Testing linearity in partial functional linear quantile regression model based on regression rank scores

Abstract

This paper investigates the hypothesis test of the parametric component in partial functional linear quantile regression model in which the dependent variable is related to both a vector of finite length and a function-valued random variable as predictor variables. A quantile rank score test based on functional principal component analysis is developed. Under mild conditions, we establish the consistency of the proposed test statistic, and show that the proposed test can detect Pitman local alternatives converging to the null hypothesis at the usual parametric rate. A simulation study shows that the proposed test procedure has good size and power with finite sample sizes. Finally, an illustrative example is given through fitting the Berkeley growth data and testing the effect of gender on the height of kids.

Appendix 1

In order to provide the proofs of the theorems, we first define some notations and give some preliminary results.

Write ${\hat{V}}_k(g)=\sum _{j=1}^m\frac{\langle {\hat{C}}_{z_kX},{\hat{v}}_j\rangle \langle {\hat{v}}_j,g\rangle }{{\hat{\lambda }}_j}$, $V_k(g)=\sum _{j=1}^{\infty }\frac{\langle C_{z_kX},v_j\rangle \langle v_j,g\rangle }{\lambda _j}$ for $g\in L^2[0,1]$, $\hat{\varvec{\Delta }}={\hat{C}}_{\varvec{z}}-\{{\hat{V}}_k({\hat{C}}_{z_{l}X})\}_{k,l=1,\ldots ,p}$. Then it is easy to show that $\varvec{\Delta }$ defined in condition C6 can be expressed as $\varvec{\Delta }=C_{\varvec{z}}-\{{V_k(C_{z_{l}X})}\}_{k,l=1,\ldots ,p}$, and that $\hat{\varvec{\Delta }}=\frac{1}{n}\varvec{Z}^T(\varvec{I}-\varvec{P})\varvec{Z}=\frac{1}{n}{\varvec{Z}^*}^T{\varvec{Z}^*}$ if ${\hat{\lambda }}_1>\cdots>{\hat{\lambda }}_n>0$ holds. Let $\varvec{S}^*_n=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\varvec{z}_i^*\psi _\tau ({\epsilon }_i)$, $R_i=\int _{0}^{1}\gamma _0(t)X_i(t)dt-\varvec{U}_i^T\varvec{\gamma }_0$, $\delta =n^{-\frac{2b-1}{2(a+2b)}}$.

The following Lemma 1, which is from He and Shao (2000, Lemma 3.2), is presented here for easy reference.

Lemma 1

Let $\{v_i(t), t\in R^q\},\;1\le i\le n$, be independent $R^d$-valued random variables with $E\{v_i(t)\}=0$ for all t. Assume that there exist $r_1>0$ and $r_2>0$ such that for every $s\in R^q$, $0<\delta ^*\le 1$, $1\le i\le n$

$$\begin{aligned} E\sup _{\Vert t-s\Vert \le \delta ^*}\bigg \Vert v_i(t)-v_i(s)\bigg \Vert \le n^{r_1}{\delta ^*}^{r_2}. \end{aligned}$$

(13)

Let

$$\begin{aligned} A_n(t,s)=\bigg (\sum _{i=1}^{n}E\big \Vert v_i(t)-v_i(s)\big \Vert ^2\bigg )^{1/2} \end{aligned}$$

and

$$\begin{aligned} B_n(t,s)=\bigg (\sum _{i=1}^{n}\big \Vert v_i(t)-v_i(s)\big \Vert ^2\bigg )^{1/2}. \end{aligned}$$

Then

$$\begin{aligned} \sup _{\Vert t\Vert \le n^{r_3},\Vert s\Vert \le n^{r_3}}\frac{\Big \Vert \sum _{i=1}^{n}\big (v_i(t)-v_i(s)\big )\Big \Vert }{n^{-2}+A_n(t,s)+B_n(t,s)}=O_p\Big (\big (q\log (n+q)\big )^{1/2}\Big ). \end{aligned}$$

for every $r_3\ge 0$.

