Automatic Structure Identification of Semiparametric Spatial Autoregressive Model Based on Smooth-Threshold Estimating Equation

Abstract

Issues concerning spatial dependence among cross-sectional units in econometrics have received more and more attention, while in statistical modeling, rarely can the analysts have a priori knowledge of the dependency relationship of the response variable with respect to independent variables. This paper proposes an automatic structure identification and variable selection procedure for semiparametric spatial autoregressive model, based on the generalized method of moments and the smooth-threshold estimating equations. The novel method is easily implemented without solving any convex optimization problems. Model identification consistency is theoretically established in the sense that the proposed method can automatically separate the linear and zero components from the varying ones with probability approaching to one. Detailed issues on computation and turning parameter selection are discussed. Some Monte Carlo simulations are conducted to demonstrate the finite sample performance of the proposed procedure. Two empirical applications on Boston housing price data and New York leukemia data are further considered.

Appendix

The following regularity conditions are required for technical proof.

(C1):

All diagonal elements of the weight matrix W are zeros. The matrix $I_n-\rho W$ is nonsingular for all $\rho$ in its compact support $\left( -1/|\lambda _{min}(W)|, 1/\lambda _{max}(W) \right)$, where $\lambda _{min}(W)$ and $\lambda _{max}(W)$, respectively, denote the minimum and maximum eigenvalues of W. Besides, the matrices W and $(I_n-\rho _0 W)^{-1}$ are uniformly bounded in absolute value in both row and column sums.

(C2):

The elements of matrix $H_n$ are uniformly bounded, and the regressors $x_i$, $i=1,\ldots ,n$ are nonstochastic with bounded support.

(C3):

The density function $f_u(\cdot )$ of random variable u is positive and has a continuous second derivative. The matrix $\varGamma (u)$ is nonsingular and Lipschitz continuous.

(C4):

The functions $\beta _{0\,s}(\cdot )$, $l=1,\ldots ,q$ are rth continuously differentiable over the interval (0, 1) with $r\ge 2$.

(C5):

The kernel function $K(\cdot )$ is a symmetric density function with a compact support $[-1,1]$ and a bounded first-order derivative.

(C6):

The random errors $\varepsilon _i$’s are independent, and $E (|\varepsilon _i|^{2+\varpi })<\infty$ for some $\varpi >0$.

(C7):

At least one nonconstant regressor in X must have significant effect on the response variable, or $\beta _0(u)\ne 0_q$ over at least one nonempty interval.

Note that conditions (C1) and (C2) are frequently postulated in spatial econometric literature including [16, 19, 34, 39], etc. Particularly, the uniform boundedness of W and $(I_n-\rho W)^{-1}$ in condition (C1) aims to limit the spatial correlation among spatial units to a manageable degree, which plays an important role in the asymptotic properties of estimators. When W is row-normalized, condition (C1) can be satisfied with $\rho \in (-1,1)$, see [18] for particular interpretations. Conditions (C3)–(C5) are regular conditions assumed in the local polynomial regression of VCPLM such as [4, 15] and the references therein. Condition (C6) is a necessary condition for estimation consistency. Condition (C7) is postulated to avoid a pure spatial autoregressive model so as to ensure that the valid instrumental variables can be generated and the proposed local linear GMM estimators are consistent. One may refer to [19] and the references therein for deeper discussions on this issue.

To proceed with the proofs of our theoretical properties, we first quote the following lemma which will be frequently used in the sequel.

Lemma 6.1

Let $(X_1,Y_1),...,(X_n,Y_n)$ be i.i.d. random vectors, where ${Y_i}'s$ are scalar random variables, and f denotes the joint density of (X, Y). Let K be a bounded positive function with bounded support, satisfying Lipschitz condition. Further assume that $\sup _{\textbf{x}} \int |y|^r f(\textbf{x},y)dy < \infty$ and $E|Y|^r < \infty$. Then,

$$\begin{aligned} \mathop {\sup }\limits _{\textbf{x}\in D} \left| \frac{1}{n} \sum \limits _{i = 1}^n{\{K_h(X_i-\textbf{x})Y_i-E[K_h(X_i-\textbf{x})Y_i]\}}\right| = O_p \left( \frac{\log ^{1/2}(1/h)}{\sqrt{nh}}\right) , \end{aligned}$$

provided that $n^{2\varepsilon -1}h \rightarrow \infty$ for some $\varepsilon < 1-r^{-1}$.

