Robust estimation for a general functional single index model via quantile regression

Abstract

This paper studies the estimation of a general functional single index model, in which the conditional distribution of the response depends on the functional predictor via a functional single index structure. We find that the slope function can be estimated consistently by the estimation obtained by fitting a misspecified functional linear quantile regression model under some mild conditions. We first obtain a consistent estimator of the slope function using functional linear quantile regression based on functional principal component analysis, and then employ a local linear regression technique to estimate the conditional quantile function and establish the asymptotic normality of the resulting estimator for it. The finite sample performance of the proposed estimation method is studied in Monte Carlo simulations, and is illustrated by an application to a real dataset.

Appendix

1.1 A.1 Proof of Proposition 1

It suffices to prove that, for any constant $u \in \mathbb {R}$, there exists a constant c such that $\mathcal {L}_{\tau }(u, \beta )\ge \mathcal {L}_{\tau }(u, c\beta _{0})$. Specifically,

$$\begin{aligned} \mathcal {L}_{\tau }(u, \beta )&=\mathbb {E}\left[ \mathbb {E}\left\{ \rho _{\tau }\left( Y-u-\langle X,\beta \rangle \right) \big |\langle X,\beta _{0}\rangle , \varepsilon \right\} \right] \nonumber \\&\ge \mathbb {E}\left[ \rho _{\tau }\left\{ \mathbb {E}\left( Y|\langle X,\beta _{0}\rangle , \varepsilon \right) -u-\mathbb {E}\left( \langle X,\beta \rangle \big |\langle X,\beta _{0}\rangle , \varepsilon \right) \right\} \right] \nonumber \\&= \mathbb {E}\left[ \rho _{\tau }\left\{ Y-u-\mathbb {E}\left( \langle X,\beta \rangle \big |\langle X,\beta _{0}\rangle \right) \right\} \right] \nonumber \\&= \mathbb {E}\left\{ \rho _{\tau }\left( Y-u-c\langle X,\beta _{0}\rangle \right) \right\} , \end{aligned}$$

where $c=\langle K(\beta _{0}), \beta \rangle /\langle K(\beta _{0}), \beta _{0}\rangle$ and the inequality follows from the convexity of $\rho _{\tau }$ and Jensen’s inequality, and the last equality holds true by invoking the linearity condition (7). This completes the proof of Proposition 1.

1.2 A.2 Proof of Theorem 1

We begin by defining some notations to be used in the proofs. Let $\xi _{i}=\left( 1,\xi _{i1},\ldots ,\xi _{im_{n}}\right) ^{\top }$, ${\widehat{\xi }}_{i}=\left( 1,{\widehat{\xi }}_{i1},\ldots ,{\widehat{\xi }}_{im_{n}}\right) ^{\top }$, ${\widetilde{\theta }}=\left( u_{\tau },\alpha _{\tau 1}, \ldots , \alpha _{\tau m_{n}}\right) ^{\top }$, $A=\left( u, \alpha _{1}, \ldots , \alpha _{m_{n}}\right) ^{\top }$ and  $\zeta _{i}=\sum _{j=m_{n}+1}^{\infty }\alpha _{\tau j}\xi _{ij}$. Then the objective function in (9) can be rewritten as

$$\begin{aligned} \sum _{i=1}^{n}\rho _{\tau }\left( \xi _{i}^{\top }{\widetilde{\theta }}-{\widehat{\xi }}_{i} ^{\top }A+\zeta _{i}+\varepsilon _{i\tau }\right) . \end{aligned}$$

(14)

Let

$$\begin{aligned} W_{i}=\left( \xi _{i}-{\widehat{\xi }}_{i}\right) ^{\top }\cdot \left( 0,\alpha ^{\top }\right) ^{\top }+\zeta _{i}, \end{aligned}$$

(15)

$\Lambda =diag \left( 1, \lambda _{1},\ldots ,\lambda _{m_{n}}\right)$, $\theta =\left( n^{1/2}(u-u_\tau ), U^{\top }\right) ^{\top }$, ${\widetilde{\xi }}_{i}=n^{-1/2}\Lambda ^{-1/2}{\widehat{\xi }}_{i}$, where

$$\begin{aligned} diag \left( 1, \lambda _{1},\ldots ,\lambda _{m_{n}}\right) =\begin{pmatrix} 1 &{} &{} &{} \\ &{}\lambda _{1} &{} &{} \\ &{} &{} \ddots &{} \\ &{} &{} &{}\lambda _{m_{n}}\\ \end{pmatrix}, \end{aligned}$$

$U=(u_{1},\ldots , u_{m_{n}})^{\top }=n^{1/2}\Lambda _{1}^{1/2}(b-\alpha )$, $b=\left( \alpha _{1}, \ldots , \alpha _{m_{n}}\right) ^{\top }$, and $\Lambda _{1}=diag \left( \lambda _{1},\ldots ,\lambda _{m_{n}}\right) .$ Let

$$\begin{aligned} S_{n}(\theta )=\sum _{i=1}^{n}\left\{ \rho _{\tau }\left( W_{i}+\varepsilon _{i\tau }-{\widetilde{\xi }}_{i}^{\top }\theta \right) - \rho _{\tau }\left( W_{i}+\varepsilon _{i\tau }\right) \right\} . \end{aligned}$$

(16)

It’s easy to see that minimizing (14) with respect to A is equivalent to minimizing (16) over $\theta .$ Let $\psi _{\tau }(t)=\tau -\mathbf {1}(t<0),$ $\Omega =\{X_{1}, \ldots , X_{n}\}$, $S_{ni}=\rho _{\tau }\left( W_{i}+\varepsilon _{i\tau }-{\widetilde{\xi }}_{i}^{\top }m_{n}^{1/2}\theta \right) - \rho _{\tau }\left( W_{i}+\varepsilon _{i\tau }\right)$, $\Gamma _{ni}=\mathbb {E}(S_{ni}|\Omega )$, $R_{ni}=S_{ni}-\Gamma _{ni}+m_{n}^{1/2}{\widetilde{\xi }}_{i}^{\top }\theta \psi _{\tau }\left( \varepsilon _{i\tau }\right)$, $S_{n}=\sum _{i=1}^{n}S_{ni}$, $\Gamma _{n}=\sum _{i=1}^{n}\Gamma _{ni}$, $R_{n}=\sum _{i=1}^{n}R_{ni}$. Then

$$\begin{aligned} S_{n}\left( m_{n}^{1/2}\theta \right) =\Gamma _{n}-m_{n}^{1/2}\sum _{i=1}^{n}{\widetilde{\xi }}_{i}^{\top }\theta \psi _{\tau }\left( \varepsilon _{i\tau }\right) +R_{n}. \end{aligned}$$

Lemma 1

Let $Z_{1}, \ldots , Z_{n}$ be arbitrary scalar random variables such that $\mathop {max }\limits _{1\le i \le n}\mathbb {E}\left( |Z_{i}|^{L_{0}}\right) <\infty$ for some $L_{0} \ge 1$. Then, we have

$$\begin{aligned} \mathbb {E}\left( \mathop {max }\limits _{1\le i \le n}|Z_{i}|\right) \le n^{1/L_{0}}\mathop {max }\limits _{1\le i \le n}\left\{ \mathbb {E}\left( |Z_{i}|^{L_{0}}\right) \right\} ^{1/L_{0}}. \end{aligned}$$

