Quantile regression and variable selection for partially linear model with randomly truncated data

Abstract

This paper focuses on the problem of estimation and variable selection for quantile regression (QR) of partially linear model (PLM) where the response is subject to random left truncation. We propose a three-stage estimation procedure for parametric and nonparametric parts based on the weights which are random quantities and determined by the product-limit estimates of the distribution function of truncated variable. The estimators obtained in the second and third stages are more efficient than the initial estimators in the first stage. Furthermore, we propose a variable selection procedure for the QR of PLM by combining the estimation method with the smoothly clipped absolute deviation penalty to get sparse estimation of the regression parameter. The oracle properties of the variable selection approach are established. Simulation studies are conducted to examine the performance of our estimators and variable selection method.

Appendix

Before we present the proofs of the theorems, we first state some regularity conditions. They are also assumed in Zhou (2011) and Kai et al. (2011). Let $\delta _n=\Big (\frac{\log (1/h)}{nh}\Big )^{1/2}$.

• (C1) The kernel function $K(\cdot )$ is a symmetry continuous density function with bounded tight support, satisfies one-order Lipschitz condition, and $\int ^{\infty }_{-\infty }u^2K(u)du<\infty$, $\int ^{\infty }_{-\infty }u^jK^2(u)du<\infty ,j=0,1,2.$

• (C2) F, G are continuous and $a_G\le a_F$.

• (C3) The random variable W has bounded support $\mathcal {W}$ and its density function $f_W(\cdot )$ is positive and has a second derivative.

• (C4) $F_{\varepsilon }(0|X,W)=\tau$ for all (X, W) with a continuous and uniformly bounded derivative and $f_\varepsilon (\cdot |X,W)$ is bounded away from zero.

• (C5) The matrixes $C_2(w) ~\text {and} ~A$ are non-singular for all $w\in \mathcal {W}$.

• (C6) The function $g(\cdot )$ has a continuous and bounded second derivative.

Lemma A.1

Let $(X_1,Y_1), \ldots , (X_n,Y_n)$ be independent and identically distributed random vectors. Assume that $E|Y|^s<\infty , sup_x\int |y|^sf(x,y)dy<\infty$, where f denotes the joint density of (X, Y). Let K be a bounded positive function with a bounded support, satisfying a Lipschitz condition. Then

$$\begin{aligned} \sup _{x} \Bigg |\frac{1}{n}\sum ^n_{i=1}\big [K_h(X_i-x)Y_i-E(K_h (X_i-x)Y_i)\big ]\Bigg |=O_p\Bigg (\frac{\log ^{1/2}(1/h)}{\sqrt{nh}}\Bigg ), \end{aligned}$$

provided that $n^{2\varepsilon -1}h\rightarrow \infty$ for some $\varepsilon <1-s^{-1}$.

Lemma A.1 follows from the result by Mack and Silverman (1982).

Lemma A.2

(Lv et al. 2015) Suppose $A_n(s)$ is convex and can be represented as $\frac{1}{2}s^TVs+U_n^Ts+C_n+r_n(s)$, where V is symmetric and positive definite, $U_n$ is stochastically bounded, $C_n$ is arbitrary and $r_n(s)$ goes to zero in probability for each s. Then $\alpha _n$, the argmin of $A_n$, is only $o_p(1)$ away from $\beta _n=-V^{-1}U_n$, the argmin of $\frac{1}{2}s^TVs+U_n^Ts+C_n$. If also $U_n \mathop {\rightarrow }\limits ^{\mathcal {D}}U$, then $\alpha _n\mathop {\rightarrow }\limits ^{\mathcal {D}}-V^{-1}U$.

Proof of Theorem 2.1

For given w, then $\tilde{g}_\tau (w), \tilde{g}'_\tau (w), \tilde{\beta }$ minimizes

$$\begin{aligned} \sum _{i=1}^n\frac{1}{G_n(Y_i)}\rho _\tau \big [Y_i-X_i^T \beta -a-b(W_i-w)\big ]K_h(W_i-w). \end{aligned}$$

