Rank method for partial functional linear regression models

Abstract

In this paper, we consider rank estimation for partial functional linear regression models based on functional principal component analysis. The proposed rank-based method is robust to outliers in the errors and highly efficient under a wide range of error distributions. The asymptotic properties of the resulting estimators are established under some regularity conditions. A simulation study conducted to investigate the finite sample performance of the proposed estimators shows that the proposed rank method performs well. Furthermore, the proposed methodology is illustrated by means of an analysis of the Berkeley growth data.

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Acknowledgements

Cao’s work is partly supported by the National Natural Science Foundation of China (No. 11701020). Xie’s work is supported by the National Natural Science Foundation of China (No. 11771032, 11571340, 11971045), Beijing Natural Science Foundation under Grant No. 1202001, the Science and Technology Project of Beijing Municipal Education Commission (KM201710005032, KM201910005015), and the International Research Cooperation Seed Fund of Beijing University of Technology (No. 006000514118553).

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Correspondence to Tianfa Xie.

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Appendix

Appendix

In order to prove the theorem, we make a linear approximation to \(|\epsilon _i-\epsilon _j-t|\) by \(D_{ij}=2I_{\{\epsilon _i>\epsilon _j\}}-1\). One intuitive interpretation of \(D_{ij}\) is that \(D_{ij}\) can be treated as the first derivative of \(|\epsilon _i-\epsilon _j-t|\) at \(t=0\) (Pollard 1991). Let \(R_i=\sum _{j=1}^m\gamma _{0j}{\hat{\xi }}_{ij}-\int _{0}^{1}\beta _0(t)X_i(t)dt=\varvec{U}_i^T\varvec{\gamma }_0-\int _{0}^{1}\beta _0(t)X_i(t)dt\). By Condition C2, one has \(E(D_{ij})=0, \text {Var}(D_{ij})=1\). Define

$$\begin{aligned} \begin{aligned} R_{ij}(\varvec{u})&= \left[ \left| \epsilon _i-\epsilon _j-\frac{1}{\sqrt{n}}(\varvec{z}_i-\varvec{z}_j)^T{\varvec{u}_1}-\frac{1}{\sqrt{n\lambda _m}}(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2-R_i+R_j\right| \right. \\&\quad -\,\left. \left| \epsilon _i-\epsilon _j\right| \right] -D_{ij} \left[ \frac{1}{\sqrt{n}}{(\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_1}+\frac{1}{\sqrt{n\lambda _m}}(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2 \right] , \\ \varvec{W}_{n1}&=\frac{1}{n\sqrt{n}}\sum _{i=1}^n\sum _{j=1}^nD_{ij}(\varvec{z}_i-\varvec{z}_j), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \varvec{W}_{n2}=\frac{1}{n\sqrt{n}}\sum _{i=1}^n\sum _{j=1}^nD_{ij}(\varvec{U}_{i}-\varvec{U}_{j}), \end{aligned}$$

where \(\varvec{u} =(\varvec{u}_1^T,\varvec{u}_2^T)^T\) and \(\varvec{W}_{n}=(\varvec{W}_{n1}^T,\varvec{W}_{n2}^T)^T.\)

Lemma 1

Denote

$$\begin{aligned} \begin{aligned} G_{n}(\varvec{u})&=\frac{1}{n}\sum \limits _{i=1}^n\sum \limits _{j=1}^n \left[ |\epsilon _i-\epsilon _j -\frac{1}{\sqrt{n}}(\varvec{z}_i-\varvec{z}_j)^T{ \varvec{u}_1}-\frac{1}{\sqrt{n\lambda _m}}(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2\right. \\&\quad \left. -\,R_i+R_j|-|\epsilon _i-\epsilon _j| \right] . \end{aligned} \end{aligned}$$

