Message/Author 

Anonymous posted on Wednesday, September 28, 2005  6:11 pm



I am analyzing shortterm longitudinal data. I have 4 observation points sample size of 200 individuals. I have the model developed, but I continue to get an error message that the psi (latent variable) covariance matrix is not positive definite. After researching this issue, beyond consulting the Wothke article (I'm having trouble getting my library to order this article), what may I do to remedy this issue? I believe that the data are salvagable so any tricks of the trade are well appreciated. Thanks in advance. 


If you are using Version 3.13 it should point to which variable has the problem. You may have a negative residual variance or a correlation of one somewhere. You can see the negative residual variance in the results and you can ask for TECH4 to see if any variables have a correlation greater than one. Otherwise, you may have a dependency among two or more of your variables. If it is a small negative residual variance, you could fix it to zero. Otherwise, you would need to change your model. If you can't figure this out with this information, please send your input, data, output, and license number to support@statmodel.com. 


Hi, I am a beginner with Mplus and I am trying to understand growth models. I have used this syntax: MODEL: ID SD  D1@0 D2@1 D3@2 D4@3; IN SN  NEG1@0 NEG2@1 NEG3@2 NEG4@3; IH SH  H1@0 H2@1 H3@2 H4@3; IP SP  POS1@0 POS2@1 POS3@2 POS4@3; but in the output there is a warning message: WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE. THIS COULD INDICATE A NEGATIVE/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 IN. In the tech4 section I have found that there are correlations higher than 1. I don't understand exactly what does it mean and I don't know how to solve the problem. Could you help? thank you very much, francesca 


I think this will be solved if you include residual covariances among the outcomes at each time point, for example, d1 WITH neg1 etc. Note that you should fit each process separately prior to putting them together in the same model. 

Liu Xiao posted on Friday, August 17, 2007  2:41 am



I am doing a latent growth modeling, and I meet the same problem. The syntax: agg2 by zcpagg2 ztrext2 (1); agg3 by zcpagg3 ztrext3 (1); agg4 by zcpagg4 ztrext4 (1); agg5 by zcpagg5 ztrext5 (1); agg6 by zcpagg6 ztrext6 (1); [zcpagg2 zcpagg3 zcpagg4 zcpagg5 zcpagg6] (2); [ztrext2 ztrext3 ztrext4 ztrext5 ztrext6] (3); i1 s1  agg2@0 agg3@1 agg4@2 agg5@3 agg6@4; zcpagg6 with zcpagg5; zcpagg4 with zcpagg5; 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 AGG2. I check the tech4, there are correlations higher than 1. In residual variance, there are agg2, agg3,agg4, agg6, lower than 0. How can I fix this problem? Can I use the MODEL CONSTRAINT to constraint the residual variance larger than 0 and smaller than 1? Thanks for your help. 


Having only two indicators for your factors can be problematic as the factor at each time point is not identified without borrowing information from other parts of the model. Try running the model without measurement invariance and without the growth model to see if the problem occurs with that model. If so, the model is probably not correct for the data. If it occurs only after imposing measurement invariance, it may be that the factors are not invariant across time. 


Hello, I have a quick question. I'm using a latent growth model to model math achievement across 3 time points. I've gotten the same message that others have written about (PSI matrix is not positivedefinite). The problem involves the variable "linear" (ie, the latent variable representing linear growth). There are no negative residual variances, although the residual variance for "linear" is 0.000. I've checked my variables for collinearity and do not believe that to be a factor. I've also made sure that the variables have similar variances. One of my math variables has very high kurtosis (39). I've tried transformations (square root, natural log, and inverse; then I reflected the variable and tried all 3 again)the transformations only make the kurtosis worse. Does anyone have any suggestions for how to fix this problem? Could it be due to the kurtosis, and if so, how does this affect the accuracy of the estimates that Mplus calculates? 


You say the residual variance for the linear slope is 0. That could be a cause for the message, for example if the slope is regressed on the intercept. Or the slope and the intercept are correlated 1.0  which you can see in the TECH4 output. To avoid this you may want to center at a different time point. 


