Empirical likelihood based tests for detecting the presence of significant predictors in marginal quantile regression

Abstract

This article investigates detecting the presence of significant predictors in marginal quantile regression. The main idea comes from the connection between the quantile correlation and the slope parameter of the marginal quantile regression, which is quite different from other methods. By introducing the local linear model and the plug-in empirical likelihood method, consistent asymptotic distribution and its adjusted version are obtained. We not only circumvent the non-regularity encountered by post-model-selected estimators but also make the results more concise. Two adaptive resampling test procedures are proposed in practice by comparing the t-statistics with a threshold to decide whether to use the traditional centered percentile bootstrap or otherwise adapt to the asymptotic distribution under the local model. Simulation studies compare these two resampling tests with other competing methods in several cases. Results show that the approaches proposed are more robust for each quantile level and can control type I error well. Two real datasets from Forbes magazine and the HIV drug resistance database are also applied to illustrate the new methods.

Technical proofs of main results

To prove Theorem 1, we need the following two lemmas.

Lemma 1

If Assumptions 1–4 hold, we have

Lemma 2

If assumptions in Theorem 1 hold, we have $\frac{1}{n}\sum _{i=1}^n(\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )^2{\mathop {\longrightarrow }\limits ^{p}}{\left\{ \begin{array}{ll} \sigma _1^2 &{} \text {if}~\varvec{\beta }_0\ne \varvec{0}\\ \sigma _2^2 &{} \text {if}~\varvec{\beta }_0=\varvec{0} \end{array}\right. }$.

Proof of Lemma 1

On the one hand, when $\varvec{\beta }_0\ne \varvec{0}$, it is clear that $\varvec{\beta }_n\rightarrow \varvec{\beta }_0$. From Li et al. (2015, Theorem 1) we have

$$\begin{aligned}&(\widehat{\text {qcor}}_\tau \{Y,X_1\},\ldots ,\widehat{\text {qcor}}_\tau \{Y,X_p\})^T\\&\qquad {\mathop {\longrightarrow }\limits ^{p}}(\text {qcor}_\tau \{Y,X_1\},\ldots ,\text {qcor}_\tau \{Y,X_p\})^T, \quad k=1,\ldots ,p. \end{aligned}$$

Since

$$\begin{aligned} \hat{k}_n=\arg \max _{k=1,\ldots ,p}\widehat{\text {qcor}}_\tau \{Y,X_k\},~ k_0=\arg \max _{k=1,\ldots ,p}\text {qcor}_\tau \{Y,X_k\}, \end{aligned}$$

we have $\hat{k}_n{\mathop {\rightarrow }\limits ^{p}}k_0$. Then from Li et al. (2015, Theorem 1),

$$\begin{aligned}&\sqrt{n}(\widehat{\text {qcov}}_\tau -\text {qcov}_\tau )\\&\quad =\sqrt{n}(\widehat{\text {qcov}}_\tau \{Y,X_{\hat{k}_n}\}-\text {qcov}_\tau \{Y,X_{k_0}\})\\&\quad =\sqrt{n}(\widehat{\text {qcov}}_\tau \{Y,X_{k_0}\}-\text {qcov}_\tau \{Y,X_{k_0}\})+o_p(1)\\&\quad =\frac{1}{\sqrt{n}}\sum _{i=1}^n[\psi _\tau (Y_i-Q_{\tau ,Y})(X_{i,k_0}-\mu _{X_{k_0}|Y})-\text {qcov}_\tau \{Y,X_{k_0}\}]+o_p(1)\\&\quad =\frac{1}{\sqrt{n}}\sum _{i=1}^n[\psi _\tau (\varvec{X}_i^T\varvec{\beta }_n+\epsilon -Q_{\tau ,\varvec{X}_i^T\varvec{\beta }_n+\epsilon })(X_{i,k_0}-\mu _{X_{k_0}|Y})\\&\quad \quad -\text {qcov}_\tau \{\varvec{X}^T\varvec{\beta }_n+\epsilon ,X_{k_0}\}]+o_p(1)\\&\quad =\frac{1}{\sqrt{n}}\sum _{i=1}^n[\psi _\tau (\varvec{X}_i^T\varvec{\beta }_0+\epsilon -Q_{\tau ,\varvec{X}_i^T\varvec{\beta }_0+\epsilon })(X_{i,k_0}-\mu _{X_{k_0}|Y})\\&\quad \quad -\text {qcov}_\tau \{\varvec{X}^T\varvec{\beta }_0+\epsilon ,X_{k_0}\}]+o_p(1). \end{aligned}$$

Hence if $\varvec{\beta }_0\ne \varvec{0}$,

$$\begin{aligned} \sqrt{n}(\widehat{\text {qcov}}_\tau -\text {qcov}_\tau ){\mathop {\longrightarrow }\limits ^{d}}M_{k_0}(\varvec{\beta }_0). \end{aligned}$$

