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sharon su posted on Saturday, July 30, 2016 - 6:36 am
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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 ~ |
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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. |
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sharon su posted on Sunday, July 31, 2016 - 7:58 pm
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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 ~ |
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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. |
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sharon su posted on Monday, August 08, 2016 - 5:02 am
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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? |
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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. |
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sharon su posted on Friday, September 02, 2016 - 1:58 am
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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: F2 WITH F1 0.260 0.016 16.405 0.000 the STDYX result: F2 WITH 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? |
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Your factor variances were probably not generated with variances one. |
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sharon su posted on Friday, September 02, 2016 - 8:00 pm
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I've checked the data, the variances were about 1, covariance about 0.25. MPLUS code: ANALYSIS: TYPE = MIXTURE; ALGORITHM = INTEGRATION; ITERATIONS = 1000; MODEL: %OVERALL% f1 BY i1-i30; f2 BY i31-i60; %c#1% [i1-i60@0]; 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? |
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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. |
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sharon su posted on Monday, September 26, 2016 - 8:18 pm
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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 : VARIABLE:NAMES ARE u1-u6; CATEGORICAL ARE u1-u6; Would it be linear link funcion assumpution? Thanks ~ |
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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. |
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sharon su posted on Tuesday, September 27, 2016 - 9:41 pm
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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 ~ |
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Link = probit; in the Analysis command. |
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sharon su posted on Wednesday, September 28, 2016 - 11:00 pm
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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 ~ |
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7.17 uses MLR and all maximum-likelihood estimators use Link=logit as the default. |
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sharon su posted on Saturday, October 01, 2016 - 12:49 am
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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? |
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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 |
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sharon su posted on Monday, November 14, 2016 - 4:19 am
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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 ~ ANALYSIS: TYPE = MIXTURE; MODEL: %OVERALL% f1 BY i1-i5; f2 BY i6-i10; %c#1% [i1-i10@0]; f1 with f2; %c#2% [i1-i10@0]; f1 with f2; |
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Your input gives equal loadings. To get group-varying residual variances, just add i1-i10; in both groups. I don't know why you fix the intercepts to zero. |
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sharon su posted on Monday, November 14, 2016 - 9:51 pm
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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 F1 BY I1 0.843 I2 0.797 I3 0.836 I4 0.797 I5 0.805 Latent Class 2 F1 BY 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? |
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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. |
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sharon su posted on Monday, November 21, 2016 - 6:47 am
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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 ~ |
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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. |
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sharon su posted on Monday, December 05, 2016 - 11:43 pm
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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 ~ |
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We need to see the two outputs to answer your question - send to Support along with your license number. |
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