Message/Author 


I am following Example 22.4 of the Mplus manual to set up a multiple indicator growth model. I have 2 indicators (each transformed to a zscore) at each of four yearly assessments. (I am ignoring missing data at this point and using only cases with complete data.)My concern about the output is that several of the correlations among the latent variables exceed 1. I am not sure why this occurred. I assumed (as in the example that residuals were not correlated), although this is unlikely with longitudinal data. I allowed for a simple autocorrelation structure (time12, 23, amd 34) in another run, but the correlations among the latent variables still exceeded 1. Any help would be greatly appreciated. Larry Kurdek Here is the program code and selected output. Mplus VERSION 2.1 MUTHEN & MUTHEN 08/29/2002 3:26 PM INPUT INSTRUCTIONS TITLE: Linear model on h self only DATA: FILE IS datso2; FORMAT IS 5x,12F6.2/1X,8F6.2/5X,12F6.2/1X,8F6.2; VARIABLE: NAMES ARE h1npa h2npa h3npa h4npa h1npb h2npb h3npb h4npb h1nsa h2nsa h3nsa h4nsa h1nsb h2nsb h3nsb h4nsb h1dis h2dis h3dis h4dis w1npa w2npa w3npa w4npa w1npb w2npb w3npb w4npb w1nsa w2nsa w3nsa w4nsa w1nsb w2nsb w3nsb w4nsb w1dis w2dis w3dis w4dis; USEVARIABLES ARE h1nsa h2nsa h3nsa h4nsa h1nsb h2nsb h3nsb h4nsb; MISSING ARE ALL (9); ANALYSIS: TYPE = MEANSTRUCTURE; ITERATIONS=70; H1ITERATIONS=70; MODEL: h1ns BY h1nsa h1nsb (1); h2ns BY h2nsa h2nsb (1); h3ns BY h3nsa h3nsb (1); h4ns BY h4nsa h4nsb (1); [h1nsa h2nsa h3nsa h4nsa] (2); [h1nsb h2nsb h3nsb h4nsb] (3); hins BY h1ns h2ns h3ns h4ns@1; hlns BY h1ns@0 h2ns@1 h3ns@2 h4ns@3; [h1nsh4ns@0 hins@0 hlns]; OUTPUT: tech1 tech4; INPUT READING TERMINATED NORMALLY Linear model on h self only SUMMARY OF ANALYSIS Number of groups 1 Number of observations 272 Number of yvariables 8 Number of xvariables 0 Number of continuous latent variables 6 Observed variables in the analysis H1NSA H2NSA H3NSA H4NSA H1NSB H2NSB H3NSB H4NSB Continuous latent variables in the analysis H1NS H2NS H3NS H4NS HINS HLNS Estimator ML Maximum number of iterations 70 Convergence criterion 0.500D04 Maximum number of steepest descent iterations 20 Input data file(s) datso2 Input data format (5X,12F6.2,/,1X,8F6.2,/,5X,12F6.2,/,1X,8F6.2) THE MODEL ESTIMATION TERMINATED NORMALLY TESTS OF MODEL FIT ChiSquare Test of Model Fit Value 390.119 Degrees of Freedom 23 PValue 0.0000 ChiSquare Test of Model Fit for the Baseline Model Value 1222.879 Degrees of Freedom 28 PValue 0.0000 CFI/TLI CFI 0.693 TLI 0.626 Loglikelihood H0 Value 2481.700 H1 Value 2286.641 Information Criteria Number of Free Parameters 21 Akaike (AIC) 5005.400 Bayesian (BIC) 5081.122 SampleSize Adjusted BIC 5014.536 (n* = (n + 2) / 24) RMSEA (Root Mean Square Error Of Approximation) Estimate 0.242 90 Percent C.I. 0.221 0.264 Probability RMSEA <= .05 0.000 SRMR (Standardized Root Mean Square Residual) Value 0.137 MODEL RESULTS Estimates S.E. Est./S.E. H1NS BY H1NSA 1.000 0.000 0.000 H1NSB 1.315 0.074 17.680 H2NS BY H2NSA 1.000 0.000 0.000 H2NSB 1.315 0.074 17.680 H3NS BY H3NSA 1.000 0.000 0.000 H3NSB 1.315 0.074 17.680 H4NS BY H4NSA 1.000 0.000 0.000 H4NSB 1.315 0.074 17.680 HINS BY H1NS 1.000 0.000 0.000 H2NS 1.108 0.064 17.428 H3NS 1.108 0.065 17.068 H4NS 1.000 0.000 0.000 HLNS BY H1NS 0.000 0.000 0.000 H2NS 1.000 0.000 0.000 H3NS 2.000 0.000 0.000 H4NS 3.000 0.000 0.000 HLNS WITH HINS 0.014 0.007 2.138 Means HINS 0.000 0.000 0.000 HLNS 0.011 0.010 1.088 Intercepts H1NSA 0.077 0.039 1.976 H2NSA 0.077 0.039 1.976 H3NSA 0.077 0.039 1.976 H4NSA 0.077 0.039 1.976 H1NSB 0.034 0.046 0.736 H2NSB 0.034 0.046 0.736 H3NSB 0.034 0.046 0.736 H4NSB 0.034 0.046 0.736 H1NS 0.000 0.000 0.000 H2NS 0.000 0.000 0.000 H3NS 0.000 0.000 0.000 H4NS 0.000 0.000 0.000 Variances HINS 0.212 0.033 6.430 HLNS 0.005 0.004 1.361 Residual Variances H1NSA 0.479 0.045 10.603 H2NSA 0.616 0.057 10.816 H3NSA 0.649 0.060 10.772 H4NSA 0.759 0.070 10.763 H1NSB 0.371 0.045 8.155 H2NSB 0.471 0.055 8.613 H3NSB 0.476 0.057 8.330 H4NSB 0.599 0.070 8.610 H1NS 0.040 0.021 1.865 H2NS 0.089 0.023 3.914 H3NS 0.079 0.024 3.246 H4NS 0.109 0.031 3.476 TECHNICAL 4 OUTPUT ESTIMATES DERIVED FROM THE MODEL ESTIMATED MEANS FOR THE LATENT VARIABLES H1NS H2NS H3NS H4NS HINS ________ ________ ________ ________ ________ 1 0.000 0.011 0.022 0.033 0.000 ESTIMATED MEANS FOR THE LATENT VARIABLES HLNS ________ 1 0.011 ESTIMATED COVARIANCE MATRIX FOR THE LATENT VARIABLES H1NS H2NS H3NS H4NS HINS ________ ________ ________ ________ ________ H1NS 0.173 H2NS 0.250 0.208 H3NS 0.264 0.318 0.265 H4NS 0.255 0.312 0.341 0.233 HINS 0.212 0.250 0.264 0.255 0.212 HLNS 0.014 0.021 0.026 0.029 0.014 ESTIMATED COVARIANCE MATRIX FOR THE LATENT VARIABLES HLNS ________ HLNS 0.005 ESTIMATED CORRELATION MATRIX FOR THE LATENT VARIABLES H1NS H2NS H3NS H4NS HINS ________ ________ ________ ________ ________ H1NS 1.000 H2NS 1.316 1.000 H3NS 1.235 1.354 1.000 H4NS 1.274 1.416 1.372 1.000 HINS 1.109 1.187 1.113 1.149 1.000 HLNS 0.495 0.652 0.713 0.859 0.446 ESTIMATED CORRELATION MATRIX FOR THE LATENT VARIABLES HLNS ________ HLNS 1.000 

bmuthen posted on Thursday, August 29, 2002  1:08 pm



You want to change your statement hins BY h1ns h2ns h3ns h4ns@1; to hins BY h1nsh4ns@1; to avoid estimating loadings for h2ns h3ns (this is not a growth model otherwise). Let us know if your correlation problem persists. 