Under the Conditions C1 and C2, it has

$$\begin{aligned} G_{n}(\varvec{u})=A_n(\varvec{u})+ \varvec{W}_{n}^T\varvec{u}+O_p\left( \lambda _m^{-1/2}\right) +O_{p}\left( n^{\frac{1}{a+2b}}\right) , \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} A_{n}(\varvec{u})&=\frac{\tau }{n^2}\varvec{u}^T_1\left( \sum _{i=1}^n\sum \limits _{j=1}^n (\varvec{z}_i-\varvec{z}_j)(\varvec{z}_i-\varvec{z}_j)^T\right) \varvec{u}_1\\&\quad +\,\frac{\tau }{\lambda _mn^2}\varvec{u}^T_2\left( \sum \limits _{i=1}^n\sum \limits _{j=1}^n (\varvec{U}_{i}-\varvec{U}_{j})(\varvec{U}_{i}-\varvec{U}_{j})^T\right) \varvec{u}_2\\&\quad +\,\frac{\tau }{\sqrt{\lambda }_mn^2}\varvec{u}^T_1\left( \sum \limits _{i=1}^n\sum \limits _{j=1}^n (\varvec{z}_{i}-\varvec{z}_{j})(\varvec{U}_{i}-\varvec{U}_{j})^T\right) \varvec{u}_2, \\ \varvec{W}_{n}^T\varvec{u}&=\varvec{W}_{n1}^T\varvec{u}_1+\varvec{W}_{n2}^T\varvec{u}_2, \end{aligned} \end{aligned}$$

where \(\varvec{W}_{n} =( \varvec{W}_{n1}^T,\varvec{W}_{n2}^T)^T,\)\(\tau =\int f^2(t)dt\) and f(t) is the density function of the random error.

Proof of Lemma 1

Let \(M(t)=E(|\epsilon _i-\epsilon _j-t|-|\epsilon _i-\epsilon _j|)\). With Condition C2, it is easy to show that M(t) has a unique minimizer at zero, and its Taylor expansion at origin has the form \(M(t)=\tau t^2+o(t^2)\). Denote \({\mathcal {F}}_n=\{(\varvec{z}_1,\varvec{X}_1),\ldots ,(\varvec{z}_n,\varvec{X}_n)\}.\) Hence, for larger n, we have

$$\begin{aligned} E(G_{n}(\varvec{u})|{\mathcal {F}}_n)=A_n(\varvec{u})+B_n(\varvec{u})+o_p(A_n(\varvec{u}))+o_p(B_n(\varvec{u})), \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} B_{n}(\varvec{u})&=\frac{\tau }{n}\sum _{i=1}^n\sum \limits _{j=1}^n (R_i-R_j)^2+\frac{\tau }{\sqrt{\lambda _m}n^{3/2}}\varvec{u}^T_2 \left( \sum \limits _{i=1}^n\sum \limits _{j=1}^n (\varvec{U}_{i}-\varvec{U}_{j})(R_i-R_j) \right) \\&\quad +\frac{\tau }{n^{3/2}}\varvec{u}^T_1 \left( \sum \limits _{i=1}^n\sum \limits _{j=1}^n (\varvec{z}_{i}-\varvec{z}_{j})(R_i-R_j) \right) \\&\triangleq B_{n1}+B_{n2}(\varvec{u})+B_{n3}(\varvec{u}) \end{aligned} \end{aligned}$$

Via Condition C1 and Cauchy–Schwarz inequality, it has

$$\begin{aligned} A_n(\varvec{u})=O_p(1). \end{aligned}$$

Note that

$$\begin{aligned} \begin{aligned} |R_i|^2&= \left| \sum _{j=1}^m\gamma _{0j}{\hat{\xi }}_{ij}-\int _{0}^{1}\beta _0(t)X_i(t)dt \right| ^2\\&\le 2 \left| \sum _{j=1}^{m}\langle X_i,{\hat{v}}_j-{v}_j\rangle \gamma _{0j} \right| ^2+2 \left| \sum _{j=m+1}^{ \infty }\langle X_i,v_j\rangle \gamma _{0j} \right| ^2\\&\triangleq 2\text {A}_1+2\text {A}_2. \end{aligned} \end{aligned}$$