On the other hand, because $\varvec{\beta }_n=\varvec{\beta }_0+\frac{1}{\sqrt{n}}\varvec{b}_0$, we have

$$\begin{aligned} \begin{aligned}&\sqrt{n}\widehat{\text {qcov}}_\tau \{Y,X_k\}\\&\quad =\sqrt{n}\bigg [\frac{1}{n}\sum _{i=1}^n\psi _\tau \bigg (\varvec{X}_i^T(\varvec{\beta }_0+\frac{1}{\sqrt{n}}\varvec{b}_0)+\epsilon _i-\hat{Q}_{\tau ,\varvec{X}^T(\varvec{\beta }_0+\frac{1}{\sqrt{n}}\varvec{b}_0)+\epsilon }\bigg )(X_{i,k}-\bar{X}_k)\bigg ].\\ \end{aligned} \end{aligned}$$

Hence if $\varvec{\beta }_0=\varvec{0}$,

$$\begin{aligned} \begin{aligned}&\sqrt{n}\widehat{\text {qcov}}_\tau \{Y,X_k\}\\&\quad =\sqrt{n}\bigg [\frac{1}{n}\sum _{i=1}^n\psi _\tau \left( \frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0+\epsilon _i-\hat{Q}_{\tau ,\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0+\epsilon }\right) (X_{i,k}-\bar{X}_k)\bigg ]\\&\quad =\sqrt{n}\bigg \{\bigg [\frac{1}{n}\sum _{i=1}^n\psi _\tau \left( \frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0+\epsilon _i-\hat{Q}_{\tau ,\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0+\epsilon }\right) (X_{i,k}-\bar{X}_k)\\&\qquad -\frac{1}{n}\sum _{i=1}^n\psi _\tau (\epsilon _i-\hat{Q}_{\tau ,\epsilon })(X_{i,k}-\bar{X}_k)\bigg ]+\bigg [\frac{1}{n}\sum _{i=1}^n\psi _\tau (\epsilon _i-\hat{Q}_{\tau ,\epsilon })(X_{i,k}-\bar{X}_k)\\&\qquad -E\psi _\tau (\epsilon -Q_{\tau ,\epsilon })(X_k-EX_k)\bigg ]\bigg \}. \end{aligned} \end{aligned}$$

The last step is obtained by the fact that $E\psi _\tau (\epsilon -Q_{\tau ,\epsilon })(X_k-EX_k)=0$. Thus we can divide $\sqrt{n}\widehat{\text {qcov}}_\tau \{Y,X_k\}$ into two parts. The first part is

$$\begin{aligned} \begin{aligned}&\sqrt{n}\bigg \{\frac{1}{n}\sum _{i=1}^n\left[ \psi _\tau \left( \frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0+\epsilon _i-\hat{Q}_{\tau ,\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0+\epsilon }\right) (X_{i,k}-\bar{X}_{k})\right] \\&\qquad -\frac{1}{n}\sum _{i=1}^n\bigg [\psi _\tau (\epsilon _i-\hat{Q}_{\tau ,\epsilon })(X_{i,k}-\bar{X}_{k})\bigg ]\bigg \} \end{aligned} \end{aligned}$$

(A.1)

and the second part is

$$\begin{aligned} \begin{aligned}&\sqrt{n}\bigg \{\big \{\frac{1}{n}\sum _{i=1}^n\big [\psi _\tau (\epsilon _i-\hat{Q}_{\tau ,\epsilon })(X_{i,k}-\bar{X}_{k})\big ]-E\big [\psi _\tau (\epsilon -Q_{\tau ,\epsilon })(X_{k}-EX_{k})\big ]\big \}\bigg \}. \end{aligned} \end{aligned}$$

(A.2)

We replace $Q_{\tau ,\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0+\epsilon }$, $\hat{Q}_{\tau ,\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0+\epsilon }$ with $Q_{\tau ,\varvec{Z}}$, $\hat{Q}_{\tau ,\varvec{Z}}$ respectively for the sake of simplicity.

For the first term,

$$\begin{aligned} \begin{aligned}&\sqrt{n}\bigg \{\frac{1}{n}\sum _{i=1}^n\left[ \psi _\tau \left( \frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0+\epsilon _i-\hat{Q}_{\tau ,\varvec{Z}}\right) (X_{i,k}-\bar{X}_{k})\right] \\&\qquad -\frac{1}{n}\sum _{i=1}^n\big [\psi _\tau (\epsilon _i-\hat{Q}_{\tau ,\epsilon })(X_{i,k}-\bar{X}_{k})\big ]\bigg \}\\&\quad =\sqrt{n}\bigg \{\frac{1}{n}\sum _{i=1}^n\left[ \psi _\tau \left( \frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0+\epsilon _i-\hat{Q}_{\tau ,\varvec{Z}})(X_{i,k}-\bar{X}_{k}\right) \right] \\&\qquad -\frac{1}{n}\sum _{i=1}^n\big [\psi _\tau (\epsilon _i-\hat{Q}_{\tau ,\varvec{Z}})(X_{i,k}-\bar{X}_{k})\big ]\bigg \}+o_p(1)\\&\quad =\sqrt{n}\left\{ \frac{1}{n}\sum _{i=1}^n\left[ \psi _\tau \left( \frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0+\epsilon _i-\hat{Q}_{\tau ,\varvec{Z}}\right) -\psi _\tau (\epsilon _i-\hat{Q}_{\tau ,\varvec{Z}})\right] (X_{i,k}-\bar{X}_{k})\right\} \\&\qquad +o_p(1). \end{aligned} \end{aligned}$$

