Spatial-Error Models and Correlated Shocks

Why Model Spatial Dependence in the Disturbance?

Outcomes in nearby locations can move together even when the model does not say that one location’s outcome directly changes another’s. A regional shock, an omitted local condition, or a measurement disturbance may affect connected locations in related ways. A spatial-error model represents this pattern by allowing disturbances to be spatially dependent.

That distinction matters for interpretation. A spatial-error model specifies dependence in unobserved determinants of the outcome. It does not, by itself, say that an outcome in one location spills over into another. The parameter governing error dependence therefore has a different meaning from the spatial-lag parameter in a model with outcome feedback.

Definition: For $n$ spatial units, a spatial-error model is $$ y=X\beta+u,\qquad u=\lambda Wu+\varepsilon. $$ Here $y$ is the $n\times 1$ outcome vector, $X$ is a matrix of observed regressors, $\beta$ is the coefficient vector, $u$ is the disturbance vector, $W$ is a specified $n\times n$ spatial weights matrix, $\lambda$ is the spatial-error parameter, and $\varepsilon$ is a vector of uncorrelated innovations. The disturbance $u$ is spatially dependent because it depends on its spatially weighted neighbours, $Wu$.

For the standard covariance result, take $W$ and $X$ as fixed, assume $E[\varepsilon\mid X]=0$, and let $\operatorname{Var}(\varepsilon\mid X)=\sigma^2 I_n$, where $\sigma^2>0$. Thus, the innovations have common variance and are uncorrelated across units. Normality is not required for the covariance derivation. A common convention is that $W$ has a zero diagonal, so a unit is not its own neighbour; the result does not depend on that convention.

The equation for $u$ is a simultaneous system: each disturbance may depend on other disturbances, which in turn may depend on it. The matrix $I_n-\lambda W$ must therefore be invertible. A sufficient condition is $|\lambda|\rho(W)<1$, where $\rho(W)$ is the largest absolute eigenvalue of $W$. This condition also permits the familiar power-series representation used below.

Reduced Form and Implied Covariance

Rearrange the disturbance equation to isolate $u$. Writing $A=(I_n-\lambda W)^{-1}$ gives the reduced form $u=A\varepsilon$. The observed outcome consequently satisfies $y=X\beta+A\varepsilon$. Since $A$ is fixed conditional on $W$, the conditional mean and covariance follow directly from those of $\varepsilon$.

E[u\mid X,W]=0,\qquad \operatorname{Var}(u\mid X,W) =A\operatorname{Var}(\varepsilon\mid X,W)A' =\sigma^2(I_n-\lambda W)^{-1}(I_n-\lambda W')^{-1}.

The transpose in the last expression is important: in general, $W$ need not be symmetric, and the covariance matrix is formed as $AA'$, not $A^2$. The resulting covariance matrix is symmetric and positive definite when $\sigma^2>0$ and $A$ is invertible. Its off-diagonal entries can be nonzero, even though the innovations $\varepsilon_i$ are uncorrelated.

Key result: Under the stated assumptions, spatial-error dependence changes the covariance of the disturbance from $\sigma^2 I_n$ to $$ \Sigma_u=\sigma^2(I_n-\lambda W)^{-1}(I_n-\lambda W')^{-1}. $$ If the innovations instead have covariance $\Omega$, the corresponding result is $\Sigma_u=A\Omega A'$. The spatial-error specification describes how innovations are propagated into $u$; it does not require the innovations themselves to be spatially correlated.

When $|\lambda|\rho(W)<1$, the inverse has the convergent expansion $(I_n-\lambda W)^{-1}=I_n+\lambda W+\lambda^2W^2+\cdots$. Thus $u=\varepsilon+\lambda W\varepsilon+\lambda^2W^2\varepsilon+\cdots$. The first term is the local innovation; later terms represent its propagation over increasingly long paths in the weights network. This is propagation in the disturbance process, not a sequence of outcome responses to neighbouring outcomes.

