Generalized partial linear models with nonignorable dropouts

Abstract

In the presence of longitudinal data with nonignorable dropouts, we propose improved estimators for generalized partial linear models that accommodate both the within-subject correlations and nonignorable missing data. To address the identifiability problem, an instrumental covariate, which is related to the response variable but unrelated to the propensity given the response variable and other covariates, is used to construct sufficient instrumental estimating equations. Subsequently, the nonparametric function is approximated by B-spline basis functions and then we construct bias-corrected generalized estimating equations based on the inverse probability weighting. In order to incorporate the within-subject correlations under an informative working correlation structure, we borrow the idea of quadratic inference function and hybrid-GEE to construct the improved empirical likelihood procedures. Under some regularity conditions, we establish asymptotic normality of the proposed estimators for the parametric components and convergence rate of the estimators for the nonparametric functions. The finite-sample performance of the proposed estimators is studied through simulations and an application to HIV-CD4 data set is also presented.

Appendix

• (C1) The covariate vectors are fixed and the first four moments of $y_{ij}$ exist. Also, for each i, ${m_i}$ is a bounded sequence of positive integers. The random error ${\varvec{\varepsilon }}_i=\varvec{y}_i-{{\varvec{\mu }}}_i$ satisfies that $E({\varvec{\varepsilon }}_i{\varvec{\varepsilon }}_i^T)=\varvec{V}_i$, $\mathop {sup}\nolimits _{i}\Vert \varvec{V}_i\Vert <\infty$ and there exists a positive constant $\delta _1$ such that $\mathop {sup}\nolimits _{i}E\Vert {\varvec{\varepsilon }}_i\Vert ^{2+\delta _1}<\infty$, where $\Vert \varvec{V}_i\Vert$ and $\Vert {\varvec{\varepsilon }}_i\Vert$ denote the modulus of the largest singular value of matrix $\varvec{V}_i$ and vector ${\varvec{\varepsilon }}_i$, respectively.

• (C2) All marginal variances $\varvec{A}_i$ is nonsingular and $\mathop {sup}\nolimits _i\Vert \varvec{A}_i\Vert <\infty$. The link function $h(\cdot )$ has bounded second derivative and $E\{\big (h(\cdot )\big )^{2+\delta _2}\}<\infty$ for some $\delta _2>2$.

• (C3) The function $f(\cdot )$ is r times continuously differentiable on (0, 1) with $r \ge 2$. The inner knots $\{a_i, i=1,\ldots , k_n\}$ satisfy

  $$\begin{aligned} \mathop {\max }\limits _{1 \le i \le {k_n}}|\kappa _{i+1}-\kappa _i|=O(k_n^{-1}),~~~~~~\frac{{\mathop {\max }\nolimits _{1 \le i \le {k_n}} \kappa _i}}{{\mathop {\min }\nolimits _{1 \le i \le {k_n}} \kappa _i }} \le C_0, \end{aligned}$$

  where $\kappa _i=a_i-a_{i-1}$ and $C_0$ is a positive constant.

• (C4) The joint distribution function $Q_{jl}(t,s)$ of any pair of $t_{ij}$ and $t_{il}$, the marginal distribution function $Q_j(t)$ of $t_{ij}$ have positive continuous density functions $q_{jl}(t,s)$ and $q_j(t)$ on $[0,1]\times [0, 1]$ and [0, 1], respectively.

• (C5) The probability function $\pi _{ij}({{\varvec{\vartheta }}}_j)$ satisfies (a) it is twice differentiable with respect to ${{\varvec{\vartheta }}}_j$; (b) $0<C_1<\pi _{ij}({{\varvec{\vartheta }}}_j)<1$ for a positive constant $C_1$; (c) $\partial \pi _{ij}({{\varvec{\vartheta }}}_j)/\partial {{\varvec{\vartheta }}}_j$ is uniformly bounded.

• (C6) There is a unique ${\varvec{\theta }}_0 \in {\mathcal {B}}$ satisfying $E({\hat{{\varvec{g}}}}_i({\varvec{\theta }}))=0$ and $E(\hat{{\varvec{\varrho }}}_i({\varvec{\theta }}))=0$, where ${\mathcal {B}}$ is the parameter space.

• (C7) Assume that $n^{-1}{\sum \nolimits _{i = 1}^n {\varvec{g}}_{0i}({{\varvec{\theta }}}_0,{\varvec{\gamma }}_0){\varvec{g}}^T_{0i}({{\varvec{\theta }}}_0,{\varvec{\gamma }}_0)}$ and $n^{-1}\sum \nolimits _{i = 1}^n {\varvec{\varrho }}_{0i}({{\varvec{\theta }}}_0,{\varvec{\gamma }}_0){\varvec{\varrho }}^T_{0i}({{\varvec{\theta }}}_0,{\varvec{\gamma }}_0)$ converge almost surely to positive definite matrixes ${\varvec{\Lambda }}_g$ and $\varvec{\Lambda }_{{\varvec{\varrho }}}$, respectively. Further, assume that $n^{-1}{\sum \nolimits _{i = 1}^n cov({\hat{{\varvec{g}}}}_{0i}({{\varvec{\theta }}}_0,\varvec{{\hat{\gamma }}})) }$ and $n^{-1}\sum \nolimits _{i = 1}^n cov(\hat{{\varvec{\varrho }}}_{0i}({{\varvec{\theta }}}_0,\varvec{{\hat{\gamma }}}))$ converges almost surely to positive definite matrixes ${\varvec{\Sigma }}_g$ and $\varvec{\Sigma }_{{\varvec{\varrho }}}$, respectively.

• (C8) Assume that ${\partial ^2 {\varvec{g}}_{0i}({{\varvec{\theta }}},{\varvec{\gamma }}_0)}/{\partial {{\varvec{\theta }}}\partial {{\varvec{\theta }}}^T}$ and ${\partial ^2 {\varvec{\varrho }}_{0i}({{\varvec{\theta }}},{\varvec{\gamma }}_0)}/{\partial {{\varvec{\theta }}}\partial {{\varvec{\theta }}}^T}$ is continuous in a neighborhood of ${{\varvec{\theta }}}_0$, $\Vert {\partial ^2 {\varvec{g}}_{0i}({{\varvec{\theta }}},{\varvec{\gamma }}_0)}/{\partial {{\varvec{\theta }}}\partial {{\varvec{\theta }}}^T}\Vert$ and $\Vert {\partial ^2 {\varvec{\varrho }}_{0i}({{\varvec{\theta }}},{\varvec{\gamma }}_0)}/{\partial {{\varvec{\theta }}}\partial {{\varvec{\theta }}}^T}\Vert$ can be bounded by some integrable functions in the neighborhood of ${{\varvec{\theta }}}_0$.

Further, suppose that ${\varvec{\xi }}_{j}^0$ is the unique solution to $E\{{\varvec{s}}_{j}(\varvec{y}_i, \varvec{v}_i, {\varvec{r}}_i, {{\varvec{\xi }}}_{j})\}=0$ and the proposed model holds, ${{\varvec{\Omega }}}_{j} =E\{{\varvec{s}}_{j}(\varvec{y}_i, \varvec{v}_i, {\varvec{r}}_i, {{\varvec{\xi }}}_{j}^0){\varvec{s}}_{j}(\varvec{y}_i, \varvec{v}_i, {\varvec{r}}_i, {{\varvec{\xi }}}_{j}^0)^T\}$ is positive definite and the matrix ${\varvec{\Upsilon }}_{j} =E[\partial {\varvec{s}}_{j}(\varvec{y}_i, \varvec{v}_i, {\varvec{r}}_i, {{\varvec{\xi }}}_{j}^0)/\partial {{\varvec{\xi }}}_{j}]$ is of full rank. As $n \rightarrow \infty$, $\sqrt{n}(\hat{{{\varvec{\xi }}}}_{j}-{{\varvec{\xi }}}_{j}^0) {\mathop {\longrightarrow }\limits ^{{{d}}}}N(0, ({\varvec{\Upsilon }}_{j}^T{{\varvec{\Omega }}}_{j}{\varvec{\Upsilon }}_{j})^{-1})$ and $\sqrt{n}(\hat{{\varvec{\gamma }}}-{\varvec{\gamma }}_0) {\mathop {\longrightarrow }\limits ^{{{d}}}}N(0,{\varvec{\Sigma }})$ in distribution. In the following we mainly derive the case in $\hat{{{\varvec{g}}}}_{i}({{\varvec{\theta }}})$, and the proof for $\hat{{{\varvec{\varrho }}}}_{i}({{\varvec{\theta }}})$ can be obtained using similar arguments.