Proof of Lemma 1

This inequality follows from the observation that

$$\begin{aligned}&\mathbb {E}\left( \mathop {max }\limits _{1\le i \le n}|Z_{i}|\right) \le \left\{ \mathbb {E}\left( \mathop {max }\limits _{1\le i \le n}|Z_{i}|^{L_{0}}\right) \right\} ^{1/L_{0}}\nonumber \\&\le \left\{ \sum _{i=1}^{n}\mathbb {E}\left( |Z_{i}|^{L_{0}}\right) \right\} ^{1/L_{0}}\le n^{1/L_{0}}\mathop {max }\limits _{1\le i \le n}\left\{ \mathbb {E}\left( |Z_{i}|^{L_{0}}\right) \right\} ^{1/L_{0}}. \end{aligned}$$

$\square$

Lemma 2

Under conditions (C1)–(C3) and (C5), we have

$$\begin{aligned} \max _{1 \le i \le n} \left\| {\widetilde{\xi }}_{i} \right\| _{2} = o_{p} \left\{ m_{n}^{-1/2}(\log n)^{-1} \right\} , \end{aligned}$$

where $\left\| {\widetilde{\xi }}_{i} \right\| _{2}=\left( {\widetilde{\xi }}_{i}^{\top } {\widetilde{\xi }}_{i}\right) ^{1/2}$.

Proof of Lemma 2

It is easy to show

$$\begin{aligned} \left\| {\widetilde{\xi }}_{i} \right\| _{2}^{2} \le&\frac{2}{n}\sum _{j=1}^{m_{n}}\lambda _{j}^{-1}\left\{ \left( {\widehat{\xi }}_{ij}-\xi _{ij}\right) ^{2}+ \xi _{ij}^{2}\right\} +\frac{1}{n} \nonumber \\ \le&\frac{2}{n}||X_{i}||^{2}\sum _{j=1}^{m_{n}}\lambda _{j}^{-1}\left\| {\widehat{\phi }}_{j}-\phi _{j}\right\| ^{2}+ \frac{2}{n}\sum _{j=1}^{m_{n}}\lambda _{j}^{-1} \xi _{ij}^{2}+\frac{1}{n}. \end{aligned}$$

By (5.21) and (5.22) in Hall et al. (2007), we have $\left\| {\widehat{\phi }}_{j}-\phi _{j}\right\| ^{2}=O_{p}(j^{2}/n)$, uniformly in $j \in \{ 1, \ldots , m_{n}\}$. Combining $\mathbb {E}\left( ||X||^{4}\right) <\infty$ and Lemma 1, we have $\mathop {max }\limits _{1\le i \le n} ||X_{i}||^{2}=O_{p}\left( n^{1/2}\right)$. Hence, $\mathop {max }\limits _{1\le i \le n}2||X_{i}||^{2}\sum _{j=1}^{m_{n}}\lambda _{j}^{-1}\left\| {\widehat{\phi }}_{j}-\phi _{j}\right\| ^{2}\le O_{p}\left( n^{-1/2}\right) \sum _{j=1}^{m_{n}}j^2/\lambda _{j}\le O_{p}\left( m_{n}^{\nu +3}n^{-1/2}\right)$. By assumption (C2) and Lemma 1, we have $\mathop {max }\limits _{1\le i \le n}\sum _{j=1}^{m_{n}}2\lambda _{j}^{-1} \xi _{ij}^{2}=O_{p}\left( m_{n}n^{1/2}\right)$. Since $\nu >1$ and $\zeta >\nu /2+1$, there exists a small constant $C_0> 0$ (depending on $\nu$ and $\zeta$) such that $\mathop {max }\limits _{1\le i \le n}2||X_{i}||^{2}\sum _{j=1}^{m_{n}}\lambda _{j}^{-1}\left\| {\widehat{\phi }}_{j}-\phi _{j}\right\| ^{2}=O_p\left\{ n^{-C_0}\left( n/m_{n}\right) \right\}$ and $\mathop {max }\limits _{1\le i \le n}\sum _{j=1}^{m_{n}}2\lambda _{j}^{-1} \xi _{ij}^{2}=O_p\left\{ n^{-C_0}\left( n/m_{n}\right) \right\}$, and complete the proof of Lemma 2. $\square$

Lemma 3

Under conditions (C1)–(C5), we have

$$\begin{aligned} \max _{1 \le i \le n}|W_{i}|=o_{p}(1), \end{aligned}$$

where $W_{i}$ are defined in (15).

Proof of Lemma 3

Following conditions (C1), (C4) and $\left\| {\widehat{\phi }}_{j}-\phi _{j}\right\| ^{2}=O_{p}\left( j^{2}/n\right)$, uniformly in $j \in \{ 1, \ldots , m_{n}\}$, we have

$$\begin{aligned}&\max _{1 \le i \le n}\left| \sum _{j=1}^{m_{n}}\left( \xi _{ij}-{\widehat{\xi }}_{ij}\right) \alpha _{\tau j}\right| \nonumber \\&\le \max _{1 \le i \le n}\Vert X_{i}\Vert \sum _{j=1}^{m_{n}}\left\| {\widehat{\phi }}_{j}-\phi _{j}\right\| \left| c\alpha _{0j}\right| =O_{p}\left( n^{-1/4}\sum _{j=1}^{m_{n}}j^{1-\zeta }\right) \nonumber \\&={\left\{ \begin{array}{ll} O_{p}\left( n^{-1/4}m_{n}^{2-\zeta }\right) &{} { if \,\, 3/2<\zeta <2, } \nonumber \\ O_{p}\left( n^{-1/4}\log m_{n}\right) &{} { if \,\, \zeta =2, } \nonumber \\ O_{p}\left( n^{-1/4}\right) &{} { if \,\, \zeta >2. } \end{array}\right. } \end{aligned}$$

