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 daniel rodriguez posted on Saturday, June 28, 2003 - 2:03 pm
I have been running GGMM with an overall model (e.g., linear), and providing start values for the growth factors level and trend, as suggested. However, my variances are always equal between classes. I am modeling with continuous dependent variables. Should I get unique variance for each class, instead of equal variance for all calsses?
 Linda K. Muthen posted on Saturday, July 05, 2003 - 9:03 am
The default is equal variances for all classes. If you want variances specific to each class, you need to mention the variances in the class specific part of the model command.
 Nathan Stein posted on Friday, January 20, 2012 - 12:22 pm
Hi Linda,

I would like the variances to be specific to each class. Could you provide the syntax for how to mention the variances in the class specific part of the model command? Thank you.

If it helps, here is my current syntax:
MODEL:
%OVERALL%
i l q | sxsevpre@0 sxsev_2@1 sxsev_4@2 sxsev_6@3 sxsev_8@4 sxsev_10@5 sxsev_12@6 sxsevpst@7 sxsev3m@8 sxsev6m@9 ;

i l q on sangpma rgcogpma;

c on sangpma rgcogpma;
 Linda K. Muthen posted on Friday, January 20, 2012 - 2:10 pm
%c#1%
i l q;

If you have more than two classes, add the other classes.
 Nathan Stein posted on Wednesday, February 01, 2012 - 9:54 am
Great. Thank you. We used the i, l, and q parameters to graph a latent class trajectory. Now we would like to add two additional curves depicting within class variance, one that is 1 standard deviation above and one that is 1 standard deviation below the trajectory. How would we do this? Could you provide us with the steps that would enable us to calculate the parameter estimates for these two additional curves?
 Linda K. Muthen posted on Wednesday, February 01, 2012 - 1:25 pm
Mplus does not provide this plot. You would need to compute that values and plot them in another program, for example, Excel or R.
 Nathan Stein posted on Tuesday, March 13, 2012 - 1:22 pm
Would I simply add (or subtract) a standard deviation from each parameter estimate. For example,

If the intercept of a line is 28.71 (variance = 52.99) and the slope is -2.87 (variance = 1.20), would the line that is one standard deviation below this line have an intercept of 21.43 (28.71 - 52.99^.5) and a slope of -3.97 (-2.87 - 1.20^.5)?
 Linda K. Muthen posted on Wednesday, March 14, 2012 - 2:46 pm
You need to compute the predicted value of y at each time point. See Slide 45 in the Topic 3 course handout to see how to do this. This is where the values of your growth curve come from. Then you need to get the standard deviation at each time point and add and subtract to the predicted value at each time point. This gives the other curves. Note that the standard deviation is the square root of the variance.
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