Lemma 1

Assume the regularity conditions in Theorem 1 hold. As $n \rightarrow \infty$,

$$\begin{aligned}&(1)~~ \frac{1}{\sqrt{n}}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({\varvec{\theta }}_0){\mathop {\longrightarrow }\limits ^{{{d}}}}N(0,\varvec{\Sigma }_{\varvec{g}}),~~~ (2)~~\frac{1}{n}\sum \limits _{i=1}^n{\hat{{\varvec{g}}}}_i({\varvec{\theta }}_0)=O_p(n^{-1/2}),~~\\&(3)~~ \frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{\varvec{\theta }}}_0)\hat{{{\varvec{g}}}}_{i}({{\varvec{\theta }}}_0)^{T}{\mathop {\longrightarrow }\limits ^{p}}\varvec{\Lambda }_{{\varvec{g}}},~~ ~~(4)~~ \max _{i}\Vert \hat{{{\varvec{g}}}}_{i}({{\varvec{\theta }}}_0)\Vert =o_{p}(n^{1/2}). \end{aligned}$$

Proof of Lemma 1

Consider the kth component of ${\hat{{\varvec{g}}}}_i({{\varvec{\theta }}}_0,\hat{{\varvec{\gamma }}})$,

$$\begin{aligned} \frac{1}{n}\sum \limits _{i=1}^n {{{\hat{{\varvec{g}}}}}_{ik}}( {{{\varvec{\theta }}}_0,\hat{{\varvec{\gamma }}} } )&=\frac{1}{n}\sum \limits _{i=1}^n \dot{{\varvec{\mu }}} _i^T \varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}{{\hat{\varvec{W}}}_i}( {{\varvec{y}_i} - {{\varvec{\mu }} _{0i}}})\\&=\frac{1}{n}\sum \limits _{i=1}^n \varvec{d}_i^T \varvec{H}'( {{{\varvec{\eta }} _{0i}}} )\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}{{\hat{\varvec{W}}}_i}( {{\varvec{y}_i} - {{\varvec{\mu }} _{0i}}}), \end{aligned}$$

where ${\varvec{\eta }}_{0i}=\varvec{x}_i{{\varvec{\beta }}}_0+\varvec{m}_i{{\varvec{\alpha }}}_0$, ${{\varvec{\mu }}}_{0i}=h(\varvec{x}_i{{\varvec{\beta }}}_0+\varvec{m}_i{{\varvec{\alpha }}}_0)$ and $\varvec{H}'( {{{\varvec{\eta }}_{0i}}} )\,\mathrm{{= diag}}( {h'( {{{\varvec{\eta }}_{0i1}}} ),\dots , h'( {{{\varvec{\eta }} _{0im}}} )} )$, $h'(t)={{dh( t )} / {dt}}$. Note that

$$\begin{aligned} {\varvec{y}_i} - {{\varvec{\mu }} _{0i}} =h(\varvec{x}_i{{\varvec{\beta }}}_0+f_0(\varvec{t}_i))-h(\varvec{x}_i{{\varvec{\beta }}}_0+\varvec{m}_i{{\varvec{\alpha }}}_0)+ {{\varvec{\varepsilon }} _i}. \end{aligned}$$

(11)

Applying the Taylor expansion to the first two terms in (11) around $( {{{\varvec{\beta }} _0},{{\varvec{\alpha }} _0}})$, we have

$$\begin{aligned} \begin{aligned} {\varvec{y}_i} - {{\varvec{\mu }} _{0i}}=&{{\varvec{\varepsilon }} _i} +\varvec{H}'( {{{\varvec{\eta }}_{0i}}}){\varvec{R}}( {{\varvec{t}_i}}) + \frac{1}{2}\varvec{H}''( {{\varvec{w}_{i}}} ){\varvec{R}^*}( {{\varvec{t}_i}} ), \end{aligned} \end{aligned}$$

(12)

where $\varvec{w}_{i}$ is between ${\varvec{\eta }}_{0i}$ and ${\varvec{x}_i}{{\varvec{\beta }} _0} + {f_0}( {{\varvec{t}_i}})$ and ${\varvec{R}}( {{\varvec{t}_i}} ) = {( {{R}( {{t_{i1}}} ),\dots ,{R}( {{t_{im}}} )} )^T}$, ${\varvec{R}^*}( {{\varvec{t}_i}} ) = {( {{R^2}( {{t_{i1}}} ),\dots ,{R^2}( {{t_{im}}} )} )^T}$, ${R}( {{t_{ij}}} ) = {f_0}( {{t_{ij}}} ) - {\varvec{m} ^T}( {{t_{ij}}} ){{\varvec{\alpha }} _0},$ $\varvec{H}''( {{\varvec{w}_i}} )\mathrm{{= diag}}( h''( {{\varvec{w} _{i1}}} ),\dots , h''( {{\varvec{w} _{im}}} ) ),$ $h''(t)={{d^2h( t )} / {dt^2}}.$

From conditions (C3)–(C4) and Corollary 6.21 in Schumaker (1981), we obtain $\Vert {{R}( {{t_{ij}}} )} \Vert = O_p( {k_n^{- r}})$ and $\Vert {\varvec{m}( {{t_{ij}}})}\Vert = O_p( 1 )$. By substituting (12) into (11), we have

$$\begin{aligned} \frac{1}{n}\sum \limits _{i = 1}^n {{{\hat{{\varvec{g}}}}}_{ik}}( {{{\varvec{\theta }}_0},\hat{{\varvec{\gamma }}} } )&= \frac{1}{n}\sum \limits _{i = 1}^n {\dot{{\varvec{\mu }} _i}^T\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}{{\hat{\varvec{W}}}_i}} {{\varvec{\varepsilon }} _i}\\&\quad + \frac{1}{n}\sum \limits _{i = 1}^n {\dot{{\varvec{\mu }}} _i^T\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}{{\hat{\varvec{W}}}_i}} \varvec{H}'( {{{\varvec{\eta }} _{0i}}} ){\varvec{R}}( {{\varvec{t}_i}} )\\&\quad + \frac{1}{{2n}}\sum \limits _{i = 1}^n {\dot{{\varvec{\mu }}} _i^T\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}{{\hat{\varvec{W}}}_i}} \varvec{H}''( {{\varvec{w} _{i}}} ){\varvec{R}^*}( {{\varvec{t}_i}} )\\&=: \varvec{I}_1+\varvec{I}_2+\varvec{I}_3. \end{aligned}$$

Under conditions (C3)–(C4) and through a simple calculation, we derive that $\varvec{I}_2=O_p(n^{-1}n^{1/2}{k_n^{-r}})=o_p(n^{-1/2})$ and $\varvec{I}_3=o_p(n^{-1/2})$. Thus,

$$\begin{aligned} \frac{1}{n}\sum \limits _{i = 1}^n{\hat{{\varvec{g}}}}_i({{\varvec{\theta }}}_0,\hat{{\varvec{\gamma }}})=\frac{1}{n}\sum \limits _{i = 1}^n {\hat{{\varvec{g}}}}_{0i}({{\varvec{\theta }}}_0,\hat{{\varvec{\gamma }}})+o_p(n^{-1/2}), \end{aligned}$$

where ${\hat{{\varvec{g}}}}_{0ik}({{\varvec{\theta }}}_0,\hat{{\varvec{\gamma }}})={\dot{{\varvec{\mu }} _i}^T\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}{{\hat{\varvec{W}}}_i}} {{\varvec{\varepsilon }} _i}$ and ${\hat{{\varvec{g}}}}_{0i}({{\varvec{\theta }}}_0,\hat{{\varvec{\gamma }}})=({\hat{{\varvec{g}}}}^T_{0i1},\ldots ,{\hat{{\varvec{g}}}}^T_{0is})^T$. Obviously, it leads to $E({\hat{{\varvec{g}}}}_{0ik}({{\varvec{\theta }}}_0,\hat{{\varvec{\gamma }}}))=0$. Applying the Taylor expansion to ${\hat{{\varvec{g}}}}_{0ik}({{\varvec{\theta }}}_0,\hat{{\varvec{\gamma }}})$ around ${\varvec{\gamma }}_0$, it leads to

$$\begin{aligned} {{\hat{{\varvec{g}}}}_{0ik}( {{\varvec{\theta }}}_0,\hat{{\varvec{\gamma }}})}&= {{\varvec{g}}_{0ik}({{\varvec{\theta }}}_0,{{\varvec{\gamma }}_0} ) + {\frac{{\partial {\varvec{g}}_{0ik}( {{\varvec{\theta }}}_0,{{\varvec{\gamma }}_0} )}}{{\partial {{\varvec{\gamma }}} }}( {\hat{{\varvec{\gamma }}} - {\varvec{\gamma }}_0 } )} } + {o_p}( 1 )\\&= :{\varvec{I}_{n1}} + {\varvec{I}_{n2}}, \end{aligned}$$

where $\varvec{I}_{n1}= {{\varvec{g}}_{0ik}( {{{\varvec{\theta }} _0},{{\varvec{\gamma }} _0}})}=\dot{{\varvec{\mu }} _i}^T\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}{\varvec{W}}_i {\varvec{\varepsilon }}_i$ and $\varvec{I}_{n2}$ is the rest part of the above equation. Under (C1)–(C2) and (C5), it can be seen that

