Appendix
Proof of Lemma 6.1
From (2.1), we have \(\frac{1}{n_l}\sum _{i=1}^{n_l}\varphi _i(\widehat{\varvec{\rho }}_{0\,l})=0\). By Taylor expansion, under (C1)(1)–(2), then
$$ 0=\frac{1}{n_l}\sum _{i=1}^{n_l}\varphi _i(\widehat{\varvec{\rho }}_{0\,l}) =\frac{1}{n_l}\sum _{i=1}^{n_l}\varphi _i(\varvec{\rho }_{0\,l}) + \frac{1}{n_l}\sum _{i=1}^{n_l} \triangledown \varphi _i(\varvec{\rho }_{0\,l})(\widehat{\varvec{\rho }}_{0\,l}-\varvec{\rho }_{0\,l}) + o_p(\Vert \widehat{\varvec{\rho }}_{0\,l}-\varvec{\rho }_{0\,l}\Vert ). $$
From (C1)(3) and central-limit theorem, it can be easily checked that \(\widehat{\varvec{\rho }}_{0\,l}-\varvec{\rho }_{0\,l}=O_p(n_l^{-1/2})\) and
$$\begin{aligned} \sqrt{n_l}(\widehat{\varvec{\rho }}_{0l}-\varvec{\rho }_{0l})=&-\Big [\frac{1}{n_l}\sum _{i=1}^{n_l}\triangledown \varphi _i(\varvec{\rho }_{0l})\Big ]^{-1}\frac{1}{\sqrt{n_l}}\sum _{i=1}^{n_l} \varphi _i(\varvec{\rho }_{0l})+o_p(1) \xrightarrow {d}N(0,I_l^{-1}(\varvec{\rho }_{0l})). \end{aligned}$$
\(\square \)
Proof of Lemma 6.2
It is sufficient to show that
$$\frac{1}{\sqrt{n_l}} \sum _{i=1}^{n_l} \frac{\widehat{\delta }_{li}}{1-{\widehat{G}}_l(Z_{li})}\big [\Lambda _l\big (\frac{F_l^{-1}(\tau ) -Z_{li}}{a_l}\big )-\tau \big ]\xrightarrow {d} N(0,\Xi _l).$$
To prove this, under (C1)–(C4), we first prove
$$\begin{aligned}&\frac{1}{\sqrt{n_l}} \sum _{i=1}^{n_l} \frac{\widehat{\delta }_{li}}{1-{\widehat{G}}_l(Z_{li})} \Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau \Big ]\nonumber \\&\quad =\frac{1}{\sqrt{n_l}} \sum _{i=1}^{n_l} \frac{\xi _{li}\delta _{li}+(1-\xi _{li})m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})}{1-G_l(Z_{li})}\Big [\Lambda _l \Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau \Big ]\nonumber \\&\qquad +\frac{1}{n_l^{3/2}}\sum _{j< k} \Big \{\frac{\xi _{lj}\delta _{lj}+(1-\xi _{lj})m_l(Z_{lj},\textbf{X}_{lj};\varvec{\rho }_{0l})}{1-G_l(Z_{lj})}\Big [\Lambda _l \Big (\frac{F_l^{-1}(\tau )-Z_{lj}}{a_l}\Big )-\tau \Big ]\Gamma _l(Z_{lk},\textbf{X}_{lk},\xi _{lk},\xi _{lk}\delta _{lk};Z_{lj})\nonumber \\&\qquad +\frac{\xi _{lk}\delta _{lk}+(1-\xi _{lk})m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})}{1-G_l(Z_{lk})}\Big [\Lambda _l \Big (\frac{F_l^{-1}(\tau )-Z_{lk}}{a_l}\Big )-\tau \Big ]\Gamma _l(Z_{lj},\textbf{X}_{lj},\xi _{lj},\xi _{lj}\delta _{lj};Z_{lk})\Big \}\nonumber \\&\qquad +\alpha _l^{\mathrm T}I_l^{-1}(\varvec{\rho }_{0l})\frac{1}{\sqrt{n_l}} \sum _{i=1}^{n_l} \frac{\xi _{li}[\delta _{li}-m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})]\triangledown m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})}{m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})[1-m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})]} + o_p(1)\nonumber \\&\;\; := E_{l1}+E_{l2}+E_{l3}+o_p(1). \end{aligned}$$
(8.1)
In fact,
$$\begin{aligned}&\frac{1}{\sqrt{n_l}}\sum _{i=1}^{n_l}\frac{\widehat{\delta }_{li}}{1-{\widehat{G}}_l(Z_{li})} \Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau \Big ]\\&\quad =\frac{1}{\sqrt{n_l}}\sum _{i=1}^{n_l}\frac{\xi _{li}\delta _{li}+(1-\xi _{li})m_l(Z_{li},\textbf{X}_{li}; \varvec{\rho }_{0l})}{1-G_l(Z_{li})}\Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau \Big ]\\&\qquad +\frac{1}{\sqrt{n_l}}\sum _{i=1}^{n_l}\frac{\xi _{li}\delta _{li}+(1-\xi _{li})m_l(Z_{li},\textbf{X}_{li}; \varvec{\rho }_{0l})}{[1-G_l(Z_{li})]^2}[{\widehat{G}}_l(Z_{li})-G_l(Z_{li})]\Big [\Lambda _l\Big (\frac{F_l^{-1} (\tau )-Z_{li}}{a_l}\Big )-\tau \Big ]\\&\qquad +\frac{1}{\sqrt{n_l}}\sum _{i=1}^{n_l}\frac{(1-\xi _{li})[m_l(Z_{li},\textbf{X}_{li}; \widehat{\varvec{\rho }}_{0l})-m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})]}{1-G_l(Z_{li})} \Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau \Big ]\\&\qquad +\frac{1}{\sqrt{n_l}}\sum _{i=1}^{n_l}\frac{\xi _{li}\delta _{li}+(1-\xi _{li})m_l(Z_{li},\textbf{X}_{li}; \varvec{\rho }_{0l})}{[1-G_l(Z_{li})]^2[1-{\widehat{G}}_l(Z_{li})]}[{\widehat{G}}_l(Z_{li})-G_l(Z_{li})]^2 \Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau \Big ]\\&\qquad +\frac{1}{\sqrt{n_l}}\sum _{i=1}^{n_l}\frac{(1-\xi _{li})[m_l(Z_{li},\textbf{X}_{li}; \widehat{\varvec{\rho }}_{0l})-m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})] [{\widehat{G}}_l(Z_{li})-G_l(Z_{li})]}{[1-G_l(Z_{li})][1-{\widehat{G}}_l(Z_{li})]} \Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau \Big ]\\&\;\; :=S_1+S_2+S_3+S_4+S_5. \end{aligned}$$
