A semiparametric dynamic higher-order spatial autoregressive model

Abstract

Conventional higher-order spatial autoregressive models assume that all regression coefficients are constant, which ignores dynamic feature that may exist in spatial data. In this paper, we introduce a semiparametric dynamic higher-order spatial autoregressive model by allowing regression coefficients in classical higher-order spatial autoregressive models to smoothly vary with a continuous explanatory variable, which enables us to explore dynamic feature in spatial data. We develop a sieve two-stage least squares method for the proposed model and derive asymptotic properties of resulting estimators. Furthermore, we develop two testing methods to check appropriateness of certain linear constraint condition on the spatial lag parameters and stationarity of the regression relationship, respectively. Simulation studies show that the proposed estimation and testing methods perform quite well in finite samples. The Boston house price data are finally analyzed to demonstrate the proposed model and its estimation and testing methods.

Appendix

In this appendix, we give detailed technical proofs of Theorems 1–5 in Sects. 2 and 3. The following three facts are frequently used in our proofs.

Fact 1. If the row and column sums of $n{\times }n$ matrices ${\textbf{A}}_{n1}$ and ${\textbf{A}}_{n2}$ are uniformly bounded in absolute value, then the row and column sums of ${\textbf{A}}_{n1}{\textbf{A}}_{n2}$ and ${\textbf{A}}_{n2}{\textbf{A}}_{n1}$ are also uniformly bounded in absolute value.

Fact 2. The largest eigenvalue of an idempotent matrix is at most one.

Fact 3. For any $n{\times }n$ matrix ${\textbf{B}}_{n}$, its spectral radius is bounded by ${\textrm{max}}_{1{\le }i{\le }n} \sum _{j=1}^{n}|b_{n,ij}|$, where $b_{n,ij}$ is the (i, j)th element of ${\textbf{B}}_{n}$.

Proof of Theorem 1

By Eq. (5) and noticing $({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{D}}_{n}={\textbf{0}}$, we obtain

$$\begin{aligned} \widehat{\varvec{\rho }}-{\varvec{\rho }}_{0}= & {} (\widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{Z}}_{n})^{-1} \widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{Y}}_{n}-{\varvec{\rho }}_{0}\nonumber \\= & {} (\widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{Z}}_{n})^{-1} \widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n})({\textbf{Z}}_{n}{\varvec{\rho }}_{0}+ {\textbf{D}}_{n}{\varvec{\gamma }}_{0}+{\textbf{R}}_{n}+{\varvec{\varepsilon }}_{n})-{\varvec{\rho }}_{0}\nonumber \\= & {} (\widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{Z}}_{n})^{-1} \widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{R}}_{n}+ (\widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{Z}}_{n})^{-1} \widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\varvec{\varepsilon }}_{n}, \end{aligned}$$

(A.1)

where ${\textbf{R}}_{n}={\textbf{M}}_{n}-{\textbf{D}}_{n}{\varvec{\gamma }}_{0}$ and $\widetilde{\textbf{R}}_{n}=({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{R}}_{n}$.

First, we consider $\widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{Z}}_{n}$. Let ${\overline{\varvec{\varepsilon }}}_{n}=({\textbf{G}}_{n1}{\varvec{\varepsilon }}_{n}, \ldots ,{\textbf{G}}_{nr}{\varvec{\varepsilon }}_{n})$, where ${\textbf{G}}_{nj}={\textbf{G}}_{nj}({\varvec{\rho }}_{0})$ ($j=1,\ldots ,r$). Then, ${\textbf{Z}}_{n}={\overline{\textbf{Z}}}_{n}+{\overline{\varvec{\varepsilon }}}_{n}$. Thus, $\widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{Z}}_{n}$ can be decomposed into

$$\begin{aligned} \widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{Z}}_{n}= & {} {\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{Z}}_{n}\\= & {} (\overline{\textbf{Z}}_{n}+\overline{\varvec{\varepsilon }}_{n})^{\textrm{T}} ({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}) (\overline{\textbf{Z}}_{n}+\overline{\varvec{\varepsilon }}_{n})\\= & {} \overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n})\overline{\textbf{Z}}_{n} +\overline{\varvec{\varepsilon }}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n})\overline{\varvec{\varepsilon }}_{n}+\\{} & {} \overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n})\overline{\varvec{\varepsilon }}_{n} +\overline{\varvec{\varepsilon }}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n})\overline{\textbf{Z}}_{n}\\= & {} n{\varvec{\varSigma }}_{n,1}+B_{n1}+B_{n2}+B_{n3}, \end{aligned}$$

where $B_{n1}=\overline{\varvec{\varepsilon }}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n})\overline{\varvec{\varepsilon }}_{n}$, $B_{n2}=\overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n})\overline{\varvec{\varepsilon }}_{n}$ and $B_{n3}=\overline{\varvec{\varepsilon }}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n})\overline{\textbf{Z}}_{n}$.

For $i,j=1,\ldots ,r$, it follows from Assumption 1.3 and Fact 1 that the row sums of ${\textbf{G}}_{nj}{\textbf{G}}_{ni}^{\textrm{T}}$ are uniformly bounded in absolute value. Hence, we obtain ${\eta }_{\max }({\textbf{G}}_{nj}{\textbf{G}}_{ni}^{\textrm{T}})={{O}}(1)$ by Fact 3. This together with Fact 2 yields

$$\begin{aligned}{} & {} {\textrm{E}}({\varvec{\varepsilon }}_{n}^{\textrm{T}}{\textbf{G}}_{ni}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{G}}_{nj}{\varvec{\varepsilon }}_{n})\\{} & {} \quad ={\sigma }_{0}^{2}{\textrm{tr}}({\textbf{G}}_{ni}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{G}}_{nj})\\{} & {} \quad ={\sigma }_{0}^{2}{\textrm{tr}}({\textbf{G}}_{nj}{\textbf{G}}_{ni}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}))\\{} & {} \quad {\le }{\sigma }_{0}^{2}{\eta }_{\max }({\textbf{G}}_{nj}{\textbf{G}}_{ni}^{\textrm{T}}){\textrm{tr}}(({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}))\\{} & {} \quad {\le }{\sigma }_{0}^{2}{\eta }_{\max }({\textbf{G}}_{nj}{\textbf{G}}_{ni}^{\textrm{T}}){\textrm{tr}}({\textbf{A}}_{n})\\{} & {} \quad ={\sigma }_{0}^{2}{\eta }_{\max }({\textbf{G}}_{nj}{\textbf{G}}_{ni}^{\textrm{T}}){\textrm{tr}}({\textbf{V}}_{n}({\textbf{V}}_{n}^{\textrm{T}}{\textbf{V}}_{n})^{-1} {\textbf{V}}_{n}^{\textrm{T}})\\{} & {} \quad ={\sigma }_{0}^{2}{\eta }_{\max }({\textbf{G}}_{nj}{\textbf{G}}_{ni}^{\textrm{T}}){\textrm{tr}}({\textbf{I}}_{s})\\{} & {} \quad ={{o}}(1). \end{aligned}$$

