[R] unexpected result in glm (family=poisson) for data with an only zero response in one factor
vito muggeo
vmuggeo at dssm.unipa.it
Wed Sep 13 12:49:59 CEST 2006
Dear Antonin,
It is a statistical problem: the well-known monotone likelihood.
In this case ML estimate does not exist (or equals infinity) and Wald
approximations (ob which SE are based) does not hold.
However LRT is valid and provides reliable results.
As far as I know, the only software dealing with monotone likelihood
problems in loglinear models is LogXact by cytel corporation.
best,
vito
Antonin Ferry wrote:
> Dear members,
> here is my trouble: My data consists of counts of trapped insects in different attractive traps. I usually use GLMs with a poisson error distribution to find out the differences between my traitments (and to look at other factor effects). But for some dataset where one traitment contains only zeros, GLM with poisson family fail to find any difference between this particular traitment and anyother one (even with traitment that have trapped a lot of insects). GLMs with gaussian family does not seem to have the same problem but GLMs with binomial family does.
> I'm not sure if it is a statistical problem or if it comes from R... in the latter case I think some solution exists (perhaps in the options of the glm() function ?).
> Thank you for your help.
>
>
> Here I figure out an exemple to past in the console:
>
> ## START ##############################################################################
> # Take a data set of counts for two traitments, one containing only zeros
> A=c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
> B=c(1,0,0,0,2,1,0,0,1,2,0,0,0,1,2,2,0,1,1,0,1,0,2,1,1,0,1,2,0,1,0,1,1,1,0,1,1,1,0,1)
> traitment=c(rep("A",40),rep("B",40))
> response=c(A,B)
> mydata=data.frame(traitment ,response)
>
>
> # Make a GLM on this dataset , with "family=poisson"
>
> g=glm(response~traitment, data=mydata, family=poisson)
> anova.glm(g,test="Chisq")
> # There is an effect of the traitment ...
>
> summary(g)
> # But traitment A does not differ from traitment B ! ! ! (the pvalue is always close from 1 in such cases)
>
> # Now if you replace only one zero of the A reponse to 1, the GLM works properly:
> mydata[1,2]=1
> g=glm(response~traitment, data=mydata, family=poisson)
> anova.glm(g,test="Chisq")
> summary(g)
> ##################################################################################### END ##
>
>
>
> Antonin Ferry (PhD)
>
> "Laboratoire d'Ecobiologie des Insectes Parasitoides"
> http://www.parasitoides.univ-rennes1.fr
> Université de Renes1, FRANCE
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
====================================
Vito M.R. Muggeo
Dip.to Sc Statist e Matem `Vianelli'
Università di Palermo
viale delle Scienze, edificio 13
90128 Palermo - ITALY
tel: 091 6626240
fax: 091 485726/485612
More information about the R-help
mailing list