By (C4), from Step 2 in the proof of Proposition 2.1, it follows that
$$\max _{1\le i\le n_l}|{\widehat{G}}_l(Z_{li})-G_l(Z_{li})|\le \sup _{0\le t\le b_l}|{\widehat{G}}_l(t)-G_l(t)|=O_p(\sqrt{\log n_l/n_l})$$
and \(S_4=O_p\big (\log n_l/\sqrt{n_l}\big )=o_p(1)\). In addition, \(S_5=O_p\big (\sqrt{\log n_l/{n_l}}\big )=o_p(1)\) by the fact \(\max _{1\le i\le n_l}|m_l(Z_{li},\textbf{X}_{li};\widehat{\varvec{\rho }}_{0\,l})-m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0\,l})|=O_p(n_l^{-1/2})\). By Proposition 2.1, we have
$$\begin{aligned} S_2\overset{a.s.}{=}\ {}&\,\frac{1}{{n_l}^{1/2}}\sum _{i=1}^{n_l}\frac{\xi _{li}\delta _{li}+(1-\xi _{li})m_l(Z_{li},\textbf{X}_{li}; \varvec{\rho }_{0l})}{1-G_l(Z_{li})}\Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau \Big ] \\ {}&\quad \times\Big [\frac{1}{n_l}\sum _{j=1}^{n_l}\psi _{lj}(Z_{li})+R_{G_l}(Z_{li})\Big ]\\ =&\,\frac{1}{n_l^{3/2}}\sum _{j< k} \Big \{\frac{\xi _{lj}\delta _{lj}+(1-\xi _{lj})m_l(Z_{lj},\textbf{X}_{lj}; \varvec{\rho }_{0l})}{1-G_l(Z_{lj})}\Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{lj}}{a_l}\Big )-\tau \Big ]\\&\quad \times\Gamma _l(Z_{lk},\textbf{X}_{lk},\xi _{lk},\xi _{lk}\delta _{lk};Z_{lj})\\&+\frac{\xi _{lk}\delta _{lk}+(1-\xi _{lk})m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})}{1-G_l(Z_{lk})} \Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{lk}}{a_l}\Big )-\tau \Big ]\\&\quad \times\Gamma _l(Z_{lj},\textbf{X}_{lj},\xi _{lj}, \xi _{lj}\delta _{lj};Z_{lk})\Big \}\\&-\frac{1}{{n_l}}\sum _{i=1}^{n_l}\frac{\xi _{li}\delta _{li}+(1-\xi _{li})m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})}{1-G_l(Z_{li})}\Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big ) -\tau \Big ]\\&\quad \times\mu _l^{\mathrm T}(Z_{li})I_l^{-1}(\varvec{\rho }_{0l})\frac{1}{n_l^{1/2}}\sum _{j=1}^{n_l}\varphi _j(\varvec{\rho }_{0l}) \\ +o_p(1) := E_{l2}+E_{l3,1}+o_p(1), \end{aligned}$$
and the \(o_p(1)\) in above equation is obtained based on the fact
$$\begin{aligned} \frac{1}{{n_l}^{1/2}}\sum _{i=1}^{n_l}\frac{\xi _{li}\delta _{li}+(1-\xi _{li})m_l(Z_{li},\textbf{X}_{li}; \varvec{\rho }_{0l})}{1-G_l(Z_{li})}\Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau \Big ]R_{G_l}(Z_{li})=o_p(1) \end{aligned}$$
and \(\frac{1}{n_l^{3/2}}\sum _{j=1}^{n_l}\frac{\xi _{lj}\delta _{lj}+(1-\xi _{lj})m_l(Z_{lj},\textbf{X}_{lj};\varvec{\rho }_{0\,l})}{1-G_l(Z_{lj})}\big [\Lambda _l\big (\frac{F_l^{-1}(\tau ) -Z_{lj}}{a_l}\big )-\tau \big ]\Gamma _l(Z_{lj},\textbf{X}_{lj},\xi _{lj},\xi _{lj}\delta _{lj};Z_{lj})=o_p(1)\). From Lemma 6.1 and using the Taylor expansion, it follows
$$\begin{aligned} S_3 =&\frac{1}{n_l}\sum _{i=1}^{n_l}\frac{(1-\xi _{li})\bigtriangledown m_l^{\mathrm T}(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})}{1-G_l(Z_{li})}\Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big ) -\tau \Big ]\Big [I_l^{-1}(\varvec{\rho }_{0l})\frac{1}{n_l^{1/2}}\sum _{j=1}^{n_l}\varphi _j(\varvec{\rho }_{0l})\\&+o_p(1)\Big ]+O_p\Big (\frac{1}{\sqrt{n_l}}\Big )\\ =&\frac{1}{n_l}\sum _{i=1}^{n_l}\frac{(1-\xi _{li})\bigtriangledown m_l^{\mathrm T}(Z_{li},\textbf{X}_{li}; \varvec{\rho }_{0l})}{1-G_l(Z_{li})}\Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big ) -\tau \Big ]I_l^{-1}(\varvec{\rho }_{0l})\frac{1}{n_l^{1/2}}\sum _{j=1}^{n_l}\varphi _j(\varvec{\rho }_{0l})+o_p(1)\\ :=&E_{l3,2}+o_p(1), \end{aligned}$$
and the remainder term \(O_p(n_l^{-1/2})\) is obtained by (C1)(2) and the fact \(\widehat{\varvec{\rho }}_{0\,l}-\varvec{\rho }_{0\,l}=O_p(n_l^{-1/2})\). Further
$$\begin{aligned}&E_{l3,1}+E_{l3,2}\\&\quad =\frac{1}{n_l}\sum _{i=1}^{n_l}\Big \{\Big [\frac{(1-\xi _{li})\bigtriangledown m_l^{\mathrm T}(Z_{li},\textbf{X}_{li}; \varvec{\rho }_{0l})}{1-G_l(Z_{li})}-\frac{\xi _{li}\delta _{li}+(1-\xi _{li})m_l(Z_{li},\textbf{X}_{li}; \varvec{\rho }_{0l})}{1-G_l(Z_{li})}\mu _l^{\mathrm T}(Z_{li})\Big ]\\&\qquad \times \Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau \Big ]\Big \}I_l^{-1}(\varvec{\rho }_{0l}) \frac{1}{n_l^{1/2}}\sum _{j=1}^{n_l}\varphi _j(\varvec{\rho }_{0l})\\&\quad =E\Big \{\Big [\frac{(1-\xi _{l})\bigtriangledown m_l^{\mathrm T}(Z_{l},\textbf{X}_{l};\varvec{\rho }_{0l})}{1-G_l(Z_{l})}-\frac{\xi _{l}\delta _{l}+(1-\xi _{l})m_l(Z_{l},\textbf{X}_{l};\varvec{\rho }_{0l})}{1-G_l(Z_{l})} \mu _l^{\mathrm T}(Z_{l})\Big ]\Big [\Lambda _{l}\Big (\frac{F_l^{-1}(\tau )-Z_l}{a_l}\Big )-\tau \Big ]\Big \}\\&\qquad \times I_l^{-1}(\varvec{\rho }_{0l})\frac{1}{n_l^{1/2}}\sum _{j=1}^{n_l}\varphi _j(\varvec{\rho }_{0l})+o_p(1):=E_{l3,3} I_l^{-1}(\varvec{\rho }_{0l})\frac{1}{n_l^{1/2}}\sum _{j=1}^{n_l}\varphi _j(\varvec{\rho }_{0l})+o_p(1). \end{aligned}$$