lkurdek posted on Thursday, August 29, 2002  5:58 pm



Thanks so much for the quick reply. I made the change you suggested, but the correlations still exceed 1. See output: Mplus VERSION 2.1 MUTHEN & MUTHEN 08/29/2002 8:53 PM INPUT INSTRUCTIONS TITLE: Linear model on h self only DATA: FILE IS datso2; FORMAT IS 5x,12F6.2/1X,8F6.2/5X,12F6.2/1X,8F6.2; VARIABLE: NAMES ARE h1npa h2npa h3npa h4npa h1npb h2npb h3npb h4npb h1nsa h2nsa h3nsa h4nsa h1nsb h2nsb h3nsb h4nsb h1dis h2dis h3dis h4dis w1npa w2npa w3npa w4npa w1npb w2npb w3npb w4npb w1nsa w2nsa w3nsa w4nsa w1nsb w2nsb w3nsb w4nsb w1dis w2dis w3dis w4dis; USEVARIABLES ARE h1nsa h2nsa h3nsa h4nsa h1nsb h2nsb h3nsb h4nsb; MISSING ARE ALL (9); ANALYSIS: TYPE = MEANSTRUCTURE; ITERATIONS=70; H1ITERATIONS=70; MODEL: h1ns BY h1nsa h1nsb (1); h2ns BY h2nsa h2nsb (1); h3ns BY h3nsa h3nsb (1); h4ns BY h4nsa h4nsb (1); [h1nsa h2nsa h3nsa h4nsa] (2); [h1nsb h2nsb h3nsb h4nsb] (3); hins BY h1nsh4ns@1; hlns BY h1ns@0 h2ns@1 h3ns@2 h4ns@3; [h1nsh4ns@0 hins@0 hlns]; OUTPUT: tech1 tech4; INPUT READING TERMINATED NORMALLY Linear model on h self only SUMMARY OF ANALYSIS Number of groups 1 Number of observations 272 Number of yvariables 8 Number of xvariables 0 Number of continuous latent variables 6 Observed variables in the analysis H1NSA H2NSA H3NSA H4NSA H1NSB H2NSB H3NSB H4NSB Continuous latent variables in the analysis H1NS H2NS H3NS H4NS HINS HLNS Estimator ML Maximum number of iterations 70 Convergence criterion 0.500D04 Maximum number of steepest descent iterations 20 Input data file(s) datso2 Input data format (5X,12F6.2,/,1X,8F6.2,/,5X,12F6.2,/,1X,8F6.2) THE MODEL ESTIMATION TERMINATED NORMALLY TESTS OF MODEL FIT ChiSquare Test of Model Fit Value 394.860 Degrees of Freedom 25 PValue 0.0000 ChiSquare Test of Model Fit for the Baseline Model Value 1222.879 Degrees of Freedom 28 PValue 0.0000 CFI/TLI CFI 0.690 TLI 0.653 Loglikelihood H0 Value 2484.071 H1 Value 2286.641 Information Criteria Number of Free Parameters 19 Akaike (AIC) 5006.141 Bayesian (BIC) 5074.651 SampleSize Adjusted BIC 5014.408 (n* = (n + 2) / 24) RMSEA (Root Mean Square Error Of Approximation) Estimate 0.233 90 Percent C.I. 0.213 0.254 Probability RMSEA <= .05 0.000 SRMR (Standardized Root Mean Square Residual) Value 0.140 MODEL RESULTS Estimates S.E. Est./S.E. H1NS BY H1NSA 1.000 0.000 0.000 H1NSB 1.317 0.075 17.635 H2NS BY H2NSA 1.000 0.000 0.000 H2NSB 1.317 0.075 17.635 H3NS BY H3NSA 1.000 0.000 0.000 H3NSB 1.317 0.075 17.635 H4NS BY H4NSA 1.000 0.000 0.000 H4NSB 1.317 0.075 17.635 HINS BY H1NS 1.000 0.000 0.000 H2NS 1.000 0.000 0.000 H3NS 1.000 0.000 0.000 H4NS 1.000 0.000 0.000 HLNS BY H1NS 0.000 0.000 0.000 H2NS 1.000 0.000 0.000 H3NS 2.000 0.000 0.000 H4NS 3.000 0.000 0.000 HLNS WITH HINS 0.013 0.007 1.829 Means HINS 0.000 0.000 0.000 HLNS 0.012 0.010 1.189 Intercepts H1NSA 0.076 0.039 1.943 H2NSA 0.076 0.039 1.943 H3NSA 0.076 0.039 1.943 H4NSA 0.076 0.039 1.943 H1NSB 0.032 0.046 0.704 H2NSB 0.032 0.046 0.704 H3NSB 0.032 0.046 0.704 H4NSB 0.032 0.046 0.704 H1NS 0.000 0.000 0.000 H2NS 0.000 0.000 0.000 H3NS 0.000 0.000 0.000 H4NS 0.000 0.000 0.000 Variances HINS 0.238 0.034 7.091 HLNS 0.007 0.004 1.833 Residual Variances H1NSA 0.479 0.045 10.600 H2NSA 0.616 0.057 10.815 H3NSA 0.649 0.060 10.769 H4NSA 0.758 0.070 10.759 H1NSB 0.370 0.046 8.128 H2NSB 0.470 0.055 8.587 H3NSB 0.476 0.057 8.321 H4NSB 0.600 0.070 8.597 H1NS 0.049 0.022 2.221 H2NS 0.083 0.022 3.712 H3NS 0.074 0.024 3.080 H4NS 0.119 0.031 3.826 TECHNICAL 4 OUTPUT ESTIMATES DERIVED FROM THE MODEL ESTIMATED MEANS FOR THE LATENT VARIABLES H1NS H2NS H3NS H4NS HINS ________ ________ ________ ________ ________ 1 0.000 0.012 0.024 0.036 0.000 ESTIMATED MEANS FOR THE LATENT VARIABLES HLNS ________ 1 0.012 ESTIMATED COVARIANCE MATRIX FOR THE LATENT VARIABLES H1NS H2NS H3NS H4NS HINS ________ ________ ________ ________ ________ H1NS 0.189 H2NS 0.251 0.188 H3NS 0.264 0.290 0.243 H4NS 0.277 0.310 0.342 0.256 HINS 0.238 0.251 0.264 0.277 0.238 HLNS 0.013 0.020 0.026 0.033 0.013 ESTIMATED COVARIANCE MATRIX FOR THE LATENT VARIABLES HLNS ________ HLNS 0.007 ESTIMATED CORRELATION MATRIX FOR THE LATENT VARIABLES H1NS H2NS H3NS H4NS HINS ________ ________ ________ ________ ________ H1NS 1.000 H2NS 1.331 1.000 H3NS 1.232 1.360 1.000 H4NS 1.257 1.412 1.373 1.000 HINS 1.121 1.188 1.099 1.121 1.000 HLNS 0.369 0.557 0.655 0.797 0.329 ESTIMATED CORRELATION MATRIX FOR THE LATENT VARIABLES HLNS ________ HLNS 1.000 Beginning Time: 20:53:57 Ending Time: 20:53:58 Elapsed Time: 00:00:01 MUTHEN & MUTHEN 11965 Venice Blvd., Suite 407 Los Angeles, CA 90066 Tel: (310) 3919971 Fax: (310) 3918971 Web: www.StatModel.com Support: Support@StatModel.com Copyright (c) 19982002 Muthen & Muthen 

bmuthen posted on Thursday, August 29, 2002  6:37 pm



You have some negative residual variances for the factor at each time point, which suggests a model misspecification. Also, your inadmissible factor correlations may suggest that some of the observedvariable correlations should not be channeled through the factors but through residual correlations. You can start by trying to add correlations between the measurement errors for the same indicator across adjacent time points, i.e. h1nsa with h2nsa and h1nsab with h2nsb, etc for later time points. 


