Composite quantile estimation in partial functional linear regression model with dependent errors

Abstract

In this paper, we consider composite quantile estimation for the partial functional linear regression model with errors from a short-range dependent and strictly stationary linear processes. The functional principal component analysis method is employed to estimate the slope function and the functional predictive variable, respectively. Under some regularity conditions, we obtain the optimal convergence rate of the slope function, and the asymptotic normality of the parameter vector. Simulation studies demonstrate that the proposed new estimation method is robust and works much better than the least squares based method when there are outliers in the dataset or the autoregressive error distribution follows a heavy-tailed distribution. Finally, we apply the proposed methodology to electricity consumption data.

Proofs of the Theorems

Proof of Theorem 1

Let $\delta _n=n^{-\frac{2b-1}{2(a+2b)}}$, $\varvec{S}_n=\delta _n^{-1}(\varvec{\hat{\alpha }}-\varvec{\alpha }_0)$, $\varvec{V}_n=\delta _n^{-1}(\varvec{\hat{\gamma }}-\varvec{\gamma }_0)$, $W_{nk}=\delta _n^{-1}(\hat{b}_k-b_{0k})$, $\varvec{W}_n=(W_{n1},\ldots , W_{nK})^T$, $r_i=\int _{0}^{1}\beta _0(t)X_i(t)dt-{\hat{\varvec{U}}}_i^T\varvec{\gamma }_0$, $\mathcal {F}_n=\Big \{(\varvec{S}_n,\varvec{V}_n,\varvec{W}_n){:}\,\big \Vert (\varvec{S}_n^T,\varvec{V}_n^T, \varvec{W}_n^T)^T\big \Vert =L\Big \}$, where L is a large enough constant, ${\varLambda }_n=\text {diag}(\hat{\lambda }_1,\ldots , \hat{\lambda }_m)$, $\varvec{{\varPi }}=E(\varvec{z}_i\varvec{z}_i^T)$, $T_n=\Big \{\left( \varvec{z}_1,X_1(\cdot )\right) ,\ldots , \left( \varvec{z}_n,X_n(\cdot )\right) \Big \}$. We next show that, for any given $\eta >0$, there exists a sufficient large constant $L=L_\eta$ such that

$$\begin{aligned}&P\bigg \{\inf _{(\varvec{S}_n,\varvec{V}_n,\varvec{W}_n)\in \mathcal {F}_n}Q_n(\varvec{\alpha }_0+\delta _n\varvec{S}_n, \varvec{\gamma }_0+\delta _n\varvec{V}_n, \varvec{b}_0+\delta _n\varvec{W}_n)>Q_n(\varvec{\alpha }_0, \varvec{\gamma }_0, \varvec{b}_0)\bigg \} \nonumber \\&\quad \ge 1-\eta . \end{aligned}$$

(15)

This implies with the probability at least $1-\eta$ that there exists a local minimizer $\varvec{\hat{\alpha }}$ and $\varvec{\hat{\gamma }}$ in the ball $\Big \{(\varvec{S}_n,\varvec{V}_n,\varvec{W}_n){:}\,\big \Vert (\varvec{S}_n^T,\varvec{V}_n^T, \varvec{W}_n^T)^T\big \Vert \le L\Big \}$ such that $\Vert \varvec{\hat{\alpha }}-\varvec{\alpha }_0\Vert =O_p(\delta _n)$ and $\Vert \varvec{\hat{\gamma }}-\varvec{\gamma }_0\Vert =O_p(\delta _n)$, which is exactly what we want to show.

