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Mplus Discussion > Latent Variable Mixture Modeling >
 sharon su posted on Saturday, July 30, 2016 - 6:36 am
Hello ~
I am trying to analyze data using ex.7.27 syntax.
1. What's the difference if I choose continuous or categorical observed variables? Will the link function change from linear to logit one?

2. Do the two types have their approriate algorithm?

3. How can I detect the label switching phenomenon?

Thanks a lot ~
 Bengt O. Muthen posted on Saturday, July 30, 2016 - 4:19 pm
1. Yes.

2. Yes.

3. Label switching is an issue with Bayesian estimation of mixtures, not ML - unless you are doing Monte Carlo studies over many replications.
 sharon su posted on Sunday, July 31, 2016 - 7:58 pm
1.When I try to figure out FMM with figures,I get stuck with the arrow from c to f. According to "latent variable hybrids", there's no arrow because the factor means can be standerized, but in Lubke & Muthen(2005, p.28), there does have an arrow from c to f. How can I explain the difference? Will standerized or not influence the model?

2. Would algorithm = intergration be suitable both for linear and logit situation?

Thank you ~
 Bengt O. Muthen posted on Monday, August 01, 2016 - 4:29 pm
1. c to f means that the means of f vary across the c classes. When c points to the indicators their means/thresholds vary across classes and therefore factor mean differences across classes cannot also be identified.

2. Yes.
 sharon su posted on Monday, August 08, 2016 - 5:02 am
Well, I generate two classes of data.
The fisrt one with two factors mean (0,0), SD equals to 1, the second data with two factors mean (1.2,1.2).
Then I combine the two data sets, and try to use ex 7.17 to analyze data. However, I can only get the item intercept deifference.

What can I do to get the factor mean difference? Does the factor mean have to equal to zero for identification?
 Bengt O. Muthen posted on Monday, August 08, 2016 - 3:29 pm
The factor mean has to be zero in one class but should be free in the other class - just like is shown in 7.17.
 sharon su posted on Friday, September 02, 2016 - 1:58 am
I gnenerate a set of data with 2 factors, their covariance is about 0.25, factors loadings are 0.8. When I use FMM analysis dealing with the data, the general result output showed results as:

F1 0.260 0.016 16.405 0.000

the STDYX result:
F1 0.743 0.018 42.427 0.000

If I choose the STDYX result,why the covariance is not near the true value, while the unstanderdized result is?
 Bengt O. Muthen posted on Friday, September 02, 2016 - 10:15 am
Your factor variances were probably not generated with variances one.
 sharon su posted on Friday, September 02, 2016 - 8:00 pm
I've checked the data, the variances were about 1, covariance about 0.25.

MPLUS code:
f1 BY i1-i30;
f2 BY i31-i60;
After rerun
The model result
F2 WITH F1 0.270
STDYX result
F2 WITH F1 0.764

I am not sure what happpened? But the estimation value(correalation?) is far away from my data.Do I miss anything?
 Bengt O. Muthen posted on Saturday, September 03, 2016 - 9:29 am
To give you good advice I have to see your full outputs for both the data-generation part and the analysis part. Please send to Support along with your license number.
 sharon su posted on Monday, September 26, 2016 - 8:18 pm
I think it's because my data is binary response. When I use continuous indicators, the estimation result seems closer to the true value.
Can I still use ex7.17 but change command to :
Would it be linear link funcion assumpution?
Thanks ~
 Bengt O. Muthen posted on Tuesday, September 27, 2016 - 12:14 pm
No, it doesn't have to do with binary response,

When you declare a variable as Categorical, a probit or logit link function is used.
 sharon su posted on Tuesday, September 27, 2016 - 9:41 pm
You mean I can imitate ex 7.27?
If I want to distinguish the probit(linear) from logit(nonlinear) link function ? How can I revise the command?
Thanks ~
 Bengt O. Muthen posted on Wednesday, September 28, 2016 - 1:09 pm
Link = probit;

in the Analysis command.
 sharon su posted on Wednesday, September 28, 2016 - 11:00 pm
If I still use the same binary data,
what's the difference between ex 7.17 and ex 7.27(link= probit)?
I am kind of confused because they both seem to provide estimation result.
Thanks ~
 Bengt O. Muthen posted on Thursday, September 29, 2016 - 8:34 am
7.17 uses MLR and all maximum-likelihood estimators use Link=logit as the default.
 sharon su posted on Saturday, October 01, 2016 - 12:49 am
1. When facing ex 7.17 & ex 7.27, they both use logit link finction as default to deal with binary data, how can I choose between them?