Denote

$$\begin{aligned} \tilde{\xi }=\sqrt{nh}\left( \begin{array}{ccc} \tilde{g}_\tau (w)-g_\tau (w)\\ h(\tilde{g}'_\tau (w)-g'_\tau (w))\\ \tilde{\beta }_\tau -\beta _\tau \end{array} \right) , \xi =\sqrt{nh}\left( \begin{array}{ccc} a-g_\tau (w)\\ h(b-g'_\tau (w))\\ \beta -\beta _\tau \end{array} \right) , N_i= \left( \begin{array}{c} 1\\ \frac{W_i-w}{h}\\ X_i \end{array} \right) , \end{aligned}$$

$\tilde{r}_i(w)=-g_\tau (W_i)+g_\tau (w)+g'_\tau (w)(W_i-w), K_i(w)=K\big (\frac{W_i-w}{h}\big )$, then $\tilde{\xi }$ will be the minimizer of

$$\begin{aligned} Q_n(\xi )=\sum _{i=1}^n\frac{K_i(w)}{G_n(Y_i)}\left[ \rho _\tau \left( \varepsilon _i-\tilde{r}_i(w)-N_i^T\xi /\sqrt{nh}\right) -\rho _\tau (\varepsilon _i-\tilde{r}_i(w))\right] . \end{aligned}$$

Following the identity by Knight (1998),

$$\begin{aligned} \rho _\tau (u-v)-\rho _\tau (u)=-v\psi _\tau (u)+\int _0^v\big [I(u\le s)-I(u\le 0)\big ]ds, \end{aligned}$$

where $\psi _\tau (u)=\tau -I(u\le 0)$, then we obtain

$$\begin{aligned} Q_n(\xi )&=\sum _{i=1}^n\frac{K_i(w)}{G_n(Y_i)}\bigg [ -\frac{N_i^T\xi }{\sqrt{nh}}\psi _\tau (\varepsilon _i-\tilde{r}_i(w))\\&\quad +\,\int _{0}^{N_i^T\xi /\sqrt{nh}}\big [I(\varepsilon _i-\tilde{r}_i(w)\le s)-I(\varepsilon _i-\tilde{r}_i(w)\le 0)\big ]ds\bigg ]\\&=-\frac{1}{\sqrt{nh}}\sum _{i=1}^n\frac{K_i(w)}{G_n(Y_i)}N_i^T \xi \psi _\tau (\varepsilon _i-\tilde{r}_i(w))\\&\quad ~~+\sum _{i=1}^n\frac{K_i(w)}{G_n(Y_i)}\int _{0}^{N_i^T\xi / \sqrt{nh}}\big [I(\varepsilon _i-\tilde{r}_i(w)\le s)-I( \varepsilon _i-\tilde{r}_i(w)\le 0)\big ]ds\\&:=-Q_{1n}^T\xi +Q_{2n}(\xi ). \end{aligned}$$

In the following, we prove $E[Q_{2n}(\xi )]=\frac{1}{2}\xi ^T\frac{f_W(w)}{\theta }C_1(w)\xi$.

Denote $\tilde{Q}_{2n}(\xi )=\sum _{i=1}^n\frac{K_i(w)}{G(Y_i)} \int _{0}^{N_i^T\xi /\sqrt{nh}}\big \{I(\varepsilon _i\le s+\tilde{r}_i(w))-I(\varepsilon _i\le \tilde{r}_i(w))\big \}ds$, $\Delta$ and $\tilde{r}_i(\mathbf{w})$ are equal to $N_i^T\xi /\sqrt{nh}$ and $\tilde{r}_i(w)$ which $X_i, W_i$ are replaced by $x,\mathbf{w}$. Since $\tilde{Q}_{2n}(\xi )$ is a summation of i.i.d. random variables of the kernel form, according to Lemma A.1, we have

$$\begin{aligned} \tilde{Q}_{2n}(\xi )=E[\tilde{Q}_{2n}(\xi )]+O_p(\delta _n). \end{aligned}$$