Lemma 2

Let $u_i(\varvec{\gamma },\varvec{\gamma }_0)=\psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma })\varvec{z}_i^*-\psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma }_0)\varvec{z}_i^* -E\psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma })\varvec{z}_i^*+E\psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma }_0)\varvec{z}_i^*$. Under the regularity conditions of Theorem 1, for any $L>0$, it has

$$\begin{aligned} \sup _{\Vert \varvec{\gamma }-\varvec{\gamma }_0\Vert \le L\delta }\frac{1}{n}\bigg \Vert \sum _{i=1}^{n}u_i(\varvec{\gamma },\varvec{\gamma }_0)\bigg \Vert =o_p(n^{-1/2}). \end{aligned}$$

(14)

Proof

We prove this lemma using Lemma 1. First, we verify the condition in (13), that is, there exist $\nu$ and $r>0$ such that

$$\begin{aligned} E\bigg [\sup _{\Vert \varvec{\gamma }-\varvec{\gamma }_0\Vert \le L\delta } \Vert u_i(\varvec{\gamma },\varvec{\gamma }_0)\Vert \bigg ]\le Kn^{\nu }\delta ^{r}. \end{aligned}$$

(15)

Noting that $I(\epsilon \le \cdot )$ and $F(\cdot )$ are monotone increasing functions, $\psi _\tau (u)=\tau - I(u<0)$ is a step function with a jump at $u=0$, $\Vert \varvec{z}_i^*\Vert =O_p(1)$ and $\Vert \varvec{U}_i\Vert =O_p(1)$, a simple calculation yields

$$\begin{aligned} \begin{aligned}&E\bigg [\sup _{\Vert \varvec{\gamma }-\varvec{\gamma }_0\Vert \le L\delta } \Vert u_i(\varvec{\gamma },\varvec{\gamma }_0)\Vert ^2 \bigg ]\\&\le E\bigg [\sup _{\Vert \varvec{\gamma }-\varvec{\gamma }_0\Vert \le L\delta }\Vert \varvec{z}_i^*\Vert ^2\Big |I(\epsilon _i\le \varvec{U}_i^T(\varvec{\gamma } -\varvec{\gamma }_0)-R_i)-I(\epsilon _i\le -R_i)\\&\quad -F\big (\varvec{U}_i^T(\varvec{\gamma }-\varvec{\gamma }_0)-R_i\big )+F(-R_i)\Big |^2\bigg ]\\&\le E\bigg [\Vert \varvec{z}_i^*\Vert ^2\Big |I(\epsilon _i\le \Vert \varvec{U}_i\Vert \delta -R_i)-I(\epsilon _i\le -R_i)\\&-F(-\Vert \varvec{U}_i\Vert \delta -R_i)+F(\Vert \varvec{U}_i\Vert \delta -R_i)\Big |^2\bigg ]\\&\le K \delta , \end{aligned} \end{aligned}$$

which implies (15).

Second, it holds that

$$\begin{aligned} \begin{aligned}&\Big \Vert \psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma })\varvec{z}_i^*-\psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma }_0)\varvec{z}_i^*\Big \Vert ^2\\&\quad \le \Vert \varvec{z}_i^*\Vert ^2 \left[ I (-R_i \le \epsilon _i< |\varvec{U}_i^T(\varvec{\gamma }-\varvec{\gamma }_0)|-R_i) + I(-|\varvec{U}_i^T(\varvec{\gamma }-\varvec{\gamma }_0)|-R_i \le \epsilon _i < -R_i)\right] ^2, \end{aligned} \end{aligned}$$

thus, we have

$$\begin{aligned} E\sum _{i=1}^{n}\Big \Vert u_i(\varvec{\gamma },\varvec{\gamma }_0)\Big \Vert ^2\le K E\sum _{i=1}^{n}\Vert \varvec{z}_i^*\Vert ^2\big |\varvec{U}_i^T(\varvec{\gamma }-\varvec{\gamma }_0)\big |, \end{aligned}$$

which implies

$$\begin{aligned} A_n=\sup _{\Vert \varvec{\gamma }-\varvec{\gamma }_0\Vert \le L\delta }\sum _{i=1}^{n}E\Big \Vert u_i(\varvec{\gamma },\varvec{\gamma }_0) \Big \Vert ^2=O(n\delta )=O\left( n^{\frac{2a+2b+1}{(2a+4b)}}\right) . \end{aligned}$$