The slope and intercept were actually not correlated very highly. I did try centering at a different time point, but it didn't help. I finally fixed the "linear" residual variance to .001 (based on other posts & responses, it seems that this is acceptable), and stopped getting the error message. If this is not a good solution, I'd appreciate hearing about it, although it seems that others have done similar things. Thanks for the help. 


Hi Virginia, What may also be happening in your data is something that seems to occur more in achievement data than any other type of data: insufficient changes in rankorder for modeling *variability* in growth (i.e., *very* high correlations between repeated measures). I ran into a similar problem as yours in working with some early reading achievement data (i.e., prekindergarten through 4th grade)  and (according to some developmental psychologist friends of mine) for some (but not all) types of achievement outcomes this phenomenon seems consistent with literature that suggests that, after a certain point kids tend to maintain their rank ordering in achievement. I'd still wait to see what Bengt suggests you do about it  but you may want to peek at the sample correlations (and see if they are really high) to help diagnose where this is coming from...... 


It's hard to say more without further information. Please send your input, data, output, and license number to support@statmodel.com. 

pm posted on Monday, November 12, 2007  12:52 pm



Hello, We are looking at academic achievement data using the same measure at three time points. We have four outcome variables: math, vocabulary, listening comprehension, and alphabet knowledge. We first ran the model for math, and it worked just fine. We then tried to run the models for the other three outcomes and we received the following message: WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE...... Upon reviewing the output, we notice we have a correlation between baseline and linear that is greater than 1. We have tried centering the variables at each time point and we've checked for collinearity. However, we still get the same warning message. We do not have enough df to do all of the with statements and therefore, that is not an option for us. We are not sure why the intercept and growth terms are so highly correlated and what we can do about it at this point. Thank you for your help. 


Is the slope variance significant? 

pm posted on Thursday, November 15, 2007  1:00 pm



Hello, The slope variance is not significant (.105). Thank you. 


Then I would fix the slope variance at zero, and then there would be no interceptslope covariance and no Psi problem. 


Hello, I'm doing a unconditional LGGM with alcohol use data. When I assumed 3 classes I also received above mentioned warning. Looking at tech4 revealed that I have a negative variance in my Intercept. Some colleagues mentioned that it helps to fix variances of the slope and/or intercept to zero to avoid the warning. But the variances of my intercept and slope are significant. Is fixing the variance of the slope or intercept to zero really the right way to handle that problem? Additional information: 1)My Data is skewed and has a preponderance of zeros and i haven't done two part yet,is this problem related to psiwarning?, (I'm doing LGMM with MLR instead, to get a clue about LGMM, which is a new topic for me). 2)I have some outliers at time1 (which is the intercept). Would deleting them be helpful? 3) are there any model restrictions (in LGGM default by mplus 4.21) which may foster the psiproblem? Can I free/fix a parameter (beside above mentioned variances) to solve the problem? 


If you have a significant negative residual variance, the model is not appropriate for the data. Outliers and a preponderance of zeros should be considered. Please send your input, data, output, and license number to support@statmodel.com for further comments. 

Jungeun Lee posted on Sunday, December 16, 2007  5:51 pm



Hi! Related to the post 'pm posted on Monday, November 12, 2007  12:52 pm' and 'Bengt O. Muthen posted on Monday, November 12, 2007  1:06 pm', can we trust variance estimates got from the model that we get some warnings like 'nonpositive PSI', 'parameters were fixed to avoid singularity of the information matrix'? 


This depends on the situation. Please send the input, data, output, and your license number to support@statmodel.com. 


Hi! I am working on a growth mixture model. When I have more than 2 classes without constraining withinclass variances, I meet the same problem (PSI is not positive definit). Whenever I have this problem, I am pretty puzzled about what I can do about it. Could you give us a general guidance in what we can do about it to fix this problem? OR, solutions to this problem depends on the situation? 