The expectation of $\tilde{Q}_{2n}(\xi )$:

$$\begin{aligned} E(\tilde{Q}_{2n}(\xi ))&=\sum _{i=1}^nE\bigg [\frac{K_i(w)}{G(Y_i)} \int _{0}^{N_i^T\xi /\sqrt{nh}}\big \{I(\varepsilon _i\le s+\tilde{r}_i(w))-I(\varepsilon _i\le \tilde{r}_i(w))\big \}ds\bigg ]\\&=\sum _{i=1}^n\int \int \int \frac{1}{G(y)}K(\frac{\mathbf{w}-w}{h}) \int _{0}^{\Delta }\big [I(y\le x^T\beta +g(\mathbf{w})+s+\tilde{r}_i(\mathbf{w}))\\&~~~~-I(y\le x^T\beta +g(\mathbf{w})+\tilde{r}_i(\mathbf{w}))\big ] dsf^{*} (x,y,\mathbf{w})dxdyd\mathbf{w}\\&=\frac{1}{\theta }\sum _{i=1}^n\mathbb {E}\bigg \{K_i(w) \int _{0}^{N_i^T\xi /\sqrt{nh}}\big \{I(\varepsilon _i\le s+\tilde{r}_i(w))-I(\varepsilon _i\le \tilde{r}_i(w))\big \} ds\bigg \}\\&=\frac{1}{\theta }\sum _{i=1}^n\mathbb {E}\bigg \{K_i(w)\mathbb {E} \big \{\int _{0}^{N_i^T\xi /\sqrt{nh}}\big [\big \{I(\varepsilon _i \le s+\tilde{r}_i(w))\\&\qquad -I(\varepsilon _i\le \tilde{r}_i(w))\big \}|X,W\big ]ds\big \}\bigg \}\\&=\frac{1}{\theta }\sum _{i=1}^n\mathbb {E}\bigg \{K_i(w) \int _{0}^{N_i^T\xi /\sqrt{nh}}\big [F_{\varepsilon }(s+\tilde{r}_i(w))-F_{\varepsilon }(\tilde{r}_i(w))\big ]ds\bigg \}\\&=\frac{1}{\theta }\sum _{i=1}^n\mathbb {E}\bigg \{K_i(w) \int _{0}^{N_i^T\xi /\sqrt{nh}}\big [sf_{\varepsilon }(\tilde{r}_i(w)|X,W)+o(1)\big ]ds\bigg \}\\&=\frac{1}{2\theta }\xi ^T\mathbb {E}\bigg \{\frac{1}{nh} \sum _{i=1}^nK_i(w)f_{\varepsilon }(\tilde{r}_i(w)|X,W)N_iN_i^T\bigg \}\xi +O_p(\delta _n). \end{aligned}$$

Similarly, we can obtain $\text {Var}[{\tilde{Q}_{2n}(\xi )}]=o(1).$ Then $\tilde{Q}_{2n}(\xi )=\frac{1}{2}\xi ^T\frac{f_W(w)}{\theta }C_1(w) \xi +O_p(\delta _n)$, where $C_1(w)=\mathbb {E}[f_{\varepsilon } (0|X,W)(1,(W-w) /h,X^T)^T(1, (W-w)/h,X^T)|W=w]$. Further, according to Lemma 5.2 in Liang and Baek (2016), we have

$$\begin{aligned} \text {sup}_y|G_n(y)-G(y)|=O_p(n^{-1/2}), \end{aligned}$$

(A.1)

and by some calculations, we have

$$\begin{aligned} | Q_{2n}(\xi )- \tilde{Q}_{2n}(\xi )|=O_p(h^{\frac{1}{2}})=o_p(1). \end{aligned}$$

(A.2)

Thus,

$$\begin{aligned} Q_{n}(\xi )&=-Q_{1n}^T\xi +E[Q_{2n}(\xi )]+O_p(\delta _n)\\&=-Q_{1n}^T\xi +\frac{1}{2}\xi ^T\frac{f_W(w)}{\theta }C_1(w) \xi +O_p(\delta _n). \end{aligned}$$

According to the Lemma A.2, the minimizer of $Q_{n}(\xi )$ can be expressed as

$$\begin{aligned} \tilde{\xi }=\theta f^{-1}_W(w)C^{-1}_1(w)Q_{1n}+o_p(1). \end{aligned}$$

(A.3)

Therefore,

$$\begin{aligned} \sqrt{nh}\Big ( \begin{array}{ccc} \tilde{g}_\tau (w)-g_\tau (w)\\ \tilde{\beta }_\tau -\beta _\tau \end{array} \Big )=\theta f^{-1}_W(w)C^{-1}_2(w)Q_{1n,1}+o_p(1). \end{aligned}$$