Combining this with Markov’s inequality yields

$$\begin{aligned} {\varvec{\varepsilon }}_{n}^{\textrm{T}}{\textbf{G}}_{ni}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{G}}_{nj}{\varvec{\varepsilon }}_{n} ={{O}}_{P}(1),\,\,i,j=1,\ldots ,r. \end{aligned}$$

(A.2)

Thus, we have $B_{n1}={{O}}_{P}(1)$.

For $i=1,\ldots ,r$, it follows from Assumption 1.3 and Facts 1 and 3 that ${\eta }_{\max }({\textbf{G}}_{ni}{\textbf{G}}_{ni}^{\textrm{T}})={{O}}(1)$. This together with Fact 2 and Assumption 3.4 yields

$$\begin{aligned}{} & {} {\textrm{E}}(\Vert \overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{G}}_{ni}{\varvec{\varepsilon }}_{n}\Vert ^2)\\= & {} {\textrm{E}}({\varvec{\varepsilon }}_{n}^{\textrm{T}}{\textbf{G}}_{ni}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n})\overline{\textbf{Z}}_{n}\overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{G}}_{ni}{\varvec{\varepsilon }}_{n})\\= & {} {\sigma }_{0}^{2}{\textrm{tr}}({\textbf{G}}_{ni}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n})\overline{\textbf{Z}}_{n}\overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{G}}_{ni})\\\le & {} {\sigma }_{0}^{2}{\eta }_{\max }({\textbf{G}}_{ni}{\textbf{G}}_{ni}^{\textrm{T}}) {\textrm{tr}}(\overline{\textbf{Z}}_{n}\overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}))\\\le & {} {\sigma }_{0}^{2}{\overline{c}}_{Z}{\eta }_{\max }({\textbf{G}}_{ni}{\textbf{G}}_{ni}^{\textrm{T}}) {\textrm{tr}}({\textbf{A}}_{n})\\= & {} {{O}}(1). \end{aligned}$$

This means that

$$\begin{aligned} \overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{G}}_{ni}{\varvec{\varepsilon }}_{n} ={{O}}_{P}(1),\,\,i=1,\ldots ,r. \end{aligned}$$

(A.3)

Therefore, we have $B_{n2}={{O}}_{P}(1)$. Similarly, we have $B_{n3}={{O}}_{P}(1)$. By combining the convergence rates of $B_{n1}$, $B_{n2}$ and $B_{n3}$, we obtain

$$\begin{aligned} n^{-1}\widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{Z}}_{n} ={\varvec{\varSigma }}_{n,1}+{{O}}_{P}(n^{-1}) ={\varvec{\varSigma }}_{n,1}+{{o}}_{P}(1). \end{aligned}$$

(A.4)

Next, we consider $\widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{R}}_{n}$. Obviously,

$$\begin{aligned}\widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{R}}_{n}= & {} {\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{R}}_{n}\\= & {} \overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{R}}_{n}+ \overline{\varvec{\varepsilon }}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{R}}_{n}\\= & {} B_{n4}+B_{n5}, \end{aligned}$$

where $B_{n4}=\overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{R}}_{n}$ and $B_{n5}=\overline{\varvec{\varepsilon }}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{R}}_{n}$. We first show that $B_{n4}={{O}}({\sqrt{n}}K^{-{\delta }})$. Let ${{R}}_{n,i}$ be the ith element of ${\textbf{R}}_{n}$, then ${{R}}_{n,i}=\sum _{j=1}^{q}x_{n,ij}[{\beta }_{0j}({U}_{n,i})-{\textbf{B}}({U}_{n,i})^{\textrm{T}} {\varvec{\gamma }}_{j0}]$. It follows from Assumption 3.1 that there exists a constant $c_{X}>0$ such that $\max _{1{\le }i{\le }n,1{\le }j{\le }q}|x_{n,ij}|{\le }c_{X}$ for all $n{\ge }1$. This together with Assumption 4.1 yields

$$\begin{aligned}|{{R}}_{n,i}|= & {} |\sum _{j=1}^{q}x_{n,ij}[{\beta }_{0j}({U}_{n,i})-{\textbf{B}}({U}_{n,i})^{\textrm{T}} {\varvec{\gamma }}_{j0}]|\\{} & {} \quad {\le }\sum _{j=1}^{q}|x_{n,ij}|{\cdot }|{\beta }_{0j}({U}_{n,i})-{\textbf{B}}({U}_{n,i})^{\textrm{T}} {\varvec{\gamma }}_{j0}|\\{} & {} \quad {\le } qc_{X}K^{-{\delta }}. \end{aligned}$$

This yields

$$\begin{aligned} \Vert {\textbf{R}}_{n}\Vert ^2=\sum _{i=1}^{n}{{R}}_{n,i}^{2} {\le }n{\cdot }\max _{1{\le }i{\le }n}{{R}}_{n,i}^{2} {\le }n{\cdot }\left( \max _{1{\le }i{\le }n}|{{R}}_{n,i}|\right) ^{2} ={{O}}(nK^{-2{\delta }}). \end{aligned}$$

(A.5)

Similar to the proof of $B_{n2}$, we have

$$\begin{aligned}\Vert B_{n4}\Vert ^2= & {} {\textbf{R}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n})\overline{\textbf{Z}}_{n}\overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{R}}_{n}\\\le & {} {\overline{c}}_{Z}{\textbf{R}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{R}}_{n}\\\le & {} {\overline{c}}_{Z}{\textbf{R}}_{n}^{\textrm{T}}{\textbf{R}}_{n}\\= & {} {{O}}(nK^{-2{\delta }}). \end{aligned}$$