Let

$$\begin{aligned}&g_\tau (\varvec{X}_i^T\varvec{b}_0,\epsilon _i,\hat{Q}_{\tau ,\varvec{Z}})\\&\quad =\psi _\tau \left( \frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0+\epsilon _i-\hat{Q}_{\tau ,\varvec{Z}})-\psi _\tau (\epsilon _i-\hat{Q}_{\tau ,\varvec{Z}}\right) \\&\quad =I\left( -\frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0<\epsilon _i-\hat{Q}_{\tau ,\varvec{Z}}<0)-I(-\frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0>\epsilon _i-\hat{Q}_{\tau ,\varvec{Z}}>0\right) \end{aligned}$$

and let

$$\begin{aligned} \zeta _n=\frac{1}{\sqrt{n}}\sum _{i=1}^n\{g_\tau (\varvec{X}_i^T\varvec{b}_0,\epsilon _i,\hat{Q}_{\tau ,\varvec{Z}})-E[g_\tau (\varvec{X}_i^T\varvec{b}_0,\epsilon _i,\hat{Q}_{\tau ,\varvec{Z}})|\varvec{X}_i]\}X_{i,k}, \end{aligned}$$

where

$$\begin{aligned}&E[g_\tau (\varvec{X}_i^T\varvec{b}_0,\epsilon _i,\hat{Q}_{\tau ,\varvec{Z}})|\varvec{X}_i]\\&\quad =E\left. \left[ I\left( -\frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0<\epsilon _i-\hat{Q}_{\tau ,\varvec{Z}}<0\right) -I\left( -\frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0>\epsilon _i-\hat{Q}_{\tau ,\varvec{Z}}>0\right) \right| \varvec{X}_i\right] \\&\quad =\int _{-\frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0+\hat{Q}_{\tau ,\varvec{Z}}}^{\hat{Q}_{\tau ,\varvec{Z}}}f_{\epsilon |\varvec{X}_i}(t)dt. \end{aligned}$$

It is clear that $E\zeta _n=0$. From Hölder’s inequality,

$$\begin{aligned} \begin{aligned} E[\zeta _n]^2\le&E[g_\tau (\varvec{X}_i^T\varvec{b}_0,\epsilon _i,\hat{Q}_{\tau ,\varvec{Z}})X_{i,k}]^2\\ \le&[P(|\epsilon _i-\hat{Q}_{\tau ,\varvec{Z}}|<1/\sqrt{n}\varvec{X}_i^T\varvec{b}_0)]^{1/2}[EX_{i,k}^4]^{1/2}\\ =&o(1). \end{aligned} \end{aligned}$$

(A.3)

Similar to the proof of Theorem 1 in Li et al. (2015), replace $\varvec{X}_i^T\varvec{b}_0$ by $v$ and denote $g_\tau (v)=g_\tau (v,\epsilon _i,\hat{Q}_{\tau ,\varvec{Z}})$, $\zeta _n(v)=\frac{1}{\sqrt{n}}\sum _{i=1}^n\{g_\tau (v)-E[g_\tau (v)|\varvec{X}_i]\}X_{i,k}$, we have

$$\begin{aligned}&\sup _{|v_1-v|<\delta }|\zeta _n(v_1)-\zeta _n(v)|\\&\quad \le \sup _{|v_1-v|<\delta }\frac{1}{\sqrt{n}}\sum _{i=1}^n\bigg \{|\{g_\tau (v_1)-g_\tau (v)\}\varvec{X}_i|+E\big [|\{g_\tau (v_1)-g_\tau (v)\}\varvec{X}_i|\big |\varvec{X}_i\big ]\bigg \}\\&\quad =\frac{1}{\sqrt{n}}\sum _{i=1}^n\bigg \{|\{g_\tau (v_1^*)-g_\tau (v)\}\varvec{X}_i|+E\big [|\{g_\tau (v_1^*)-g_\tau (v)\}\varvec{X}_i|\big |\varvec{X}_i\big ]\bigg \}, \end{aligned}$$

in which $v_1^*$ equals to $v+\delta$ or $v-\delta$. Then

$$\begin{aligned} \begin{aligned}&E\sup _{|v_1-v|<\delta }\big |\zeta _n(v_1)-\zeta _n(v)\big |\\&\quad \le 2\sqrt{n}E\big |\{g_\tau (v_1^*)-g_\tau (v)\}X_{i,k}\big |\\&\quad =2\sqrt{n}E\bigg |\int _{\hat{Q}_{\tau ,\varvec{Z}}+n^{-1/2}v}^{\hat{Q}_{\tau ,\varvec{Z}}+n^{-1/2}v_1^*}f_{\epsilon |\varvec{X}}(t)dtX_{i,k}\bigg |\\&\quad \le \delta \cdot 2E\big [\sup _{|t|<\pi }f_{\epsilon |\varvec{X}_i}(\hat{Q}_{\tau ,\varvec{Z}}+t)|X_{i,k}|\big ]. \end{aligned} \end{aligned}$$