$$\begin{aligned} E(\varvec{I}_{n1}\varvec{I}_{n1}^T)&=E[{\dot{{\varvec{\mu }} _i}^T\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}\varvec{W}_i{\varvec{\varepsilon }}_i{\varvec{\varepsilon }}_i^T\varvec{W}_i\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}\dot{{\varvec{\mu }} _i}}]\\&=E[E[{\dot{{\varvec{\mu }} _i}^T\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}\varvec{W}_i{\varvec{\varepsilon }}_i{\varvec{\varepsilon }}_i^T\varvec{W}_i\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}\dot{{\varvec{\mu }} _i}}|{\varvec{x}}_i]]\\&={\dot{{\varvec{\mu }} _i}^T\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}\varvec{T}_i\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}\dot{{\varvec{\mu }} _i}},\\ E(\varvec{I}_{n2}\varvec{I}_{n2}^T)&=E\big [ {{{\partial {{\varvec{g}}_{0ik}}\big ( {{{{\varvec{\theta }}}_0},{{\varvec{\gamma }} _0}} \big )} \big / {\partial {\varvec{\gamma }} }}} \big ]{\varvec{\Sigma }} {E^T}\big [ {{{\partial {{\varvec{g}}_{0ik}}\big ( {{{{\varvec{\theta }}}_0},{{\varvec{\gamma }} _0}} \big )} \big / {\partial {{\varvec{\gamma }}}}}} \big ],\\ E(\varvec{I}_{n1} \varvec{I}_{n2})&=O_p(n^{-1/2}),~~~~~~~Cov(\varvec{I}_{n1},\varvec{I}_{n2})=o_p(1), \end{aligned}$$

where $\varvec{T}_i$ is a symmetric matrix with its $(j, j')$ element $(j\le j')$ as $\varepsilon _{ij}\varepsilon _{ij'}/\pi _{ij}$. Therefore,

$$\begin{aligned} Cov({\hat{{\varvec{g}}}}_{0ik}({\varvec{\theta }}_0,\hat{{\varvec{\gamma }}}))&={\dot{{\varvec{\mu }} _i}^T\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}\varvec{T}_i\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}\dot{{\varvec{\mu }} _i}}\\&\quad +E\big [ {{{\partial {{\varvec{g}}_{0ik}}\big ( {{{{\varvec{\theta }}}_0},{{\varvec{\gamma }} _0}} \big )} \big / {\partial {\varvec{\gamma }} }}} \big ]{\varvec{\Sigma }} {E^T}\big [ {{{\partial {{\varvec{g}}_{0ik}}\big ( {{{{\varvec{\theta }}}_0},{{\varvec{\gamma }} _0}} \big )} \big / {\partial {{\varvec{\gamma }}}}}} \big ], \end{aligned}$$

and

$$\begin{aligned}Cov({\hat{{\varvec{g}}}}_{0i}({\varvec{\theta }}_0,\hat{{\varvec{\gamma }}}))={\left( \begin{array}{ccc} \varvec{N}_{11} &{} \ldots &{} \varvec{N}_{1s} \\ \vdots &{} \ddots &{} \vdots \\ \varvec{N}_{s1} &{} \ldots &{} \varvec{N}_{ss} \\ \end{array} \right) }+ {\left( \begin{array}{ccc} \varvec{U}_{11} &{} \ldots &{} \varvec{U}_{1s} \\ \vdots &{} \ddots &{} \vdots \\ \varvec{U}_{s1} &{} \ldots &{} \varvec{U}_{ss} \\ \end{array} \right) },\end{aligned}$$

where $\varvec{N}_{lk}=\dot{{\varvec{\mu }} _i}^T\varvec{A}_i^{- 1/2}{\varvec{B}_l}\varvec{A}_i^{- 1/2}\varvec{T}_i\varvec{A}_i^{- 1/2}{\varvec{B}_k}\varvec{A}_i^{- 1/2}\dot{{\varvec{\mu }} _i}$ and $\varvec{U}_{lk}=E\big [ {\partial {{\varvec{g}}_{0il}}\big ( {{{\varvec{\theta }} _0},{{\varvec{\gamma }} _0}} \big )} /{\partial {\varvec{\gamma }} }\big ] {\varvec{\Sigma }} {E^T}\big [ {{{\partial {{\varvec{g}}_{0ik}}\big ( {{{\varvec{\theta }} _0},{{\varvec{\gamma }} _0}} \big )} / {\partial {{\varvec{\gamma }}}}}} \big ]$ for $l, k=1,\ldots ,s$. Under the conditions (C6)–(C7), it can be verified that

$$\begin{aligned} n^{-1}\sum \limits _{i = 1}^nCov({\hat{{\varvec{g}}}}_{0i}({\varvec{\theta }}_0,\hat{{\varvec{\gamma }}})){\mathop {\longrightarrow }\limits ^{{{p}}}}{\varvec{\Sigma }}_g, \end{aligned}$$

in probability and then

$$\begin{aligned} \frac{1}{{\sqrt{n} }}\sum \limits _{i = 1}^n {{{{\hat{{\varvec{g}}}}}_i}} ( {{{\varvec{\theta }} _0,\varvec{{\hat{\gamma }}}}} ) = \frac{1}{{\sqrt{n} }}\sum \limits _{i = 1}^n {{{{\hat{{\varvec{g}}}}}_{0i}}}( {{{\varvec{\theta }} _0,\varvec{{\hat{\gamma }}}}} ) + o_p(1){\mathop {\longrightarrow }\limits ^{{{d}}}}N(0,{\varvec{\Sigma }}_g), \end{aligned}$$

in distribution. Moreover, we can obtain $n^{-1}\sum \nolimits _{i = 1}^n{\hat{{\varvec{g}}}}_i({\varvec{\theta }}_0,\hat{{\varvec{\gamma }}})=O_p(n^{-1/2})$. Noting that $\sqrt{n}(\varvec{{\hat{\gamma }}}-{\varvec{\gamma }}_0)=O_p(1)$ and ${\partial {\varvec{g}}_{0ik}( {{\varvec{\theta }}}_0,{{\varvec{\gamma }}_0} ) \big / \partial {{\varvec{\gamma }}}}=O_p(1)$ from (C5) and (C8), we can obtain

$$\begin{aligned}\begin{aligned} \frac{1}{n}\sum \limits _{i = 1}^n {{{{\hat{{\varvec{g}}}}}_{il}}} ( {{{\varvec{\theta }} _0,\varvec{{\hat{\gamma }}}}} ){{{{\hat{{\varvec{g}}}}}_{ik}^T}} ( {{{\varvec{\theta }} _0,\varvec{{\hat{\gamma }}}}} ) =\frac{1}{n}\sum \limits _{i = 1}^n {{{{\varvec{g}}}_{0il}}} ( {{{\varvec{\theta }} _0,\varvec{\gamma _0}}} ){{{{\varvec{g}}}_{0ik}^T}} ( {{{\varvec{\theta }} _0,\varvec{\gamma _0}}} )+o_p(1). \end{aligned}\end{aligned}$$

Hence

$$\begin{aligned} \begin{aligned} \frac{1}{n}\sum \limits _{i = 1}^n {{{{\hat{{\varvec{g}}}}}_{i}}} ( {{{\varvec{\theta }} _0,\varvec{{\hat{\gamma }}}}} ){{{{\hat{{\varvec{g}}}}}_{i}^T}} ( {{{\varvec{\theta }} _0,\varvec{{\hat{\gamma }}}}} )&=\frac{1}{n}\sum \limits _{i = 1}^n {{{{\varvec{g}}}_{0i}}} ( {{{\varvec{\theta }} _0,\varvec{\gamma _0}}} ){{{{\varvec{g}}}_{0i}^T}} ( {{{\varvec{\theta }} _0,{{\varvec{\gamma }}_0}}} )+o_p(1)\\&=\frac{1}{n}\sum \limits _{i=1}^n{\left( \begin{array}{ccc} \varvec{N}_{11} &{} \ldots &{} \varvec{N}_{1s} \\ \vdots &{} \ddots &{} \vdots \\ \varvec{N}_{s1} &{} \ldots &{} \varvec{N}_{ss} \\ \end{array} \right) }+o_p(1){\mathop {\longrightarrow }\limits ^{{{p}}}}{\varvec{\Lambda }}_{g}. \end{aligned}\end{aligned}$$