Thus, we have $B_{n4}={{O}}({\sqrt{n}}K^{-{\delta }})$. For $i=1,\ldots ,r$, similar to the proof of (A.2), we have

$$\begin{aligned}{} & {} {\textrm{E}}(\Vert {\varvec{\varepsilon }}_{n}^{\textrm{T}}{\textbf{G}}_{ni}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{R}}_{n}\Vert ^2)\\= & {} {\textrm{E}}({\varvec{\varepsilon }}_{n}^{\textrm{T}}{\textbf{G}}_{ni}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{R}}_{n}{\textbf{R}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{G}}_{ni}{\varvec{\varepsilon }}_{n})\\= & {} {\sigma }_{0}^{2}{\textrm{tr}}({\textbf{G}}_{ni}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{R}}_{n}{\textbf{R}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{G}}_{ni})\\\le & {} {\sigma }_{0}^{2}{\eta }_{\max }({\textbf{G}}_{ni}{\textbf{G}}_{ni}^{\textrm{T}}) {\textrm{tr}}({\textbf{R}}_{n}{\textbf{R}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}))\\\le & {} {\sigma }_{0}^{2}{\eta }_{\max }({\textbf{G}}_{ni}{\textbf{G}}_{ni}^{\textrm{T}}) {\textrm{tr}}({\textbf{R}}_{n}{\textbf{R}}_{n}^{\textrm{T}})\\= & {} {{O}}(nK^{-2{\delta }}). \end{aligned}$$

This implies that

$$\begin{aligned} {\varvec{\varepsilon }}_{n}^{\textrm{T}}{\textbf{G}}_{ni}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{R}}_{n} ={{O}}_{P}({\sqrt{n}}K^{-{\delta }}),\,i=1,\ldots ,r. \end{aligned}$$

Therefore, we have $B_{n5}={{O}}_{P}({\sqrt{n}}K^{-{\delta }})$. By combining the convergence rates of $B_{n4}$ and $B_{n5}$, we obtain

$$\begin{aligned} \widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{R}}_{n} ={{O}}_{P}({\sqrt{n}}K^{-{\delta }}). \end{aligned}$$

(A.6)

Last, we consider $\widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\varvec{\varepsilon }}_{n}$. It directly follows from (A.2) that

$$\begin{aligned} \Vert {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{G}}_{ni}{\varvec{\varepsilon }}_{n}\Vert ^{2} ={{O}}_{P}(1),\,i=1,\ldots ,r. \end{aligned}$$

(A.7)

By an analogous proof to that of (A.2), we can show that

$$\begin{aligned} \Vert {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\varvec{\varepsilon }}_{n}\Vert ^{2} ={{O}}_{P}(1). \end{aligned}$$

(A.8)

By combining (A.7), (A.8) and Cauchy–Schwarz inequality, we obtain

$$\begin{aligned}{\varvec{\varepsilon }}_{n}^{\textrm{T}}{\textbf{G}}_{ni}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\varvec{\varepsilon }}_{n}\le & {} \Vert {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\textbf{G}}_{ni}{\varvec{\varepsilon }}_{n}\Vert {\cdot } \Vert {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\varvec{\varepsilon }}_{n}\Vert \\= & {} {{O}}_{P}(1),\,i=1,\ldots ,r. \end{aligned}$$

This implies that $\overline{\varvec{\varepsilon }}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\varvec{\varepsilon }}_{n}={{O}}_{P}(1)$. Thus, we have

$$\begin{aligned} \widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\varvec{\varepsilon }}_{n}= & {} {\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\varvec{\varepsilon }}_{n}\nonumber \\= & {} \overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\varvec{\varepsilon }}_{n}+ \overline{\varvec{\varepsilon }}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\varvec{\varepsilon }}_{n}\nonumber \\= & {} \overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\varvec{\varepsilon }}_{n}+{{O}}_{P}(1). \end{aligned}$$

(A.9)

By combining (A.4), (A.6) and (A.9), we obtain

$$\begin{aligned}{} & {} \sqrt{n}(\widehat{\varvec{\rho }}-{\varvec{\rho }}_{0})\\= & {} (n^{-1}\widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{Z}}_{n})^{-1} (n^{-1/2}\widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\varvec{\varepsilon }}_{n} +n^{-1/2}\widetilde{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{A}}_{n}\widetilde{\textbf{R}}_{n})\\= & {} [{\varvec{\varSigma }}_{n,1}+{{o}}_{P}(1)]^{-1} [n^{-1/2}\overline{\textbf{Z}}_{n}^{\textrm{T}}({\textbf{I}}_{n}-{\textbf{P}}_{n}) {\textbf{A}}_{n}({\textbf{I}}_{n}-{\textbf{P}}_{n}){\varvec{\varepsilon }}_{n}+{{o}}_{P}(1)]. \end{aligned}$$

Invoking the central limit theorem and Slutsky’s Lemma, we have

$$\begin{aligned} {\sqrt{n}}(\widehat{\varvec{\rho }}-{\varvec{\rho }}_{0}) \overset{\text {D}}{\longrightarrow }N({\textbf{0}},{\varvec{\varSigma }}). \end{aligned}$$

Thus, we complete the proof of Theorem 1. $\square$

Proof of Theorem 2

First, we consider the convergence rate of ${\widehat{\varvec{\gamma }}}$. By simple calculation, we have

$$\begin{aligned}{\widehat{\varvec{\gamma }}}-{\varvec{\gamma }}_{0}= & {} ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1} {\textbf{D}}_{n}^{\textrm{T}}({\textbf{Y}}_{n}-{\textbf{Z}}_{n}\widehat{\varvec{\rho }}) -{\varvec{\gamma }}_{0}\\= & {} ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1} {\textbf{D}}_{n}^{\textrm{T}}({\textbf{Z}}_{n}{\varvec{\rho }}_{0}+ {\textbf{D}}_{n}{\varvec{\gamma }}_{0}+{\textbf{R}}_{n}+{\varvec{\varepsilon }}_{n} -{\textbf{Z}}_{n}\widehat{\varvec{\rho }})-{\varvec{\gamma }}_{0}\\= & {} ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\textbf{Z}}_{n} ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }}) +({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\textbf{R}}_{n} +({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n}\\= & {} B_{n6}+B_{n7}+B_{n8}, \end{aligned}$$

where $B_{n6}=({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\textbf{Z}}_{n} ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})$, $B_{n7}=({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\textbf{R}}_{n}$ and $B_{n8}=({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n}$.