Moreover, we can get

$$\begin{aligned} B_n=\sup _{\Vert \varvec{\gamma }-\varvec{\gamma }_0\Vert \le L\delta }\sum _{i=1}^{n}\Big \Vert u_i(\varvec{\gamma },\varvec{\gamma }_0) \Big \Vert ^2=O_p\left( n^{\frac{2a+2b+1}{(2a+4b)}}\right) . \end{aligned}$$

Invoking Lemma 1 and Condition C2(b), we have

$$\begin{aligned} \begin{aligned} \sup _{\Vert \varvec{\gamma }-\varvec{\gamma }_0\Vert \le L\delta }\frac{1}{n}\bigg \Vert \sum _{i=1}^{n}u_i(\varvec{\gamma },\varvec{\gamma }_0)\bigg \Vert&=\frac{1}{n}O_p\Big ([m\log (n)]^{1/2}(n^{-2}+A_n^{1/2}+B_n^{1/2})\Big )\\&=O_p\left( n^{-\frac{2a+6b-3}{4a+8b}}\right) (\log (n))^{1/2}=o_p(n^{-1/2}). \end{aligned} \end{aligned}$$

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

Lemma 3

Under the regularity conditions of Theorem 1, for any $L>0$, it has

$$\begin{aligned} \sup _{\Vert \varvec{\gamma }-\varvec{\gamma }_0\Vert \le L\delta }n^{-1/2}\bigg \Vert \sum _{i=1}^{n}E\psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma })\varvec{z}_i^*-E\psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma }_0)\varvec{z}_i^*\bigg \Vert =o(1). \end{aligned}$$

(16)

Proof

Since $E\psi _\tau (\epsilon )=\tau -F(0)=0$, where F is the conditional distribution function of $\epsilon$. Using the Taylor expansion, we have

$$\begin{aligned} \begin{aligned}&E\psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma })\varvec{z}_i^*-E\psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma }_0)\varvec{z}_i^*\\&\quad =f(0)\varvec{z}_i^*\varvec{U}_i^T(\varvec{\gamma }-\varvec{\gamma }_0)+\frac{1}{2}f'(0)\varvec{z}_i^*(\varvec{\gamma }-\varvec{\gamma }_0)^T\varvec{U}_i\varvec{U}_i^T(\varvec{\gamma }-\varvec{\gamma }_0) +o\left( \frac{1}{\sqrt{n}}\right) . \end{aligned} \end{aligned}$$

The remainder is $o\big (1/\sqrt{n}\big )$ because $\Vert \varvec{\gamma }-\varvec{\gamma }_0\Vert \le L\delta$ and $b>\frac{a}{2}+1$. Then, we can obtain that

$$\begin{aligned}\begin{aligned}&\sup _{\Vert \varvec{\gamma }-\varvec{\gamma }_0\Vert \le L\delta }n^{-1/2}\bigg \Vert \sum _{i=1}^{n}E\psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma })\varvec{z}_i^*-E\psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma }_0)\varvec{z}_i^*\bigg \Vert \\&\quad =\sup _{\Vert \varvec{\gamma }-\varvec{\gamma }_0\Vert \le L\delta }n^{-1/2}\bigg \Vert bE{\varvec{Z}^*}^T\varvec{U}(\varvec{\gamma }-\varvec{\gamma }_0)+ c\sum _{i=1}^{n}E\varvec{z}_i^*(\varvec{\gamma }-\varvec{\gamma }_0)^T\varvec{U}_i\varvec{U}_i^T(\varvec{\gamma }-\varvec{\gamma }_0) \bigg \Vert +o(1) \end{aligned} \end{aligned}$$

