mylmeMF1 <- lme( y ~ time*sex, data=DAT, random=~time*sex | id, method="ML")
I meant to put "sex" into both "fixed" and "random". It should work fine, provided the configuration of the data will honestly support estimating everything.
hope this helps. spencer graves
Jacob Wegelin wrote:
Dear Mr. Graves:
Thank you. You're right -- I should have said method="ML".
Putting sex into the model as a fixed effect enables us to see if the relationship between time and y differs between the sexes. But it doesn't do anything to the covariance of the random effects.
Jake
On Mon, 12 Jul 2004, Spencer Graves wrote:
Have you tried something like the following:
mylmeMF0 <- lme( y ~ time, data=DAT, random=~time | id, method="ML")
mylmeMF1 <- lme( y ~ time*sex, data=DAT, random=~time | id, method="ML")
anova(mylemFM0, mylmeMF1)
Please check Pinheiro and Bates regarding "method". They explain when "ML" and "REML" are appropriate. I don't have the book handy, and I can't remember for sure, but I believe that "REML" is would only be apropriate here if "sex" were NOT in the "fixed" model.
hope this helps. spencer graves
Jacob Wegelin wrote:
How does one implement a likelihood-ratio test, to test whether the variances of the random effects differ between two groups of subjects?
Suppose your data consist of repeated measures on subjects belonging to two groups, say boys and girls, and you are fitting a linear mixed-effects model for the response as a function of time. The within-subject errors (residuals) have the same variance in both groups. But the dispersion of the random effects differs between the groups. The boys' random effects -- say, the intercepts -- have greater variance than the girls'. One can see this by partitioning the data by sex and fitting two separate models.
The model for the girls,
library("nlme") mylmeF0 <- lme( y ~ time, data=DAT, random=~time | id, subset=sex=="F")
yields a variance of about one for the random intercepts:
StdDev Corr (Intercept) 0.9765052 (Intr) time 0.1121913 -0.254 Residual 0.1806528
whereas in the model for the boys, the corresponding variance is ten times that amount:
mylmeM0 <- lme( y ~ time, data=DAT, random=~time | id, subset=sex=="M")
StdDev Corr (Intercept) 10.1537946 (Intr) time 0.1230063 -0.744 Residual 0.1298910
I would like to use a likelihood ratio to test this difference. The smaller ("null") model would be
mylme0 <- lme( y ~ time, data=DAT, random=~time | id ) .
This model forces the random intercepts for both boys and girls to come
from a single normal distribution.
The larger model would allow the boys' and girls' random intercepts (or more generally their random effects) to come from separate normal distributions with possibly unequal variances.
There must be some straightforward obvious way to fit the larger model, but I do not see it.
Pinheiro and Bates, chapter 5.2, show how to model unequal *residual* ("within-group") variances for the two groups using varIdent. They also tantalizingly say, "The single-level variance function model (5.10) can be generalized to multilevel models" (page 206), which seems to suggest that a solution to the current problem might exist.
The pdMat classes provide a way to *constrain* the dispersion matrix of the random effects, not make it more general.
Of course, one way to test for unequal variances is to apply an F-test for equal variances to the random intercepts. If the data are as shown at the bottom of this email, the test can be implemented as follows:
stuff<-as.data.frame(summary(mylme0)$coefficients$random$id) stuff$sex<-factor(substring(row.names(stuff), 1,1)) mysplit<-split(stuff[,"(Intercept)"], stuff[,"sex"]) ns<-sapply(mysplit, length) vars<-sapply(mysplit, var) p<- 1-pf( vars["M"]/vars["F"], ns["M"]-1, ns["F"]-1)
Alternatively, one could implement a permutation test for the ratio of the variances of the random intercepts--these variances derived from the two halves of the partitioned data.
But surely there's a direct, model-based way to do this?
Thanks for any suggestions
Jake
P.S. Here is the code by which the "data" were generated.
nb<-1 ntimepts<-3 girls<-data.frame( y= rep(-nb:nb , each=ntimepts) , id=rep( paste("F", 1:(2*nb+1), sep=""), each=ntimepts) , time=rep(1:(2*nb+1), length=ntimepts) ) boys <-data.frame( y= rep(10*(-nb:nb) , each=ntimepts) , id=rep( paste("M", 1:(2*nb+1), sep=""), each=ntimepts) , time=rep(1:(2*nb+1), length=ntimepts) ) DAT<-rbind(girls,boys) DAT$y<-DAT$y + rnorm(nrow(DAT))/5 DAT$sex<-factor(substring( as.character(DAT[,"id"]), 1,1)) row.names(DAT)<-paste( DAT[,"id"], DAT[,"time"], sep=".")
Jacob A. Wegelin Assistant Professor Division of Biostatistics, School of Medicine University of California, Davis One Shields Ave, TB-168 Davis CA 95616-8638 USA http://wegelin.ucdavis.edu/ [EMAIL PROTECTED]
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