Robust confidence regions for the index and functional coefficients in the single-index varying coefficients regression model

Abstract

A single-index varying coefficients regression model is considered, from which robust confidence regions/intervals of the regression parameters are derived. Considering the rank-based estimating equations, empirical likelihood objective functions of the index and functional coefficients objective functions are defined and their asymptotic properties are established under mild regularity conditions. The performance of the proposed approach is demonstrated via extensive Monte Carlo simulation experiments. The simulation results are compared with those obtained from a normal approximation alternative. Also the proposed method is compared with the least squares and least absolute deviations alternatives. Finally, a real data example is given to illustrate the method.

Keywords:

• Rank pseudo-norm
• empirical likelihood
• asymptotic distributions
• confidence regions

2000 AMS Subject Classifications:

• Primary 62J02
• 62G05
• Secondary 62F12
• 62G20

Disclosure statement

No potential conflict of interest was reported by the author(s).

Appendix

The following lemma is necessary in the proof of Theorem 3.2

Lemma 4.1

Set $P_{\mathrm{jn}}\left(G\right(t_0\left)\right)=\frac{1}{n}\sum _{i=1}^n\left[\mathrm{\nabla G}\right(t_{0i}\left){]}^{\tau }{\mathbf{z}}_i\phi \right({\tilde{F}}_{\mathrm{ijn}}\left(v_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)\left)v_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)w_i\right(t_{0i})$, where ${\tilde{F}}_{\mathrm{ijn}}$ is the cumulative distribution function of $v_{\mathrm{ij}}\left(G\right(t_0\left)\right)$. Under $\left(I_1\right)-\left(I_8\right)$, we have $\underset{n\to \mathrm{\infty }}{lim}\underset{G\left(t_0\right)}{sup}\Vert J_{\mathrm{jn}}\left(G\right(t_0\left)\right)-P_{\mathrm{jn}}\left(G\right(t_0\left)\right)\Vert =0a.s.$

Proof.

Following [32], it is not hard to show that $R\left(v_{\mathrm{ij}}\right(G\left(t_0\right)\left)\right)/(n+1)-{\tilde{F}}_{\mathrm{ijn}}\left(v_{\mathrm{ij}}\right(G\left(t_0\right)\left)\right)\to 0a.s.,\text{for all}i,j=1,2.$By continuity of φ together with the generalized continuous mapping theorem [33], we have ${\displaystyle \underset{1\le i\le n}{max}|\phi \left(R\right(v_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)\left)/\right(n+1\left)\right)-\phi \left({\tilde{F}}_{\mathrm{ijn}}\right(v_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)\left)\right)|\to 0a.s.}$ On the other hand, $\begin{array}{rl}& \Vert J_{\mathrm{jn}}\left(G\right(t_0\left)\right)-P_{\mathrm{jn}}\left(G\right(t_0\left)\right)\Vert \\ & \le \frac{1}{n}\sum _{i=1}^n\Vert \mathrm{\nabla G}\left(t_{0i}\right)\Vert \Vert {\mathbf{z}}_i\Vert |\phi \left(\frac{R\left(v_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)}{n+1}\right)-\phi ({\tilde{F}}_{\mathrm{ijn}}\left(v_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)\left)\right||w_i\left(t_{0i}\right)|.\end{array}$Note that by definition, $|w_i\left(t_0\right)|\le 1$ and by assumption $\left(I_3\right)$, $\Vert \mathrm{\nabla G}\left(t_0\right)\Vert \le \sum _{k=0}^q{J}_k\left(\mathbf{x}\right)$. Then, by Cauchy–Schwarz inequality, we have $\begin{array}{rl}& \Vert J_{\mathrm{jn}}\left(G\right(t_0\left)\right)-P_{\mathrm{jn}}\left(G\right(t_0\left)\right)\Vert \\ & \le \frac{1}{n}\sum _{i=1}^n\left\{\sum _{k=0}^q{J}_k\right({\mathbf{x}}_i\left)\right\}\Vert {\mathbf{z}}_i\Vert |\phi \left(\frac{R\left(v_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)}{n+1}\right)-\phi ({\tilde{F}}_{\mathrm{ijn}}\left(v_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)\left)\right|\\ & \le {[\frac{1}{n}\sum _{i=1}^n{\left\{\sum _{k=0}^q{J}_k\right({\mathbf{x}}_i\left)\right\}}^2\Vert {\mathbf{z}}_i{\Vert }^2]}^{1/2}\\ & \times {\left[\underset{1\le i\le n}{max}{|\phi \left(\frac{R\left(v_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)}{n+1}\right)-\phi ({\tilde{F}}_{\mathrm{ijn}}\left(v_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)\left)\right|}^2\right]}^{1/2}.\end{array}$By $\left(I_3\right)$ and $\left(I_6\right)$, $E\left[\right\{\sum _{k=0}^q{J}_k\left({\mathbf{x}}_i\right){\}}^2\Vert {\mathbf{z}}_i{\Vert }^2]<\mathrm{\infty }$. So, by the Strong Law of Large Numbers, ${\displaystyle \frac{1}{n}\sum _{i=1}^n\left\{\sum _{k=0}^q{J}_k\right({\mathbf{x}}_i){\}}^2\Vert {\mathbf{z}}_i{\Vert }^2\to E[\left\{\sum _{k=0}^q{J}_k\right(\mathbf{x}){\}}^2\Vert \mathbf{z}{\Vert }^2]<\mathrm{\infty }a.s.}$ Moreover, as $n\to \mathrm{\infty }$, $\underset{1\le i\le n}{max}{|\phi \left(\frac{R\left(v_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)}{n+1}\right)-\phi ({\tilde{F}}_{\mathrm{ijn}}\left(v_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)\left)\right|}^2\to 0a.s.$Thus, ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\underset{G\left(t_0\right)}{sup}\Vert J_{\mathrm{jn}}\left(G\right(t_0\left)\right)-P_{\mathrm{jn}}\left(G\right(t_0\left)\right)\Vert =0a.s.}$

