FPCA-based estimation for generalized functional partially linear models

Abstract

In real data analysis, practitioners frequently come across the case that a discrete response will be related to both a function-valued random variable and a vector-value random variable as the predictor variables. In this paper, we consider the generalized functional partially linear models (GFPLM). The infinite slope function in the GFPLM is estimated by the principal component basis function approximations. Then, we consider the theoretical properties of the estimator obtained by maximizing the quasi likelihood function. The asymptotic normality of the estimator of the finite dimensional parameter and the rate of convergence of the estimator of the infinite dimensional slope function are established, respectively. We investigate the finite sample properties of the estimation procedure via Monte Carlo simulation studies and a real data analysis.

Appendix: Proofs

Before we present the proof of Theorem 1, we give some useful lemmas and their proofs. For any matrix $\varvec{A}$, denote $L_2$ norm for $\varvec{A}$ as $\Vert \varvec{A}\Vert _2 = \sup \limits _{\Vert \varvec{x}\Vert \ne 0}\frac{ \Vert \varvec{A}\varvec{x}\Vert }{\Vert \varvec{x}\Vert }.$ Let $\eta _{0i}=\varvec{Z}_i^T\varvec{\alpha }_0+\int _0^1X_i(t)\beta _0(t)dt,$ and $\tilde{\eta }_i=\varvec{Z}_i^T{\varvec{\alpha }}_0+\varvec{U}_i^T\varvec{\gamma }_0.$

Lemma 1

Let $R_i=\int _0^1X_i(t)\beta _0(t)dt-\varvec{U}_i^T\varvec{\gamma }_0$, for $i=1,2,\dots ,n$. Under conditions C1–C7, it has

$$\begin{aligned} \Vert R_i\Vert ^2=\Vert \eta _{0i}-\tilde{\eta }_i \Vert ^2=O_p\left( n^{-(2b+a-1)/(a+2b)}\right) . \end{aligned}$$

Furthermore, one has $\eta _{0i}-\tilde{\eta }_i=O_p\left( n^{-(2b+a-1)/2(a+2b)}\right) .$

Proof of Lemma 1

By the Karhunen–Loève representation and the fact that sequence $\{\phi _j\}$ forms an orthonormal basis in $L^2([0, 1])$, we have

$$\begin{aligned} R_i= & {} \int _0^1X_i(t)\beta _0(t)dt-\varvec{U}_i^T\varvec{\gamma }_0\\= & {} \int _0^1\sum \limits _{j=1}^\infty \sum \limits _{k=1}^\infty \langle X_i,\phi _j\rangle \langle \beta _0,\phi _k\rangle \phi _j(t)\phi _k(t)dt-\sum \limits _{k=1}^m\langle X_i,\hat{\phi }_k\rangle \langle \beta _0,\phi _k\rangle \\= & {} \sum \limits _{k=1}^\infty \langle X_i,\phi _k\rangle \langle \beta _0,\phi _k\rangle -\sum \limits _{k=1}^m\langle X_i,\hat{\phi }_k\rangle \langle \beta _0,\phi _k\rangle \\= & {} \sum \limits _{k=1}^m\langle X_i,\phi _k\rangle \langle \beta _0,\phi _k\rangle -\sum \limits _{k=1}^m\langle X_i,\hat{\phi }_k\rangle \langle \beta _0,\phi _k\rangle +\sum \limits _{k=1+m}^\infty \langle X_i,\phi _k\rangle \langle \beta _0,\phi _k\rangle \\= & {} \sum \limits _{k=1}^m \langle X_i,\phi _k-\hat{\phi }_k\rangle \gamma _k+\sum \limits _{k=1+m}^\infty x_{ik}\gamma _k\\= & {} I_1+I_2, \end{aligned}$$

where

$$\begin{aligned} I_1=\sum \limits _{k=1}^m \langle X_i,\phi _k-\hat{\phi }_k\rangle \gamma _k \end{aligned}$$

and

$$\begin{aligned} I_2= \sum \limits _{k=1+m}^\infty x_{ik}\gamma _k. \end{aligned}$$

First, we consider $I_1.$ By triangle inequality, C1 and C2, we have

$$\begin{aligned} \Vert I_1\Vert ^2= & {} \Vert \sum \limits _{k=1}^m \langle X_i,\phi _k-\hat{\phi }_k\rangle \gamma _k \Vert ^2\\\le & {} m \sum \limits _{k=1}^m \Vert \langle X_i,\phi _k-\hat{\phi }_k\rangle \gamma _k \Vert ^2 \\\le & {} m\sum \limits _{k=1}^m \Vert \phi _k-\hat{\phi }_k\Vert ^2|\gamma _k |^2\\\le & {} O_p(m)\sum \limits _{k=1}^m \frac{k^2}{ n}k^{-2b} \\\le & {} O_p\left( \frac{m}{ n }\right) \sum \limits _{k=1}^m k^{-2b+2}\\\le & {} O_p\left( \frac{m}{n}\right) . \end{aligned}$$