Therefore, by condition (C5), we obtain $\max _{1 \le i \le n}\left| \sum _{j=1}^{m_{n}}\left( \xi _{ij}-{\widehat{\xi }}_{ij}\right) \alpha _{\tau j}\right| =o_{p}(1)$. Following conditions (C2)–(C5) and Lemma 1, we have

$$\begin{aligned} \begin{aligned} \mathbb {E}\left( \max _{1 \le i \le n}\left| \zeta _{i}\right| \right)&\le \mathbb {E}\left( \sum _{j=m_{n}+1}^{\infty }|\alpha _{\tau j}|\max _{1 \le i \le n}|\xi _{ij}|\right) \le \sum _{j=m_{n}+1}^{\infty } C n^{1 / 4}\left| \alpha _{0j}\right| \lambda _{j}^{1 / 2} \\&\le C n^{1 / 4} \sum _{j=m_{n}+1}^{\infty } j^{-(\zeta +v / 2)} =C n^{1 / 4} \sum _{j=m_{n}+1}^{\infty } \int _{j-1}^{j}j^{-(\zeta +v/2)}dx \\&\le C n^{1 / 4} \sum _{j=m_{n}+1}^{\infty } \int _{j-1}^{j}x^{-(\zeta +v/2)}dx=C n^{1/4}\int _{m_{n}}^{\infty }x^{-(\zeta +v/2)}dx\\&=C n^{1 / 4}\frac{1}{\zeta +v/2-1}m_{n}^{-(\zeta +v/2)+1}=\frac{C}{\zeta +v/2-1}n^{1 / 4}n^\frac{-(\zeta +v/2)+1}{v+2\zeta }\\&=\frac{C}{\zeta +v/2-1}n^{\frac{1}{v+2\zeta }-\frac{1}{4}}=o(1), \end{aligned} \end{aligned}$$

where $v>1$ and $\zeta >v/2+1.$ Thus, we have

$$\begin{aligned} \max _{1 \le i \le n}\left| W_{i}\right| \le \max _{1 \le i \le n}\left| \sum _{j=1}^{m_{n}}\left( \xi _{ij}-{\widehat{\xi }}_{ij}\right) \alpha _{\tau j}\right| +\max _{1 \le i \le n}|\zeta _{i}|=o_{p}(1). \end{aligned}$$

Lemma 4

For $k \in \left\{ 1,\ldots , m_{n}\right\}$, define the following expressions:

$$\begin{aligned} \vartheta ^{(1)}=\sum _{i=1}^{n}\sum _{s=m_{n}+1}^{\infty }\xi _{is}\alpha _{\tau s}\Lambda ^{-1/2}\xi _{i}, \,\,\, \vartheta ^{(2)}_{k}=\sum _{i=1}^{n}\left( {\widehat{\xi }}_{ik}-\xi _{ik}\right) /\sqrt{\lambda _{k}}\sum _{s=m_{n}+1}^{\infty }\xi _{is}\alpha _{\tau s} , \end{aligned}$$

$$\begin{aligned} \vartheta ^{(3)}=\sum _{i=1}^{n}\sum _{s=1}^{m_{n}}\left( {\widehat{\xi }}_{is}-\xi _{is}\right) \alpha _{\tau s}\Lambda ^{-1/2}\xi _{i},\,\,\vartheta ^{(4)}_{k}=\sum _{i=1}^{n}\sum _{s=1}^{m_{n}}\alpha _{\tau s}\left( {\widehat{\xi }}_{is}-\xi _{is}\right) \left( {\widehat{\xi }}_{ik}-\xi _{ik}\right) /\sqrt{\lambda _{k}}. \end{aligned}$$

Under conditions (C1)–(C5), we have

$$\begin{aligned} \left\| \vartheta ^{(1)}\right\| _{2}=o_{p}\left\{ \left( m_{n}n\right) ^{1/2}\right\} ,\,\,\, \vartheta ^{(2)}_{k}=o_{p}\left( n^{1/2}\right) , \end{aligned}$$

$$\begin{aligned} \left\| \vartheta ^{(3)}\right\| _{2}=O_{p}\left\{ \left( m_{n}n\right) ^{1/2}\right\} ,\,\,\, \vartheta ^{(4)}_{k}=O_{p}\left( k^{\nu /2+1}\right) , \end{aligned}$$

where $o_{p}$ and $O_{p}$ are uniform for $k \in \{1,\ldots , m_{n}\}.$

Proof of Lemma 4

Please refer to the proof of Lemma 2 of Kong et al. (2016). $\square$

proof of Theorem 1

By simple calculation, we have

$$\begin{aligned} \left| S_{ni}+m_{n}^{1/2}{\widetilde{\xi }}_{i}^{\top }\theta \psi _{\tau }\left( \varepsilon _{i\tau }\right) \right| \le 2\left| m_{n}^{1/2}{\widetilde{\xi }}_{i}^{\top }\theta \right| \mathbf {1}\left\{ \left| \varepsilon _{i\tau }\right| \le \left| m_{n}^{1/2}{\widetilde{\xi }}_{i}^{\top }\theta -W_{i}\right| \right\} . \end{aligned}$$

Therefore,

$$\begin{aligned}&\mathbb {E}\left( R_{n}\right) ^{2}\nonumber \\&=\mathbb {E}\left[ \mathbb {E}\left\{ \left( \sum _{i=1}^{n}R_{ni}^{2}+2\sum _{i=1}^{n}\sum _{k: i<k}R_{ni}R_{nk}\right) \bigg |\Omega \right\} \right] \nonumber \\&=\mathbb {E}\left[ \mathbb {E}\left\{ \left( \sum _{i=1}^{n}R_{ni}^{2}\right) \bigg |\Omega \right\} \right] \nonumber \\&=\sum _{i=1}^{n}\mathbb {E}\left( \mathbb {E}\left[ \left\{ S_{ni}-\Gamma _{ni}+m_{n}^{1/2}{\widetilde{\xi }}_{i}^{\top }\theta \psi _{\tau }\left( \varepsilon _{i\tau }\right) \right\} ^{2}\Big |\Omega \right] \right) \nonumber \\&\le \sum _{i=1}^{n}\mathbb {E}\left( \mathbb {E}\left[ \left\{ S_{ni}+m_{n}^{1/2}{\widetilde{\xi }}_{i}^{\top }\theta \psi _{\tau }\left( \varepsilon _{i\tau }\right) \right\} ^{2}\Big |\Omega \right] \right) \nonumber \\&\le 2m_{n}\mathbb {E}\left[ \sum _{i=1}^{n}\left( {\widetilde{\xi }}_{i}^{\top }\theta \right) ^{2}\mathbb {E}\left\{ \mathbf {1}\left( \left| \varepsilon _{i\tau _{j}}\right| \le \max _{1 \le i \le n}\left| m_{n}^{1/2}{\widetilde{\xi }}_{i}^{\top }\theta -W_{i}\right| \right) \Big |\Omega \right\} \right] \nonumber \\&\le 2m_{n}\left[ \mathbb {E}\left\{ \sum _{i=1}^{n}\left( {\widetilde{\xi }}_{i}^{\top }\theta \right) ^{2}\right\} ^{2}\right] ^{1/2}\mathbb {E}\left\{ \mathbf {1}\left( \left| \varepsilon _{i\tau }\right| \le \max _{1 \le i \le n}\left| m_{n}^{1/2}{\widetilde{\xi }}_{i}^{\top }\theta -W_{i}\right| \right) \right\} .\nonumber \end{aligned}$$

