43
$\begingroup$

I recently realized that a mixed-model with only subject as a random factor and the other factors as fixed factors is equivalent to an ANOVA when setting the correlational structure of the mixed model to compound symmetry.

Therefore I would like to know what does compound symmetry mean in the context of a mixed (i.e., split-plot) ANOVA, at best explained in plain English.

Besides compound symmetry lme offers other types of correlational structures, such as

corSymm general correlation matrix, with no additional structure.

or different types of spatial correlation.

Therefore, I have the related question on which other types of correlational structures may be advisable to use in the context of designed experiments (with between- and within-subjects factors)?

It would be great if answers could point to some references for different correlational structures.

$\endgroup$
2
  • 1
    $\begingroup$ Since it would be hard for me to explain CS in plain English, just a comment: I like chapter 7 "Examining the Multilevel's Error Covariance Structure" in Singer/Willett's (2003) "Applied Longitudinal Data Analysis". It gives a great overview. $\endgroup$ Commented Sep 2, 2011 at 12:18
  • $\begingroup$ I'll second the advice of getting a good textbook. Singer/Willett is good; I also like Weiss (2005) "Modeling Longitudinal Data"; chapter 8 "Modeling the Covariance Matrix" has this specific information. $\endgroup$ Commented Sep 2, 2011 at 13:16

3 Answers 3

44
$\begingroup$

Compound symmetry is essentially the "exchangeable" correlation structure, except with a specific decomposition for the total variance. For example, if you have mixed model for the subject $i$ in cluster $j$ response, $Y_{ij}$, with only a random intercept by cluster

$$ Y_{ij} = \alpha + \gamma_{j} + \varepsilon_{ij} $$

where $\gamma_{j}$ is the cluster $j$ random effect with variance $\sigma^{2}_{\gamma}$ and $\varepsilon_{ij}$ is the subject $i$ in cluster $j$ "measurement error" with variance $\sigma^{2}_{\varepsilon}$ and $\gamma_{j}, \varepsilon_{ij}$ are independent. This model implicitly specifies the compound symmetry covariance matrix between observations in the same cluster:

$$ {\rm cov}(Y_{ij}, Y_{kj}) = \sigma^{2}_{\gamma} + \sigma^{2}_{\varepsilon} \cdot \mathcal{I}(k = i) $$

Note that the compound symmetry assumption implies that the correlation between distinct members of a cluster is $\sigma^{2}_{\gamma}/(\sigma^{2}_{\gamma} + \sigma^{2}_{\varepsilon})$.

In "plain english" you might say this covariance structure implies that all distinct members of a cluster are equally correlated with each other and the total variation, $\sigma^{2} = \sigma^{2}_{\gamma} + \sigma^{2}_{\varepsilon}$, can be partitioned into the "shared" (within a cluster) component, $\sigma^{2}_{\gamma}$ and the "unshared" component, $\sigma^{2}_{\varepsilon}$.

Edit: To aid understanding in the "plain english" sense, consider an example where individuals are clustered within families so that $Y_{ij}$ denotes the subject $i$ in family $j$ response. In this case the compound symmetry assumption means that the total variation in $Y_{ij}$ can be partitioned into the variation within a family, $\sigma^{2}_{\varepsilon}$, and the variation between families, $\sigma^{2}_{\gamma}$.

$\endgroup$
3
  • 1
    $\begingroup$ I think you mean "where $\gamma_j$ is the cluster $j$ random effect"... What's the bit that goes $I(k=i)$? $\endgroup$ Commented Nov 22, 2012 at 13:07
  • $\begingroup$ Thank you Kyle! Btw, @Jack, the $\mathcal{I}(k=i)$ bit was just a compact way of writing that, if you're talking about the same individual, then you have perfect correlation (i.e. the covariance is equal to the total variance); i.e. you have $\sigma_{\varepsilon}^2 + \sigma_{\gamma}^2$ down the diagonal and $\sigma_{\gamma}^2$ everywhere else. Does this clarify? $\endgroup$
    – Macro
    Commented Dec 1, 2012 at 22:56
  • 1
    $\begingroup$ This suggests, that compound symmetry has only non-negative correlations. It is not true in general. To be positive-definite, it needs 1>ρ>−1/(n−1) so it CAN be negative. That's why random-intercept is not fully equivalent to compound symmetry. Please consider editing your answer to make a note for readers, that may come here searching for this term by the occasion of mixed models. That's why the results given by GLS and LMM may differ if negative correlations occur. LMM will zero them, GLS will handle them. Source - Pineiro, Bates Mixed Effect models in S and S PLUS. $\endgroup$
    – Bastian
    Commented Aug 4, 2022 at 11:49
2
$\begingroup$

Compound Symmetry just means that all the variances are equal and all the covariances are equal. So the same variance and covariance are used for all subjects. If you think this applies to the factors in your ANOVA model, compound symmetry is a good covariance structure to use because of its simple structure.

$\endgroup$
2
$\begingroup$

For me, the best answer I've seen about compound symmetry is from David Howell (here):

In other words, the correlation between trial 1 and trial 2 is equal to the correlation between trial 1 and trial 4 or trial 3 and trial 4, etc. But a more direct way to think about compound symmetry is to say that it requires that all subjects in each group change in the same way over trials. In other words the slopes of the lines regressing the dependent variable on time are the same for all subjects. Put that way it is easy to see that compound symmetry can really be an unrealistic assumption. If some of your subjects improve but others don't, you do not have compound symmetry and you make an error if you use a solution that assumes that you do.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.