Firstly, by $\Vert v_j-\hat{v}_j\Vert ^2=O_p(n^{-1}j^2)$ (see e.g., Shin 2009; Yu et al. 2016a), one has

$$\begin{aligned} \begin{aligned} |r_i|^2&=\bigg |\int _{0}^{1}\beta _0(t)X_i(t)dt-{\hat{\varvec{U}}}_i^T\varvec{\gamma }_0\bigg |^2\\&\le 2 \bigg |\sum _{j=1}^{m}\langle X_i,\hat{v}_j-{v}_j\rangle \gamma _{0j}\bigg |^2+2\bigg |\sum _{j=m+1}^{ \infty }\langle X_i,v_j\rangle \gamma _{0j}\bigg |^2\\&\triangleq 2\text {A}_1+2\text {A}_2. \end{aligned} \end{aligned}$$

For $\text {A}_1$, by conditions C1, C2 and the Hölder inequality, it is obtained

$$\begin{aligned} \begin{aligned} \text {A}_1&=\left| \sum _{j=1}^{m}\langle X_i,{v}_j-\hat{v}_j\rangle \gamma _{0j}\right| ^2 \le cm\sum _{j=1}^{m}\Vert {v}_j-\hat{v}_j\Vert ^2|\gamma _{0j}|^2\\&\le cm\sum _{j=1}^{m}O_p(n^{-1}j^{2-2b}) =O_p(n^{-\frac{a+4b-4}{a+2b}})=o_p(\delta _n^2). \end{aligned} \end{aligned}$$

As for $\text {A}_2$, due to

$$\begin{aligned} \begin{aligned}&E\left\{ \sum _{j=m+1}^{ \infty }\langle X_i,v_j\rangle \gamma _{0j} \right\} =0,\\&\text {Var}\left\{ \sum _{j=m+1}^{ \infty }\langle X_i,{v}_j\rangle \gamma _{0j}\right\} =\sum _{j=m+1}^{ \infty }\lambda _j{\gamma _{0j}}^2 \le c\sum _{j=m+1}^{ \infty }j^{-(a+2b)} =O(n^{-\frac{a+2b-1}{a+2b}}), \end{aligned} \end{aligned}$$

one has

$$\begin{aligned} A_2=O_p\left( n^{-\frac{a+2b-1}{a+2b}}\right) =o_p(\delta _n^2). \end{aligned}$$

Taking these together, we have

$$\begin{aligned} |r_i|^2=O_p\left( n^{-\frac{a+2b-1}{a+2b}}\right) =o_p(\delta _n^2). \end{aligned}$$

(16)

Let

$$\begin{aligned} P_n(\varvec{S}_n,\varvec{V}_n,\varvec{W}_n)=Q_n(\varvec{\alpha }_0+\delta _n\varvec{S}_n, \varvec{\gamma }_0+\delta _n\varvec{V}_n, \varvec{b}_0+\delta _n\varvec{W}_n)-Q_n(\varvec{\alpha }_0, \varvec{\gamma }_0, \varvec{b}_0).\nonumber \\ \end{aligned}$$

(17)

By the Knight identity (1998)

$$\begin{aligned} | r-s|-| r|=-s\left( I{(r>0)}-I{(r<0)}\right) +2\int _{0}^{s}\left[ I{(r\le t)}-I{(r\le 0)}\right] dt, \end{aligned}$$

we have

$$\begin{aligned} \rho _\tau (r-s)-\rho _\tau (r)=s(I{(r<0)}-\tau )+\int _{0}^{s}\left[ I{(r\le t)}-I{(r\le 0)}\right] dt. \end{aligned}$$

(18)

Then we can write $P_n(\varvec{S}_n,\varvec{V}_n,\varvec{W}_n)$ as follows:

$$\begin{aligned}&P_n(\varvec{S}_n,\varvec{V}_n,\varvec{W}_n)\nonumber \\&\quad =\sum _{k=1}^{K}\sum _{i=1}^{n}\delta _n(\varvec{z}_i^T\varvec{S}_n+{\hat{\varvec{U}}}_i^T\varvec{V}_n+W_{nk})\left[ I(e_i<r_i+b_{0k})-\tau _k\right] \nonumber \\&\qquad +\,\sum _{k=1}^{K}\sum _{i=1}^{n}\int _{0}^{\delta _n(\varvec{z}_i^T\varvec{S}_n+{\hat{\varvec{U}}}_i^T\varvec{V}_n+W_{nk})}\left[ I(e_i\le t+r_i+b_{0k})-I(e_i\le r_i+b_{0k})\right] dt\nonumber \\&\quad \triangleq \sqrt{n}\delta _n(\varvec{A}_n^T\varvec{S}_n+\varvec{B}_n^T\varvec{V}_n+\varvec{C}_n^T\varvec{W}_{n})+\sum _{k=1}^{K}\varvec{D}_n^{(k)}, \end{aligned}$$