Worked Example: Two Connected Units

Consider two hypothetical locations with equal reciprocal weights, $W=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, spatial-error parameter $\lambda=0.2$, and independent innovations with variance $\sigma^2=9$ outcome-units squared. The eigenvalues of $W$ are $1$ and $-1$, so $|\lambda|\rho(W)=0.2<1$; the reduced form exists. The assumptions also specify zero-mean, uncorrelated innovations with equal variance.

The inverse is $$ (I_2-0.2W)^{-1} =\frac{1}{1-0.2^2} \begin{pmatrix}1&0.2\\0.2&1\end{pmatrix} =\frac{1}{0.96} \begin{pmatrix}1&0.2\\0.2&1\end{pmatrix}. $$

Using $\Sigma_u=9AA'$, the product of the unscaled matrix with its transpose is $\begin{pmatrix}1.04&0.4\\0.4&1.04\end{pmatrix}$. Therefore $$ \Sigma_u =\frac{9}{0.96^2} \begin{pmatrix}1.04&0.4\\0.4&1.04\end{pmatrix} = \begin{pmatrix} 10.1563&3.9063\\ 3.9063&10.1563 \end{pmatrix}. $$

The two disturbances have variance $10.1563$ outcome-units squared and covariance $3.9063$ outcome-units squared. Their correlation is $3.9063/10.1563\approx0.3846$. The positive covariance does not mean that one location’s outcome causes the other’s outcome to rise. It arises because each disturbance contains contributions from innovations at both locations.

Reading the Spatial-Error Parameter

The parameter $\lambda$ controls the strength and direction of the spatial dependence in the disturbance equation, conditional on the chosen $W$. It is not a direct measure of a change in $y_i$ caused by a change in $y_j$. Its numerical meaning also depends on the scale and normalization of $W$: changing the weights changes the system in which $\lambda$ operates.

For nonnegative weights and positive $\lambda$, the power-series terms transmit an innovation along spatial paths with the same sign. A negative $\lambda$ alternates signs across successive powers. These are descriptions of the disturbance representation. They do not alone establish a causal pathway, and a covariance’s sign depends on the entire matrix product, not just on the sign of $\lambda$.

Worked Example: Covariance on a Three-Location Path

Suppose three hypothetical locations form a path: location 1 is connected to 2, and location 2 to 3. Use row-standardized weights $W=\begin{pmatrix}0&1&0\\0.5&0&0.5\\0&1&0\end{pmatrix}$, $\lambda=0.3$, and independent zero-mean innovations with variance $4$ outcome-units squared. The eigenvalues of this $W$ are $0$ and $\pm 1$, so $|\lambda|\rho(W)=0.3<1$. The inverse exists and the stipulated innovation assumptions support the covariance formula.

Direct inversion gives $$ A=(I_3-0.3W)^{-1} =\frac{1}{0.91} \begin{pmatrix} 0.955&0.3&0.045\\ 0.15&1&0.15\\ 0.045&0.3&0.955 \end{pmatrix}. $$

To show the multiplication, let $B$ denote the numerator matrix above, so that $\Sigma_u=4BB'/0.91^2$. The first diagonal element of $BB'$ is $0.955^2+0.3^2+0.045^2=1.00405$. The middle diagonal element is $0.15^2+1^2+0.15^2=1.045$. The covariance numerator for locations 1 and 2 is $0.955(0.15)+0.3(1)+0.045(0.15)=0.45$; for locations 1 and 3 it is $0.955(0.045)+0.3(0.3)+0.045(0.955)=0.17595$. Hence $$ \Sigma_u =\frac{4}{0.91^2} \begin{pmatrix} 1.00405&0.45&0.17595\\ 0.45&1.045&0.45\\ 0.17595&0.45&1.00405 \end{pmatrix} = \begin{pmatrix} 4.8499&2.1737&0.8499\\ 2.1737&5.0477&2.1737\\ 0.8499&2.1737&4.8499 \end{pmatrix}. $$

For example, the covariance of locations 1 and 2 is $4(0.45)/(0.91^2)=2.1736505$, which rounds to $2.1737$ outcome-units squared. The covariance between locations 1 and 3 is $4(0.17595)/(0.91^2)\approx0.8499$ outcome-units squared. The endpoints are not directly connected in $W$, yet their disturbances can covary because the inverse incorporates indirect paths through the middle location.

