SAS Proc Mixed random statement

[deleted_user]

Hi there,

I tried to do a mixed model with AR(1)+RE variance structure. I did it in two ways, as below. Here id is the subject id. There are two covariates: treatment (trt) and month (continuous).

proc mixed data=one;
class id;
model y= trt month /s;
repeated / type=ar(1) sub=id;
random id;
run;

proc mixed data=one;
class id;
model y= trt month /s;
random id / type=ar(1);
run;

However, they produced very different results. I checked the book "SAS for Mixed Models" Second Edition. It seems the first one is correct. But I wonder why the second one failed, though it converged with 3 covariance parameters: variance, AR(1) and residual. Does anyone know what is the actual covariance structure for the second model? Thanks a lot!

Lei

[Dale]

The second model states (more or less) that, for ordered IDs, the person-specific mean values follow an AR(1) structure. That is, suppose that you had 5 individuals who had the following ID's and mean values:

       ID                mean    
       243              18.21
       129              18.96
       141              19.44
       227              26.33
       313              22.71


Now, if we order this table by ID, we have

       ID                mean    
       129              18.96
       141              19.44
       227              26.33
       243              18.21
       313              22.71


The ordered mean values are 18.96, 19.44, 26.33, 18.21, and 22.71. There is a variance to these mean values. That is what the first of the three covariance parameters is estimating. The AR(1) parameter is estimating a covariance structure in which individuals who have closer ID values would have stronger covariance of the mean values - an assumption that I doubt you want to make. The last of the three covariance parameters, the residual variance, is estimating the variability of person-specific values about their mean values.