2. Do the two examples both set constraint factor means = 0, factor variances =1 for identification? Do I have to give more restrictions?

3. When c were more than 2, is it enough to give constraint of foctor means = 0 and factor variances =1 for only one group? or both two groups need to have constraint?
 Bengt O. Muthen posted on Monday, October 03, 2016 - 7:09 am
See my paper on our website:

Muthén, B. (2008). Latent variable hybrids: Overview of old and new models. In Hancock, G. R., & Samuelsen, K. M. (Eds.), Advances in latent variable mixture models, pp. 1-24. Charlotte, NC: Information Age Publishing, Inc. Click here for information about the book.
download paper contact author
 sharon su posted on Monday, November 14, 2016 - 4:19 am
If I want to have different resisual variances but have the same factor loadings on the two groups,how can I revise my code? Thanks ~


f1 BY i1-i5;
f2 BY i6-i10;
f1 with f2;

f1 with f2;
 Bengt O. Muthen posted on Monday, November 14, 2016 - 5:09 pm
Your input gives equal loadings. To get group-varying residual variances, just add


in both groups.

I don't know why you fix the intercepts to zero.
 sharon su posted on Monday, November 14, 2016 - 9:51 pm
1. I change the code, only the model result shows equal factor loadings, but not in STANDARDIZED MODEL RESULTS, why?

STDYX Standardization
Latent Class 1
I1 0.843
I2 0.797
I3 0.836
I4 0.797
I5 0.805
Latent Class 2
I1 0.798
I2 0.780
I3 0.784
I4 0.793
I5 0.798

2. I give the assumption that the two groups have the same intercepts equal to zero when generating the data, so I give the constraint?

3. I generate the data with factor loadings 0.8, residual variances are 0.36, I choose the STDYX Standardization, is it appropriate?
 Bengt O. Muthen posted on Tuesday, November 15, 2016 - 5:27 pm
1. I assume it is because your variances are not equal across groups and they are used to standardize.

2. It is ok to generate by 0, but don't fix them to 0 in the analysis - you never have that knowledge in real data.

3. If you have a factor variance of 1 the total variance is 1 as you seem to assume here.
 sharon su posted on Monday, November 21, 2016 - 6:47 am
1. When I choose ANALYSIS: TYPE = MIXTURE to analyze data of two factors and two classes, why do I get different factor means in .out file and SAVE = FSCORES file.
For example, in .out file for c1 and c2
f1 mean 0.768, f2 mean 0.736 (c2 are set to zero) of STDYX.
However, in fscore file, I got the result for c1:
f1 0.819523732, f2 0.802319149
for c2:
f1-0.381434447, f2 -0.362233933
I am not sure what happened?
2. I have results of F1 and C_F1. They are both eta results, what's the difference?
Thanks ~
 Bengt O. Muthen posted on Monday, November 21, 2016 - 4:53 pm
1. Estimated factor scores don't have the same means and variances as true scores.

2. F is the overall value, mixed over classes and C_F is the class-specific value.
 sharon su posted on Monday, December 05, 2016 - 11:43 pm
I am trying to find the more suitable factor mixture model for the simulated data. I have two assumptions: 1factor/1class¡B1factor/2class. However, in the 1factor/2class situation, person belonged to the second class were zero. The output still had different Loglikelihood and Information Criteria values compared with 1factor/1class.
Why the estimation can be computed when one class were 0?and different with 1factor/1class?
Thanks ~
 Bengt O. Muthen posted on Tuesday, December 06, 2016 - 6:11 pm
We need to see the two outputs to answer your question - send to Support along with your license number.
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