Correlated Disturbances Are Not Outcome Spillovers

The distinction becomes clear by comparing conditional means. In the spatial-error model, $y=X\beta+u$ and $E[u\mid X,W]=0$, so $E[y\mid X,W]=X\beta$. The spatial structure changes the covariance of outcomes around that conditional mean; it does not multiply $X\beta$ by a spatial multiplier.

By contrast, a spatial-lag model has the form $y=\rho Wy+X\beta+\varepsilon$. When $I_n-\rho W$ is invertible and $E[\varepsilon\mid X,W]=0$, its conditional mean is $E[y\mid X,W]=(I_n-\rho W)^{-1}X\beta$. That reduced form transmits changes in regressors through the outcome equation. The spatial-error model does not imply this outcome feedback.

Worked Example: Same Regressor, Different Mean Implications

Take two hypothetical locations with $W=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, and suppose $X\beta=\begin{pmatrix}10\\0\end{pmatrix}$ outcome-units. In a spatial-error model with $\lambda=0.2$, assume the innovations have zero conditional mean and finite variance as required above; $|\lambda|\rho(W)=0.2<1$. Then $E[y\mid X,W]=X\beta=\begin{pmatrix}10\\0\end{pmatrix}$. Error dependence affects the covariance, not this conditional mean.

For comparison, in a spatial-lag model with $\rho=0.2$, the same $W$ has $|\rho|\rho(W)=0.2<1$, so its mean multiplier is defined: $$ E[y\mid X,W] =(I_2-0.2W)^{-1} \begin{pmatrix}10\\0\end{pmatrix} = \begin{pmatrix}10.4167\\2.0833\end{pmatrix}. $$

The second mean includes an induced mean outcome of $2.0833$ units at location 2. This contrast is a consequence of the different equations, not evidence that spatial-error dependence itself creates an outcome spillover.

Assumptions, Scope, and a Common Pitfall

The covariance formula is conditional on the chosen weights matrix and on the innovation covariance. If the innovations are heteroskedastic or correlated, replacing their covariance with $\Omega$ yields $A\Omega A'$; the simpler $\sigma^2AA'$ formula would then be inappropriate. If $I_n-\lambda W$ is singular, the reduced-form covariance is not defined by this representation. In applications, the validity of the weights specification and the adequacy of the disturbance assumptions are substantive questions, not automatic consequences of writing down the model.

A common mistake is to read a positive estimate of $\lambda$ as proof that a unit’s outcome raises the outcomes of its neighbours. That reasoning confuses a dependence parameter in $u=\lambda Wu+\varepsilon$ with a coefficient on neighbouring outcomes in an outcome-lag equation. The correction is to identify what appears on the right-hand side: spatially weighted disturbances imply correlated shocks; spatially weighted outcomes imply outcome feedback. Neither specification alone establishes causation.

Key takeaway: A spatial-error model produces $u=(I_n-\lambda W)^{-1}\varepsilon$ and, under independent equal-variance innovations, $\Sigma_u=\sigma^2(I_n-\lambda W)^{-1}(I_n-\lambda W')^{-1}$. This is a model for spatial dependence in unobserved disturbances, distinct from spillovers in the conditional mean of outcomes.

Check Your Understanding

Answer each question using the model and assumptions stated in the tutorial.

Why does the covariance matrix of $u$ involve $AA'$ rather than $A^2$ when $W$ is not necessarily symmetric?
What condition on $I_n-\lambda W$ is needed to write the reduced form, and what convenient sufficient condition was given?
If $\operatorname{Var}(\varepsilon\mid X,W)=\Omega$, what is the implied covariance of $u$?
In the spatial-error model, what is $E[y\mid X,W]$ when $E[\varepsilon\mid X,W]=0$?
Explain why nonzero covariance between two disturbances does not by itself establish an outcome spillover or causal effect.