(A.4)

where $C_2(w)=\mathbb {E}\{f_{\varepsilon }(0|X,W)(1,X^T)^T (1,X^T)|W=w\}$, $Q_{1n,1}=\frac{1}{\sqrt{nh}}\sum _{i=1}^n \frac{K_i(w)}{G_n(Y_i)}\psi _\tau (\varepsilon _i-\tilde{r}_i)(1,X_i^T)^T$. In the following, consider $Q_{1n,1}$. Denote $Q^{*}_{1n,1}=\frac{1}{\sqrt{nh}}\sum _{i=1}^n\frac{K_i(w)}{G(Y_i)}(1,X_i^T)^T\psi _\tau (\varepsilon _i)$,

$$\begin{aligned} E(Q^{*}_{1n,1})&=\frac{1}{\sqrt{nh}}\sum _{i=1}^nE\big \{ \frac{K_i(w)}{G(Y_i)}(1,X_i^T)^T\psi _\tau (\varepsilon _i)\big \}\\&=\frac{1}{\theta \sqrt{nh}}\sum _{i=1}^n\mathbb {E}\big [K_i(w) (1,X_i^T)^T\mathbb {E}\{(\tau -I(\varepsilon _i\le 0))|X,W\}\big ]\\&=\frac{1}{\theta \sqrt{nh}}\sum _{i=1}^n\mathbb {E}\big [K_i (w)(1,X_i^T)^T\big (\tau -F_{\varepsilon }(0|X,W)\big )\big ]=0, \end{aligned}$$

and Var$(Q^{*}_{1n,1})\rightarrow \frac{\tau (1-\tau )f_W(w)\nu _0}{\theta }D_2(w)$, where $D_2(w)=\mathbb {E}[\frac{1}{G(Y)}(1,X^T)^T(1,X^T)|W=w]$. By the Cramér-Wald theorem and the central limit theorem, we have

$$\begin{aligned} Q^{*}_{1n,1}\mathop {\rightarrow }\limits ^{\mathcal {D}}N\left( 0,\frac{\tau (1-\tau )f_W(w)\nu _0}{\theta }D_2(w) \right) . \end{aligned}$$

Define $\tilde{Q}_{1n,1}=\frac{1}{\sqrt{nh}}\sum _{i=1}^n\frac{K_i(w)}{G(Y_i)}(1,X_i^T)^T\psi _\tau (\varepsilon _i-\tilde{r}_i),$ we have

$$\begin{aligned}&\text {Var}(\tilde{Q}_{1n,1}-Q^{*}_{1n,1})\\&\quad \le \frac{C}{\theta G(a_F)nh}\sum _{i=1}^nK^2_i(w)(1,X_i^T)^T(1,X_i^T)max\{F_{\varepsilon } (|\tilde{r}_i|)-F_{\varepsilon }(0)\}=o_p(1). \end{aligned}$$

Thus

$$\begin{aligned} \text {Var}(\tilde{Q}_{1n,1}-Q^{*}_{1n,1})=o(1). \end{aligned}$$

By Slutsky’s theorem, conditioning on X, W, we have

$$\begin{aligned} \tilde{Q}_{1n,1}-E(\tilde{Q}_{1n,1})\mathop {\rightarrow }\limits ^{\mathcal {D}}N(0,\frac{\tau (1-\tau )f_W(w)\nu _0}{\theta }D_2(w) ). \end{aligned}$$

(A.5)

Note that

$$\begin{aligned} Q_{1n,1}=Q_{1n,1}-\tilde{Q}_{1n,1}+(\tilde{Q}_{1n,1}-E\tilde{Q}_{1n,1})+E\tilde{Q}_{1n,1}. \end{aligned}$$

(A.6)

Similar to the proof of (A.2), we have $Q_{1n,1}-\tilde{Q}_{1n,1}=o_p(1)$. Thus,

$$\begin{aligned} Q_{1n,1}-E\tilde{Q}_{1n,1}=(\tilde{Q}_{1n,1}-E\tilde{Q}_{1n,1})+o_p(1). \end{aligned}$$

(A.7)

Next we calculate the mean of $\tilde{Q}_{1n,1}$.

