\varvec{\theta }_0}\right)= & {} -\frac{1}{n}\sum _{i=1}^nE\left\{ \varvec{\Psi }(\varvec{\beta }_0,\text{ Z }_i,\alpha _0)^{\otimes 2}\right\} =-\text{ J },\\ \frac{1}{n}E\left( \frac{\partial \text{ U}_2}{\partial \varvec{\beta }^T}|_{\varvec{\theta }=\varvec{\theta }_0}\right)= & {} \frac{1}{n}\sum _{i=1}^nE\left\{ \frac{\partial \varvec{\Psi }(\varvec{\beta },\text{ Z }_i,\alpha )}{\partial \varvec{\beta }^T}|_{\varvec{\theta }=\varvec{\theta }_0}\right\} {\mathop {=}\limits ^{\wedge }}\text{ B }_2,\\ \frac{1}{n}E\left( \frac{\partial \text{ U}_2}{\partial \alpha }|_{\varvec{\theta }=\varvec{\theta }_0}\right)= & {} \frac{1}{n}\sum _{i=1}^nE\left( \frac{\partial \varvec{\Psi }(\varvec{\beta },\text{ Z }_i,\alpha )}{\partial \alpha }|_{\varvec{\theta }=\varvec{\theta }_0}\right) {\mathop {=}\limits ^{\wedge }}\text{ B }_3,\\ \frac{1}{n}E\left( \frac{\partial \mathrm{U_3}}{\partial \varvec{\beta }^T}|_{\varvec{\theta }=\varvec{\theta }_0}\right)= & {} {\mathbf {0}}\\ \frac{1}{n}E\left( \frac{\partial \mathrm{U_4}}{\partial \alpha }|_{\varvec{\theta }=\varvec{\theta }_0}\right)= & {} 0. \end{aligned}$$
Put these results together, we have
$$\begin{aligned} \text{ B }=-\frac{1}{n}E\left\{ \dfrac{\partial \text{ U }(\varvec{\theta })}{\partial \varvec{\theta }^T}|_{\varvec{\theta }=\varvec{\theta }_0}\right\} = \begin{pmatrix} \varvec{\Sigma }&{} {\mathbf {0}} &{} -\text{ B }_1 &{} {\mathbf {0}} \\ {\mathbf {0}} &{} \text{ J }&{} -\text{ B }_2 &{} -\text{ B }_3 \\ -\text{ B }_1^T &{} -\text{ B }_2^T &{} {\mathbf {0}} &{} {\mathbf {0}} \\ {\mathbf {0}} &{} -\text{ B }_3^T &{} {\mathbf {0}} &{} 0 \end{pmatrix}. \end{aligned}$$
Next, we derive an expression for the inverse of \(\text{ B }\). For this purpose, we write
$$\begin{aligned} {\widetilde{\text{ B }}}_1=(-\text{ B }_1,{\mathbf {0}}), \quad {\widetilde{\text{ B }}}_2=(-\text{ B }_2,-\text{ B }_3), \end{aligned}$$
then the 4-by-4 block matrix \(\text{ B }\) could be rewritten as a 3-by-3 block matrix
$$\begin{aligned} \text{ B }=\begin{pmatrix} \varvec{\Sigma }&{} {\mathbf {0}} &{} {\widetilde{\text{ B }}}_1 \\ {\mathbf {0}} &{} \text{ J }&{} {\widetilde{\text{ B }}}_2 \\ {\widetilde{\text{ B }}}_1^T &{} {\widetilde{\text{ B }}}_2^T &{} {\mathbf {0}} \end{pmatrix}. \end{aligned}$$
Observe that