Thanks again. Indeed, allowing for lag1 errors eliminated all negative variances and brought all factor correlations below 1.00. 

bmuthen posted on Friday, August 30, 2002  10:13 pm



Great. 

Anonymous posted on Tuesday, October 05, 2004  7:53 am



I am fitting a Multiple Indicator LGC model. I have the same latent factor measured by six items at each time point. All observed items are measured on a 15 point scale and I have used the default option of setting the unstandardised loading of the first variable to 1 to set the scale of the latent variable. When I estimate the growth parameters, everything seems fine apart from the intercept is estimated as 6.49. Is this possible or is my model misspecified? Thank you. 


I would have to see your full output to answer your question. Please send it to support@statmodel.com. 

Anonymous posted on Friday, October 15, 2004  1:24 am



Linda thank you for pointing out that my error was in failing to constrain the intercept mean to zero. The model now works fine. However, my followup question concerns the interpretation of the intercept parameter  is there no way of getting a handle on the starting point of the growth trajectory (assuming the model specifies the intercept to be average trajectory at time 1? Patrick 

Anonymous posted on Friday, October 15, 2004  1:28 am



Dear Bengt and Linda If I have a latent growth curve model with 5 waves of data and I specify the trajectory as linear, what use am I making of the intervening 3 waves. Would my model be any different if I used only the first and last waves? 


Re: October 15  1:24am  That would be the mean of the intecept growth factor. 


Re: October 15  10:13am  First of all, the model would not be identified with only two waves. In addition, interveing timepoints enhances the power, that is, reducing the standard errors of the parameter estimates. 

Jon Heron posted on Friday, October 22, 2004  12:39 am



Hi, I have a question about Multiple Indicator example 6.14 in the Mplus3 manual. Page 93 states that the default parameterization is to fix the intercepts of f_i to zero. Surely these f_i are what the growth model is fitted to, and hence centring them would be like fitting a growth model to observed zscores which I doubt would yield very much at all. Have I lost the plot here? many thanks Jon 


In a multiple indicator growth model, f takes the role of the observed outcome in a regular growth model for a single outcome at each timepoint. In the regular growth model, the intercept of the outcome is fixed at zero, allowing the outcome means to vary as a function of the growth factor means across time. In other words, fixing the intercepts of fi to zero does not imply that the means of fi are zero at each timepoint. You might find it helpful to look at the table in Chapter 16 which shows the new growth model language in comparison to using the regular Mplus language. 

Jon Heron posted on Tuesday, October 26, 2004  6:05 am



Thanks Linda, so why is it necessary to fix the mean of i to zero in the MI model but not the regular model. Other than just to avoid the warning: THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL. PROBLEM INVOLVING PARAMETER 24. 


In the regular model, you fix the intercepts of the outcome to zero and estimate the means of i and s. In the MI model, the intercepts of the factor indicators are held equal, the mean of i is fixed, and the mean of s is estimated. These are two different parameterizations. I believe that Bengt teaches the multiple indicator model is his web class which can be accessed at: http://www.statmodel.com/trainhandouts.html This will give a complete picture. It's difficult to understand a model a parameter at a time. 

Anonymous posted on Tuesday, July 05, 2005  7:26 pm



I have run a multiple indicator growth model and estimated a single class solution. I am wondering if it is possible to extend the multiple indicator growth model to examine classes of growth. If so, are the model specifications for multiple classes the same as in growth models using observed data? Thanks. 


Yes, this model can be extended to multiple classes. As a first step, just change the CLASSES statement to 2 instead of 1. Then you can see which parameters are constrained to be equal across classes and make adjustments as you wish. 


Dear All, I wonder whether you have a reference or paper which explains in details the meaning and interpretation of the parameters in the the "Multiple Indicator Growth Model for categorical outcomes" along with some practical examples. I am also interested to a reference describing the model formulation and estimation. Many thanks, A. Oulhaj 

finnigan posted on Monday, January 19, 2009  2:16 pm



Linda/Bengt I am considering a power analysis to establish a sample size for a multiple indicator growth model covering five measurement occasions. All constructs are measured using likert scales. Data will be collected from individuals within organisations within each occasion. The study is exploratory and there is practically no longitudinal research in this field to provide any input for parametre values Is there any MPLUS example that you are aware of that addresses this. thanks 


The Monte Carlo counterpart for Example 6.15 is a good place to start. I don't know of any papers that cover this topic. 

finnigan posted on Tuesday, January 20, 2009  4:49 am



Thanks,Linda. I have a few more questions: I will be using the LMACS framework which involves: 1 use of factor analysis 2 establishing invariance 3 using multigroup analysis 4 adding predictors 5 regressing growth parametres on each other. Would a separate montecarlo analysis need to be considered for each of these steps or would the one montecarlo analysis suffice that looks like ex 6.15? What modifications would be needed to the MC example to amend it to consider clustered data and a longitudinal study ie individual within organisation over five measurement occasions. Given that there is very little empirical estimates of parametre values available in the my field for the MC analysis , is there any other way to estimate them in the MC analysis? Finally,what would be the difficulties if one was to adopt Cohen's approach,(typified in the free software), to undertake the steps above in the context of longitudinal study with clustered data? Thanks 


I would do a single MC analysis so that the full model is in place to assess power. See chapter 11 of the UG where a twolevel growth model (3level analysis) is given as an example (ex 11.4). Second to last question  no. Last question  I don't know of a Cohen approach that could cover this situation. 


Hi Bengt and Linda, I am running multiple indicator growth model. With full measurement invariance not satisfied what is the next step? Assuming i have partial measurement invariance (say for example 50 % of the loadings and intercepts are invariant)is it still appropriate to run growth models? How does that affect my results and conclusion? Your advice on this is greatly appreciated. thanks a lot 


The question is how the noninvariance shows up and how interpretable it is. If you have only 50% of the measurement parameters invariant across adjacent time points that is much worse than 50% across several time points. Ultimately, the interpretability of the non invariance is key. Generally speaking, in my view only a minority of the measurement parameters can be noninvariant while still being able to make a plausible claim that you are studying growth for the same latent variable construct. If your construct changes over time (as evidenced by measurement noninvariance), perhaps a single growth model should be replaced by two or more related processes. 


Thanks Bengt. The noninvariance shows mostly in intercepts which are more important than noninvariance in factor loadings in my caseam I right? I am also asking the subject matter experts to make judgments on the clinical significance of the observed differences in intercepts. Thanks 


I think intercept noninvariance is more common than factor loading noninvariance but I don't think one is more important than the other. 


thank you. When checking measurement invariance across time is it appropriate and also interpretable to control for other covariates(For example gender impacting on one of the indicators)? Thanks. 


