Convergence problems in multi-group m... PreviousNext
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 Michelle posted on Friday, January 20, 2006 - 2:36 pm
I am trying to run a multi-group path model with all observed variables.
When I constrain the parameters to be equal, the model runs fine. But when I try to free the parameters across groups, the model will not converge. Could you please tell me what is wrong with my code? (note, I have also tried it without the corr between inv and genn). thanks

USEVARIABLES ARE cluster weight inv
use gen t1achvt t2achvt;

CATEGORICAL IS inv;

CLUSTER is cluster;
WEIGHT is weight;

MISSING IS .;

GROUPING IS sex (0 = male 1 = female);

analysis:
type=complex;

model:
inv on use;
inv on t1achvt;
inv with gen;
gen on t1achvt;
t2achvt on inv;
t2achvt on t1achvt;
t2achvt on gen;

model male:
inv on use;
inv on t1achvt;
inv with gen;
gen on t1achvt;
t2achvt on inv;
t2achvt on t1achvt;
t2achvt on gen;

output: standardized ;
 Linda K. Muthen posted on Friday, January 20, 2006 - 3:15 pm
I would suggest you run the model separately for each group and see if they each converge. If they do, and the multiple group run does not converge, there are some suggestions in the user's guide about convergence problems. If these do not work, send the outputs for each group, the output for the multiple group run, the data, the input for the multiple group run, and your license number to support@statmodel.com
 Alex Zammit posted on Friday, July 18, 2008 - 4:56 pm
I have the same problem as Michelle. I have a 3 factor model. I am comparing two groups and when I try to freely estimate the loadings across the two groups, the model will not converge, even after 500000 iterations. I have tried assigning starting values to the loadings in the second group but this was unsuccessful. However, when I constrain the loadings in the second group, the model converges.

Oddly, If I reverse the order of the groups in the 2 group comparison, the model will converge.

Is there any information available from the work done on Michelle's problem?
 Linda K. Muthen posted on Friday, July 18, 2008 - 5:14 pm
I would suggest doing each group separately. With multiple group analysis, the first step is to determine whether a model with the same number of factors fits the data for each group. You can start with EFA or CFA. If each group has the same number of factors, then see if the same model fits the data for each group. Only then would I put the groups together to test for measurement invariance. Also, see the suggestions in the user's guide for convergence problems. If you continue to have problem, send your input, data, output, and license number to support@statmodel.com.
 Alex Zammit posted on Friday, July 18, 2008 - 5:39 pm
Linda,

The model has good fit on each group separately.

Unfortunately, the ethics approval for my project prohibits sending data to external parties. This is why I was hoping that the recommendations that were made to solve Michelle's problem - which sounded exactly like my problem - might be available to give me a starting point to solve my problem.

Thanks,
 Linda K. Muthen posted on Friday, July 18, 2008 - 5:41 pm
I would see suggestions in the user's guide. Convergence problems can be caused by many things. Send your output and license number to support and be sure to include sample statistics in the output.
 Rachael Robnett posted on Wednesday, April 25, 2018 - 10:59 am
Hello! I'm conducting a multiple-group LPA.

I get a message about the model not converging, bu the best LL value was replicated. Can I proceed with invariance testing?

The messages are below:

736 perturbed starting value run(s) did not converge.

THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES MAY NOT BE TRUSTWORTHY FOR SOME PARAMETERS DUE TO A NON-POSITIVE DEFINITE
FIRST-ORDER DERIVATIVE PRODUCT MATRIX. THIS MAY BE DUE TO THE STARTING VALUES BUT MAY ALSO BE AN INDICATION OF MODEL NONIDENTIFICATION. THE CONDITION NUMBER IS -0.233D-15. PROBLEM INVOLVING THE FOLLOWING PARAMETER: Parameter 18, %CG#1.C#3%: SERIOUS (equality/label)

And here is the model input pertaining to the above warning:

%OVERALL%
c#1 on cg#1; c#2 on cg#1;
G By common* serious;
G@1;
[G@0];

%cg#1.c#3%
G by common* serious;
[common serious] (m5-m6);
common serious (v5-v6);
 Bengt O. Muthen posted on Wednesday, April 25, 2018 - 5:21 pm
Having only 2 indicators is not enough to identify a factor model; you need at least 3.
 Rachael Robnett posted on Wednesday, April 25, 2018 - 7:50 pm
Thanks so much for the quick reply! I included the factor model because local independence is violated in the LPA.

Is there a workaround such as constraining the factor loadings to equality? Or should I do away with the factor model altogether and just run a standard LPA?
 Bengt O. Muthen posted on Thursday, April 26, 2018 - 4:49 pm
Just run a standard LPA.
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