[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),
Toeplitz.
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
nested?
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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