Next, we consider $I_2$. Invoking the fact that $x_{ik}=\langle X_i,\phi _k\rangle$ are uncorrelated random variables, we have

$$\begin{aligned} \text {E}\left( \Vert I_2\Vert ^2\right)= & {} \text {E}\left( \sum \limits _{k=1+m}^\infty x_{ik}\gamma _k\right) ^2\\= & {} \sum \limits _{k=1+m}^\infty \sum \limits _{j=1+m}^\infty \gamma _k\gamma _j \text {E}\left( x_{ik} x_{ij}\right) \\= & {} \sum \limits _{k=1+m}^\infty \gamma _k^2\text {E}\left( x_{ik}^2\right) \\= & {} \sum \limits _{k=1+m}^\infty \gamma _k^2\lambda _k\\\le & {} C\sum \limits _{k=1+m}^\infty k^{-2b} k^{-a} \\\le & {} C\sum \limits _{k=1+m}^\infty k^{-(2b+a)} \\= & {} O\left( m^{-(2b+a-1)}\right) . \end{aligned}$$

By C3, we have $I_2=O_p\left( m^{-(2b+a-1)/2}\right) =O_p\left( n^{-\frac{(2b+a-1)}{2(a+2b)}}\right) .$ Combining the convergence rate of $I_1$ with $I_2$’s, we have

$$\begin{aligned} R_i= & {} O_p\left( \sqrt{\frac{m}{n}}\right) +O_p\left( n^{-\frac{(2b+a-1)}{2(a+2b)}}\right) \\= & {} O_p\left( n^{-\frac{(2b+a-1)}{2(a+2b)}}\right) . \end{aligned}$$

By direct calculation and the fact $\Vert R_i\Vert ^2=\Vert \eta _{0i}-\tilde{\eta }_i \Vert ^2=O_p\left( n^{-(2b+a-1)/(a+2b)}\right)$, we have

$$\begin{aligned} \eta _{0i}-\tilde{\eta }_i=O_p\left( n^{-(2b+a-1)/2(a+2b)}\right) . \end{aligned}$$

Thus Lemma 1 is proved. $\square$

Let

$$\begin{aligned} \check{\varvec{\alpha }}= & {} \arg \max \limits _{\varvec{\alpha }} \sum _{i=1}^n Q\left( Y_i,g^{-1}\left( (\varvec{Z}_i- {\varvec{h}}(X_i))^T\varvec{\alpha }+\varvec{U}_i^T\varvec{\gamma }_0+ {\varvec{h}}(X_i)\varvec{\alpha }_0\right) \right) \\= & {} \sum _{i=1}^n Q\left( Y_i,g^{-1}\left( \tilde{\varvec{Z}}_i ^T\varvec{\alpha }+\varvec{U}_i^T\varvec{\gamma }_0+ {\varvec{h}}(X_i)\varvec{\alpha }_0\right) \right) , \end{aligned}$$

where $\tilde{ \varvec{Z}}_i =\varvec{Z}_i- {\varvec{h}}(X_i)$ and $\varvec{\gamma }_0=( \langle \beta _0, \hat{\phi }_1\rangle ,\langle \beta _0, \hat{\phi }_2\rangle ,\dots ,\langle \beta _0, \hat{\phi }_m\rangle )^T.$ The following lemma gives the asymptotic distribution of $\check{\varvec{\alpha }}.$

Lemma 2

Under conditions C1–C7, it has

$$\begin{aligned} \sqrt{n}(\check{\varvec{\alpha }}- {\varvec{\alpha }}_0) \rightarrow N\left( 0, \Sigma _{1}^{-1} \Sigma _2 \Sigma _{1}^{-1}\right) , \end{aligned}$$

where $\Sigma _{1} =E[\rho _2(\eta _0)\tilde{\varvec{Z}}\tilde{ \varvec{Z}}^T]$, $\Sigma _2 =E[q_1^2(\eta _0)\tilde{\varvec{Z}} \tilde{\varvec{Z}}^T]$ and $\tilde{ \varvec{Z}}=\varvec{Z}-\varvec{h}(X)$.