$$\begin{aligned}&\begin{pmatrix} \text{ I }&{} {\mathbf {0}} &{} {\mathbf {0}} \\ {\mathbf {0}} &{} \text{ I }&{} {\mathbf {0}}\\ -{\widetilde{\text{ B }}}_1^T\varvec{\Sigma }^{-1} &{} -{\widetilde{\text{ B }}}_2^T\text{ J }^{-1} &{} \text{ I }\end{pmatrix} \begin{pmatrix} \varvec{\Sigma }&{} {\mathbf {0}} &{} {\widetilde{\text{ B }}}_1 \\ {\mathbf {0}} &{} \text{ J }&{} {\widetilde{\text{ B }}}_2 \\ {\widetilde{\text{ B }}}_1^T &{} {\widetilde{\text{ B }}}_2^T &{} {\mathbf {0}} \end{pmatrix} \begin{pmatrix} \text{ I }&{} {\mathbf {0}} &{} -\varvec{\Sigma }^{-1}{\widetilde{\text{ B }}}_1 \\ {\mathbf {0}} &{} \text{ I }&{} -\text{ J }^{-1}{\widetilde{\text{ B }}}_2 \\ {\mathbf {0}} &{} {\mathbf {0}} &{} \text{ I }\end{pmatrix}\\&=\begin{pmatrix} \varvec{\Sigma }&{} {\mathbf {0}} &{} {\mathbf {0}} \\ {\mathbf {0}} &{} \text{ J }&{} {\mathbf {0}} \\ {\mathbf {0}} &{} {\mathbf {0}} &{} {\widetilde{\varvec{\Gamma }}} \end{pmatrix}, \end{aligned}$$
where \({\widetilde{\varvec{\Gamma }}}=-(\text{ B }_1^T\varvec{\Sigma }^{-1}\text{ B }_1+\text{ B }_2^T\text{ J }^{-1}\text{ B }_2)\), we have
$$\begin{aligned}&\text{ B }^{-1}=\text{ A }=\begin{pmatrix} \text{ I }&{} {\mathbf {0}} &{} -\varvec{\Sigma }^{-1}{\widetilde{\text{ B }}}_1 \\ {\mathbf {0}} &{} \text{ I }&{} -\text{ J }^{-1} {\widetilde{\text{ B }}}_2 \\ {\mathbf {0}} &{} {\mathbf {0}} &{} \text{ I }\end{pmatrix} \begin{pmatrix} \varvec{\Sigma }^{-1} &{} {\mathbf {0}} &{} {\mathbf {0}} \\ {\mathbf {0}} &{} \text{ J }^{-1} &{} {\mathbf {0}} \\ {\mathbf {0}} &{} {\mathbf {0}} &{} {\widetilde{\varvec{\Gamma }}}^{-1} \end{pmatrix} \begin{pmatrix} \text{ I }&{} {\mathbf {0}} &{} {\mathbf {0}} \\ {\mathbf {0}} &{} \text{ I }&{} {\mathbf {0}} \\ -{\widetilde{\text{ B }}}_1^T\varvec{\Sigma }^{-1} &{} -{\widetilde{\text{ B }}}_2^T\text{ J }^{-1} &{} \text{ I }\end{pmatrix}\\&\quad =\begin{pmatrix} \varvec{\Sigma }^{-1}+\varvec{\Sigma }^{-1}{\widetilde{\text{ B }}}_1{\widetilde{\varvec{\Gamma }}}^{-1}{\widetilde{\text{ B }}}_1^T\varvec{\Sigma }^{-1} &{} \varvec{\Sigma }^{-1}{\widetilde{\text{ B }}}_1\varvec{\Gamma }^{-1}{\widetilde{\text{ B }}}_2^T\text{ J }^{-1} &{} -\varvec{\Sigma }^{-1}{\widetilde{\text{ B }}}_1{\widetilde{\varvec{\Gamma }}}^{-1}\\ \text{ J }^{-1}{\widetilde{\text{ B }}}_2{\widetilde{\varvec{\Gamma }}}^{-1}{\widetilde{\text{ B }}}_1^T\varvec{\Sigma }^{-1} &{} \text{ J }^{-1}+\text{ J }^{-1}{\widetilde{\text{ B }}}_2{\widetilde{\varvec{\Gamma }}}^{-1}{\widetilde{\text{ B }}}_2^T\text{ J }^{-1} &{} -\text{ J }^{-1}{\widetilde{\text{ B }}}_2{\widetilde{\varvec{\Gamma }}}^{-1} \\ -{\widetilde{\varvec{\Gamma }}}^{-1}{\widetilde{\text{ B }}}_1^T\varvec{\Sigma }^{-1} &{} -{\widetilde{\varvec{\Gamma }}}{\widetilde{\text{ B }}}_2^T\text{ J }^{-1} &{} {\widetilde{\varvec{\Gamma }}}^{-1} \end{pmatrix}\\&\quad {\mathop {=}\limits ^{\wedge }}\begin{pmatrix} \text{ A }_{11} &{} \text{ A }_{12} &{} -\varvec{\Sigma }^{-1}{\widetilde{\text{ B }}}_1{\widetilde{\varvec{\Gamma }}}^{-1} \\ \text{ A }_{21} &{} \text{ A }_{22} &{} -\text{ J }^{-1}{\widetilde{\text{ B }}}_2{\widetilde{\varvec{\Gamma }}}^{-1} \\ -{\widetilde{\varvec{\Gamma }}}^{-1}{\widetilde{\text{ B }}}_1^T\varvec{\Sigma }^{-1} &{} -{\widetilde{\varvec{\Gamma }}}{\widetilde{\text{ B }}}_2^T\text{ J }^{-1} &{} {\widetilde{\varvec{\Gamma }}}^{-1} \end{pmatrix}. \end{aligned}$$
Now we calculate the detailed expression of \({\widetilde{\varvec{\Gamma }}}^{-1}\), \(-{\widetilde{\varvec{\Gamma }}}^{-1}{\widetilde{\text{ B }}}_1^T\varvec{\Sigma }^{-1}\) and \(-{\widetilde{\varvec{\Gamma }}}{\widetilde{\text{ B }}}_2^T\text{ J }^{-1}\) appeared in \(\text{ A }^{-1}\).
First, we calculate \({\widetilde{\varvec{\Gamma }}}^{-1}\). Note that
$$\begin{aligned} {\widetilde{\varvec{\Gamma }}}= & {} -({\widetilde{\text{ B }}}_1^T\varvec{\Sigma }^{-1}{\widetilde{\text{ B }}}_1+{\widetilde{\text{ B }}}_2^T\text{ J }^{-1}{\widetilde{\text{ B }}}_2)\\= & {} -\begin{pmatrix} -\text{ B }_1^T \\ {\mathbf {0}} \end{pmatrix} \varvec{\Sigma }^{-1} \begin{pmatrix} -\text{ B }_1&{\mathbf {0}} \end{pmatrix} - \begin{pmatrix} -\text{ B }_2^T \\ -\text{ B }_3^T \end{pmatrix} \text{ J }^{-1} \begin{pmatrix} -\text{ B }_2&-\text{ B }_3 \end{pmatrix}\\= & {} -\begin{pmatrix} \text{ B }_1^T\varvec{\Sigma }^{-1}\text{ B }_1+\text{ B }_2^T\text{ J }^{-1}\text{ B }_2 &{} \text{ B }_2^T\text{ J }^{-1}\text{ B }_3 \\ \text{ B }_3^T\text{ J }^{-1}\text{ B }_2 &{} \text{ B }_3^T\text{ J }^{-1}\text{ B }_3 \end{pmatrix}{\mathop {=}\limits ^{\wedge }}-{\bar{\varvec{\Gamma }}}. \end{aligned}$$
Using algebraic transformation, we have
$$\begin{aligned}&\begin{pmatrix} \text{ I }&{} {\mathbf {0}} \\ -(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1} &{} \text{ I }\end{pmatrix} \begin{pmatrix} \text{ B }_1^T\varvec{\Sigma }^{-1}\text{ B }_1+\text{ B }_2^T\text{ J }^{-1}\text{ B }_2 &{} \text{ B }_2^T\text{ J }^{-1}\text{ B }_3 \\ \text{ B }_3^T\text{ J }^{-1}\text{ B }_2 &{} \text{ B }_3^T\text{ J }^{-1}\text{ B }_3 \end{pmatrix} \begin{pmatrix} \text{ I }&{} -\text{ H }^{-1}(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3) \\ {\mathbf {0}} &{} \text{ I }\end{pmatrix}\\&\quad \quad =\begin{pmatrix} \text{ H }&{} {\mathbf {0}} \\ {\mathbf {0}} &{} \mathrm{Q} \end{pmatrix}, \end{aligned}$$
where \(\text{ H }=\text{ B }_1^T\varvec{\Sigma }^{-1}\text{ B }_1+\text{ B }_2^T\text{ J }^{-1}\text{ B }_2\) and \(\mathrm{Q} =\text{ B }_3^T\text{ J }^{-1}\text{ B }_3-(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1}(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3)\). Thus