(A.4)

When $n\rightarrow \infty$, $|n^{-1/2}v|<\pi$, $|n^{-1/2}v_1^*|<\pi$. Combine (A.3) and (A.4) with assumptions, for any $M>0$ and $E\sup _{|v|\le M}|\zeta _n(v)|=o(1)$ we have

$$\begin{aligned} \begin{aligned}&\sqrt{n}\bigg \{\frac{1}{n}\sum _{i=1}^ng_\tau (\varvec{X}_i^T\varvec{b}_0,\epsilon _i,\hat{Q}_{\tau ,\varvec{Z}})X_{i,k}\bigg \}\\&\quad =\zeta _n+\sqrt{n}\bigg \{\frac{1}{n}\sum _{i=1}^nE[g_\tau (\varvec{X}_i^T\varvec{b}_0,\epsilon _i,\hat{Q}_{\tau ,\varvec{Z}})|\varvec{X}_i]X_{i,k}\bigg \}\\&\quad =\sqrt{n}\bigg \{\frac{1}{n}\sum _{i=1}^n\int _{-\frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0+\hat{Q}_{\tau ,\varvec{Z}}}^{\hat{Q}_{\tau ,\varvec{Z}}}f_{\epsilon |\varvec{X}_i}(t)dtX_{i,k}\bigg \}+o_p(1)\\&\quad =\frac{1}{n}\sum _{i=1}^nf_{\epsilon |\varvec{X}_i}(\xi ')\varvec{X}_i^T\varvec{b}_0X_{i,k}+o_p(1),~\xi '\in (-\frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0+\hat{Q}_{\tau ,\varvec{Z}},\hat{Q}_{\tau ,\varvec{Z}}). \end{aligned} \end{aligned}$$

With similar argument we have

$$\begin{aligned} \begin{aligned}&\sqrt{n}\bigg \{\frac{1}{n}\sum _{i=1}^ng_\tau (\varvec{X}_i^T\varvec{b}_0,\epsilon _i,\hat{Q}_{\tau ,\varvec{Z}})\bar{X}_{k}\bigg \}\\&\quad =\sqrt{n}\bigg \{\frac{1}{n}\sum _{i=1}^n\int _{-\frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0+\hat{Q}_{\tau ,\varvec{Z}}}^{\hat{Q}_{\tau ,\varvec{Z}}}f_{\epsilon |\varvec{X}_i}(t)dt\bar{X}_{k}\bigg \}+o_p(1)\\&\quad =\frac{1}{n}\sum _{i=1}^nf_{\epsilon |\varvec{X}_i}(\xi ')\varvec{X}_i^T\varvec{b}_0\bar{X}_{k}+o_p(1),~\xi '\in (-\frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0+\hat{Q}_{\tau ,\varvec{Z}},\hat{Q}_{\tau ,\varvec{Z}}). \end{aligned} \end{aligned}$$

Hence (A.1) can be written as

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^nf_{\epsilon |\varvec{X}_i}(\xi ')\varvec{X}_i^T\varvec{b}_0(X_{i,k}-\bar{X}_{k})+o_p(1),~\xi '\in (-\frac{1}{\sqrt{n}}\varvec{X}_i^T\varvec{b}_0+\hat{Q}_{\tau ,\varvec{Z}},\hat{Q}_{\tau ,\varvec{Z}}). \end{aligned}$$

Because $\frac{1}{\sqrt{n}}\varvec{b}_0\rightarrow 0$ when $n\rightarrow \infty$, we may take $\xi '=\hat{Q}_{\tau ,\epsilon }$. Since $\hat{Q}_{\tau ,\epsilon }{\mathop {\longrightarrow }\limits ^{p}}Q_{\tau ,\epsilon }$, we have $\xi '{\mathop {\longrightarrow }\limits ^{p}}Q_{\tau ,\epsilon }$. Thus the first term can be written as

$$\begin{aligned} E[f_{\epsilon |\varvec{X}}(Q_{\tau ,\epsilon })\varvec{X}^T(X_{k}-EX_{k})]\varvec{b}_0+o_p(1). \end{aligned}$$

(A.5)

For the second term (A.2), it is same to the form in Li et al. (2015, Theorem 1). Combining their theorem with (A.5) yields that $\sqrt{n}\widehat{\text {qcov}}_\tau \{Y,X_k\}$ can be written as

$$\begin{aligned} M_k(\varvec{0})+E[f_{\epsilon |\varvec{X}}(Q_{\tau ,\epsilon })\varvec{X}^T(X_k-EX_k)]\varvec{b}_0+o_p(1). \end{aligned}$$

(A.6)

Under assumptions of Theorem 1, we know that for any $j\ne k$ a.s.

