Message/Author 

Lee_50 posted on Friday, December 30, 2005  12:46 pm



I use the Mplus User's Guide and then have some questions. 1. Some papers write down the status facotr, it is the same the intercept growth factor, isn't it? ( I think it is same but I want to make sure.) 2. Fix the model, consist of two basic randomeffects growth factors: an overall status factor and a linear growth facotr...so overall status factor is also the same in (1). 3. Want to use: Fullinformation maximumlikelihood estimation under missingness, so I search it but the option='estimator', it is not find...could you tell me? just default or ML( is the FIML) 4. Want to use: satorrabentler corrections to chisquare tests of model fit and parameter standard errors, but if I just use the Multilevel model, I got chisquare test value. is this same?? 5. Use the other kind of modelin, I got the Pvalue from output, but the mulitlevel model result didn't give this part(pvalue) how to calcuate pvalue? or it is possible in Mplus? Some questions is not difficult. But I really want to make sure. Cuz, each book and paper..they use the different word even if the same meanning..this kinds of thing it is not good me..confuse. Thank you so much for helping. It is too help to me. 


12. When the slope growth factor has the zero time score at the first timepoint, the intercept growth factor describes initial status. If the zero time score is at the last timepoint, the intercept growth factor describes final status. Perhaps this is what you mean. 3. Any maximum likelihood estimator will give you fullinformation maximum likelihood. 4. MLM gives the SatorraBentler chisquare. 5. I think you mean you are not getting chisquare and the pvalue for chisquare. When means, variances, and covariances are not sufficient statistics for model estimation, chisquare is not available. 


I have run a latent growth curve model to analyze the development of ADH problems. I have analyzed whether the development of these problems is different in the control and intervention condition, by regressing the intercept and the slope of the growth curve model on the intervention status (cond). There is a significant effect of the intervention status on the slope (not on the intercept). QUESTION: I would like to visualize the growth of ADH problems in the control and intervention condition. Can this be done in Mplus? I have already tried this, but I cannot find a way to visualize the growth in the two groups: Iadh Sadh  ADHD1@0 ADHD2@.5 ADHD3@1 ADHD4@1.5; Iadh Sadh; Iadh with Sadh; Iadh on male cond; Sadh on male cond; output: sampstat modindices(3) stand cinterval TECH4 TECH1; PLOT: TYPE IS PLOT3; SERIES is adhd1  adhd4 (Sadh); 


Look at the adjusted means plot. I think that will give you what you want. 


Hi Linda, I am running a latent growth model with 4 time points, and received this message: THE MODEL ESTIMATION TERMINATED NORMALLY WARNING: 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 SLOPE. Do you see anything wrong with my code below? I tried loosening starting values for loadings to slope, and tried providing a positive starting value for the variance of SLOPE, since the message suggested that there may be a negative variance for this factorI did indeed see this in the TECH4. Can you give me suggestions, or may I send you my data? Thank you! USEVARIABLES ARE ALDH2DI YEAR1 YEAR2 YEAR3 YEAR4; MISSING IS .; ANALYSIS: TYPE = MEANSTRUCTURE; MODEL: INTERCEPT BY YEAR1@1 YEAR2@1 YEAR3@1 YEAR4@1; SLOPE BY YEAR1@0 YEAR2*.33 YEAR3*.67 YEAR4@1; INTERCEPT ON ALDH2DI; SLOPE ON ALDH2DI; INTERCEPT WITH SLOPE; OUTPUT: SAMP STAND RES MODINDICES(0) tech1 tech4; 


Part of the problem may be that you are not specifying the growth model correctly. If you use BY, you must fix the intercepts to zero and free the growth factor means. It would be easier if you use the growth language: intercept slope  YEAR1@0 YEAR2*.33 YEAR3*.67 YEAR4@1; Then everything is done correctly as the default and the default starting values are better than those for the BY statements. I would also fit the growth model before adding ON statements. 


Thanks, Linda! It worked! I have one more question: I saw both variance of latent intercept and slope, and RESIDUAL variance for latent intercept and slope reported in Bengt Muthen's writeup of a similar analysis, where he modeled the effect of predictors on latent intercept and latent slope: http://gseis.ucla.edu/faculty/muthen/Papers/Article_080.pdf In my output, I do see the residual variances, including for latent intercept and latent slope, but not the variances of latent intercept and slope themselves. Where do I find variance of latent intercept and latent slope in the Mplus output (not residual variances)? Thank you so much! 


You can find these in TECH4. 