If you plan on making comparisons across gender, you should check for measurement invariance across gender. With gender as a covariate, you can check intercept invariance. If you want to also check factor loading invariance, you would need to do a multiple group analysis. 


thank you Linda! to clarify things my interest is in longitudinal invariance. i do have invariance in factor loading longitudinally. i do not satisfy longitudinal measurement invariance in intercepts of some items. I suspect gender plays a role in this and would like to control for gender when checking MI longitudinally. my question was can I control for gender by allowing gender to be a covariate(impacting on the items) when checking MI longitudinally? Thanks, Selahadin 


Yes. This would test for intercept invariance not factor loading invariance. 


I estimate a linear growth model on a continuous indicator at different time points. I did not impose any restrictions on the model other than the time scores required for a linear growth model. The estimate of the mean of the intercept is not equal to the estimate of the mean of the indicator for which the time score was set to zero. I thought they should be equal. I also checked example6.1 from user's guide, and the mean of the intercept did not turn out to be equal to the mean of the indicator where the time score is zero. For the example 6.1 the difference between the two is small, but for my application, the difference is about 50%. Should not they be equal? Thank you. 


They would only be equal if the model fits perfectly. If the model is correct, then in the population the mean of the 1st outcome is equal to [i]. 


Dear Drs. Muthén, I am using two wave panel data. My goal is to model parallel developmental processes of 3 latent variables. My previous results are based on latent difference score models. Now I tried to do a multigroup LDS model, but I get the error, that the model may not be identified. Then I did a 2Wave LGM (fixing the resid. variances of the latent var equal to 0), what I think is exactly the same as the LDS. I got the same Chi²(df) and also the same estimates. But in this case the mutiple group analysis is possible. Why is the multigroup LDS not possible? My syntax for the growth part is. LGM: i s  exb2@0 exb3@1; [i s]; exb2@0; exb3@0; LDS: delta by exb3@1; exb3 on exb2@1; exb3@0; [exb2 delta]; Thanks Christoph Weber 


Please send the two outputs and your license number to support@statmodel.com. 


Dear Dr. Muthén, I've allready found the difference. Using the LDS approach Mplus estimates the intercept of exb3 for the second group. After fixing it = 0 the LGM and LDS show the same results. thanks christoph Weber 

dlsp posted on Wednesday, March 31, 2010  7:21 am



Dear Dr. Muthén, I am also running a multiple indicator LGM with continous outcomes across three time points and five indicators at every time point. The mean of the Intercept is fixed to zero as the default. Is there a possibility to estimate the Intercept freely or can I calculate the Intercept by hand? 


The model would not be identified if you free the mean of the intercept growth factor unless you place other restrictions on the intercepts of the observed factor indicators. 

Yen M. To posted on Thursday, July 29, 2010  10:53 am



Dear Drs. Muthén & Muthén, Thank you in advance for your time. I am using data from a restricted dataset and I am attempting to perform a multiple indicator growth model for continuous outcomes with Mplus v.5. I am examining a theoretical model of parental involvement across time. The research indicates that parental involvement is determined by engagement (E), accessibility (A), and responsibility (R). From the dataset, I have identified several factor indicators that reliably fit into each component. After reviewing the examples from the short course handouts and the user’s guide I still have a few inquires I was hoping you could address… 1) Do factors always have to deal with time? That is, is it possible for me to have factors that are unrelated to each other temporally? 2) Is it possible for me to have a different set of indicators for each factor? I noticed that in the examples there are a different number of variables, but I also notice that these variables are the same in each factor. Can I have a different number of variables and the variables be qualitatively different too? If so, can you direct me to an example of the syntax? 3) Is it possible for the indicators of a factor to have differing scales, such that one may be continuous and the others categorical? Thank you for any guidance you can give me. Respectfully, Yen To 


You can have a crosssectional model where factors measured at the same time are related to each other or you can have one or more factors for which the factor indicators have been measured repeatedly across time. In a crosssectional study, you can compare the parameters of the same factors across groups. In a growth model, you can compare parameters of the same factors across time. In both cases, the key for making the comparisons is that the factors being compared have been tested for measurement invariance. If at least partial measurement invariance does not exist, it does not make sense to compare factors across groups or time. In most cases, the same factor will have the same factor indicators. 

Yen M. To posted on Friday, July 30, 2010  1:44 pm



Ah, I see now that I was conceptualizing it all wrong. Thank you for your help! YT 

Hemant Kher posted on Wednesday, May 25, 2011  8:44 pm



Dear Professor Muthen and Muthen, I am writing about the multiple indicator growth model; in particular I have some questions on the course handout, Topic 4, slide #82. 1. I find point #s 2 and 3 on slide 82 very interesting; point 2 suggests examining growth in each indicator separately and in the sum of indicators, and the suggestion in point 3 is that growth models must be the same. Can you please elaborate on point 3 – are you suggesting that the slopes of all these models should be about the same (roughly parallel)? 2. Can we interpret point #3 to imply that, if some of the slopes are vastly different (e.g. suppose we have 4 indicators, and that the slopes for 3 indicators are about the same, but the slope for 4th indicator is twice as large as others) then the result will be lack of invariance? 3. I have read a few papers on the impact of invariance on a multiple indicator growth model, but they do not use points 2 and 3 – have you written about this (or are aware of a paper that uses this approach)? Thank you as always for your time, Hemant 


12. The same type of model should fit each indicator. For example, they should all be linear or all quadratic. 3. I don't know of a paper that describes this. We have not written about it. We have only taught about it. 

Vanessa posted on Wednesday, September 28, 2011  8:37 pm



Hi. I am trying to model a multiple indicator LGCM. I have 3 tasks as established markers of latent factor, each assessed at 4 time points. I have modelled each indicator separately, to examine their growth models. Task 1 has sig linear +quad means; Task 2 has near sig (.07) lin + quad means; while Task 3 has a cubic shape. Task 3 is cubic because a 'parallel' form was used at T2 and T4 and this form was harder than that used at T1 and T3: Because of this I have released the intercepts for T2 and T4. Is this an appropriate way to account for the change in task means (due to a more difficult version of task being used) and is this then fine to include this in a multipleindicator LGC with the other two indicators, since it's different shape has an explanation due to different forms (of the same task) being used at the different time points? Many thanks in advance 


Typically, you need the same growth shape for all factor indicators in order for growth modeling of a factor to be well motivated. You seem to say that Task 1 and Task 2 are perhaps too easy, or that Task 3 is too hard, and this is the reason for the different shapes. If that means ceiling effects for Task 1 and 2 and/or floor effects for Task 3 I can imagine that the growth modeling can work, if those effects are handled by say censorednormal modeing. But that's quite advanced modeling. 

Vanessa posted on Thursday, September 29, 2011  4:56 pm



Thanks for your reply above. Sorry, it seems I wasn't quite clear. I have 3 indicators for a multiple indicator LCG, each measured at 4 timepoints. Two indicators show the same growth shape (increasing but linear and quadratic). The other indicator goes up (time 2), then down (time 3), then up again (time 4). This is because for this indicator the same parallel form was used at Time 1 and 3 (hence similar means) and a different, apparently easier form at times 2 and 4 (hence the same, and higher than time 1 and 3, means at these times). My question is, is it valid to free the intercepts for this indicator at time 2 and 4, to account for the different means due to a different form being used? Could I then include this indicator with the others in a LGCM? Thanks again 


You can do that if you think the intercept differences across time for that indicator accounts for its growth shape being different from the other two indicators. You still have two indicators with timeinvariant measurement parameters. And the third indicator contributes by adding information about the factor. 

