Likelihood ratios
By "complexity-corrected", I simply meant a likelihood ratio that takes into account potential differences in the number of parameters between models, penalizing the more complex model. In the Glover & Dixon paper I cited in my original post, they compute a likelihood ratio based on ANOVA sums of squares, then apply a complexity correction factor (they show formulae for both AIC and BIC) to yield a final ratio that I've been calling a "complexity-corrected likelihood ratio". Sorry for causing confusion by using my idiosyncratic nomenclature!
On Tue, Jun 1, 2010 at 8:25 PM, Ben Bolker <bolker at ufl.edu> wrote:
?Yes, or exp(logLik(fit2,REML=FALSE)-logLik(fit1,REML=FALSE)) ? What is a "complexity-corrected" likelihood ratio? ? I'm very nervous about mixing in the AIC penalty terms with the rest of the likelihood ratio -- and, as I think has been pointed out, you have to be careful to set REML=FALSE ... fengsj at mail.utexas.edu wrote:
Is there someting similar for glmer models? Thanks! Quoting Mike Lawrence <Mike.Lawrence at dal.ca>:
After posting this, I thought to contact Pete Dixon himself and indeed
it seems he already coded the functions to obtain a likelihood ratio
comparing two lmer models:
AIC_lmer = function(x){
? ? ?require(lme4)
? ? ?print(formula(attr(x,"call")))
? ? ?summary(x)@AICtab
}
LR_lmer = function(m0,m1){
? ? ?exp((AIC_lmer(m0)[[1]]-AIC_lmer(m1)[[1]])/2)
}
#example usage:
LR_lmer( my_fit1 , my_fit2 )
On Tue, Jun 1, 2010 at 1:50 PM, Mike Lawrence <Mike.Lawrence at dal.ca> wrote:
oops, I guess that should be: LR = exp( anova( fit1 , fit2 )$Chisq[2] / -2 ) On Tue, Jun 1, 2010 at 1:28 PM, Mike Lawrence <Mike.Lawrence at dal.ca> wrote:
Hi folks, I have 2 lmer fits, one (fit1) nested in the other (fit2), and I'd like to compute the likelihood ratio comparing the models so I can say something like "there is X times more evidence for fit1 than for fit2" (as in Glover & Dixon, 2004, www.ncbi.nlm.nih.gov/pubmed/15732688). I know I can use anova(fit1,fit2) to obtain a null-hypothesis significance test of the fits, and I suspect the output also contains the information I need to make my evidentiary statement, but I'm not confident of what I'm doing here. Is it correct that the reported value of chi-square from anova() is simply the D of the likelihood ratio test (http://en.wikipedia.org/wiki/Likelihood_ratio_test)? If so, does it sound right that I can simply derive the complexity-corrected likelihood ratio as: LR = exp( -2 * anova( fit1 , fit2 )$Chisq[2] ) ? Mike -- Mike Lawrence Graduate Student Department of Psychology Dalhousie University Looking to arrange a meeting? Check my public calendar: http://tr.im/mikes_public_calendar ~ Certainty is folly... I think. ~
-- Mike Lawrence Graduate Student Department of Psychology Dalhousie University Looking to arrange a meeting? Check my public calendar: http://tr.im/mikes_public_calendar ~ Certainty is folly... I think. ~
-- Mike Lawrence Graduate Student Department of Psychology Dalhousie University Looking to arrange a meeting? Check my public calendar: http://tr.im/mikes_public_calendar ~ Certainty is folly... I think. ~
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-- Ben Bolker Associate professor, Biology Dep't, Univ. of Florida *** NEW E-MAIL ADDRESSES: *** ? bbolker at gmail.com , bolker at math.mcmaster.ca bolker at ufl.edu / people.biology.ufl.edu/bolker GPG key: people.biology.ufl.edu/bolker/benbolker-publickey.asc
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Mike Lawrence Graduate Student Department of Psychology Dalhousie University Looking to arrange a meeting? Check my public calendar: http://tr.im/mikes_public_calendar ~ Certainty is folly... I think. ~