where $b=f(0)$ and $c=f'(0)$. Note that $\varvec{U}$ and $\varvec{Z}^*$ are orthogonal to each other, that is, $\varvec{U}^T\varvec{Z}^*=0$. Moreover,

$$\begin{aligned} \begin{aligned} \sup _{\Vert \varvec{\gamma }-\varvec{\gamma }_0\Vert \le L\delta }n^{-1/2}\bigg \Vert \sum _{i=1}^{n}E\varvec{z}_i^*(\varvec{\gamma }-\varvec{\gamma }_0)^T\varvec{U}_i\varvec{U}_i^T(\varvec{\gamma }-\varvec{\gamma }_0)\bigg \Vert = O\left( n^{-\frac{2b-a-2}{2(a+2b)}}\right) =o(1). \end{aligned} \end{aligned}$$

Thus completes the proof of Lemma 3. $\square$

Proof of Theorem 1

First, note that

$$\begin{aligned} \begin{aligned} |R_i|^2&\le 2 \bigg |\sum _{j=1}^{m}\langle X_i,{\hat{v}}_j-{v}_j\rangle \gamma _{0j}\Big |^2+2\Big |\sum _{j=m+1}^{ \infty }\langle X_i,v_j\rangle \gamma _{0j}\bigg |^2\\&\equiv 2\text {A}_1+2\text {A}_2. \end{aligned} \end{aligned}$$

For $\text {A}_1$, by condition C1 and the Hölder inequality, it follows

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

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

$$\begin{aligned} E\Big \{\sum _{j=m+1}^{ \infty }\langle X_i,v_j\rangle \beta _{0j}\Big \}=0, \end{aligned}$$

and

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

one has

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

Taking these together, we have

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

(17)

According to Theorem 3.2 of Kato (2012), we can get the fact that $|\hat{\varvec{\gamma }}-\varvec{\gamma }_0|=O_p(\delta )$. Therefore, we can derive from Lemmas 2 and  3 that

$$\begin{aligned} n^{-1/2}\bigg \Vert \sum _{i=1}^{n}\psi _\tau (Y_i-\varvec{U}_i^T\varvec{{\hat{\gamma }}})\varvec{z}_i^*-\psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma }_0)\varvec{z}_i^*\bigg \Vert =o_p(1). \end{aligned}$$

(18)

Moreover, invoking Eq. (17) and following the similar arguments used for Eq. (18), a simple calculation yields

$$\begin{aligned} \begin{aligned} n^{-1/2}\bigg \Vert \sum _{i=1}^{n}\psi _\tau (Y_i-\varvec{U}_i^T\varvec{\gamma }_0)\varvec{z}_i^*-\psi _\tau (\epsilon _i)\varvec{z}_i^*\bigg \Vert =o_p(1). \end{aligned} \end{aligned}$$

(19)

Combining (18) and (19), we have

$$\begin{aligned} n^{-1/2}\bigg \Vert \sum _{i=1}^{n}\psi _\tau (Y_i-\varvec{U}_i^T\varvec{{\hat{\gamma }}})\varvec{z}_i^*-\psi _\tau (\epsilon _i)\varvec{z}_i^*\bigg \Vert =o_p(1). \end{aligned}$$

(20)

In addition, it’s easy to see that

$$\begin{aligned} \text {Cov}(\varvec{S}_n^*)=\frac{1}{n}\sum _{i=1}^{n}\text {Cov}\bigg (\varvec{z}_i^*\psi _\tau (\epsilon _i)\bigg )=\frac{1}{{n}}\sum _{i=1}^{n}\varvec{z}_i^*{\varvec{z}_i^*}^T\tau (1-\tau )=\varvec{V}_n. \end{aligned}$$

By the central limit theorem, we have

$$\begin{aligned} \varvec{V}_n^{-1/2}\varvec{S}_n^*\overset{d}{\longrightarrow }N(0,\varvec{I}_p). \end{aligned}$$