Proof of Lemma 2

Let $\varvec{u}=\sqrt{n}( \varvec{\alpha }- {\varvec{\alpha }}_0)$ and $\check{\varvec{u}}= \sqrt{n}(\check{\varvec{\alpha }}- {\varvec{\alpha }}_0).$ Since that $\check{\varvec{\alpha }}$ maximizes $\sum _{i=1}^n Q\left( Y_i,g^{-1}\left( \tilde{ \varvec{Z}}_i^T\varvec{\alpha }+\varvec{U}_i^T\varvec{\gamma }_0+{\varvec{h}}(X_i)\varvec{\alpha }_0\right) \right)$, we have $\check{\varvec{u}}$ maximizes

$$\begin{aligned} M_n(\varvec{u})= & {} \sum _{i=1}^n Q\left( Y_i,g^{-1}( \tilde{\varvec{Z}}_i^T{\varvec{\alpha }}_0+\varvec{U}_i^T\varvec{\gamma }_0+{\varvec{h}}(X_i)\varvec{\alpha }_0+ \tilde{ \varvec{Z}}_i^T\varvec{u}/\sqrt{n})\right) \\&- \sum _{i=1}^n Q\left( Y_i,g^{-1}(\varvec{Z}_i^T{\varvec{\alpha }}_0+\varvec{U}_i^T\varvec{\gamma }_0)\right) . \end{aligned}$$

Invoking Taylor expansion, one has

$$\begin{aligned} M_n(\varvec{u})= & {} \frac{1}{\sqrt{n}}\sum _{i=1}^n q_1\left( Y_i,\tilde{\eta }_i\right) \tilde{ \varvec{Z}}_i^T\varvec{u}+\frac{1}{2} \varvec{u}^T \Sigma _n \varvec{u}, \end{aligned}$$

where

$$\begin{aligned} \Sigma _n= & {} \frac{1}{n}\sum _{i=1}^nq_2\left( Y_i,\tilde{\eta }_i+\zeta _i\right) \tilde{ \varvec{Z}}\tilde{ \varvec{Z}}_i^T\\= & {} \frac{1}{n}\sum _{i=1}^n[(Y_i-g^{-1}(\tilde{\eta }_i+\zeta _i))\rho '_1(\tilde{\eta }_i+\zeta _i)-\rho _2(\tilde{\eta }_i+\zeta _i)] \tilde{ \varvec{Z}}_i \tilde{ \varvec{Z}}_i^T \end{aligned}$$

with $\zeta _i$ lies between 0 and $\tilde{ \varvec{Z}}_i^T\varvec{u}/\sqrt{n}$ for $i=1,2,\dots ,n.$

Denote $\mathcal {F}_n=\{ q_2(y,u): y\in R, | u |\le C\},$ where C is a constant. By Theorem 2.7.1 of van der Vaart and Wellner (1996), one has that $\mathcal {F}_n$ is a P-Donsker class of measurable functions. Thus, by Theorem 19.24 of van der Vaart (1998), we have

$$\begin{aligned} \Sigma _n=-\frac{1}{n}\sum _{i=1}^n\rho _2(\eta _{0i}) \tilde{ \varvec{Z}}_i\tilde{ \varvec{Z}}_i^T+o_p(1)=-\text {E}[\rho _2(\eta _{0}) \tilde{\varvec{Z}}\tilde{\varvec{Z}}^T]+o_p(1) =-\Sigma _{1}+o_p(1). \end{aligned}$$

(5)

In addition,

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{i=1}^n q_1\left( Y_i,\tilde{\eta }_i\right) \tilde{ \varvec{Z}}_i= & {} \frac{1}{\sqrt{n}}\sum _{i=1}^n q_1\left( Y_i, \eta _{0i}\right) \tilde{ \varvec{Z}}_i+\frac{1}{\sqrt{n}}\sum _{i=1}^n q_2\left( Y_i, \eta _{0i}\right) \tilde{ \varvec{Z}}_i(\tilde{\eta }_i-\eta _{0i})\\&+ O_p(\sqrt{n} n^{-(2b+a-1)/(a+2b)})\\= & {} \frac{1}{\sqrt{n}}\sum _{i=1}^n q_1\left( Y_i, \eta _{0i}\right) \tilde{ \varvec{Z}}_i+\frac{1}{\sqrt{n}}\sum _{i=1}^n q_2\left( Y_i, \eta _{0i}\right) \tilde{ \varvec{Z}}_i(\tilde{\eta }_i-\eta _{0i})\\&+o_p(1)\\= & {} I_{3}+I_{4}+ o_p(1), \end{aligned}$$

where

$$\begin{aligned} I_{3}=\frac{1}{\sqrt{n}}\sum _{i=1}^n q_1\left( Y_i, \eta _{0i}\right) \tilde{ \varvec{Z}}_i \end{aligned}$$

and

$$\begin{aligned} I_{4}=\frac{1}{\sqrt{n}}\sum _{i=1}^n q_2\left( Y_i, \eta _{0i}\right) \tilde{ \varvec{Z}}_i(\tilde{\eta }_i-\eta _{0i}). \end{aligned}$$