Thanks, Linda. I see this in the TECH4: ESTIMATED COVARIANCE MATRIX FOR THE LATENT VARIABLES INTERCEP SLOPE ETHNIC INTERCEP 1.677 SLOPE 0.318 1.117 ETHNIC 0.033 0.082 0.250 But, what are the standard errors and significance levels? I saw that Bengt provided these along with the latent intercept and slope variances in his aforementioned writeup. Thanks again. 


Linda, can you tell me whether standard errors and significance levels are available for these variances? 


To get the SEs of thes TECH4 quantities, you would have to express the quantities in Model Constraint as new parameters. Then the SEs are obtained automcatically using the delta method. 


Thanks, Bengt. Can you tell me what I am doing wrong here (see code below)? Is it possible to estimate both the variance and residual variance of a latent variable using the delta method? I get the message: THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL. PROBLEM INVOLVING PARAMETER 14. THE CONDITION NUMBER IS 0.000D+00. USEVARIABLES ARE ETHNIC YEAR1 YEAR2 YEAR3 YEAR4; MISSING IS .; DEFINE: ETHNIC = ETHNIC1; ANALYSIS: TYPE = MEANSTRUCTURE; MODEL: INTERCEPT SLOPE  YEAR1@0 YEAR2* YEAR3* YEAR4@1; INTERCEPT ON ETHNIC; SLOPE ON ETHNIC; INTERCEPT WITH SLOPE; MODEL CONSTRAINT: NEW(INTERCEPT SLOPE); OUTPUT: SAMP STAND RES MODINDICES(0) tech1 tech4; 


You have to define New parameters in Model Constraint  you haven't done that. For an example, see UG ex 5.20. 


Bengt, I apologize for any oversight I am making, but I can't think of how to define the variances of latent intercept and latent slope using other terms in the model, except R2 + residual variance (i.e., variance explained plus variance unexplained). However, R2 and residual variance are not have explicitly named terms in the model; these are just statistics produced automatically by Mplus. What I am looking for are the standard errors of the variances for latent intercept and latent slope in my latent growth model, as you reported in your writeup here: http://gseis.ucla.edu/faculty/muthen/Papers/Article_080.pdf I see the variances in the TECH4 output, but not the standard errors of the variances. Can you tell me how you would write these model constraints? 


I apologize: In my previous post, I meant to write, "However, R2 and residual variance are not explicitly named terms in the model; these are just statistics produced automatically by Mplus." So, I am just not sure how to write the model constraints. 


TECH4 does not have standard errors. For a wellfitting model, you can obtain standard errors for the variances by running the unconditional model where variances are estimated. This is probably the best solution in your case where residual variances at not model parameters. 


Hello again Linda. If a predictor GENDER with females coded as 1 affects a latent growth factor negatively, is it more appropriate to say that females showed LESS growth from the starting point (reflected by score on Initial Status) or showed SLOWER growth from the starting point? I have heard velocity of latent growth. How does one measure velocity of latent growth? Is this something different than what is reflected by the loadings onto the latent growth factor, or the effect of a predictor such as gender on the latent growth factor? Thank you for your knowledge, Lisa 


Linda, my quesrtion is: How does one measure velocity of latent growth? Is this something different than what is reflected by the loadings onto the latent growth factor, or the effect of a predictor such as gender on the latent growth factor? Is it reflected in how much freely estimated loadings onto the latent growth factor jump from one time point to another? For example, if you have several loadings: .00, *, *, *, *, 1.00 (with * meaning it is freely estimated), and the jump between loadings between time 1 and time 2 is a bigger jump than the loadings between time 4 and time 5, could you say that the first part of the trajectory has greater VELOCITY of growth? Is this velocity quantifiable? 


I am not familiar with the concept of velocity of growth  perhaps you can google for references that describe it. 


Hi, Dr. Muthen! I think Lisa M. Yarnell may be referring to the linear term in a growth model. I've heard the linear term referred to as velocity and the quadratic term as acceleration. I think the mean of the linear term in the LGM is what she's looking for. 


