Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|improve this question
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. – Bernd Weiss Sep 2 '11 at 12:18
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. – Aaron Sep 2 '11 at 13:16
up vote 27 down vote accepted

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, $\sigma^{2}_{\gamma}$.

share|improve this answer
(+1) Of possible interest also: An introduction to sphericity. – chl Nov 22 '12 at 10:37
I think you mean "where $\gamma_j$ is the cluster $j$ random effect"... What's the bit that goes $I(k=i)$? – Jack Tanner Nov 22 '12 at 13:07
Yes, thank you for the correction, @Jack. – Macro Dec 1 '12 at 20:32
Great answer, @Macro! – Kyle. Dec 1 '12 at 21:18
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? – Macro Dec 1 '12 at 22:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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