As for \(E_{l3,3}\), we have
$$\begin{aligned} E_{l3,3} =&\,E\Big \{\Big [\frac{(1-\xi _{l})\bigtriangledown m_l^{\mathrm T}(Z_{l},\textbf{X}_{l}; \varvec{\rho }_{0l})}{1-G_l(Z_{l})}-\frac{\delta _l}{1-G_l(Z_{l})}\mu _l^{\mathrm T}(Z_{l})\Big ] \big [I(Z_l\le F_l^{-1}(\tau ))-\tau \big ]\Big \}\\&+E\Big \{\Big [\frac{(1-\xi _{l})\bigtriangledown m_l^{\mathrm T}(Z_{l},\textbf{X}_{l}; \varvec{\rho }_{0l})}{1-G_l(Z_{l})}-\frac{\delta _l}{1-G_l(Z_{l})}\mu _l^{\mathrm T}(Z_{l})\Big ] \\&\quad \times \Big [\Lambda _{l}\Big (\frac{F_l^{-1}(\tau )-Z_l}{a_l}\Big )-I(Z_l\le F_l^{-1}(\tau ))\Big ]\Big \} :=\alpha _l^{\mathrm T}+E_{l3,4}. \end{aligned}$$
Next, we show \(E_{l3,4}=O(a_l)\). It is sufficient to show that
$$\begin{aligned}&E_{l3,4}^{(1)}:=E\Big \{\frac{(1-\xi _{l})\bigtriangledown m_l^{\mathrm T}(Z_{l},\textbf{X}_{l};\varvec{\rho }_{0l})}{1-G_l(Z_{l})}\Big [\Lambda _{l}\Big (\frac{F_l^{-1}(\tau )-Z_l}{a_l}\Big )-I(Z_l\le F_l^{-1}(\tau ))\Big ]\Big \}=O(a_l),\\&E_{l3,4}^{(2)}:=E\Big \{\frac{\delta _l}{1-G_l(Z_{l})}\mu _l^{\mathrm T}(Z_{l}) \Big [\Lambda _{l}\Big (\frac{F_l^{-1}(\tau )-Z_l}{a_l}\Big )-I(Z_l\le F_l^{-1}(\tau ))\Big ]\Big \}=O(a_l). \end{aligned}$$
By (C3), we know that \(w_l(\cdot )\) are density functions with compact supports. Without loss of generality, assume that the support of \(w_l(\cdot )\) is \([\underline{c}_l,{\overline{c}}_l]\), then
$$\begin{aligned} \Lambda _l\Big (\frac{F_l^{-1}(\tau )-y}{a_l}\Big )=\left\{ \begin{array}{ll} 1, &{} \text{ if }~ y< F_l^{-1}(\tau )-a_l{\overline{c}}_l,\\ \Lambda _l\Big (\frac{F_l^{-1}(\tau )-y}{a_l}\Big ), &{} \text{ if }~ F_l^{-1}(\tau ) -a_l{\overline{c}}_l\le y\le F_l^{-1}(\tau )-a_l\underline{c}_l,\\ 0, &{} \text{ if }~ y>F_l^{-1}(\tau )-a_l\underline{c}_l.\\ \end{array}\right. \end{aligned}$$
There are three cases for the choices of \(\{\underline{c}_l,{\overline{c}}_l\}\): (i) \(\underline{c}_l\ge 0, {\overline{c}}_l\ge 0\); (ii) \(\underline{c}_l< 0, {\overline{c}}_l\ge 0\); (iii) \(\underline{c}_l\le 0, {\overline{c}}_l\le 0\). Here, we give the proof of the conclusion when \(\underline{c}_l\ge 0, {\overline{c}}_l\ge 0\), and others can be proved similarly.
$$\begin{aligned} \Vert E_{l3,4}^{(2)}\Vert =&\,\Big \Vert E\Big \{\mu _l^{\mathrm T}(Y_{l})\Big [\Lambda _{l}\Big (\frac{F_l^{-1}(\tau )-Y_l}{a_l}\Big )-I(Y_l\le F_l^{-1}(\tau ))\Big ]\Big \}\Big \Vert \\ =&\,\Big \Vert \Big [\int _{-\infty }^{F_l^{-1}(\tau )-a_l{\overline{c}}_l}+\int _{F_l^{-1}(\tau ) -a_l{\overline{c}}_l}^{F_l^{-1}(\tau )-a_l\underline{c}_l}+\int _{F_l^{-1}(\tau ) -a_l\underline{c}_l}^{F_l^{-1}(\tau )}+\int _{F_l^{-1}(\tau )}^{\infty }\Big ]\\&\times \mu _l^{\mathrm T}(y)\Big [\Lambda _{l}\Big (\frac{F_l^{-1}(\tau )-y}{a_l}\Big )-I(y\le F_l^{-1}(\tau ))\Big ]f_l(y)dy\Big \Vert \\ =&\,\Big \Vert \int _{F_l^{-1}(\tau )-a_l{\overline{c}}_l}^{F_l^{-1}(\tau )}\mu _l^{\mathrm T}(y) \Big [\Lambda _{l}\Big (\frac{F_l^{-1}(\tau )-y}{a_l}\Big )-I(y\le F_l^{-1}(\tau ))\Big ]f_l(y)dy\Big \Vert =O(a_l). \end{aligned}$$
Similarly to the proof as for \(E_{l3,4}^{(2)}\), we have
$$\begin{aligned} \Vert E_{l3,4}^{(1)}\Vert \le&\, C E\Big |\Lambda _{l}\Big (\frac{F_l^{-1}(\tau )-Z_l}{a_l}\Big )-I(Z_l\le F_l^{-1}(\tau ))\Big |\\ \le&\, C \Big [\int _{-\infty }^{F_l^{-1}(\tau )-a_l{\overline{c}}_l}+\int _{F_l^{-1}(\tau ) -a_l{\overline{c}}_l}^{F_l^{-1}(\tau )-a_l\underline{c}_l}+\int _{F_l^{-1}(\tau ) -a_l\underline{c}_l}^{F_l^{-1}(\tau )}+\int _{F_l^{-1}(\tau )}^{\infty }\Big ] \\&\times \Big |\Lambda _l\Big (\frac{F_l^{-1}(\tau )-z}{a_l}\Big )-I(z\le F_l^{-1}(\tau ))\Big |h_l(z)dz\\ =&\,C \int _{F_l^{-1}(\tau )-a_l{\overline{c}}_l}^{F_l^{-1}(\tau )} \Big |\Lambda _l\Big (\frac{F_l^{-1}(\tau )-z}{a_l}\Big )-I(z\le F_l^{-1}(\tau ))\Big |h_l(z)dz=O(a_l), \end{aligned}$$
where \(h_l(z)=f_l(z)(1-G_l(z))+(1-F_l(z))g_l(z)\) is the density of the observable variable \(Z_l\). Hence, \(E_{l3,4}=E_{l3,4}^{(1)}-E_{l3,4}^{(2)}=O(a_l)\) and \(E_{l3,3}=\alpha _l^{\mathrm T}+O(a_l)\). Then, one can obtain (8.1) easily.
