Nonpositive definite psi matrix PreviousNext
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 daniel adkins posted on Tuesday, November 28, 2006 - 10:39 am
output indicates the psi matrix is nonpos def. tech4 output indicates that the issue is that the standardized corr between the intercept and linear slope component factors is greater than one. however, the raw, unstandardized corr is not problematic. is this an issue? can these parameter estimates be considered accurate? the estimates are consistent with earlier model runs using a very similar measure which estimated cleanly.
 Linda K. Muthen posted on Tuesday, November 28, 2006 - 1:38 pm
I suspect that you have covariates in your model and that what you are looking at in the results section of the output is a residual covaraince and a standardized residual covariance. TECH4 provides a correlation. If this is greater than one, then the model has a problem.
 daniel adkins posted on Tuesday, November 28, 2006 - 2:04 pm
there are no covariates in the model, it is an unconditional quadratic LCM w/ 6 repeated measures. the corr btwn the growth factors is given at 2 places in the output--in the standard output (under the growth factor cfa's) and in the tech4 output. the numbers in the tech4 growth factor corr matrix correspond to the corrs given in the standardized columns (ie- columns 4 and 5) of the regular output. does that make sense? if so, does the model have a problem? if so, any remedial advice?
 Linda K. Muthen posted on Tuesday, November 28, 2006 - 2:35 pm
I thought you said that the correlations in TECH4 and the standardized values were different. If they are the same, that makes sense, and if they are one or greater that is a problem. I would need you to send the input, data, output, and your license number to support@statmodel.com to see if I can suggest any model modifications.
 Carolin posted on Wednesday, June 08, 2011 - 2:22 am
Dear Mr and Mrs Muthen,

Iím new to Mplus and GMM and have some questions.
Iím analyzing a GMM with 4 timepoints without covariates. I have a negative variance of the intercept, that is not significant (I get the warning PSI Matrix is not positive definite).
1) How can this happen, what does this mean?

Then, I tried to fix i@0, with this I donít get the warning message anymore.
2) Can I do this? This means, that I donít allow the intercepts to vary around the mean, right? Do you have other suggestions to solve the problem with the negative variance?

3) Can I still put covariates in my model with this restraint because there is no variance to explain when I fix it to zero.

Thanks a lot for your help!
 Linda K. Muthen posted on Wednesday, June 08, 2011 - 6:03 am
If you have a negative variance for the intercept growth factor, the model is inadmissible. If it is small and not significant, you can fix it at zero. This means that the intercept is a fixed rather than a random effect. It is still possible to use covariates.
 Carolin posted on Wednesday, June 08, 2011 - 6:21 am
Thanks a lot for your answer! I have two further questions:

Can you imagine different reasons how it can happen that a model is inadmissable?

Will the covariates that I use still be able to have influence when the intercept is fixed,is there still variance?
 Linda K. Muthen posted on Wednesday, June 08, 2011 - 1:48 pm
If you have a significant negative variance or residual variance, it means the model is not correct for the data.

You can still regress the intercept growth factor on covariates even if you fix the variance of the intercept growth factor to zero. Bringing the covariates into the model can increase the power to detect that the variance is significant.
 Carolin posted on Thursday, June 09, 2011 - 5:53 am
Thanks a lot again, you are sp helpful!!!

You wrote the model is not correct for the data with "significant negative variance" - is this the same for non significant negative variance?

I assume that I still fix the variance in the model with covariates (as in that one without covariates), right?

When I write a paper I think I have to tell the fact that I fixed the variance. Do you have any experiences if that would be a problem?
 Linda K. Muthen posted on Thursday, June 09, 2011 - 6:53 am
If the negative variance is small and not significant, you can fix it at zero. I think this is commonly done. I would try letting the variance be free when I added to covariates to see if that changes things and fix it only if needed.
 Anna Radkovsky posted on Tuesday, February 12, 2013 - 3:46 am
Hello,
I am currently running a bivariate linear latent growth curve analysis with four time points, and I am getting some convergence problems due to nonpositive definite PSI matrix. Considering the TECH4 Output, the correlation of slopes is greater than one. Because I am interested in the slope-slope relationship, setting the variance of the slope to one is not satisfactory to me. Therefore, I have two questions:
1) Why is the correlation of slopes greater than one?
2) Is there anything I can do to modify the model in a way it still involves the slope-slope correlation?
You can find the input model Output below.
Thanks a lot for your help!

MODEL:
iD sD | BDI_MT1@0 BDI_MT2@1 BDI_MT3@2 BDI_MT4@3;
iER sER | S27_9_t1@0 S27_9_t2@1 S27_9_t3@2 S27_9_t4@3;
 Linda K. Muthen posted on Tuesday, February 12, 2013 - 6:13 am
You may need to add residual covariances of the two outcomes at each time point, for example,

bdi_mt1 WITH s27_9_t1;

If you have not already done so, I would fit each process separately before I put them together.
 Anna Radkovsky posted on Tuesday, February 12, 2013 - 9:03 am
Thank you very much for your quick reply.
I have already done the analyses for the univariate LGCs, and both models fit very well.

I added residual covariances, as you proposed, but this didn't change anything. So in addition I tried setting the residuals equal within each process. Luckily this solved the problem and now the model fits well.

But I still wonder how correlations greater than one could occur in growth curve modeling. Do you have any hints for me?

Thanks again,
Anna
 Linda K. Muthen posted on Wednesday, February 13, 2013 - 3:56 pm
The needs for residual covariances across processes is the main reason I have seen.
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