(21)

Then, Theorem 1 follows immediately from (20) and (21). $\square$

Lemma 4

Define

$$\begin{aligned} {\varvec{{\bar{S}}}}_n=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\varvec{z}_i^*\psi _\tau \Big (Y_i-\varvec{U}_i^T\varvec{{\hat{\gamma }}}\Big ), \end{aligned}$$

where $\hat{\varvec{\gamma }}=\text {argmin}_{{\varvec{\gamma }}\in R^m}\sum _{i=1}^{n}\rho _\tau (Y_i-\varvec{U}_i\varvec{\gamma }-\varvec{z}_i^T\frac{\varvec{\beta }_0}{\sqrt{n}})$. Then under ${\widetilde{H}}_A$ and Conditions C1–C7, we have

$$\begin{aligned} {\varvec{{\bar{S}}}}_n={\varvec{{S}}}_n^*+\frac{1}{n}f(0)\sum _{i=1}^{n}{\varvec{z}_i^*}\varvec{z}_i^T\varvec{\beta }_0+o_p(1). \end{aligned}$$

(22)

Proof

Since $E\psi _\tau (\epsilon )=\tau -F(0)$ and

$$\begin{aligned} \begin{aligned}&E\big [\varvec{z}_i^*\big (\psi _\tau (Y_i-\varvec{U}_i^T\varvec{{\hat{\gamma }}})-\psi _\tau (\epsilon _i)\big )\big ]\\&\quad =E\Big [\varvec{z}_i^*\big ( \psi _\tau \big (R_i+\varvec{U}_i^T(\varvec{\gamma }_0-\varvec{{\hat{\gamma }}})+\varvec{z}_i^T\frac{\varvec{\beta }_0}{\sqrt{n}}+\epsilon _i\big )-\psi _\tau (\epsilon _i)\big )\Big ]. \end{aligned} \end{aligned}$$

Then, invoking (17) and the orthogonality between $\varvec{U}$ and $\varvec{Z}^*$, we have

$$\begin{aligned}\begin{aligned} E\big ({\varvec{{\bar{S}}}}_n-{\varvec{{S}}}_n^*\big )&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\varvec{z}_i^*\Big [F\Big (R_i+\varvec{U}_i^T(\varvec{\gamma }_0-\varvec{{\hat{\gamma }}})+\varvec{z}_i^T\frac{\varvec{\beta }_0}{\sqrt{n}}\Big )-F(0)\Big ]\\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}f(0)\varvec{z}_i^*\left( R_i+\varvec{U}_i^T(\varvec{\gamma }_0-\varvec{{\hat{\gamma }}})+\varvec{z}_i^T\frac{\varvec{\beta }_0}{\sqrt{n}}\right) +o_p(1)\\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}f(0)\varvec{z}_i^*\varvec{z}_i^T\frac{\varvec{\beta }_0}{\sqrt{n}}+o_p(1). \end{aligned} \end{aligned}$$

This completes the proof of Lemma 4. $\square$

Proof of Theorem 2

According to Lemma 1 in Yu et al. (2016), we have

$$\begin{aligned} \hat{\varvec{\Delta }}\overset{p}{\longrightarrow }\varvec{\Delta }, \quad {\varvec{V}_n}\overset{p}{\longrightarrow }\tau (1-\tau )\varvec{\Delta }, \quad {\text{ as } \ \ n \rightarrow \infty }. \end{aligned}$$

(23)

Note that ${\varvec{{S}}}_n^*$ is the summation of independent entries. It follows from the Lindberg-Feller central limit theorem that

$$\begin{aligned} \varvec{V}_n^{-1/2}\varvec{S}_n^*\overset{d}{\longrightarrow }N(0,\varvec{I}_p). \end{aligned}$$

(24)

The combination of (23), (24), Lemma 4 and the definition of non-central Chi-squared distribution allow us to finish the proof of Theorem 2. $\square$