Based on (8.1), by the central limit theorem of U-statistics, it follows \(E_{l2}\xrightarrow {d} N(0,\Omega _{l2})\). In addition, with the aid of central limit theorem, we have \(E_{l1}\xrightarrow {d} N(0,\Omega _{l1})\), \(E_{l3}\xrightarrow {d} N(0,\Omega _{l3})\). Under the MAR assumption, it can be checked that \(\textrm{Cov}(E_{l1},E_{l2}) \rightarrow \Omega _{l12}\), \(\textrm{Cov}(E_{l1},E_{l3}) \rightarrow \Omega _{l13}\), \(\textrm{Cov}(E_{l2},E_{l3}) \rightarrow \Omega _{l23}.\) Hence, the conclusion holds. \(\square \)
Proof of Lemma 6.3
We only prove (a) for \(l=2\). In fact,
$$\begin{aligned}&\frac{1}{n_2} \sum _{j=1}^{n_2} {\widehat{W}}^2_{2j}(\theta ,\Delta )\\&\quad =\frac{1}{n_2}\sum _{j=1}^{n_2} \frac{\xi _{2j}\delta _{2j}+(1-\xi _{2j})m_2^2(Z_{2j},\textbf{X}_{2j};\varvec{\rho }_{02})}{[1-G_2(Z_{2j})]^2}\Big [\Lambda _2\Big (\frac{\Delta -Z_{2j}}{a_2}\Big )-\tau \Big ]^2\\&\qquad +\frac{1}{n_2}\sum _{j=1}^{n_2}\frac{[\xi _{2j}\delta _{2j}+(1-\xi _{2j})m_2^2(Z_{2j},\textbf{X}_{2j};\varvec{\rho }_{02})][(1-G_2(Z_{2j}))^2-(1-{\widehat{G}}_2(Z_{2j}))^2]}{[1-{\widehat{G}}_2(Z_{2j})]^2[1-G_2(Z_{2j})]^2}\\&\qquad \times \Big [\Lambda _2\Big (\frac{\Delta -Z_{2j}}{a_2}\Big )-\tau \Big ]^2\\&\qquad +\frac{1}{n_2}\sum _{j=1}^{n_2}\frac{(1-\xi _{2j})[m_2^2(Z_{2j},\textbf{X}_{2j};\widehat{\varvec{\rho }}_{02})-m_2^2(Z_{2j},\textbf{X}_{2j};\varvec{\rho }_{02})]}{[1-{\widehat{G}}_2(Z_{2j})]^2}\Big [\Lambda _2\Big (\frac{\Delta -Z_{2j}}{a_2}\Big )-\tau \Big ]^2\\&\;\; := R_1+R_2+R_3. \end{aligned}$$
From Proposition 2.1 and Lemma 6.1, it is obvious that \(R_2\le C\max _{1\le j\le n_2} |{\widehat{G}}_2(Z_{2j})-G_2(Z_{2j})|=o_p(1)\) and
$$R_3\le C\max _{1\le j\le n_2} |{m}_2(Z_{2j},\textbf{X}_{2j};\widehat{\varvec{\rho }}_{02})-m_2(Z_{2j},\textbf{X}_{2j};{\varvec{\rho }}_{02})|=o_p(1)$$
. Now, we show \(R_1=\Omega _{21}+o_p(1)\). In fact,
$$\begin{aligned} R_1=&\,E\Big \{\frac{\xi _{2}\delta _{2}+(1-\xi _{2})m_2^2(Z_{2},\textbf{X}_{2};\varvec{\rho }_{02})}{[1-G_2(Z_{2})]^2}\Big [\Lambda _2\Big (\frac{\Delta -Z_{2}}{a_2}\Big ) -\tau \Big ]^2\Big \}+o_p(1)\\ =&\,E\Big \{\frac{\xi _{2}\delta _{2}+(1-\xi _{2})m_2^2(Z_{2},\textbf{X}_{2};\varvec{\rho }_{02})}{[1-G_2(Z_{2})]^2}\big [I(Z_2\le F_2^{-1}(\tau ))-\tau \big ]^2\Big \}\\&+E\Big \{\frac{\xi _{2}\delta _{2}+(1-\xi _{2})m_2^2(Z_{2},\textbf{X}_{2};\varvec{\rho }_{02})}{[1-G_2(Z_{2})]^2}\Big \{\Big [\Lambda _2\Big (\frac{\Delta -Z_{2}}{a_2}\Big )-\tau \Big ]^2 -\big [I(Z_2\le F_2^{-1}(\tau ))-\tau \big ]^2\Big \}\Big \}+o_p(1)\\ :=&\,\Omega _{21}+R_1^{(2)}+o_p(1). \end{aligned}$$
Next, we prove \(R_1^{(2)}=O(a_2)\), whose proof is similar to \(E_{l3,4}^{(2)}\). Note that when \(\Theta \in \{\Theta :\Vert \Theta - \Theta _0\Vert \le \varepsilon _n\}\), we have \(|\Delta -F_2^{-1}(\tau )|\le \varepsilon _n=o(a_2)\) (see condition (C5)). Then
$$\begin{aligned} R_1^{(2)} \le&\, E\Big \{\frac{\delta _2}{[1-G_2(Z_{2})]^2}\Big |\Big [\Lambda _2\Big (\frac{\Delta -Z_{2}}{a_2}\Big )-\tau \Big ]^2 -\big [I(Z_2\le F_2^{-1}(\tau ))-\tau \big ]^2\Big |\Big \}\\ =&\, E\Big \{\frac{1}{1-G_2(Y_{2})}\Big |\Big [\Lambda _2\Big (\frac{\Delta -Y_{2}}{a_2}\Big )-\tau \Big ]^2 -\big [I(Y_2\le F_2^{-1}(\tau ))-\tau \big ]^2\Big |\Big \}\\ =&\, \Big [\int _{-\infty }^{F_2^{-1}(\tau )-a_2{\overline{c}}_2}+\int _{F_2^{-1}(\tau )-a_2{\overline{c}}_2}^{F_2^{-1}(\tau ) -a_2\underline{c}_2}+\int _{F_2^{-1}(\tau )-a_2\underline{c}_2}^{F_2^{-1}(\tau )}+\int _{F_2^{-1}(\tau )}^{\infty }\Big ]\\&\times \frac{f_2(y)}{1-G_2(y)}\Big |\Big [\Lambda _2\Big (\frac{\Delta -y}{a_2}\Big )-\tau \Big ]^2 -\big [I(y\le F_2^{-1}(\tau ))-\tau \big ]^2\Big |dy\\ =&\, \int _{F_2^{-1}(\tau )-a_2{\overline{c}}_2}^{F_2^{-1}(\tau )} \frac{f_2(y)}{1-G_2(y)}\Big |\Big [\Lambda _2\Big (\frac{\Delta -y}{a_2}\Big )-\tau \Big ]^2-\big [I(y\le F_2^{-1}(\tau ))-\tau \big ]^2\Big |dy=O(a_2). \end{aligned}$$
Hence, \( \frac{1}{n_2} \sum _{j=1}^{n_2} {\widehat{W}}^2_{2j}(\theta ,\Delta ) =\Omega _{21}+o_p(1) \), and the first claim is proved.