These problems depend on the situation. You can send your input, data, output, and license number to support@statmodel.com is you want further information. 


hi, i am working on a growth curve model with an intercept and a slope factor. outcome variable is the summarized annual rate in selfreported delinquency for 15 offenses. the rate is transformed to its natural logarithm. besides that i included a range of TVCs (modelled as markovchains) and three TICs. i have five timepoints to analyze. the slope factor loadings are fixed to 0 and 1 for time point 1 ad 2 resp. the rest is freely estimated. all constructs (except the TICs) are treated as latent variables. here comes the problem: when estimating the model under listwise deletion conditions everything is fine. using the option of full information estimation leads to a correlation greater 1 between intercept and slope. i am wondering why? and what can be done? fixing the variance at a certain value? fixing the growth factor means at zero (loss of information)? thanks for your time. daniel 


The data are quite different when you use listwise deletion versus TYPE=MISSING; which suggests the listwise deletion sample is selective. To better understand why you get growth factors correlating greater than one, I would first fit the growth model without covariates. Then I would add timeinvariant covariates. After that I would add timevarying covariates. 


Dear Prof Muthen me again. Is there any possibility to change something in the model or any way out this problem? THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE. That is the modell: W1_TA by w1_taet2 w1_taet4 w1_taet5 w1_taet6; W2_TA by w2_taet2 w2_taet4 w2_taet5 w2_taet6; W3_TA by w3_taet2 w3_taet4 w3_taet5 w3_taet6; W1_OP by w1_opf2 w1_opf3 w1_opf4 w1_opf5; W2_OP by w2_opf2 w2_opf3 w2_opf4 w2_opf5; W3_OP by w3_opf2 w3_opf3 w3_opf4 w3_opf5; W2_TA on w1_OP w1_TA; w2_OP on w1_OP w1_TA; w3_TA on w2_OP w2_TA; w3_OP on w2_OP w2_TA; w1_OP with w1_TA; w2_OP with w2_TA; w3_OP with w3_TA; Thank you, Bettina 


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


Hello, I am doing a parallel process model using the following syntax; i1 s1 q1pa4@0 pa6@1 pa8@2 pa0@3 pa2@4 paA@5 paB@6; i2 s2sk4@0 sk6@1 sk8@2 sk0@3 sk2@4 skA@5 skB@6; s1 ON i2; s2 ON i1; q1 ON i2; s1 WITH S2; After running this syntax I get the following message: WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE. THIS COULD INDICATE A NEGATIVE/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 S2. After checking TECH4, I see that in the estimated covariance matrix I have negative values and values greater than 1. How can I solve this problem? In the the estimated correlation matrix, I have negative values. This should not be a problem, should it? Thanks Wayne 


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

Carolin posted on Monday, August 01, 2011  2:46 am



Dear Mrs and Mr Muthén, I have fixed i@0 in the linear three class model (GMM) because of a nonpositive definite psi matrix (variance of i was small and not significant). My question: do I have to fix i@0 in the quadratic model or the 4 class model too for better comparability or parsimony? Thanks a lot 


I would only fix it in the quadratic model if it is estimated as a small negative value. 

Alan Gow posted on Wednesday, August 24, 2011  5:01 am



I'm running a growth curve model with 3 latent factors (each defined by 4 measures). The warning message 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 I. appears. Checking TECH4 shows a correlation greater than 1 with the first factor and a covariate. How can I remedy this? 


You may need a direct effect from the covariate to a factor indicator. Check the modification indices. 

Till posted on Tuesday, September 13, 2011  6:09 am



Dear Mrs. and Mr. Muthén, I'm conducting a latent growth curve analysis with the factors F1 F2 F3 as latent exogenous variables. I'am interested in the parameter loadings between these factors and the slope. Slope and intercept are assessed at nine time points. In one sample, MPlus warns me that the latent variable covariance matrix is not positive definit which seems to be due to a negative residualvariance and a negative variance of the slope. Is there a way to solve that problem, for example by fixing the residualvariance to 0.01 or would that mean to suppress the variance of the slope which I'am mainly interested in? Could I, as an Alternative, just report the unstandardized values of the parameters? Please find my Input data below: variable: names= g1 e1 n1 g2 e2 n2 g3 e3 n3 l01 l02 l03 l04 l05 l06 l07 l08 l09; usevariable=all; missing=all(99); model: i s  l01@0 l02 l03 l04 l05@1 l06 l07 l08 l09; F1 by n1 n2 n3; F2 by e1 e2 e3; F3 by g1 g2 g3; i s on F1 F2 F3; Analysis: Estimator=MLR; output: samp standardized tech4; Thank you very much in advance Kind regards Till 

