# Endogenous Spatial Lags for the Linear Regression Model

Over the past number of years, I have noted that spatial econometric methods have been gaining popularity. This is a welcome trend in my opinion, as the spatial structure of data is something that should be explicitly included in the empirical modelling procedure. Omitting spatial effects assumes that the location co-ordinates for observations are unrelated to the observable characteristics that the researcher is trying to model. Not a good assumption, particularly in empirical macroeconomics where the unit of observation is typically countries or regions.

Starting out with the prototypical linear regression model: $y = X \beta + \epsilon$, we can modify this equation in a number of ways to account for the spatial structure of the data. In this blog post, I will concentrate on the spatial lag model. I intend to examine spatial error models in a future blog post.

The spatial lag model is of the form: $y= \rho W y + X \beta + \epsilon$, where the term $\rho W y$ measures the potential spill-over effect that occurs in the outcome variable if this outcome is influenced by other unit’s outcomes, where the location or distance to other observations is a factor in for this spill-over. In other words, the neighbours for each observation have greater (or in some cases less) influence to what happens to that observation, independent of the other explanatory variables $(X)$. The $W$ matrix is a matrix of spatial weights, and the $\rho$ parameter measures the degree of spatial correlation. The value of $\rho$ is bounded between -1 and 1. When $\rho$ is zero, the spatial lag model collapses to the prototypical linear regression model.

The spatial weights matrix should be specified by the researcher. For example, let us have a dataset that consists of 3 observations, spatially located on a 1-dimensional Euclidean space wherein the first observation is a neighbour of the second and the second is a neighbour of the third. The spatial weights matrix for this dataset should be a $3 \times 3$ matrix, where the diagonal consists of 3 zeros (you are not a neighbour with yourself). Typically, this matrix will also be symmetric. It is also at the user’s discretion to choose the weights in $W$. Common schemes include nearest k neighbours (where k is again at the users discretion), inverse-distance, and other schemes based on spatial contiguities. Row-standardization is usually performed, such that all the row elements in $W$ sum to one. In our simple example, the first row of a contiguity-based scheme would be: [0, 1, 0]. The second: [0.5, 0, 0.5]. And the third: [0, 1, 0].

While the spatial-lag model represents a modified version of the basic linear regression model, estimation via OLS is problematic because the spatially lagged variable $(Wy)$ is endogenous. The endogeneity results from what Charles Manski calls the ‘reflection problem’: your neighbours influence you, but you also influence your neighbours. This feedback effect results in simultaneity which renders bias on the OLS estimate of the spatial lag term. A further problem presents itself when the independent variables $(X)$ are themselves spatially correlated. In this case, completely omitting the spatial lag from the model specification will bias the $\beta$ coefficient values due to omitted variable bias.

Fortunately, remedying these biases is relatively simple, as a number of estimators exist that will yield an unbiased estimate of the spatial lag, and consequently the coefficients for the other explanatory variables—assuming, of course, that these explanatory variables are themselves exogenous. Here, I will consider two: the Maximum Likelihood estimator (denoted ML) as described in Ord (1975), and a generalized two-stage least squares regression model (2SLS) wherein spatial lags, and spatial lags lags (i.e. $W^{2} X$) of the explanatory variables are used as instruments for $Wy$. Alongside these two models, I also estimate the misspecified OLS both without (OLS1) and with (OLS2) the spatially lagged dependent variable.

To examine the properties of these four estimators, I run a Monte Carlo experiment. First, let us assume that we have 225 observations equally spread over a $15 \times 15$ lattice grid. Second, we assume that neighbours are defined by what is known as the Rook contiguity, so a neighbour exists if they are in the grid space either above or below or on either side. Once we create the spatial weight matrix we row-standardize.