For \(\text {A}_1\), by Condition C1 and Hölder inequality, it is obtained

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

As for \(\text {A}_2\), since

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

and

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

one has

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

Combining above discussions, it follows

$$\begin{aligned} |R_i|^2=O_p\left( n^{-\frac{a+2b-1}{a+2b}}\right) . \end{aligned}$$
(5)

Moreover, it has

$$\begin{aligned} \begin{aligned} B_{n1}&=\frac{\tau }{n}\sum _{i=1}^{n}\sum \limits _{j=1}^{n} (R_i-R_j)^2\\&=nO_{p}\left( n^{-\frac{2b+a-1}{a+2b}}\right) \\&=O_{p}\left( n^{\frac{1}{a+2b}}\right) . \end{aligned} \end{aligned}$$

Taking advantage of the stochastic order of \(B_{n1}\) and \(A_n(\varvec{u})\) and Cauchy–Schwarz inequality, one has

$$\begin{aligned} B_{n2} (\varvec{u}) =B_{n3}(\varvec{u}) =O_{p}\left( n^{\frac{1}{a+2b}}\right) . \end{aligned}$$

Therefore, it follows

$$\begin{aligned} B_{n}=O_{p}\left( n^{\frac{1}{a+2b}}\right) . \end{aligned}$$

Furthermore, we obtain

$$\begin{aligned} E(G_{n}(\varvec{u})|{\mathcal {F}}_n)=A_n(\varvec{u})+O_{p}\left( n^{\frac{1}{a+2b}}\right) . \end{aligned}$$

Thus,

$$\begin{aligned} G_n(\varvec{u})=E[G_n(\varvec{u})|{\mathcal {F}}_n]+\varvec{W}_{n}^T\varvec{u}+\frac{1}{n}\sum \limits _{j=1}^n\sum \limits _{i=1}^n\left[ R_{ij}(\varvec{u}) -E(R_{ij}(\varvec{u}))\right] . \end{aligned}$$

By elementary calculation, one has

$$\begin{aligned} \begin{aligned} |R_{ij}(\varvec{u})|&\le \left| \frac{1}{\sqrt{n}}(\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_1+\frac{1}{\sqrt{n\lambda _m}}(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2+R_i-R_j \right| \\&\quad \times I_{\left\{ \epsilon _i-\epsilon _j\le \left| \frac{1}{\sqrt{n}}{ (\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_1}+\frac{1}{\sqrt{n\lambda _m}}(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2+R_i-R_j \right| \right\} }. \end{aligned} \end{aligned}$$

With cross product terms being cancellated, it follows

$$\begin{aligned} \begin{aligned} E \left( \frac{1}{n}\sum \limits _{i=1}^n\sum \limits _{j=1}^n\left[ R_{ij}(\varvec{u}) -E(R_{ij}(\varvec{u}))\right] \right) ^2&\le \frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^nE\left[ R_{ij}(\varvec{u}) -E(R_{ij}(\varvec{u}))\right] ^2\\&\le I_1(\varvec{u})+I_2(\varvec{u})+I_3(\varvec{u}), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} I_1(\varvec{u})&=\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n\varvec{u}_1^T(\varvec{z}_i-\varvec{z}_j)(\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_1\\&\quad \times EI_{\left\{ \epsilon _i-\epsilon _j\le \left| \frac{1}{\sqrt{n}}\frac{1}{\sqrt{n}}(\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_1-\frac{1}{\sqrt{n\lambda _m}}(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2+R_i-R_j \right| \right\} },\\ I_2(\varvec{u})&=\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n\frac{1}{ {n\lambda _m}}\varvec{u}_2^T(\varvec{U}_{i}-\varvec{U}_{j})(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2\\&\quad \times EI_{\left\{ \epsilon _i-\epsilon _j\le \left| \frac{1}{\sqrt{n}}\frac{1}{\sqrt{n}}(\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_1-\frac{1}{\sqrt{n\lambda _m}}(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2+R_i-R_j \right| \right\} } \end{aligned} \end{aligned}$$

and

$$\begin{aligned} I_3(\varvec{u})=\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n (R_{i }-R_{j })^2 EI_{\left\{ \epsilon _i-\epsilon _j\le \left| \frac{1}{\sqrt{n}}\frac{1}{\sqrt{n}}(\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_1-\frac{1}{\sqrt{n\lambda _m}}(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2+R_i-R_j \right| \right\} }. \end{aligned}$$