$$\begin{aligned}&{\widetilde{\varvec{\Gamma }}}^{-1} = -\begin{pmatrix} \text{ I }&{} -\text{ H }^{-1}(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3) \\ {\mathbf {0}} &{} \text{ I }\end{pmatrix} \begin{pmatrix} \text{ H }^{-1} &{} {\mathbf {0}} \\ {\mathbf {0}} &{} \mathrm{Q^{-1}} \end{pmatrix} \begin{pmatrix} \text{ I }&{} {\mathbf {0}} \\ -(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1} &{} \text{ I }\end{pmatrix}\\&\quad =-\begin{pmatrix} \text{ H }^{-1}+\text{ H }^{-1}(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3)\mathrm{Q^{-1}}(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1} &{} -\text{ H }^{-1}(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3)\mathrm{Q^{-1}}\\ -\mathrm{Q^{-1}}(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1} &{} \mathrm{Q^{-1}} \end{pmatrix}\\&\quad {\mathop {=}\limits ^{\wedge }} \begin{pmatrix} \text{ A }_{33} &{} \text{ A }_{34} \\ \text{ A }_{43} &{} \mathrm{A_{44}} \end{pmatrix}. \end{aligned}$$
Similarly, we have
$$\begin{aligned}&-{\widetilde{\varvec{\Gamma }}}^{-1}{\widetilde{\text{ B }}}_1^T\varvec{\Sigma }^{-1}\\&\quad =\begin{pmatrix} \text{ H }^{-1}+\text{ H }^{-1}(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3)\mathrm{Q^{-1}}(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1} &{} -\text{ H }^{-1}(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3)\mathrm{Q^{-1}} \\ -\mathrm{Q^{-1}}(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1} &{} \mathrm{Q^{-1}} \end{pmatrix} \begin{pmatrix} -\text{ B }_1^T \\ {\mathbf {0}} \end{pmatrix} \varvec{\Sigma }^{-1} \\&\quad = \begin{pmatrix} -[\text{ H }^{-1}+\text{ H }^{-1}(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3)\mathrm{Q^{-1}}(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1}]\text{ B }_1^T\varvec{\Sigma }^{-1} \mathrm{Q^{-1}}(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1}\text{ B }_1^T\varvec{\Sigma }^{-1} \end{pmatrix}\\&\quad {\mathop {=}\limits ^{\wedge }}\begin{pmatrix} \text{ A }_{31}\\ \text{ A }_{41} \end{pmatrix}, \end{aligned}$$
and
$$\begin{aligned}&-{\widetilde{\varvec{\Gamma }}}^{-1}{\widetilde{\text{ B }}}_2^T\text{ J }^{-1}\\&\quad =\begin{pmatrix} \text{ H }^{-1}+\text{ H }^{-1}(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3)\mathrm{Q^{-1}}(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1} &{} -\text{ H }^{-1}(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3)\mathrm{Q^{-1}} \\ -\mathrm{Q^{-1}}(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1} &{} \mathrm{Q^{-1}} \end{pmatrix} \begin{pmatrix} -\text{ B }_2^T \\ -\text{ B }_3^T \end{pmatrix} \text{ J }^{-1} \\&\quad = \begin{pmatrix} -[\text{ H }^{-1}+\text{ H }^{-1}(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3)\mathrm{Q^{-1}}(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1}]\text{ B }_2^T\text{ J }^{-1}+\text{ H }^{-1}(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3) \mathrm{Q^{-1}}\text{ B }_3^T\text{ J }^{-1} \\ \mathrm{Q^{-1}}(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1}\text{ B }_2^T\text{ J }^{-1}-\mathrm{Q^{-1}}\text{ B }_3^T\text{ J }^{-1} \end{pmatrix}\\&\quad {\mathop {=}\limits ^{\wedge }}\begin{pmatrix} \text{ A }_{32}\\ \text{ A }_{42} \end{pmatrix}. \end{aligned}$$
So the inverse of matrix \(\text{ B }\) can be expressed as
$$\begin{aligned} \text{ A }=\begin{pmatrix} \text{ A }_{11} &{} \text{ A }_{12} &{} \text{ A }_{31}^T &{} \text{ A }_{41}^T \\ \text{ A }_{21} &{} \text{ A }_{22} &{} \text{ A }_{32}^T &{} \text{ A }_{42}^T \\ \text{ A }_{31} &{} \text{ A }_{32} &{} \text{ A }_{33} &{} \text{ A }_{34} \\ \text{ A }_{41} &{} \text{ A }_{42} &{} \text{ A }_{43} &{} \mathrm{A_{44}} \end{pmatrix}. \end{aligned}$$
From Taylor expansion, we have
$$\begin{aligned} \sqrt{n}({\widehat{\varvec{\theta }}}-\varvec{\theta }_0)=\text{ A }\times n^{-1/2}\text{ U }_{0}+o_p(1), \end{aligned}$$
where \(\text{ U }_0=(\text{ U }_{01}^T,\text{ U }_{02}^T,\text{ U }_{03}^T,\mathrm{U_{04}})^T\). It follows that \(\sqrt{n}({\widehat{\varvec{\theta }}}-\varvec{\theta }_0)\) is asymptotically normal with zero mean and variance-covariance matrix \(\text{ A }\varvec{\Omega }\text{ A }^T\). Specifically, the variance-covariance matrix of \(\sqrt{n}({\widehat{\varvec{\beta }}}-\varvec{\beta }_0)\) is
$$\begin{aligned} \varvec{\Sigma }_0= & {} \text{ A }_{31}\varvec{\Sigma }\text{ A }_{31}^T+\text{ A }_{32}\text{ J }\text{ A }_{32}^T\\= & {} \text{ H }^{-1}+\text{ H }^{-1}(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3)\mathrm{Q^{-1}}(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)\text{ H }^{-1}\\= & {} [\text{ H }-(\text{ B }_2^T\text{ J }^{-1}\text{ B }_3){(\text{ B }_3^T\text{ J }^{-1}\text{ B }_3)}^{-1}(\text{ B }_3^T\text{ J }^{-1}\text{ B }_2)]^{-1}. \end{aligned}$$
This completes the proof. \(\square \)
1.2 Subgroup survival probability in the simulation study
In this subsection, we derived the specific formula for computing the subgroup survival probability in the simulation study. For illustration, we only calculated the \(t^*\)-year survival probability for subgroup \(\Omega _1\) of Model 1 in the simulation study. Other calculations follow similar procedures.