By routine calculation, we have $I_{4}= O_p\left( n^{-(2b+a-1)/2(a+2b)}\right) =o_p(1).$

Therefore, we have

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{i=1}^n q_1\left( Y_i,\tilde{\eta }_i\right) \tilde{ \varvec{Z}}_i = \frac{1}{\sqrt{n}}\sum _{i=1}^n q_1\left( Y_i, \eta _{0i} \right) \tilde{ \varvec{Z}}_i + o_p(1). \end{aligned}$$

(6)

Let $\varvec{W}_n=\frac{1}{\sqrt{n}}\sum _{i=1}^n q_1\left( Y_i, \eta _{0i} \right) \tilde{\varvec{Z}}_i.$ By the Lindeberg-Feller central limit theorem, we have

$$\begin{aligned} \varvec{W}_n \rightarrow N\left( 0, \Sigma _2 \right) , \end{aligned}$$

(7)

where $\Sigma _2 =E[q_1^2(Y,\eta _0)\tilde{\varvec{Z}} \tilde{\varvec{Z}}^T].$ Combining (5) with (7), we have

$$\begin{aligned} M_n(\varvec{u})= & {} \varvec{W}_n^T \varvec{u}-\frac{1}{2} \varvec{u}^T \Sigma _{1} \varvec{u}+o_p(1). \end{aligned}$$

Here, the epi-convergence results of Geyer (1994) and Convexity Lemma of Pollard (1991) imply that

$$\begin{aligned} \check{\varvec{u}}= \Sigma _{1}^{-1}\varvec{W}_n +o_p(1). \end{aligned}$$

(8)

By Slutsky Lemma, we have

$$\begin{aligned} \sqrt{n}(\check{\varvec{\alpha }}- {\varvec{\alpha }}_0) \rightarrow N\left( 0, \Sigma _{1}^{-1} \Sigma _2 \Sigma _{1}^{-1}\right) . \end{aligned}$$

Thus, the proof for Lemma 2 is completed. $\square$

Denote $\hat{\varvec{\theta }}=(\hat{\varvec{\alpha }}^T,\hat{\varvec{\gamma }}^T)^T$ and $\check{{\varvec{\theta }}}=( {\check{\varvec{\alpha }}}^T, {\varvec{\gamma }_0}^T)^T.$ We further have the following lemma.

Lemma 3

Under conditions C1–C7, it has

$$\begin{aligned} \Vert \hat{\varvec{\theta }}-\check{{\varvec{\theta }}}\Vert =O_p\left( n^{-(2b-1)/2(a+2b) }\right) . \end{aligned}$$

Proof of Lemma 3

Note that

$$\begin{aligned} \frac{\partial L(\varvec{\theta })}{\partial \varvec{\theta }}\large \mid _{\varvec{\theta }=\hat{\varvec{\theta }}}- \frac{\partial L(\varvec{\theta })}{\partial \varvec{\theta }} \mid _{\varvec{\theta }= \check{{\varvec{\theta }}}}= \frac{\partial ^2 L(\varvec{\theta })}{\partial \varvec{\theta }\partial \varvec{\theta }^T} \mid _{\varvec{\theta }= \bar{\varvec{\theta }}}(\hat{\varvec{\theta }}-\check{{\varvec{\theta }}}), \end{aligned}$$

where $\bar{\varvec{\theta }}=\nu \hat{\varvec{\theta }}+(1-\nu )\check{{\varvec{\theta }}}$, and $0\le \nu \le 1.$ By the fact that $\frac{\partial L(\hat{\varvec{\theta }})}{\partial \varvec{\theta }}=0,$ we have

$$\begin{aligned} \hat{\varvec{\theta }}-\check{{\varvec{\theta }}}=-\left( \frac{\partial ^2 L(\varvec{\theta })}{\partial \varvec{\theta }\partial \varvec{\theta }^T} \mid _{\varvec{\theta }= \bar{\varvec{\theta }}}\right) ^{-1} \frac{\partial L(\varvec{\theta })}{\partial \varvec{\theta }} \mid _{\varvec{\theta }= \check{{\varvec{\theta }}}}. \end{aligned}$$