$$\begin{aligned} \frac{1}{\sqrt{nh}}E(\tilde{Q}_{1n,1})&=\frac{1}{nh} \sum _{i=1}^nE\bigg \{\frac{K_i(w)}{G(Y_i)}\psi _\tau (\varepsilon _i- \tilde{r}_i(w))(1,X_i^T)^T\bigg \}\nonumber \\&=-\frac{1}{h\theta }\mathbb {E}\big [K_i(w)\mathbb {E}\{(I(\varepsilon _i -\tilde{r}_i(w)\le 0)-\tau )|X,W\}(1,X_i^T)^T\big ]\nonumber \\&=-\frac{1}{h\theta }\mathbb {E}\big [K_i(w)\{F_{\varepsilon }(\tilde{r}_i(w)|X,W)-F_{\varepsilon }(0|X,W)\}(1,X_i^T)^T\big ]\nonumber \\&=-\frac{1}{h\theta }\mathbb {E}\big \{K_i(w)\tilde{r}_i(w)f_{\varepsilon }(0|X,W)\big (1+o(1)\big )(1,X_i^T)^T\big \}\nonumber \\&=\frac{\mu _2h^2}{2\theta }f_W(w)C_2(w)\bigg ( \begin{array}{ccc} {g''}_{\!\tau }(w)\\ 0 \end{array} \bigg )+o_p(h^2). \end{aligned}$$

(A.8)

Combining (A.4), (A.5), (A.6), (A.7) and (A.8), the proof of the Theorem 2.1 is completed.

Proof of Theorem 2.2

Given $\tilde{g}_\tau (W_i)$, then

$$\begin{aligned} \hat{\beta }_\tau =\underset{\beta }{\arg \min }\sum _{i=1}^n\frac{1}{G_n(Y_i)}\rho _\tau \big (Y_i-\tilde{g}_\tau (W_i)-X_i^T\beta \big ). \end{aligned}$$

Denote $r_i=\tilde{g}_\tau (W_i)-g_\tau (W_i)$, $\gamma =\sqrt{n} (\beta -\beta _\tau )$ and $\hat{\gamma }=\sqrt{n}(\hat{\beta }-\beta _\tau )$. Then $\hat{\gamma }$ will be the minimizer of the

$$\begin{aligned} V_n(\gamma )=\sum _{i=1}^n\frac{1}{G_n(Y_i)}\big [\rho _\tau (\varepsilon _i- r_i-X_i^T\gamma /\sqrt{n})-\rho _\tau (\varepsilon _i- r_i)\big ]. \end{aligned}$$

(A.9)

Using the identity by Knight (1998),

$$\begin{aligned} \rho _\tau (u-v)-\rho _\tau (u)=-v\psi _\tau (u)+\int _0^v\big [I(u\le s)-I(u\le 0)\big ]ds, \end{aligned}$$

(A.9) can be written as

(A.10)

Consider $V_{2n}(\gamma )$ firstly. Denote

$$\begin{aligned} V^*_{2n}(\gamma )=\sum _{i=1}^n\frac{1}{G(Y_i)}\int _{r_i}^{ r_i+X_i^T\gamma /\sqrt{n}}\big [I(\varepsilon _i\le s)-I(\varepsilon _i\le 0)\big ]ds \end{aligned}$$

The conditional expectation of $V^*_{2n}(\gamma )$:

By some calculations, we have $\text {Var}[V^*_{2n}(\gamma )]=o(1).$ Thus,

Denote $\tilde{R}_n(\gamma )=V^*_{2n}(\gamma )-E^*(V_{2n}(\gamma )|X,W)$, it is easy to get that $\tilde{R}_n(\gamma )=o_p(1)$. Note that $V_{2n}(\gamma ) =V^*_{2n}(\gamma )+o_p(1)$, then we have

Next, consider $V_{1n}$. Define $V^*_{1n}=\frac{1}{\sqrt{n}} \sum _{i=1}^n \frac{1}{G(Y_i)}X_i\psi _\tau (\varepsilon _i)$. According to (A.1), we have $V_{1n}=V^*_{1n}+o_p(1)$. Hence,

$$\begin{aligned} V_n(\gamma )= & {} -\bigg [\frac{1}{\sqrt{n}}\sum _{i=1}^n\frac{1}{G(Y_i)}X_i\psi _\tau (\varepsilon _i)\bigg ]^T\gamma +\frac{1}{2} \gamma ^TA_n\gamma \nonumber \\&\quad +\,\bigg [\frac{1}{\sqrt{n}}\sum _{i=1}^n\frac{1}{G(Y_i)} f_{\varepsilon }(0|X,W)r_iX_i\bigg ]^T\gamma +o_p(1). \end{aligned}$$