By Condition C2 and the weak law of large numbers, one has

$$\begin{aligned}&\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n\varvec{u}_1^T(\varvec{z}_i-\varvec{z}_j)(\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_1=O_p(1), \end{aligned}$$
(6)
$$\begin{aligned}&\max \limits _{i=1,2,\ldots ,n}\left| \frac{1}{\sqrt{n}}(\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_2-\frac{1}{\sqrt{n\lambda _m}}{ (\varvec{U}_{i}-\varvec{U}_{j})^T } \right| =o_p(1), \end{aligned}$$
(7)

and

$$\begin{aligned} || R_i-R_j||^2=o_p(1). \end{aligned}$$
(8)

Therefore, combining (6) and (7) with (8), we can conclude that \(I_1(\varvec{u})=o_p(1)\).

Next, consider the stochastic order of \(I_{2}(\varvec{u})\).

$$\begin{aligned} \begin{aligned} I_2(\varvec{u})&=\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n\frac{1}{ {n\lambda _m}}\varvec{u}_2^T(\varvec{U}_{i}-\varvec{U}_{j})(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2\\&\quad \times EI_{\left\{ \epsilon _i-\epsilon _j\le \left| \frac{1}{\sqrt{n}}\frac{1}{\sqrt{n}}(\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_1-\frac{1}{\sqrt{n\lambda _m}}(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2+R_i-R_j \right| \right\} }\\&=\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n\frac{1}{ {n\lambda _m}}\varvec{u}_2^T\varvec{U}_{i}\varvec{U}_{i}^T\varvec{u}_2EI_{\left\{ \epsilon _i-\epsilon _j\le \left| \frac{1}{\sqrt{n}}\frac{1}{\sqrt{n}}(\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_1-\frac{1}{\sqrt{n\lambda _m}}(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2+R_i-R_j \right| \right\} }\\&\quad +\,\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n\frac{1}{ {n\lambda _m}}\varvec{u}_2^T\varvec{U}_{j}\varvec{U}_{j}^Tu_2EI_{\left\{ \epsilon _i-\epsilon _j\le \left| \frac{1}{\sqrt{n}}\frac{1}{\sqrt{n}}(\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_1-\frac{1}{\sqrt{n\lambda _m}}(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2+R_i-R_j \right| \right\} }\\&\quad -\,2\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n\frac{1}{ {n\lambda _m}}\varvec{u}_2^T\varvec{U}_{i}\varvec{U}_{j}^T\varvec{u}_2EI_{\left\{ \epsilon _i-\epsilon _j\le \left| \frac{1}{\sqrt{n}}\frac{1}{\sqrt{n}}(\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_2-\frac{1}{\sqrt{n\lambda _m}}(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}+R_i-R_j \right| \right\} }\\&\triangleq I_{21}(\varvec{u})+I_{21}(\varvec{u})-2I_{23}(\varvec{u}). \end{aligned} \end{aligned}$$

By Lemma 2 of Yu et al. (2016b) and (8), one has

$$\begin{aligned} \begin{aligned} I_{21}(\varvec{u})&={ \frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n\frac{1}{ {n\lambda _m}}\varvec{u}_2^T\varvec{U}_{i}\varvec{U}_{i}^T\varvec{u}_2EI_{\left\{ \epsilon _i-\epsilon _j\le \left| \frac{1}{\sqrt{n}}\frac{1}{\sqrt{n}}(\varvec{z}_i-\varvec{z}_j)^T\varvec{u}_1-\frac{1}{\sqrt{n\lambda _m}}(\varvec{U}_{i}-\varvec{U}_{j})^T\varvec{u}_2+R_i-R_j \right| \right\} }}\\&=o_p(1) \sum \limits _{i=1}^n \frac{1}{ {n\lambda _m}}\varvec{u}_2^T\varvec{U}_{i}\varvec{U}_{i}^T\varvec{u}_2 \\&= o_p(1) \frac{1}{ { \lambda _m}} \varvec{u}_2^T \varvec{\Lambda }\varvec{u}_2 \\&= O_p( { \lambda _m^{-1}}), \end{aligned} \end{aligned}$$

