Spatial-Lag Models and Simultaneous Feedback

Why a Spatial Lag Is a Simultaneous Relationship

When an outcome in one location depends on outcomes in nearby locations, those neighboring outcomes may themselves depend on the first location’s outcome. The resulting feedback is not a sequence in which one unit’s outcome is fixed before the others are determined. Instead, all outcomes are jointly determined. This distinction matters because the spatially lagged outcome is then generally correlated with the structural disturbance, even if the observed regressors are exogenous.

Let there be \(n\) spatial units. The known \(n\times n\) matrix \(W\) records the chosen spatial weights: its entry \(w_{ij}\) describes the weight assigned to unit \(j\) in the spatial lag for unit \(i\). The vector \(y\) contains the outcomes, \(X\) is an \(n\times k\) matrix of regressors, \(\beta\) is a \(k\)-vector of coefficients, and \(\varepsilon\) is the vector of structural disturbances. The scalar \(\rho\) measures the strength and direction of dependence on the spatially lagged outcome.

A common specification has \(w_{ii}=0\), so a unit does not receive a direct weight on its own outcome. This does not remove feedback: unit \(i\) can affect a neighbor whose outcome then contributes to the spatial lag for unit \(i\). Whether weights are row-standardized or otherwise normalized depends on how \(W\) was constructed. Here \(W\) is treated as fixed and known; the algebra does not require a particular normalization.

Solving the Simultaneous System

Move the spatial-lag term to the left-hand side. If the matrix \(I-\rho W\) is nonsingular, the simultaneous equations have the unique solution

This expression is an algebraic solution conditional on nonsingularity; it does not by itself establish that a chosen specification is stable or appropriate. The solution shows that every outcome can depend on the covariates and disturbances across the system, not only on those of the same unit. For the disturbance component, define \(A=(I-\rho W)^{-1}\), so that \(y=AX\beta+A\varepsilon\). The spatial lag is consequently \(Wy=WAX\beta+WA\varepsilon\).

Why the Spatial Lag Is Endogenous

To make the source of endogeneity explicit, suppose the disturbances satisfy the conditional exogeneity condition \(E[\varepsilon\mid X,W]=0\). For a simple covariance calculation, also suppose

This covariance assumption means that, conditional on \(X\) and \(W\), disturbances have the same variance and zero covariance across distinct units. It does not require normality. Since \(A\), \(W\), and \(X\) are fixed conditional on \(X,W\), the solution gives \(Wy=WAX\beta+WA\varepsilon\). The first term is fixed under that conditioning, while the second transmits disturbances into the spatial lag. In particular,

Thus the spatial lag will generally be correlated with the disturbance vector. For the usual regression moment involving the spatial-lag regressor, the corresponding conditional expected cross-product is

The equality follows by writing \(Wy\) as its conditional mean plus \(WA\varepsilon\): the fixed conditional-mean term has zero expected product with \(\varepsilon\), and \(E[\varepsilon'(WA)'\varepsilon\mid X,W]=\sigma^2\operatorname{tr}((WA)')=\sigma^2\operatorname{tr}(WA)\). For a standard zero-diagonal weights matrix and nonzero feedback, this trace is typically nonzero. The exact covariance depends on \(W\) and \(\rho\), so “generally endogenous” does not mean that every conceivable configuration must produce a nonzero moment.

Exogenous Regressors Do Not Make \(Wy\) Exogenous

The condition \(E[\varepsilon\mid X,W]=0\) says that the structural disturbances have conditional mean zero given the observed regressors and the weights matrix. It supports the claim that the included covariates are exogenous under the model. But \(Wy\) is not simply another observed covariate: it contains outcomes generated by the system, including outcomes affected by \(\varepsilon\). The condition on \(X\) therefore does not imply \(E[(Wy)'\varepsilon\mid X,W]=0\).

A short comparison makes the role of feedback clear. With the two-unit matrix above, if \(\rho=0\), then \(A=I\) and \(WA=W\). Because \(W\) has zero diagonal, \(\operatorname{tr}(WA)=\operatorname{tr}(W)=0\): under the stipulated uncorrelated disturbances, each unit’s spatial lag has no same-unit disturbance component in this no-feedback case. At \(\rho=0.4\), however, the calculation above gives \(\operatorname{tr}(WA)=0.9524\). Feedback creates a route back to the unit’s own disturbance. This comparison concerns the expected regression cross-product; it does not claim that every finite sample has a zero cross-product when \(\rho=0\).

Common Pitfall and Practical Importance

Incorrect reasoning: “The regressors \(X\) are exogenous and the disturbance has mean zero, so every right-hand-side variable is uncorrelated with the disturbance. Therefore ordinary regression of \(y\) on \(Wy\) and \(X\) is justified by exogeneity.”

Correction: Exogeneity of \(X\) does not extend automatically to \(Wy\). The latter is constructed from \(y\), which itself contains \(\varepsilon\) through the simultaneous system. Under the stated homoskedastic, uncorrelated-disturbance assumptions, \(\operatorname{Cov}(Wy,\varepsilon\mid X,W)=\sigma^2W(I-\rho W)^{-1}\); its diagonal feedback terms, and typically the associated cross-product moment, are nonzero. A regression method must account for this endogeneity rather than treating \(Wy\) as an exogenous covariate.

The key modeling distinction is between spatially patterned outcomes and simultaneous outcome feedback. The model \(y=\rho Wy+X\beta+\varepsilon\) specifically says that neighboring outcomes enter the outcome equation. Its solution makes the feedback structure explicit and shows why the spatial-lag regressor requires special econometric treatment. This is separate from the question of whether disturbances themselves are spatially correlated: here the disturbance covariance was taken to be \(\sigma^2I\), yet feedback alone generated endogeneity in \(Wy\).