The proof of Lemma 6.1 can be referred to [27].

Proof of Theorem 2.2

The asymptotic normality of $\hat{\beta }(u;\hat{\rho })$ can be directly obtained from [48], so we focus on the part of $\hat{\beta }^{\prime }(u;\hat{\rho })$. Adopting the notations in Sect. 2, it follows from Eq. (2.1) with h replaced by $h_1$ that

$$\begin{aligned} \hat{\beta }^{\prime }(u,\hat{\rho })= & {} ({\textbf {0}}_p,I_p)\left( D_u^T K_u D_u \right) ^{-1}D_u^T K_u (Y-\hat{\rho }Y) \\= & {} (0_p,I_p)\left( D_u^T K_u D_u \right) ^{-1}D_u^T K_u \left\{ M_0+\varepsilon +(\rho _0-\hat{\rho })WY \right\} , \end{aligned}$$

where ${\textbf {0}}_p$ means a $p\times p$ matrix with all components being equal to 0. Thus, we have

$$\begin{aligned} \sqrt{nh_1}(\hat{\beta }^{\prime }(u,\hat{\rho })-\beta ^{\prime }_0(u))= & {} \sqrt{nh_1}\left\{ ({\textbf {0}}_p,I_p)\left( D_u^T K_u D_u \right) ^{-1}D_u^T K_u (Y-\hat{\rho }Y)- \beta ^{\prime }_0(u) \right\} \nonumber \\= & {} R_{n1}(u) + R_{n2}(u) + R_{n3}(u), \end{aligned}$$

(6.1)

where

$$\begin{aligned}{} & {} R_{n1}(u)=\sqrt{nh_1} \left\{ ({\textbf {0}}_p,I_p)\left( D_u^T K_u D_u \right) ^{-1}D_u^T K_u M_0-\beta ^{\prime }_0(u) \right\} , \\{} & {} R_{n2}(u)=\sqrt{nh_1} ({\textbf {0}}_p,I_p)\left( D_u^T K_u D_u \right) ^{-1}D_u^T K_u \varepsilon , \\{} & {} R_{n3}(u)=\sqrt{nh_1} ({\textbf {0}}_p,I_p)\left( D_u^T K_u D_u \right) ^{-1}D_u^T K_u(\rho _0-\hat{\rho })WY. \end{aligned}$$

Next, we mainly consider $R_{n1}(u)$ since the properties of $R_{n2}(u)$ and $R_{n3}(u)$ can be similarly derived from [48]. Note that

$$\begin{aligned} D_u^TK_uD_u = \left( \begin{array}{cc} \sum _{i=1}^n{ Z_iZ_i^TK_{h_1}(U_i-u) } &{} \sum _{i=1}^n{ Z_iZ_i^T (U_i-u) K_{h_1}(U_i-u) } \\ \sum _{i=1}^n{ Z_iZ_i^T (U_i-u) K_{h_1}(U_i-u) } &{} \sum _{i=1}^n{ Z_iZ_i^T (U_i-u)^2 K_{h_1}(U_i-u) } \end{array} \right) . \end{aligned}$$

Each element of the above matrix is in the form of a kernel regression. Hence, by Lemma 6.1, we have

$$\begin{aligned} \frac{1}{n}D_u^TK_uD_u = f_U(u)\left( \begin{array}{cc} \varGamma (u) &{} 0 \\ 0 &{} \mu _2h_1^2 \varGamma (u) \end{array} \right) \left\{ 1+O_p(c_n) \right\} , \end{aligned}$$

(6.2)

where $c_n= \{\log (1/h_1)/ nh_1 \}^{1/2}+h_1^2$.

In addition, since the functions $\beta _{0}(\cdot )$ are presumed to be second-order continuously differentiable, it follows from Taylor expansion that

$$\begin{aligned} M_0-D_u g_0(u)= & {} \left( \begin{array}{c} Z_1^T[\beta _{0}(U_1)-\beta _{0}(u)-\beta _{0}^{\prime }(u)(U_1-u)] \\ \vdots \\ Z_n^T[\beta _{0}(U_n)-\beta _{0}(u)-\beta _{0}^{\prime }(u)(U_n-u)] \end{array} \right) \\= & {} \left( \begin{array}{c} \frac{1}{2}Z_1^T\beta _{0}^{\prime \prime }(u)(U_1-u)^2 \{ 1+o((U_1-u)^2)\} \\ \vdots \\ \frac{1}{2}Z_n^T\beta _{0}^{\prime \prime }(u)(U_n-u)^2 \{ 1+o((U_n-u)^2)\} \end{array} \right) , \end{aligned}$$