According to (3),

$$\begin{aligned}\frac{{{n^{- 1}}{{( {\mathop {\max }\limits _{1 \le i \le n} \Vert {{{{\hat{{\varvec{g}}}}}_{i}}} ( {{{\varvec{\theta }} _0,\varvec{{\hat{\gamma }}}}} ) \Vert } )}^2}}}{{{n^{- 1}}\sum \limits _{i = 1}^n {{{{{\hat{{\varvec{g}}}}}_{i}}} ( {{{\varvec{\theta }} _0,\varvec{{\hat{\gamma }}}}} ){{{{\hat{{\varvec{g}}}}}^T_{i}}} ( {{{\varvec{\theta }} _0,\varvec{{\hat{\gamma }}}}} )} }} \rightarrow 0,\end{aligned}$$

which leads to ${\mathop {\max }\nolimits _{1 \le i \le n} \Vert {{{{\hat{{\varvec{g}}}}}_{i}}} ( {{{\varvec{\theta }} _0,\varvec{{\hat{\gamma }}}}} ) \Vert } =o_p(n^{1/2})$. The proof of Lemma 1 is completed. $\square$

Lemma 2

Assume the regularity conditions in Theorem 1 hold. We have

$$\begin{aligned} n^{-1}\hat{R}_Q({{\varvec{\theta }}}_0)= {\hat{{\varvec{g}}}}({{\varvec{\theta }}}_0)^T\varvec{S}^{-1}_n({{\varvec{\theta }}}_0){\hat{{\varvec{g}}}}({{\varvec{\theta }}}_0)+o_p(1), \end{aligned}$$

where ${\hat{{\varvec{g}}}}({{\varvec{\theta }}}_0)=n^{-1}\sum \nolimits _{i = 1}^n{\hat{{\varvec{g}}}}_i({{\varvec{\theta }}}_0)$ and $\varvec{S}_n( {{{{\varvec{\theta }}}_0}} ) = {n^{- 1}}\sum \nolimits _{i = 1}^n {{{{\hat{{\varvec{g}}}}}_i}( {{{{\varvec{\theta }}}_0}})} {{{{\hat{{\varvec{g}}}}}_i}( {{{{\varvec{\theta }}}_0}})}^T$.

Proof of Lemma 2

Noting that the Lagrange multiplier method leads to the empirical log-likelihood ratio function for ${{\varvec{\theta }}}_0$ as

$$\begin{aligned} \hat{R}_Q({{\varvec{\theta }}}_0)=2\sum \limits _{i=1}^n \log \left\{ 1+{{\varvec{\lambda }}}^{T}({{\varvec{\theta }}}_0)\hat{{{\varvec{g}}}}_{i}({{\varvec{\theta }}}_0)\right\} , \end{aligned}$$

where vector ${{\varvec{\lambda }}}$ is the solution to the equation

$$\begin{aligned} D({\varvec{\lambda }})=:\frac{1}{n}\sum \limits _{i = 1}^n {\frac{{{{{\hat{{\varvec{g}}}}}_i}\big ({{\varvec{\theta }}}_0 \big )}}{{1 + {{\varvec{\lambda }} ^T}{{{\hat{{\varvec{g}}}}}_i}\big ( {{\varvec{\theta }}}_0 \big )}}} = 0. \end{aligned}$$

Denote ${{\varvec{\lambda }}} = \rho {\varvec{{\varvec{u}}}}$ where $\rho \ge 0$, ${\varvec{u}}\in {R^p}$ and $\big \Vert {\varvec{u}}\big \Vert = 1$ is an unit vector. Substituting ${1 \big / {\big ( {1 + {{\varvec{\lambda }} ^T}{{{\hat{{\varvec{g}}}} }_{i}}\big ( {{{{\varvec{\theta }}}}} \big )} \big )}} = 1 - {{{{\varvec{\lambda }} ^T}{{{\hat{{\varvec{g}}}} }_{i}}\big ( {{{{\varvec{\theta }}}}} \big )} \big / {\big ( {1 + {{\varvec{\lambda }} ^T}{{{\hat{{\varvec{g}}}} }_{i}}\big ( {{{{\varvec{\theta }}}}} \big )} \big )}}$ and ${{\varvec{\lambda }}} = \rho {\varvec{{\varvec{u}}}}$ into the above equation, we have

$$\begin{aligned} 0&= \frac{1}{n}\sum \limits _{i = 1}^n {\frac{{{{{\hat{{\varvec{g}}}}}_i}\big ( {{{{\varvec{\theta }}}_0}} \big )}}{{1 + {{\varvec{\lambda }} ^T}{{{\hat{{\varvec{g}}}}}_i}\big ( {{{{\varvec{\theta }}}_0}} \big )}}} = \frac{1}{n}\sum \limits _{i = 1}^n {{\varvec{u}^T}{{{\hat{{\varvec{g}}}}}_i}({{{\varvec{\theta }}}_0}) - \rho } \frac{1}{n}\sum \limits _{i = 1}^n {\frac{{{{({{\varvec{u}}^T}{{{\hat{{\varvec{g}}}}}_i}({{{\varvec{\theta }}}_0}))}^2}}}{{1 + \rho {{\varvec{u}}^T}{{{\hat{{\varvec{g}}}}}_i}({{{\varvec{\theta }}}_0})}}} \\&\le {{\varvec{u}}^T}\frac{\mathrm{{1}}}{n}\sum \limits _{i = 1}^n {{{{\hat{{\varvec{g}}}}}_i}\big ( {{{{\varvec{\theta }}}_0}} \big )} - \frac{\rho }{{1 + \rho \mathop {\max }\limits _{1 \le i \le n} \big \Vert {{{{\hat{{\varvec{g}}}}}_i}({{{\varvec{\theta }}}_0})} \big \Vert }}{{\varvec{u}}^T}\varvec{S}_n\big ( {{{{\varvec{\theta }}}_0}} \big ){\varvec{u}}, \end{aligned}$$

where $\varvec{S}_n( {{{{\varvec{\theta }}}_0}} ) = {n^{- 1}}\sum \nolimits _{i = 1}^n {{{{\hat{{\varvec{g}}}}}_i}( {{{{\varvec{\theta }}}_0}})} {{{{\hat{{\varvec{g}}}}}_i}( {{{{\varvec{\theta }}}_0}})}^T={\varvec{\Lambda }}_g(1+o_p(1))$ and the inequality follows from the positivity of $(1+\rho \varvec{u}^T{\hat{\varvec{g}}_i({{\varvec{\theta }}}_0)})$. Therefore, it can be shown that

$$\begin{aligned} \rho {{\varvec{u}}^T}\varvec{S}_n\big ( {{{{\varvec{\theta }}}_0 }} \big ){\varvec{u}}\le \{{1+\rho \mathop {\max }\limits _{1 \le i \le n} ||{{{\hat{{\varvec{g}}}} }_{i}}({{{\varvec{\theta }}}_0 })||}\} \Vert {{\varvec{u}}^T}\frac{\mathrm{{1}}}{n}\sum \limits _{i = 1}^n {{{{\hat{{\varvec{g}}}} }_{i}}\big ( {{{{\varvec{\theta }}}_0 }} \big )}\Vert . \end{aligned}$$

Noting that $\max \nolimits _{1\le i\le n}|| \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0}) | |=o_p(n^{1/2})$ and according to the conclusion provided in the proof of Lemma 1, we have

$$\begin{aligned} \rho [\varvec{u}^{T} \varvec{\Lambda }_{{\varvec{g}}} \varvec{u}+o_{p}(1)]\le O_p(n^{-1/2})\{1+\rho o_p(n^{1/2})\}, \end{aligned}$$

which leads to $\rho =O_{p}(n^{-1/2}),$ i.e., $|| {{\varvec{\lambda }}}||=O_{p}(n^{-1/2}).$ Naturally,

$$\begin{aligned}\max \limits _{1\le i\le n}| |{{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0}) | |\le || {{\varvec{\lambda }}}|| \max \limits _{1\le i\le n}|| \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0}) | |=O_{p}(n^{-1/2}) o_p(n^{1/2})=o_{p}(1).\end{aligned}$$

Expanding the equation $D({\varvec{\lambda }})$, we get

$$\begin{aligned} 0&= \frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\Big \{1-{{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0}) +\frac{[{{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})]^2 }{(1+\xi _i)^3}\Big \} \\&= \frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0}) - \frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})^{T}{{\varvec{\lambda }}}+ \frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\frac{[{{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})]^2 }{(1+\xi _i)^3}, \end{aligned}$$

where $\xi _i\in (0, {{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})).$ Using the fact that $\max \nolimits _{1\le i\le n}|| {{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0}) | |=o_{p}(1),$ then $|\xi _i| =o_{p}(1).$ Since

$$\begin{aligned} \Big \Vert \frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\frac{[{{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})]^2 }{(1+\xi _i)^3} \Big \Vert&\le \frac{\max \limits _{1\le i\le n}|| \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0}) | |}{1-\max \limits _{1\le i\le n}|\xi _i| } \Big \Vert {{\varvec{\lambda }}}^{T} \frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})^{T} {{\varvec{\lambda }}}\Big \Vert \\&= o(n^{1/2})O_{p}(n^{-1}) \\&= o_{p}(n^{-1/2}), \end{aligned}$$

thus

$$\begin{aligned} {{\varvec{\lambda }}}&= \Big [\frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})^{T}\Big ]^{-1}\Big [\frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\Big ]+ {\varvec{\zeta }}, \end{aligned}$$