(19)

where

$$\begin{aligned} \begin{aligned}&\varvec{A}_n=n^{-1/2}\sum _{i=1}^{n}\varvec{z}_i\sum _{k=1}^{K}\left[ I(e_i<r_i+b_{0k})-\tau _k\right] ,\\&\varvec{B}_n=n^{-1/2}\sum _{i=1}^{n}\hat{\varvec{U}}_i\sum _{k=1}^{K}\left[ I(e_i<r_i+b_{0k})-\tau _k\right] ,\\&{C}_{nk}=n^{-1/2}\sum _{i=1}^{n}\left[ I(e_i<r_i+b_{0k})-\tau _k\right] ,\quad \varvec{C}_{n}=({C}_{n1},\ldots , {C}_{nK})^T,\\&\varvec{D}_n^{(k)}=\sum _{i=1}^{n}\int _{0}^{\delta _n(\varvec{z}_i^T\varvec{S}_n+\hat{\varvec{U}}_i^T\varvec{V}_n+W_{nk})}\left[ I(e_i\le t+r_i+b_{0k})-I(e_i\le r_i+b_{0k})\right] dt. \end{aligned} \end{aligned}$$

Note that, by conditions C2 and C4, respectively, one has $\Vert {\varLambda }_n\Vert =O(1)$, $\Vert \varvec{{\varPi }}\Vert =O(1)$, and $\lim _{n\rightarrow \infty }E\varvec{B}_n^T\varvec{V}_n=0$, $E\big \{(\varvec{B}_n^T\varvec{V}_n)^2\big \}=\varvec{V}_n^TE(\varvec{B}_n\varvec{B}_n^T)\varvec{V}_n=O(\Vert \varvec{V}_n\Vert ^2)$. Then we have $\varvec{B}_n^T\varvec{V}_n=O_p(\Vert \varvec{V}_n\Vert )$. Similarly, using condition C4, we get $\varvec{A}_n^T\varvec{S}_n=O_p(\Vert \varvec{S}_n\Vert )$. This combined with (19) leads to

$$\begin{aligned} P_n(\varvec{S}_n,\varvec{V}_n,\varvec{W}_n)=\sum _{k=1}^{K}\varvec{D}_n^{(k)}+o_p(n\delta _n^2)\Vert \varvec{S}_n\Vert +o_p(n\delta _n^2)\Vert \varvec{V}_n\Vert . \end{aligned}$$

(20)

Invoking condition C9, a simple calculation yields

$$\begin{aligned}&E(\varvec{D}_n^{(k)}|T_n)\\&\quad =E\left[ \sum _{i=1}^{n}\int _{0}^{\delta _n(\varvec{z}_i^T\varvec{S}_n+\hat{\varvec{U}}_i^T\varvec{V}_n+W_{nk})}\left( I(e_i\le t+r_i+b_{0k})-I(e_i\le r_i+b_{0k})\right) dt|T_n\right] \\&\quad =\sum _{i=1}^{n}\int _{0}^{\delta _n(\varvec{z}_i^T\varvec{S}_n+\hat{\varvec{U}}_i^T\varvec{V}_n+W_{nk})}\left[ F(t+r_i+b_{0k})-F(r_i+b_{0k})\right] dt\\&\quad =\sum _{i=1}^{n}\int _{0}^{\delta _n(\varvec{z}_i^T\varvec{S}_n+\hat{\varvec{U}}_i^T\varvec{V}_n+W_{nk})}\left[ f(r_i+b_{0k})t(1+o_p(1))\right] dt\\&\quad \le \frac{3f(b_{0k})}{2}n\delta _n^2\left( W^2_{nk}+\varvec{V}_n^T{\varLambda }_n\varvec{V}_n+\varvec{S}_n^T{\varPi }\varvec{S}_n\right) (1+o_p(1)). \end{aligned}$$