where \(\varvec{\Lambda }= \text {diag}( \lambda _1,\lambda _2,\ldots ,\lambda _m).\) Similar to \(I_{21}(\varvec{u})\), we have \(I_{22}(\varvec{u})=O_p( { \lambda _m^{-1}}).\) Then, invoking Cauchy–Schwarz inequality, one has \(I_{23}(\varvec{u})=O_p( { \lambda _m^{-1}}).\) Thus, \(I_{2}(\varvec{u})=O_p( { \lambda _m^{-1}}).\) Now, we consider \(I_3(\varvec{u})\). Similar to \(B_n(\varvec{u})\), we can show that \(I_3(\varvec{u})=o_{p}\left( n^{\frac{1}{a+2b}}\right) .\)

Considering the stochastic order of \(I_{1}(\varvec{u})\), \(I_{2}(\varvec{u})\) and \(I_{3}(\varvec{u})\), one has

$$\begin{aligned} G_{n}(\varvec{u})=A_n(\varvec{u})+\varvec{W}_{n}^T\varvec{u}+O_p\left( \lambda _m^{-1/2}\right) +O_{p}\left( n^{\frac{1}{a+2b}}\right) . \end{aligned}$$

Thus completes the proof of Lemma 1. \(\square\)

Proof of Theorem 1

Write \(\varvec{U}_{i}=({\hat{\xi }}_{i1},\ldots ,{\hat{\xi }}_{im})^T, \varvec{\theta }=(\varvec{\alpha }^T,\varvec{\gamma }^T)^T\). Note that

$$\begin{aligned} \hat{\varvec{\theta }}=\arg \min \limits _{\varvec{\theta }}\sum \limits _{i=1}^n\sum \limits _{j=1}^n \frac{1}{n}|Y_i-Y_j-\varvec{z}_i^T\varvec{\alpha }-\varvec{U}_{i}^T\varvec{\gamma } + \varvec{z}_j^T\varvec{\alpha }+\varvec{U}_{j}^T\varvec{\gamma } |, \end{aligned}$$

so \(\hat{\varvec{\theta }}\) minimizes

$$\begin{aligned} \begin{aligned} Q_{n}(\varvec{\theta })&=\sum \limits _{i=1}^n\sum \limits _{j=1}^n \frac{1}{n}|Y_i-Y_j-\varvec{z}_i^T\varvec{\alpha }-\varvec{U}_{i}^T\varvec{\gamma } + \varvec{z}_j^T\varvec{\alpha }+\varvec{U}_{j}^T\varvec{\gamma } |-\frac{1}{n}\sum \limits _{i=1}^n\sum \limits _{j=1}^n|\epsilon _i-\epsilon _j| \\&=\frac{1}{n}\sum \limits _{i=1}^n\sum \limits _{j=1}^n \left| \epsilon _i-\epsilon _j-(\varvec{z}_i-\varvec{z}_j)^T(\varvec{\alpha }-\varvec{\alpha }_0) -(\varvec{U}_{i}-\varvec{U}_{j})^T(\varvec{\gamma }-\varvec{\gamma }_{0})-R_i-R_j \right| \\&\quad -\,\frac{1}{n}\sum \limits _{i=1}^n\sum \limits _{j=1}^n|\epsilon _i-\epsilon _j|{,} \end{aligned} \end{aligned}$$

where \(R_{i}=\sum _{j=1}^{m}{\hat{\xi }}_{ij}\gamma _{0j}-\int _{0}^{1}\beta _0(t)x_i(t)dt.\)