Using \(\log n_2/(n_2a_2)\rightarrow 0\), from (C1)–(C4), one can deduce
$$\begin{aligned}&\frac{1}{n_2}\sum _{j=1}^{n_2} {\widehat{W}}'_{2j}(\theta ,\Delta ) =\frac{1}{n_2}\sum _{j=1}^{n_2} \frac{\widehat{\delta }_{2j}}{1-{\widehat{G}}_2(Z_{2j})} \frac{1}{a_2}w_2\Big (\frac{\Delta -Z_{2j}}{a_2}\Big )\\&\quad =\frac{1}{n_2} \sum _{j=1}^{n_2} \frac{\xi _{2j}\delta _{2j}+(1-\xi _{2j})m_2(Z_{2j}, \textbf{X}_{2j};\varvec{\rho }_{02})}{1-G_2(Z_{2j})}\frac{1}{a_2}w_2\Big (\frac{\Delta -Z_{2j}}{a_2}\Big )\\&\qquad +\frac{1}{n_2}\sum _{j=1}^{n_2} \frac{(1-\xi _{2j})[m_2(Z_{2j},\textbf{X}_{2j}; \widehat{\varvec{\rho }}_{02})-m_2(Z_{2j},\textbf{X}_{2j};\varvec{\rho }_{02})]}{1-G_2(Z_{2j})} \frac{1}{a_2}w_2\Big (\frac{\Delta -Z_{2j}}{a_2}\Big )\\&\qquad +\frac{1}{n_2}\sum _{j=1}^{n_2} \frac{\xi _{2j}\delta _{2j}+(1-\xi _{2j})m_2(Z_{2j},\textbf{X}_{2j}; \widehat{\varvec{\rho }}_{02})}{[1-G_2(Z_{2j})][1-{\widehat{G}}_2(Z_{2j})]} [{\widehat{G}}_2(Z_{2j})-G_2(Z_{2j})]\frac{1}{a_2}w_2\Big (\frac{\Delta -Z_{2j}}{a_2}\Big )\\&\quad =\frac{1}{n_2}\sum _{j=1}^{n_2} \frac{\xi _{2j}\delta _{2j}+(1-\xi _{2j})m_2(Z_{2j},\textbf{X}_{2j}; \varvec{\rho }_{02})}{1-G_2(Z_{2j})}\frac{1}{a_2}w_2\Big (\frac{\Delta -Z_{2j}}{a_2}\Big ) +O_p\Big (\sqrt{\frac{\log n_2}{n_2}}\Big )\\&\quad =f_2(F_2^{-1}(\tau ))+o_p(1). \end{aligned}$$
Finally,
$$\begin{aligned}&\frac{1}{n_2}\sum _{j=1}^{n_2} {\widehat{W}}_{2j}(\theta ,\Delta ) = \frac{1}{n_2}\sum _{j=1}^{n_2} \frac{\widehat{\delta }_{2j}}{1-{\widehat{G}}_2(Z_{2j})} \Big [\Lambda _2\Big (\frac{\Delta -Z_{2j}}{a_2}\Big )-\tau \Big ]\\&\quad =\frac{1}{n_2}\sum _{j=1}^{n_2} \frac{\xi _{2j}\delta _{2j}+(1-\xi _{2j})m_2(Z_{2j},\textbf{X}_{2j}; \varvec{\rho }_{02})}{1-G_2(Z_{2j})}\Big [\Lambda _2\Big (\frac{\Delta _\tau -Z_{2j}}{a_2}\Big )-\tau \Big ]\\&\qquad + \frac{1}{n_2}\sum _{j=1}^{n_2}\frac{\xi _{2j}\delta _{2j}+(1-\xi _{2j})m_2(Z_{2j},\textbf{X}_{2j}; \varvec{\rho }_{02})}{1-G_2(Z_{2j})}\frac{1}{a_2}w_2\Big (\frac{\Delta ^*-Z_{2j}}{a_2}\Big )(\Delta -\Delta _\tau ) +O_p\Big (\sqrt{\frac{\log n_2}{n_2}}\Big ) \\&\;\; := I_1+I_2+O_p\Big (\sqrt{\frac{\log n_2}{n_2}}\Big ), \end{aligned}$$
where \(\Delta ^*\) is between \(\Delta \) and \(\Delta _\tau \). By (C2) and law of large numbers, we have \(I_2=[f_2(F_2^{-1}(\tau ))+o_p(1)](\Delta -\Delta _\tau )\). By (C1) and \(n_2a_2^{2r_2}\rightarrow 0\), it is easy to see that, for sufficiently large C,
$$\begin{aligned} P(|I_1|>Cn_2^{-1/2}) \le \frac{n_2a_2^{2r_2}+C_0}{C} \rightarrow 0, \end{aligned}$$
where \(C_0=\Omega _{21}+o(1)\), so \(I_1=O_p(n_2^{-1/2})\). Thus, the third claim is proved. \(\square \)
Proof of Lemma 6.4
We only show the conclusions hold when \(l=2\). In fact,
$$\begin{aligned}&0=\Big |\frac{1}{n_2}\sum _{j=1}^{n_2} \frac{{\widehat{W}}_{2j}(\theta ,\Delta )}{1+\lambda _2 {\widehat{W}}_{2j}(\theta ,\Delta )}\Big |\\&\quad \ge |\lambda _2|\frac{1}{n_2}\sum _{j=1}^{n_2} \frac{{\widehat{W}}^2_{2j}(\theta ,\Delta )}{1+|\lambda _2|\max _{1\le j\le n_2}|{\widehat{W}}_{2j}(\theta ,\Delta )|}-\Big |\frac{1}{n_2}\sum _{j=1}^{n_2}{\widehat{W}}_{2j}(\theta ,\Delta )\Big |. \end{aligned}$$
Note that under (C3)–(C4), combined with Proposition 2.1, we have
$$\begin{aligned} \max _{1\le j\le n_2}|{\widehat{W}}_{2j}(\theta ,\Delta )|=\max _{1\le j \le n_2}\Big |\frac{\widehat{\delta }_{2j}}{1-{\widehat{G}}_2(Z_{2j})} \Big [\Lambda _2\Big (\frac{\Delta -Z_{2j}}{a_2}\Big )-\tau \Big ]\Big |=O_p(1), \end{aligned}$$
and combined with Lemma 6.3, \(\lambda _2=O_p(\varepsilon _n)\) follows. In addition,
$$\begin{aligned} 0=\frac{1}{n_2}\sum _{j=1}^{n_2} {\widehat{W}}_{2j}(\theta ,\Delta ) - \lambda _2\frac{1}{n_2}\sum _{j=1}^{n_2}{\widehat{W}}_{2j}^2(\theta ,\Delta ) +O_p(\varepsilon _n^2), \end{aligned}$$
which yields \(\lambda _2 = \big [\frac{1}{n_2}\sum _{j=1}^{n_2}{\widehat{W}}^2_{2j}(\theta ,\Delta )\big ]^{-1}\frac{1}{n_2} \sum _{j=1}^{n_2}{\widehat{W}}_{2j}(\theta ,\Delta )+O_p(\varepsilon _n^2)\). \(\square \)
Proof of Lemma 6.5
To prove this Lemma, it is sufficient to show that for \(\theta -\theta _0=\varepsilon _n u_1\), \(\Delta -\Delta _\tau =\varepsilon _n u_2\) and \(u_1^2+u_2^2=1\), we have \(-l_1(\theta ,\Delta ) > -l_1(\theta _0,\Delta _\tau )\) and \(l_2(\theta ,\Delta )\ge -l_2(\theta _0,\Delta _\tau )\). Here, we only show the second claim. When \(u_2=0\), \(-l_2(\theta ,\Delta )=-l_2(\theta _0,\Delta _\tau )\). When \(u_2\ne 0\), by the Taylor expansion, one can get
$$\begin{aligned} -l_2(\theta ,\Delta )=&\,\frac{n_2}{2}\Big [\frac{1}{n_2}\sum _{j=1}^{n_2} {\widehat{W}}_{2j}^2(\theta ,\Delta )\Big ]^{-1}\Big [\frac{1}{n_2} \sum _{j=1}^{n_2}{\widehat{W}}_{2j}(\theta ,\Delta )\Big ]^2+O_p(n_2\varepsilon _n^3)\\ =&\, \frac{n_2}{2} \Big [\frac{1}{n_2}\sum _{j=1}^{n_2}{\widehat{W}}_{2j}^2(\theta ,\Delta )\Big ]^{-1} \Big [\frac{1}{n_2} \sum _{j=1}^{n_2}{\widehat{W}}_{2j}(\theta _0,\Delta _\tau )+\frac{1}{n_2} \sum _{j=1}^{n_2}{\widehat{W}}'_{2j}(\theta _0,\Delta _\tau ^*) (\Delta -\Delta _\tau )\Big ]^2\\&\quad +O_p(n_2\varepsilon _n^3)\\ =&\, \frac{n_2}{2} \Big [\frac{1}{n_2}\sum _{j=1}^{n_2} {\widehat{W}}^2_{2j}(\theta _0,\Delta _\tau )\Big ]^{-1} \Big [\frac{1}{n_2}\sum _{j=1}^{n_2} {\widehat{W}}_{2j}(\theta _0,\Delta _\tau )\Big ]^2+Cn_2\varepsilon _n^2\\ >&\, \frac{n_2}{2} \Big [\frac{1}{n_2}\sum _{j=1}^{n_2} {\widehat{W}}_{2j}^2(\theta _0,\Delta _\tau )\Big ]^{-1} \Big [\frac{1}{n_2}\sum _{j=1}^{n_2}{\widehat{W}}_{2j}(\theta _0,\Delta _\tau )\Big ]^2 +O_p(n_2\varepsilon _n^3) =-l_2(\theta _0,\Delta _\tau ), \end{aligned}$$
where \(\Delta _\tau ^*\) is between \(\Delta \) and \(\Delta _\tau \). By similar arguments, one can obtain \(-l_1(\theta ,\Delta ) > -l_1(\theta _0,\Delta _\tau )\). Moreover, \(l_n(\theta ,\Delta )\) is continuous. Hence, \(Q_{in}(\widehat{\theta }_0,\widehat{\Delta }_\tau )=0\) for \(i=1,\ldots ,4\). \(\square \)
Proof of Lemma 6.6
To prove the first part of this Lemma, it suffices to show \(-l_1(\theta _0,\Delta _\tau \pm \varepsilon _n) > -l_1(\theta _0,\Delta _\tau )\) and \(-l_2(\theta _0,\Delta _\tau \pm \varepsilon _n) > -l_2(\theta _0,\Delta _\tau )\), which can be proved similarly to Lemma 6.5.
