Latent growth modeling of binary outc... PreviousNext
Mplus Discussion > Growth Modeling of Longitudinal Data >
 Girmay Medhin posted on Wednesday, February 24, 2010 - 1:15 pm
I am almost new in using Mplus in general and for growth modelling in particular.

I have two binary outcomes (stunting and underweight) of infants measured at 2, 4, 6, 9, 12 and 18 months of age. 4 month time point has large number of missing values although ather time points also have few missing values

I am using WLSMV estimator following the recommenadtion for binary outcome and I have also specified H1

I am not getting measures of global fit although the modeling ended normally and parameters are generated

If I use ML as an estimator it generates some but not all fit indices

Any one who might give me an advice

 Linda K. Muthen posted on Wednesday, February 24, 2010 - 1:38 pm
It sounds like the model does not converge. Please send the output and your license number to
 Girmay Medhin posted on Monday, March 01, 2010 - 4:00 pm
I am still working with LGM of stunting and underweight to identify the best fitting unconditional model (linear, quadratic or feely estimated non-linear) .

You were right in your response and the model was not converging. The whole problem was something to do with constraints. I did consult Mplus uses Guide and two books ( Bollen K etal 2006, Duncan TE, etal 2006)

Three questions for now
1. Under Theta parameterization is it meaningful to constrain all variances of repeated measure to be equal? (eg y1 y2 y3 (1);

2. Unlike ordered categorical variable binary outcome has only one threshold and the advice is to constrain it to Zero across measurement time with nean or variance of the latent variable. Is it also meaningful to let the mean of threshold be estimated in the model but fix it overtime ?
Mplus command: [ y1$1-y5$1] (1); in place of [ y1$1-y5$1@0] ;

3. My data collection time points are 2, 4, 6, 9, 12, and 18 months of age. If the quadric model fits well and resulted, say, intercept = -3.4, slope = 1.2, quadratic = -0.8 -- all statistically significant how do I interpret these parameters? I just couldn’t get around the probit-probability link when it comes to the coefficients { any reference would also be helpful)

thank you
 Bengt O. Muthen posted on Tuesday, March 02, 2010 - 1:21 pm
1. Yes

2. That's an alternative parameterization where you have to use


3. First, you want to have a zero time score say at the first time point (so use 0, 2, ... instead of 2, 4, ...). Second, the linear and quadratic signs imply a shape of an increasing function which then levels off. The intercept is the negative threshold, so your threshold is positive. A high threshold implies a small probability of u=1 as opposed to u=0.
 Girmay Medhin posted on Saturday, March 13, 2010 - 6:30 pm
After obtaining best fitting quadratic unconditional LGM I am trying to identify factors that might affect the parameters of the latent growth curve within conditional LGM. Main interest being maternal mental health (CMD) other factors may also affect both CMD and the three growth parameters.

1. Is it correct to alter (remove or add covariance) while fitting conditional LGM (covariances between the three parameters) based on MI values?
2. For 0/1 coded binary outcome Mplus reports mean values of the threshold that corresponds to the category 0. Does that make any difficulty in the interpretation of the probit regression coefficients or the probability of getting 1?

3. Covariances between growth parameters are not significant. But their presence (freely estimated) significantly affects the effect of CMD on growth parameters. Should I have them in the model or fix them to zero? While fitting unconditional LGM it was necessary to have i with s@0; i with q@0) to make the model converge.

4. Is it correct to interpret the standardized probit regression coefficients as the expected change of an outcome (probit(u)) in response to a unit change on the risk factor? Or should we interpret it in terms of the expected effect on the probability?

Thank you
 Girmay Medhin posted on Saturday, March 13, 2010 - 6:33 pm
One related question

CMD (as a binary variable) is hypothesized to be influenced by other factors and also influence latent growth parameters. If we use a continuous measure of CMD (i.e. score) should we use the same method of estimation?

Thank you
 Linda K. Muthen posted on Sunday, March 14, 2010 - 10:26 am
1. and 3. Always include the residual covariances of the growth factors in the model.

2. To understand thresholds, see the Topic 2 course handout.

3. This is the change in u* in standard deviation units for a one standard deviation unit of x.

The same method of estimation is used if CMD is categorical or continuous if there are other categorical dependent variables in the model.
 Girmay Medhin posted on Monday, March 15, 2010 - 12:15 am
Thank you Linda for the response. It was very helpful

Few more points:
1. One of the unconditioal models does not converge if the covariance of interecept factor and quadratic facor are included. But, it converges if I constrain it to Zero Is it then wrong to have i with Q@0?

2. In the conditional LGM is it alowerd to constrain one or more of the three variances (i, l, q)to zero? (this is again to force the model convergence. If YES what are the covariates which are significant predictors of that particular factor predicting?

Many thanks
 Linda K. Muthen posted on Monday, March 15, 2010 - 9:45 am
If a model does not converge, it is best to try to find the reason why. Fixing a value arbitrarily is not a good idea. Please send your output and license number to
 Girmay Medhin posted on Tuesday, March 16, 2010 - 5:30 pm
Thank you for helpful advice. I am Back to the basics

-now trying to understand best fitting function of which freely loading non-linear function one of them
-Thetha parametrization because I will be conditioning later
i by u1@1 u2@1 u3@1 u4@1 u5@1 u6@1;
s by u1@0 u2* u3* u4* u5* u6@1;
s i ; s with i; [i@0 s ]; [u1$1-u6$1] (1) ;
u1@1 u2 u3 u4 u5 u6;
resulted in unidenified model. The complaint on the outpute is variance of s
is it logical to fix u6 to u6@1 on the last command because the time is bounded by u1@0 and u6@1 in the s by command?

thank you
 Linda K. Muthen posted on Tuesday, March 16, 2010 - 5:39 pm
Please send your full output and license number to
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