By Lemma 1, with simple calculation, it yields

$$\begin{aligned} Q_{n}(\varvec{\theta })=V_{n}(\varvec{\theta })+\sqrt{ n}\varvec{W}_{n1}^T(\varvec{\alpha }-\varvec{\alpha }_0)+\sqrt{ n}\varvec{W}_{n2}^T(\varvec{\gamma }-\varvec{\gamma }_0)+O_p\left( \lambda _m^{-1/2}\right) +O_{p}\left( n^{\frac{1}{a+2b}}\right) , \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} V_{n}(\varvec{\theta })&=\frac{\tau }{n}(\varvec{\alpha }-\varvec{\alpha }_0)^T \left( \sum _{i=1}^n\sum \limits _{j=1}^n (\varvec{z}_i-\varvec{z}_j)(\varvec{z}_i-\varvec{z}_j)^T\right) (\varvec{\alpha }-\varvec{\alpha }_0) \\&\quad +\,\frac{\tau }{\lambda _mn}(\varvec{\gamma }-\varvec{\gamma }_{0})^T\left( \sum \limits _{i=1}^n\sum \limits _{j=1}^n (\varvec{U}_{i}-\varvec{U}_{j})(\varvec{U}_{i}-\varvec{U}_{j})^T\right) (\varvec{\gamma }-\varvec{\gamma }_{0})\\&\quad +\,\frac{\tau }{\sqrt{\lambda }_mn}(\varvec{\alpha }-\varvec{\alpha }_0)^T\left( \sum \limits _{i=1}^n\sum \limits _{j=1}^n (\varvec{z}_{i}-\varvec{z}_{j})(\varvec{U}_{i}-\varvec{U}_{j})^T\right) (\varvec{\gamma }-\varvec{\gamma }_{0}){.} \end{aligned} \end{aligned}$$

By the definition of \(\hat{\varvec{\theta }}\) and Lemma 1, we can conclude that

$$\begin{aligned} \frac{\partial V_{n}(\hat{\varvec{\theta }})}{\partial \varvec{\theta }}+\sqrt{n} \varvec{W}_n=O_p\left( \lambda _m^{-1/2}\right) +O_{p}\left( n^{\frac{1}{a+2b}}\right) . \end{aligned}$$
(9)

And with simple calculation, it yields

$$\begin{aligned} \begin{aligned}&\frac{2\tau }{\lambda _mn} \left( \sum \limits _{i=1}^n\sum \limits _{j=1}^n (\varvec{U}_{i}-\varvec{U}_{j})(\varvec{U}_{i}-\varvec{U}_{j})^T \right) (\hat{\varvec{\gamma }}-\varvec{\gamma }_{0})\\&\qquad + \frac{\tau }{\sqrt{\lambda }_mn}\left( \sum \limits _{i=1}^n\sum \limits _{j=1}^n (\varvec{z}_{i}-\varvec{z}_{j})(\varvec{U}_{i}-\varvec{U}_{j})^T\right) (\hat{\varvec{\alpha }}-\varvec{\alpha }_0)+\sqrt{n}\varvec{W}_{n2}\\&\quad =O_p\left( \lambda _m^{-1/2}\right) +O_{p}\left( n^{\frac{1}{a+2b}}\right) , \end{aligned} \end{aligned}$$
(10)

and

$$\begin{aligned} \begin{aligned}&\frac{2\tau }{\lambda _mn} \left( \sum \limits _{i=1}^n\sum \limits _{j=1}^n (\varvec{z}_{i}-\varvec{z}_{j})(\varvec{z}_{i}-\varvec{z}_{j})^T \right) (\hat{\varvec{\alpha }}-\varvec{\alpha }_{0})\\&\qquad + \frac{\tau }{\sqrt{\lambda }_mn}\left( \sum \limits _{i=1}^n\sum \limits _{j=1}^n (\varvec{z}_{i}-\varvec{z}_{j})(\varvec{U}_{i}-\varvec{U}_{j})^T\right) (\hat{\varvec{\gamma }}-\varvec{\gamma }_{0})+\sqrt{n}\varvec{W}_{n1}\\&\quad =O_p\left( \lambda _m^{-1/2}\right) +O_{p}\left( n^{\frac{1}{a+2b}}\right) . \end{aligned} \end{aligned}$$
(11)

