Spatial Simultaneous Autoregressive Models for Compositional Data: Application to Land Use

Abstract

Econometric land use models study determinants of land use shares of different classes: “agriculture”, “forest”, “urban” and “other” for example. Land use shares have a compositional nature as well as an important spatial dimension. We compare two compositional regression models with a spatial autoregressive nature in the framework of land use. We study the impact of the choice of coordinate space and prove that a choice of coordinate representation does not have any impact on the parameters in the simplex as long as we do not impose further restrictions. We discuss parameters interpretation taking into account the non-linear structure as well as the spatial dimension. In order to assess the explanatory variables impact, we compute and interpret the semi-elasticities of the shares with respect to the explanatory variables and the spatial impact summary measures.

6 Appendix

1.1 6.1 Matrix Product Expression Using Alr Transformation

Let us consider the $\text {alr}_D$ transformation $\text {alr}_D(\mathbf {u})= \mathbf {F}_D\ln (\mathbf {u})$ for a vector $\mathbf {u} \in \mathbf {S}^D$ and with $\mathbf {F}_D= \left[ \mathbf {I}_{D-1} \, -\mathbf {j}_{D-1}\right]$. The result can be generalized to any $\text {alr}_m$ transformation.

$$\begin{aligned} \text {alr}_D\left( \mathbf {B} \boxdot \mathbf {u}\right) =\mathbf {F}_D\ln \left( \mathbf {B} \boxdot \mathbf {u}\right) = \mathbf {F}_D\ln \left( \text {ilr}^{-1}_V\left( \mathbf {B}^*_{V} \text {ilr}_V(\mathbf {u})\right) \right) , \end{aligned}$$

where $\mathbf {V}\mathbf {B}^*_V=\mathbf {B}\mathbf {V}$. Since $\mathbf {V} \mathbf {V}^T=\mathbf {I}_D -\mathbf {j}_D \mathbf {j}_D^T/D$ and $\mathbf {F}_D \mathbf {j}_D=\mathbf {0}_D$, we have

$$\begin{aligned} \text {alr}_D\left( \mathbf {B} \boxdot \mathbf {u}\right) =\mathbf {F}_D\ln \left( \mathcal {C}\left( \exp (\mathbf {V}\mathbf {B}^*_V \text {ilr}_V(\mathbf {u})\right) \right) = \mathbf {F}_D \mathbf {V} \mathbf {B}^*_V \text {ilr}_V(\mathbf {u}) = \mathbf {F}_D \mathbf {B} \ln (\mathbf {u}). \end{aligned}$$

1.2 6.2 Writing the Model in Reduced Form in Coordinate Space

Recognizing that the matrix form of the term $\sum _{m \in S_l^{\mathbf {WY}^*}} \mathbf {R}^*_{ml}\mathbf {W} \mathbf {Y}_{.m}^*$ is $\mathbf {W}\mathbf {Y}^*\mathbf {R}^*$ , we get easily that the matrix formulation of the model can be written

$$\begin{aligned} \mathbf {Y}^* = \mathbf {W}\mathbf {Y}^*\mathbf {R}^* + \begin{bmatrix} \mathbf {X}_{S_1^{\mathbf {X}}}\pmb {\beta }_{S_1^\mathbf {X}}^*&\ldots&\mathbf {X}_{S_{D-1}^{\mathbf {X}}}\pmb {\beta }^*_{S^\mathbf {X}_{D-1}} \end{bmatrix} + \pmb {\epsilon }^*. \end{aligned}$$

We are going to use the following property: if we have three matrices $\mathbf {A},$ $\mathbf {B}$ and $\mathbf {C}$ such that the number of columns of $\mathbf {A}$ is equal to the number of rows of $\mathbf {B}$ and the number of columns of $\mathbf {B}$ is equal to the number of rows of $\mathbf {C},$ then

$$\begin{aligned} \text {vec}_c(\mathbf {ABC}) = (\mathbf {C}^T\otimes \mathbf {A})\text {vec}_c(\mathbf {B}). \end{aligned}$$

(23)

