# [R-sig-ME] chi-square mixtures for random effects LRTs

Douglas Bates bates at stat.wisc.edu
Sun Aug 10 10:20:27 CEST 2008

```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
>
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>

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