Following

$$\begin{aligned} \frac{1}{n} \sum _{i=1}^{n} {\widehat{\xi }}_{ik} {\widehat{\xi }}_{ij}=\iint _{\mathcal {I}^{2}} {{\widehat{K}}}(s, t) {\widehat{\phi }}_{ k}(s) {\widehat{\phi }}_{j}(t) d s d t=\left\{ \begin{array}{ll}{{\widehat{\lambda }}_{k}} &{} {if \,\,\, k=j,} \\ {0} &{} {if \,\,\, k\ne j}\end{array}\right. \end{aligned}$$

and (5.2) in Hall et al. (2007), we get

$$\begin{aligned}&\sum _{i=1}^{n}\left( {\widetilde{\xi }}_{i}^{\top }\theta \right) ^{2}\nonumber \\&= \sum _{i=1}^{n}\left( \theta ^{\top }\left( n^{-1/2}, 0_{m_{n}}^{\top }\right) ^{\top }+\theta ^{\top }\left( 0, \left\{ n^{-1/2}\Lambda _{1}^{-1/2}\left( {\widehat{\xi }}_{i1},\ldots ,{\widehat{\xi }}_{im_{n}}\right) ^{\top }\right\} ^{\top }\right) ^{\top }\right) ^{2}\nonumber \\&= \theta _{1}^{2}+\theta ^{\top }\text {diag}\left( 0, {\widehat{\lambda }}_{1}/\lambda _{1}, \ldots , {\widehat{\lambda }}_{m_{n}}/\lambda _{m_{n}}\right) \theta \nonumber \\&=\Vert \theta \Vert _{2}^{2}+\theta ^{\top }\text {diag}\left( 0, \left( {\widehat{\lambda }}_{1}-\lambda _{1}\right) /\lambda _{1}, \ldots , \left( {\widehat{\lambda }}_{m_{n}}-\lambda _{m_{n}}\right) /\lambda _{m_{n}}\right) \theta ,\\&\le \left\| \theta \right\| _{2}^{2}+\Vert \theta \Vert _{2}^{2}{\widehat{\Delta }}/\lambda _{m_{n}},\nonumber \end{aligned}$$

(17)

where $\theta _{1}$ is the first component of $\theta$, $0_{m_{n}}$ represents an $m_{n}$-dimensional vector whose components are all zero, and ${\widehat{\Delta }}=\left\{ \displaystyle {\iint _{\mathcal {I}^{2}}\left( {{\widehat{K}}}-K\right) ^{2}}\right\} ^{1/2}.$ By (5.9) in Hall et al. (2007) and conditions (C3)–(C5), we get $\mathbb {E}\left( {\widehat{\Delta }}^{2}/\lambda _{m_{n}}^{2}\right) =o(1).$ Therefore,

$$\begin{aligned} \mathbb {E}\left\{ \sum _{i=1}^{n}\left( {\widetilde{\xi }}_{i}^{\top }\theta \right) ^{2}\right\} ^{2}&\le 2\left\| \theta \right\| _{2}^4\left\{ 1+\mathbb {E} \left( {\widehat{\Delta }}^{2}/\lambda _{m_{n}}^{2}\right) \right\} =C\left\| \theta \right\| _{2}^4. \end{aligned}$$

(18)

Suppose that $L_{1}$ is a positive constant. For $\Vert \theta \Vert _{2}\le L_{1},$ following Lemmas 2 and 3, and $\varepsilon _{i\tau }=O_{p}(1)$, we have

$$\begin{aligned}&\mathbb {E}\left\{ \mathbf {1}\left( \left| \varepsilon _{i\tau }\right| \le \max _{1 \le i \le n}\left| m_{n}^{1/2}{\widetilde{\xi }}_{i}^{\top }\theta -W_{i}\right| \right) \right\} \nonumber \\&=\mathbb {P}\left( \left| \varepsilon _{i\tau }\right| \le \max _{1 \le i \le n}\left| m_{n}^{1/2}{\widetilde{\xi }}_{i}^{\top }\theta -W_{i}\right| \right) =o(1). \end{aligned}$$

(19)

Combining (18) and (19), we obtain

$$\begin{aligned} \mathbb {E}\left\{ \left( R_{n}\right) ^{2}\right\}&=o\left( m_{n}\left\| \theta \right\| _{2}^2\right) . \end{aligned}$$

(20)

It is easy to show

$$\begin{aligned} \mathbb {E}\left\{ m_{n}^{1/2}\sum _{i=1}^{n}{\widetilde{\xi }}_{i}^{\top }\theta \psi _{\tau }\left( \varepsilon _{i\tau }\right) \right\} ^{2}&=\mathbb {E}\left( \mathbb {E}\left[ \left\{ m_{n}^{1/2}\sum _{i=1}^{n}{\widetilde{\xi }}_{i}^{\top }\theta \psi _{\tau }\left( \varepsilon _{i\tau }\right) \right\} ^{2}\bigg |\Omega \right] \right) \nonumber \\&=m_{n}\sum _{i=1}^{n}\mathbb {E}\left( \mathbb {E}\left[ {\left\{ {\widetilde{\xi }}_{i}^{\top }\theta \psi _{\tau }\left( \varepsilon _{i\tau }\right) \right\} }^{2}\Big |\Omega \right] \right) \nonumber \\&\le m_{n}\mathbb {E}\left\{ \sum _{i=1}^{n}\left( {\widetilde{\xi }}_{i}^{\top }\theta \right) ^{2}\right\} =O\left( m_{n}\Vert \theta \Vert _{2}^{2}\right) . \end{aligned}$$

(21)

Since $\mathbb {E}\left( {\widehat{\Delta }}^{2}/\lambda _{m_{n}}^{2}\right) =o(1),$ we have

$$\begin{aligned}&\left| \theta ^{\top }\text {diag}\left( 0, \left( {\widehat{\lambda }}_{1}-\lambda _{1}\right) /\lambda _{1}, \ldots , \left( {\widehat{\lambda }}_{m_{n}}-\lambda _{m_{n}}\right) /\lambda _{m_{n}}\right) \theta \right| \nonumber \\&\le \Vert \theta \Vert _{2}^{2}{\widehat{\Delta }}/\lambda _{m_{n}}=o_{p}\left( \Vert \theta \Vert _{2}^{2}\right) . \end{aligned}$$

Hence, by (17), we get

$$\begin{aligned}&\sum _{i=1}^{n}\left( m_{n}^{1/2}{\widetilde{\xi }}_{i}^{\top }\theta \right) ^{2}\nonumber \\&=m_{n}\Vert \theta \Vert _{2}^{2}+m_{n}\theta ^{\top }\text {diag}\left( 0, \left( {\widehat{\lambda }}_{1}-\lambda _{1}\right) /\lambda _{1}, \ldots , \left( {\widehat{\lambda }}_{m_{n}}-\lambda _{m_{n}}\right) /\lambda _{m_{n}}\right) \theta \nonumber \\&=m_{n}\Vert \theta \Vert _{2}^{2}\{1+o_{p}(1)\}. \end{aligned}$$

(22)

It is easy to prove

$$\begin{aligned}&\sum _{i=1}^{n}m_{n}^{1/2}W_{i}{\widetilde{\xi }}_{i}^{\top }\theta \nonumber \\&=\left( m_{n}/n\right) ^{\frac{1}{2}}\left( \theta ^{\top }\vartheta ^{(3)}+\sum _{s=1}^{m_{n}}\vartheta _{s}^{(4)}u_{s}+\theta ^{\top }\vartheta ^{(1)}+\sum _{s=1}^{m_{n}}\vartheta _{s}^{(2)}u_{s}\right) . \end{aligned}$$