Taking our spatial weights matrix as defined, we want to simulate the following linear model: $y = \rho Wy + \beta_{1} + \beta_{2}x_{2} + \beta_{3}x_{3} + \epsilon$, where we set $\rho=0.4$ , $\beta_{1}=0.5$, $\beta_{2}=-0.5$, $\beta_{3}=1.75$. The explanatory variables are themselves spatially autocorrelated, so our simulation procedure first simulates a random normal variable for both $x_{2}$ and $x_{3}$ from: $N(0, 1)$, then assuming a autocorrelation parameter of $\rho_{x}=0.25$, generates both variables such that: $x_{j} = (1-\rho_{x}W)^{-1} N(0, 1)$ for $j \in \{ 1,2 \}$. In the next step we simulate the error term $\epsilon$. We introduce endogeneity into the spatial lag by assuming that the error term $\epsilon$ is a function of a random normal $v$ so $\epsilon = \alpha v + N(0, 1)$ where $v = N(0, 1)$ and $\alpha=0.2$, and that the spatial lag term includes $v$. We modify the regression model to incorporate this: $y = \rho (Wy + v) + \beta_{1} + \beta_{2}x_{2} + \beta_{3}x_{3} + \epsilon$. From this we can calculate the reduced form model: $y = (1 - \rho W)^{-1} (\rho v + \beta_{1} + \beta_{2}x_{2} + \beta_{3}x_{3} + \epsilon)$, and simulate values for our dependent variable $y$.

Performing 1,000 repetitions of the above simulation permits us to examine the distributions of the coefficient estimates produced by the four models outlined in the above. The distributions of these coefficients are displayed in the graphic in the beginning of this post. The spatial autocorrelation parameter $\rho$ is in the bottom-right quadrant. As we can see, the OLS model that includes the spatial effect but does not account for simultaneity (OLS2) over-estimates the importance of spatial spill-overs. Both the ML and 2SLS estimators correctly identify the $\rho$ parameter. The remaining quadrants highlight what happens to the coefficients of the explanatory variables. Clearly, the OLS1 estimator provides the worst estimate of these coefficients. Thus, it appears preferable to use OLS2, with the biased autocorrelation parameter, than the simpler OLS1 estimator. However, the OLS2 estimator also yields biased parameter estimates for the $\beta$ coefficients. Furthermore, since researchers may want to know the marginal effects in spatial equilibrium (i.e. taking into account the spatial spill-over effects) the overestimated $\rho$ parameter creates an additional bias.

To perform these calculations I used the spdep package in R, with the graphic created via ggplot2. Please see the R code I used in the below.

library(spdep) ; library(ggplot2) ; library(reshape)

rm(list=ls())
n = 225
data = data.frame(n1=1:n)
# coords
data$lat = rep(1:sqrt(n), sqrt(n)) data$long = sort(rep(1:sqrt(n), sqrt(n)))
# create W matrix
wt1 = as.matrix(dist(cbind(data$long, data$lat), method = "euclidean", upper=TRUE))
wt1 = ifelse(wt1==1, 1, 0)
diag(wt1) = 0
# row standardize
rs = rowSums(wt1)
wt1 = apply(wt1, 2, function(x) x/rs)
lw1 = mat2listw(wt1, style="W")

rx = 0.25
rho = 0.4
b1 = 0.5
b2 = -0.5
b3 = 1.75
alp = 0.2

inv1 = invIrW(lw1, rho=rx, method="solve", feasible=NULL)
inv2 = invIrW(lw1, rho=rho, method="solve", feasible=NULL)

sims = 1000
beta1results = matrix(NA, ncol=4, nrow=sims)
beta2results = matrix(NA, ncol=4, nrow=sims)
beta3results = matrix(NA, ncol=4, nrow=sims)
rhoresults = matrix(NA, ncol=3, nrow=sims)

for(i in 1:sims){
u1 = rnorm(n)
x2 = inv1 %*% u1
u2 = rnorm(n)
x3 = inv1 %*% u2
v1 = rnorm(n)
e1 = alp*v1 + rnorm(n)
data1 = data.frame(cbind(x2, x3),lag.listw(lw1, cbind(x2, x3)))
names(data1) = c("x2","x3","wx2","wx3")

