Robust empirical likelihood for partially linear models via weighted composite quantile regression

Abstract

In this paper, we investigate robust empirical likelihood inferences for partially linear models. Based on weighted composite quantile regression and QR decomposition technology, we propose a new estimation method for the parametric components. Under some regularity conditions, we prove that the proposed empirical log-likelihood ratio is asymptotically chi-squared, and then the confidence intervals for the parametric components are constructed. The resulting estimators for parametric components are not affected by the nonparametric components, and then it is more robust, and is easy for application in practice. Some simulations analysis and a real data application are conducted for further illustrating the performance of the proposed method.

Appendix: Proof of theorems

To facilitate the proof, firstly, we list some lemmas.

Lemma 1

Suppose that the regularity conditions C1–C5 hold, and the number of knots $\kappa$ satisfies $\kappa =O(n^{1/(2r+1)})$. Then, we have that

$$\begin{aligned} \hat{b}_k-b_k=O_P(n^{-1/2}). \end{aligned}$$

Proof

Similar to the proof of Theorem 1 in Zou and Yuan (2008), this lemma can be easily derived. $\square$

Lemma 2

Suppose that the regularity conditions C1–C5 hold, and the number of knots $\kappa$ satisfies $\kappa =O(n^{1/(2r+1)})$. Then, we have that

$$\begin{aligned} \frac{1}{\sqrt{n-L}}\sum _{i=1}^{n-L} \hat{\eta }_{i}(\beta ){\mathop {\longrightarrow }\limits ^\mathcal{L}}N(0,\varLambda ), \end{aligned}$$

(14)

where $\varLambda =G\omega ^{T}\varOmega \omega =\sum _{k=1}^K \sum _{l=1}^K\omega _{k}\omega _{l}(\min (\tau _{k},\tau _{l})-\tau _{k}\tau _{l})E(X^{*}X^{*T})$.

Proof

Let $B_{ik}=I(\varepsilon _{i}^{*} \le b_k)-I(\varepsilon _{i}^{*} \le \hat{b}_k)$, then it is easy to show that

$$\begin{aligned} \frac{1}{\sqrt{n-L}}\sum _{i=1}^{n-L} \hat{\eta }_{i}(\beta )= & {} \frac{1}{\sqrt{n-L}}\sum _{i=1}^{n-L} \sum _{k=1}^{K}\omega _{k}X_{i}^{*}(\tau _k-I(\varepsilon _i \le b_k))\nonumber \\&+\frac{1}{\sqrt{n-L}}\sum _{i=1}^{n-L} \sum _{k=1}^K B_{ik}\omega _{k}X_{i}^{*} \nonumber \\\equiv & {} \frac{1}{\sqrt{n-L}}\sum _{i=1}^{n-L} A_{i1}+\frac{1}{\sqrt{n-L}}\sum _{i=1}^{n-L} A_{i2}. \end{aligned}$$

(15)

Note that

$$\begin{aligned} E\{A_{i1}\}=E\left\{ \sum _{k=1}^K \omega _{k}X_{i}^{*}(\tau _k-I(\varepsilon _{i}^{*} \le b_k))\right\} =0, \end{aligned}$$

then by some calculations, we can get

$$\begin{aligned} \begin{array}{lll} Var\{A_{i1}\}&{}=&{}\displaystyle Var\left\{ \sum _{k=1}^K\omega _{k}X_{i}^{*}(\tau _k-I(\varepsilon _{i}^{*} \le b_k))\right\} \\ &{}=&{}\displaystyle E\left\{ \left( \sum _{k=1}^K\omega _{k}X_{i}^{*}(\tau _k-I(\varepsilon _{i}^{*} \le b_k))\right) \left( \sum _{k=1}^K\omega _{k}X_{i}^{*}(\tau _k-I(\varepsilon _{i}^{*} \le b_k))\right) ^{T}\right\} \\ &{}=&{}\displaystyle E\left\{ \sum _{k=1}^K\sum _{l=1}^K \omega _{k}\omega _{l}(\tau _k-I(\varepsilon _{i}^{*} \le b_k))(\tau _{l}-I(\varepsilon _{i}^{*} \le b_{l}))X_{i}^{*}X_{i}^{*T}\right\} . \end{array} \end{aligned}$$