Using (23), we get

$$\begin{aligned} \text {vec}_c ( \mathbf {W}\mathbf {Y}^*\mathbf {R}^*)= \left( (\mathbf {R}^*)^T\otimes \mathbf {W}\right) \text {vec}_c(\mathbf {Y}^*). \end{aligned}$$

Therefore the c-vectorization of the whole model is

$$\begin{aligned} \text {vec}_c \mathbf {Y}^* = \left( (\mathbf {R}^*)^T\otimes \mathbf {W}\right) \text {vec}_c(\mathbf {Y}^*) + \begin{bmatrix} \mathbf {X_{S_1}} \pmb {\beta }_1^*\\ . \\ \mathbf {X_{S_{D-1}}}\pmb {\beta }_{D-1}^* \end{bmatrix}+\text {vec}_c\pmb {\epsilon }^* \end{aligned}$$

and the reduced form of the model in vectorized form is

$$\begin{aligned} \text {vec}_c \mathbf {Y}^* = (\mathbf {I}_{n(D-1)} - ((\mathbf {R}^*)^T\otimes \mathbf {W}))^{-1}\begin{bmatrix} \mathbf {X_{S_1}} \pmb {\beta }_1^*\\ . \\ \mathbf {X_{S_{D-1}}}\pmb {\beta }_{D-1}^* \end{bmatrix} +\text {vec}_c\pmb {\epsilon }^*. \end{aligned}$$

1.3 6.3 Lemma Used in the Proof of Theorem 1

Lemma 1

Let $\mathbf {j}_D$ (resp. $\pmb {0}_D$) denote the D-dimensional column vector of ones (resp. zeros) and $\mathbf {I}_D$ be the $D \times D$ identity matrix. Let $\mathbf {K}=\begin{bmatrix} \mathbf {I}_{D-1} - \mathbf {j}_{D-1}\mathbf {j}_{D-1}^T/D\\ -\mathbf {j}_{D-1}^T/D \end{bmatrix}$ a $D \times (D-1)$ matrix and $\mathbf {F}= \left[ \mathbf {I}_{D-1} \, -\mathbf {j}_{D-1}\right]$   a   $(D-1) \times D$ matrix. Let $\mathbf {B}$ be a $D \times D$ matrix such that $\mathbf {B} \mathbf {j}_D = \pmb {0}_D$ and $\mathbf {B}^T\mathbf {j}_D= \pmb {0}_D$ and $\mathbf {B}^*$ be a $(D-1) \times (D-1)$ matrix. We have that $\mathbf {F}\mathbf {B}=\mathbf {B}^*\mathbf {F}$ is equivalent to $\mathbf {B}=\mathbf {K} \mathbf {B}^* \mathbf {F}$.

We have

$$ \mathbf {K}= \begin{bmatrix} 1-\frac{1}{D} &{} -\frac{1}{D} &{}-\frac{1}{D} &{}\ldots &{} -\frac{1}{D} \\ -\frac{1}{D} &{} 1-\frac{1}{D} &{}-\frac{1}{D} &{}\ldots &{}-\frac{1}{D} \\ -\frac{1}{D} &{}-\frac{1}{D} &{} 1-\frac{1}{D} &{}\ldots &{}-\frac{1}{D} \\ \vdots &{}\vdots &{}\vdots &{}\ddots &{}\vdots \\ -\frac{1}{D} &{}-\frac{1}{D} &{} \frac{1}{D} &{}\ldots &{} 1-\frac{1}{D} \\ -\frac{1}{D} &{}-\frac{1}{D} &{} -\frac{1}{D} &{}\ldots &{}-\frac{1}{D} \end{bmatrix} \;\; \text { and } \;\; \mathbf {F} = \begin{bmatrix} 1 &{} 0 &{} 0 &{}\ldots &{}0 &{}-1 \\ 0 &{} 1 &{} 0 &{} \ldots &{} 0 &{}-1 \\ 0 &{} 0 &{} 1 &{} \ldots &{} 0 &{}-1 \\ \vdots &{} \vdots &{} &{}\ddots &{}0 &{} -1\\ 0 &{} 0 &{} 0 &{} &{} 1 &{} -1. \end{bmatrix} $$

