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08-03-2010 11:04 PM

Hello all,

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

For my example s^2=8.8771 and p=.04977

Now my R matrix is a 2x2 matrix [{8.8771, 3.4493},{3.4493,8.8771}]

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?

Is cov(i,j)=s^2*p^|t(i)-t(j)| ????

Thanks

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

For my example s^2=8.8771 and p=.04977

Now my R matrix is a 2x2 matrix [{8.8771, 3.4493},{3.4493,8.8771}]

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?

Is cov(i,j)=s^2*p^|t(i)-t(j)| ????

Thanks

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Posted in reply to trekvana

08-04-2010 07:31 AM

According to the PROC MIXED documentation,

cov(i,j) = s^2*p^(d(i,j)) where d(i,j) is the Euclidean distance between i and j

d(i,j) = sqrt( sum [over m from 1 to k](c(m,i) - c(m,j)^2).

So for one dimensional data, like time series, your expression should be correct.

Steve Denham

cov(i,j) = s^2*p^(d(i,j)) where d(i,j) is the Euclidean distance between i and j

d(i,j) = sqrt( sum [over m from 1 to k](c(m,i) - c(m,j)^2).

So for one dimensional data, like time series, your expression should be correct.

Steve Denham