Now we prove the second part of conclusion. For \(Q_{in}(\theta _0,{\widetilde{\Delta }}_\tau )=0\) (\(i=1,2\)) and \(Q_{3n}(\theta _0,{\widetilde{\Delta }}_\tau )+\frac{n_2}{n_1} Q_{4n}(\theta _0,{\widetilde{\Delta }}_\tau )=0\), by Taylor expansion, combined with Lemma 6.3, we have
$$\begin{aligned} \left( \begin{array}{c} 0\\ 0\\ 0 \end{array}\right) = \left( \begin{array}{c} \frac{1}{n_1}\sum _{i=1}^{n_1}{\widehat{W}}_{1i}(\theta _0,\Delta _\tau )\\ \frac{1}{n_2}\sum _{j=1}^{n_2}{\widehat{W}}_{2j}(\theta _0,\Delta _\tau )\\ 0 \end{array}\right) + S \left( \begin{array}{c} {\widetilde{\Delta }}_\tau - \Delta _\tau \\ {\widetilde{\lambda }}_1\\ {\widetilde{\lambda }}_2 \end{array}\right) + o_p(T_n), \end{aligned}$$
where \(T_n=|{\widetilde{\Delta }}_\tau -\Delta _\tau |+|{\widetilde{\lambda }}_1|+|{\widetilde{\lambda }}_2|\) and \(S=\left( \begin{array}{ccc} f_1(F_1^{-1}(\tau )) &{} -\Omega _{11} &{} 0\\ f_2(F_2^{-1}(\tau )) &{} 0 &{} -\Omega _{21}\\ 0 &{} f_1(F_1^{-1}(\tau )) &{} \gamma f_2(F_2^{-1}(\tau )) \end{array}\right) \). From Lemma 6.2, it follows that \(T_n=O_p(n^{-1/2})\) and
$$\begin{aligned} {\widetilde{\lambda }}_1&= \frac{\gamma f_2^2(F_2^{-1}(\tau ))n_1^{-1} \sum _{i=1}^{n_1} {\widehat{W}}_{1i}(\theta _0,\Delta _\tau ) -\gamma f_1(F_1^{-1}(\tau )) f_2(F_2^{-1}(\tau ))n_2^{-1}\sum _{j=1}^{n_2} {\widehat{W}}_{2j}(\theta _0,\Delta _\tau )}{\det } \\&\quad +o_p(n^{-1/2}),\\ {\widetilde{\lambda }}_2&= \frac{-f_1(F_1^{-1}(\tau )) f_2(F_2^{-1}(\tau ))n_1^{-1}\sum _{i=1}^{n_1} {\widehat{W}}_{1i}(\theta _0,\Delta _\tau )+f_1^2(F_1^{-1}(\tau )) n_2^{-1}\sum _{j=1}^{n_2} {\widehat{W}}_{2j}(\theta _0,\Delta _\tau )}{\det } \\&\quad +o_p(n^{-1/2}). \end{aligned}$$
Therefore, \({\widetilde{\lambda }}_1 = -\gamma \frac{f_2(F_2^{-1}(\tau ))}{f_1(F_1^{-1}(\tau ))}{\widetilde{\lambda }}_2 + o_p(n^{-1/2})\). Applying Lemma 6.2, the second conclusion can be verified. \(\square \)
Proof of Lemma 6.7
The proof is similar to the proof of Lemma 6.4, and we only prove the conclusions when \(l=2\). Let \(\lambda _2^A=\Vert \lambda _2^A\Vert \eta _2\), then
$$\begin{aligned} 0&=\eta _2^{\mathrm T} \frac{1}{n_2}\sum _{j=1}^{n_2}\frac{{\widehat{W}}_{2j}^A(\theta ,\Delta )}{1+\lambda _2^{A\mathrm T}{\widehat{W}}_{2j}^A(\theta ,\Delta )}\\&\ge \Vert \lambda _2^A\Vert \frac{\eta _2^{\mathrm T}\frac{1}{n_2}\sum _{j=1}^{n_2} {\widehat{W}}_{2j}^A(\theta ,\Delta ){\widehat{W}}_{2j}^{A\mathrm T}(\theta ,\Delta )\eta _2}{1+\Vert \lambda _2^A\Vert \max _{j}|\eta _2^{\mathrm T}{\widehat{W}}_{2j}^A(\theta ,\Delta )|}-\big |\eta _2^{\mathrm T} \frac{1}{n_2}\sum _{j=1}^{n_2}{\widehat{W}}_{2j}^A(\theta ,\Delta )\big |. \end{aligned}$$
Under (A1), one can check that \(\frac{1}{n_2}\sum _{j=1}^{n_2}{\widehat{W}}_{2j}^A(\theta ,\Delta ){\widehat{W}}_{2j}^{A^\mathrm T}(\theta ,\Delta )= \left( \begin{array}{cc} \Omega _{21} &{} \sigma _{\beta _2}^{\mathrm T}\\ \sigma _{\beta _2} &{} \sigma _{\beta _2\beta _2} \end{array}\right) +o_p(1)\). Combined with (C3)–(C4), (A2) and Lemma 6.5, we have \(\Vert \lambda _2^A\Vert =O_p(\varepsilon _n)\). Furthermore
$$\begin{aligned} \textbf{0}&=\frac{1}{n_2}\sum _{j=1}^{n_2}\frac{{\widehat{W}}_{2j}^A(\theta ,\Delta )}{1+\lambda _2^{A\mathrm T}{\widehat{W}}_{2j}^A(\theta ,\Delta )}\\&=\frac{1}{n_2}\sum _{j=1}^{n_2}{\widehat{W}}_{2j}^A(\theta ,\Delta )-\frac{1}{n_2} \sum _{j=1}^{n_2}{\widehat{W}}_{2j}^A(\theta ,\Delta ) {\widehat{W}}_{2j}^{A\mathrm T}(\theta ,\Delta )\lambda _2^A+O_p(\varepsilon _n^2), \end{aligned}$$
therefore, \(\lambda _2^A=\big [\frac{1}{n_2}\sum _{j=1}^{n_2}{\widehat{W}}_{2j}^A(\theta ,\Delta ) {\widehat{W}}_{2j}^{A\mathrm T}(\theta ,\Delta )\big ]^{-1} \frac{1}{n_2}\sum _{j=1}^{n_2}{\widehat{W}}_{2j}^A(\theta ,\Delta )+O_p(\varepsilon _n^2)\). Similarly, one can verify the conclusions for \(l=1\). \(\square \)
Proof of Lemma 6.7
The proof is similar to Lemma 6.5. To prove this Lemma, it is sufficient to show that for \(\theta -\theta _0=\varepsilon _n u_1\), \(\Delta -\Delta _\tau =\varepsilon _n u_2\) and \(u_1^2+u_2^2=1\), we have
$$\begin{aligned} -{{\overline{l}}}_1^A(\theta ,\Delta ) > -{{\overline{l}}}_1^A(\theta _0,\Delta _\tau )~~\text{ and }~~-{{\overline{l}}}_2^A(\theta ,\Delta )\ge -{{\overline{l}}}_2^A(\theta _0,\Delta _\tau ), \end{aligned}$$
where \({{\overline{l}}}_1^A(\theta ,\Delta )=-\sum _{i=1}^{n_1}\log (1+\lambda _1^{A\mathrm T}{\widehat{W}}_{1i}^A(\theta ,\Delta ))\), \({{\overline{l}}}_2^A(\theta ,\Delta )=-\sum _{j=1}^{n_2}\log (1+\lambda _2^{A\mathrm T}{\widehat{W}}_{2j}^A(\theta ,\Delta ))\).