where $g_0(u)=\left( \beta _0(u)^T,\beta ^{\prime }_0(u)^T \right) ^T$. Based on some direct calculations and Lemma 6.1, it follows

$$\begin{aligned}{} & {} n^{-1}D_u^TK_u [M_0-D_u g_0(u)] \nonumber \\= & {} \left( \begin{array}{c} \frac{1}{2} f_U(u) \mu _2 h_1^2 \varGamma (u)\alpha _{0}^{\prime \prime }(u) \\ 0 \end{array} \right) (1+o(h_1^2))(1+O_p(c_n)). \end{aligned}$$

(6.3)

Therefore, applying (6.2) and (6.3) to $R_{n1}(u)$ yields

$$\begin{aligned} R_{n1}(u)= & {} \sqrt{nh_1} ({\textbf {0}}_p,I_p)\left( D_u^T K_u D_u \right) ^{-1}D_u^T K_u \left\{ M_0- D_u g_0(u) \right\} \nonumber \\= & {} \sqrt{nh_1} ({\textbf {0}}_p,I_p)\left( \frac{1}{n}D_u^T K_u D_u \right) ^{-1}\frac{1}{n}D_u^T K_u \left\{ M_0- D_u g_0(u) \right\} \nonumber \\= & {} o\big \{ (nh_1^5)^{1/2} \big \}(1+O_p(c_n))=o_p(1), \end{aligned}$$

(6.4)

where the last equality holds due to the bandwidth assumption $h_1=O(n^{-1/5})$.

For the terms $R_{n2}(u)$ and $R_{n3}(u)$, exactly following the same line as done in the proof of Theorem 2 in [48], wherein the 2-dimensional index variable ${\textbf {s}}$ is replaced by scalar one u and $(I_p,{\textbf {0}}_{2p})$ in $\varDelta _{n2}({\textbf {s}}),\varDelta _{n3}({\textbf {s}})$ is replaced by $({\textbf {0}}_{p},I_p)$, we can obtain that

$$\begin{aligned} R_{n2}(u) ~~\mathop \rightarrow \limits ^d~~ N\left( 0, h_1^{-2}\nu _2\sigma ^2(u)\left\{ \mu _2^2 f_U(u)\varGamma (u)\right\} ^{-1}\right) , ~~~~R_{n3}(u)=o_p(1).\nonumber \\ \end{aligned}$$

(6.5)

As a consequence, we derive the following conclusion by uniting results (6.1), (6.4) and (6.5), that is,

$$\begin{aligned} \sqrt{nh_1^3}\left( \hat{\beta }^{\prime }(u;\hat{\rho })-\beta _0^{\prime }(u)\right) ~~\mathop \rightarrow \limits ^d~~ N\left( 0, \nu _2\sigma ^2(u)\left\{ \mu _2^2 f_U(u)\varGamma (u)\right\} ^{-1}\right) . \end{aligned}$$

This completes the proof. $\square$

Proof of Theorem 3.1

From Theorem 2.2 and the assumption $h_1=O(n^{-1/5})$, we should always keep in mind the fact, which plays a crucial role in this proof, that $\hat{\beta }(u)$ and $\hat{\beta }^{\prime }(u)$ are consistent estimators of $\beta _0(u)$ and $\beta _0^{\prime }(u)$ with convergence rates $n^{2/5}$ and $n^{1/5}$, respectively. Suppose $\eta >0$ be an arbitrary positive number, we now present the proof by the following steps.