Vanessa posted on Monday, October 03, 2011  6:25 pm



Thanks for your reply. Just to clarify your response, does this mean that if I don't hold the intercepts invariant for one of the indicators, then this indicator will not be contributing information about change over time, to the factor means (as it will be accounted for by the varying intercepts)? An additional question regarding assessing the shape of individual indicators, I am basing this on the significance level of the means for linear and quadratic slopes in LGCMs. If eg. the quadratic mean is significant under the unstandardised solution, but not the standardised solution (while linear sig under both), how would you interpret the shape? Thanks a lot for your time. 


Right. Although it does contribute information about the factor. Fine. I would typically trust the unstandardized significance more. Ratios can have nonnormal distributions (which can be checked out by Bayes). 

Vanessa posted on Monday, October 03, 2011  8:34 pm



Many thanks for your responses! 

Vanessa posted on Monday, October 03, 2011  9:04 pm



I have one additional and related question. I am testing the effect of an intervention, modelling it as a covariate (in the first instance) in multiple indicator models. I am just wondering what the impact of mismodelling the shape of the change could have? Ie. The indicators (for the control group) all show significant linear and quadratic means. (One indicator in TX group showed only linear). When I model only a linear slope, the impact of Group on slope is sig (and not only just); if I model linear and quad slopes, the effect of Group on both slopes is notsignificant. Thanks in advance for any insight 

Vanessa posted on Monday, October 03, 2011  9:42 pm



Following on from your response at October 03, 2011  8:23 pm, regarding determining the shape of the curve for an indicator... (and related to above) is there an optimum way of doing this? I have an example where, for 4 timepoints, if I model a quadratic and linear slope, both are significant. But when I model a cubic, in addition to linear and quad, only the linear is significant. Would I take the shape as linear or quadratic? Is it best to model all potential slopes (L+q+c) first in just one model, and then take the shape as following only the significant means in this model?? (ie. basing it only on one model) Or test for sig. of slopes incrementally in different models, starting with the simplest curve, and then...?? 


There is not an optimum way to do this. As a general rule you want to choose a starting point for your modeling that is as close as possible to the best model  but you don't know what that is. I would work stepwise, going from a simple model with linear slope to one with both linear and quadratic. For 4 time points as you have, I wouldn't bother with cubic. I would keep the quadratic if either its mean or variance was significant. Regarding your earlier question on effects of the tx dummy on growth factors, that is tricky with a quadratic. See how we describe and resolve this problem in http://www.statmodel.com/download/Article_0832.pdf 

Vanessa posted on Wednesday, October 19, 2011  10:46 pm



Hi, I have some followon questions regarding testing for interevention effects, following Muthen & Curran 97... I have multiple indicator models, and 4 equally spaced time points, the intervention starting directly after T0. How should one determine the shape of the additional treatment slope? Is it based on the shape of the curve when examining a LGCM in treatment group only (eg. as above, a significant linear mean in Tx group)? In the miultigroup model, if I free the middle time scores for the added treatment slope, (ie. T1@0 T2* T3* T4@1) does the significance value for the mean of the t slope now refer only to the change between T1 and T4? If I'm also examining numerous interactions, including between initial status and Tx slope [T], as well as regressing T on covariates, how is the intercept of T now interpreted? Is the significance of T equivalent to testing the main effect now? Many thanks in advance 


Q1. The treatmentinduced change may not follow the original growth shape, but depends on the intervention. You want to keep it simple, so linear is a good start and probably all you need. Q2. Yes. But you can evaluate significance of change between any points. All you have to do is to create a "lambda*mu" NEW parameter in Model Constraint, where lambda is the estimated time score and mu is the slope growth factor mean. Q3. Yes. 

Vanessa posted on Thursday, October 20, 2011  6:26 pm



Following on from your response above 1) If linear is a good start, how can one determine whether it is sufficient? In a model testing only the main treatment effect, when I free two time points for the Tslope (eg. T1@0 T2* T3* T4@1) Chi2 diff test indicates this is sign. better fit than the linear Tx slope (the linear also needs/has the T@0 while the nonlinear does not). Furthermore, the linear Txslope mean is not sig, while the nonlinear Tx slope mean is sig p < .01 significant... Are there any ways of evaluating which is the more appropriate Treatment growth shape? 2) With such a nonlinear Tx slope (t1@0 t2* t3* t4@1) would you consider it necessary to evaluate the significance of change between each time point, or simply the change between T1T4? 3) Just to clarify, even when T is regressed on multiple covariates in the one model, the intercept of T is still interpreted as the mean of T (and significance indicates significant main Tx effect)? I wasn't sure whether this was only the case when T on i is the only 'covariate' in the model, because the mean of i is set at zero... many thanks 


1) chi2 testing is a good approach as you are doing. 2) No 3) I would center the covariates so that the intercept is the mean of T, and then consider its intercept as the main effect conditional on the covariate means. 

Vanessa posted on Wednesday, October 26, 2011  5:17 pm



Thanks for your reply. Just to clarify your 2) No, response. When evaluating the significance of a treatment effect (following M & C, 97), if the additional Tx slope is nonlinear (t1@0 t2* t3* t4@1), simply assessing the sig. of the change from T1 to T4 is all that is necessary? 


Yes, I think so. 

Vanessa posted on Thursday, October 27, 2011  5:21 pm



Thanks  just a theoretical query then: In the M&C, 97 multigroup modelling of interventions, do you think it is possible , if both Tx and Ctrl group graphs look linear (over 4 time points), and linear is determined as the normative slope (based on significance of slope in Ctrl group LGCM), that the added treatment slope could still be nonlinear?? 


Yes, because the treatment effect may be of different strength at different time points. 

Vanessa posted on Wednesday, November 02, 2011  8:29 pm



Still on the M&C 97 approach of testing for intervention effects, is there an optimum way of investigating whether covariates like gender etc. have an impact on the treatment slope, before actually modelling them? Are there any arguments for leaving in covariates that do not significantly impact on Tx slope? Thanks in advance 


Q1. I can't think of a useful way to in a simple way do some preprocessing that would single out such covariates. Q2. Usually you would try to let theory guide you in selecting a set of covariates and then report on the model that includes all of them, even if some are insignificant. 


I would like to to a multigroup multiple indicator growth model for continuous outcomes. I would like to see whether the intercept and slope means are different across groups. In an earlier post you indicate that restrictions need to be placed on the intercepts of the observed variable indicators. I would like to ask what these should be? (intercepts of the same indicator at different timepoints are now equal within and across groups, but the model is still not identified) thank you very much in advance 


The intercepts of the outcomes should be equal across time and groups. The factor intercepts should be zero across time and groups. The intercept growth factor mean should be zero in one group and free in the others. The slope growth factor mean should be free in all groups. 


Dear, I understand that the intercept of the LGM with multiple indicators is automatically set to zero. How do we interpret the slopemean in this case? LGM with three measurement points in 5 years. Means of the latent concepts (4point scale): Time 1: 2.47 Time 2: 2.45 Time 3: 2.58 Slope = 0.069 If the slope represents the average growth for each unit increase in time, that means that latent concept increases with 5*0.069 = 0.345 over the fiveyear period. Yet, there is only a growth of 2.582.47 = 0.11 over time. How is this possible, or am I completely wrong in interpreting the slope here? 