Similarly,

$$\begin{aligned} \begin{aligned}&\text {Var}(\varvec{D}_n^{(k)}|T_n)\\&\quad =\text {Var}\left[ \sum _{i=1}^{n}\int _{0}^{\delta _n(\varvec{z}_i^T\varvec{S}_n+\hat{\varvec{U}}_i^T\varvec{V}_n+W_{nk})}\left( I(e_i\le t+r_i+b_{0k})-I(e_i\le r_i+b_{0k})\right) dt|T_n\right] \\&\quad \le \sum _{i=1}^{n}E\left[ \left( \int _{0}^{\delta _n(\varvec{z}_i^T\varvec{S}_n+\hat{\varvec{U}}_i^T\varvec{V}_n+W_{nk})}\left[ I(e_i\le t+r_i+b_{0k})-I(e_i\le r_i+b_{0k})\right] dt\right) ^2|T_n\right] \\&\quad \le \sum _{i=1}^{n}\int _{0}^{|\delta _n(\varvec{z}_i^T\varvec{S}_n+\hat{\varvec{U}}_i^T\varvec{V}_n+W_{nk})|}\int _{0}^{|\delta _n(\varvec{z}_i^T\varvec{S}_n+\varvec{U}_i^T\varvec{V}_n+W_{nk})|}\\&\qquad \times \left[ F(|\delta _n(\varvec{z}_i^T\varvec{S}_n+\hat{\varvec{U}}_i^T\varvec{V}_n+W_{nk})|+r_i+b_{0k})-F(r_i+b_{0k})\right] dt_1dt_2\\&\quad \le o\left( \sum _{i=1}^{n}|\delta _n(\varvec{z}_i^T\varvec{S}_n+\hat{\varvec{U}}_i^T\varvec{V}_n+W_{nk})|^2\right) \\&\quad =o_p(n\delta _n^2)\left( \Vert \varvec{S}_n\Vert ^2+\Vert \varvec{V}_n\Vert ^2+\Vert \varvec{W}_n\Vert ^2\right) . \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} \varvec{D}_n^{(k)}\le \frac{3f(b_{0k})}{2}n\delta _n^2\left( W^2_{nk}+\varvec{V}_n^T{\varLambda }_n\varvec{V}_n+\varvec{S}_n^T{\varPi }\varvec{S}_n\right) (1+o_p(1)). \end{aligned}$$

Then, we can obtain

$$\begin{aligned} \begin{aligned} P_n(\varvec{S}_n,\varvec{V}_n,\varvec{W}_n)&\le \frac{3f(b_{0k})}{2}n\delta _n^2\left( W^2_{nk}+\varvec{V}_n^T{\varLambda }_n\varvec{V}_n+\varvec{S}_n^T{\varPi }\varvec{S}_n\right) (1+o_p(1))\\&\quad +\,o_p(n\delta _n^2)\left( \Vert \varvec{S}_n\Vert ^2+\Vert \varvec{V}_n\Vert ^2+\Vert \varvec{W}_n\Vert ^2\right) . \end{aligned} \end{aligned}$$

(21)

Taking these together, we can obtain that $P_n(\varvec{S}_n,\varvec{V}_n,\varvec{W}_n)$ is dominated by the positive quadratic term $n\delta _n^2\sum _{k=1}^{K}f(b_{0k})\left( W^2_{nk}+\varvec{V}_n^T{\varLambda }_n\varvec{V}_n+\varvec{S}_n^T{\varPi }\varvec{S}_n\right)$ as long as L is large enough. Hence, Eq. (15) holds, and there exists local minimizer $\hat{\varvec{\gamma }}$ such that

$$\begin{aligned} \Vert \varvec{\hat{\gamma }}-\varvec{\gamma }_0\Vert =O_p(\delta _n). \end{aligned}$$

(22)