First, we consider $\frac{\partial L(\varvec{\theta })}{\partial \varvec{\theta }} \mid _{\varvec{\theta }= \check{{\varvec{\theta }}}}.$

$$\begin{aligned} \frac{\partial L(\varvec{\theta })}{\partial \varvec{\theta }} \mid _{\varvec{\theta }= \check{{\varvec{\theta }}}}= & {} \left\{ \left( \frac{\partial L(\check{{\varvec{\theta }}})}{\partial \varvec{\alpha }}\right) ^T, \left( \frac{\partial L(\check{{\varvec{\theta }}})}{\partial \varvec{\gamma }}\right) ^T\right\} ^T\\= & {} \sum _{i=1}^n q_1\left( Y_i, \check{\eta }_{i}\right) (\varvec{Z}_i^T,\varvec{U}_i^T)^T , \end{aligned}$$

where $\frac{\partial L(\check{{\varvec{\theta }}})}{\partial \varvec{\alpha }}= \sum _{i=1}^n q_1\left( Y_i, \check{\eta }_{i}\right) \varvec{Z}_i$, $\frac{\partial L(\check{{\varvec{\theta }}})}{\partial \varvec{\gamma }} = \sum _{i=1}^n q_1\left( Y_i, \check{\eta }_{i}\right) \varvec{U}_i.$ and $\check{{\eta _i}}=\varvec{Z}_i^T{\check{\alpha }}+\varvec{U}_i^T\varvec{\gamma }_0,\ i=1,\ldots ,n$. Note that

$$\begin{aligned} \frac{\partial L(\check{{\varvec{\theta }}})}{\partial \varvec{\alpha }}= & {} \sum _{i=1}^n q_1\left( Y_i, \check{\eta }_{i}\right) \varvec{Z}_i\\= & {} \sum _{i=1}^n q_1\left( Y_i, \eta _{0i}\right) \varvec{Z}_i+ \sum _{i=1}^n q_2\left( Y_i, { \eta }^*_{i}\right) (\varvec{Z}_i^T(\check{\varvec{\alpha }}- {\varvec{\alpha }}_0) +R_i)\varvec{Z}_i \end{aligned}$$

and

$$\begin{aligned} \frac{\partial L(\check{{\varvec{\theta }}})}{\partial \varvec{\gamma }}= & {} \sum _{i=1}^n q_1\left( Y_i, \check{\eta }_{i}\right) \varvec{U}_i\\= & {} \sum _{i=1}^n q_1\left( Y_i, \eta _{0i}\right) \varvec{U}_i+ \sum _{i=1}^n q_2\left( Y_i, { \eta }^*_{i}\right) (\varvec{Z}_i^T(\check{\varvec{\alpha }}- {\varvec{\alpha }}_0) +R_i)\varvec{U}_i, \end{aligned}$$

where ${ \eta }^*_{i}=\nu \eta _{0i}+(1-\nu )\check{\eta }_{i},$ $0\le \nu \le 1$ for $i=1,2,\dots ,n.$ By routine calculation, we have

$$\begin{aligned} E\left( \left\| \sum _{i=1}^n q_1\left( Y_i, \eta _{0i}\right) \varvec{Z}_i\right\| \right) = O(\sqrt{n}). \end{aligned}$$

(9)

In addition, by condition C2 and the fact $\Vert R_i\Vert ^2=O_p(n^{-(2b+a-1)/(a+2b)})$, we have

$$\begin{aligned} \left\| \sum _{i=1}^n q_2\left( Y_i, { \eta }^*_{i}\right) (\varvec{Z}_i^T(\check{\varvec{\alpha }}- {\varvec{\alpha }}_0) +R_i)\varvec{Z}_i \right\|= & {} O_p(\sqrt{n})+O_p\left( n\cdot n^{-(2b+a-1)/2(a+2b) }\right) \nonumber \\= & {} O_p\left( n\cdot n^{-(2b+a-1)/2(a+2b) }\right) . \end{aligned}$$

(10)

Combining (9) with (10), one has $\frac{\partial L(\check{{\varvec{\theta }}})}{\partial \varvec{\alpha }}= \sum _{i=1}^n q_1\left( Y_i, \check{\eta }_{i}\right) \varvec{Z}_i=O_p\left( n n^{-(2b+a-1)/2(a+2b) }\right) .$

Similarly, we have

$$\begin{aligned} E\left( \left\| \sum _{i=1}^n q_1\left( Y_i, \eta _{0i}\right) \varvec{U}_i\right\| ^2\right) =O(nm). \end{aligned}$$