Drs. Muthen: If in a 4 time point LGM, I set error variance to be the same across time points (which I see demonstrated in the Mplus manual), is it also possible to correlate errors for subsequent time points? Given that it is the same scale administered at multiple time points, it makes sense that scores would be correlated. However, does this introduce some sort of dependency issue in the modeleither due to the errors being set to be equal across the time points, or otherwise? The model below runs well without the correlated errors, but crashes when I correlate them in either of the patterns shown below (in the correlations between YEAR1YEAR4). MODEL: INTERCEPT SLOPE  YEAR1@0 YEAR2@0.385 YEAR3@0.692 YEAR4@1; INTERCEPT ON GENDER ETHNIC; SLOPE ON ALLELE; INTERCEPT WITH SLOPE; YEAR1YEAR4 (1); YEAR1 WITH YEAR2; YEAR2 WITH YEAR3; YEAR3 WITH YEAR4; !YEAR1 WITH YEAR2 YEAR3 YEAR4; !YEAR2 WITH YEAR3 YEAR4; !YEAR3 WITH YEAR4; 


The fact that the same item is repeatedly measured is what the growth models takes into account. There may also be a need for residual covariances across time. You can look at modification indices to check this. You can hold the residual variances equal across time or not. There are generally held equal in multilevel modeling programs but that is not necessary in Mplus. If you want me to look at your output, please send it and your license number to support@statmodel.com. 


I am running a conditional LGCM with multiple indicators at each time point; each indicator is an ordered categorical. In my model I have timevarying (TVC) and timeinvariant covariates (TIC). TVC and TIC are included as sets of dummies, as they are all categorical too. The model runs well, however I loose 50% of the cases compared with the unconditional LGCM. Then I tried to run the same conditional model after restating the names of the TIC and TVC, as we do for other models when we want Mplus to use all the available cases (fullinformation ML), for example: construct1991 0n educ1 educ2; educ1 educ2; ... In this second case I only loose around 15% of cases, but unfortunately the model does not converge (message: number of iterations exceeded). Please, can I do anything to avoid dropping all those cases and get my model to run? If not, would it be correct to compare results, e.g. growth factors' estimates, between the unconditional and conditional models if the number of cases varies so much between the two? Thank you. 


Here's a trick that we suggest when you have missing on tics and the missingness for the tic at time t implies that the outcomes y at time t are missing. Score the tics at values not designated as missing data symbols. Then subjects with missing on tics will be included but not affect the likelihood due to missing on outcomes. 

ywang posted on Wednesday, March 26, 2014  12:56 pm



Dear Drs. Muthen, We worked on a latent class growth model. The outcome variable is clearly very skewed. One reviewer mentioned that it has implications for erroneously identifying latent classes in a population when we treated it as a continuous variable. What can we do for the skewed outcome variable in latent class growth modeling (and the parallel latent growth modeling linked to another outcome)? Thanks a lot! 


Two issues are relevant in this situation. One is whether the classes make substantive sense based on theory. The other is whether external validity can be demonstrated using a distal outcome. See the following paper which is available on the website: Muthén, B. (2003). Statistical and substantive checking in growth mixture modeling. Psychological Methods, 8, 369377. 


Dear Drs. Muthens, I have developed a quadratic LGCM with 5 time points (varying times of observation). I notice that when I add a time invariant covariate (dichotomous) to the model, it will not converge, stating that there is a "nonpositive definite fisher information matrix". The output also states: PROBLEM INVOLVING PARAMETER 9. Using TECH 1 I see that this parameter is the correlation between my covariate and the intercept. By chance I have managed to resolve the problem by defining time differently (time = time/24). 1. However, I am unclear as to what the initial problem was. Are you able to tell me what this might have been? 2. Also, unfortunately, this new time scale does not make much sense in my model, is there any other way of resolving this problem without changing the time scale? Thank you very much for your help! 


Since you mention a correlation, it sounds like you have both i ON x; and i WITH x; which would not be identified. But maybe I am not understanding your situation, in which case you should send the problematic output to support along with your license number. 


Dear Bengt, Thank you very much for your reply. I believe I have solved this problem. However, I have a remaining issue, which is that the Loglikelhood of the unrestricted model (H1) appears to be more negative than the restricted model (H0  my model). This appears to imply that the my model fits the data better than this unrestricted model, which doesn't seem to make sense. This also provides a negative chisquare when comparing the model. I am wondering if you have any idea why this might be the case, and what I might be able to do to rectify this? Thank you very much for all your help. 


Please send this output to Support along with your license number. 


Dear Linda or Bengt! I have run a latent growth model where I regress the intercept and slope growth factors on a continuous predictor X. X has a negative effect on the intercept and a positive effect on the slope while the correlation between the intercept and the slope is negative. The suspicion arises that the positive effect of X on the slope might be confounded by the negative association between X and the intercept (those high in X tend to have lower starting values and therefore more room to increase in the outcome). Would it make sense to let the slope regress on both X and the intercept? Would this give me the effect of X on the slope while controlling for the intercept? 