data1$y1 = inv2 %*% (b1 + b2*x2 + b3*x3 + rho*v1 + e1) data1$wy1 = lag.listw(lw1, data1$y1) data1$w2x2 = lag.listw(lw1, data1$wx2) data1$w2x3 = lag.listw(lw1, data1$wx3) data1$w3x2 = lag.listw(lw1, data1$w2x2) data1$w3x3 = lag.listw(lw1, data1$w2x3) m1 = coef(lm(y1 ~ x2 + x3, data1)) m2 = coef(lm(y1 ~ wy1 + x2 + x3, data1)) m3 = coef(lagsarlm(y1 ~ x2 + x3, data1, lw1)) m4 = coef(stsls(y1 ~ x2 + x3, data1, lw1)) beta1results[i,] = c(m1[1], m2[1], m3[2], m4[2]) beta2results[i,] = c(m1[2], m2[3], m3[3], m4[3]) beta3results[i,] = c(m1[3], m2[4], m3[4], m4[4]) rhoresults[i,] = c(m2[2],m3[1], m4[1]) } apply(rhoresults, 2, mean) ; apply(rhoresults, 2, sd) apply(beta1results, 2, mean) ; apply(beta1results, 2, sd) apply(beta2results, 2, mean) ; apply(beta2results, 2, sd) apply(beta3results, 2, mean) ; apply(beta3results, 2, sd) colnames(rhoresults) = c("OLS2","ML","2SLS") colnames(beta1results) = c("OLS1","OLS2","ML","2SLS") colnames(beta2results) = c("OLS1","OLS2","ML","2SLS") colnames(beta3results) = c("OLS1","OLS2","ML","2SLS") rhoresults = melt(rhoresults) rhoresults$coef = "rho"
rhoresults$true = 0.4 beta1results = melt(beta1results) beta1results$coef = "beta1"
beta1results$true = 0.5 beta2results = melt(beta2results) beta2results$coef = "beta2"
beta2results$true = -0.5 beta3results = melt(beta3results) beta3results$coef = "beta3"
abline(v=1)


# Monty Hall Meets Monte Carlo

In 1990, Parade magazine columnist Marilyn vos Savant caused a giant kerfuffle over an answer she gave to a question posed by a reader. The problem, as stated by the omnipotent Wikipedia, goes like this:

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

Vos Savant’s answer was you should always switch. The probability of winning the star prize (a brand new Rover Vitesse Fastback) is 2/3 if you switch, and only 1/3 if you stick. However, this answer outraged many people, and 10,000 furious souls dabbed their quills in some ink and wrote strongly worded, and possibly impolite, letters to Vos Savant refuting her analysis. It was Vos Savant who would have the last laugh though, as her answer was indeed correct (LOL).

That people would get this puzzle wrong is understandable. I know I did the first time I was faced with it. The reason is deeply rooted in behavioural psychology. Humans are not good at dealing with conditional probabilities. If only Vos Savant’s irate readers knew how to use Bayes theorem, they could have saved themselves the ink and the emotional turmoil by deriving the analytic solution to the problem (although, on reflection, doing this would have also led to some ink usage).

What if you’re too lazy or unsure how to derive an analytic solution to the problem? Is there a way for you, you feckless badger? Why yes Kent, there is – use simulation. In the below, I have included some simple R code which simulates the problem. As you can see you get close enough to the Vos Savant’s correct answer.

The code below is an example of how it is possible to use simulation techniques to derive solutions when you do not want to, or are too stupid to do the mathematics. The beauty of computer simulation is that it can be used in much more complex settings where analytic solutions are not feasible. Extracting results via computer simulation is crucial in Bayesian statistical analysis, and also in frequentist calculations of model uncertainty like bootstrapping.

# clean
rm(list=ls())
# outcome matrix
cars <- matrix(NA,nrow=10000,ncol=2)
for(i in 1:10000){
# initial choices
choices <- 1:3
cardoor <- sample(choices,1)
goatdoor <- choices[-cardoor]
contest <- sample(choices,1)
choice2 <- c(contest,NA)
choice2[2] <- ifelse(contest==cardoor,sample(goatdoor,1),cardoor)
# the guy who sticks
cars[i,1] <- ifelse(contest==cardoor,1,0)
# the guy who changes
cars[i,2] <- ifelse(choice2[choice2!=contest]==cardoor,1,0)
}

c(mean(cars[,1]),mean(cars[,2]))