$$\begin{aligned}&\frac{M_j(\varvec{0})+E[f_{\epsilon |\varvec{X}}(Q_{\tau ,\epsilon })\varvec{X}^T(X_j-EX_j)]\varvec{b}_0)}{\sqrt{(\tau -\tau ^2)\hat{\sigma }_{X_j}^2}}\\&\qquad \ne \frac{M_k(\varvec{0})+E[f_{\epsilon |\varvec{X}}(Q_{\tau ,\epsilon })\varvec{X}^T(X_k-EX_k)]\varvec{b}_0}{\sqrt{(\tau -\tau ^2)\hat{\sigma }_{X_k}^2}}, \end{aligned}$$

so $K$ is unique a.s.. Let $\hat{\varvec{q}}=(\widehat{\text {qcov}}_\tau \{Y,X_1\},\ldots ,\widehat{\text {qcov}}_\tau \{Y,X_p\})^T$ and $\hat{\varvec{R}}=(\widehat{\text {qcor}}_\tau \{Y,X_1\},\ldots ,\widehat{\text {qcor}}_\tau \{Y,X_p\})^T$, then we have if $\varvec{\beta }_0=\varvec{0}$,

$$\begin{aligned} \sqrt{n}\hat{\varvec{R}}{\mathop {\longrightarrow }\limits ^{d}}&\bigg (\frac{M_1(\varvec{0})+E[f_{\epsilon |\varvec{X}}(Q_{\tau ,\epsilon })\varvec{X}^T(X_1-EX_1)]\varvec{b}_0}{\sqrt{(\tau -\tau ^2)\hat{\sigma }_{X_1}^2}},\\&\ldots ,\frac{M_p(\varvec{0})+E[f_{\epsilon |\varvec{X}}(Q_{\tau ,\epsilon })\varvec{X}^T(X_p-EX_p)]\varvec{b}_0}{\sqrt{(\tau -\tau ^2)\hat{\sigma }_{X_p}^2}}\bigg )^T. \end{aligned}$$

Similar to McKeague and Qian (2015, Theorem 1), let

$$\begin{aligned} h(\varvec{t})=(\varvec{1}_{\arg \max _kt_k=1},\ldots ,\varvec{1}_{\arg \max _kt_k=p})^T, \end{aligned}$$

where $\varvec{t}=(t_1,\ldots ,t_p)^T\in \mathbb {R}^p$. Notice that if $\arg \max _kt_k$ is unique, $h$ is continuous at $\varvec{t}$ and

$$\begin{aligned} \sqrt{n}\widehat{\text {qcov}}=\sqrt{n}\hat{\varvec{q}}^Th(\sqrt{n}\hat{\varvec{R}}). \end{aligned}$$

Hence

$$\begin{aligned} K=\arg \max _{k=1,\ldots ,p}\frac{M_k(\varvec{0})+E[f_{\epsilon |\varvec{X}}(Q_{\tau ,\epsilon })\varvec{X}^T(X_k-EX_k)]\varvec{b}_0}{\sqrt{(\tau -\tau ^2)\hat{\sigma }_{X_k}^2}}. \end{aligned}$$

(A.7)

Combining (A.6) with (A.7) and by the continuous mapping theorem we have if $\varvec{\beta }_0=\varvec{0}$,

$$\begin{aligned} \begin{aligned}&\sqrt{n}\widehat{\text {qcov}}_\tau {\mathop {\longrightarrow }\limits ^{d}}\\&M_K(\varvec{0})+\{E[f_{\epsilon |\varvec{X}}(Q_{\tau ,\epsilon })\varvec{X}^T(X_K-EX_K)]-E[f_{\epsilon |\varvec{X}}(Q_{\tau ,\epsilon })\varvec{X}^TX_{\bar{k}(\varvec{b}_0)}]\}\varvec{b}_0. \end{aligned} \end{aligned}$$

(A.8)

Because $\frac{1}{\sqrt{n}}\varvec{b}_0\rightarrow \varvec{0}$ when $n\rightarrow \infty$, we have $\hat{Q}_{\tau ,\varvec{X}^T(\varvec{\beta }_0+\frac{1}{\sqrt{n}}\varvec{b}_0)+\epsilon }\rightarrow \hat{Q}_{\tau ,\varvec{X}^T\varvec{\beta }_0+\epsilon },Q_{\tau ,\varvec{X}^T(\varvec{\beta }_0+\frac{1}{\sqrt{n}}\varvec{b}_0)+\epsilon }\rightarrow Q_{\tau ,\varvec{X}^T\varvec{\beta }_0+\epsilon }$. Notice the fact that $E\psi _\tau (\epsilon -Q_{\tau ,\epsilon })(X_{\bar{k}(\varvec{b}_0)}-EX_{\bar{k}(\varvec{b}_0)})=0$, then $-\sqrt{n}\text {qcov}_\tau$ can be written as

$$\begin{aligned} \begin{aligned}&\sqrt{n}\bigg \{-\big \{E\big [\psi _\tau (\varvec{X}^T(\varvec{\beta }_0+\frac{1}{\sqrt{n}}\varvec{b}_0)+\epsilon -Q_{\tau ,\varvec{X}^T(\varvec{\beta }_0+\frac{1}{\sqrt{n}}\varvec{b}_0)+\epsilon })(X_{\bar{k}(\varvec{b}_0)}-EX_{\bar{k}(\varvec{b}_0)})\big ]\\&-E\big [\psi _\tau (\epsilon -Q_{\tau ,Z})(X_{\bar{k}(\varvec{b}_0)}-EX_{\bar{k}(\varvec{b}_0)})\big ]\big \}\bigg \}+o_p(1). \end{aligned} \end{aligned}$$

