chi-square mixtures for random effects LRTs
On Sat, Aug 9, 2008 at 12:54 PM, Daniel Ezra Johnson
<danielezrajohnson at gmail.com> wrote:
Pinheiro & Bates (2000:86-87) discuss, but in the end do not recommend, using mixed chi-squared distributions for random effects likelihood ratio tests. Their recommendation is to use the conservative 'naive' df, equal to the difference in the number of non-redundant parameters in the model. They discuss two examples (updated to lmer notation below):
Example 1: fm1Machine <- lmer(score~Machine+(1|Worker),data=Machines) fm2Machine <- lmer(score~Machine+(1|Worker/Machine),data=Machines)
Example 2: fm1OrthF <- lmer(distance~age+(1|Subject),data=Orthodont[Orthodont$Sex=="Female",]) fm2OrthF <- lmer(distance~age+(age|Subject),data=Orthodont[Orthodont$Sex=="Female",])
In Example 1, the difference between the models is the addition of a (nested) random intercept, therefore the number of parameters increases by one. In Example 2, the difference is a random slope, which also generates a correlation parameter, so the number of parameters increases by two.
Therefore, P&B recommend using a conservative df of 1 for the test in Example 1, and df of 2 in Example 2.
My first question is, if the difference in your model was the addition of a crossed random effect intercept, not discussed in P&B:
Example 3: imaginary1 <- lmer(score~sex+(1|subject)) imaginary2 <- lmer(score~sex+(1|subject)+(1|item))
No correlation term would be generated here, so would this pattern just like Example 1? That is, would df=1 be the naive and conservative choice?
Yes.
A second question: if one did wish to employ Stram and Lee's correction using a mixed chi-squared distribution -- between the df given above and one less degree of freedom, e.g. Mix(0,1) in Examples 1 (and 3?), and Mix(1,2) in Example 2 -- how would this be done?
Parameters for which 0 is not the boundary value, e.g. correlations, count for 1 degree of freedom. Parameters for which 0 is the boundary value, e.g. standard deviations or variances, contribute a mixture of 0 and 1 degree of freedom. In example 3 the degrees of freedom for reference distribution would be Mix(0,1) as you suspect. That is relatively easy to implement in that the effective p-value is 1/2 the p-value that is calculated for 1 degree of freedom. In general if you have a mixture that is 50% chi-squared J and 50% chi-squared K then you calculate the p-value for J and the p-value for K and average them. It happens that the p-value for a chi-squared with 0 degrees of freedom is 0 for any positive likelihood ratio.
P&B implement it somewhere in plot(simulate.lme()) but I cannot find the code for it. Is it as simple as: mean(pchisq(LRTS,df=c(0,1),lower.tail=F)) # Example 1 (and 3?) aka pchisq(LRTS,df=1,lower.tail=F)/2 in this special case mean(pchisq(LRTS,df=c(1,2),lower.tail=F)) # Example 2 If this is correct, it would seem reasonably easy to use Stram & Lee's df correction as long as the models being compared differ minimally, as in these examples. Figure 2.4 in P&B shows that the correction still leaves a conservative result in the ML case for Example 1, but it still looks better than the 'naive' df. So I'm a bit puzzled why P&B don't in the end recommend the chi-squared mixture adjustment. Thanks a lot, Daniel
_______________________________________________ R-sig-mixed-models at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models