I'm wondering whether setting, say, a compound symmetry covariance structure between random effects and setting the residual covariance to 0 is effectively the same as not using the random effects G covariance matrix and choosing CS in the residual one?

My goal is to model the true compound symmetry only via residual covariance structure, but I would like to be able to use the benefits of the lme4 or glmmTMB, rather than nlme::gls(). lme4 is the best equipped package, but doesn't allow to specify the residual covariance structure. glmmTMB allows, but I'm not sure whether it's done via G or R matrix.

I know, that if I use the G (random effects) covariance matrix, I have to "zero" the R one (which is, probably, by default, either the heterogeneous diagonal matrix or just homogeneous sigma*I), to avoid problems.

  • In lme4 I can use control=list(sigma=1e-8))
  • In glmmTMB I can use dispformula = ~0
  • In nlme::gls() I don't have to do anything.

But is setting the structure via G and zeroing R the same as using the structure exclusively in R?

In other words, is (1 + factor | ID) equivalent to cs(0 + factor | ID) with zeroed R? And is this the same as using corCompSymm() in nlme::gls()?


1 Answer 1


A couple of notes:

  • In glmmTMB() and for a normally distributed outcome, specifying the random-effects structure as (1 + factor | ID) will be equivalent to us(0 + factor | ID) provided that dispformula = ~ 0.
  • However, this is not equivalent to compound symmetry. The compound symmetry structure typically assumes that the covariance is the same for all pairs of measurements. The unstructured covariance matrix, on the other hand, postulates a different covariance for each pair of measurements.

EDIT: If you instead specify cs(0 + factor | ID), then you postulate indeed a compound symmetry structure for the levels of factor. This should be equivalent to corCompSymm(form = ~ 1 | ID).

Note however that the compound symmetry structure has a more clear interpretation in the case of normally distributed outcomes. For other types, it only refers to the structure you assume for the random-effects covariance matrix, which will not directly translate to correlations for the outcomes themselves.

  • $\begingroup$ Thank you for your answer. I made a mistake. There should be cs, not us, the compound symmetry. I know it is not equivalent to the random intercept model, but was wondering if it could be achieved with zeroed R in lme4 and glmmTMB. $\endgroup$
    – Damasco
    Commented Jan 22, 2020 at 23:14

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