Proof of Theorem 3.1

By definition, $\mathit{\lambda }$ satisfies $\frac{1}{n}\sum _{i=1}^n\frac{z_{\mathrm{ij}}\left({\text{β}}_0\right)}{1+{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right)}=0.$Now, by assumption $\left(I_1\right)$ and $\left(I_3\right)$, one can obtain in a straightforward manner that ${\displaystyle \underset{1\le i\le n}{max}\Vert z_{\mathrm{ij}}\left({\text{β}}_0\right)\Vert =o_p\left(n^{1/2}\right)}$. Also, $n^{-1}\sum _{i=1}^n\Vert z_{\mathrm{ij}}\left({\text{β}}_0\right){\Vert }^3=o_p\left(n^{1/2}\right)$. Following the same argument as in [34], we obtain that $\Vert \mathit{\lambda }\Vert =O_p\left(n^{-1/2}\right)$, and by Theorem 2.3 of [23], $\Vert S_{\mathrm{jn}}\left({\text{β}}_0\right)\Vert =O_p\left(n^{1/2}\right)$. From this, a Taylor expansion to the right hand side of Equation (8(8) $\frac{1}{n}\sum _{i=1}^n\frac{z_{\mathrm{ij}}\left({\text{β}}_0\right)}{1+{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right)}=\mathbf{0}.$(8) ) gives $R_j\left({\text{β}}_0\right)=2\sum _{i=1}^n\left[{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\right({\text{β}}_0)-\frac{1}{2}{\left({\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\right({\text{β}}_0\left)\right)}^2]+{\mathit{\gamma }}_n,$where $\Vert {\mathit{\gamma }}_n\Vert =O_p\left(1\right)\sum _{i=1}^n|{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right){|}^3=O_p\left(1\right)\Vert \mathit{\lambda }{\Vert }^3\sum _{i=1}^n\Vert z_{\mathrm{ij}}\left({\text{β}}_0\right){\Vert }^3=o_p\left(1\right)$. On the other hand, note that solving Equation (8(8) $\frac{1}{n}\sum _{i=1}^n\frac{z_{\mathrm{ij}}\left({\text{β}}_0\right)}{1+{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right)}=\mathbf{0}.$(8) ) is equivalent to solving $\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\left({\text{β}}_0\right)(1-{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}({\text{β}}_0\left)\right)+\frac{1}{n}\sum _{i=1}{z}_{\mathrm{ij}}\left({\text{β}}_0\right)\frac{{\left({\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\right({\text{β}}_0\left)\right)}^2}{1+{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right)}=0,$which in turn can be rewritten as (17) $\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\left({\text{β}}_0\right)-\left(\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\right({\text{β}}_0\left)z_{\mathrm{ij}}^{\tau }\right({\text{β}}_0\left)\right)\mathit{\lambda }+\frac{1}{n}\sum _{i=1}{z}_{\mathrm{ij}}\left({\text{β}}_0\right)\frac{{\left({\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\right({\text{β}}_0\left)\right)}^2}{1+{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right)}=0.$(17) This implies that $\begin{array}{rl}\mathit{\lambda }& ={\left(\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\right({\text{β}}_0\left)z_{\mathrm{ij}}^{\tau }\right({\text{β}}_0\left)\right)}^{-1}\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\left({\text{β}}_0\right)+{\left(\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\right({\text{β}}_0\left)z_{\mathrm{ij}}^{\tau }\right({\text{β}}_0\left)\right)}^{-1}\\ & \times \frac{1}{n}\sum _{i=1}{z}_{\mathrm{ij}}\left({\text{β}}_0\right)\frac{{\left({\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\right({\text{β}}_0\left)\right)}^2}{1+{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right)}\end{array}$from