By (A.3),

$$\begin{aligned}&=\frac{1}{\sqrt{n}}\sum _{i=1}^n\frac{1}{G(Y_i)}f_{\varepsilon } (0|X,W)r_iX_i\\&=\frac{1}{\sqrt{n}}\sum _{i=1}^n\frac{1}{G(Y_i)}\frac{f_{ \varepsilon }(0|X,W)}{f_W(w)}\theta X_i\Big ( \begin{array}{ccc} 1\\ \mathbf 0 \end{array} \Big )^T C^{-1}_2(W_i)\\&\quad ~\times \, \left\{ \frac{1}{nh}\sum _{j=1}^n \frac{K_j}{G_n(Y_j)}\Big ( \begin{array}{ccc} 1\\ X_j \end{array}\Big )\psi _\tau (\varepsilon _j-\tilde{r}_j)\right\} \\&\quad ~+O_p(h^{\frac{3}{2}}+\log ^{\frac{1}{2}}(1/h)/\sqrt{nh^2})\\&= \frac{1}{\sqrt{n}}\sum _{j=1}^n\frac{1}{G_n(Y_j)}\psi _\tau ( \varepsilon _j)\delta (X_j,W_j)+O_p(n^{\frac{1}{2}}h^2+\log ^{ \frac{1}{2}}(1/h)/\sqrt{nh^2})\\&=\frac{1}{\sqrt{n}}\sum _{i=1}^n\frac{1}{G(Y_i)}\psi _\tau ( \varepsilon _i)\delta (X_i,W_i)+o_p(1), \end{aligned}$$

where $\delta (X_i,W_i)=\mathbb {E}\big [f_{\varepsilon }(0| X,W)X(1,\mathbf 0 ^T)\big ]C^{-1}_2(W_i)(1,X_i^T)^T$. Thus,

$$\begin{aligned} V_n(\gamma )&=\frac{1}{2}\gamma ^TA_n\gamma -\bigg [\frac{1}{\sqrt{n}}\sum _{i=1}^n\frac{1}{G(Y_i)}\psi _\tau (\varepsilon _i) \{X_i-\delta (X_i,W_i)\}\bigg ]^T\gamma +o_p(1)\\&{:=}\frac{1}{2}\gamma ^TA_n\gamma -B^T_n\gamma +o_p(1). \end{aligned}$$

Observe that $A_n=EA_n+o_p(1)=\frac{1}{\theta } \mathbb {E} \big \{f_{\varepsilon }(0|X,W)XX^T\big \}+o_p(1):=\frac{1}{ \theta }A+o_p(1)$, hence

$$\begin{aligned} V_n(\gamma )=\frac{1}{2\theta }\gamma ^TA\gamma -B^T_n\gamma +o_p(1). \end{aligned}$$

According to the Lemma A.2, we have

$$\begin{aligned} \hat{\gamma }=\theta A^{-1}B_n+o_p(1). \end{aligned}$$

(A.11)

Further, according to the Cramér-Wald device and central limit theorem, we have

$$\begin{aligned} B_n\mathop {\rightarrow }\limits ^{\mathcal {D}}N\left( 0,\frac{\tau (1-\tau )}{\theta } B\right) , \end{aligned}$$

(A.12)

where $B=\mathbb {E}\{X-\delta (X,W)\}^{\otimes 2}.$ Combining (A.11) with (A.12), we accomplish the proof of Theorem 2.2.

Proof of Theorem 2.3

The asymptotic normality of $\hat{g}_\tau (w)$ can be obtained by following the ideas in the proof of Theorem 2.1, we omit the details here.

Proof of Theorem 2.4

Denote $\zeta =\sqrt{n}(\beta -\beta _\tau ),$$\hat{\zeta }=\sqrt{n}(\hat{\beta }^\lambda -\beta _\tau ),$$\hat{\zeta }_1=\sqrt{n}(\hat{\beta }_1^\lambda -\beta _{1\tau })$ and $r_i=\hat{g}_\tau (W_i)-g_\tau (W_i)$, then $\hat{\beta }^\lambda$ be the minimizer of the following penalized target function:

$$\begin{aligned} \sum ^n_{i=1}\frac{1}{G_n(Y_i)}\rho _\tau \{Y_i-X_i^T\beta -\hat{g}_\tau (W_i)\}+n\sum ^q_{j=1}p^{\prime }_\lambda (|\beta ^{(0)}_{j}|) \text {sgn}(\beta ^{(0)}_j)(\beta _{j}-\beta _{\tau j}). \end{aligned}$$