By Assumption 4.4, Fact 2 and ${\textbf{Z}}_{n}=\overline{\textbf{Z}}_{n}+\overline{\varvec{\varepsilon }}_{n}$, we have

$$\begin{aligned}\Vert B_{n6}\Vert ^{2}= & {} ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})^{\textrm{T}} {\textbf{Z}}_{n}^{\textrm{T}}{\textbf{D}}_{n}({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1} ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\textbf{Z}}_{n} ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})\\\le & {} n^{-1}{\eta }_{\min }^{-1}(n^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n}) ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})^{\textrm{T}} {\textbf{Z}}_{n}^{\textrm{T}}{\textbf{D}}_{n}({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1} {\textbf{D}}_{n}^{\textrm{T}}{\textbf{Z}}_{n}({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})\\\le & {} {\underline{c}}_{D}^{-1}({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})^{\textrm{T}} (n^{-1}{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{Z}}_{n}) ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})\\= & {} {\underline{c}}_{D}^{-1}({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})^{\textrm{T}} (n^{-1}\overline{\textbf{Z}}_{n}^{\textrm{T}}\overline{\textbf{Z}}_{n}) ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }}) +{\underline{c}}_{D}^{-1}({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})^{\textrm{T}} (n^{-1}\overline{\varvec{\varepsilon }}_{n}^{\textrm{T}}\overline{\varvec{\varepsilon }}_{n}) ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})\\{} & {} \,+\, 2{\underline{c}}_{D}^{-1}({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})^{\textrm{T}} (n^{-1}\overline{\textbf{Z}}_{n}^{\textrm{T}}\overline{\varvec{\varepsilon }}_{n}) ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})\\= & {} B_{n61}+B_{n62}+B_{n63}, \end{aligned}$$

where $B_{n61}={\underline{c}}_{D}^{-1}({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})^{\textrm{T}} (n^{-1}\overline{\textbf{Z}}_{n}^{\textrm{T}}\overline{\textbf{Z}}_{n}) ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})$, $B_{n62}={\underline{c}}_{D}^{-1}({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})^{\textrm{T}} (n^{-1}\overline{\varvec{\varepsilon }}_{n}^{\textrm{T}}\overline{\varvec{\varepsilon }}_{n}) ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})$ and $B_{n63}=2{\underline{c}}_{D}^{-1}({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})^{\textrm{T}} (n^{-1}\overline{\textbf{Z}}_{n}^{\textrm{T}}\overline{\varvec{\varepsilon }}_{n}) ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})$.

From Theorem 1, we obtain $\Vert \widehat{\varvec{\rho }}-{\varvec{\rho }}_{0}\Vert ^{2}= {{O}}_{P}(n^{-1})$. This together with Assumption 3.4 yields

$$\begin{aligned}B_{n61}= & {} {\underline{c}}_{D}^{-1}({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})^{\textrm{T}} (n^{-1}\overline{\textbf{Z}}_{n}^{\textrm{T}}\overline{\textbf{Z}}_{n}) ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})\\\le & {} {\underline{c}}_{D}^{-1}{\eta }_{\max }(n^{-1}\overline{\textbf{Z}}_{n}^{\textrm{T}} \overline{\textbf{Z}}_{n})\Vert \widehat{\varvec{\rho }}-{\varvec{\rho }}_{0}\Vert ^{2}\\= & {} {{O}}_{P}(n^{-1}). \end{aligned}$$

For $i,j=1,\ldots ,r$, it follows from Assumption 1.3 and Facts 1 and 3 that ${\textrm{E}}(n^{-1}{\varvec{\varepsilon }}_{n}^{\textrm{T}}{\textbf{G}}_{ni}^{\textrm{T}}{\textbf{G}}_{nj}{\varvec{\varepsilon }}_{n}) =n^{-1}{\sigma }_{0}^{2}{\textrm{tr}}({\textbf{G}}_{ni}^{\textrm{T}}{\textbf{G}}_{nj}) =n^{-1}{\sigma }_{0}^{2}{\textrm{tr}}({\textbf{G}}_{nj}{\textbf{G}}_{ni}^{\textrm{T}}) {\le }{\sigma }_{0}^{2}{\eta }_{\max }({\textbf{G}}_{nj}{\textbf{G}}_{ni}^{\textrm{T}}) ={{O}}(1).$ This implies that $n^{-1}{\varvec{\varepsilon }}_{n}^{\textrm{T}}{\textbf{G}}_{ni}^{\textrm{T}}{\textbf{G}}_{nj}{\varvec{\varepsilon }}_{n} ={\textrm{O}}_{P}(1)$. Thus, we have

$$\begin{aligned}B_{n62}={\underline{c}}_{D}^{-1}({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})^{\textrm{T}} (n^{-1}\overline{\varvec{\varepsilon }}_{n}^{\textrm{T}}\overline{\varvec{\varepsilon }}_{n}) ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})={{O}}_{P}(n^{-1}). \end{aligned}$$

By using Cauchy–Schwarz inequality and the orders of $B_{n61}$ and $B_{n62}$, we have

$$\begin{aligned}B_{n63}= & {} 2{\underline{c}}_{D}^{-1}({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})^{\textrm{T}} (n^{-1}\overline{\textbf{Z}}_{n}^{\textrm{T}}\overline{\varvec{\varepsilon }}_{n}) ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})\\\le & {} 2\left| \left[ {\underline{c}}_{D}^{-1/2}n^{-1/2}({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})^{\textrm{T}} \overline{\textbf{D}}_{n}^{\textrm{T}}\right] {\cdot }\left[ {\underline{c}}_{D}^{-1/2}n^{-1/2}\overline{\varvec{\varepsilon }}_{n} ({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})\right] \right| \\\le & {} 2(B_{n61}B_{n62})^{1/2}\\= & {} {{O}}_{P}(n^{-1}). \end{aligned}$$

Combining the orders of $B_{n61}$, $B_{n62}$ and $B_{n63}$, we obtain $\Vert B_{n6}\Vert ={{O}}_{P}(n^{-1/2})$.

By Assumption 4.4, Fact 2 and (A.5), we obtain

$$\begin{aligned}\Vert B_{n7}\Vert ^{2}= & {} {\textbf{R}}_{n}^{\textrm{T}}{\textbf{D}}_{n}({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1} ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\textbf{R}}_{n}\\\le & {} n^{-1}{\eta }_{\min }^{-1}(n^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n}){\textbf{R}}_{n}^{\textrm{T}} {\textbf{D}}_{n}({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\textbf{R}}_{n}\\\le & {} n^{-1}{\underline{c}}_{D}^{-1}\Vert {\textbf{R}}_{n}\Vert ^{2}\\= & {} {{O}}(K^{-2{\delta }}). \end{aligned}$$

This means that $\Vert B_{n7}\Vert ={{O}}(K^{-{\delta }})$.

