Variogram Paradox

On 04/08/2013 08:12 PM, Saman Monfared wrote:

Dear All,
We know that the estimation of covariance parameters is an important
problem for spatial processes because the variogram shows the spatial
variation. In many cases to select the best variogram model some
parametric models considered and some criterions such as mean
prediction error, mean square error, correlation between the observed
and predicted values and correlation between the predicted and the
residual values in cross validation method uses to select the best
variogram model.

Below codes get an example whit two variogram models which are have
very different parameters (sill, range and nugget) but values of
mentioned criterions are approximately equal for them.
Why?
What is the role of variogram?
What is the role of empirical variogram when a variogram function
which is so far away than it can has approximately equaled cross
validation results.

library(gstat)
data(meuse)
coordinates(meuse)<-~x+y
v<-variogram(log(zinc)~1,meuse)
v.f<-fit.variogram(v,vgm(.205,"Mat",700,0.008,kappa=1))
plot(v,v.f)
v.ff<-fit.variogram(v,vgm(.205,"Mat",700,0.008,kappa=1)
,fit.sills =F, fit.ranges =F)
plot(v,v.ff)
k1<-krige.cv(log(zinc)~1,meuse,v.f)
k2<-krige.cv(log(zinc)~1,meuse,v.ff)
mean(k1$residual)
mean(k2$residual)
mean(k1$residual^2)
mean(k2$residual^2)
cor(k1$var1.pred,k1$observed)
cor(k2$var1.pred,k2$observed)
cor(k1$var1.pred,k1$residual)
cor(k2$var1.pred,k2$residual)

Best,
Saman.

Because the ratio of the two models,

v1 = variogramLine(v.f, 500)
v2 = variogramLine(v.ff, 500)
plot(v1[,1], v1[,2]/v2[,2])

is fairly constant.

If a variogram model gets multiplied by a positive constant, the kriging
predictions will remain identical:

v.fff = v.ff
v.ff$psill = v.ff$psill * 1234 # or any pos. number

summary(k3$var1.pred-k2$var1.pred)

all.equal(k3$var1.pred, k2$var1.pred)

k3 <- krige.cv(log(zinc)~1,meuse,v.fff)
summary(k3$var1.pred-k2$var1.pred)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
      0       0       0       0       0       0