(A.9)

If $\varvec{\beta }_0=\varvec{0}$, the term (A.9) can be expressed by

$$\begin{aligned}&-\sqrt{n}\bigg \{E\big [\psi _\tau (\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0+\epsilon -Q_{\tau ,\varvec{Z}})(X_{\bar{k}(\varvec{b}_0)}-EX_{\bar{k}(\varvec{b}_0)})\big ]\\&-E\big [\psi _\tau (\epsilon -Q_{\tau ,\varvec{Z}})(X_{\bar{k}(\varvec{b}_0)}-EX_{\bar{k}(\varvec{b}_0)})\big ]\bigg \}+o_p(1)\\ =&-\sqrt{n}E\bigg \{E\big [\psi _\tau (\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0+\epsilon -Q_{\tau ,\varvec{Z}})(X_{\bar{k}(\varvec{b}_0)}-EX_{\bar{k}(\varvec{b}_0)})\big |\varvec{X}\big ]\\&-E\big [\psi _\tau (\epsilon -Q_{\tau ,\varvec{Z}})(X_{\bar{k}(\varvec{b}_0)}-EX_{\bar{k}(\varvec{b}_0)})\big |\varvec{X}\big ]\bigg \}+o_p(1)\\ =&-\sqrt{n}E\bigg \{E\big [\psi _\tau (\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0+\epsilon -Q_{\tau ,\varvec{Z}})-\psi _\tau (\epsilon -Q_{\tau ,\varvec{Z}})\big |\varvec{X}\big ](X_{\bar{k}(\varvec{b}_0)}-EX_{\bar{k}(\varvec{b}_0)})\bigg \}\\&+o_p(1), \end{aligned}$$

where

$$\begin{aligned}&E\big [\psi _\tau (\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0+\epsilon -Q_{\tau ,\varvec{Z}})-\psi _\tau (\epsilon -Q_{\tau ,\varvec{Z}})\big |\varvec{X}\big ]\\&\quad =E\big [-[I(\epsilon -Q_{\tau ,\varvec{Z}}<-\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0)-I(\epsilon -Q_{\tau ,\varvec{Z}}<0)]\big |\varvec{X}\big ]\\&\quad =E\big [[I(-\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0<\epsilon -Q_{\tau ,\varvec{Z}}<0)-I(-\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0>\epsilon -Q_{\tau ,\varvec{Z}}>0)]\big |\varvec{X}\big ]\\&\quad =\int _{-\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0+Q_{\tau ,\varvec{Z}}}^{Q_{\tau ,\varvec{Z}}}f_{\epsilon |\varvec{X}}(t)dt\\&\quad =f_{\epsilon |\varvec{X}}(\xi )\cdot \frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0,~\xi \in (-\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0+Q_{\tau ,\varvec{Z}},Q_{\tau ,\varvec{Z}}). \end{aligned}$$

The last equation is obtained by the assumptions and mean value theorem for definite integrals. Hence (A.9) can be written as $-E[f_{\epsilon |\varvec{X}}(\xi )\varvec{X}^T(X_{\bar{k}(\varvec{b}_0)}-EX_{\bar{k}(\varvec{b}_0)})]\varvec{b}_0+o_p(1),\xi \in (-\frac{1}{\sqrt{n}}\varvec{X}^T\varvec{b}_0+Q_{\tau ,\varvec{Z}},Q_{\tau ,\varvec{Z}})$. Since $\frac{1}{\sqrt{n}}\varvec{b}_0\rightarrow \varvec{0}$ as $n\rightarrow \infty$, we may take $\xi =Q_{\tau ,\epsilon }$. Then $-\sqrt{n}\text {qcov}_\tau$ can be written as

$$\begin{aligned} -E[f_{\epsilon |\varvec{X}}(Q_{\tau ,\epsilon })\varvec{X}^T(X_{\bar{k}(\varvec{b}_0)}-EX_{\bar{k}(\varvec{b}_0)})]\varvec{b}_0+o_p(1). \end{aligned}$$

(A.10)

Aggregating (A.8), (A.10) and Slutsky’s lemma we finish the proof of Lemma 1.