(13)

where $| |{\varvec{\zeta }}| |=o_{p}(n^{-1/2}).$ A Taylor expansion of $\hat{R}_Q({{\varvec{\theta }}}_0)$ yields

$$\begin{aligned} \hat{R}_Q({{\varvec{\theta }}}_0)&= 2 \sum \limits _{i=1}^n \Big \{ {{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})-\frac{1}{2}[{{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})]^2 +\frac{1}{3}\frac{[{{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})]^3}{(1+\xi _i)^3} \Big \}\\&= 2 {{\varvec{\lambda }}}^{T} \sum \limits _{i=1}^n\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})-\sum \limits _{i=1}^n {{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})^{T}{{\varvec{\lambda }}}+\frac{2}{3}\sum \limits _{i=1}^n\frac{[{{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})]^3}{(1+\xi _i)^3}. \end{aligned}$$

Similarly,

$$\begin{aligned} \Big \Vert \sum \limits _{i=1}^n \frac{[{{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})]^3 }{(1+\xi _i)^3} \Big \Vert&\le \frac{\max \limits _{1\le i\le n}|| {{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0}) | |}{1-\max \limits _{1\le i\le n}|\xi _i| } \Big \Vert {{\varvec{\lambda }}}^{T} \frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})^{T} {{\varvec{\lambda }}}\Big \Vert \\&= o_{p}(1)nO_{p}(n^{-1}) \\&= o_{p}(1), \end{aligned}$$

therefore,

$$\begin{aligned} \hat{R}_Q({{\varvec{\theta }}}_0)&= 2 {{\varvec{\lambda }}}^{T} \sum \limits _{i=1}^n\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})-\sum \limits _{i=1}^n {{\varvec{\lambda }}}^{T}\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})^{T}{{\varvec{\lambda }}}+o_{p}(1). \end{aligned}$$

(14)

Substituting (13) in (14), it holds that

$$\begin{aligned} \hat{R}_Q({{\varvec{\theta }}}_0)&=n\Big [\frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})^{T}\Big ]\Big [\frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})^{T}\Big ]^{-1}\Big [\frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\Big ]\\&\quad -n{\varvec{\zeta }}^{T}\frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})^{T}{\varvec{\zeta }}+o_{p}(1). \end{aligned}$$

A simple calculation shows that

$$\begin{aligned} n{\varvec{\zeta }}^{T}\frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})^{T}{\varvec{\zeta }}&=no_{p}(n^{-1/2})o_{p}(n^{-1/2})=o_{p}(1). \end{aligned}$$

Therefore,

$$\begin{aligned} \hat{R}_Q({{\varvec{\theta }}}_0)&=\Big [\frac{1}{\sqrt{n}}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})^{T}\Big ]\Big [\frac{1}{n}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})^{T}\Big ]^{-1}\Big [\frac{1}{\sqrt{n}}\sum \limits _{i=1}^n \hat{{{\varvec{g}}}}_{i}({{{\varvec{\theta }}}_0})\Big ]+o_{p}(1). \end{aligned}$$

Since ${\hat{{{\varvec{\theta }}}}}_Q=\mathop {\arg \min }\nolimits _{{{\varvec{\theta }}}\in {\mathcal {B}}} {\hat{R}_Q({{\varvec{\theta }}})}$, we have two properties of ${\hat{R}}_Q({{\varvec{\theta }}}_0)$ as follows:

$$\begin{aligned} \frac{1}{n}\hat{R}'_Q({{\varvec{\theta }}}_0)&=2{\hat{{\varvec{g}}}}'({{\varvec{\theta }}}_0)^T\varvec{S}^{-1}_n({{\varvec{\theta }}}_0){\hat{{\varvec{g}}}}({{\varvec{\theta }}}_0)-R_{n1}\\ \frac{1}{n}\hat{R}''_Q({{\varvec{\theta }}}_0)&=2{{\hat{{\varvec{g}}}}'}({{\varvec{\theta }}}_0)^T\varvec{S}^{-1}_n({{\varvec{\theta }}}_0){\hat{{\varvec{g}}}}'({{\varvec{\theta }}}_0)+R_{n2}, \end{aligned}$$

where

$$\begin{aligned} R_{n1}&={\hat{{\varvec{g}}}}^{T}({{\varvec{\theta }}}_0)\varvec{S}^{-1}_n({{\varvec{\theta }}}_0)\{\varvec{S}'_n({{\varvec{\theta }}}_0)\}^{-1}\varvec{S}^{-1}_n({{\varvec{\theta }}}_0){\hat{{\varvec{g}}}}({{\varvec{\theta }}}_0),\\ R_{n2}&=2{\hat{{\varvec{g}}}}''({{\varvec{\theta }}}_0)^T\varvec{S}^{-1}_n({{\varvec{\theta }}}_0){\hat{{\varvec{g}}}}({{\varvec{\theta }}}_0)-4n{\hat{{\varvec{g}}}}'({{\varvec{\theta }}}_0)^{T}\varvec{S}^{-1}_n({{\varvec{\theta }}}_0)\{\varvec{S}'_n({{\varvec{\theta }}}_0)\}^{-1}\varvec{S}^{-1}_n({{\varvec{\theta }}}_0){\hat{{\varvec{g}}}}({{\varvec{\theta }}}_0)\\&\quad +2{\hat{{\varvec{g}}}}^{T}({{\varvec{\theta }}}_0)\varvec{S}^{-1}_n({{\varvec{\theta }}}_0)\{\varvec{S}'_n({{\varvec{\theta }}}_0)\}^{-1}\varvec{S}^{-1}_n({{\varvec{\theta }}}_0)\varvec{\{}\varvec{S}'_n({{\varvec{\theta }}}_0)\}^{-1}\varvec{S}^{-1}_n({{\varvec{\theta }}}_0){\hat{{\varvec{g}}}}({{\varvec{\theta }}}_0)\\&\quad -{\hat{{\varvec{g}}}}^{T}({{\varvec{\theta }}}_0)\varvec{S}^{-1}_n({{\varvec{\theta }}}_0)\{\varvec{S}''_n({{\varvec{\theta }}}_0)\}^{-1}\varvec{S}^{-1}_n({{\varvec{\theta }}}_0){\hat{{\varvec{g}}}}({{\varvec{\theta }}}_0). \end{aligned}$$

Since $n^{-1}\sum \nolimits _{i = 1}^n{\hat{{\varvec{g}}}}_i({\varvec{\theta }}_0,\hat{{\varvec{\gamma }}})=O_p(n^{-1/2})$, together with conditions (C5)–(C8), we have $R_{n1}=O_p(n^{-1})$ and $R_{n2}=o_p(1)$. The proof of Lemma 2 is completed. $\square$

Proof of Theorem 1

Similar to the proof of Lemma 1 in Qin and Lawless (1994), under conditions (C3)–(C4) and (C6)–(C7), we have $\hat{{{\varvec{\theta }}}}_Q {\mathop {\longrightarrow }\limits ^{{{p}}}}{{\varvec{\theta }}}_0$ in probability. This implies $\Vert {\hat{{{\varvec{\beta }}}}}_Q-{{\varvec{\beta }}}_0\Vert =O_p(n^{-1/2})$ and $\Vert {\hat{{{\varvec{\alpha }}}}}_Q-{{\varvec{\alpha }}}_0\Vert =O_p(n^{-1/2})$. Under (C3)–(C4),

$$\begin{aligned}&\mathop {lim}_{n\rightarrow \infty }\frac{1}{n}\sum \limits _{i=1}^n({\hat{f}}_Q(\varvec{t}_i)-f_0(\varvec{t}_i))^T({\hat{f}}_Q(\varvec{t}_i)-f_0(\varvec{t}_i))=\int _0^1 {({\hat{f}}_Q(t)-f_0( t))^2 dt}\\&\quad =\int _0^1\big (\varvec{m}^T(t)({\hat{{{\varvec{\alpha }}}}}_Q-{{\varvec{\alpha }}}_0)+R(t)\big )^2dt\\&\quad \le 2\int _0^1\big (\varvec{m}^T(t)({\hat{{{\varvec{\alpha }}}}}_Q-{{\varvec{\alpha }}}_0)\big )^2dt+2\int _0^1 R(t)^2 dt\\&\quad = 2 ({\hat{{{\varvec{\alpha }}}}}_Q-{{\varvec{\alpha }}}_0)^T\int _0^1 \varvec{m}(t)\varvec{m}^T(t) dt ({\hat{{{\varvec{\alpha }}}}}_Q-{{\varvec{\alpha }}}_0)+ 2\int _0^1 R(t)^2 dt. \end{aligned}$$

Since $\Vert R(t_{ij})\Vert =O_p(k_n^{-r})=O_p(n^{-r/(2r+1)})$, we obtain

$$\begin{aligned} \int _0^1 {({\hat{f}}_Q(t)-f_0( t))^2 dt}=O_p(n^{-2r/(2r+1)}). \end{aligned}$$