(11)

and

$$\begin{aligned} \Vert \sum _{i=1}^n q_2\left( Y_i, { \eta }^*_{i}\right) (\varvec{Z}_i^T(\check{\varvec{\alpha }}- {\varvec{\alpha }}_0) +R_i)\varvec{U}_i \Vert =O_p(\sqrt{nm}). \end{aligned}$$

(12)

Combining (11) with (12), one has $\frac{\partial L(\check{{\varvec{\theta }}})}{\partial \varvec{\gamma } }= \sum _{i=1}^n q_1\left( Y_i, {\eta }^*_{i}\right) \varvec{U}_i= O_p\left( n n^{-(2b+a-1)/2(a+2b) }\right) .$ Therefore, we have

$$\begin{aligned} \left\| \frac{\partial L(\varvec{\theta })}{\partial \varvec{\theta }} \mid _{\varvec{\theta }= \check{{\varvec{\theta }}}}\right\| = O_p\left( n n^{-(2b+a-1)/2(a+2b) }\right) . \end{aligned}$$

(13)

Following Lemma A.3 of Wang et al. (2011), we have

$$\begin{aligned} \left\| \left( \frac{1}{n}\frac{\partial ^2 L(\varvec{\theta })}{\partial \varvec{\theta }\partial \varvec{\theta }^T} \mid _{\varvec{\theta }= \bar{\varvec{\theta }}}\right) ^{-1}\right\| =O_p\left( \frac{1}{\sqrt{\lambda _m}}\right) =O_p\left( \sqrt{m^{a}}\right) . \end{aligned}$$

(14)

By (13) and (14), we have

$$\begin{aligned} \Vert \hat{\varvec{\theta }}-\check{{\varvec{\theta }}}\Vert\le & {} \left\| \left( \frac{1}{n}\frac{\partial ^2 L(\varvec{\theta })}{\partial \varvec{\theta }\partial \varvec{\theta }^T} \mid _{\varvec{\theta }= \bar{\varvec{\theta }}}\right) ^{-1} \right\| \left\| \frac{1}{n} \frac{\partial L(\varvec{\theta })}{\partial \varvec{\theta }} \mid _{\varvec{\theta }= \check{{\varvec{\theta }}}}\right\| \\= & {} O_p\left( \sqrt{m^{a}}\right) O_p\left( n^{-(2b+a-1)/2(a+2b) }\right) \\= & {} O_p\left( n^{-(2b-1)/2(a+2b) }\right) . \end{aligned}$$

$\square$

Proof of Theorem 1

Similar to Lu et al. (2014), one has

$$\begin{aligned} \Vert \hat{\beta }-\beta _{0}\Vert ^{2}= & {} \left\| \sum _{j=1}^{m}\hat{\gamma }_{j}\hat{\phi }_{j}-\sum _{j=1}^{\infty }\gamma _{j}\phi _{j}\right\| ^{2}\\\le & {} 2\left\| \sum _{j=1}^{m}\hat{\gamma }_{j}\hat{\phi }_{j}-\sum _{j=1}^{m}\gamma _{j}\phi _{j}\right\| ^{2} +2\left\| \sum _{j=m+1}^{\infty }\gamma _{j}\phi _{j}\right\| ^{2}\\\le & {} 4\left\| \sum _{j=1}^{m}(\hat{\gamma }_{j}-\gamma _{j})\hat{\phi }_{j}\right\| ^{2} +4\left\| \sum _{j=1}^{m}\gamma _{j}(\hat{\phi }_{j}-\phi _{j})\right\| ^{2} +2\sum _{j=m+1}^{\infty }\gamma _{j}^{2}\\\le & {} 4\left\| \hat{\varvec{\gamma }} -\varvec{\gamma }_{0} \right\| ^{2} +4\left\| \sum _{j=1}^{m}\gamma _{j}(\hat{\phi }_{j}-\phi _{j})\right\| ^{2} +2\sum _{j=m+1}^{\infty }\gamma _{j}^{2}\\= & {} 4\left\| \hat{\varvec{\gamma }} -\varvec{\gamma }_{0} \right\| ^{2}+4I_{5}+2I_{6}. \end{aligned}$$