Here the Mplus code of the previous question: ANALYSIS: ESTIMATOR IS MLR; MODEL: etno2006 by etno11* (1) etno12* (2)etno13* (3); etno2008 by etno21* (1) etno22* (2) etno23* (3); etno2011 by etno31* (1) etno32* (2) etno33* (3); [etno11* etno21* etno31*] (4); [etno12* etno22* etno32*] (5); [etno13* etno23*] (6); [etno33*]; etno11* etno21* etno31*; etno12* etno22* etno32*; etno13* etno23* etno33*; etno2006@1 etno2008* etno2011*; etno11 with etno21* etno31*; etno21 with etno31*; etno12 with etno22* etno32*; etno22 with etno32*; etno13 with etno23* etno33*; etno23 with etno33*; int by etno2006@1 etno2008@1 etno2011@1; slp by etno2006@0 etno2008@2 etno2011@5; [etno2006@0 etno2008@0 etno2011@0]; [int@0 slp]; Thank you very much! 


Hi, I am wondering if you can run a multiple indictor growth model with more then one latent factor? For example, if CFA tells me that my indicators of internalizing are best modeled as 2 factors  depression and anxiety, can I then run a single latent growth curve model using both latent factors? OR... do I use each set of latent factors to run separate growth models, and then also model a higher order growth model. Thanks 


You can have more than one latent factor in a multiple indicator growth model. You would have one growth curve for depression and one for anxiety. You can examine the relationship between the two curves. 

WenHsu Lin posted on Sunday, January 20, 2013  5:33 pm



Dear I am trying to run a multiple indicator LGM. My model is identical as Example 6.14. So, I set up my syntax as it is written for EXP 6.14. However, I got the following message: "THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL. PROBLEM INVOLVING PARAMETER 21. THE CONDITION NUMBER IS 0.214D13." Would you mind to help. My syntax is as follow: model: swb1 by shappy1 sar1(1) w1dep(2); swb2 by shappy2 sar2(1) w2dep(2); swb3 by shappy3 sar3(1) w3dep(2); [sar1 sar2 sar3]; [shappy1 shappy2 shappy3]; [w1dep w2dep w3dep]; i sswb1@0 swb2@1 swb3@2; 


In your bracket statements for the intercepts you need to add ...(1); ...(2); ...(3); in order to hold them equal across time. 

WenHsu Lin posted on Monday, January 21, 2013  11:03 pm



Hi, professor Muthen, a follow up question is that if I can allow correlations among indicators? Thank you. 


Yes, these can be allowed as long as they are identified. 

WenHsu Lin posted on Wednesday, January 23, 2013  3:56 am



Hi, Professors: A final follow up. Is it reasonable and correct to allow indicators of the latent variable to correlated across time? 


It seems reasonable that these parameters could be part of a growth model. 

WenHsu Lin posted on Thursday, January 31, 2013  6:38 pm



Hi, Professors, I am running a a model and is indicated by the following syntax. Although the model seemed to fit the data marginally, the MI indicated that if I free the mean of one of the three time point, it will improve the fit. Is that reasonable, although I have some theoretical reason to back me up. Thank you. swb1 by shappy1 sar1(1) w1dep(2); swb2 by shappy2 sar2(1) w2dep(2); swb3 by shappy3 sar3(1) w3dep(2); [sar1 sar2 sar3](3); [shappy1 shappy2 shappy3](4); [w1dep w2dep w3dep](5); i sswb1@0 swb2@1 swb3@2; sar1 with sar2 sar3; sar2 with sar3; shappy1 with shappy2 shappy3; shappy2 with shappy3; w1dep with w2dep w3dep; w2dep with w3dep; Chisquare = 240(24) RMSEA = .058 CFI = .964 TLI = .945 


Are you talking about the intercepts for swb1, swb2, and swb3? 

WenHsu Lin posted on Monday, February 04, 2013  11:38 pm



Hi, Pro. Muthen: No. I am talking about the indicators of swb latent variables. 


These are held equal as part of the growth model parameterization. They should remain that way. 


Hi, I am running a multipleindicator latent growth curve model with categorical outcomes. The problems that I have found are: 1) Integration does not support models with scale factors 2) The crosssweep correlations between the indicators are not computed 3)Even when I cancel the scale factor, as well as the correlations in (2), the message I get is "FATAL ERROR THERE IS NOT ENOUGH MEMORY SPACE TO RUN Mplus ON THE CURRENT INPUT FILE. ... NOTE THAT THE OPERATING SYSTEM HAS NO RESTRICTION ON MEMORY USAGE FOR Mplus 64BIT." I have already run the same model with WLSMV estimator, got not very ideal fit indices (CFI and TLI=.938; RMSEA=0.069); also I have tried with WEAK invariance constraints and WLSMV and the fit definitely improves. Thus, my questions are, please: 1) Do you think my model is too complicated for the MLR estimation option? 2) Is there anything I can do to let this MLR model run? 3) If nothing can be done with MLR, am I allowed to use a STRONG invariance model to run a conditional growth curve with both timevarying and time invariant covariates, even if the WEAK invariance model fits better for the unconditional case? I apologise for the many issues presented and potential confusion, but hope you are able to help. Please, let me know should you need me to specify futher my points above. Thank you in advance, Gabriella 


The feasibility of MLR depends on how many dimensions of integration it says that your case has. With say less than 8, you can try integration=montecarlo(5000); You should also read Muthen & Asparouhov (2013). Item response modeling in Mplus: A multidimensional, multilevel, and multitimepoint example. and well as Web Note 17, both of which are on the Mplus website, for the alternative of using Bayesian analysis. I would try to relax the full invariance model, either by partial invariance or approximate invariance. 


Thank you very much for your answer, I will try with Bayesian analysis too. A further question for a parallel project under development, please: what kind of approach to longitudinal multiple indicator analysis would you suggest if (some of) the indicators change across time points? This is a problem common in developmental studies, such as cohort studies. McArdle et al. (2009) proposed a multilevel longitudinal IRT analysis based on Rasch parameterisation that only assumes measurement invariance (as it models Rasch factor scores). Could you please suggest how to deal with this issue in Mplus, as well as how to test for measurement invariance for timevarying indicators... if possible? Many thanks again. 


You do this in a wide approach  as we show for multipleindicator growth  where you let different items at different time points have differences among their measurement parameters (so only partial invariance). 


Thank you. 


Hello, I am struggling with a multiple indicator growth model. I have set my model up following example 6.14 from the user's guide (although I am working with couple data, so have also allowed partners' residuals to correlate). Two waves are missing one of the indicators, so I have constrained repeated measure residuals to be time invariant to allow model identification. Additionally, due to relatively high nonlinearity, I have included a treatment variable as a timevarying predictor rather than build a multigroup model. I receive a warning that the psi matrix is not positive definite. Looking in tech 4, one of the latent variables has correlations slightly larger than 1 with a few of the other latent variables (perhaps because, after accounting for treatment, there is not much change, and this measure is taken within a few weeks of other timepoints?). Otherwise, model fit statistics look acceptable. Any advice would be much appreciated. Thanks so much for your time. Best, Matt 


Please send the output and your license number to support@statmodel.com. 

RuoShui posted on Thursday, December 12, 2013  6:50 pm



Dear Dr. Muthen, I know that in LGCM with multiple indicators, the intercept growth factor is fixed and not estimated. But just as you said in the above posts "In other words, fixing the intercepts of fi to zero does not imply that the means of fi are zero at each timepoint." Is there any way I can get an actual number for the initial status since the real initial status is not zero? Thank you so much! 


Fixing the intercept growth factor mean at zero is not a limitation. You can't identify both it and the intercepts of the indicators. You pick any indicator and fix its intercept to zero (at all time points) and then free the intercept growth factor mean, but that doesn't buy you anything. 