When \(u_2=0\), \(-{{\overline{l}}}_2^A(\theta ,\Delta )=-{{\overline{l}}}_2^A(\theta _0,\Delta _\tau )\). When \(u_2\ne 0\), by Taylor expansion, similar to Lemma 6.5, we have \( -{{\overline{l}}}_2^A(\theta ,\Delta )=Cn\varepsilon _n^2>-{{\overline{l}}}_2^A(\theta _0,\Delta _\tau )=o_p(n\varepsilon _n^2). \) Similarly, one can obtain that \(-{{\overline{l}}}_1^A(\theta ,\Delta )>-{{\overline{l}}}_1^A(\theta _0,\Delta _\tau )\). Therefore, the conclusion holds. \(\square \)
Proof of Lemma 6.9
The proof of the first part is similar to that of Lemmas 6.6 and 6.8. Now we prove the second conclusion. Applying Taylor expansion to \(Q_{in}^A(\theta _0,{\widetilde{\Delta }}_\tau ^A)=\textbf{0}\) for \(i=1,2\), and \(Q_{3n}^A(\theta _0,{\widetilde{\Delta }}_\tau ^A)+\frac{n_2}{n_1}Q_{4n}^A(\theta _0,{\widetilde{\Delta }}_\tau ^A)=0\), one can obtain
$$\begin{aligned} \left( \begin{array}{c} \textbf{0}\\ \textbf{0}\\ 0 \end{array}\right) =\left( \begin{array}{c} \frac{1}{n_1}\sum _{i=1}^{n_1}{\widehat{W}}_{1i}^A(\theta _0,\Delta _\tau )\\ \frac{1}{n_2}\sum _{j=1}^{n_2}{\widehat{W}}_{2j}^A(\theta _0,\Delta _\tau )\\ 0 \end{array}\right) +S^A\left( \begin{array}{c} {\widetilde{\Delta }}_\tau ^A-\Delta _\tau \\ {\widetilde{\lambda }}_1^A\\ {\widetilde{\lambda }}_2^A \end{array}\right) +o_p(T_n^A), \end{aligned}$$
where \(T_n^A=|{\widetilde{\Delta }}_{\tau }^A-\Delta _\tau |+\Vert {\widetilde{\lambda }}_1^A\Vert +\Vert {\widetilde{\lambda }}_2^A\Vert \) and
$$\begin{aligned} S^A=\left( \begin{array}{ccccc} f_1(F_1^{-1}(\tau )) &{} -\Omega _{11} &{} -\sigma _{\beta _1}^{\mathrm T} &{} 0 &{} \textbf{0}\\ \textbf{0} &{} -\sigma _{\beta _1} &{} -\sigma _{\beta _1\beta _1} &{} \textbf{0} &{} \textbf{0}\\ f_2(F_2^{-1}(\tau )) &{} 0 &{} \textbf{0} &{}-\Omega _{21} &{} -\sigma _{\beta _2}^{\mathrm T}\\ \textbf{0} &{} \textbf{0} &{} \textbf{0} &{} -\sigma _{\beta _2} &{} -\sigma _{\beta _2\beta _2}\\ 0 &{} f_1(F_1^{-1}(\tau )) &{} \textbf{0} &{} \gamma f_2(F_2^{-1}(\tau )) &{}\textbf{0} \end{array} \right) . \end{aligned}$$
From Lemma 6.2, we have \(T_n^A=O_p(n^{-1/2})\). Based on knowledge of block matrices, we have
$$\begin{aligned} {\widetilde{\Delta }}_\tau ^A - \Delta _\tau =&-\frac{1}{\det ^A} \Big \{\frac{f_1(F_1^{-1}(\tau ))}{\Omega _{11}^{A}} \frac{1}{n_1} \sum _{i=1}^{n_1}\big [{\widehat{W}}_{1i}(\theta _0,\Delta _\tau )-\sigma _{\beta _1}^{\mathrm T}\sigma _{\beta _1\beta _1}^{-1}{\overline{W}}_{1i}^A\big ]\\&+\frac{\gamma f_2(F_2^{-1}(\tau ))}{\Omega _{21}^{A}} \frac{1}{n_2} \sum _{j=1}^{n_2}\big [{\widehat{W}}_{2j}(\theta _0,\Delta _\tau )-\sigma _{\beta _2}^{\mathrm T} \sigma _{\beta _2\beta _2}^{-1}{\overline{W}}_{2j}^A\big ]\Big \}+o_p(n^{-1/2}), \end{aligned}$$
where \(\det ^A=\frac{f_1^2(F_1^{-1}(\tau ))}{\Omega _{11}^{A}}+\frac{\gamma f_2^2(F_2^{-1}(\tau ))}{\Omega _{21}^{A}}\). \(\square \)
Proof of Lemma 6.10
The proof is similar to the proof of Lemma 6.2. In fact,
$$\begin{aligned}&\frac{1}{\sqrt{n_l}} \sum _{i=1}^{n_l} \big [{\widehat{W}}_{li}(\theta _0,\Delta _{\tau })-\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}{\overline{W}}_{li}^A\big ]\\&\quad =\frac{1}{\sqrt{n_l}}\sum _{i=1}^{n_l} \frac{\xi _{li}\delta _{li}+(1-\xi _{li})m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})}{1-{G}_l(Z_{li})}\Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_{li})\Big ]\\&\qquad + \frac{1}{\sqrt{n_l}}\sum _{i=1}^{n_l} \frac{\xi _{li}\delta _{li}+(1-\xi _{li})m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})}{1-{G}_l(Z_{li})}\frac{{\widehat{G}}_l(Z_{li})-G_l(Z_{li})}{1-G_l(Z_{li})}\\&\qquad \times \Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_{li})\Big ]\\&\qquad +\frac{1}{\sqrt{n_l}} \sum _{i=1}^{n_l} \frac{(1-\xi _{li})[m_l(Z_{li},\textbf{X}_{li};\widehat{\varvec{\rho }}_{0l})-m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})]}{1-G_l(Z_{li})}\Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_{li})\Big ]\\&\qquad +\frac{1}{\sqrt{n_l}}\sum _{i=1}^{n_l}\frac{[\xi _{li}\delta _{li}+(1-\xi _{li})m_l({Z_{li}},\textbf{X}_{li};\varvec{\rho }_{0l})][{\widehat{G}}_l(Z_{li})-G_l(Z_{li})]^2}{[1-G_l(Z_{li})]^2[1-{\widehat{G}}_l(Z_{li})]}\\&\qquad \times \Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_{li})\Big ]\\&\qquad +\frac{1}{\sqrt{n_l}}\sum _{i=1}^{n_l}\frac{(1-\xi _{li})[m_l(Z_{li},\textbf{X}_{li};\widehat{\varvec{\rho }}_{0l})-m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})][{\widehat{G}}_l(Z_{li})-G_l(Z_{li})]}{[1-G_l(Z_{li})][1-{\widehat{G}}_l(Z_{li})]}\\&\qquad \times \Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_{li})\Big ] :=D_1^A+D_2^A+D_3^A+D_4^A+D_5^A. \end{aligned}$$
