Local influence analysis in general spatial models

Abstract

We study the local influence in the general spatial model which includes the spatial autoregressive model and the spatial error model as two special cases. The stepwise local influence procedure is employed in our diagnostic analysis. We derive the local diagnostic measures in the general spatial model under three perturbation schemes, namely, the variance perturbation, dependent variable perturbation and explanatory variable perturbation schemes. A simulation example and two real-data examples are analysed in detail and they show that the stepwise local influence analysis is effective in identifying influential observations and is a powerful tool for uncovering masking effects.

Appendix: Proofs

1.1 1.   Proof of equation (15)

Using Eq. (10), we have

$$\begin{aligned} \frac{\partial ^{2}{L}}{\partial {\rho }\partial {\rho }}|_{\omega =\mathbf {1}_{n}}&=-tr(C_{1}^{2})-\frac{1}{\sigma ^{2}}y^TW^T_{1}B^TBW_{1}y\\ \frac{\partial ^{2}{L}}{\partial {\rho }\partial {\lambda }}|_{\omega =\mathbf {1}_{n}}&=\frac{\partial ^{2}{L}}{\partial {\lambda }\partial {\rho }}|_{\omega =\mathbf {1}_{n}} =-\frac{1}{\sigma ^{2}}e^TW^T_{2}BW_{1}y-\frac{1}{\sigma ^{2}}e^TB^TW_{2}W_{1}y\\ \frac{\partial ^{2}{L}}{\partial {\rho }\partial {\sigma ^{2}}}|_{\omega =\mathbf {1}_{n}}&=\frac{\partial ^{2}{L}}{\partial {\sigma ^{2}}\partial {\rho }}|_{\omega =\mathbf {1}_{n}} =-\frac{1}{\sigma ^{4}}e^TB^TBW_{1}y\\ \frac{\partial ^{2}{L}}{\partial {\rho }\partial {\beta ^T}}|_{\omega =\mathbf {1}_{n}}&=\left( \frac{\partial ^{2}{L}}{\partial {\beta }\partial {\rho }}\right) ^T|_{\omega =\mathbf {1}_{n}} =-\frac{1}{\sigma ^{2}}y^TW^T_{1}B^TBX\\ \frac{\partial ^{2}{L}}{\partial {\lambda }\partial {\lambda }}|_{\omega =\mathbf {1}_{n}}&=-tr(C_{2}^{2})-\frac{1}{\sigma ^{2}}e^TW^T_{2}W_{2}e,\ \frac{\partial ^{2}{L}}{\partial {\lambda }\partial {\sigma ^{2}}}|_{\omega =\mathbf {1}_{n}} =\frac{\partial ^{2}{L}}{\partial {\sigma ^{2}}\partial {\lambda }}|_{\omega =\mathbf {1}_{n}}\\&=-\frac{1}{\sigma ^{4}}e^TB^TW_{2}e\\ \frac{\partial ^{2}{L}}{\partial {\lambda }\partial {\beta ^T}}|_{\omega =\mathbf {1}_{n}}&=\left( \frac{\partial ^{2}{L}}{\partial {\beta }\partial {\lambda }}\right) ^T|_{\omega =\mathbf {1}_{n}} =-\frac{1}{\sigma ^{2}}e^TB^TW_{2}X-\frac{1}{\sigma ^{2}}e^TW^T_{2}BX\\ \frac{\partial ^{2}{L}}{\partial {\sigma ^{2}}\partial {\sigma ^{2}}}|_{\omega =\mathbf {1}_{n}}&=\frac{n}{2\sigma ^{4}}-\frac{1}{\sigma ^{6}}e^TB^TBe, \ \frac{\partial ^{2}{L}}{\partial {\beta }\partial {\beta ^T}}|_{\omega =\mathbf {1}_{n}} =-\frac{X^TB^TBX}{\sigma ^{2}}\\ \frac{\partial ^{2}{L}}{\partial {\sigma ^{2}}\partial {\beta ^T}}|_{\omega =\mathbf {1}_{n}}&=\left( \frac{\partial ^{2}{L}}{\partial {\beta }\partial {\sigma ^{2}}}\right) ^T|_{\omega =\mathbf {1}_{n}} =-\frac{1}{\sigma ^{4}}e^TB^TBX \end{aligned}$$