RuoShui posted on Saturday, December 14, 2013  6:36 pm



Dear Bengt, Thank you very much! I understand that fixing the intercept growth factor mean at zero is certainly not a limitation. One confusion I have is that when it comes to having different classes, if the last class intercept growth factor mean is fixed at zero while the other class intercept growth factor mean is estimated (for example 0.5), can I say that the two classes are different on their slope, but not necessarily on where they start? Will the class with intercept growth factor means fixed at zero actually start higher than 0.5? And how can I know? Please let me know if understand this all wrong. Thank you so much! 


With the last class intercept growth factor mean fixed at zero and the other class (say first class) having it estimated as 0.5, the first class has a higher intercept growth factor mean. The zscore ratio 0.5/SE tells you if it is significantly higher. 

RuoShui posted on Tuesday, December 17, 2013  3:43 am



Thank you very much Bengt! 


Hi, I am using Multiple indicator linear growth model for continuous outcomes (Example 6.14) with three indicators (Nss, Ass, Iss) across 4 timpoints. Since these three indicators present different perspectives of disabilities, could you please advice how to interpret S=1.009? I cannot decide the "unit" of the S. For example,can I say "in the past 4 years, people will increase 1.009 (Nss, Ass, or Iss)disabilities per year?" Please advise how to interpret the slop and the unit of the slop (by which variable). Thank you very much!! MODEL RESULTS S  T1 1.000 0.000 999.000 999.000 T2 1.490 0.033 45.290 0.000 T3 2.797 0.057 48.658 0.000 T4 4.000 0.000 999.000 999.000 Means I 0.000 0.000 999.000 999.000 S 1.009 0.041 24.705 0.000 


Please send your output and license number to support@statmodel.com. 

Dan Cloney posted on Tuesday, July 01, 2014  7:59 pm



Hi, I am having trouble producing PVs for 3 simultaneous multiple indicator linear growth models. Specification is similar to EX 6.15 (UGv7, p137). Observed indicator ('u11u610') thresholds and loadings of the factor indicators ('f1f10') are constrained to equality over time. The intercept and slope factors of the three LGMs are related so that: i1 WITH i2, i3, s1, s2@0, s3@0; i2 WITH i3, s2, s1@0, s3@0; i3 WITH s3, s1@0, s2@0; s1 WITH s2, s3; s2 WITH s3; The model converges and I get adequate model fit and parameter estimes. I can extract EAP factor scores. When I try and extract PVs I get an initial error message: "*** FATAL ERROR THE VARIANCE COVARIANCE MATRIX IS NOT SUPPORTED. ONLY FULL VARIANCE COVARIANCE BLOCKS ARE ALLOWED. USE ALGORITHM=GIBBS(RW) TO RESOLVE THIS PROBLEM." When I use 'ALGORITHM IS GIBBS(RW);' I get a new error message: "THE CONVERGENCE CRITERION IS NOT SATISFIED. INCREASE THE MAXIMUM NUMBER OF ITERATIONS OR INCREASE THE CONVERGENCE CRITERION." I have tried to play around, e.g., 'BCONVERGENCE = 0.1;' but not sure if I'm on the right track. Any advice about simple problems i may be missing? 


Hello, I am trying to fit a LGC to my data but the model is not estimating a mean score for the growth intercept. My data is weighted and ordinal and I'm using the WLSMV estimator. I have imposed a series of constraints based on factorial invariance testing thus all indicator factor loadings and some thresholds were constrained to be equal across time. Having seen your advice above, I've tried your suggestion of constraining one variable to be zero across time to allow the estimation of the mean, but it hasn't worked, I'm still not having the mean of the growth intercept estimated: Means I 0.000 0.000 999.000 999.000 S 0.061 0.003 21.940 0.000 Can you suggest anything? 


Simone: With categorical variables, the growth model parametrization used holds the thresholds equal across time and fixes the mean of the intercept growth factor to zero. See pages 676677 of the user's guide. 


Dan I would get rid of this line i1 WITH i2, i3, s1, s2@0, s3@0; i2 WITH i3, s2, s1@0, s3@0; i3 WITH s3, s1@0, s2@0; s1 WITH s2, s3; s2 WITH s3; and let all the latent variables correlate freely and remove ALGORITHM IS GIBBS(RW). If the above constraints are very important then using biter=100000 or thin=10, i.e., increasing the number of mcmc iterations should eventually lead to convergence. 

fred posted on Friday, October 17, 2014  4:39 am



Hello, I am trying to estimate a growth curve with 3 measurments (at baseline, 1 year and 2 years). I have also three indicators (x1, x4, x10) for a factor at each assessment (aFactor, dFactor, fFactor). Here is my code: USEVARIABLES = aX1 aX4 ax10 dx1 dx4 dx10 fx1 fx4 fx10; MISSING = ALL(999); MODEL: aFactor BY ax1 ax4 ax10(12) ; dfactor BY dX1 dX4 dX10(12) ; fFactor BY fX1 fX4 fX10(12); [aX1 dX1 fX1 ] (3); [aX4 dX4 fX4] (4); [aX10 dX10 fX10] (5); [aFactor@0 dFactor fFactor]; i s  aFactor@0 dFactor@1 fFactor@2; as the following output it looks like I need to make changes to make the model identified, but I dont know what needs to be fixed or freed! THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL. PROBLEM INVOLVING PARAMETER 17. THE CONDITION NUMBER IS 0.733D17. Also, I have earlier assessed liongtidunal invariance and it looks like it holds for intercepts as well. What is the problem with my code? Thanks Fred 

Jon Heron posted on Friday, October 17, 2014  4:46 am



I think you need to centre all of your first order factors. Currently the slope factor is attempting to capture change but you have already explained change by allowing those factor means to be estimated. 


See the V7 UG, page 681 for the parametrization Jon refers to. 

fred posted on Friday, October 17, 2014  12:13 pm



Thanks, does this mean I need to remove this line [aFactor@0 dFactor fFactor]; As far as I can see from the example at page 681 that is the difference. Am I right? 


You have to change that line to [aFactor@0 dFactor@0 fFactor@0]; The means of the factors are then orchestrated by the means of the growth factors as in regular observedoutcome growth modeling. 

fred posted on Monday, October 20, 2014  1:23 am



Thanks, Two things hapens when I use [aFactor@0 dFactor@0 fFactor@0]. First it looks like the intercept of LGC is not estimated: Means I 0.000 0.000 999.000 999.000 S 0.008 0.013 0.631 0.528 Also there are negative covarainces showing up! (From tech4) ESTIMATED COVARIANCE MATRIX FOR THE LATENT VARIABLES includes S cov with Factor = 0.023 S cov with I = 0.023 Isnt it possible to estimate the mean of the intercept in this model? and what would you trace the negative covaraince to? Thanks fred 

Jon Heron posted on Monday, October 20, 2014  4:22 am



Hi Fred that's right, the mean of the intercept is not estimated. I can recommend a couple of good sources of info for these models. The first is a paper in "teacher's corner" in the SEM journal http://www.tandfonline.com/doi/pdf/10.1207/S15328007SEM0803_7 And the second, perhaps better but not as readily accessible is Sayers and Cumsille's chapter in the book "new method for the analysis of change". As for the negative covariance, there's nothihg wrong with that per se. A negative association between intercept and slope implies the data are "fanning in", i.e. variability is decreasing. cheers, Jon 