EFried posted on Saturday, January 21, 2012  12:27 pm



We are using a simple growth model at the moment before including the other covariates. i s  y0@0 y1@1 y2@2 y3@3 y4@4; We get the PSI warning that people have mentioned in the thread. TECH4 shows a value of 1.004 between S and I in the correlation matrix, I guess that's the reason. So I included y0 y1 y2 y3 PWITH y1 y2 y3 y4; as seen in one of your videos. This changes the correlation to 0.028, the PSI warning disappears, and the MI also get smaller (highest one from 240 to 190, but still high). However, since the MI indices are very high and the fit is bad, I freed the last 3 timescores (y0@0 y1@1 y2*2 y3*3 y4*4;). The fit indices get wonderful and the MI indices much better, but the correlation between I and S goes up again to 3.9 again and the PSI warning appears. The same happens in models with several classes as soon as I move from a LCGA model (is@0;) to a GMM model (!is@0): good fit, but correlation between I and S >1. (Negative residual variances seem not to be a problem.) Thank you for suggestions Torvon 

EFried posted on Saturday, January 21, 2012  1:03 pm



(Update: in some class solutions I do actually have negative residual variances in the biggest class, usually on I. In the so far best 2 class solution the negative residual variance of I = .6, too big to just fix it to zero ) 


I would run the model with one class as a first step. Try centering the time scores in the middle. 

EFried posted on Sunday, January 22, 2012  11:11 pm



Linda, I ran the models with only one class as first step and encounter the problem, as described above. I also tried now to center the time scores in the middle and still get the warning. No negative residual variances, but a correlation higher than 1 between Intercept and Slope in class2. What else can I do to fix this? Thank you! Torvon 


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


Hi I have run this model F11 by x11 x21 x31 F21 by F11 y1 z1 F12 by x12 x22 x32 F22 by F2 y2 z2 [...] F1n by x1n x2n x3n F2n by F1n yn zn F3 by f21 f22 ... f2n ______________________________________ All observed variables x y z are categorical. We have 7 years of measure for all observed dimentions. If I run every year independently, every single year model return a good fit. When I add the latent variable F3 to capture the persistence of simultaneous superior F1, y and x, some correlation problems emerge. When I use 3 years, the model return negative residual variance around 0,5 0,75. When I use all years, tetrachoric correlation between different year observation can,t be computed (acce1 and acce2 for example). The problem pertain to high correlation between latent variables. But since these high correlation relate to the dependence between each parallel year observation, how should i fix this. 


Too much of the correlations may be forced to go through the factors. Ask for modification indices and see if there are correlations needed among the observed variables. 

rgrove posted on Thursday, March 15, 2012  3:22 am



Hi, I am receiving the same nonpositive definite psi matrix error when running a CFA. My model is COG BY cog_emp* EYES KDEF; EMOT BY comt_emp* AQ_soc; COG@1; EMOT@1; EMOT WITH COG; I have no negative residuals. Is it that the correlation between my variables is 1.05? How do I account for this in my model? Thanks for your help 


Yes, a correlation greater than one will generate the message. This makes your model inadmissible. You must change your model. 


Hi, I'm trying to run a straightforward growth curve model of a single observed variable over 5 time points, and I am getting the nonpositive definite PSI matrix error. The slope term is causing a problem: according to the TECH4 output the correlations of S with S, and of S with I, are both 999. I had previously tried running it with a larger data set, when the correlation of S with I was estimated as above 3, and most of the other parameters seemed to be estimated normally. There is nothing strange about the data as far as I can tell  all five variables are roughly normally distributed, with correlations between the time points ranging from 0.43 to 0.59. My model syntax is as follows (which works fine for different variables): i s  errs07@0 errs08@1 errs09@2 errs10@3 errs11@4; Many thanks in advance for any suggestions you might have! Jeremy 