(A.13)

Minimizing (A.13) is equivalent to minimizing

$$\begin{aligned} L_n(\zeta )&=\sum ^n_{i=1}\frac{1}{G_n(Y_i)}\big \{\rho _\tau ( \varepsilon _i-r_i-X_i^T\zeta /\sqrt{n})-\rho _\tau (\varepsilon _i -r_i)\big \} \\&\quad +n\sum ^q_{j=1}p^{'}_\lambda (|\beta ^{(0)}_{j}|) \text {sgn}(\beta ^{(0)}_j)(\beta _{j}-\beta _{\tau j}). \end{aligned}$$

The second term above can be expressed as

$$\begin{aligned} n\sum ^q_{j=1}p'_\lambda \text {sgn}(\beta ^{(0)}_j)(\beta _{j}-\beta _{\tau j})\mathop {\rightarrow }\limits ^{\mathcal {P}}\bigg \{ \begin{array}{ll} 0,&{}\quad \text {if}~~ \beta _2=\beta _{2\tau },\\ \infty , &{}\quad \text {otherwise}. \end{array} \end{aligned}$$

Therefore, we obtain $\hat{\beta }_2^{\lambda }\mathop {\rightarrow }\limits ^{\mathcal {P}}0$.

Denote $B_{n,11}$ is upper-left $s\times s$ submatrix of $B_n$. Due to $\hat{\zeta }$ is the minimizer of the $L_n(\zeta )$ and $L_n(\zeta )$ can be written asymptotically as

$$\begin{aligned} L_n\big ((\zeta _1^T, \mathbf 0 ^T)^T\big )= & {} \frac{1}{2}(\zeta _1^T, \mathbf 0 ^T)\frac{A}{\theta }(\zeta _1^T,\mathbf 0 ^T)^T-B_n^T( \zeta _1^T, \mathbf 0 ^T)^T\\&+\,n\sum ^q_{j=1}p^{'}_\lambda (|\beta ^{ (0)}_{j}|)\text {sgn}(\beta ^{(0)}_j)(\beta _{j}-\beta _{\tau j})+o_p(1)\\\rightarrow & {} L(\zeta _1)=\frac{1}{2}\zeta _1^T\frac{\Sigma _1}{\theta } \zeta _1-B_{n,11}^T\zeta _1. \end{aligned}$$

Note that $L_n(\zeta )$ is a convex function of $\zeta$ and $L(\zeta _1)$ has a unique minimizer, the epiconvergence theory of Geyer (1994) imply that

$$\begin{aligned} \arg \min L_n(\zeta )=\sqrt{n}(\hat{\beta }^\lambda -\beta _\tau )\mathop {\rightarrow }\limits ^{ \mathcal {D}}\arg \min L(\zeta _1), \end{aligned}$$

which establishes the asymptotic normality part.

To prove the consistency property of model selection, we need only to show $\hat{\beta }_2^{\lambda }=0$ with probability tending to 1. It is equivalent to prove that if $\beta _{\tau j}=0$, then $P(\hat{\beta }_j^{ \lambda }\ne 0)\rightarrow 0$. Recall the fact that $|\frac{\rho _{ \tau } (t_2)-\rho _{\tau }(t_1)}{t_2-t_1}|\le \mathrm {max}(\tau ,1-\tau )<1$, if $\hat{\beta }_j^{\lambda }\ne 0,$ then we have $\sqrt{n}p^{\prime }_\lambda ( |\beta ^{(0)}_{j}|)<n^{-1}\sum ^n_{i=1}\frac{1}{G_n(Y_i)}|X_{ij}|$. Therefore, $P(\hat{\beta }_j^{\lambda }\ne 0)\le P\big (\sqrt{n} p^{\prime }_\lambda (|\beta ^{(0)}_{j}|)<n^{-1}\sum ^n_{i=1}\frac{1}{G_n(Y_i)} |X_{ij}|\big )$, which together with $\sqrt{n}p^{\prime }_\lambda (|\beta ^{(0)}_{j}|) \rightarrow \infty$ yields that $P(\hat{\beta }_j^{\lambda }\ne 0)\rightarrow 0$.