Let $\mathbf {B}$ be a $D \times D$ matrix such that $\mathbf {B} \mathbf {j}_D = \pmb {0}_D$ and $\mathbf {B}^T\mathbf {j}_D= \pmb {0}_D$ and $\mathbf {B}^*$ be a $(D-1) \times (D-1)$ matrix. First, let us prove that if $\mathbf {B}=\mathbf {K} \mathbf {B}^* \mathbf {F}$, then $\mathbf {F}\mathbf {B}=\mathbf {B}^*\mathbf {F}$. It is easy to show that $\mathbf {F}\mathbf {K}= \mathbf {I}_{D-1}$. Thus, if $\mathbf {B} = \mathbf {KB}^*\mathbf {F}$,

$$\begin{aligned} \mathbf {FB} = \mathbf {FKB}^*\mathbf {F}= \mathbf {I}_{D-1}\mathbf {B}^*\mathbf {F} = \mathbf {B}^*\mathbf {F}. \end{aligned}$$

Let us now prove the converse. We have $\mathbf {B} \mathbf {j}_D= \mathbf {B}^T\mathbf {j}_D= \pmb {0}_D$ and we assume $\mathbf {FB}=\mathbf {B}^*\mathbf {F}$. We write $\mathbf {B}$ in blocks as follows:

$$\begin{aligned} \mathbf {B} = \begin{bmatrix} \mathbf {B}_1 &{}\mathbf {b}_2\\ \mathbf {b}_3^T &{} b_4, \end{bmatrix} \end{aligned}$$

where $\mathbf {B}_1$ is a $D-1 \times D-1$ matrix, $\mathbf {b}_2$ and $\mathbf {b}_3^T\in \mathbf {R}^{D-1}$ and $\mathbf {b}_4 \in \mathbf {R}$. Using the fact that $\mathbf {B} \mathbf {j}_D= \mathbf {B}^T\mathbf {j}_D= \pmb {0}_D$, we get

$$\begin{aligned} {\left\{ \begin{array}{ll} \mathbf {b}_2 =- \mathbf {B}_1\mathbf {j}_{D-1}\\ \mathbf {b}_3 = -\mathbf {B}_1^T\mathbf {j}_{D-1}\\ \mathbf {b}_4 = \mathbf {j}_{D-1}^T \mathbf {B}_1\mathbf {j}_{D-1} \end{array}\right. }. \end{aligned}$$

(24)

To find $\mathbf {B}$, we only need to find $\mathbf {B}_1$. Using (24), we write the $(D-1) \times D$ matrix $\mathbf {FB}$ as a function of $\mathbf {B}_1$:

$$\begin{aligned} \mathbf {FB}&= [(\mathbf {I}_{D-1}+ \mathbf {j}_{D-1}\mathbf {j}_{D-1}^T)\mathbf {B}_1 \ \ \ \ -(\mathbf {I}_{D-1} + \mathbf {j}_{D-1}\mathbf {j}_{D-1}^T)\mathbf {B}_1\mathbf {j}_{D-1}]. \end{aligned}$$

Furthermore, $\mathbf {B}^*\mathbf {F} = [\mathbf {B}^*\ \ -\mathbf {B}^*\mathbf {j}_{D-1}]$. So, $\mathbf {FB} = \mathbf {B}^*\mathbf {F}$ implies

$$\begin{aligned} \left( \mathbf {I}_{D-1}+ \mathbf {j}_{D-1}\mathbf {j}_{D-1}^T\right) \mathbf {B}_1 = \mathbf {B}^*. \end{aligned}$$

The inverse matrix of $\left( \mathbf {I}_{D-1}+ \mathbf {j}_{D-1}\mathbf {j}_{D-1}^T\right)$ is $\left( \mathbf {I}_{D-1}+ \mathbf {j}_{D-1}\mathbf {j}_{D-1}^T/D\right)$. Thus,

$$\begin{aligned} \mathbf {B}_1 = (\mathbf {I}_{D-1}+ \mathbf {j}_{D-1}\mathbf {j}_{D-1}^T/D)\mathbf {B}^*. \end{aligned}$$

(25)

Using (24) and (25), it is now easy to check that $\mathbf {B}=\mathbf {K} \mathbf {B}^* \mathbf {F}$.