where $C_{1}=W_{1}A^{-1}$, $C_{2}=W_{2}B^{-1}$, $e=Ay-X\beta$, $A=I_{n}-\rho W_{1}$, $B=I_{n}-\lambda W_{2}$. From (3) and (4), we have the following equivalent relations:

$$\begin{aligned}&tr({\hat{C}}_{1})=\frac{\displaystyle {1}}{\displaystyle {{\hat{\sigma }}^{2}}}{\hat{e}}^T{\hat{V}}W_{1}y,\quad {\hat{\sigma }}^{2}=\frac{\displaystyle {1}}{\displaystyle {n}}{\hat{e}}^T{\hat{V}}{\hat{e}},\quad \end{aligned}$$

(20)

$$\begin{aligned}&tr({\hat{C}}_{2})=\frac{\displaystyle {1}}{\displaystyle {{\hat{\sigma }}^{2}}}{\hat{e}}^T{\hat{B}}^TW_{2}{\hat{e}},\quad X^T{\hat{V}}{\hat{e}}=\mathbf {0}. \end{aligned}$$

(21)

where ${\hat{V}}={\hat{B}}^T{\hat{B}}$, ${\hat{C}}_{1}=W_{1}{\hat{A}}^{-1}$, ${\hat{C}}_{2}=W_{2}{\hat{B}}^{-1}$, ${\hat{e}}={\hat{A}}y-X{\hat{\beta }}$, ${\hat{A}}=I_{n}-\hat{\rho } W_{1}$ and ${\hat{B}}=I_{n}-\hat{\lambda } W_{2}$. Therefore matrix $\ddot{L}$ defined in Eq. (5) can be obtained to be

$$\begin{aligned} \ddot{L}=-\frac{1}{{\hat{\sigma }}^{2}}\cdot \begin{pmatrix} {\hat{\sigma }}^{2}tr({\hat{C}}_{1}^{2})+{\hat{\eta }}^T_{1}{\hat{\eta }}_{1} &{}\quad {\hat{\eta }}^T_{1}{\hat{\eta }}_{2} &{}\quad tr({\hat{C}}_{1}) &{}\quad {\hat{\eta }}^T_{1}{\hat{B}}X\\ {\hat{\eta }}^T_{1}{\hat{\eta }}_{2} &{}\quad {\hat{\sigma }}^{2}tr({\hat{C}}_{2}^{2})+{\hat{e}}^TW^T_{2}W_{2}{\hat{e}} &{}\quad tr({\hat{C}}_{2}) &{}\quad {\hat{\eta }}^T_{2}{\hat{B}}X\\ tr({\hat{C}}_{1}) &{}\quad tr({\hat{C}}_{2}) &{}\quad \frac{\displaystyle {n}}{\displaystyle {2{\hat{\sigma }}^{2}}} &{}\quad \mathbf {0}\\ X^T{\hat{B}}^T{\hat{\eta }}_{1} &{} X^T{\hat{B}}^T{\hat{\eta }}_{2} &{}\quad \mathbf {0} &{}\quad X^T{\hat{V}}X \end{pmatrix}, \end{aligned}$$

(22)

where ${\hat{\eta }}_{1}={\hat{B}}W_{1}y$, ${\hat{\eta }}_{2}={\hat{B}}^{T-1}({\hat{B}}^TW_{2} +W^T_{2}{\hat{B}}){\hat{e}}$.

