[R-sig-ME] nested residual covariance matrices longitudinal data

Ben Pelzer b.pelzer at maw.ru.nl
Thu Sep 28 14:04:11 CEST 2017

Dear list,

I would like to ask a few questions about comparing longitudinal model 
results. For longitudinal data with say 4 fixed occasions, my model 
looks like this:

Y = b0 + b1*occ1 + b2*occ2 + b3*occ3 + e

where occ1, occ2 and occ3 are 0-1 dummy variables for the first three 
occasions and "e" is the error term. The data are in "long" format with 
four records for each person. When using "gls" from the nlme package, I 
can choose among many different covariance structures for the residuals 
"e". In would like to confine my question to four of these:

unstructered (UN), compound symmetry (CS), autoregressive 1 (AR1), 

As far as these different structures are nested, one can use likelihood 
ratio tests. E.g. one could compare Toeplitz with AR1, since AR1 is a 
special case of Toeplitz. Am I right here? The anova command in R allows 
doing this comparison for Toeplitz and AR1 and produces a chisquare 
test, but does it indeed make sense, i.e. are these two models indeed 

For the comparison of CS and AR1, however, anova does not produce a 
chisquare test to compare these models, I guess since the two are NOT 
nested. Is this correct? IMHO they are indeed not nested, but I'm not 
completely sure...

For comparing CS, AR1 and Toeplitz with UN,  anova nicely produces 
chisquares. Does it mean that (as I would expect) the three are all 
nested within UN?

And finally the last question. In a book (can't remember which one...) I 
once found a general remark about using likelihood ratio tests (LRT) to 
compare models  with different covariance structures for residuals but 
identical fixed effects. A warning was given that one should be cautious 
to use LRT's in case the restricted model arises by putting variances 
equal to zero in the full model. In such situation the H0 hypothesized 
value of the variance lies "on the boundary of the parameter space" and 
consequently the difference in deviance of the two models will not be 
chisquare distributed. Subsequently, one uses AIC and BIC instead of 
model deviances to select the "better" model.
     However, for the above comparisons I do not see that variances are 
set equal to zero. E.g. comparing Toeplitz and CS comes down to assuming 
EQUAL covariances instead of just ONE covariance. Would you agree that 
therefore, when pairwise comparing the models mentioned (as far as 
nested) this "boundary" problem does not exist?

Thanks a lot for any help!!!

Ben Pelzer.

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