spdep (version 0.1-10)

predict.sarlm: Prediction for spatial simultaneous autoregressive linear model objects

Description

predict.sarlm() calculates predictions as far as is at present possible for for spatial simultaneous autoregressive linear model objects, using Haining's terminology for decomposition into trend, signal, and noise --- see reference.

Usage

predict.sarlm(object, newdata = NULL, listw = NULL, zero.policy = FALSE, ...)
print.sarlm.pred(x, ...)

Arguments

object

sarlm object returned by lagsarlm or errorsarlm

newdata

Data frame in which to predict --- if NULL, predictions are for the data on which the model was fitted

listw

a listw object created for example by nb2listw

zero.policy

if TRUE assign zero to the lagged value of zones without neighbours, if FALSE (default) assign NA - causing lagsarlm() to terminate with an error

x

the object to be printed

...

further arguments passed through

Value

• predict.sarlm() returns a vector of predictions with two attribute vectors of trend and signal values with class sarlm.pred. print.sarlm.pred is a print function for this class, printing and returning a data frame with columns: "fit", "trend" and "signal".

Details

In the following, the trend is the non-spatial smooth, the signal is the spatial smooth, and the noise is the residual. The fit returned is the sum of the trend and the signal.

The function approaches prediction first by dividing invocations between those with or without newdata. When no newdata is present, the response variable may be reconstructed as the sum of the trend, the signal, and the noise (residuals). Since the values of the response variable are known, their spatial lags are used to calculate signal components. For the error model, trend = $X \beta$, and signal = $\lambda W y - \lambda W X \beta$. For the lag and mixed models, trend = $X \beta$, and signal = $\rho W y$.

When however newdata is used for prediction, no observations of the response variable being predicted are available. Consequently, while the trend components are the same, the signal cannot take full account of the spatial smooth. In the error model, the signal is set to zero, since the spatial smooth is expressed in terms of the error: $(I - \lambda W)^{-1} \varepsilon$.

In the lag model, the signal can be expressed in the following way:

$$(I - \rho W) y = X \beta + \varepsilon$$, $$y = (I - \rho W)^{-1} X \beta + (I - \rho W)^{-1} \varepsilon$$

giving a feasible signal component of:

$$\rho W y = \rho W (I - \rho W)^{-1} X \beta$$

setting the error term to zero. This also means that predictions of the signal component for lag and mixed models require the inversion of an n-by-n matrix.

Because the outcomes of the spatial smooth on the error term are unobservable, this means that the signal values for newdata are incomplete. In the mixed model, the spatially lagged RHS variables influence both the trend and the signal, so that the root mean square prediction error in the examples below for this case with newdata is smallest, although the model was not the best fit

References

Haining, R. 1990 Spatial data analysis in the social and environmental sciences, Cambridge: Cambridge University Press, p. 258.

See Also

errorsarlm, lagsarlm

Examples

data(oldcol)
COL.lag.eig <- lagsarlm(CRIME ~ INC + HOVAL, data=COL.OLD, nb2listw(COL.nb))
COL.mix.eig <- lagsarlm(CRIME ~ INC + HOVAL, data=COL.OLD, nb2listw(COL.nb),
  type="mixed")
COL.err.eig <- errorsarlm(CRIME ~ INC + HOVAL, data=COL.OLD, nb2listw(COL.nb))
print(p1 <- predict(COL.mix.eig))
print(p2 <- predict(COL.mix.eig, newdata=COL.OLD, listw=nb2listw(COL.nb)))
AIC(COL.mix.eig)
sqrt(deviance(COL.mix.eig)/length(COL.nb))
sqrt(sum((COL.OLD$CRIME - as.vector(p1))^2)/length(COL.nb))
sqrt(sum((COL.OLD$CRIME - as.vector(p2))^2)/length(COL.nb))
AIC(COL.err.eig)
sqrt(deviance(COL.err.eig)/length(COL.nb))
sqrt(sum((COL.OLD$CRIME - as.vector(predict(COL.err.eig)))^2)/length(COL.nb))
sqrt(sum((COL.OLD$CRIME - as.vector(predict(COL.err.eig, newdata=COL.OLD,
  listw=nb2listw(COL.nb))))^2)/length(COL.nb))
AIC(COL.lag.eig)
sqrt(deviance(COL.lag.eig)/length(COL.nb))
sqrt(sum((COL.OLD$CRIME - as.vector(predict(COL.lag.eig)))^2)/length(COL.nb))
sqrt(sum((COL.OLD$CRIME - as.vector(predict(COL.lag.eig, newdata=COL.OLD,
  listw=nb2listw(COL.nb))))^2)/length(COL.nb))