That's a perfectly fine approach. 


I am trying to estimate the unconditional model for a piecewise latent growth curve model using the following syntax: Alpha_C by Crit_1@1 Crit_2@1 Crit_3@1 Crit_4@1 Crit_5@1; Beta1_C by Crit_1@0 Crit_2@1 Crit_3@2 Crit_4@2 Crit_5@2; Beta2_C by Crit_1@0 Crit_2@0 Crit_3@0 Crit_4@1 Crit_5@2; [Crit_1@0 Crit_2@0 Crit_3@0 Crit_4@0 Crit_5@0]; [Alpha_C Beta1_C Beta2_C]; Alpha_C with Beta1_C Beta2_C; Beta1_C with Beta2_C; I am getting the following message: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE...PROBLEM INVOLVING VARIABLE BETA2_C. I have looked at the regular and TECH4 output and am unable to figure out why I can getting this message. Any ideas on how to remedy this problem? 


Please send the output and your license number to support@statmodel.com. 


I am trying to estimate a conditional model for a piecewise latent growth curve model using the following truncated syntax which does not show the syntax for estimating the latent intercepts (alphas) and slopes (betas): Alpha_P with Beta1_P Beta2_P; Beta1_P with Beta2_P; Beta1_P Beta2_P on Alpha_C; Beta1_C with Beta2_C; Alpha_C with Beta1_C Beta2_C; Beta1_P with Beta1_C; Beta2_P with Beta2_C; Panic_M with Alpha_P Beta1_P Beta2_P; Crit_M with Alpha_C Beta1_C Beta2_C; Panic_7@0; Panic_M with Crit_M; I am receiving the following message: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE...PROBLEM INVOLVING VARIABLE BETA2_C. I have checked the regular and tech outputs as well as correlations and variances/res. variances. Any ideas on how to diagnose the problem? 


With regard to the model in the previous post, I also receive this message: THE STANDARD ERRORS FOR H1 ESTIMATED SAMPLE STATISTICS COULD NOT BE COMPUTED. THIS MAY BE DUE TO LOW COVARIANCE COVERAGE. THE ROBUST CHISQUARE COULD NOT BE COMPUTED. Does this message mean that parameter estimates for this model should not be trusted? Also, how does one determine model fit if the chisquare cannot be computed? 


Please send the output and your license number to support@statmodel.com. 


I'm running a 2group latent growth model of longitudinal model to find support for invariance (the slope loadings and residual variances at each time point) between the two treatment groups in my sample. I want to justify collapsing both groups into one sample to run subsequent analysis. When I first ran the two groups together, the model of random linear and random quadratic slope fit well, but the variance of each growth factor was not significant.. When running the model as a 2group, the variance of each growth factor within the each group is now significant! Can this be due to increased power with a 2group approach? 


Check that each group in the 2grp run has the same number of parameters as each group's run. Perhaps invariance of some parameters is defaulted in which case more power is obtained. 


Thank you for your response. I have a followup question. If variance of a growth parameter is found to be initially not significant, (1) are you justified in adding a covariate (i.e. treatment arm) to the model? (2) Given the above results, what does it mean if the effect of the covariate on the growth factor is found to be significant? Many thanks 


(1) Yes, because you have a different amount of power to detect the variation when a covariate is included. (2) It means that there is significant variation in the growth factor. 


Hi Dr. Muthen, I am running a quadratic latent growth model with four time points, and the intercept fixed at time 1. The unconditional model yieled a nonsignificant negative residual variance for the linear slope term, so this term was fixed to 0. When various predictors of growth parameters are added in the conditional model, I am receiving an error that there is a nonpositive definite firstorder derivative product matrix, and the error indicates that the problematic parameter is the variance of the latent intercept in the psi matrix. I have tried a number of solutions, including fixing the residual variance, changing the starting values, etc., but nothing seems to solve the problem. An additional issue seems to be that the conditional model has worse fit than the unconditional model Do you have any suggestions for how to proceed? Are the estimates for prediction of the growth parameters from the conditional model interpretable? Thank you. 


Perhaps you have a binary covariate that has its mean or variance mentioned in which case the message can be ignored. If not, send output to Support along with license number. The conditional model imposes more restrictions and therefore has more ways of misspecification. 


Thanks for the quick response! There are not any binary covariates, but there are some Likerttype covariates that are positively or negatively skewed for which the means are estimated. Could this result in the same situation as the binary covariate? 


No. You should send your output to Support along with your license number so we can diagnose. 

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