Following Lemma 4 and condition (C5), we conclude that

$$\begin{aligned} (m_{n}/n)^{\frac{1}{2}}\theta ^{\top }\vartheta ^{(3)}&=O_{p}(m_{n})\left\| \theta \right\| _{2}, \,\,\,\, (m_{n}/n)^{\frac{1}{2}}\theta ^{\top }\vartheta ^{(1)}=o_{p}(m_{n})\left\| \theta \right\| _{2},\nonumber \\&(m_{n}/n)^{\frac{1}{2}}\sum _{s=1}^{m_{n}}\vartheta _{s}^{(2)}u_{s}=o_{p}(m_{n})\left\| \theta \right\| _{2},\nonumber \\ (m_{n}/n)^{\frac{1}{2}}\sum _{s=1}^{m_{n}}\vartheta _{s}^{(4)}u_{s}&=O_{p}\left\{ (m_{n}/n)^{\frac{1}{2}}\left( \sum _{s=1}^{m_{n}}s^{\nu +2}\right) ^{1/2}\right\} \left\| \theta \right\| _{2}\nonumber \\&=O_{p}\left\{ \left( m_{n}/n\right) ^{\frac{1}{2}}m_{n}^{(\nu +3)/2}\right\} \left\| \theta \right\| _{2} =o_{p}\left( m_{n}\right) \left\| \theta \right\| _{2}. \end{aligned}$$

Thus,

$$\begin{aligned} \sum _{i=1}^{n}m_{n}^{1/2}W_{i}{\widetilde{\xi }}_{i}^{\top }\theta =O_{p}\left( m_{n}\right) \Vert \theta \Vert _{2}. \end{aligned}$$

(23)

By the proof of Lemma 2 of Yao et al. (2017) and condition (C6),

$$\begin{aligned} \Gamma _{n}&=\sum _{i=1}^{n}\mathbb {E}\left\{ \rho _{\tau }\left( W_{i}+\varepsilon _{i\tau }-{\widetilde{\xi }}_{i}^{\top }m_{n}^{1/2}\theta \right) \Big |\Omega \right\} - \mathbb {E}\left\{ \rho _{\tau }\left( W_{i}+\varepsilon _{i\tau }\right) \big |\Omega \right\} \nonumber \\&=\frac{1}{2}\sum _{i=1}^{n}f_{\tau }(0|X_{i})\left\{ \left( m_{n}^{1/2}{\widetilde{\xi }}_{i}^{\top }\theta \right) ^{2}-2m_{n}^{1/2}W_{i}{\widetilde{\xi }}_{i}^{\top }\theta \right\} \left\{ 1+o_{p}(1)\right\} \end{aligned}$$

Following condition (C6), (22) and (23), we get

$$\begin{aligned} \Gamma _{n}=m_{n}\Vert \theta \Vert _{2}^{2}\{1+o_{p}(1)\}+O_{p}\left( m_{n}\right) \Vert \theta \Vert _{2}. \end{aligned}$$

(24)

Combining (20), (21) and (24), for sufficiently large $L_{1}$, we have

$$\begin{aligned} \inf _{\Vert \theta \Vert _{2}=L_{1}}S_{n}\left( m_{n}^{1 / 2} \theta \right) \geqslant c_{0} L_{1}^{2} m_{n}\{1+o_{p}(1)\}, \end{aligned}$$

where $c_{0}$ is a positive constant. This implies that

$$\begin{aligned} \mathbb {P}\left( \inf _{\Vert \theta \Vert _{2} \geqslant L_{1}}\left[ \sum _{i=1}^{n}\left\{ \rho _{\tau }\left( W_{i}+\varepsilon _{i\tau }-m_{n}^{1/2}{\widetilde{\xi }}_{i}^{\top }\theta \right) - \rho _{\tau }\left( W_{i}+\varepsilon _{i\tau }\right) \right\} \right] >0\right) \rightarrow 1 \end{aligned}$$

as $n\rightarrow \infty .$ Hence, $\mathbb {P}\left( \left\| {\widehat{\theta }}\right\| _{2}\le L_{1}m_{n}^{1/2}\right) \rightarrow 1$ as $n\rightarrow \infty ,$ where ${\widehat{\theta }}$ is the minimizer of (16). Hence, $\left\| {\widehat{\theta }}\right\| _{2}=O_{p}\left( m_{n}^{1/2}\right) .$ Therefore, we have

$$\begin{aligned} \left\| n^{1/2}\Lambda _{1}^{1/2}\left( {\widehat{\alpha }}-\alpha \right) \right\| _{2}=O_{p}\left( m_{n}^{1/2}\right) . \end{aligned}$$

(25)

By some straightforward calculations, we get

$$\begin{aligned}&\left\| {\widehat{\beta }}_{\tau }-\beta _{\tau }\right\| ^{2}\nonumber \\&=\Bigg |\Bigg |\sum \limits _{j=1}^{m_{n}}{\widehat{\alpha }}_{\tau j}{\widehat{\phi }}_{j}-\sum \limits _{j=1}^{\infty }\alpha _{\tau j}\phi _{j}\Bigg |\Bigg |^{2} \nonumber \\&\le 2\Bigg |\Bigg |\sum \limits _{j=1}^{m_{n}}{\widehat{\alpha }}_{\tau j}{\widehat{\phi }}_{j}-\sum \limits _{j=1}^{m_{n}}\alpha _{\tau j}\phi _{j}\Bigg |\Bigg |^{2}+2 \Bigg |\Bigg |\sum \limits _{j=m_{n}+1}^{\infty }\alpha _{\tau j}\phi _{j}\Bigg |\Bigg |^{2} \nonumber \\&\le 4\Bigg |\Bigg |\sum \limits _{j=1}^{m_{n}}\left( {\widehat{\alpha }}_{\tau j}-\alpha _{\tau j}\right) {\widehat{\phi }}_{j}\Bigg |\Bigg |^{2}+4\Bigg |\Bigg |\sum \limits _{j=1}^{m_{n}}\alpha _{\tau j}\left( {\widehat{\phi }}_{j}-\phi _{j}\right) \Bigg |\Bigg |^{2}+2\sum \limits _{j=m_{n}+1}^{\infty }\alpha _{\tau j}^{2} \nonumber \\&\le 4\sum \limits _{j=1}^{m_{n}}\left( {\widehat{\alpha }}_{\tau j}-\alpha _{\tau j}\right) ^{2}+4m_{n}\sum \limits _{j=1}^{m_{n}}\alpha _{\tau j}^{2}\left\| {\widehat{\phi }}_{j}-\phi _{j}\right\| ^{2}+2\sum \limits _{j=m_{n}+1}^{\infty }\alpha _{\tau j}^{2} \nonumber \\&\le 4(n\lambda _{m_{n}})^{-1}\left\| n^{1/2}\Lambda _{1}^{1/2}\left( {\widehat{\alpha }}-\alpha \right) \right\| _{2}^{2}+4m_{n}\sum \limits _{j=1}^{m_{n}}\alpha _{\tau j}^{2}\left\| {\widehat{\phi }}_{j}-\phi _{j}\right\| ^{2}+2\sum \limits _{j=m_{n}+1}^{\infty }\alpha _{\tau j}^{2}. \end{aligned}$$