The mean of the intercept growth factor is not identified because you already have parameters that cover the means at the first time point, namely the measurement intercepts of the factor indicators. You can certainly choose to fix the measurement intercept of say the first indicator to zero, but nothing is gained by this  the model fits the same and you have just moved a parameter from one array to another. There is not need for a nonzero intercept growth factor mean. 

fred posted on Tuesday, October 21, 2014  11:31 pm



Thank you Bengt and Jon! I am also wondering whether embeding the factorial model in the LGC somehow inherently addersses the issue of longitundanal invariance, I mean would I gain further information by allowing i.e. auroregressive correlationsanalys between the error terms etc.? 

fred posted on Tuesday, October 21, 2014  11:54 pm



I mean of course autoregressive "correlations" 

Jon Heron posted on Wednesday, October 22, 2014  2:23 am



Yes, you would require at least partial longitudinal invariance to fit this model. I found slide 82 very useful http://www.statmodel.com/download/Topic4v.pdf and yes, you may find the need to correlate the errors for each item repeated across time, or perhaps introduce some method factors to capture this same feature. Christian Geiser uses the method factor approach in his Mplus book. best, Jon 

RuoShui posted on Tuesday, February 24, 2015  11:10 pm



Dear Drs. Muthen, I am conducting a study using multiple indicator growth curve model. I am wondering when testing for measurement invariance over time, is it ok to impute the missing values of the items? Or will imputation violate any assumption of measurement invariance? Thank you! 


Why impute?  ML under MAR ("FIML") handles the missing data directly and gives the same results in large samples. Imputation does not give any gain here as far as I can see. 

RuoShui posted on Wednesday, February 25, 2015  9:51 pm



Thank you very much Dr. Muthen. 


Dear Drs. Muthén, I am trying to run a multiple indicator growth model for categorical indicators with uneqal spaced time points. My idea was to combine the codes from examples 6.12 and 6.15 in the Mplus user guide. Unfortunately, this is not working and I was wondering what I am doing wrong. When I am using scaling factors as in 6.15 I get the following error: ALGORITHM=INTEGRATION does not support models with scale factors. When I am using Theta parameterization as on page 682 I get these errors: PARAMETERIZATION=THETA is not allowed for TYPE=MIXTURE or ALGORITHM=INTEGRATION. Setting is ignored. Variances for categorical outcomes can only be specified using PARAMETERIZATION=THETA with estimators WLS, WLSM, or WLSMV. This is my input code (with which I have already tested measurement invariance): VARIABLE: ... tscores=FS1 FS2 FS3; ANALYSIS: ESTIMATOR = WLSMV; TYPE= random; !PARAMETERIZATION=THETA; MODEL: F1 BY i0101 i0201i0901(L2L12) i1301; F2 BY i0102...; F3 BY i0103...; [i0101$1 i0102$1 i0103$1](I1); ... {i0101i1301@1 i0102i1703}; i s  F1 F2 F3 AT FS1 FS2 FS3; Thanks, Susanne 


If you want the random slopes of UG ex 6.22 you need to use ML. If you want to have the flexibility of scale factors you need to use WLSMV. I would recommend ML so not use scale factors. That gives better missing data handling (MAR). 


Dear Drs. Muthén, I have been trying repeatedly (with various measures) to estimate a multiple indicator latent growth curve model. I am able to test for measurement invariance and to estimate a nogrowth model, but I keep getting an error when I try to estimate linear growth. I'm including the model command in a separate post. Here is the type of error I get when I include the latent slope: THE MODEL ESTIMATION TERMINATED NORMALLY THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL. PROBLEM INVOLVING THE FOLLOWING PARAMETER: Parameter 35, [ S ] THE CONDITION NUMBER IS 0.107D15. Is there a constraint that I'm missing? This happens even if I fix the slope variance to zero. Thank you! Liz 


Here is the model: !FACTORS PUBy3 BY M1y3@1 M2y3(b) M3y3(c) M4y3(d) M5y3(e); PUBy4 BY M1y4@1 M2y4(b) M3y4(c) M4y4(d) M5y4(e); PUBy5 BY M1y5@1 M2y5(b) M3y5(c) M4y5(d) M5y5(e); PUBy6 BY M1y6@1 M2y6(b) M3y6(c) M4y6(d) M5y6(e); I S  puby3@0 puby4@1 puby5@2 puby6@3 ; !RESIDS M1y3 (er_a) M2y3 (er_b) M3y3 (er_c) M4y3 (er_d) M5y3 (er_e) M1y4 (er_a) M2y4 (er_b) M3y4 (er_c) M4y4 (er_d) M5y4 (er_e) M1y5 (er_a) M2y5 (er_b) M3y5 (er_c) M4y5 (er_d) M5y5 (er_e) M1y6 (er_a) M2y6 (er_b) M3y6 (er_c) M4y6 (er_d) M5y6 (er_e) ; 


Please disregard my prior two messages. I figured out that I had neglected to constrain the intercepts of the indicators to be equal over time. Thank you! 


Dear Drs. Muthén, I'm trying to test the longitudinal invariance of my 5factors measure and I need some help. I have 5 factors measured at 5 time point. Each factor is measured by 6 items. I defined the model specifying the factor structure for each time point A_T1 BY item1_T1 item2_T1 item3_T1... B_T1 BY... C T1 BY... D_T1 BY... E_T1 BY... A_T2 BY item1_T2 item2_T2 item3_T2... B_T2 BY... ETC.. Then I allowed errors for each item within different time points to correlate (i.e., item1_t1 with item1_t2 item1_t3 item1_t4 item1_t5 etc.) finally I run this model, to establish configural invariance. The problem is that the analysis seems to require a huge amount of time and I did not get any result. Is there any mistake in my procedure? How can I solve my probem? Thanks for your help Grace 


It sounds like you are analyzing 5 x 6 x 5 = 150 variables. This can take a long time using ML. As a first step, you should analyze the 5 factors at each time point separately, then each factor separately at all time points. You may then find that you need to simplify your model, for example drop some factors. 


Dear Dr. Muthén, Thank you very much for your reply. I did what you suggested and I found that the 5 factors structure holds at all time points. Similarly, I obtained good fit indices analyzing each factor separately at all time points. These results were expected, as the structure of the measure has been studied in different researches, all supporting the 5 factors solution. How do you suggest to go on? 


Try Estimator = uls; Y 


Dear Dr. Muthén, Thank you again for your reply. I have missing values in my data file, I recieve a warning message saying that uls cannot be used with missing values. Is there any alternative? Thanks for your help 


You can build your model gradually. Run one factor for all time point separately. A_T1, A_T2, ..., A_T5. Use this analysis to study longitudinal invariance for that factor and most importantly reduce the number of item specific correlations. Given that the factors are correlated across time there is no need to include all item specific correlations (these contribute a lot to the slow estimation). Keep only those that are statistically significant. Do this for all factors and use the ouput:svalues command to get good starting values for the final model that combines all factors across all time points. You might also find that this command improves the computation: output:nochi; This eliminates the H1 model estimation that will be slow with many variables. Instead of chisquare you can use likelihood ratio test to evaluate the longitudinal invariance. 


Thank you very much for your detailed answer, Dr. Asparouhov. Do you suggest to use starting values for all my parameters? Am I supposed to modify them in some cases? 


I would recommend using the staring values you can get through output:svalues; from simpler models. You don't need to modify them and you don't need to give starting values for all parameters, especially if you don't have such starting values (for parameters not available in the simpler models). Starting values will help reduce computational time a little bit but they are not that important. 

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