It follows from Assumption 4.4 that

$$\begin{aligned}{\textrm{E}}(\Vert B_{n8}\Vert ^{2})= & {} {\textrm{E}}({\varvec{\varepsilon }}_{n}^{\textrm{T}}{\textbf{D}}_{n}({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1} ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n})\\\le & {} n^{-1}{\eta }_{\min }^{-1}(n^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n}) {\textrm{E}}({\varvec{\varepsilon }}_{n}^{\textrm{T}} {\textbf{D}}_{n}({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n})\\= & {} n^{-1}{\underline{c}}_{D}^{-1}{\sigma }_{0}^{2} {\textrm{tr}}({\textbf{D}}_{n}({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}})\\= & {} {{O}}(K/n). \end{aligned}$$

This implies that $\Vert B_{n8}\Vert ={{O}}_{P}({\sqrt{K/n}})$. By triangle inequality and the orders of $B_{n6}$, $B_{n7}$ and $B_{n8}$, we have

$$\begin{aligned} \Vert {\widehat{\varvec{\gamma }}}-{\varvec{\gamma }}_{0}\Vert {\le }\Vert B_{n6}\Vert +\Vert B_{n7}\Vert +\Vert B_{n8}\Vert ={{O}}_{P}({\sqrt{K/n}}+K^{-{\delta }}). \end{aligned}$$

Next, we consider the uniform convergence rate of ${\widehat{\varvec{\beta }}}(u)$. By the definition of ${\widehat{\varvec{\beta }}}(u)$, convergence rate of ${\widehat{\varvec{\gamma }}}$, and Assumptions 4.1 and 4.3, we obtain

$$\begin{aligned}{} & {} {\sup }_{{{u}}{\in }{{\mathcal {U}}}}|{\widehat{{\beta }}}_{j}(u)-{{\beta }}_{0j}(u)|\\{} & {} \quad ={\sup }_{{{u}}{\in }{{\mathcal {U}}}}|{\textbf{B}}(u)^{\textrm{T}}{\widehat{\varvec{\gamma }}}_{j} -{{\beta }}_{0j}(u)|\\{} & {} \quad ={\sup }_{{{u}}{\in }{{\mathcal {U}}}}|{\textbf{B}}(u)^{\textrm{T}}({\widehat{\varvec{\gamma }}}_{j}- {{\varvec{\gamma }}}_{j0})+{\textbf{B}}(u)^{\textrm{T}}{{\varvec{\gamma }}}_{0j} -{{\beta }}_{0j}(u)|\\{} & {} \quad {\le } {\sup }_{{{u}}{\in }{{\mathcal {U}}}}|{\textbf{B}}(u)^{\textrm{T}}({\widehat{\varvec{\gamma }}}_{j}- {{\varvec{\gamma }}}_{j0})| +{\sup }_{{{u}}{\in }{{\mathcal {U}}}}|{{\beta }}_{0j}(u)-{\textbf{B}}(u)^{\textrm{T}}{{\varvec{\gamma }}}_{j0}|\\{} & {} \quad {\le } \Vert {\widehat{\varvec{\gamma }}}- {{\varvec{\gamma }}}_{0}\Vert {\cdot }{\sup }_{{{u}}{\in }{{\mathcal {U}}}}\Vert {\textbf{B}}({\textbf{u}})\Vert + {\sup }_{{{u}}{\in }{{\mathcal {U}}}}|{{\beta }}_{0j}(u)-{\textbf{B}}(u)^{\textrm{T}}{{\varvec{\gamma }}}_{j0}|\\{} & {} \quad {\le } {\zeta }(K)\Vert {\widehat{\varvec{\gamma }}}- {{\varvec{\gamma }}}_{0}\Vert +{{O}}(K^{-{\delta }})\\{} & {} \quad ={{O}}_{P}({\zeta }(K)({\sqrt{K/n}}+K^{-{\delta }})). \end{aligned}$$

This yields

$$\begin{aligned} {\sup }_{{{u}}{\in }{{\mathcal {U}}}}\Vert {\widehat{\varvec{\beta }}}(u)-{\varvec{\beta }}_{0}(u)\Vert ={{O}}_{P}({\zeta }(K)({\sqrt{K/n}}+K^{-{\delta }})). \end{aligned}$$

(A.10)

Finally, we consider the limiting distribution of ${\widehat{\varvec{\beta }}}(u)$. By the definition of ${\widehat{\varvec{\beta }}}(u)$ and Assumption 4.1, we have

$$\begin{aligned}{\widehat{\varvec{\beta }}}(u)-{\varvec{\beta }}_{0}(u)= & {} {\varvec{\varGamma }}(u)({\widehat{\varvec{\gamma }}}- {{\varvec{\gamma }}}_{0})+{\varvec{\varGamma }}(u){{\varvec{\gamma }}}_{0} -{\varvec{\beta }}_{0}(u)\\= & {} {\varvec{\varGamma }}(u)(B_{n6}+B_{n7}+B_{n8}) +{{O}}(K^{-{\delta }})\\= & {} {\varvec{\varGamma }}(u)(B_{n6}+B_{n7}+B_{n8}) +{{o}}(1). \end{aligned}$$

By combining the convergence rates of $B_{n6}$ and $B_{n7}$ and Assumption 4.3, we have

$$\begin{aligned} {\varvec{\varGamma }}(u)B_{n6}={{O}}_{P}(n^{-1/2}{\zeta }(K))\,\,\, {\textrm{and}}\,\,\, {\varvec{\varGamma }}(u)B_{n7}={{O}}({\zeta }(K)K^{-{\delta }}). \end{aligned}$$

It follows from Assumption 4.5 and ${\zeta }(K){\rightarrow }{\infty }$ as $n{\rightarrow }{\infty }$ that $n^{-1/2}{\zeta }(K)={{o}}(1)$. This together with ${\sqrt{n}}K^{-{\delta }}={\textrm{o}}(1)$ yields ${\zeta }(K)K^{-{\delta }}=(n^{-1/2}{\zeta }(K))({\sqrt{n}}K^{-{\delta }}) ={{o}}(1)$. Thus, we have

$$\begin{aligned} {\widehat{\varvec{\beta }}}(u)-{\varvec{\beta }}_{0}(u)={\varvec{\varGamma }}(u) ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n} +{{o}}_{P}(1). \end{aligned}$$