By conditions (C3) and (C5), and (25), we obtain

$$\begin{aligned} \left( n\lambda _{m_{n}}\right) ^{-1}\left\| n^{1/2}\Lambda _{1}^{1/2}\left( {\widehat{\alpha }}-\alpha \right) \right\| _{2}^{2}\le O_{p}\left( n^{-1}m_{n}^{v+1}\right) =O_{p}\left\{ n^{-(2\zeta -1)/(\nu +2\zeta )}\right\} . \end{aligned}$$

(26)

Since $\left\| {\widehat{\phi }}_{j}-\phi _{j}\right\| ^{2}=O_{p}\left( j^{2}/n\right)$, uniformly in $j \in \{ 1, \ldots , m_{n}\}$, we have

$$\begin{aligned}&m_{n}\sum \limits _{j=1}^{m_{n}}\alpha _{\tau j}^{2}\left\| {\widehat{\phi }}_{j}-\phi _{j}\right\| ^{2} =m_{n}\sum \limits _{j=1}^{m_{n}}j^{-2\zeta }O_{p}\left( n^{-1}j^{2}\right) =\frac{m_{n}}{n}\, O_{p}\left( \sum \limits _{j=1}^{m_{n}}j^{2-2\zeta }\right) \nonumber \\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,={\left\{ \begin{array}{ll} O_{p}\left( \frac{m_{n}}{n}\right) &{} {if \,\, 2-2\zeta <-1, } \nonumber \\ O_{p}\left( \frac{m_{n}\log m_{n}}{n}\right) &{} {if \,\, 2-2\zeta =-1, } \nonumber \\ O_{p}\left( \frac{m_{n}^{4-2\zeta }}{n}\right) &{} {if \,\, 2-2\zeta >-1. } \end{array}\right. } \end{aligned}$$

Since $\zeta >\nu /2+1$ and $\nu >1$, we conclude that

$$\begin{aligned} m_{n}\sum _{j=1}^{m_{n}}\alpha _{\tau j}^{2}\left\| {\widehat{\phi }}_{j}-\phi _{j}\right\| ^{2}=O_{p}( {m_{n}}/{n})=o_{p}\left\{ n^{-(2\zeta -1)/(\nu +2\zeta )}\right\} . \end{aligned}$$

(27)

By conditions (C4) and (C5), we get

$$\begin{aligned} \sum \limits _{j=m_{n}+1}^{\infty }\alpha _{\tau j}^{2} \le C \sum \limits _{j=m_{n}+1}^{\infty }j^{-2\zeta }=O\left\{ m_{n}^{-(2\zeta -1)}\right\} =O\left\{ n^{-(2\zeta -1)/(\nu +2\zeta )}\right\} . \end{aligned}$$

(28)

Combining (26)–(28), we complete the proof of Theorem 1.

1.3 A.3 Proof of Theorem 2

Quadratic Approximation Lemma Pollard (1991): suppose $\mathcal {L}_{n}(\delta )$ is convex and can be represented as $\delta ^{\top }B\delta /2+u_{n}^{\top }\delta +a_{n}+R_{n}(\delta ),$ where B is symmetric and positive definite, $u_{n}$ is stochastically bounded, $a_{n}$ is arbitrary, and $R_{n}(\delta )$ goes to zero in probability for each $\delta$. Then ${\widehat{\delta }},$ the minimizer of $\mathcal {L}_{n}(\delta ),$ is only $o_{p}(1)$ away $\mathrm {from}-B^{-1} u_{n} .$ If $u_{n}$ converges in distribution to u,  then ${\widehat{\delta }}$ converges in distribution to $-B^{-1}u.$

We use the Quadratic Approximation Lemma to complete the proof. Recall the notations $z=\langle X_{0}$, $\beta _{\tau }\rangle$, ${{\widehat{z}}}=\langle X_{0}$, ${\widehat{\beta }}_{\tau }\rangle$, $Z_{i}=\langle X_{i}$, $\beta _{\tau }\rangle$, ${{\widehat{Z}}}_{i}=\langle X_{i}$, ${\widehat{\beta }}_{\tau }\rangle$. Let $a_{0}=Q_{\tau }(z)=Q_{\tau }(\langle X_{0}$, $\beta _{\tau }\rangle )$, $b_{0}=Q'_{\tau }(z)$, ${{\widehat{K}}}_{i}=K\left\{ \left( {{\widehat{Z}}}_{i}-{{\widehat{z}}}\right) /h\right\}$, $K_{i}=K\left\{ \left( Z_{i}-z\right) /h\right\}$ and $K'_{i}=K'\{(Z_{i}-z)/h\}$ for $i=1, \ldots , n$. We first show that

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^{n}\left[ \rho _{\tau }\left\{ Y_{i}-a-b\left( {{\widehat{Z}}}_{i}-{{\widehat{z}}}\right) \right\} {{\widehat{K}}}_{i}-\rho _{\tau }\left\{ Y_{i}-a-b\left( Z_{i}-z\right) \right\} K_{i}\right] =O_{p}\left( n^{-\frac{2\zeta -1}{2(\nu +2\zeta )}}\right) . \end{aligned}$$

(29)

Define

$$\begin{aligned} R_{11}&=\frac{1}{n}\sum _{i=1}^{n}K_{i}\left[ \rho _{\tau }\left\{ Y_{i}-a-b\left( {{\widehat{Z}}}_{i}-{{\widehat{z}}}\right) \right\} -\rho _{\tau }\left\{ Y_{i}-a-b\left( Z_{i}-z\right) \right\} \right] , \nonumber \\ R_{12}&=\frac{1}{n}\sum _{i=1}^{n}\left( {{\widehat{K}}}_{i}-K_{i}\right) \rho _{\tau }\left\{ Y_{i}-a-b\left( Z_{i}-z\right) \right\} , \nonumber \\ R_{13}&=\frac{1}{n}\sum _{i=1}^{n}\left[ \rho _{\tau }\left\{ Y_{i}-a-b\left( {{\widehat{Z}}}_{i}-{{\widehat{z}}}\right) \right\} -\rho _{\tau }\left\{ Y_{i}-a-b\left( Z_{i}-z\right) \right\} \right] \left( {{\widehat{K}}}_{i}-K_{i}\right) . \end{aligned}$$

Then

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^{n}\left[ \rho _{\tau }\left\{ Y_{i}-a-b\left( {{\widehat{Z}}}_{i}-{{\widehat{z}}}\right) \right\} {{\widehat{K}}}_{i}-\rho _{\tau }\left\{ Y_{i}-a-b\left( Z_{i}-z\right) \right\} K_{i}\right] =R_{11}+R_{12}+R_{13}. \end{aligned}$$