Proof of Theorem 2

Denote $\varvec{L}_{1n}({{\varvec{\beta }}},{{\varvec{\alpha }}})$ and $\varvec{L}_{2n}({{\varvec{\beta }}},{{\varvec{\alpha }}})$ as the first derivatives of ${\hat{R}}_Q({{\varvec{\theta }}})$ with respect to ${{\varvec{\beta }}}$ and ${{\varvec{\alpha }}}$ respectively. Obviously, $\varvec{L}_{1n}({\hat{{{\varvec{\beta }}}}}_Q,{\hat{{{\varvec{\alpha }}}}}_Q)=0$ and $\varvec{L}_{2n}({\hat{{{\varvec{\beta }}}}}_Q,{\hat{{{\varvec{\alpha }}}}}_Q)=0$. Applying Taylor expansion to $\varvec{L}_{1n}$ around $({{\varvec{\beta }}}_0,{{\varvec{\alpha }}}_0)$, it follows that

$$\begin{aligned} \varvec{L}_{1n}({\hat{{{\varvec{\beta }}}}}_Q,{\hat{{{\varvec{\alpha }}}}}_Q)&=\varvec{L}_{1n}({{\varvec{\beta }}}_0,{{\varvec{\alpha }}}_0)+\frac{{\partial {\varvec{L}}_{1n}({{\varvec{\beta }}}_0,{{\varvec{\alpha }}}_0)}}{{\partial {{\varvec{\beta }}}}}({\hat{{{\varvec{\beta }}}}}_Q-{{\varvec{\beta }}}_0)+ \frac{{\partial {\varvec{L}}_{1n}({{\varvec{\beta }}}_0,{{\varvec{\alpha }}}_0)}}{{\partial {{\varvec{\alpha }}}}}({\hat{{{\varvec{\alpha }}}}}_Q-{{\varvec{\alpha }}}_0)\\&\quad +\frac{1}{2}({\hat{{{\varvec{\theta }}}}}_Q-{{\varvec{\theta }}}_0)^T\frac{{\partial ^2 {\varvec{L}}_{1n}({\tilde{{{\varvec{\beta }}}}}_0,{\tilde{{{\varvec{\alpha }}}}}_0)}}{{\partial {{\varvec{\theta }}}^2 }}({\hat{{{\varvec{\theta }}}}}_Q-{{\varvec{\theta }}}_0), \end{aligned}$$

where $({\tilde{{{\varvec{\beta }}}}}_0^T,{\tilde{{{\varvec{\alpha }}}}}_0^T)^T$ is between $({{\varvec{\beta }}}_0^T,{{\varvec{\alpha }}}_0^T)^T$ and $({\hat{{{\varvec{\beta }}}}}_Q^T,{\hat{{{\varvec{\alpha }}}}}_Q^T)^T$. Since

$$\begin{aligned}n^{-1}\hat{R}'_Q({{\varvec{\theta }}}_0)=2{\hat{{\varvec{g}}}}'({{\varvec{\theta }}}_0)^T\varvec{S}^{-1}_n({{\varvec{\theta }}}_0){\hat{{\varvec{g}}}}({{\varvec{\theta }}}_0)+O_p(n^{-1}),\end{aligned}$$

we have

$$\begin{aligned} \frac{1}{n}\varvec{L}_{1n}({{\varvec{\beta }}}_0,{{\varvec{\alpha }}}_0)=-\frac{2}{n^2}&\sum \limits _{l_1,l_2 = 1}^n\sum \limits _{k_1,k_2 = 1}^s \{\varvec{x}_{l_1}^T\varvec{H}'({\varvec{\eta }}_{0l_1})\varvec{A}_{l_1}^{-1/2} \varvec{B}_{k_1}\varvec{A}_{l_1}^{-1/2}\hat{\varvec{W}}_{l_1}\varvec{H}'({\varvec{\eta }}_{0l_1})\varvec{d}_{l_1}\\&\varvec{S}^{-1}_{n,k_1k_2}\varvec{d}_{l_2}^T\varvec{H}'({\varvec{\eta }}_{0l_2})\varvec{A}_{l_2}^{-1/2}\varvec{B}_{k_2}\varvec{A}_{l_2}^{-1/2}\hat{\varvec{W}}_{l_2}(\varvec{y}_{l_2}-{{\varvec{\mu }}}_{0l_2})\} +o_p(n^{-1/2})\\ =-\frac{2}{n^2}&\sum \limits _{l_1,l_2= 1}^n\tilde{\varvec{x}}_{l_1}\hat{{\varvec{\tau }}}_{l_1l_2}\{\tilde{\varvec{R}}(\varvec{t}_{l_2})+{\varvec{\varepsilon }}_{l_2}\} +o_p(n^{-1/2}), \end{aligned}$$

where $\varvec{S}^{-1}_{n,k_1k_2}$ represents the $(k_1,k_2)$ part of $\varvec{S}_n^{-1}$. Let $\tilde{\varvec{x}}_{l_1}=\varvec{H}'({\varvec{\eta }}_{0i})\varvec{x}_{l_1}$, $\tilde{\varvec{R}}(\varvec{t}_{l_2})=\varvec{H}'({\varvec{\eta }}_{0{l_2}})\varvec{R}(\varvec{t}_{l_2})$ and for simplicity,

$$\begin{aligned}\hat{{\varvec{\tau }}}_{l_1l_2}=\sum \limits _{k_1,k_2 = 1}^s \varvec{A}_{l_1}^{-1/2} \varvec{B}_{k_1}\varvec{A}_{l_1}^{-1/2}\hat{\varvec{W}}_{l_1}\varvec{H}'({\varvec{\eta }}_{0l_1})\varvec{d}_{l_1} \varvec{S}^{-1}_{n,k_1k_2}\varvec{d}_{l_2}^T\varvec{H}'({\varvec{\eta }}_{0l_2})\varvec{A}_{l_2}^{-1/2}\varvec{B}_{k_2}\varvec{A}_{l_2}^{-1/2}\hat{\varvec{W}}_{l_2}, \end{aligned}$$

and

$$\begin{aligned} {{\varvec{\tau }}_{l_1l_2}}=\sum \limits _{k_1,k_2 = 1}^s \varvec{A}_{l_1}^{-1/2} \varvec{B}_{k_1}\varvec{A}_{l_1}^{-1/2}{\varvec{W}}_{l_1}\varvec{H}'({\varvec{\eta }}_{0l_1})\varvec{d}_{l_1} {\varvec{\Lambda }}^{-1}_{g,k_1k_2}\varvec{d}_{l_2}^T\varvec{H}'({\varvec{\eta }}_{0l_2})\varvec{A}_{l_2}^{-1/2}\varvec{B}_{k_2}\varvec{A}_{l_2}^{-1/2}{\varvec{W}}_{l_2}. \end{aligned}$$

Similar with the proof of Lemma 1, we have $\hat{{\varvec{\tau }}}_{l_1l_2}={\varvec{\tau }}_{l_1l_2}(1+o_p(1))$. Therefore, we have

$$\begin{aligned}&\frac{1}{n}\frac{{\partial {\varvec{L}}_{1n}({{\varvec{\beta }}}_0,{{\varvec{\alpha }}}_0)}}{{\partial {{\varvec{\beta }}}}}=-\frac{2}{n^2}\sum \limits _{l_1,l_2 = 1}^n\tilde{\varvec{x}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\tilde{\varvec{x}}_{l_2}+o_p(n^{-1/2}),\\&\frac{1}{n}\frac{{\partial {\varvec{L}}_{1n}({{\varvec{\beta }}}_0,{{\varvec{\alpha }}}_0)}}{{\partial {{\varvec{\alpha }}}}}=-\frac{2}{n^2}\sum \limits _{l_1,l_2= 1}^n\tilde{\varvec{x}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\tilde{\varvec{m}}_{l_2}+o_p(n^{-1/2}), \end{aligned}$$

where $\tilde{\varvec{m}}_{l_2}=\varvec{H}'({\varvec{\eta }}_{0l_2})\varvec{m}_{l_2}$. Thus,

$$\begin{aligned} \begin{aligned} \frac{1}{n}\varvec{L}_{1n}({\hat{{{\varvec{\beta }}}}}_Q,{\hat{{{\varvec{\alpha }}}}}_Q)&=-\frac{2}{n^2}\sum \limits _{l_1,l_2 = 1}^n\tilde{\varvec{x}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\{\tilde{\varvec{x}}_{l_2}({\hat{{{\varvec{\beta }}}}}_Q-{{\varvec{\beta }}}_0)+\tilde{\varvec{m}}_{l_2}({\hat{{{\varvec{\alpha }}}}}_Q-{{\varvec{\alpha }}}_0)\}\\&\quad -\frac{2}{n^2}\sum \limits _{l_1,l_2 = 1}^n\tilde{\varvec{x}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\{\tilde{\varvec{R}}(\varvec{t}_{l_2})+{\varvec{\varepsilon }}_{l_2}\}+o_p(n^{-1/2}). \end{aligned} \end{aligned}$$