Observe that

$$\begin{aligned} \begin{aligned} \Vert \hat{\beta }(t)-\beta _0(t)\Vert ^2&=\bigg \Vert \sum _{j=1}^{m}\hat{\gamma }_{j}\hat{v}_j-\sum _{j=1}^{\infty }{\gamma }_{0j}{v}_j\bigg \Vert ^2\\&\le 2\bigg \Vert \sum _{j=1}^{m}\hat{\gamma }_{j}\hat{v}_j-\sum _{j=1}^{m}{\gamma }_{0j}{v}_j\bigg \Vert ^2+ 2\bigg \Vert \sum _{j=m+1}^{\infty }{\gamma }_{0j}{v}_j\bigg \Vert ^2\\&\le 4\bigg \Vert \sum _{j=1}^{m}(\hat{\gamma }_j-\gamma _{0j})\hat{v}_j\bigg \Vert ^2+ 4\bigg \Vert \sum _{j=1}^{m}{\gamma }_{0j}(\hat{v}_j-v_j)\bigg \Vert ^2+2\sum _{j=m+1}^{\infty }{\gamma }_{0j}^2\\&\triangleq 4D_1+4D_2+2D_3. \end{aligned} \end{aligned}$$

Invoking Eq. (16), condition C2, the orthogonality of $\{\hat{v}_j\}$ and $\Vert v_j-\hat{v}_j\Vert ^2=O_p(n^{-1}j^2)$, one has

$$\begin{aligned} D_1&=\bigg \Vert \sum _{j=1}^{m}(\hat{\gamma }_j-\gamma _{0j})\hat{v}_j\bigg \Vert ^2 \le \sum _{j=1}^{m}\big |\hat{\gamma }_j-\gamma _{0j}\big |^2\nonumber \\&=\Vert \hat{\varvec{\gamma }}-\varvec{\gamma }_0\Vert ^2= O_p(\delta _n^2).\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \end{aligned}$$

(23)

$$\begin{aligned} D_2&=\bigg \Vert \sum _{j=1}^{m}{\gamma }_{0j}(\hat{v}_j-v_j)\bigg \Vert ^2\le m\sum _{j=1}^{m}\parallel \hat{v}_j-v_j\parallel ^2{\gamma }_{0j}^2\le \frac{m}{n}O_p\left( \sum _{j=1}^{m}j^2{\gamma }_{0j}^2\right) \nonumber \\&=O_p\left( n^{-1}m\sum _{j=1}^{m}j^{2-2b}\right) =O_p(n^{-1}m)=o_p(n^{-\frac{2b-1}{a+2b}})=o_p(\delta _n^2).\end{aligned}$$

(24)

$$\begin{aligned} D_3&=\sum _{j=m+1}^{\infty }\gamma _{0j}^2\le C\sum _{j=m+1}^{\infty }j^{-2b}=O(n^{-\frac{2b-1}{a+2b}})=O(\delta _n^2).\qquad \qquad \;\;\;\; \end{aligned}$$

(25)

Then, combining Eqs. (23)–(25), we can complete the proof of Theorem 1. $\square$

Proof of Theorem 2

According to Theorem 1, we know that, as $n\rightarrow \infty$, with probability tending to 1, $Q_n(\varvec{\alpha }, \varvec{\gamma }, \varvec{b})$ attains the minimal value at $(\hat{\varvec{\alpha }},\hat{\varvec{\gamma }},\hat{\varvec{b}})$. Then, we have the following score equations

$$\begin{aligned} \begin{aligned}&\frac{1}{n}\sum _{k=1}^{K}\sum _{i=1}^{n}\varvec{z}_i^T\psi _{\tau _k}\left( Y_i-b_{0k}-\varvec{z}_i^T\varvec{\hat{\alpha }}-\hat{\varvec{U}}_i^T \hat{\varvec{\gamma }}\right) =0, \end{aligned} \end{aligned}$$

(26)

$$\begin{aligned} \begin{aligned}&\frac{1}{n}\sum _{k=1}^{K}\sum _{i=1}^{n}\hat{\varvec{U}}_i^T\psi _{\tau _k}\left( Y_i-b_{0k}-\varvec{z}_i^T\varvec{\hat{\alpha }}-\hat{\varvec{U}}_i^T \hat{\varvec{\gamma }}\right) =0, \end{aligned} \end{aligned}$$