Now, we consider $I_{5}.$ Note that $\Vert \phi _{j}-\hat{\phi }_{j}\Vert ^2=O_p(n^{-1}j^2)$, one has

$$\begin{aligned} I_{5}\le & {} m\sum _{j=1}^{m}\Vert \hat{\phi }_{j}-\phi _{j}\Vert ^{2}\gamma _{j}^{2}\\\le & {} O_p(1) n^{-1}m\sum _{j=1}^{m}j^{2}\gamma _{j}^{2}\\= & {} O_{p}\left( n^{-1}m\right) \\= & {} O_{p}\left( n^{-\frac{a+2b-1}{a+2b}}\right) . \end{aligned}$$

Next, we consider $I_{6}.$

$$\begin{aligned} I_{6}\le & {} C\sum _{j=m+1}^{\infty }\gamma _{j}^{2}\\\le & {} C\sum _{j=m+1}^{\infty }j^{-2b}\\= & {} O\left( m^{-(2b-1)}\right) \\= & {} O\left( n^{-\frac{2b-1}{a+2b}}\right) . \end{aligned}$$

Therefore, combining these with Lemma 3, we have

$$\begin{aligned} \Vert \hat{\beta }-\beta _0\Vert ^{2}=O_p\left( n^{-(2b-1)/(a+2b) }\right) . \end{aligned}$$

Next, we prove the asymptotic normality of $\hat{\varvec{\alpha }}$. Let $\hat{\eta }_{i}=\varvec{Z}_i^T\hat{\varvec{\alpha }}+\varvec{U}_i^T\hat{\varvec{\gamma }}.$ For any $\varvec{v}\in R^p$, define

$$\begin{aligned} \hat{\eta }(\varvec{v})= & {} \varvec{Z}^T\hat{\varvec{\alpha }}+\varvec{U}^T\hat{\varvec{\gamma }}+(\varvec{Z}-h(X))^T\varvec{v} \\= & {} \varvec{Z}^T(\hat{\varvec{\alpha }}+\varvec{v})+\varvec{U}^T\hat{\varvec{\gamma }}- h(X)^T\varvec{v} . \end{aligned}$$

Note that when $\varvec{v} = 0,$ $\hat{\eta }(\varvec{v})$ maximizes $L(\hat{\eta }(\varvec{v}))= \sum _{i=1}^nQ\left( Y_i,g^{-1}\left( \hat{\eta }_i(\varvec{v})\right) \right) .$ Thus, one has

$$\begin{aligned} \frac{\partial L(\hat{\eta }(\varvec{v}))}{ \partial \varvec{v}}\large \mid _{\varvec{v}=0}=0. \end{aligned}$$

By Lemma 3, we have

$$\begin{aligned} 0=\frac{\partial L(\hat{\eta }(\varvec{v}))}{ \partial \varvec{v}}\large \mid _{\varvec{v}=0}= & {} \sum _{i=1}^n q_1\left( Y_i, \hat{\eta }_{i}\right) \tilde{ \varvec{Z}}_i\\= & {} \sum _{i=1}^n \left( Y_i-g^{-1}(\hat{\eta }_{i})\right) \rho _1(\hat{\eta }_{i}) \tilde{\varvec{Z}}_i\\= & {} \sum _{i=1}^n \left( Y_i-g^{-1}( {\eta }_{0i})\right) \rho _1({\eta }_{0i}) \tilde{\varvec{Z}}_i\\&+ \sum _{i=1}^n \varepsilon _i(\rho _1(\hat{\eta }_{i})-\rho _1({\eta }_{0i})) \tilde{\varvec{Z}}_i\\&+ \sum _{i=1}^n \left( g^{-1}( {\eta }_{0i})-g^{-1}(\hat{\eta }_{i})\right) \rho _1(\hat{\eta }_{i})\tilde{\varvec{Z}}_i\\= & {} A_1+A_2+A_3. \end{aligned}$$

We first consider $A_2.$ Obviously, $E(A_2)=0.$ By Lemma 1, we have

$$\begin{aligned} E(A_2^2)= & {} \sum _{i=1}^n E\left( V(\eta _i)(\rho _1(\hat{\eta }_{i})-\rho _1({\eta }_{0i}))^2 \tilde{\varvec{Z}}_i\tilde{\varvec{Z}}_i^T\right) \\\le & {} C \sum _{i=1}^n E\left( (\hat{\eta }_{i}-{\eta }_{i})^2 \tilde{\varvec{Z}}_i\tilde{\varvec{Z}}_i^T\right) \\= & {} O(m). \end{aligned}$$