which, it follows that $\mathit{\lambda }={\left(\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\right({\text{β}}_0\left)z_{\mathrm{ij}}^{\tau }\right({\text{β}}_0\left)\right)}^{-1}\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\left({\text{β}}_0\right)+o_p\left(1\right),$since $\left(\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\right({\text{β}}_0\left)z_{\mathrm{ij}}^{\tau }\right({\text{β}}_0\left){)}^{-1}\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\right({\text{β}}_0)\frac{\left({\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\right({\text{β}}_0){)}^2}{1+{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right)}=o_p\left(1\right)$. Also, premultiplying Equation (17(17) $\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\left({\text{β}}_0\right)-\left(\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\right({\text{β}}_0\left)z_{\mathrm{ij}}^{\tau }\right({\text{β}}_0\left)\right)\mathit{\lambda }+\frac{1}{n}\sum _{i=1}{z}_{\mathrm{ij}}\left({\text{β}}_0\right)\frac{{\left({\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\right({\text{β}}_0\left)\right)}^2}{1+{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right)}=0.$(17) ) by $\mathit{\lambda }$, yields $\frac{1}{n}\sum _{i=1}^n{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right)-\frac{1}{n}\sum _{i=1}^n{\left({\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\right({\text{β}}_0\left)\right)}^2+\frac{1}{n}\sum _{i=1}^n\frac{{\left({\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\right({\text{β}}_0\left)\right)}^3}{1+{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right)}=0,$which in turn results in $\sum _{i=1}^n{\left({\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\right({\text{β}}_0\left)\right)}^2=\sum _{i=1}^n{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right)+o_p\left(1\right)$since ${\displaystyle \sum _{i=1}^n\frac{\left({\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\right({\text{β}}_0){)}^3}{1+{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right)}=o_p\left(1\right)}$. Hence, $\begin{array}{rl}{R}_j\left({\text{β}}_0\right)& =\sum _{i=1}^n{\mathit{\lambda }}^{\tau }{z}_{\mathrm{ij}}\left({\text{β}}_0\right)+o_p\left(1\right)\\ & ={\left(\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\right({\text{β}}_0\left)\right)}^{\tau }{\left(\sum _{i=1}^n{z}_{\mathrm{ij}}\right({\text{β}}_0\left)z_{\mathrm{ij}}^{\tau }\right({\text{β}}_0\left)\right)}^{-1}\left(\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\right({\text{β}}_0\left)\right)+o_p\left(1\right)\end{array}$From this, note that by Theorem 2.3 of [23], we have $\frac{1}{n}\sum _{i=1}^n{z}_{\mathrm{ij}}\left({\text{β}}_0\right)z_{\mathrm{ij}}^{\tau }\left({\text{β}}_0\right)\to {\mathbf{V}}_{j}a.s.$Thus $R_j\left({\text{β}}_0\right)\approx {\left({\mathbf{V}}_j^{-1/2}\sqrt{n}{S}_{\mathrm{jn}}\right({\text{β}}_0\left)\right)}^{\tau }\left({\mathbf{V}}_j^{-1/2}\sqrt{n}{S}_{\mathrm{jn}}\right({\text{β}}_0\left)\right)+o_p\left(1\right).$Hence, noting that ${\mathbf{V}}_j^{-1/2}\sqrt{n}{S}_{\mathrm{nj}}\left({\text{β}}_0\right)⟹\mathcal{D}{N}_p(\mathbf{0},\mathbf{I})$, where $\mathbf{I}$ is a $p\times p$ identity matrix, $R_j\left({\text{β}}_0\right)⟹\mathcal{D}{\chi }_p^2$ and the proof is complete.