Following

$$\begin{aligned}&\rho _{\tau }(x-y)-\rho _{\tau }(x)=y\{\mathbf {1}(x\le 0)-\tau \}+\int _0^{y}\{\mathbf {1}(x\le t)-\mathbf {1}(x\le 0)\}\,dt, \end{aligned}$$

(30)

we have $|\rho _{\tau }(x-y)-\rho _{\tau }(x)|\le 2|y|$. By Theorem 1, we get

$$\begin{aligned} \left| R_{11}\right| \le&\frac{1}{n}\sum _{i=1}^{n}K_{i}\left| \rho _{\tau }\left\{ Y_{i}-a-b\left( {{\widehat{Z}}}_{i}-{{\widehat{z}}}\right) \right\} -\rho _{\tau }\left\{ Y_{i}-a-b\left( Z_{i}-z\right) \right\} \right| \nonumber \\ \le&2|b|\mathop {sup }\limits _{t}\left| K(t)\right| \frac{1}{n}\sum _{i=1}^{n}\left( \left| {{\widehat{Z}}}_{i}-Z_{i}\right| +\left| {{\widehat{z}}}-z\right| \right) \nonumber \\ =&2|b|\mathop {sup }\limits _{t}\left| K(t)\right| \frac{1}{n}\sum _{i=1}^{n}\left( \left| \langle X_{i}, {\widehat{\beta }}_{\tau }-\beta _{\tau }\rangle \right| +\left| \langle X_{0}, {\widehat{\beta }}_{\tau }-\beta _{\tau }\rangle \right| \right) =O_{p}\left( n^{-\frac{2\zeta -1}{2(\nu +2\zeta )}}\right) . \end{aligned}$$

Hence, we have $R_{11}=O_{p}\left( n^{-\frac{2\zeta -1}{2(\nu +2\zeta )}}\right)$. It is easy to show that

$$\begin{aligned}& |R_{12}|\\& \quad \le (nh)^{-1}\sum _{i=1}^{n}\left |K{^{\prime}}\left( \frac{Z_{i}-z}{h}\right) \rho _{\tau }\left\{ Y_{i}-a-b\left( Z_{i}-z\right) \right\} {\langle } X_{i}-X_{0}, {\widehat{\beta }}_{\tau }-\beta _{\tau } {\rangle }\{1+o_{p}(1)\}\right |. \end{aligned}$$

Let $Z_{i0}=(X_i-X_{0})/h$. Then

$$\begin{aligned}&(nh)^{-1}\sum _{i=1}^{n}\left| K'\left( \frac{Z_{i}-z}{h}\right) \rho _{\tau }\left\{ Y_{i}-a-b\left( Z_{i}-z\right) \right\} \langle X_{i}-X_{0}, {\widehat{\beta }}_{\tau }-\beta _{\tau }\rangle \right| \nonumber \\&=\frac{1}{n}\sum _{i=1}^{n}\Big |K'(\langle Z_{i0}, \beta _{\tau }\rangle )\rho _{\tau }\left\{ Y_{i}-a-bh\langle Z_{i0}, \beta _{\tau }\rangle \right\} \langle Z_{i0}, {\widehat{\beta }}_{\tau }-\beta _{\tau }\rangle \Big | \nonumber \\&=O_{p}\left( n^{-\frac{2\zeta -1}{2(\nu +2\zeta )}}\right) . \end{aligned}$$

(31)

This implies that $R_{12}=O_{p}\left( n^{-\frac{2\zeta -1}{2(\nu +2\zeta )}}\right)$. Following similar arguments, we have $R_{13}=O_{p}\left( n^{-\frac{2\zeta -1}{2(\nu +2\zeta )}}\right) .$ Thus (29) holds. Following (29) and $n^{\frac{\nu +1}{\nu +2\zeta }}h\rightarrow 0$, we get

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^{n}\rho _{\tau }\left\{ Y_{i}-a-b\left( {{\widehat{Z}}}_{i}-{{\widehat{z}}}\right) \right\} {{\widehat{K}}}_{i}=\frac{1}{n}\sum _{i=1}^{n}\rho _{\tau }\left\{ Y_{i}-a-b\left( Z_{i}-z\right) \right\} K_{i}+o_{p}\left\{ \left( nh\right) ^{-1/2}\right\} . \end{aligned}$$

(32)

This means that the estimate of $Q_{\tau }(\cdot )$ based on $\left\{ \left( \langle X_{i}, {\widehat{\beta }}_{\tau }\rangle , Y_{i}\right) , i=1,2, \ldots , n\right\}$ is asymptotically as efficient as that based on $\left\{ \left( \langle X_{i}, \beta _{\tau }\rangle , Y_{i}\right) , i=1,2, \ldots , n\right\} .$

Let $\eta _{i}=(a-a_{0})+(b-b_{0})(Z_{i}-z)$. Following (30), we obtain

$$\begin{aligned}&\sum _{i=1}^{n}\rho _{\tau }\left\{ Y_{i}-a-b\left( Z_{i}-z\right) \right\} K_{i}-\sum _{i=1}^{n}\rho _{\tau }\left\{ Y_{i}-a_{0}-b_{0}\left( Z_{i}-z\right) \right\} K_{i} \nonumber \\&=\sum _{i=1}^{n} \left[ \eta _{i}\{\mathbf {1}(Y_{i}-a_{0}-b_{0}(Z_{i}-z)\le 0)-\tau \} \right. \\&\left. \,\,\,+ \int _{0}^{\eta _{i}}\{\mathbf {1}(Y_{i}-a_{0}-b_{0}(Z_{i}-z)\le t)-\mathbf {1}(Y_{i}-a_{0}-b_{0}(Z_{i}-z)\le 0)\}dt\right] K_{i}. \nonumber \end{aligned}$$

(33)

Let $\varrho =(nh)^{1/2}\{a-a_{0}, h(b-b_{0})\}^\top$. In the sequel, we show the first quantity in (33) has approximately the form $W_{n}^{\top }\varrho$, and the second quantity has the form $\frac{1}{2}\varrho ^{\top }S_{n}\varrho$. Denote by $F_{Y}(\cdot |Z)$ the conditional distribution of Y given $\langle X, \beta _{\tau } \rangle =Z$. Then

$$\begin{aligned}&\sum _{i=1}^{n}\eta _{i}\{\mathbf {1}(Y_{i}-a_{0}-b_{0}(Z_{i}-z)\le 0)-\tau \}K_{i} \nonumber \\&=\sum _{i=1}^{n}\eta _{i}\{F_{Y_{i}}(a_{0}+b_{0}(Z_{i}-z)|Z_{i})-\tau \}K_{i} \\&\,\,\,\,\,+\sum _{i=1}^{n}\eta _{i}\{\mathbf {1}(Y_{i}-a_{0}-b_{0}(Z_{i}-z)\le 0)-F_{Y_{i}}(a_{0}+b_{0}(Z_{i}-z)|Z_{i})\}K_{i}. \nonumber \end{aligned}$$

(34)