In addition, note that $E\{I(\varepsilon _{i}^{*} \le b_k)\}=\tau _k$, $E\{I(\varepsilon _{i}^{*} \le b_{l})\}=\tau _{l}$ and $E\{I(\varepsilon _{i}^{*} \le b_k)I(\varepsilon _{i}^{*} \le b_{l})\}=\min \{\tau _k, \tau _{l}\}$, we have

$$\begin{aligned} \begin{array}{lll} \displaystyle E\{(\tau _k-I(\varepsilon _{i}^{*} \le b_k))(\tau _{l}-I(\varepsilon _{i}^{*} \le b_{l}))\}\\ ~~~\displaystyle =E\{\tau _k\tau _{l}-\tau _kI(\varepsilon _{i}^{*} \le b_{l})-I(\varepsilon _{i}^{*} \le b_k)\tau _{l}+I(\varepsilon _{i}^{*} \le b_k)I(\varepsilon _{i}^{*} \le b_{l})\}\\ ~~~\displaystyle =\min \{\tau _k, \tau _{l}\}-\tau _k\tau _{l}. \end{array} \end{aligned}$$

Hence, we get that

$$\begin{aligned} Var\{A_{i1}\}=\sum _{k=1}^{K}\sum _{l=1}^{K}\omega _{k}\omega _{l}(\min (\tau _{k},\tau _{l})-\tau _{k}\tau _{l})E(X^{*}X^{*T}). \end{aligned}$$

Then, using central limit theorem, we have

$$\begin{aligned} \frac{1}{\sqrt{n-L}}\sum _{i=1}^{n-L} A_{i1}{\mathop {\longrightarrow }\limits ^\mathcal{L}}N(0,\varLambda ). \end{aligned}$$

(16)

Next, we prove $\frac{1}{\sqrt{n-L}}\sum _{i=1}^{n-L} A_{i2}=o_{p}(1)$. Let $\sum _{k=1}^{K}\omega _{k}B_{ik}=\delta _{i}$, $A_{i2,j}$ be the jth component of $A_{i2}$ and $X_{ij}^{*}$ be the jth component of $X_{i}^{*}$. Then by Lemma 2 in Zhao and Xue (2010), we have that

$$\begin{aligned} \max _{ s}\left| \sum _{i=1}^{s}X_{ij}^{*}\right| =O_{p}(\sqrt{n}\log n). \end{aligned}$$

(17)

In addition, by Lemma 1 we have that

$$\begin{aligned} B_{ik}=I(\varepsilon _i^{*} \le b_k)-I(\varepsilon _i^{*} \le \hat{b}_k)=O_{p}(n^{-1/2}). \end{aligned}$$

(18)

Then combing (17) and (18), and using the Abel inequality, it follows that

$$\begin{aligned} \begin{array}{lll} \displaystyle \left| \frac{1}{\sqrt{n-L}}\sum _{i=1}^{n-L} A_{i2,j}\right| &{}=&{}\displaystyle \frac{1}{\sqrt{n-L}}\left| \sum _{i=1}^{n-L}\delta _{i}X_{ij}^{*}\right| \\ &{}\le &{} \displaystyle \frac{1}{\sqrt{n-L}}\max _{ i}|\delta _{i}|\max _{ s}\left| \sum _{i=1}^{s}X_{ij}^{*}\right| =o_{p}(1). \end{array} \end{aligned}$$

Thus, we have

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{i=1}^{n} A_{i2}=o_{p}(1). \end{aligned}$$

(19)

Then, the proof of this lemma is completed by invoking (15), (16) and (19). $\square$

Lemma 3

Suppose that the regularity conditions C1–C5 hold, and the number of knots $\kappa$ satisfies $\kappa =O(n^{1/(2r+1)})$. Then, we have that

$$\begin{aligned} \frac{1}{ n-L}\sum _{i=1}^{n-L} \hat{\eta }_i(\beta )\hat{\eta }_i^{T}(\beta ){\mathop {\longrightarrow }\limits ^\mathcal{P}}\varLambda , \end{aligned}$$

where “${\mathop {\longrightarrow }\limits ^\mathcal{P}}$” means the convergence in probability.

Proof

We also use the notations in the proof of Lemma 2. Then we have

$$\begin{aligned} \frac{1}{ n-L}\sum _{i=1}^{n-L} \hat{\eta }_i(\beta )\hat{\eta }_i^{T}(\beta )= & {} \frac{1}{ n-L}\sum _{i=1}^{n-L} A_{i1}A_{i1}^{T}+\frac{1}{ n-L}\sum _{i=1}^{n-L} A_{i1}A_{i2}^{T}\\&+\frac{1}{ n-L}\sum _{i=1}^{n-L} A_{i2}A_{i1}^{T}+ \frac{1}{ n-L}\sum _{i=1}^{n-L} A_{i2}A_{i2}^{T}\\\equiv & {} B_{1}+B_{2}+B_{3}+B_{4}. \end{aligned}$$