It is easy to see that \(D_4^A=O_p(\log n_l/\sqrt{n_l})=o_p(1)\) and \(D_5^A=O_p(\sqrt{\log n_l/n_l})=o_p(1)\). By Proposition 2.1 and Lemma 6.1, we have
$$\begin{aligned} D_2^A=&\frac{1}{n_l^{3/2}}\sum _{j<k}\Big \{\frac{\xi _{lj}\delta _{lj}+(1-\xi _{lj})m_l(Z_{lj},\textbf{X}_{lj};\varvec{\rho }_{0l})}{1-G_l(Z_{lj})}\Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{lj}}{a_l}\Big )-\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_{lj})\Big ]\\&\times \Gamma _l(Z_{lk},\textbf{X}_{lk},\xi _{lk},\xi _{lk}\delta _{lk};Z_{lj})+\frac{\xi _{lk}\delta _{lk}+(1-\xi _{lk})m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})}{1-G_l(Z_{lk})}\\&\times \Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{lk}}{a_l}\Big )-\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_{lk})\Big ]\Gamma _l(Z_{lj},\textbf{X}_{lj},\xi _{lj},\xi _{lj}\delta _{lj};Z_{lk})\Big \}\\&-E\Big [\frac{\delta _l}{1-G_l(Z_l)}[I(Z_l\le F_l^{-1}(\tau ))-\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_l)]\mu _l^{\mathrm T}(Z_l)\Big ]I_l^{-1}(\varvec{\rho }_{0l})\\&\times \frac{1}{\sqrt{n_l}}\sum _{k=1}^{n_l} \frac{\xi _{lk}[\delta _{lk}-m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})]\nabla m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})}{m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})[1-m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})]}+o_p(1), \end{aligned}$$
and
$$\begin{aligned} D_3^A=&E\Big [\frac{(1-\xi _l)[I(Z_l\le F_l^{-1}(\tau )) -\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_l)]\nabla m_l^{\mathrm T}(Z_{l},\textbf{X}_{l}; \varvec{\rho }_{0l})}{1-G_l(Z_l)}\Big ]I_l^{-1}(\varvec{\rho }_{0l})\\&\times \frac{1}{\sqrt{n_l}}\sum _{k=1}^{n_l} \frac{\xi _{lk}[\delta _{lk}-m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})]\nabla m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})}{m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})[1-m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})]}+o_p(1). \end{aligned}$$
Therefore
$$\begin{aligned}&\frac{1}{\sqrt{n_l}} \sum _{i=1}^{n_l} \big [{\widehat{W}}_{li}(\theta _0,\Delta _{\tau }) -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}{\overline{W}}_{li}^A\big ]\\&\quad\quad =\frac{1}{\sqrt{n_l}}\sum _{i=1}^{n_l} \frac{\xi _{li}\delta _{li}+(1-\xi _{li})m_l(Z_{li},\textbf{X}_{li};\varvec{\rho }_{0l})}{1-{G}_l(Z_{li})}\Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{li}}{a_l}\Big )-\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_{li})\Big ]\\&\qquad +\frac{1}{n_l^{3/2}}\sum _{j<k}\Big \{\frac{\xi _{lj}\delta _{lj}+(1-\xi _{lj})m_l(Z_{lj},\textbf{X}_{lj};\varvec{\rho }_{0l})}{1-G_l(Z_{lj})}\Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{lj}}{a_l}\Big )-\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_{lj})\Big ]\\&\qquad \times \Gamma _l(Z_{lk},\textbf{X}_{lk},\xi _{lk},\xi _{lk}\delta _{lk};Z_{lj})+\frac{\xi _{lk}\delta _{lk}+(1-\xi _{lk})m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})}{1-G_l(Z_{lk})}\\&\qquad \times \Big [\Lambda _l\Big (\frac{F_l^{-1}(\tau )-Z_{lk}}{a_l}\Big )-\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_{lk})\Big ]\Gamma _l(Z_{lj},\textbf{X}_{lj},\xi _{lj},\xi _{lj}\delta _{lj};Z_{lk})\Big \}\\&\qquad +E\Big \{\frac{(1-\xi _l)[I(Z_l\le F_l^{-1}(\tau ))-\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_l)]\nabla m_l^{\mathrm T}(Z_{l},\textbf{X}_{l};\varvec{\rho }_{0l})}{1-G_l(Z_l)}\\&\qquad -\frac{\delta _l}{1-G_l(Z_l)}[I(Z_l\le F_l^{-1}(\tau ))-\tau -\sigma _{\beta _l}^{\mathrm T}\sigma _{\beta _l\beta _l}^{-1}\beta _l(Z_l)]\mu _l^{\mathrm T}(Z_l) \Big \}I_l^{-1}(\varvec{\rho }_{0l})\\&\qquad \times \frac{1}{\sqrt{n_l}}\sum _{k=1}^{n_l} \frac{\xi _{lk}[\delta _{lk}-m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})]\nabla m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})}{m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})[1-m_l(Z_{lk},\textbf{X}_{lk};\varvec{\rho }_{0l})]}+o_p(1)\\&\;\; :=E_{l1}^A+E_{l2}^A+E_{l3}^A+o_p(1). \end{aligned}$$
By the central limit theorem of U-statistics, it follows \(E_{l2}^A\xrightarrow {d} N(0,\Omega _{l2}^A)\). In addition, with the aid of central limit theorem, we have \(E_{l1}^A\xrightarrow {d} N(0,\Omega _{l1}^A)\), \(E_{l3}^A\xrightarrow {d} N(0,\Omega _{l3}^A)\). Under the MAR assumption, it can be checked that \(\textrm{Cov}(E_{l1}^A,E_{l2}^A) \rightarrow \Omega _{l12}^A\), \(\textrm{Cov}(E_{l1}^A,E_{l3}^A) \rightarrow \Omega _{l13}^A\), \(\textrm{Cov}(E_{l2}^A,E_{l3}^A) \rightarrow \Omega _{l23}^A.\) Hence, the conclusion holds. \(\square \)