Step 1: If $k \in {\mathcal {A}}_{v}$, we have $\Vert n^{-1/2}{\hat{a}}_k\Vert >0$ and $\Vert n^{-1/2}{\hat{b}}_k\Vert >0$. Thus,

$$\begin{aligned}{} & {} P \left( \frac{\lambda _1}{\Vert n^{-1/2}{\hat{a}}_k\Vert ^{1+\gamma _1}}> n^{-2/5}\eta \right) \\= & {} P \left( \lambda _1 n^{2/5}/\eta> \Vert n^{-1/2}{\hat{a}}_k\Vert ^{1+\gamma _1} \right) \\\le & {} P \left( \left( \lambda _1 n^{2/5}/\eta \right) ^{1/(1+\gamma _1)} > \min _{k\in {\mathcal {A}}_{v}} \Vert n^{-1/2}{\hat{a}}_k\Vert - O_p(n^{-2/5}) \right) \\\rightarrow & {} 0 \end{aligned}$$

by condition $n^{2/5}\lambda _1 \rightarrow 0$, and

$$\begin{aligned}{} & {} P \left( \frac{\lambda _2}{\Vert n^{-1/2}{\hat{b}}_k\Vert ^{1+\gamma _2}}> n^{-1/5}\eta \right) \\= & {} P \left( \lambda _2 n^{1/5}/\eta> \Vert n^{-1/2}{\hat{b}}_k\Vert ^{1+\gamma _2} \right) \\\le & {} P \left( \left( \lambda _2 n^{1/5}/\eta \right) ^{1/(1+\gamma _2)} > \min _{k\in {\mathcal {A}}_{v}} \Vert n^{-1/2}{\hat{b}}_k\Vert - O_p(n^{-2/5}) \right) \\\rightarrow & {} 0 \end{aligned}$$

by condition $n^{1/5}\lambda _2 \rightarrow 0$. This implies,

$$\begin{aligned}{} & {} \hat{\delta }_{1k}=\min \left\{ 1,\frac{\lambda _1}{|n^{-1/2}{\hat{a}}_k|^{1+\gamma _1}} \right\} = o_p(n^{-2/5}),\\{} & {} \hat{\delta }_{2k}=\min \left\{ 1,\frac{\lambda _2}{|n^{-1/2}{\hat{b}}_k|^{1+\gamma _2}} \right\} = o_p(n^{-1/5}). \end{aligned}$$

Accordingly, we have

$$\begin{aligned} P \left( \hat{\delta }_{1k}< 1, ~\hat{\delta }_{2k}<1, ~ \text{ for } \text{ all }~ k\in {\mathcal {A}}_{v} \right) \rightarrow 1. \end{aligned}$$

(6.6)

Step 2: If $k \in {\mathcal {A}}_{v}^c$, that means $k \in {\mathcal {A}}_{c}$ or $k \in {\mathcal {A}}_{z}$.

Step 2.1: If $k \in {\mathcal {A}}_{c}$, then $\Vert n^{-1/2}{\hat{a}}_k\Vert >0$ and $\Vert n^{-1/2}{\hat{b}}_k\Vert =O_p(n^{-1/5})$. By Step 1, we have $P \left( \hat{\delta }_{1k} < 1 ~ \text{ for } \text{ all }~ k\in {\mathcal {A}}_{c} \right) \rightarrow 1$. Moreover, by condition $n^{(1+\gamma _2)/5}\lambda _2 \rightarrow \infty$,

$$\begin{aligned} P \left( \frac{\lambda _2}{\Vert n^{-1/2}{\hat{b}}_k\Vert ^{1+\gamma _2}}<1 \right) = P \left( O_p\left( \lambda _2 n^{(1+\gamma _2)/5} \right) <1 \right) \rightarrow 0, \end{aligned}$$

which is equivalent to $P \left( \hat{\delta }_{2k} = 1 ~ \text{ for } \text{ all }~ k\in {\mathcal {A}}_{c} \right) \rightarrow 1$. Consequently, we obtain

$$\begin{aligned} P \left( \hat{\delta }_{1k} < 1,~\hat{\delta }_{2k} = 1, ~ \text{ for } \text{ all }~ k\in {\mathcal {A}}_{c} \right) \rightarrow 1. \end{aligned}$$

(6.7)

Step 2.2: If $k \in {\mathcal {A}}_{z}$, then $\Vert n^{-1/2}{\hat{a}}_k\Vert =O_p(n^{-2/5})$ and $\Vert n^{-1/2}{\hat{b}}_k\Vert =O_p(n^{-1/5})$. By Step 2.1, we have $P \left( \hat{\delta }_{2k} = 1 ~ \text{ for } \text{ all }~ k\in {\mathcal {A}}_{z} \right) \rightarrow 1$. Besides, by condition $n^{2(1+\gamma _1)/5}\lambda _1 \rightarrow \infty$,