(15)

Similarly,

$$\begin{aligned} \begin{aligned} \frac{1}{n}\varvec{L}_{2n}({\hat{{{\varvec{\beta }}}}}_Q,{\hat{{{\varvec{\alpha }}}}}_Q)&=-\frac{2}{n^2}\sum \limits _{l_1,l_2 = 1}^n\tilde{\varvec{m}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\{\tilde{\varvec{x}}_{l_2}({\hat{{{\varvec{\beta }}}}}_Q-{{\varvec{\beta }}}_0)+\tilde{\varvec{m}}_{l_2}({\hat{{{\varvec{\alpha }}}}}_Q-{{\varvec{\alpha }}}_0)\}\\&\quad -\frac{2}{n^2}\sum \limits _{l_1,l_2 = 1}^n\tilde{\varvec{m}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\{\tilde{\varvec{R}}(\varvec{t}_{l_2})+{\varvec{\varepsilon }}_{l_2}\}+o_p(n^{-1/2}). \end{aligned} \end{aligned}$$

(16)

From the Eq. (16), we have

$$\begin{aligned} {\hat{{{\varvec{\alpha }}}}}_Q-{{\varvec{\alpha }}}_0=-\varvec{V}_g^{-1}\big \{\varvec{K}_g({\hat{{{\varvec{\beta }}}}}_Q-{{\varvec{\beta }}}_0)+\varvec{P}_g\big \}, \end{aligned}$$

(17)

where $\varvec{V}_g=n^{-2}{\sum \nolimits _{l_1,l_2= 1}^n\tilde{\varvec{m}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\tilde{\varvec{m}}_{l_2}}$, $\varvec{K}_g=n^{-2}\sum \nolimits _{l_1,l_2 = 1}^n\tilde{\varvec{m}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\tilde{\varvec{x}}_{l_2}$ and $\varvec{P}_g=n^{-2}\sum \nolimits _{l_1,l_2= 1}^n\tilde{\varvec{m}}_{l_1}^T \hat{{\varvec{\tau }}}_{l_1l_2} \{\tilde{\varvec{R}}(\varvec{t}_{l_2})+{\varvec{\varepsilon }}_{l_2}\}$. By the central limit theorem, it follows that $\varvec{V}_g=n^{-2}{\sum \nolimits _{l_1,l_2= 1}^n\tilde{\varvec{m}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\tilde{\varvec{m}}_{l_2}} {\mathop {\longrightarrow }\limits ^{{{p}}}}E(\tilde{\varvec{m}}^T{\varvec{\tau }}\tilde{\varvec{m}})$, $\varvec{K}_g=n^{-2}\sum \nolimits _{l_1,l_2 = 1}^n\tilde{\varvec{m}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\tilde{\varvec{x}}_{l_2} {\mathop {\longrightarrow }\limits ^{{{p}}}}E(\tilde{\varvec{m}}^T{\varvec{\tau }}\tilde{\varvec{x}})$ with $\tilde{\varvec{x}}=(\tilde{\varvec{x}}_1^T,\ldots ,\tilde{\varvec{x}}_n^T)^T$, $\tilde{\varvec{m}}=(\tilde{\varvec{m}}_1^T,\ldots ,\tilde{\varvec{m}}_n^T)^T$ and

$$\begin{aligned}{\varvec{\tau }}=\left( \begin{array}{ccc} {\varvec{\tau }}_{11} &{} \ldots &{} {\varvec{\tau }}_{1n} \\ \vdots &{} \ddots &{} \vdots \\ {\varvec{\tau }}_{n1} &{} \ldots &{} {\varvec{\tau }}_{nn} \\ \end{array} \right) .\end{aligned}$$

By substituting (17) into the Eq. (15), we obtain

$$\begin{aligned} \begin{aligned}&\frac{1}{n^2}\sum \limits _{l_1,l_2 = 1}^n\tilde{\varvec{x}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\{\tilde{\varvec{x}}_{l_2}-\tilde{\varvec{m}}_{l_2}\varvec{V}_g^{-1}\varvec{K}_g\}({\hat{{{\varvec{\beta }}}}}_Q-{{\varvec{\beta }}}_0)+o_p(n^{-1/2})\\&\quad =\frac{1}{n^2}\sum \limits _{l_1,l_2 = 1}^n\tilde{\varvec{x}}_{l_2}^T\hat{{\varvec{\tau }}}_{l_1l_2}\{\tilde{\varvec{R}}(\varvec{t}_{l_2})+{\varvec{\varepsilon }}_{l_2}-\tilde{\varvec{m}}_{l_2}\varvec{V}_g^{-1}\varvec{P}_g\}. \end{aligned} \end{aligned}$$

(18)

By a simple calculation, we can obtain

$$\begin{aligned}n^{-2}\sum \limits _{l_1,l_2= 1}^n\varvec{K}_g^T \varvec{V}_g^{-1}\tilde{\varvec{m}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\{\tilde{\varvec{x}}_{l_2}-\tilde{\varvec{m}}_{l_2}\varvec{V}_g^{-1}\varvec{K}_g\}=0,\end{aligned}$$

$$\begin{aligned}n^{-2}\sum \limits _{l_1,l_2= 1}^n\varvec{K}_g^T \varvec{V}_g^{-1}\tilde{\varvec{m}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\{{\varvec{\varepsilon }}_{l_2}+\tilde{\varvec{R}}(\varvec{t}_{l_2})-\tilde{\varvec{m}}_{l_2}\varvec{V}_g^{-1}\varvec{P}_g\}=0.\end{aligned}$$

By defining $\varvec{x}_i^*=\tilde{\varvec{x}}^T_i-\varvec{K}_g^T\varvec{V}_g^{-1}\tilde{\varvec{m}}_i^T$ and substituting it into (18), it leads to

$$\begin{aligned}&\big \{\frac{1}{n^2}\sum \limits _{l_1,l_2 = 1}^n\varvec{x}_{l_1}^*\hat{{\varvec{\tau }}}_{l_1l_2}\varvec{x}_{l_2}^{*T} +o_p(1)\big \}\sqrt{n}({\hat{{{\varvec{\beta }}}}}_Q-{{\varvec{\beta }}}_0)\\&\quad =\frac{1}{n^{{3}/{2}}}\sum \limits _{l_1,l_2 = 1}^n \varvec{x}_i^{*}\hat{{\varvec{\tau }}}_{l_1l_2}{\varvec{\varepsilon }}_{l_2} -\frac{1}{n^{3/2}}\sum \limits _{l_1,l_2= 1}^n \varvec{x}_{l_{1}}^{*} \hat{{\varvec{\tau }}}_{l_{1}l_{2}} \tilde{\varvec{m}}_{l_{1}}\varvec{V}_g^{-1}\varvec{P}_g +\frac{1}{n^{3/2}}\sum \limits _{l_1,l_2 = 1}^n \varvec{x}_{l_1}^{*}\hat{{\varvec{\tau }}}_{l_1l_2}\tilde{\varvec{R}}(\varvec{t}_{l_2})\\&\quad =:\varvec{A}_1+\varvec{A}_2+\varvec{A}_3. \end{aligned}$$

Note that

$$\begin{aligned} \begin{aligned} \varvec{A}_1&=\frac{1}{n^{3/2}}\sum \limits _{l_1,l_2 = 1}^n\varvec{x}_{l_1}^*\hat{{\varvec{\tau }}}_{l_1l_2}{\varvec{\varepsilon }}_{l_2}=\frac{1}{\sqrt{n}}\sum \limits _{l_2= 1}^n\big \{\frac{1}{n}\sum \limits _{l_1 = 1}^n\varvec{x}_{l_1}^*\hat{{\varvec{\tau }}}_{l_1l_2}{\varvec{\varepsilon }}_{l_2}\big \} =\frac{1}{\sqrt{n}}\sum \limits _{l_2= 1}^n \hat{{\varvec{z}}}_{gl_2}({{\varvec{\beta }}}_0,\hat{{\varvec{\gamma }}}), \end{aligned}\end{aligned}$$

