sorry, that escaped my attention. it was meant to be
 
y2=10+rand.int+tim+rand.int*tim+e
 
Daniel
 
-------------------------
cuncta stricte discussurus
-------------------------
 

  _____  

Von: Dennis Murphy [mailto:djmuser@gmail.com] 
Gesendet: Tuesday, November 24, 2009 5:00 PM
An: Daniel Malter
Betreff: Re: [R] random effects correlation in lmer


HI Daniel:

I liked your explanation, but in the code snippet you provided, y1 and y2
are the
same. Did you mean to put

y2 <- 10 + rand.slop + time + e   or   y2 <- 10 + rand.int * rand.slop + tim
+ e?

Dennis


On Tue, Nov 24, 2009 at 12:17 PM, Daniel Malter <daniel@umd.edu> wrote:



It implies that the random intercept is perfectly collinear with the random
slope, as you suggested. I attach an example.

The data generating process of y1 has a random intercept, but no random
slope. When you fit a model with random intercept and random slope, the
correlation between the two is estimated at -1. However, note that the
variance of the random slope is almost zero. Thus, we fit the wrong model. A
random intercept only model would have sufficed.

The data generating process of y2 includes a random slope, but those with
the higher intercepts also have the greater slopes. Random intercept and
random slope estimates are perfectly collinear. This leads to the problem
you encounter. Models 2a and 2b provide random intercept only and random
slope only estimates for comparison. It would suffice in this case to fit a
random intercept model.

Finally, the data generating process for y3 has random intercept and slope,
but both are independent, so that the problem does not occur.

library(lme4)

tim=rep(10:19,10)
id=rep(1:10,each=10)
rand.int=rep(rnorm(10,0,1),each=10)
rand.slop=rep(rnorm(10,0,1),each=10)
e=rnorm(100,0,0.5)

y1=10+rand.int+tim+e
y2=10+rand.int+tim+e
y3=10+rand.int+tim+rand.slop*tim+e

reg1=lmer(y1~tim+(tim|id))
summary(reg1)

reg2=lmer(y2~tim+(tim|id))
summary(reg2)

reg2a=lmer(y2~tim+(1|id))
summary(reg2a)

reg2b=lmer(y2~tim+(-1+tim|id))
summary(reg2b)

reg3=lmer(y3~tim+(tim|id))
summary(reg3)

HTH,
Daniel


Kurt Smith-3 wrote:
>
> I am having an issue with lmer that I wonder if someone could explain.
>
> I am trying to fit a mixed effects model to a set of longitudinal data
> over a set of individual subjects:
> (fm1 <- lmer(x ~ time + (time|ID),aa))
>
>
> I quite often find that the correlation between the random effects is 1.0:
> Linear mixed model fit by REML
> Formula: x ~ time + (time | ID)
>    Data: aa
>    AIC   BIC logLik deviance REMLdev
>  28574 28611 -14281    28561   28562
> Random effects:
>  Groups   Name        Variance Std.Dev. Corr
>  ID       (Intercept)  77.035   8.7770
>           time         10.817   3.2889  1.000
>  Residual             112.151  10.5901
> Number of obs: 3539, groups: ID, 1000
>
> Fixed effects:
>             Estimate Std. Error t value
> (Intercept)  98.7601     0.3894  253.64
> time          1.3671     0.2001    6.83
>
> Correlation of Fixed Effects:
>      (Intr)
> time -0.045
>
>
> All other parameters seem to converge as I increase the size of the
> data set, or have a reasonable distribution over several bootstrap
> samples. This suggests to me there is a singularity or something in
> solving for the random effects correlation. Does anyone have any
> insight?
> Thanks,
> Kurt Smith
>
> ______________________________________________
> R-help@r-project.org 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.
>
>

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