Using Taylor’s expansion, we obtain

$$\begin{aligned} Q_{\tau }(Z_i)=a_{0}+b_{0}(Z_{i}-z)+Q''_{\tau }(z)(Z_{i}-z)^{2}\left\{ 1/2+o_{p}(1)\right\} , \end{aligned}$$

for $|Z_{i}-z|\le h$, and

$$\begin{aligned} F_{Y_{i}}(a_{0}+b_{0}(Z_{i}-z)|Z_{i})&=F_{Y_{i}}\left( Q_{\tau }(Z_i)-Q''_{\tau }(z)(Z_{i}-z)^{2}\left\{ 1/2+o_{p}(1)\right\} \big |Z_{i}\right) \nonumber \\&=\tau -f_{Y_{i}}(Q_{\tau }(Z_i)|Z_{i})Q''_{\tau }(z)(Z_{i}-z)^{2}\left\{ 1/2+o_{p}(1)\right\} , \end{aligned}$$

where $f_{Y_{i}}(\cdot |Z_{i})$ is the conditional density function of $Y_{i}$ given $\langle X_{i}, \beta _{\tau }\rangle =Z_{i}$. Hence, we get

$$\begin{aligned}&\sum _{i=1}^{n}\eta _{i}\{F_{Y_{i}}(a_{0}+b_{0}(Z_{i}-z)|Z_{i})-\tau \}K_{i} \nonumber \\&=-\sum _{i=1}^{n}\eta _{i}f_{Y_{i}}(Q_{\tau }(Z_i)|Z_{i})Q''_{\tau }(z)(Z_{i}-z)^{2}\left\{ 1/2+o_{p}(1)\right\} K_{i}, \end{aligned}$$

which is again of the form

$$\begin{aligned}&-(nh)^{-1/2}\sum _{i=1}^{n}\varrho ^{\top }\begin{pmatrix} 1 \\ \frac{(Z_{i}-z)}{h} \\ \end{pmatrix} f_{Y_{i}}(Q_{\tau }(Z_i)|Z_{i})Q''_{\tau }(z)(Z_{i}-z)^{2}\left\{ 1/2+o_{p}(1)\right\} K_{i}. \end{aligned}$$

(35)

The second term on the right side of (36) can be approximated as

$$\begin{aligned} \sum _{i=1}^{n}\eta _{i}\{\mathbf {1}(Y_{i}\le Q_{\tau }(Z_{i}))-\tau \}K_{i}\{1+o_{p}(1)\}, \end{aligned}$$

which is again of the form

$$\begin{aligned}&(nh)^{-1/2}\sum _{i=1}^{n}\varrho ^{\top }\begin{pmatrix} 1 \\ \frac{(Z_{i}-z)}{h} \\ \end{pmatrix} \{\mathbf {1}(Y_{i}\le Q_{\tau }(Z_{i}))-\tau \}K_{i}\{1+o_{p}(1)\}. \end{aligned}$$

(36)

The second term on the right side of (33) can be approximated as

$$\begin{aligned}&\sum _{i=1}^{n}\left[ \int _{0}^{\eta _{i}}\{\mathbf {1}(Y_{i}-a_{0}-b_{0}(Z_{i}-z)\le t)-\mathbf {1}(Y_{i}-a_{0}-b_{0}(Z_{i}-z)\le 0)\}dt\right] K_{i}\nonumber \\&=\frac{1}{2}\sum _{i=1}^{n}f_{Y_{i}}(a_{0}+b_{0}(Z_{i}-z)|Z_{i})\eta ^{2}_{i}K_{i}\{1+o_{p}(1)\}\nonumber \\&=\frac{1}{2}\sum _{i=1}^{n}f_{Y_{i}}(Q_{\tau }(Z_{i})|Z_{i})\eta ^{2}_{i}K_{i}\{1+o_{p}(1)\}, \end{aligned}$$

which is now of the form

$$\begin{aligned}&\frac{1}{2}\sum _{i=1}^{n}(nh)^{-1}f_{Y_{i}}(Q_{\tau }(Z_{i})|Z_{i})\varrho ^{\top }\begin{pmatrix} 1 &{} \frac{Z_{i}-z}{h} \\ \frac{Z_{i}-z}{h} &{}\left\{ \frac{(Z_{i}-z)}{h}\right\} ^{2}\\ \end{pmatrix} \varrho K_{i}\{1+o_{p}(1)\}. \end{aligned}$$

(37)

Combining (34)–(37), we observe that (33) is now approximated by a quadratic function of $\varrho$. By the Quadratic Approximation Lemma, $\varrho$ is only $o_{p}(1)$ away from $-{S}_{n}^{-1}W_{n}$, where

$$\begin{aligned} S_{n} =&(nh)^{-1}\sum _{i=1}^{n}f_{Y_{i}}(Q_{\tau }(Z_{i})|Z_{i})\begin{pmatrix} 1 &{} \frac{Z_{i}-z}{h} \\ \frac{Z_{i}-z}{h} &{}\left( \frac{Z_{i}-z}{h}\right) ^{2}\\ \end{pmatrix} K_{i},\nonumber \\ W_{n}=&(nh)^{1/2}\left[ (nh)^{-1}\sum _{i=1}^{n} \begin{pmatrix} 1 \\ \frac{Z_{i}-z}{h} \\ \end{pmatrix} \{\mathbf {1}(Y_{i}\le Q_{\tau }(Z_{i}))-\tau \}K_{i} \nonumber \right. \\&\left. -(nh)^{-1}\sum _{i=1}^{n} \begin{pmatrix} 1 \\ \frac{Z_{i}-z}{h} \\ \end{pmatrix} f_{Y_{i}}(Q_{\tau }(Z_i)|Z_{i})Q''_{\tau }(z)(Z_{i}-z)^{2}\frac{K_{i}}{2}\right] . \end{aligned}$$

It can be easily shown that, as $n\longrightarrow \infty$, $S_{n}$ converges in probability to

$$\begin{aligned} f_{Y}(Q_{\tau }(z)|z)f_{\beta _{\tau }}(z)\begin{pmatrix} 1 &{}0 \\ 0&{}\mu _{2} \\ \end{pmatrix}, \end{aligned}$$

where $\mu _{2}=\int _{-1}^{1} t^{2}K(t)dt$. By some straightforward calculations, we can show $(nh)^{-1}\sum _{i=1}^{n}\{\mathbf {1}(Y_{i}\le Q_{\tau }(Z_{i}))-\tau \}K_{i}$ converges in distribution to a normal distribution with zero mean and variance $\tau (1-\tau )f_{\beta _{\tau }}(z)\int _{-1}^{1} K^{2}(t)dt$, and $\frac{1}{2}(nh)^{-1}\sum _{i=1}^{n}f_{Y_{i}}(Q_{\tau }(Z_i)|Z_i)Q''_{\tau }(z)(Z_{i}-z)^{2}K_{i}$ converges in probability to $\frac{1}{2}h^2 Q''_{\tau }(z)\mu _{2}f_{Y}(Q_{\tau }(z)|z)f_{\beta _{\tau }}(z).$ The proof of Theorem 2 is completed by combining the above results.