I read discussions on how the random intercept model is not equivalent to the compound symmetry. I understand, that the CS model allows for a case, where the responses are more similar across subjects than "within subjects", but there is one thing I don't understand - "parameter space". In addition to the simplified answer, it was said, that "both methods differ in parameter spaces". What does it mean? What kind of parameters is mentioned?

For example: Is it possible to fit mixed-models via gls?

Also beware of the difference in parameter spaces: the parameter space for the compound symmetry model is bigger than it is for the random intercept model. The random-effect variance is necessarily non-negative which leads to a non-negative corr but the corr in the compound symmetry model can also be negative (though not too much). So while two model fits can be equivalent (if ρ≥0) they need not be (if ρ<0) and strictly speaking the underlying models are not the same. – Rune H Christensen May 23 '18 at 6:35

  • $\begingroup$ Please reference or cite the context of "parameter space". At present you leave us guessing. Also have a look at this question/answer: stats.stackexchange.com/a/15107/182480 $\endgroup$
    – Carsten
    Mar 2, 2020 at 15:44
  • $\begingroup$ @Carsten Thank you. I updated my answer. $\endgroup$
    – dasivalana
    Mar 2, 2020 at 15:56

1 Answer 1


Let's assume that we have two measurements per subject $i$. The random intercepts model postulates that for any $j$ and $j'$, the correlation between the outcome measurements $Y_{ij}$ and $Y_{ij'}$ is

$$\mbox{cor}(Y_{ij}, Y_{ij'}) = \frac{\sigma_b^2}{\sigma_b^2 + \sigma^2},$$

where $\sigma_b^2$ is the variance of the random intercepts and $\sigma^2$ is the variance of the error terms. The important note is that this correlation cannot be negative. This is because both the numerator and denominator are non-negative because they involve variance terms that are restricted to be non-negative.

However, the general compound symmetry matrix postulates that the same correlation is

$$\mbox{cor}(Y_{ij}, Y_{ij'}) = \rho,$$

where $\rho$ can freely vary in the interval $(-1, 1)$.

Hence, if this correlation in your data is, say $-0.2$, then you can never get that with a random-intercepts model. The random-intercepts model would tell you that the correlation is zero (because this is the closest permissible value to the true value of $-0.2$).

  • $\begingroup$ Thank you very much for the answer. Tell me, please, what is the relationship to the "parameter space", as in my question? Is this that RI model calculates 2 parameters (RI variance and error variance), while the CS only one - the ρ? But it says the opposite - that the parameter space for CS is bigger... So, is this about the range of values it can take? Then yes, in CS ρ can be also negative, while in RI - only positive...? I feel the difference between the two models, but I don't understand the "parameter space" term. Space of what parameters? And why in CS it's bigger? $\endgroup$
    – dasivalana
    Mar 2, 2020 at 19:52
  • $\begingroup$ @dasivalana The parameter space is the permissible range of values for the correlation. In the random intercepts model, the parameter space for $\rho$ is the interval $(0, 1)$. In the general compound symmetry model the parameter space for $\rho$ is the interval $(-1, 1)$. $\endgroup$ Mar 2, 2020 at 20:27
  • $\begingroup$ And now it makes sense perfectly. Thank you! $\endgroup$
    – dasivalana
    Mar 2, 2020 at 20:43
  • $\begingroup$ It is not true that $\rho$ freely varies in $(-1,1)$. The $n\times n$ matrix with $1$ at each diagonal entry and $\rho$ at all other entries is positive definite if and only if $1 > \rho > - \frac{1}{n-1}$. I'm not sure how this extends to a block-diagonal matrix; maybe the lower bound on $\rho$ is $-\frac{1}{N-1}$ with $N = \max J(i)$, ,denoting by $J(i)$ the number of $y_{ij}$? cc @dasivalana $\endgroup$ Mar 2, 2020 at 21:58
  • $\begingroup$ @StéphaneLaurent you are right. I had in my mind the situations with two repeated measurements in which case the $\rho$ varies in $(-1, 1)$. I have updated the answer to mention that. $\endgroup$ Mar 3, 2020 at 10:26

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