Thus, one has $A_2=O_p\left( \sqrt{m}\right) =o_p\left( {\sqrt{n}}\right) .$ Now, we consider $A_3.$ Note that

$$\begin{aligned} A_3= & {} \sum _{i=1}^n \left( g^{-1}( {\eta }_{0i})-g^{-1}(\hat{\eta }_{i})\right) \rho _1(\hat{\eta }_{i})\tilde{ \varvec{Z}}_i\\= & {} \sum _{i=1}^n \left( g^{-1}( {\eta }_{0i})-Y_i+Y_i-g^{-1}(\hat{\eta }_{i})\right) \rho _1(\hat{\eta }_{i})\tilde{ \varvec{Z}}_i\\= & {} \sum _{i=1}^n \left( Y_i-g^{-1}(\hat{\eta }_{i})\right) \rho _1(\hat{\eta }_{i})\tilde{ \varvec{Z}}_i-\sum _{i=1}^n\left( Y_i-g^{-1}({\eta }_{0i})\right) \rho _1({\eta }_{0i})\tilde{ \varvec{Z}}_i\\&-\sum _{i=1}^n \left( Y_i-g^{-1}({\eta }_{0i})\right) (\rho _1({\hat{\eta }}_{i})-\rho _1({\eta }_{0i}))\tilde{ \varvec{Z}}_i\\= & {} \sum _{i=1}^n (q_1(\hat{\eta }_{i})-q_1({\eta }_{0i}))\tilde{ \varvec{Z}}_i-\sum _{i=1}^n \left( Y_i-g^{-1}({\eta }_{0i})\right) (\rho _1({\hat{\eta }}_{i})-\rho _1({\eta }_{0i}))\tilde{ \varvec{Z}}_i\\= & {} \sum _{i=1}^n q_2({\eta }_{0i})(\hat{\eta }_i-\eta _{0i})\tilde{ \varvec{Z}}_i+o_p(\sqrt{n})\\&-\sum _{i=1}^n \left( Y_i-g^{-1}({\eta }_{0i})\right) (\rho _1({\hat{\eta }}_{i})-\rho _1({\eta }_{0i}))\tilde{ \varvec{Z}}_i\\= & {} \sum _{i=1}^n \left( Y_i-g^{-1}({\eta }_{0i})\right) \rho _1'({\eta }_{0i})(\hat{\eta }_i-\eta _{0i})\tilde{ \varvec{Z}}_i-\sum _{i=1}^n \rho _2(\eta _{i0})(\hat{\eta }_i-\eta _{0i})\tilde{ \varvec{Z}}_i\\&-\sum _{i=1}^n \left( Y_i-g^{-1}({\eta }_{0i})\right) (\rho _1({\hat{\eta }}_{i})-\rho _1({\eta }_{0i}))\tilde{ \varvec{Z}}_i+o_p(\sqrt{n})\\= & {} -\sum _{i=1}^n \rho _2(\eta _{i0})(\hat{\eta }_i-\eta _{0i})\tilde{ \varvec{Z}}_i+o_p(\sqrt{n}). \end{aligned}$$

Invoking Lemmas 2 and 3, we have

$$\begin{aligned} \sum _{i=1}^n \rho _2(\eta _{i0})(\hat{\eta }_i-\eta _{0i})\tilde{ \varvec{Z}}_i=\sum _{i=1}^n \rho _2(\eta _{i0})\tilde{ \varvec{Z}}_i \tilde{ \varvec{Z}}^T_i(\hat{\varvec{\alpha }}-\varvec{\alpha }_0)+o_p(\sqrt{n}). \end{aligned}$$

Thus, by $A_1,$ $A_2$ and $A_3$, we have

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n \rho _2( {\eta }_{0i})\tilde{ \varvec{Z}}_i\tilde{ \varvec{Z}}_i^T(\hat{\varvec{\alpha }}-{\varvec{\alpha }_0})=\frac{1}{n}\sum _{i=1}^n \left( Y_i-g^{-1}( {\eta }_{0i})\right) \rho _1({\eta }_{0i})\tilde{\varvec{Z}}_i+o_p\left( \frac{1}{\sqrt{n}}\right) . \end{aligned}$$

Furthermore, by the central limit theorem and slutsky’s Lemma, we have

$$\begin{aligned} \sqrt{n}(\hat{\varvec{\alpha }}-{\varvec{\alpha }}_0)\rightarrow N(0,\Sigma _{1}^{-1}\Sigma _{2} \Sigma _{1}^{-1}), \end{aligned}$$

where $\Sigma _{1}=E\left( \rho _2( {\eta _0})\tilde{ \varvec{Z}}\tilde{ \varvec{Z}}^T\right)$ and $\Sigma _{2}=E\left( q_1^2( {\eta _0})\tilde{ \varvec{Z}}\tilde{\varvec{Z}}^T\right)$. $\square$