1.2 2.   Proof of Theorem 1 (I)

Using Eq. (10), we have

$$\begin{aligned}&\frac{\partial ^{2}{L}}{\partial {\rho }\partial {\omega _{i}}}|_{\omega =\mathbf {1}_{n}} =\frac{1}{\sigma ^{2}}d^T_{i}Be\cdot d^T_{i}BW_{1}y, \ \ i=1,2,\ldots ,n \\&\frac{\partial ^{2}{L}}{\partial {\lambda }\partial {\omega _{i}}}|_{\omega =\mathbf {1}_{n}} =\frac{1}{\sigma ^{2}}d^T_{i}Be\cdot d^T_{i}W_{2}e, \ \ i=1,2,\ldots ,n \\&\frac{\partial ^{2}{L}}{\partial {\sigma ^{2}}\partial {\omega _{i}}}|_{\omega =\mathbf {1}_{n}} =\frac{(d^T_{i}Be)^{2}}{2\sigma ^{4}}, \ \ \frac{\partial ^{2}{L}}{\partial {\beta }\partial {\omega _{i}}}|_{\omega =\mathbf {1}_{n}} =\frac{1}{\sigma ^{2}}d^T_{i}Be\cdot X^TB^Td_{i}, \ \ i=1,2,\ldots ,n \end{aligned}$$

where $\omega _{i}$ represents the ith element of $\omega$ for $i=1,2,\ldots ,n$. By the definition, we immediately obtain the form of $\Delta$ as given in Theorem 1 (I).

1.3 3.   Proof of Theorem 1 (II)

Using Eq. (12), we have

$$\begin{aligned}&\frac{\partial ^{2}{L}}{\partial {\rho }\partial {\omega ^{T}}}|_{\omega =\mathbf {0}_{n}} =\frac{1}{\sigma ^{2}}\left( e^TB^TBW_{1}+y^TW^T_{1}B^TBA\right) \\&\frac{\partial ^{2}{L}}{\partial {\lambda }\partial {\omega ^{T}}}|_{\omega =\mathbf {0}_{n}} =\frac{1}{\sigma ^{2}}\left( e^TW^T_{2}BA+e^TB^TW_{2}A\right) \\&\frac{\partial ^{2}{L}}{\partial {\sigma ^{2}}\partial {\omega ^{T}}}|_{\omega =\mathbf {0}_{n}} =\frac{\displaystyle {e^TB^TBA}}{\displaystyle {\sigma ^{4}}},\quad \frac{\partial ^{2}{L}}{\partial {\beta }\partial {\omega ^{T}}}|_{\omega =\mathbf {0}_{n}} =\frac{1}{\sigma ^{2}}X^TB^TBA, \end{aligned}$$

which immediately result in Theorem 1 (II).

1.4 4.   Proof of Theorem 1 (III)

Using Eq. (14), we have

$$\begin{aligned}&\frac{\partial ^{2}{L}}{\partial {\rho }\partial {\omega ^{T}}}|_{\omega =\mathbf {0}_{n}} =-\frac{1}{\sigma ^{2}}l^T_{p}\beta \cdot y^TW^T_{1}B^TB \\&\frac{\partial ^{2}{L}}{\partial {\lambda }\partial {\omega ^{T}}}|_{\omega =\mathbf {0}_{n}} =-\frac{1}{\sigma ^{2}}l^T_{p}\beta \cdot \left( e^TW^T_{2}B+e^TB^TW_{2}\right) \\&\frac{\partial ^{2}{L}}{\partial {\sigma ^{2}}\partial {\omega ^{T}}}|_{\omega =\mathbf {0}_{n}} =-\frac{\displaystyle {1}}{\displaystyle {\sigma ^{4}}}l^T_{p}\beta \cdot e^TB^TB\\&\frac{\partial ^{2}{L}}{\partial {\beta }\partial {\omega ^{T}}}|_{\omega =\mathbf {0}_{n}} =\frac{1}{\sigma ^{2}}\left( l_{p}e^TB^TB-l^T_{p}\beta \cdot X^TB^TB\right) \end{aligned}$$

where $l_{p}$ is a $k\times 1$ vector with the pth element equal to 1 and the rest equal to 0. Thus, $\Delta$ given in Theorem 1 (III) can be easily obtained.

1.5 5. Matlab code for computing diagnostics

Following is a function used to calculate the local diagnostic in stepwise local influence analysis. The method refers to Sect. 3.2 or Shi and Huang (2011).