Multiple Indicator Growth Model PreviousNext
Mplus Discussion > Growth Modeling of Longitudinal Data >
Message/Author
 Larry Kurdek posted on Thursday, August 29, 2002 - 12:50 pm
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 z-score) 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 (time1-2, 2-3, amd 3-4) 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;
[h1ns-h4ns@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 y-variables 8
Number of x-variables 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 0.500D-04
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

Chi-Square Test of Model Fit

Value 390.119
Degrees of Freedom 23
P-Value 0.0000

Chi-Square Test of Model Fit for the Baseline Model

Value 1222.879
Degrees of Freedom 28
P-Value 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
Sample-Size 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 h1ns-h4ns@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 h1ns-h4ns@1;
hlns BY h1ns@0 h2ns@1 h3ns@2 h4ns@3;
[h1ns-h4ns@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 y-variables 8
Number of x-variables 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 0.500D-04
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

Chi-Square Test of Model Fit

Value 394.860
Degrees of Freedom 25
P-Value 0.0000

Chi-Square Test of Model Fit for the Baseline Model

Value 1222.879
Degrees of Freedom 28
P-Value 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
Sample-Size 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) 391-9971
Fax: (310) 391-8971
Web: www.StatModel.com
Support: Support@StatModel.com

Copyright (c) 1998-2002 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 observed-variable 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.
 Larry Kurdek posted on Friday, August 30, 2002 - 4:58 pm
Thanks again. Indeed, allowing for lag-1 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 1-5 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 mis-specified? Thank you.
 Linda K. Muthen posted on Tuesday, October 05, 2004 - 8:54 am
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 follow-up 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?
 Linda K. Muthen posted on Friday, October 15, 2004 - 10:13 am
Re: October 15 - 1:24am - That would be the mean of the intecept growth factor.
 Linda K. Muthen posted on Friday, October 15, 2004 - 10:15 am
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 z-scores which I doubt would yield very much at all.

Have I lost the plot here?


many thanks


Jon
 Linda K. Muthen posted on Friday, October 22, 2004 - 9:36 am
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.
 Linda K. Muthen posted on Tuesday, October 26, 2004 - 8:22 am
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.
 Linda K. Muthen posted on Wednesday, July 06, 2005 - 8:53 am
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.
 Abderrahim Oulhaj posted on Thursday, April 12, 2007 - 6:47 am
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
 Linda K. Muthen posted on Monday, January 19, 2009 - 4:09 pm
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 multi-group 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
 Bengt O. Muthen posted on Tuesday, January 20, 2009 - 4:02 pm
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 two-level growth model (3-level 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.
 Selahadin Ibrahim posted on Monday, April 27, 2009 - 5:14 am
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
 Bengt O. Muthen posted on Monday, April 27, 2009 - 9:52 am
The question is how the non-invariance 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 non-invariant 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.
 Selahadin Ibrahim posted on Monday, April 27, 2009 - 1:20 pm
Thanks Bengt. The non-invariance shows mostly in intercepts -which are more important than non-invariance in factor loadings in my case-am I right?
I am also asking the subject matter experts to make judgments on the clinical significance of the observed differences in intercepts.

Thanks
 Linda K. Muthen posted on Tuesday, April 28, 2009 - 10:45 am
I think intercept non-invariance is more common than factor loading non-invariance but I don't think one is more important than the other.
 Selahadin Ibrahim posted on Thursday, April 30, 2009 - 11:24 am
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.
 Linda K. Muthen posted on Thursday, April 30, 2009 - 1:14 pm
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.
 Selahadin Ibrahim posted on Thursday, April 30, 2009 - 6:40 pm
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
 Linda K. Muthen posted on Thursday, April 30, 2009 - 6:46 pm
Yes. This would test for intercept invariance not factor loading invariance.
 Roxana Dragan posted on Friday, May 29, 2009 - 8:31 am
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.
 Bengt O. Muthen posted on Friday, May 29, 2009 - 8:56 am
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].
 Christoph Weber posted on Wednesday, September 23, 2009 - 3:35 am
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 2-Wave 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
 Linda K. Muthen posted on Wednesday, September 23, 2009 - 10:22 am
Please send the two outputs and your license number to support@statmodel.com.
 Christoph Weber posted on Thursday, September 24, 2009 - 1:01 am
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?
 Linda K. Muthen posted on Wednesday, March 31, 2010 - 10:41 am
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
 Linda K. Muthen posted on Thursday, July 29, 2010 - 4:36 pm
You can have a cross-sectional 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 cross-sectional 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
 Linda K. Muthen posted on Thursday, May 26, 2011 - 3:24 pm
1-2. 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 multiple-indicator 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
 Bengt O. Muthen posted on Thursday, September 29, 2011 - 9:01 am
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 censored-normal 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
 Bengt O. Muthen posted on Thursday, September 29, 2011 - 9:16 pm
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 time-invariant 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.
 Bengt O. Muthen posted on Monday, October 03, 2011 - 8:23 pm
Right. Although it does contribute information about the factor.

Fine.

I would typically trust the unstandardized significance more. Ratios can have non-normal 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 mis-modelling 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 not-significant.