(27)

where $\psi _{\tau _k}(u)=\rho '_{\tau _k}(u)=\tau _k-{I}(u<0)$ is score function. By Eqs. (26) and (27), we have

$$\begin{aligned}&\begin{aligned} \frac{1}{n}\sum _{k=1}^{K}\sum _{i=1}^{n}\varvec{z}_i^T\psi _{\tau _k}\left( e_i-b_{0k}-r_i-\varvec{z}_i^T(\varvec{\hat{\alpha }}-\varvec{\alpha }_0)-\hat{\varvec{U}}_i^T (\hat{\varvec{\gamma }}-\varvec{\gamma }_0)\right) =0, \end{aligned}\end{aligned}$$

(28)

$$\begin{aligned}&\begin{aligned} \frac{1}{n}\sum _{k=1}^{K}\sum _{i=1}^{n}\hat{\varvec{U}}_i^T\psi _{\tau _k}\left( e_i-b_{0k}-r_i-\varvec{z}_i^T(\varvec{\hat{\alpha }}-\varvec{\alpha }_0)-\hat{\varvec{U}}_i^T (\hat{\varvec{\gamma }}-\varvec{\gamma }_0)\right) =0. \end{aligned} \end{aligned}$$

(29)

Further, we can write Eq. (28) as

$$\begin{aligned} \begin{aligned} (28)=-\sum _{i=1}^{n}H_n+\sum _{k=1}^{K}B^{(k)}_{n1}+\sum _{k=1}^{K}B^{(k)}_{n2}, \end{aligned} \end{aligned}$$

(30)

where

$$\begin{aligned} H_n= & {} \frac{1}{n}\sum _{i=1}^{n}\varvec{z}_i\sum _{k=1}^{K}\left[ I(e_i<b_{0k}+r_i)-\tau _k\right] , \\ B^{(k)}_{n1}= & {} \frac{1}{n}\sum _{i=1}^{n}\varvec{z}_i\left[ F(b_{0k}+r_i)-F\left( b_{0k}+r_i+\varvec{z}_i^T(\varvec{\hat{\alpha }}-\varvec{\alpha }_0)+\hat{\varvec{U}}_i^T \left( \hat{\varvec{\gamma }}-\varvec{\gamma }_0\right) \right) \right] , \\ B^{(k)}_{n2}= & {} \frac{1}{n}\sum _{i=1}^{n}\varvec{z}_i\left\{ \left[ I(e_i<b_{0k}+r_i)-I(e_i<b_{0k}+r_i+\varvec{z}_i^T(\varvec{\hat{\alpha }}-\varvec{\alpha }_0) \right. \right. \\&\left. \left. \quad +\hat{\varvec{U}}_i^T (\hat{\varvec{\gamma }}-\varvec{\gamma }_0))\right] \right. \\&\left. -\left[ F(b_{0k}+r_i)-F\left( b_{0k}+r_i+\varvec{z}_i^T(\varvec{\hat{\alpha }}-\varvec{\alpha }_0)+\hat{\varvec{U}}_i^T (\hat{\varvec{\gamma }}-\varvec{\gamma }_0)\right) \right] \right\} . \end{aligned}$$

Invoking Taylor expansion and Theorem 1, a simple calculation yields

$$\begin{aligned} B^{(k)}_{n1}=-\frac{1}{n}\sum _{i=1}^{n}f(b_{0k})\left[ \varvec{z}_i\varvec{z}_i^T(\varvec{\hat{\alpha }}-\varvec{\alpha }_0)+\varvec{z}_i\hat{\varvec{U}}_i^T (\hat{\varvec{\gamma }}-\varvec{\gamma }_0)\right] (1+o(1)), \end{aligned}$$

By direct calculation of the mean and variance, we can show, as in Jiang et al. (2012), that $B^{(k)}_{n2}=o_p(\delta _n).$ Thus, we have

$$\begin{aligned}&\frac{1}{n}\sum _{i=1}^{n}\varvec{z}_i\sum _{k=1}^{K}\left[ I(e_i<b_{0k}+r_i)-\tau _k\right] \nonumber \\&\quad =\frac{1}{n}\sum _{i=1}^{n}\sum _{k=1}^{K}f(b_{0k})\left[ \varvec{z}_i\varvec{z}_i^T(\varvec{\hat{\alpha }}-\varvec{\alpha }_0)+\varvec{z}_i\hat{\varvec{U}}_i^T (\hat{\varvec{\gamma }}-\varvec{\gamma }_0)\right] (1+o(1)). \end{aligned}$$

