

Latent growth modeling with three tim... 

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Ben Luo posted on Thursday, September 08, 2011  1:49 am



Hi, I am a beginner of MPlus. I am grateful if you could help me with the following question. I tested a latent growth model with 102 subjects and three points in time. The code was as follows. Variable: NAMES ARE C_1 C_2 C_3; MISSING = ALL (99); ANALYSIS: ESTIMATOR = ML; MODEL: I S  C_1@0 C_2@1 C_3@2; OUTPUT: TECH1; TECH4; Here, C_1, C_2, and C_3 are all continuous. However, the output showed a warning message, that 'THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE. THIS COULD INDICATE A NEGATIVE VARIANCE/RESIDUAL VARIANCE FOR A LATENT VARIABLE, A CORRELATION GREATER OR EQUAL TO ONE BETWEEN TWO LATENT VARIABLES, OR A LINEAR DEPENDENCY AMONG MORE THAN TWO LATENT VARIABLES. CHECK THE TECH4 OUTPUT FOR MORE INFORMATION. PROBLEM INVOLVING VARIABLE I.' I noticed that the variances of I and S are both negative (while the residual variances are both positive). In addition, in TECH4 output, the correlation between I and S is 999.000. These are consistent with what the warning says. What might the underlying probelm be? How can I fix it? Many thanks! 


A growth model is not suitable for any data set. You should first do a Type=Basic analysis and look at the means and variances for the 3 outcomes as well as plot the observed individual curves for each subject. This will tell you if a linear model seems appropriate for the data. For instance, if the means increase over time and the variances decrease over time, a growth model might be hard to fit. Or, perhaps there is no trend so that you don't need a slope growth factor. 

Ben Luo posted on Wednesday, October 12, 2011  6:26 am



Thank you, Dr. Muthen! Your instructions are very helpful to me. Following your suggestions, I have calculated the means and variances of three variables, and found that the means have a U shape while the variances have an invertedU shape across three time points. In addition, I have also checked the individual curves for each case, and found that about half of the cases presented a U shape, one fifth linearly increase, and one fifth linearly decrease. Based on these results, can I say that there is little likelihood that a linear growth model fits well with the data? Can I model a Ushaped growth model instead with my three timepoint dataset? Is it possible that including another variable (timeinvariant or timevariant) will improve the goodness of fit for the growth model? Thank you for considering my naïve questions! Many thanks! Ben 


The problem with having only three time points is that you cannot fit a quadratic model. The linear model has only one degree of freedom. For further details, see Slide 52 of the Topic 3 course handout. 

Ben Luo posted on Monday, October 17, 2011  4:49 am



Thanks Dr. Muthen! Now I know my dataset can only possibly fit a linear model. There is one followup question that I hope you can help with. As so far the univariate linear growth model does not work for my dataset, is it possible that a more complicated, multivariate LGM still can instead fit better with the data? In other words, is univariate LGM a basis of multivariate LGM? Thanks a lot! Ben 


A univariate linear growth model is the same as a multivariate linear growth model with the exception that the residual variances are held equal across time for the univariate model. The model I refer to with one degree of freedom is a multivariate linear growth model where residual variances are estimated at each time point. 

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