According to the Cram$\mathrm {\acute{e}}$r-Wold device, it is sufficient to prove

$$\begin{aligned} {\textbf{c}}^{\textrm{T}}{\varvec{\varGamma }}(u)({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n}\overset{\text {D}}{\longrightarrow } N({{0}},{\textbf{c}}^{\textrm{T}}{\varvec{\varSigma }}(u){\textbf{c}}) \end{aligned}$$

for any nonzero $q{\times }1$ vector of constants ${\textbf{c}}$. Let ${\textbf{d}}_{n,i}=({\textbf{I}}_{q}{\otimes }{\textbf{B}}(U_{n,i})){\textbf{X}}_{n,i}$ ($i=1,\ldots ,n$), then

$$\begin{aligned}{}[{\textbf{c}}^{\textrm{T}}{\varvec{\varSigma }}(u){\textbf{c}}]^{-1/2}{\textbf{c}}^{\textrm{T}}{\varvec{\varGamma }}(u) ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n} =\sum _{i=1}^{n}{\xi }_{n,i}, \end{aligned}$$

where ${\xi }_{n,i}=[{\textbf{c}}^{\textrm{T}}{\varvec{\varSigma }}(u){\textbf{c}}]^{-1/2}{\textbf{c}}^{\textrm{T}}{\varvec{\varGamma }}(u) ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{d}}_{n,i}{\varepsilon }_{n,i}$. It follows from Assumptions 3.1 and 4.3 that

$$\begin{aligned} \Vert {\textbf{d}}_{n,i}\Vert ^{2}=\sum _{j=1}^{q}x_{n,ij}^{2}\Vert {\textbf{B}}(U_{n,i})\Vert ^{2}{\,} {\le }\left( \sup _{u{\in }{{\mathcal {U}}}}\Vert {\textbf{B}}(u)\Vert \right) ^{2}{\,}qc_{X}^{2}. \end{aligned}$$

This together with Assumptions 4.3 and 4.4 yields

$$\begin{aligned}{} & {} \max _{1{\le }i{\le }n}|[{\textbf{c}}^{\textrm{T}}{\varvec{\varSigma }}(u){\textbf{c}}]^{-1/2}{\textbf{c}}^{\textrm{T}}{\varvec{\varGamma }}(u) ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{d}}_{n,i}|\\{} & {} \quad \le [{\textbf{c}}^{\textrm{T}}{\varvec{\varSigma }}(u){\textbf{c}}]^{-1/2}{\cdot }\Vert {\textbf{c}}^{\textrm{T}}{\varvec{\varGamma }}(u) ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}\Vert {\cdot }\max _{1{\le }i{\le }n}\Vert {\textbf{d}}_{n,i}\Vert \\{} & {} \quad \le {\sqrt{q}}c_{X}{\zeta }(K)[{\textbf{c}}^{\textrm{T}}{\varvec{\varSigma }}(u){\textbf{c}}]^{-1/2} [{\textbf{c}}^{\textrm{T}}{\varvec{\varGamma }}(u)({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-2}{\varvec{\varGamma }}(u)^{\textrm{T}}{\textbf{c}}]^{1/2}\\{} & {} \quad \le {\sqrt{q}}c_{X}{\zeta }(K){\underline{c}}_{D}^{-1/2}n^{-1/2}[{\textbf{c}}^{\textrm{T}}{\varvec{\varSigma }}(u){\textbf{c}}]^{-1/2} [{\textbf{c}}^{\textrm{T}}{\varvec{\varGamma }}(u)({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\varvec{\varGamma }}(u)^{\textrm{T}}{\textbf{c}}]^{1/2}\\{} & {} \quad ={\sqrt{q}}c_{X}{\underline{c}}_{D}^{-1/2}{\sigma }_{0}^{-1}{\zeta }(K)n^{-1/2}. \end{aligned}$$

This together with Assumptions 2 and 4.5 yields

$$\begin{aligned}\sum _{i=1}^{n}{\textrm{E}}|{\xi }_{n,i}|^{3}= & {} \sum _{i=1}^{n}|[{\textbf{c}}^{\textrm{T}}{\varvec{\varSigma }}(u){\textbf{c}}]^{-1/2}{\textbf{c}}^{\textrm{T}}{\varvec{\varGamma }}(u) ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{d}}_{n,i}|^{3} {\textrm{E}}|{\varepsilon }_{n,i}|^{3}\\\le & {} Cn{\cdot }\max _{1{\le }i{\le }n}|[{\textbf{c}}^{\textrm{T}}{\varvec{\varSigma }}(u){\textbf{c}}]^{-1/2}{\textbf{c}}^{\textrm{T}}{\varvec{\varGamma }}(u) ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{d}}_{n,i}|^{3}\\= & {} {{O}}(n^{-1/2}{\zeta }(K)^{3})\\= & {} {{o}}(1). \end{aligned}$$

Combining this result with Lyapunov central limit theorem, we obtain

$$\begin{aligned}{}[{\textbf{c}}^{\textrm{T}}{\varvec{\varSigma }}(u){\textbf{c}}]^{-1/2}{\textbf{c}}^{\textrm{T}}{\varvec{\varGamma }}(u) ({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n} \overset{\text {D}}{\longrightarrow }N(0,1). \end{aligned}$$

Thus, we complete the proof of Theorem 2. $\square$

Proof of Theorem 3

First, we prove the consistency of ${\widehat{\sigma }}^{2}$. By the definition of ${\widehat{\sigma }}^{2}$, we have

$$\begin{aligned}{\widehat{\sigma }}^{2}= & {} n^{-1}\Vert {\textbf{Y}}_{n}-{\textbf{Z}}_{n}\widehat{\varvec{\rho }} -{\textbf{D}}_{n}\widehat{\varvec{\gamma }}\Vert ^{2}\\= & {} n^{-1}\Vert {\textbf{Z}}_{n}{\varvec{\rho }}_{0}+{\textbf{R}}_{n}+ {\textbf{D}}_{n}{\varvec{\gamma }}_{0}+{\varvec{\varepsilon }}_{n} -{\textbf{Z}}_{n}\widehat{\varvec{\rho }} -{\textbf{D}}_{n}\widehat{\varvec{\gamma }}\Vert ^{2}\\= & {} n^{-1}\Vert {\textbf{Z}}_{n}({\varvec{\rho }}_{0}-\widehat{\varvec{\rho }})+ {\textbf{D}}_{n}({\varvec{\gamma }}_{0}-\widehat{\varvec{\gamma }})+{\textbf{R}}_{n}+ {\varvec{\varepsilon }}_{n}\Vert ^{2}\\= & {} n^{-1}{\varvec{\varepsilon }}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n} +C_{n1}+C_{n2}+C_{n3}+C_{n4}-2C_{n5}-2C_{n6}-2C_{n7}-2C_{n8}+2C_{n9}, \end{aligned}$$