Try centering the time scores, for example, i s  errs07@2 errs08@1 errs09@0 errs10@1 errs11@2; 

Michael posted on Tuesday, May 08, 2012  6:08 am



Dear Dr.s Muthen, I´m running an growth curve model with 4 time points. When I use only an linear latent slope, the model is performing well. when I introduce an quadratic slope I get the warnings WARNING: THE RESIDUAL COVARIANCE MATRIX (THETA) IS NOT POSITIVE DEFINITE. and WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE. Indeed, I checked, the model estimates two negativ variances here´s my input i1 s1 q1 y11@0 y21@1 y31@2 y41@3 ; i2 s2 q2 y12@0 y22@1 y32@2 y42@3 ; Y11 Y21 Y31 Y41 pwith Y12 Y22 Y32 Y42 ; s2 on i1; s1 on i2; q2 on s1; q1 on s2; i1 with s1; i2 with s2; q1 with i1; q1 with s1; q2 with i2; q2 with s2; I do not understand why this warnings occur. the numer of free parameters is 36, as much as my empirical parameters. Is it an identification problem? Many thanks in advance Michael 


Thanks for the suggestion (centering the time scores)  unfortunately this doesn't work, and I get the same error message. Do you have any other ideas? Many thanks, Jeremy 


Dear Drs. Muthén, I am trying to estimate a multiple indicator linear growth model for two parallel processes for continuous outcomes and I am getting the warning that the latent variable covariance matrix (PSI) is not positive definite. When inspecting the model results and TECH4 output, it turns out that one of the slopes has a negative variance and a correlation of 999 with all other latent variables. I have tried to center the time scores, but this doesn’t solve the problem. When I fix the variance of the slope to zero I do not get the warning anymore, but the correlation between the slope and all other latent variables remains to be 999 and more importantly, it does not allow me to investigate the correlated change. Is it possible to solve this problem without fixing the variance of the slope to be zero? Thank you very much for your help and suggestions in advance! Best, Roos Hutteman 


Michael: The message is because of the negative residual variances. You can try holding them equal across time. 


Jeremy: It sounds like the data and model are not compatible. 


Roos: If the negative residual variance is small and not significant, you can fix it to zero. 


Hello, I am doing a multigroup comparison of a pretty complicated SEM model. When I run the analysis, the model converged for two of my groups but not the third. For this group, I get a not positive definite psi matrix. The problem is with the residual variance of my latent outcome. The size of the residual variance is .015 and it is highly insignificant p=.623. For the other groups, the residual variance is very small (.008 and .007) and also insignificant (.781 and .819). I am wondering if it would, therefore, be justifiable to fix the residual variance of this latent variable to zero? Thanks, Sarah 


Hello, Sorry for the previous message, I was looking at the wrong version of my model. This is a crosssectional complex model where I am doing a multigroup comparison. The model is pretty complicated. When I run it, I get a not positive definite psi matrix for all my groups. The problem is with the residual variance of my latent outcome. The size of the residual variance is .002, .004, and .024 depending on the groups and is highly insignificant p>.5 Given this, I am wondering if it would be justifiable to fix the residual variance of this latent variable to zero? Thanks, Sarah 


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


Hello Dr. Muthén, I am trying to run this GMM but am getting an error: WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IN CLASS 2 IS NOT POSITIVE DEFINITE. THIS COULD INDICATE A NEGATIVE VARIANCE/RESIDUAL VARIANCE FOR A LATENT... Is it because I'm getting a negative variance? ESTIMATED COVARIANCE MATRIX FOR THE LATENT VARIABLES ICEPT LINEAR ________ ________ ICEPT 0.082 LINEAR 0.021 0.007 How can I fix this? I read the other posts but I wasn't sure and don't want to do it incorrectly. Thank you very much, Danyel 


Yes, a negative variance gives this message. This implies that the latent classes explain all variation in the linear growth factor and there is no further withinclass variation. You can fix this variance at zero, or you can use fewer classes. 


