Ben Luo posted on Wednesday, September 07, 2011 - 7:49 pm
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@0C_2@1C_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 - 12:26 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 inverted-U 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 U-shaped growth model instead with my three time-point dataset?
Is it possible that including another variable (time-invariant or time-variant) will improve the goodness of fit for the growth model?
Thank you for considering my na´ve questions! Many thanks!
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 Sunday, October 16, 2011 - 10:49 pm
Thanks Dr. Muthen! Now I know my dataset can only possibly fit a linear model.
There is one follow-up 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?
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.