https://communities.sas.com/t5/Statistical-Procedures/Autoregressive-model-with-random-times/m-p/60109#M2781 According to the PROC MIXED documentation,<BR /> <BR /> cov(i,j) = s^2*p^(d(i,j)) where d(i,j) is the Euclidean distance between i and j<BR /> <BR /> d(i,j) = sqrt( sum [over m from 1 to k](c(m,i) - c(m,j)^2).<BR /> <BR /> So for one dimensional data, like time series, your expression should be correct.<BR /> <BR /> Steve Denham Wed, 04 Aug 2010 11:31:32 GMT SteveDenham 2010-08-04T11:31:32Z https://communities.sas.com/t5/Statistical-Procedures/Autoregressive-model-with-random-times/m-p/60108#M2780 Hello all,<BR /> <BR /> I am using type=sp(pow)(time) for an AR(1) model with random times and I am getting two parameter estimates: the residual variance s^2 and correlation coef p<BR /> <BR /> For my example s^2=8.8771 and p=.04977<BR /> <BR /> Now my R matrix is a 2x2 matrix [{8.8771, 3.4493},{3.4493,8.8771}]<BR /> <BR /> I can see why the diagonal elements which are the variances are 8.8771 but I dont understand where the 3.4493 is coming from. Is it the covariance between to adjacent observations?<BR /> <BR /> Is cov(i,j)=s^2*p^|t(i)-t(j)| ????<BR /> <BR /> Thanks Wed, 04 Aug 2010 03:04:37 GMT https://communities.sas.com/t5/Statistical-Procedures/Autoregressive-model-with-random-times/m-p/60108#M2780 trekvana 2010-08-04T03:04:37Z https://communities.sas.com/t5/Statistical-Procedures/Autoregressive-model-with-random-times/m-p/60109#M2781 According to the PROC MIXED documentation,<BR /> <BR /> cov(i,j) = s^2*p^(d(i,j)) where d(i,j) is the Euclidean distance between i and j<BR /> <BR /> d(i,j) = sqrt( sum [over m from 1 to k](c(m,i) - c(m,j)^2).<BR /> <BR /> So for one dimensional data, like time series, your expression should be correct.<BR /> <BR /> Steve Denham Wed, 04 Aug 2010 11:31:32 GMT https://communities.sas.com/t5/Statistical-Procedures/Autoregressive-model-with-random-times/m-p/60109#M2781 SteveDenham 2010-08-04T11:31:32Z