Proof of Theorem 3.2

For fixed $t_0$ and from Equation (13(13) $\sum _{i=1}^n\frac{{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)}{1+{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)}=0.$(13) ), $\mathit{\xi }$ satisfies (18) $\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n\frac{{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i})}{1+{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)}=0.$(18) Now, by Assumptions $\left(I_1\right)$ and $\left(I_3\right)$, one can obtain in a straightforward manner that ${\displaystyle \underset{1\le i\le n}{max}\Vert {\varphi }_{\mathrm{ij}}\left(t_{0i}\right)\Vert =o_p\left(\sqrt{n{h}_n^{q+1}}\right)}$. Also, $\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n\Vert {\varphi }_{\mathrm{ij}}\left(t_{0i}\right){\Vert }^3=o_p\left(\sqrt{n{h}_n^{q+1}}\right)$. Following the same argument as in the proof of Theorem 3.1, we have $\Vert \mathit{\xi }\Vert =O_p\left(1/\sqrt{n{h}_n^{q+1}}\right)$, and by Lemma 3.1, $\Vert \mathrm{\nabla }{J}_{\mathrm{jn}}\left(G\right(t_0\left)\right)\Vert =O_p\left(\sqrt{n{h}_n^{q+1}}\right)$. Once again, a direct application of the Taylor expansion up to order 2 to the right-hand side of Equation (14(14) $H_j\left(G\right(t_0)=2\sum _{i=1}^n\log (1+{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)),$(14) ) gives $H_j\left(G\right(t_0\left)\right)=2\sum _{i=1}^n\left[{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right))-\frac{1}{2}{\left({\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right))}^2]+{\mathit{\xi }}_n,$where $\Vert {\mathit{\xi }}_n\Vert =O_p\left(1\right)\sum _{i=1}^n|{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right){|}^3=O_p\left(1\right)\Vert \mathit{\xi }{\Vert }^3\sum _{i=1}^n\Vert {\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right){\Vert }^3=o_p\left(1\right)$. Following similar argument as in the proof of Theorem 3.1, solving Equation (18(18) $\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n\frac{{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i})}{1+{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)}=0.$(18) ) is equivalent to solving $\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)(1-{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}(G\left(t_{0i}\right)\left)\right)+\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)\frac{{\left({\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)}^2}{1+{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)}=0,$which in turn can be rewritten as (19) $\begin{array}{rl}& \frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)-\left(\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left){\varphi }_{\mathrm{ij}}^{\tau }\right(G\left(t_{0i}\right)\left)\right)\mathit{\xi }\\ & +\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)\frac{{\left({\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)}^2}{1+{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)}=0.\end{array}$(19) Solving this equation for $\mathit{\xi }$ gives $\begin{array}{rl}\mathit{\xi }& ={\left(\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left){\varphi }_{\mathrm{ij}}^{\tau }\right(G\left(t_{0i}\right)\left)\right)}^{-1}\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)\\ & +{\left(\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left){\varphi }_{\mathrm{ij}}^{\tau }\right(G\left(t_{0i}\right)\left)\right)}^{-1}\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)\frac{{\left({\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)}^2}{1+{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)},\end{array}$from