Thank you for replying. I fixed it to zero and still received the same message. It said to look at class 2 matrices, but don't see what it's referring to. ESTIMATED MEANS FOR THE LATENT VARIABLES ICEPT LINEAR ________ ________ 1 1.502 0.083 S.E. FOR ESTIMATED MEANS FOR THE LATENT VARIABLES ICEPT LINEAR ________ ________ 1 0.100 0.016 ESTIMATED COVARIANCE MATRIX FOR THE LATENT VARIABLES ICEPT LINEAR ________ ________ ICEPT 0.258 LINEAR 0.038 0.000 S.E. FOR ESTIMATED COVARIANCE MATRIX FOR THE LATENT VARIABLES ICEPT LINEAR ________ ________ ICEPT 0.018 LINEAR 0.002 0.000 ESTIMATED CORRELATION MATRIX FOR THE LATENT VARIABLES ICEPT LINEAR ________ ________ ICEPT 1.000 LINEAR 999.000 999.000 S.E. FOR ESTIMATED CORRELATION MATRIX FOR THE LATENT VARIABLES ICEPT LINEAR ________ ________ ICEPT 0.000 LINEAR 999.000 0.000 


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


Hello, I have fixed this variance to zero, but after reading some more on your website it seems that I should not fix this variance to zero if it significant correct? The slope variance of .007 is significant. See below for reference. Linda K. Muthen posted on Wednesday, November 28, 2007  2:41 pm If you have a significant negative residual variance, the model is not appropriate for the data. Outliers and a preponderance of zeros should be considered. Please send your input, data, output, and license number to support@statmodel.com for further comments. 


Linda's answer applies. 


Dr. Muthén, Thank you for replying. I appreciate it. Danyel 


Dr. Muthén, I'm estimating a dyadic growth model and am getting a warning about one of the latent variables, but cannot see why this warning is coming up. Can you please help? 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 SW. Thanks, Danyel 


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


One of our analyses generated the message below. The problem variable (S) has a positive variance that is not significantly different from zero. We have at least one correlation greater than 1. Is it permissible to fix the variance of S to zero, and could it resolve the problem of having correlations greater than 1? Thank you! WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IN CLASS 1 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 S. 


If the correlation >1 is between s and i, then yes that would be a possible remedy. 


The correlation between I and S was greater than 1. Strangely, the correlations between Q and every other latent variable were also greater than 1, but no warning message was generated for Q until after the variance of S was fixed to 0. In the event that this error message is generated for multiple parameters at once, should they all be constrained at once? Or is it possible that constraining one will resolve the problems with the others? Thanks for your help! 


The latter. 


Thank you! 


I am running a GMM with a quadratic and cubic term, both significant, VARIABLE: NAMES ARE b1b5; Missing are all (999); CLASSES = c (3); ANALYSIS: TYPE = MIXTURE; STARTS = 400 20; MODEL: %OVERALL% i s q u b1@0 b2@1 bi3@2 b4@3 b5@4; OUTPUT: TECH1 TECH4; I receive this message WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IN CLASS 1 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 U. I looked at tech 4, not sure what the problem is. If I drop the cubic term and leave the quadratic (some of the curves look quadratic) I get WARNING: THE RESIDUAL COVARIANCE MATRIX (THETA) IN CLASS 1 IS NOT POSITIVE DEFINITE...CHECK THE RESULTS SECTION FOR MORE INFORMATION. PROBLEM INVOLVING VARIABLE B5. for all the classes There is substantial missing data at timepoint B5. B5 has a negative residual variance. I am not sure how to improve this model? thanks 


The message is caused by the negative residual variance which makes the results inadmissible. If it is small and not significant, you can try fixing it to zero. You can also try holding the residual variances equal across time. 


