Growth curve model with binary outcome
Message/Author
 Alexander Pabst posted on Thursday, November 26, 2015 - 3:00 am
Dear all,
I am using a growth curve model in order to analyze change (i.e., treatment effect) in a binary indicator across 2 measurement occasions (Voelkle, 2007, Psychology Science, 49, 375–414). Unfortunately, the model does not provide standard errors / p-values and MPlus says “The degrees of freedom for this model are negative. The model is not identified.” I tried different options for estimator, weight, cluster, bootstrap etc. – nothing worked. Is something wrong with the model or is it underestimated due to the fact that I only have 2 time points? I’m wondering about this because a model with a continuous indicator (also two time points) can easily be estimated. In addition, when I add a third time point (+ 1 random binary variable) the model does not work either, but when I have four time points (+ 2 random binary variables) it suddenly worked fine.

The code with 2 time Points is:
MODEL:
i s | out0@0 out1@1 ;
out0@0 out1@0 ;
i s ON group ;
s ON i ;
 Bengt O. Muthen posted on Thursday, November 26, 2015 - 6:45 pm
With 2 time points and continuous outcomes you have for each group 5 pieces of information: 2 means, 2 var's, and 1 cov. With binary outcomes you have only 3 pieces. With 2 time points and binary outcomes your model has more parameters than that.

Note that you don't want to fix the residual variances at zero.

Note also that you cannot identify the variance of s with only 2 time points for either the continuous or binary case.
 Alexander Pabst posted on Friday, November 27, 2015 - 2:55 am
Dear Prof. Muthen,

Thank you very much for the information.
we are not quite sure about your explanations; as far as we understand it, the model with a continuous manifest outcome is fully saturated for two points in time if the residuals of the measurement model are fixed at zero (this is just another notation of a random anova). Please see a piece of output below.

Unfortunately, we are not able to adopt this notation if the manifest variables are binary. Moreover, we don`t understand why we only have 3 pieces of information in this case.

Thanks again for further explanations in advance.

Output for continuous model:

i s |kost0@0 kost1@1;
kost0@0 kost1@0;

Estimate
I |
KOST0 1.000
KOST1 1.000

S |
KOST0 0.000
KOST1 1.000

S WITH I -2359.906

Means
I 33.122
S 5.482

Intercepts
KOST0 0.000
KOST1 0.000

Variances
I 4330.523
S 6831.682

Residual Variances
KOST0 0.000
KOST1 0.000
 Linda K. Muthen posted on Friday, November 27, 2015 - 8:02 am
The sample statistics for binary variables are thresholds and correlations. Variances are not model parameters for binary variables. For two binary variables, you have two thresholds and one correlation, These are the three pieces of information.