I'm fitting a linear mixed model by SAS. There are 596 sectors and 8489 subjects. (each sector contains 10~15 subjects). Each subject is measured at most 6 times, so the total number of observation is 50043. SAS code are as follows.

proc mixed data=work.dat2 covtest method=ml maxfunc=1000 ;
   class group_k sectorid childuid;
   model laz=group_k x1 x2 x4 x6 x1_k x2_k x4_k x6_k 
             / solution cl outpm=out;
         random sectorid;
         repeated / subject=childuid type=cs ;

One of the result tables is as follows. The first table is without the option 'type=cs' (default type=VC), and the second table is with the option 'type=cs'. enter image description here enter image description here

I don't understand the relation between the SAS result table and the theory presented here http://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_mixed_sect003.htm

The covariance matrix for the default 'variance component' is $ \sigma_k^2 1(i=j) $ and that for the type 'compound symmetry' is $ \sigma_1 + \sigma^2 1(i=j) $ In the two tables, I don't understand which corresponds to $ \sigma_k, \sigma_1, \sigma $ respectively.

  • $\begingroup$ Was the answer below of any help? $\endgroup$ – ocram Jun 7 '14 at 5:16

The statement random sectorid stipulates that the variable sectorid enters as a random effect ($G$-side of the model). This is the only random effect for the $G$-side ($k=1$). By default, the covariance structure is "Variance Components" (VC). The variance component is given in the row called SECTORID in the "Covariance Parameter Estimates" Table. In your first case, then, $ G = 0.08433 \, I_{596}, $ with $I_{n}$ the identity matrix of dimension $n$.

The statement repeated / subject=childuid stipulates that the $R$ matrix, i.e. the covariance matrix of the residuals, is block diagonal with one block for each level of childuid. By default, the common covariance structure is "Variance Components" (VC). The corresponding variance component, $\sigma^2$, is given in the row called Residual in the "Covariance Parameter Estimates" Table. In your first case, then, each block has a $1.1341$ on the diagonal.

Specifying type= in the repeated statement changes the common block structure of the $R$ matrix. With type=cs, $\sigma^2$ is still in the row called Residual and $\sigma^2_1$ is given in the row called CS.

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