By the law of large numbers, we can derive that $B_{1}{\mathop {\longrightarrow }\limits ^\mathcal{P}}\varLambda$. We now show $B_{2}{\mathop {\longrightarrow }\limits ^\mathcal{P}}0$. Let $B_{2,rs}$ be the (r, s) component of $B_{2}$, $A_{ij,r}$ be the rth component of $A_{ij}$, $j=1,2$. Then we use the Cauchy–Schwarz inequality to get

$$\begin{aligned} |B_{2,rs}|\le \left( \frac{1}{ n-L}\sum _{i=1}^{n-L}A_{i1,r}^{2}\right) ^{1/2}\left( \frac{1}{ n-L}\sum _{i=1}^{n-L}A_{i2,s}^{2}\right) ^{1/2}. \end{aligned}$$

From the proof of Lemma 2, we can see $(n-L)^{-1}\sum _{i=1}^{n-L}A_{i1,r}^{2}=O_{p}(1)$ and $(n-L)^{-1}\sum _{i=1}^{n-L}A_{i2,s}^{2}=o_{p}(1)$. Hence $B_{2}{\mathop {\longrightarrow }\limits ^\mathcal{P}}0$. Using the similar argument, we can prove that $B_{v}{\mathop {\longrightarrow }\limits ^\mathcal{P}}0$, $v=3,4$. This completes the proof of this lemma. $\square$

Proof of Theorem 1

By the definition of $\hat{\eta }_{i}(\beta )$, and using the arguments similar to Owen (1990), we can obtain

$$\begin{aligned} \max _{ i}\Vert \hat{\eta }_{i}(\beta )\Vert =o_{p}((n-L)^{1/2}), \end{aligned}$$

(20)

and

$$\begin{aligned} \Vert \lambda \Vert =O_{p}((n-L)^{-1/2}). \end{aligned}$$

(21)

Combing (20) and (21), and using the same arguments as in the proof of Theorem 4 in Xue and Zhu (2007), we can get

$$\begin{aligned} R(\beta )=\left\{ \frac{1}{\sqrt{n-L}}\sum _{i=1}^{n-L}\hat{\eta }_{i}(\beta )\right\} ^{T}\hat{\varLambda }^{-1} \left\{ \frac{1}{\sqrt{n-L}}\sum _{i=1}^{n-L}\hat{\eta }_{i}(\beta )\right\} +o_{p}(1), \end{aligned}$$

where $\hat{\varLambda }=(n-L)^{-1}\sum _{i=1}^{n-L}\hat{\eta }_{i}(\beta )\hat{\eta }_{i}^{T}(\beta )$. This together with Lemmas 2 and 3 proves Theorem 1. $\square$

Proof of Theorem 2

Note that $\hat{\beta }-\beta =O_{p}(n^{-1/2})$ and $\hat{b}_{k}-b_{k}=O_{p}(n^{-1/2})$, $k=1,\ldots ,K$, then by the Taylor expansion, we have that

$$\begin{aligned} K\left( \frac{Y_{i}^{*}-X_{i}^{*T}\hat{\beta }-\hat{b}_{k}}{h}\right)= & {} K\left( \frac{Y_{i}^{*}-X_{i}^{*T}\beta -b_{k}}{h}+ \frac{X_{i}^{*T}(\beta -\hat{\beta })+(b_{k}-\hat{b}_{k})}{h}\right) \nonumber \\= & {} K\left( \frac{Y_{i}^{*}-X_{i}^{*T}\beta -b_{k}}{h}\right) +O_{p}\left( \frac{(\beta -\hat{\beta })+(b_{k}-\hat{b}_{k})}{h}\right) \nonumber \\= & {} K\left( \frac{\varepsilon _{i}^{*}-b_{k}}{h}\right) +o_{p}(1). \end{aligned}$$

(22)

In addition, by some calculations, we have that

$$\begin{aligned} \frac{1}{h}E\left\{ K\left( \frac{\varepsilon _{i}^{*}-b_{k}}{h}\right) \right\}= & {} \frac{1}{h}\int K\left( \frac{t-b_{k}}{h}\right) f(t)dt\nonumber \\= & {} \int K(w)f(hw+b_{k})dw\nonumber \\= & {} f(b_{k})+O_{p}(h). \end{aligned}$$

(23)

Invoking (22) and (23), we obtain $E(\hat{\varPi }_{k})=f(b_{k})\varGamma +o_{p}(1)$. Hence, by the law of large number, we derive that $\hat{\varPi }_{k}{\mathop {\longrightarrow }\limits ^\mathcal{P}}f(b_{k})\varGamma$. Then, from the definition of $\hat{\varPi }$, we have

$$\begin{aligned} \hat{\varPi }=\sum _{k=1}^{K}\omega _{k}\hat{\varPi }_{k}{\mathop {\longrightarrow }\limits ^\mathcal{P}}\sum _{k=1}^{K}\omega _{k} f(b_{k})\varGamma =\varPi . \end{aligned}$$

This completes the proof of Theorem 2. $\square$