Proof of Lemma 2

If $\varvec{\beta }_0\ne \varvec{0}$,

$$\begin{aligned}&\bigg |\frac{1}{n}\sum _{i=1}^n[\psi _\tau (Y_i-\hat{Q}_{\tau ,Y})(X_{i,\hat{k}_n}-\bar{X}_{\hat{k}_n})-\text {qcov}_\tau ]^2\\&\quad \quad -\frac{1}{n}\sum _{i=1}^n[\psi _\tau (Y_i-Q_{\tau ,Y})(X_{i,\hat{k}_n}-EX_{\hat{k}_n})-\text {qcov}_\tau ]^2\bigg |\\&\quad \le O_p(1)\frac{1}{n}\sum _{i=1}^n\bigg |\psi _\tau (Y_i-\hat{Q}_{\tau ,Y})(X_{i,\hat{k}_n}-\bar{X}_{\hat{k}_n})-\psi _\tau (Y_i-Q_{\tau ,Y})(X_{i,\hat{k}_n}-EX_{\hat{k}_n})\bigg |\\&\quad \le O_p(1) \frac{1}{n}\sum _{i=1}^n\bigg |EX_{\hat{k}_n}-\bar{X}_{\hat{k}_n}\bigg |\\&\quad =o_p(1). \end{aligned}$$

Because $\hat{k}_n{\mathop {\longrightarrow }\limits ^{p}} k_0$, from the definition of convergence in probability we know that for all $\epsilon _0>0$, there exists some integer $N_1$ such that when $n\ge N_1$, we have

$$\begin{aligned} P(|\hat{k}_n-k_0|>\frac{1}{2})<\epsilon _0. \end{aligned}$$

Since $\hat{k}_n$ takes integer value, we have

$$\begin{aligned} \{|\hat{k}_n-k_0|>\frac{1}{2}\}=\{\hat{k}_n\ne k_0\}. \end{aligned}$$

Besides, from the uniform law of large numbers (ULLN) we have

$$\begin{aligned}&\frac{1}{n}\sum _{i=1}^n[\psi _\tau (Y_i-Q_{\tau ,Y})(X_{i,k_0}-EX_{k_0})-\text {qcov}_\tau ]^2\\&\qquad {\mathop {\longrightarrow }\limits ^{p}}E[\psi _\tau (Y-Q_{\tau ,Y})(X_{k_0}-EX_{k_0})]^2=\sigma _1^2, \end{aligned}$$

which means that there exists some integer $N_2$ such that when $n\ge N_2$, we have

$$\begin{aligned} P\left( |\frac{1}{n}\sum _{i=1}^n[\psi _\tau (Y_i-Q_{\tau ,Y})(X_{i,k_0}-EX_{k_0})-\text {qcov}_\tau ]^2-\sigma _1^2|>\delta \right) <\epsilon _0, \end{aligned}$$

where $\delta >0$ is arbitrary.

Hence

$$\begin{aligned}&P\left( |\frac{1}{n}\sum _{i=1}^n[\psi _\tau (Y_i-Q_{\tau ,Y})(X_{i,\hat{k}_n}-EX_{\hat{k}_n})-\text {qcov}_\tau ]^2-\sigma _1^2|>\delta \right) \\&\quad =P\left( |\frac{1}{n}\sum _{i=1}^n \left[ \psi _\tau (Y_i-Q_{\tau ,Y})(X_{i,\hat{k}_n}-EX_{\hat{k}_n})-\text {qcov}_\tau \right] ^2\right. \\&\qquad \quad \left. -\sigma _1^2|>\delta ,|\hat{k}_n-k_0|>\frac{1}{2}\right) \\&\qquad +P\left( \!|\frac{1}{n}\!\sum _{i=1}^n[\psi _\tau (Y_i-Q_{\tau ,Y})(X_{i,\hat{k}_n}-EX_{\hat{k}_n})\!-\!\text {qcov}_\tau ]^2-\sigma _1^2|>\delta ,|\hat{k}_n-k_0|\!\le \!\frac{1}{2}\!\right) \\&\quad \le P\left( |\hat{k}_n-k_0|>\frac{1}{2}\right) \\&\quad \quad +P\left( |\frac{1}{n}\sum _{i=1}^n \left[ \psi _\tau (Y_i-Q_{\tau ,Y})(X_{i,\hat{k}_n}-EX_{\hat{k}_n})-\text {qcov}_\tau \right] ^2-\sigma _1^2|>\delta ,\hat{k}_n=k_0\right) \\&\quad = P\left( |\hat{k}_n-k_0|>\frac{1}{2}\right) \\&\quad \quad +P\left( |\frac{1}{n}\sum _{i=1}^n \left[ \psi _\tau (Y_i-Q_{\tau ,Y})(X_{i,k_0}-EX_{k_0})-\text {qcov}_\tau \right] ^2-\sigma _1^2|>\delta \right) . \end{aligned}$$

Take $N=\max \{N_1,N_2\}$, we have when $n>N$,

$$\begin{aligned} P\big (|\frac{1}{n}\sum _{i=1}^n[\psi _\tau (Y_i-Q_{\tau ,Y})(X_{i,\hat{k}_n}-EX_{\hat{k}_n})-\text {qcov}_\tau ]^2-\sigma _1^2|>\delta \big )<2\epsilon _0 \end{aligned}$$

for any $\delta >0$ and $\epsilon _0>0$, which means

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n[\psi _\tau (Y_i-Q_{\tau ,Y})(X_{i,\hat{k}_n}-EX_{\hat{k}_n})-\text {qcov}_\tau ]^2{\mathop {\longrightarrow }\limits ^{p}}\sigma _1^2. \end{aligned}$$

If $\varvec{\beta }_0=\varvec{0}$, with similar argument we have

$$\begin{aligned}&\frac{1}{n}\sum _{i=1}^n[\psi _\tau (Y_i-Q_{\tau ,Y})(X_{i,\hat{k}_n}-EX_{\hat{k}_n})-\text {qcov}_\tau ]^2\\&\qquad {\mathop {\longrightarrow }\limits ^{p}}E[\psi _\tau (Y-Q_{\tau ,Y})(X_{\bar{k}(\varvec{b}_0)}-EX_{\bar{k}(\varvec{b}_0)})]^2=\sigma _2^2. \end{aligned}$$

Combining all above yields Lemma 2.