(31)

Similarly, we have

$$\begin{aligned}&\frac{1}{n}\sum _{i=1}^{n}\varvec{U}_i\sum _{k=1}^{K}\left[ I(e_i<b_{0k}+r_i)-\tau _k\right] \nonumber \\&\quad =\frac{1}{n}\sum _{i=1}^{n}\sum _{k=1}^{K}f(b_{0k})\left[ \hat{\varvec{U}}_i\varvec{z}_i^T(\varvec{\hat{\alpha }}-\varvec{\alpha }_0)+\hat{\varvec{U}}_i\hat{\varvec{U}}_i^T (\hat{\varvec{\gamma }}-\varvec{\gamma }_0)\right] (1+o(1)). \end{aligned}$$

(32)

Let ${\varPhi }_n=\frac{1}{n}\sum _{i=1}^{n}\hat{\varvec{U}}_i\hat{\varvec{U}}_i^T$, ${\varPsi }_n=\frac{1}{n}\sum _{i=1}^{n}\hat{\varvec{U}}_i\varvec{z}_i^T$, ${\varUpsilon }_n=\frac{1}{n}\sum _{i=1}^{n}\hat{\varvec{U}}_i\left[ I(e_i<b_{0k}+r_i)-\tau _k\right]$. By Eq. (32), we have

$$\begin{aligned} \hat{\varvec{\gamma }}-\varvec{\gamma }_0=\left( {\varPhi }_n+o_p(1)\right) ^{-1}\left[ {\varUpsilon }_n+{\varPsi }_n(\varvec{\gamma }_0-\varvec{\hat{\gamma }}_0)\right] . \end{aligned}$$

(33)

Substituting Eq. (33) into Eq. (31), we can obtain that

$$\begin{aligned}&\frac{1}{n}\sum _{i=1}^{n}\sum _{k=1}^{K}f(b_{0k})\varvec{z}_i[\varvec{z}_i-{\varPsi }_n^T{\varPhi }_n^{-1}\hat{\varvec{U}}_i]^T(\hat{\varvec{\alpha }}-\varvec{\alpha }_0)+o_p(\hat{\varvec{\alpha }}-\varvec{\alpha }_0)\nonumber \\&\quad =\frac{1}{n}\sum _{i=1}^{n}\sum _{k=1}^{K}\varvec{z}_i\left( \left[ I(e_i<b_{0k}+r_i)-\tau _k\right] -f(b_{0k})\hat{\varvec{U}}_i^T{\varPhi }_n^{-1}{\varUpsilon }_n\right) \end{aligned}$$

(34)

Note that

$$\begin{aligned}&\frac{1}{n}\sum _{i=1}^{n}\sum _{k=1}^{K}f(b_{0k}){{\varPsi }}_n^T{\varPhi }_n^{-1}\hat{\varvec{U}}_i[\varvec{z}_i-{\varPsi }_n^T{\varPhi }_n^{-1}\hat{\varvec{U}}_i]^T=0,\qquad \quad \; \end{aligned}$$

(35)

$$\begin{aligned}&\frac{1}{n}\sum _{i=1}^{n}\sum _{k=1}^{K}f(b_{0k}){{\varPsi }}_n^T{\varPhi }_n^{-1}\hat{\varvec{U}}_i\Big \{\left[ I(e_i<b_{0k}+r_i)-\tau _k\right] -\hat{\varvec{U}}_i^T{\varPhi }_n^{-1}{\varUpsilon }_n\Big \}=0.\qquad \quad \end{aligned}$$

(36)