It can be rewritten as follows

$$\begin{aligned} 4\tau \sqrt{ n}({ \varvec{Z} }^T(\varvec{I}-\varvec{S}_m)\varvec{Z} ) ( \hat{\varvec{\alpha }}-\varvec{\alpha }_0 )=\frac{1}{n\sqrt{n}}\sum _{i=1}^n\sum _{j=1}^nD_{ij}(\tilde{\varvec{Z}}_i-\tilde{\varvec{Z}}_j)+o_p(1), \end{aligned}$$
(12)

where \(\varvec{Z}=[\varvec{z}_1,\ldots ,\varvec{z}_n]^T\), and \(\tilde{\varvec{Z}}_i\) is the ith column component of \((\varvec{I}-\varvec{S}_m)\varvec{Z},\) here \(\varvec{S}_m=\varvec{U}_m(\varvec{U}_m^T\varvec{U}_m)^{-1}\varvec{U}_m^T.\)

Similar to Lemma 2 of Du et al. (2018), we have

$$\begin{aligned} \frac{1}{n\sqrt{n}}\sum _{i=1}^n\sum _{j=1}^nD_{ij}(\tilde{\varvec{Z}}_i-\tilde{\varvec{Z}}_j){\longrightarrow } N \left( 0 ,\frac{4}{3}\varvec{\Sigma }\right) . \end{aligned}$$

Therefore, according to the law of large numbers, we have

$$\begin{aligned} \begin{aligned} \sqrt{n}(\hat{\varvec{\alpha }}-\varvec{\alpha }_0)&= ( 4\tau (\varvec{Z}^T(\varvec{I}-\varvec{S}_m)\varvec{Z}/n ) )^{-1} \frac{1}{n\sqrt{n}}\sum _{i=1}^n\sum _{j=1}^nD_{ij}(\tilde{\varvec{Z}}_i-\tilde{\varvec{Z}}_j)+o_p(1)\\&=( 4\tau \varvec{\Sigma })^{-1} \frac{1}{n\sqrt{n}}\sum _{i=1}^n\sum _{j=1}^nD_{ij}(\tilde{\varvec{Z}}_i-\tilde{\varvec{Z}}_j)+o_p(1)\\&\rightarrow N \left( 0 ,\frac{1}{12\tau ^2}\varvec{\Sigma }^{-1}\right) . \end{aligned} \end{aligned}$$

Furthermore, we have \(\Vert \hat{\varvec{\alpha }}-\varvec{\alpha }_0\Vert =O_p(n^{-1/2}).\) Combining this with the definition of \(\varvec{W}_{n2}\) and (10), one has

$$\begin{aligned} \Vert \hat{\varvec{\gamma }}-\varvec{\gamma }_{0}\Vert =O_p(n^{-1/2})+O_p\left( \sqrt{\frac{m^{a+1}}{n}}\right) +O_p\left( \sqrt{\frac{m }{n}}\right) =O_p\left( \sqrt{\frac{m^{a+1}}{n}}\right) . \end{aligned}$$

Using the similar argument as in Theorem 3.2 of Shin (2009) and the condition \(m\sim n^{1/(a+2b)}\), we have

$$\begin{aligned} \Vert {{\hat{\beta }}}-\beta _0\Vert =O_p\left( n^{-(2b-1)/(a+2b)}\right) . \end{aligned}$$

\(\square\)

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Cao, R., Xie, T. & Yu, P. Rank method for partial functional linear regression models. J. Korean Stat. Soc. (2020). https://doi.org/10.1007/s42952-020-00075-4

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Keywords

  • Rank estimation
  • Karhunen–Loève expansion
  • Asymptotic normality
  • Functional principal component analysis
  • Convergence rate

Mathematics Subject Classification

  • 62J05
  • 62M10