Psi matrix troubles PreviousNext
Mplus Discussion > Growth Modeling of Longitudinal Data >
Message/Author
 EFried posted on Tuesday, February 21, 2012 - 10:43 am
Dear Dr. Muthén,

I'm running a GMM with 5 equidistant measurement points, continuous dependent variable.

GMM shows much better fit than LCGA in all class solutions, and adding a quadratic term always improves the fit.

However, I get the warning that the latent variable covariance matrix (psi) is not positive definite in all classes in all models, even in a one-class solution.

The reason is usually a correlation greater than 1 in the tech4 output, usually between Q and S or I and S, with very high correlation values up to 2.1 (sometimes a negative residual variance).

Despite reading many threads on this forum for this problem and watching your tutorial videos I have not yet found a way to fix this. Help would be much appreciated.

Thank you!
EFried
 Linda K. Muthen posted on Tuesday, February 21, 2012 - 7:49 pm
Try centering the time scores.
 EFried posted on Wednesday, February 22, 2012 - 9:45 am
Thank you!

I ran the different models, in addition to the standard model
y0@0 y1@1 y2@2 y3@3 y4@4
as
y0@-4 y1@-3 y2@-2 y3@-1 y4@0
y0@-2 y1@-1 y2@0 y3@1 y4@2

I get exactly the same fit indices (LLH, BIC, class solutions, entropy, etc), and the PSI problems persist.

Allowing for residual covariances of the outcomes
(y0 y1 y2 y3 PWITH y1 y2 y3 y4)
which was recommended in your videos decreases fit and causes other conversion problems.

Any other suggestions?
Thank you!
EFried
 Linda K. Muthen posted on Wednesday, February 22, 2012 - 7:33 pm
You can try a linear model or a model with free time scores.
 EFried posted on Friday, February 24, 2012 - 8:08 am
Linear doesn't fit, freeing time scores leads to conversion problems.
Looking at the output, the variance of the quadratic growth factor was pretty similar, so I fixed it to zero which makes conversion possible now.

Would that be considered "overfitting", or does it make sense / is it common to use

I S Q | y1@0 y2@1 y3@2 y4@3;
q@0;
i s ON x1 x2 x3;
%c#1%
i s ON x1 x2 x3;
%c#2%
i s ON x1 x2 x3;

models?
 Linda K. Muthen posted on Friday, February 24, 2012 - 5:21 pm
I think it is often the case that the variance of the quadratic growth factor is zero.
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