According to Eqs. (34)–(36), it is easy to show that

$$\begin{aligned}&\left( \frac{1}{n}\sum _{i=1}^{n}\sum _{k=1}^{K}f(b_{0k})\widetilde{\varvec{z}}_i\widetilde{\varvec{z}}_i^T\right) \sqrt{n} (\hat{\varvec{\alpha }}-{\varvec{\alpha }}_0) \nonumber \\&\quad =\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\sum _{k=1}^{K}\left[ I(e_i<b_{0k}+r_i)-\tau _k\right] \widetilde{\varvec{z}}_i+o_p(1), \end{aligned}$$

(37)

where $\widetilde{\varvec{z}}_i={\varvec{z}}_i-{{\varPsi }}_n^T{\varPhi }_n^{-1}\hat{\varvec{U}}_i$. According to Lemma 1 in Yu et al. (2016b) and condition C6, as ${n \rightarrow \infty }$, one has

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^{n}\widetilde{\varvec{z}}_i\widetilde{\varvec{z}}_i^T=\hat{\varvec{{\varSigma }}}\overset{p}{\longrightarrow }\varvec{{\varSigma }}. \end{aligned}$$

Note that

$$\begin{aligned}&\eta _i-\sum _{k=1}^{K}\left[ I(e_i<b_{0k}+r_i)-\tau _k\right] =o_p(1), \\&\lim _{n\rightarrow \infty }\text {E} \left( \sum _{k=1}^{K}\left[ I(e_i<b_{0k}+r_i)-\tau _k\right] \right) =0, \\&\lim _{n\rightarrow \infty }\text {Var} \left( \sum _{k=1}^{K}\left[ I(e_i<b_{0k}+r_i)-\tau _k\right] \right) =\sum _{k,k'=1}^{K}\min (\tau _k,\tau _{k'})\left( 1-\max (\tau _k,\tau _{k'})\right) . \end{aligned}$$

As the random errors are from a stationary process, thus the correlation between $e_i$ and $e_j$ only depends on $|i-j|$, and thus, $E(\eta _i\eta _{i+s})=E(\eta _1\eta _s)$. Invoking conditions C6–C7, we have

$$\begin{aligned} \begin{aligned} \text {Var}(\widetilde{\varvec{z}}_i)&=\frac{1}{n}\widetilde{\varvec{z}}_i\widetilde{\varvec{z}}_i^T E(\eta _i^2)+2\sum _{s=1}^{n-1}\sum _{i=1}^{n-s}\widetilde{\varvec{z}}_i\widetilde{\varvec{z}}_{i+s}^TE(\eta _i\eta _{i+s})+o_p(1)\\&=E(\eta _1)^2\bigg \{\hat{\varvec{{\varSigma }}}+2\sum _{s=1}^{n-1}\varvec{{\varDelta }}_s\frac{E(\eta _1\eta _{1+S})}{E(\eta _1^2)}\bigg \}+o_p(1)\\&=\sum _{k,k'=1}^{K}\min (\tau _k,\tau _{k'})\left( 1-\max (\tau _k,\tau _{k'}\right) \Big \{\hat{\varvec{{\varSigma }}}+2\sum _{s=1}^{n-1}\varvec{{\varDelta }}_s\rho (s)\Big \}+o_p(1)\\&\rightarrow \sum _{k,k'=1}^{K}\min (\tau _k,\tau _{k'})\left( 1-\max (\tau _k,\tau _{k'}\right) \varvec{{\varXi }}. \end{aligned} \end{aligned}$$

Using the central limits theorem, we have

$$\begin{aligned} \sqrt{n}(\hat{\alpha }-\alpha _0)\overset{d}{\longrightarrow } N(0,\varvec{{\varSigma }}_{\text {CQR}}). \end{aligned}$$

(38)

where $\varvec{{\varSigma }}_{\text {CQR}}=\frac{\sum _{k,k'=1}^{K}\min (\tau _k,\tau _{k'})\left( 1-\max (\tau _k,\tau _{k'})\right) }{\left( \sum _{k=1}^{K}f({b_{0k}})\right) ^2}\varvec{{\varSigma }}^{-1}\varvec{{\varXi }}\varvec{{\varSigma }}^{-1}$. We complete the proof of Theorem 2. $\square$