where $C_{n1}=n^{-1}(\widehat{\varvec{\rho }}-{\varvec{\rho }}_{0})^{\textrm{T}}{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{Z}}_{n} (\widehat{\varvec{\rho }}-{\varvec{\rho }}_{0})$, $C_{n2}=n^{-1}(\widehat{\varvec{\gamma }}-{\varvec{\gamma }}_{0})^{\textrm{T}}{\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n} (\widehat{\varvec{\gamma }}-{\varvec{\gamma }}_{0})$, $B_{n3}=n^{-1}\Vert {\textbf{R}}_{n}\Vert ^{2}$, $C_{n4}=n^{-1}(\widehat{\varvec{\rho }}-{\varvec{\rho }}_{0})^{\textrm{T}}{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{D}}_{n} (\widehat{\varvec{\gamma }}-{\varvec{\gamma }}_{0})$, $C_{n5}=n^{-1}(\widehat{\varvec{\rho }}-{\varvec{\rho }}_{0})^{\textrm{T}}{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{R}}_{n}$, $C_{n6}=n^{-1}(\widehat{\varvec{\rho }}-{\varvec{\rho }}_{0})^{\textrm{T}}{\textbf{Z}}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n}$, $C_{n7}=n^{-1}(\widehat{\varvec{\gamma }}-{\varvec{\gamma }}_{0})^{\textrm{T}}{\textbf{D}}_{n}^{\textrm{T}}{\textbf{R}}_{n}$, $C_{n8}=n^{-1}(\widehat{\varvec{\gamma }}-{\varvec{\gamma }}_{0})^{\textrm{T}}{\textbf{D}}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n}$ and $C_{n9}=n^{-1}{\textbf{R}}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n}$.

By applying the law of large numbers for independent and identically distributed random variables, we have $n^{-1}{\varvec{\varepsilon }}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n} \overset{\text {P}}{\longrightarrow }{\sigma }_{0}^2$. Thus, to complete the proof of part (a), it suffices to show $C_{nj}\overset{\text {P}}{\longrightarrow }0$ ($j=1,\ldots ,9$).

By Theorems 1 and 2 and their proofs, we have $\Vert \widehat{\varvec{\rho }}-{\varvec{\rho }}_{0}\Vert ={{O}}_{p}(n^{-1/2})$, $\Vert \widehat{\varvec{\gamma }}-{\varvec{\gamma }}_{0}\Vert ={{O}}_{P}({\sqrt{K/n}}+K^{-{\delta }})$, $\Vert {\textbf{R}}_{n}\Vert ^{2}={{O}}(nK^{-2{\delta }})$ and $n^{-1}{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{Z}}_{n}={{O}}_{p}(1)$. Combining these results with $\Vert n^{-1/2}{\varvec{\varepsilon }}_{n}\Vert =(n^{-1}{\varvec{\varepsilon }}_{n}^{\textrm{T}}{\varvec{\varepsilon }}_{n}) ^{1/2}={{O}}_{p}(1)$, Assumption 4.4 and Cauchy–Schwarz inequality, we obtain

$$\begin{aligned}C_{n1}= & {} (\widehat{\varvec{\rho }}-{\varvec{\rho }}_{0})^{\textrm{T}}(n^{-1}{\textbf{Z}}_{n}^{\textrm{T}}{\textbf{Z}}_{n}) (\widehat{\varvec{\rho }}-{\varvec{\rho }}_{0}) ={{O}}_{p}(n^{-1})={{o}}_{p}(1),\\ C_{n2}\le & {} {\overline{c}}_{D}\Vert \widehat{\varvec{\gamma }}-{\varvec{\gamma }}_{0}\Vert ^{2} ={{O}}_{p}(K/n+K^{-2{\delta }})={{o}}_{p}(1),\\ C_{n3}= & {} n^{-1}\Vert {\textbf{R}}_{n}\Vert ^{2} ={{O}}(K^{-2{\delta }})={{o}}(1),\\ C_{n4}\le & {} \Vert n^{-1/2}{\textbf{Z}}_{n}(\widehat{\varvec{\rho }}-{\varvec{\rho }}_{0})\Vert {\cdot } \Vert n^{-1/2}{\textbf{D}}_{n}(\widehat{\varvec{\gamma }}-{\varvec{\gamma }}_{0})\Vert ={{o}}_{p}(1),\\ C_{n5}\le & {} \Vert n^{-1/2}{\textbf{Z}}_{n}(\widehat{\varvec{\rho }}-{\varvec{\rho }}_{0})\Vert {\cdot } \Vert n^{-1/2}{\textbf{R}}_{n}\Vert ={{O}}_{p}(n^{-1/2}K^{-{\delta }})={{o}}_{p}(1),\\ C_{n6}\le & {} \Vert n^{-1/2}{\textbf{Z}}_{n}(\widehat{\varvec{\rho }}-{\varvec{\rho }}_{0})\Vert {\cdot } \Vert n^{-1/2}{\varvec{\varepsilon }}_{n}\Vert ={{O}}_{p}(n^{-1/2})={\textrm{o}}_{p}(1),\\ C_{n7}\le & {} \Vert n^{-1/2}{\textbf{D}}_{n}(\widehat{\varvec{\gamma }}-{\varvec{\gamma }}_{0})\Vert {\cdot } \Vert n^{-1/2}{\textbf{R}}_{n}\Vert ={{O}}_{p}(K^{-{\delta }}({\sqrt{K/n}}+K^{-{\delta }}))={{o}}_{p}(1),\\ C_{n8}\le & {} \Vert n^{-1/2}{\textbf{D}}_{n}(\widehat{\varvec{\gamma }}-{\varvec{\gamma }}_{0})\Vert {\cdot } \Vert n^{-1/2}{\varvec{\varepsilon }}_{n}\Vert ={{O}}_{p}({\sqrt{K/n}}+K^{-{\delta }})={{o}}_{p}(1),\\ C_{n9}\le & {} \Vert n^{-1/2}{\textbf{R}}_{n}\Vert {\cdot } \Vert n^{-1/2}{\varvec{\varepsilon }}_{n}\Vert ={{O}}_{p}(K^{-{\delta }})={{o}}_{p}(1). \end{aligned}$$

Next, we prove part (b) of Theorem 3. By Theorems 1 and 2, we have $\widehat{\varvec{\rho }}\overset{\text {P}}{\longrightarrow }{\varvec{\rho }}_{0}$ and ${\widehat{\varvec{\beta }}}(u)\overset{\text {P}}{\longrightarrow } {\varvec{\beta }}_{0}(u)$. This together with part (a) yields $\widehat{\varvec{\varSigma }}\overset{\text {P}}{\longrightarrow } {\varvec{\varSigma }}$.

