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Please, help me with 'mattern' variogram‏

5 messages · Jimmy Martina, Paulo Justiniano Ribeiro Jr

Original
Jimmy Martina
# 
Hi, R-folks:
 
I have been tryin many combination of parameter to make Matern variogram to work, but I can't find the available one. I'm near to be crazy.
I tiped:
A?o2003Selg.lf<-likfit(A?o2003Selg,cov.model="matern",ini.cov.pars=c(1.5,14),kappa=2.5,fix.kappa=FALSE,nugget=0.08,lambda=0.008,fix.lambda=FALSE,hessian=TRUE)
 
the hessian shows:
 
$hessian
          [,1]           [,2]        [,3]      [,4]
[1,]  9.139986  -8.562736e+00   41.648355 -3.801615
[2,] -8.562736  3.351952e+159 -147.279881 -7.083089
[3,] 41.648355  -1.472799e+02  347.493191 -7.509035
[4,] -3.801615  -7.083089e+00   -7.509035 90.681917

Then, I put:  solve(A?o2003Selg.lf$info.minimisation.function$hessian)
 
...and it's shown as following:
 
Error in solve.default(A?o2003Selg.lf$info.minimisation.function$hessian) : 
  sistem is computationally singular: number of reciprocity condition= 1.05091e-159

on account of I work in Spanish, it's shown:
Error in solve.default(A?o2003Selg.lf$info.minimisation.function$hessian) : 
  sistema es computacionalmente singular: n?mero de condici?n rec?proco = 1.05091e-159
 
I also wander if there is any limit for kappa. Which one would be the maximun. As far as I'm concerned, it must be always major than cero, right?. I also know that its values depend on the tendency of cloud semivariogram, if it flattens slowly (Gaussian) it's major.
 
I'd thank the one available to answer my desperate question.
 		 	   		  
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Paulo Justiniano Ribeiro Jr
# 
Jimmy

can you show the result of
suummary() on your ligfit object.
I'd like to have a hunch on the parameter estimates

Looking at the diagonal of your hessian matrix
we see a very large value (a small variance for the estimator)
relatively to the other
which is probably causing some numerical 
instability

Also in general I'm not it is not always worth running the ML 
estimation on kappa. Sometimes I'd rather
run the several maximization on a grid of meaniful kappa values
seq(0.5, 5, by=0.5) for instance,
like profiling the likelihood.
Small changes is kappa usually does not make material.

Finally in my experience, in general,  it does not make much of a 
diference estimating lambda altogether with other parameters or just 
marginally


best
P.J.


Paulo Justiniano Ribeiro Jr
LEG (Laboratorio de Estatistica e Geoinformacao)
Universidade Federal do Parana
Caixa Postal 19.081
CEP 81.531-990
Curitiba, PR  -  Brasil
Tel: (+55) 41 3361 3573
VOIP: (+55) (41) (3361 3600) 1053 1066
Fax: (+55) 41 3361 3141
e-mail: paulojus AT  ufpr  br
http://www.leg.ufpr.br/~paulojus
On Wed, 17 Nov 2010, Jimmy Martina wrote:

            

      
      
      
    

  
  


  


      

    

    
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