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...??
 Bengt O. Muthen posted on Tuesday, October 04, 2011 - 9:15 pm
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 follow-on 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
 Bengt O. Muthen posted on Thursday, October 20, 2011 - 9:43 am
Q1. The treatment-induced 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 non-linear does not). Furthermore, the linear Txslope mean is not sig, while the non-linear 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 non-linear 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 T1-T4?

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
 Bengt O. Muthen posted on Friday, October 21, 2011 - 4:29 pm
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 non-linear (t1@0 t2* t3* t4@1), simply assessing the sig. of the change from T1 to T4 is all that is necessary?
 Bengt O. Muthen posted on Thursday, October 27, 2011 - 8:57 am
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 non-linear??
 Bengt O. Muthen posted on Thursday, October 27, 2011 - 6:13 pm
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
 Bengt O. Muthen posted on Thursday, November 03, 2011 - 11:12 am
Q1. I can't think of a useful way to in a simple way do some pre-processing 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.
 Alithe van den Akker posted on Wednesday, December 07, 2011 - 3:17 am
I would like to to a multi-group 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
 Linda K. Muthen posted on Wednesday, December 07, 2011 - 11:53 am
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.
 Cecil Meeusen posted on Thursday, January 05, 2012 - 6:27 am
Dear,

I understand that the intercept of the LGM with multiple indicators is automatically set to zero.
How do we interpret the slope-mean in this case?

LGM with three measurement points in 5 years.

Means of the latent concepts (4-point 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 five-year period. Yet, there is only a growth of 2.58-2.47 = 0.11 over time.
How is this possible, or am I completely wrong in interpreting the slope here?
 Cecil Meeusen posted on Thursday, January 05, 2012 - 6:43 am
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!
 Kara Thompson posted on Monday, April 23, 2012 - 2:04 pm
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
 Linda K. Muthen posted on Monday, April 23, 2012 - 4:11 pm
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.
 Wen-Hsu 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 -0.214D-13."

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 s|swb1@0 swb2@1 swb3@2;
 Bengt O. Muthen posted on Sunday, January 20, 2013 - 5:38 pm
In your bracket statements for the intercepts you need to add

...(1);

...(2);


...(3);

in order to hold them equal across time.
 Wen-Hsu 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.
 Linda K. Muthen posted on Tuesday, January 22, 2013 - 9:32 am
Yes, these can be allowed as long as they are identified.
 Wen-Hsu 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?
 Linda K. Muthen posted on Wednesday, January 23, 2013 - 12:09 pm
It seems reasonable that these parameters could be part of a growth model.
 Wen-Hsu 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 s|swb1@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;

Chi-square = 240(24)
RMSEA = .058
CFI = .964
TLI = .945
 Linda K. Muthen posted on Friday, February 01, 2013 - 9:30 am
Are you talking about the intercepts for swb1, swb2, and swb3?
 Wen-Hsu Lin posted on Monday, February 04, 2013 - 11:38 pm
Hi, Pro. Muthen:

No. I am talking about the indicators of swb latent variables.
 Linda K. Muthen posted on Tuesday, February 05, 2013 - 5:59 am
These are held equal as part of the growth model parameterization. They should remain that way.
 Gabriella Melis posted on Tuesday, July 02, 2013 - 10:00 am
Hi,

I am running a multiple-indicator 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 cross-sweep 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 64-BIT."

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 time-varying 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
 Bengt O. Muthen posted on Tuesday, July 02, 2013 - 5:57 pm
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 multi-dimensional, multi-level, and multi-timepoint 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.
 Gabriella Melis posted on Wednesday, July 03, 2013 - 10:39 am
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 multi-level 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 time-varying indicators... if possible?

Many thanks again.
 Bengt O. Muthen posted on Wednesday, July 03, 2013 - 3:03 pm
You do this in a wide approach - as we show for multiple-indicator growth - where you let different items at different time points have differences among their measurement parameters (so only partial invariance).
 Gabriella Melis posted on Wednesday, July 03, 2013 - 11:34 pm
Thank you.
 Matt Hawrilenko posted on Wednesday, September 04, 2013 - 1:43 pm
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 time-varying predictor rather than build a multi-group 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
 Linda K. Muthen posted on Wednesday, September 04, 2013 - 2:07 pm
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!
 Bengt O. Muthen posted on Friday, December 13, 2013 - 4:30 pm
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!
 Bengt O. Muthen posted on Sunday, December 15, 2013 - 10:55 am
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 z-score 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!
 Ya-Mei Chen posted on Thursday, May 15, 2014 - 6:54 pm
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
 Linda K. Muthen posted on Friday, May 16, 2014 - 9:54 am
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 ('u11-u610') thresholds and loadings of the factor indicators ('f1-f10') 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?
 Simone Croft posted on Wednesday, July 02, 2014 - 5:56 am
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?
 Linda K. Muthen posted on Wednesday, July 02, 2014 - 6:49 am
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 676-677 of the user's guide.
 Tihomir Asparouhov posted on Monday, July 07, 2014 - 11:40 am
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.
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