where

$$\begin{aligned} \hat{\varvec{z}}_{gl_2}({{\varvec{\beta }}}_0,\hat{{\varvec{\gamma }}})&=\frac{1}{n}\sum \limits _{l_1 = 1}^n\sum \limits _{k_1,k_2= 1}^n \varvec{x}_{l_1}^*\varvec{A}_{l_1}^{-1/2} \varvec{B}_{k_1}\varvec{A}_{l_1}^{-1/2}\hat{\varvec{W}}_{l_1}\varvec{H}'({\varvec{\eta }}_{0l_1})\varvec{d}_{l_1} \varvec{S}^{-1}_{n,k_1k_2}\\&\quad \times \varvec{d}_{l_2}^T\varvec{H}'({\varvec{\eta }}_{0l_2})\varvec{A}_{l_2}^{-1/2}\varvec{B}_{k_2}\varvec{A}_{l_2}^{-1/2}\hat{\varvec{W}}_{l_2}{\varvec{\varepsilon }}_{l_2}, {\varvec{z}}_{gl_2}({{\varvec{\beta }}}_0,{{\varvec{\gamma }}_0})\\&=\frac{1}{n}\sum \limits _{l_1 = 1}^n\sum \limits _{k_1,k_2= 1}^n \varvec{x}_{l_1}^*\varvec{A}_{l_1}^{-1/2} \varvec{B}_{k_1}\varvec{A}_{l_1}^{-1/2}{\varvec{W}}_{l_1}\varvec{H}'({\varvec{\eta }}_{0l_1})\varvec{d}_{l_1} {\varvec{\Lambda }}^{-1}_{g,k_1k_2}\\&\quad \times \varvec{d}_{l_2}^T\varvec{H}'({\varvec{\eta }}_{0l_2})\varvec{A}_{l_2}^{-1/2}\varvec{B}_{k_2}\varvec{A}_{l_2}^{-1/2}{\varvec{W}}_{l_2}{\varvec{\varepsilon }}_{l_2}. \end{aligned}$$

Using the similar arguments in the proof of Lemma 1, we have

$$\begin{aligned} \frac{1}{{\sqrt{n} }}\sum \limits _{l_2 = 1}^n {\hat{\varvec{z}}_{gl_2}( {{{{\varvec{\beta }}}_0},\hat{{\varvec{\gamma }}} } )\varvec{z}_{gl_2}^T} ( {{{{\varvec{\beta }}}_0},\hat{{\varvec{\gamma }}} } ) {\mathop {\longrightarrow }\limits ^{{{p}}}}{\varvec{B}_{g}}, ~~~\frac{1}{{\sqrt{n} }}\sum \limits _{l_2= 1}^n {\hat{\varvec{z}}_{gl_2}( {{{{\varvec{\beta }}}_0},\hat{{\varvec{\gamma }}} } ) {\mathop {\longrightarrow }\limits ^{d}}} N( {0,{{\varvec{\Psi }}_{g}}} ), \end{aligned}$$

where $\varvec{B}_{g} = E\big ( {{\varvec{z}_{gl_2}}( {{{{\varvec{\beta }}}_0},{{\varvec{\gamma }} _0}} )\varvec{z}_{gl_2}^T( {{{{\varvec{\beta }}}_0},{{\varvec{\gamma }} _0}} )} \big )=E\big \{[\tilde{\varvec{x}}^T{\varvec{\tau }}-\tilde{\varvec{x}}^T{\varvec{\tau }}\tilde{\varvec{m}}[\tilde{\varvec{m}}^T{\varvec{\tau }}\tilde{\varvec{m}}]^{-1}\tilde{\varvec{m}}^T{\varvec{\tau }}] {{\varvec{\varepsilon }}}\big \}^{\otimes 2}$ and ${\varvec{\Psi }}_{g}=\varvec{B}_{g}+E( {{{{\varvec{z}_{gl_2}}( {{{{\varvec{\beta }}}_0},{{\varvec{\gamma }} _0}} )} / {\partial {{\varvec{\gamma }}} }}} ){{\varvec{\Sigma }}}E( {{{{{{\varvec{z}_{gl_2}}( {{{{\varvec{\beta }}}_0},{{\varvec{\gamma }} _0}} )} / {\partial {{\varvec{\gamma }}} }}}^T}} )$. Then we have $\varvec{A}_1{\mathop {\longrightarrow }\limits ^{d}}N(0,{\varvec{\Psi }}_g)$ in distribution. Since

$$\begin{aligned}\frac{1}{n^{3/2}}\sum \limits _{l_1,l_2= 1}^n\varvec{x}_{l_1}^*\hat{{\varvec{\tau }}}_{l_1l_2}\tilde{\varvec{m}}_{l_2}=0,\end{aligned}$$

then $\varvec{A}_2=0$. Expanding ${\varvec{x}}_{l_1}^*$ in $\varvec{A}_3$, we have

$$\begin{aligned} \begin{aligned} \varvec{A}_3&=\frac{1}{n^{3/2}}\sum \limits _{l_1,l_2,=1}^n(\tilde{\varvec{x}}^T_{l_1}-\varvec{K}^T_g\varvec{V}_g^{-1}\tilde{\varvec{m}}_{l_1}^T)\hat{{\varvec{\tau }}}_{l_1l_2}\tilde{\varvec{R}}(\varvec{t}_{l_2})\\&=\frac{1}{n^{3/2}}\sum \limits _{l_1,l_2,=1}^n\big [\tilde{\varvec{x}}^T_{l_1}-\{E^T(\tilde{\varvec{m}}^T{\varvec{\tau }}\tilde{\varvec{x}})\}\{E^{-1}(\tilde{\varvec{m}}^T{\varvec{\tau }}\tilde{\varvec{m}})\}\tilde{\varvec{m}}_{l_1}^T\big ]\hat{{\varvec{\tau }}}_{l_1l_2}\tilde{\varvec{R}}(\varvec{t}_{l_2})\\&\quad -\frac{1}{n^{3/2}}\sum \limits _{l_1,l_2,=1}^n\big [\varvec{K}^T_g\varvec{V}_g^{-1}-\{E^T(\tilde{\varvec{m}}^T{\varvec{\tau }}\tilde{\varvec{x}})\}\{E^{-1}(\tilde{\varvec{m}}^T{\varvec{\tau }}\tilde{\varvec{m}})\}\big ]\tilde{\varvec{m}}_{l_1}^T\hat{{\varvec{\tau }}}_{l_1l_2}\tilde{\varvec{R}}(\varvec{t}_{l_2})\\&=:\varvec{A}_{31}+\varvec{A}_{32}. \end{aligned} \end{aligned}$$

From the definition of $\varvec{K}_g$, it is easy to verify that

$$\begin{aligned}E\big [\big \{\tilde{\varvec{x}}^T_{l_1}-E^T(\tilde{\varvec{m}}^T{\varvec{\tau }}\tilde{\varvec{x}})E^{-1}(\tilde{\varvec{m}}^T{\varvec{\tau }}\tilde{\varvec{m}})\tilde{\varvec{m}}_{l_1}^T\big \}\hat{{\varvec{\tau }}}_{l_1l_2}\tilde{\varvec{m}}_{l_2}\big ]=0,\end{aligned}$$

which implies

$$\begin{aligned}\frac{1}{n^2}\sum \limits _{l_1,l_2=1}^n \big \{\tilde{\varvec{x}}^T_{l_1}-E^T(\tilde{\varvec{m}}^T{\varvec{\tau }}\tilde{\varvec{x}})E^{-1}(\tilde{\varvec{m}}^T{\varvec{\tau }}\tilde{\varvec{m}})\tilde{\varvec{m}}_{l_1}^T\big \}\hat{{\varvec{\tau }}}_{l_1l_2}\tilde{\varvec{m}}_{l_2}=O_p(1).\end{aligned}$$

Recalling that $\Vert {\varvec{m}( {{t_{ij}}})}\Vert = O_p( 1 )$ and $\Vert {{R}( {{t_{ij}}} )} \Vert =o_p(1)$, we have $\varvec{A}_{31}=o_p(1)$. Similarly, we obtain $\varvec{A}_{32}=o_p(1)$ and then $\varvec{A}_{3}=o_p(1)$. By the law of large numbers, we have

$$\begin{aligned}\frac{1}{n^2}\sum \limits _{l_1,l_2 = 1}^n\varvec{x}_{l_1}^*\hat{{\varvec{\tau }}}_{l_1l_2}\varvec{x}_{l_2}^{*T}{\mathop {\longrightarrow }\limits ^{{{p}}}}{\varvec{\Gamma }}_g,\end{aligned}$$

in probability where ${\varvec{\Gamma }}_g=E\big \{\tilde{\varvec{x}}^T{\varvec{\tau }}\tilde{\varvec{x}}-\tilde{\varvec{x}}^T{\varvec{\tau }}\tilde{\varvec{m}}[\tilde{\varvec{m}}^T{\varvec{\tau }}\tilde{\varvec{m}}]^{-1}\tilde{\varvec{m}}^T{\varvec{\tau }}\tilde{\varvec{x}}\big \}$. Hence,

$$\begin{aligned}\sqrt{n}({\hat{{{\varvec{\beta }}}}}_Q-{{\varvec{\beta }}}_0){\mathop {\longrightarrow }\limits ^{d}}N(0,{\varvec{\Gamma }}_g^{-1}{\varvec{\Psi }}_g{\varvec{\Gamma }}_g^{-1}),\end{aligned}$$

in distribution and the proof of Theorem 2 is completed. $\square$