Recall that \(\Omega _1=\{(Z_1,Z_2)|Z_1\ge 0.5, Z_2=0\}\), with \(Z_1\) being uniformly distributed over [0, 1] and \(Z_2\) following a Bernoulli distribution with success probability of 0.5, which are statistically independent to each other, we have
$$\begin{aligned} P(T>t^*|Z_1\ge 0.5,Z_2=0)= & {} \frac{P(T>t^*,Z_1\ge 0.5|Z_2=0)}{P(Z_1\ge 0.5|Z_2=0)}\\= & {} \frac{P(T>t^*,Z_1\ge 0.5|Z_2=0)}{P(Z_1\ge 0.5)}\\= & {} 2P(T>t^*,Z_1\ge 0.5|Z_2=0). \end{aligned}$$
The conditional density of T given \(Z_1=z_1\) and \(Z_2=z_2\) is
$$\begin{aligned} p_{T|Z_1,Z_2}(t|z_1,z_2)=\lambda (t|z_1,z_2)S(t|z_1,z_2)=\lambda (t|z_1,z_2)\exp \left\{ -\int _0^t \lambda (u|z_1,z_2)du\right\} . \end{aligned}$$
In addition, the joint density of T and \(Z_1\) given \(Z_2=z_2\) is
$$\begin{aligned} p_{T,Z_1|Z_2}(t,z_1)= & {} p_{T|Z_1,Z_2}(t|z_1,z_2)p_{Z_1|Z_2}(z_1|z_2)\\= & {} p_{T|Z_1,Z_2}(t|z_1,z_2)p_{Z_1}(z_1)\\= & {} \lambda (t|z_1,z_2)\exp \left\{ -\int _0^t \lambda (u|z_1,z_2)du\right\} I(0<z_1<1), \end{aligned}$$
where \(I(0<z_1<1)\) is the usual indicator function which takes the value of 1 when \(0<z_1<1\) and 0 otherwise. For Model 1, we have \(\lambda (t|z_1,z_2)=2t+\beta _1 z_1+\beta _2 z_2\). Combining this with expressions in the last display, it yields
$$\begin{aligned} p_{T,Z_1|Z_2=0}(t,z_1)=(2t+\beta _1z_1)\exp (-t^2+\beta _1z_1t)I(0<z_1<1). \end{aligned}$$
After some algebraic manipulations, we have
$$\begin{aligned}&P(T>t^*,Z_1\ge 0.5|Z_2=0)=\int _{t^*}^{+\infty }dt\int _{\frac{1}{2}}^1 (2t+\beta _1z_1)\exp (-t^2+\beta _1z_1t)dz_1\\&=\int _{t^*}^{+\infty }\exp (-t^2)\\&\quad \left\{ \left( \frac{1}{\beta _1}+\frac{1}{2t}+\frac{1}{\beta _1t^2}\right) \exp (-\beta _1t/2)-\left( \frac{2}{\beta _1} +\frac{1}{t}+\frac{1}{\beta _1t^2}\right) \exp (-\beta _1t)\right\} dt\\&=\int _{-\infty }^{+\infty }\exp (-t^2)I\left( t>t^*\right) \\&\quad \left\{ \left( \frac{1}{\beta _1}+\frac{1}{2t}+\frac{1}{\beta _1t^2}\right) \exp (-\beta _1t/2)-\left( \frac{2}{\beta _1} +\frac{1}{t}+\frac{1}{\beta _1t^2}\right) \exp (-\beta _1t)\right\} dt. \end{aligned}$$
If we let g(t) denote the product of \(I\left( t>t^*\right) \) with the expressions between the large curly braces in the last integration, it follows that
$$\begin{aligned} P(T>t^*,Z_1\ge 0.5|Z_2=0)=\sqrt{\pi }E\{g(X)\}, \end{aligned}$$
where \(X\sim N(0,1/2)\). As a result, the \(t^*\)-year survival probability can be computed by using the Monte Carlo method. In the simulation studies, the Monte Carlo sampling size was 10, 000.