which, it follows that $\mathit{\xi }={\left(\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left){\varphi }_{\mathrm{ij}}^{\tau }\right(G\left(t_{0i}\right)\left)\right)}^{-1}\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)+o_p\left(1\right),$as $\left(\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left){\varphi }_{\mathrm{ij}}^{\tau }\right(G\left(t_{0i}\right)\left){)}^{-1}\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right))\frac{\left({\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)){)}^2}{1+{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)}=o_p\left(1\right)$. Moreover, premultiplying Equation (19(19) $\begin{array}{rl}& \frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)-\left(\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left){\varphi }_{\mathrm{ij}}^{\tau }\right(G\left(t_{0i}\right)\left)\right)\mathit{\xi }\\ & +\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)\frac{{\left({\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)}^2}{1+{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)}=0.\end{array}$(19) ) by $\mathit{\xi }$, yields $\begin{array}{rl}& \frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)-\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\left({\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)}^2+\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n\frac{{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right){)}^3}{1+{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)}\\ & =0,\end{array}$which in turn results in $\sum _{i=1}^n{\left({\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)}^2=\sum _{i=1}^n{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)+o_p\left(1\right)$since ${\displaystyle \sum _{i=1}^n\frac{\left({\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)){)}^3}{1+{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)}=o_p\left(1\right)}$. Thus $\begin{array}{rl}{H}_j\left(G\right(t_0\left)\right)& =\sum _{i=1}^n{\mathit{\xi }}^{\tau }{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right)+o_p\left(1\right)\\ & ={\left(\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)}^{\tau }{\left(\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left){\varphi }_{\mathrm{ij}}^{\tau }\right(G\left(t_{0i}\right)\left)\right)}^{-1}\\ & \times \left(\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\right(G\left(t_{0i}\right)\left)\right)+o_p\left(1\right).\end{array}$Noting that $\frac{1}{n{h}_n^{q+1}}\sum _{i=1}^n{\varphi }_{\mathrm{ij}}\left(G\right(t_{0i}\left)\right){\varphi }_{\mathrm{ij}}^{\tau }\left(G\right(t_{0i}\left)\right)={\mathrm{\Omega }}_{\mathrm{nj}}$ is symmetric, we have $\begin{array}{rl}{H}_j\left(G\right(t_{0i}\left)\right)& ={\left({\mathrm{\Omega }}_{\mathrm{nj}}^{-1/2}\right(t_0\left)\sqrt{n{h}_n^{q+1}}\mathrm{\nabla }{J}_{\mathrm{jn}}\right(G\left(t_0\right)\left)\right)}^{\tau }\left({\mathrm{\Omega }}_{\mathrm{nj}}^{-1/2}\right(t_0\left)\sqrt{n{h}_n^{q+1}}\mathrm{\nabla }{J}_{\mathrm{jn}}\right(G\left(t_0\right)\left)\right)\\ & +o_p\left(1\right).\end{array}$To this end, from Remark 3.2, ${\mathrm{\Omega }}_{\mathrm{nj}}^{-1/2}\left(t_0\right)\sqrt{n{h}_n^{q+1}}\mathrm{\nabla }{J}_{\mathrm{jn}}\left(G\right(t_0\left)\right)⟹\mathcal{D}{N}_{q+1}(\mathbf{0},\mathbf{I})$, where $\mathbf{I}$ is a $(q+1)\times (q+1)$ identity matrix. Thus $H_j\left(G\right(t_{0i}\left)\right)⟹\mathcal{D}{\chi }_{q+1}^2$, which completes the proof.