Finally, we prove part (c) of Theorem 3. Let $\widehat{{\varSigma }}_{ij}(u)$ and ${{\varSigma }}_{ij}(u)$ be (i, j)th elements of $\widehat{\varvec{\varSigma }}(u)$ and ${\varvec{\varSigma }}(u)$, respectively. Then $\widehat{{\varSigma }}_{ij}(u)={{\widehat{\sigma }}}^{2}{\textbf{e}}_{i}^{\textrm{T}}{\varvec{\varGamma }}(u)({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1} {\varvec{\varGamma }}(u)^{\textrm{T}}{\textbf{e}}_{j}$ and ${{\varSigma }}_{ij}(u)={\sigma }_{0}^{2}{\textbf{e}}_{i}^{\textrm{T}}{\varvec{\varGamma }}(u)({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1} {\varvec{\varGamma }}(u)^{\textrm{T}}{\textbf{e}}_{j}$, where ${\textbf{e}}_{i}$ is a $q{\times }1$ vector with its ith element being 1 and other elements are 0, and ${\textbf{e}}_{j}$ is similarly defined. It follows from part (a) of Theorem 3 that ${\widehat{\sigma }}^{2}-{\sigma }_{0}^2={{o}}_{p}(1)$. This together with Assumptions 4.3–4.5 yields

$$\begin{aligned}\widehat{{\varSigma }}_{ij}(u)-{{\varSigma }}_{ij}(u)= & {} ({\widehat{\sigma }}^{2}-{\sigma }_{0}^{2})({\textbf{B}}(u)^{\textrm{T}}{\otimes }{\textbf{e}}_{i}^{\textrm{T}})({\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1} ({\textbf{e}}_{j}{\otimes }{\textbf{B}}(u))\\= & {} {\textrm{o}}_{p}(1)n^{-1}({\textbf{B}}(u)^{\textrm{T}}{\otimes }{\textbf{e}}_{i}^{\textrm{T}})(n^{-1}{\textbf{D}}_{n}^{\textrm{T}}{\textbf{D}}_{n})^{-1} ({\textbf{e}}_{j}{\otimes }{\textbf{B}}(u))\\\le & {} {\underline{c}}_{D}^{-1}{\textrm{o}}_{p}(1)n^{-1}\Vert {\textbf{B}}(u)\Vert ^{2}I(i=j)\\\le & {} {\textrm{o}}_{p}({\zeta }(K)^{2}/n)I(i=j)\\= & {} {\textrm{o}}_{p}(({\zeta }(K)^{2}K/n){K}^{-1})I(i=j)\\= & {} {\textrm{o}}_{p}(1), \end{aligned}$$

where $I(\cdot )$ is the indicator function. This shows that $\widehat{\varvec{\varSigma }}(u)$ is a consistent estimator of ${\varvec{\varSigma }}(u)$. $\square$

Proof of Theorems 4 and 5

We only prove Theorem 5 because Theorem 4 is a special case of Theorem 5. By Theorem 1, it is easy to show

$$\begin{aligned} {\sqrt{n}}({\textbf{R}}\widehat{\varvec{\rho }}-{\textbf{b}}) \overset{\text {D}}{\longrightarrow }N({\sqrt{n}}({\textbf{R}}{\varvec{\rho }}_{0}-{\textbf{b}}), {\textbf{R}}{\varvec{\varSigma }}{\textbf{R}}^{\textrm{T}}). \end{aligned}$$

This implies that

$$\begin{aligned} {\sqrt{n}}({\textbf{R}}{\varvec{\varSigma }}{\textbf{R}}^{\textrm{T}})^{-1/2}({\textbf{R}}\widehat{\varvec{\rho }}-{\textbf{b}}) \overset{\text {D}}{\longrightarrow }N({\varvec{\mu }},{\textbf{I}}_{d}), \end{aligned}$$

where ${\varvec{\mu }}={\sqrt{n}}({\textbf{R}}{\varvec{\varSigma }}{\textbf{R}}^{\textrm{T}})^{-1/2} ({\textbf{R}}{\varvec{\rho }}_{0}-{\textbf{b}})$. This together with Theorem 3(b) and the Slutsky theorem yields

$$\begin{aligned} {\sqrt{n}}({\textbf{R}}\widehat{\varvec{\varSigma }}{\textbf{R}}^{\textrm{T}})^{-1/2} ({\textbf{R}}\widehat{\varvec{\rho }}-{\textbf{b}}) \overset{\text {D}}{\longrightarrow }N({\varvec{\mu }},{\textbf{I}}_{d}). \end{aligned}$$

Thus, we have

$$\begin{aligned} T_{1}=n({\textbf{R}}\widehat{\varvec{\rho }}-{\textbf{b}})^{\textrm{T}} ({\textbf{R}}\widehat{\varvec{\varSigma }}{\textbf{R}}^{\textrm{T}})^{-1} ({\textbf{R}}\widehat{\varvec{\rho }}-{\textbf{b}}) \overset{\text {D}}{\longrightarrow }{\chi }_{d}^{2}(\lambda ), \end{aligned}$$

where ${\lambda }=\lim _{n{\rightarrow }{\infty }}{\varvec{\mu }}^{\textrm{T}}{\varvec{\mu }} =\lim _{n{\rightarrow }{\infty }}n({\textbf{R}}{\varvec{\rho }}_{0}-{\textbf{b}})^{\textrm{T}} ({\textbf{R}}{\varvec{\varSigma }}{\textbf{R}}^{\textrm{T}})^{-1} ({\textbf{R}}{\varvec{\rho }}_{0}-{\textbf{b}})$. $\square$