$$\begin{aligned} P \left( \frac{\lambda _1}{\Vert n^{-1/2}{\hat{a}}_k\Vert ^{1+\gamma _1}}<1 \right) = P \left( O_p\left( \lambda _1 n^{2(1+\gamma _2)/5} \right) <1 \right) \rightarrow 0, \end{aligned}$$

which is equivalent to $P \left( \hat{\delta }_{1k} = 1 ~ \text{ for } \text{ all }~ k\in {\mathcal {A}}_{z} \right) \rightarrow 1$. As a result,

$$\begin{aligned} P \left( \hat{\delta }_{1k} = 1,~\hat{\delta }_{2k} = 1, ~ \text{ for } \text{ all }~ k\in {\mathcal {A}}_{z} \right) \rightarrow 1. \end{aligned}$$

(6.8)

Combining expressions (6.6), (6.7) and (6.8), it follows that

$$\begin{aligned} P \left( {\mathcal {A}}_{v}=\hat{{\mathcal {A}}}_{v} \right) \rightarrow 1. \end{aligned}$$

(6.9)

Step 3: If $k \in {\mathcal {A}}_{c}$, from Step 2.1, we have $P \left( \hat{\delta }_{1k} < 1,~\hat{\delta }_{2k} = 1, ~ \text{ for } \text{ all }~ k\in {\mathcal {A}}_{c} \right) \rightarrow 1$. If $k \in {\mathcal {A}}_{c}^c$, then $k \in {\mathcal {A}}_{v}$ or $k \in {\mathcal {A}}_{z}$. When $k \in {\mathcal {A}}_{v}$, it follows from Step 1 that $P \left( \hat{\delta }_{1k}< 1, ~\hat{\delta }_{2k}<1, ~ \text{ for } \text{ all }~ k\in {\mathcal {A}}_{v} \right) \rightarrow 1$, while $P \left( \hat{\delta }_{1k} = 1,~\hat{\delta }_{2k} = 1, ~ \text{ for } \text{ all }\right. \left. k\in {\mathcal {A}}_{z} \right) \rightarrow 1$ by Step 2.2 when $k \in {\mathcal {A}}_{z}$. Therefore, we have

$$\begin{aligned} P \left( {\mathcal {A}}_{c}=\hat{{\mathcal {A}}}_{c} \right) \rightarrow 1. \end{aligned}$$

(6.10)

Step 4: If $k \in {\mathcal {A}}_{z}$, from Step 2.2, we have $P \left( \hat{\delta }_{1k} = 1,~\hat{\delta }_{2k} = 1, ~ \text{ for } \text{ all }~ k\in {\mathcal {A}}_{z} \right) \rightarrow 1$. If $k \in {\mathcal {A}}_{z}^c$, then $k \in {\mathcal {A}}_{v}$ or $k \in {\mathcal {A}}_{c}$. If $k \in {\mathcal {A}}_{v}$, $P \left( \hat{\delta }_{1k}< 1, ~\hat{\delta }_{2k}<1, ~ \text{ for } \text{ all }~ k\in {\mathcal {A}}_{v} \right) \rightarrow 1$ holds from Step 1. If $k \in {\mathcal {A}}_{c}$, we have $P \left( \hat{\delta }_{1k} < 1,~\hat{\delta }_{2k} = 1, ~ \text{ for } \text{ all }~ k\in {\mathcal {A}}_{c} \right) \rightarrow 1$ from Step 2.1. Hence,

$$\begin{aligned} P \left( {\mathcal {A}}_{z}=\hat{{\mathcal {A}}}_{z} \right) \rightarrow 1. \end{aligned}$$

(6.11)

Note that the sets ${\mathcal {A}}_{v}$, ${\mathcal {A}}_{c}$ and ${\mathcal {A}}_{z}$ constitute a partition of the covariate indexes. Namely, ${\mathcal {A}}_{v}\cup {\mathcal {A}}_{c}\cup {\mathcal {A}}_{z}=\{1,2,\ldots ,p\}$ and the intersection of any two sets is empty. This together with results (6.9), (6.10) and (6.11) implies

$$\begin{aligned} P\left( \hat{{\mathcal {A}}}_{v}={\mathcal {A}}_{v},~~\hat{{\mathcal {A}}}_{c}={\mathcal {A}}_{c},~~\hat{{\mathcal {A}}}_{z}={\mathcal {A}}_{z} \right) \rightarrow 1. \end{aligned}$$

This completes the proof. $\square$