Proof of Theorem 1

Let $W_n=\max \limits _{1\le i\le n}\big |\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau \big |$. It is obvious that $W_n=O(1)$ a.s..

Let $S_n=\frac{1}{n}\sum _{i=1}^n(\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )^2$. From equation (2.4) we have

$$\begin{aligned} 0=|f(\lambda )|&=\frac{1}{n}\bigg |\sum _{i=1}^n(\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )-\lambda \sum _{i=1}^n\frac{(\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )^2}{1+\lambda (\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )}\bigg |\\&\ge \frac{|\lambda |}{n}\sum _{i=1}^n\frac{(\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )^2}{1+\lambda (\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )}-\frac{1}{n}\left| \sum _{i=1}^n(\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )\right| \\&\ge \frac{|\lambda |S_n}{1+|\lambda |W_n}-\left| \frac{1}{n}\sum _{i=1}^{n}\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau \right| . \end{aligned}$$

From Lemma 1 we obtain $\frac{|\lambda |S_n}{1+|\lambda |W_n}=O_p(n^{-1/2})$, which yields $|\lambda |=O_p(n^{-1/2})$.

Let $\gamma _i=\lambda (\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )$, we have $\max \limits _{1\le i\le n}|\gamma _i|=O_p(n^{-1/2})O(1)=O_p(n^{-1/2})$. Then equation (2.4) can be written as:

$$\begin{aligned} 0=f(\lambda )&=\frac{1}{n}\sum _{i=1}^n\left[ \left( \hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )(1-\gamma _i+\frac{\gamma _i^2}{1+\gamma _i}\right) \right] \\&=\frac{1}{n}\sum _{i=1}^n\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau -S_n\lambda +\frac{1}{n}\sum _{i=1}^n(\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )\frac{\gamma _i^2}{1+\gamma _i}. \end{aligned}$$

From discussed above the last term $\frac{1}{n}\sum _{i=1}^n(\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )\frac{\gamma _i^2}{1+\gamma _i}=O_p(n^{-3/2})$, so we obtain

$$\begin{aligned} \lambda =S_n^{-1}\left( \frac{1}{n}\sum _{i=1}^n\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )+\beta ,~\beta =O_p(n^{-1}\right) . \end{aligned}$$

Hence from Taylor’s expansion we have

$$\begin{aligned} l(\text {qcov}_\tau )=2\sum _{i=1}^n\gamma _i-\sum _{i=1}^n\gamma _i^2+2\eta _n, \end{aligned}$$

where

$$\begin{aligned} |\eta _n|\le C\sum _{i=1}^n|\lambda (\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )|^3\le C|\lambda |^3n=O_p(n^{-1/2}). \end{aligned}$$

Plugging $\lambda$ we obtain

$$\begin{aligned} l(\text {qcov}_\tau )&=2n\lambda (\frac{1}{n}\sum _{i=1}^n(\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau ))-nS_n\lambda ^2+2\eta _n\\&=\frac{n(\frac{1}{n}\sum _{i=1}^n\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )^2}{S_n}-nS_n\beta ^2\\&\qquad +2\eta _n-2S_n^{-1}\left( \frac{1}{n}\sum _{i=1}^n\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau \right) \beta n. \end{aligned}$$

So if $\varvec{\beta }_0\ne \varvec{0}$,

$$\begin{aligned} l(\text {qcov}_\tau )=\frac{n(\frac{1}{n}\sum _{i=1}^n\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )^2}{\Omega _{k_0}(\varvec{\beta }_0)}\frac{\Omega _{k_0}(\varvec{\beta }_0)}{S_n}+o_p(1){\mathop {\longrightarrow }\limits ^{p}}\frac{\Omega _{k_0}(\varvec{\beta }_0)}{\sigma _1^2}\chi _1^2. \end{aligned}$$

The last step could be deduced from Lemma 1 and Lemma 2 directly.

Similarly, if $\varvec{\beta }_0=\varvec{0}$, we have

$$\begin{aligned} l(\text {qcov}_\tau )=\frac{n(\frac{1}{n}\sum _{i=1}^n\hat{V}_{i,\hat{k}_n}-\text {qcov}_\tau )^2}{\Omega _{K}(\varvec{0})}\frac{\Omega _{K}(\varvec{0})}{S_n}+o_p(1){\mathop {\longrightarrow }\limits ^{p}}\frac{\Omega _{K}(\varvec{0})}{\sigma _2^2}\chi _1^2(\delta ). \end{aligned}$$

The proof is completed.

Proof of Theorem 2

Proof of Theorem 2 is similar to that of Theorem 1, and hence we omit it here.