Dr. Muthén, I'm estimating a dyadic growth model and am getting a warning about one of the latent variables, but cannot see why this warning is coming up. Can you please help? 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 SWM. Thanks, Danyel 


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


Hi Dr. Muthen, I have been trying to run a growth model for two parallel processes with continuous outcomes. Unfortunately I have a small sample size, only 61. However, I am trying to cover my bases. I ran the model and got the error message of a negative residual, problem with s2. I fixed s2@0 and it failed to converge. Do you have any suggestions on next steps or do you think the model is just not a good fit with the sample size? Thanks, Eddie i1 s1  zT1PDS@0 zT2PCL@1 zT3PCL@2; i2 s2  t1CSESUM@0 t2CSESUM@1 t3CSESUM@2; i1 ON s2; i2 ON s1; s2@0; 


As a first step, you should run each process separately. 


I'm trying to run a number of DGM's and I keep coming across the same error message (even after centering all variables): THE MODEL ESTIMATION TERMINATED NORMALLY WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE.... TECH4 shows that in the estimated covariance matrix I have negative values and values greater than 1, and I also have some strange values in the estimated correlation matrix. It is a similar story with some of the other DGM's with different variables (also after centering). Will this make a huge difference to the output? Or is there something else I can try? Kind Regards 


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

DavidBoyda posted on Sunday, February 28, 2016  3:38 am



Dear Support, I have a mediation model with dichotomous outcomes thus estimated the model using WLSMV but I am receiving this error: THE SAMPLE CORRELATION OF Y4 AND Y3 IS 1.000. DUE TO ONE OR MORE ZERO CELLS IN THEIR BIVARIATE TABLE. INFORMATION FROM THESE VARIABLES CAN BE USED TO CREATE ONE NEW VARIABLE. THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE. I see by Tech 4 that the covariance and correlation matrix that my 3 outcome Dvs appear to be almost near 1.00 (0.992) in association. How can I remedy this? 


Either use only one of the variables or combine them into one variable. 

DavidBoyda posted on Sunday, February 28, 2016  7:29 am



That simple! Again thank you Linda. 


Dear support, I am comparing 3class models by gradually freeing parameters and testing whether the covariates have classspecific effects on the growth factors, to find the best fitting model. When I let the slope be freely estimated within each class, I get the ‘NOT POSTIVE DEFINITE’ message. Inspecting tech 4, I found that for class two, i and s had a correlation greater than 1. Q1: Can this be remedied by fixing the variance of s to 0, for class two only? Since I have covariates in the model, does specifying s@0 actually mean that I am fixing the residual variance of s to 0? Q2: If freeing both i and s leads to a model with problems (either nonconvergence or outofrange estimates), do I still report the model’s fit indices in the table presenting the models I am comparing? Or do I exclude fit results from this model stating that it was unidentifiable, and rather give the results from revised models that worked okay? Q3: When I add covariates, the sample size drops. I read in the UG that covariates are not part of the model and thus missingness for them is not allowed. It also says, however, that covariate missingness can be modeled if the covariates are brought into the model. Q4: How do I bring the covariates into the model? Q5: Journals often require a correlation matrix for study variables. If I let the covariates be exogenous, is it then better to report bivariate correlations of the reduced sample? 


Q1. You can do that if it makes sense and doesn't worsen BIC. Yes, residual variance. Q2. I would not report on a model with inadmissible estimates. Q4. You can bring them in by mentioning their variances, but I think you will run into trouble because they predict c. You can try multiple imputation but that has its own limitations. Q5.Yes. 


Thank you for Your response, Dr. Muthen. very helpful. May I follow up With another question. I have now two models that are better than the rest, but these two are very similar with reagrds to BIC and entropy. One has the covariates' effects Equal within classes, whereas the other has the effects of covariates freely estimated within classes. They get the same entropy value of .74, and almost the same BIC value. So I can perform a difference test using their respective loglikelihood (H0) values. The more restrictive model (with covariates effects held Equal) has a logL value of 29843.64 and 28 parameters and the one With covariates freed has a LogL of 29791.45 With 52 parameters. If I understand this correctly, since the difference x 2 is 104.38 is more than the chisquare Critical value With 24 df (the difference in number of parameters), it indicates that the model With more restrictions is worse? 


You are right. 

Emma Thomas posted on Thursday, March 03, 2016  11:40 pm



Dear Drs Muthen, I am conducting autoregressive mediation models using two latent factors and one observed dependent variable, measured over three timepoints. When I run the models I get the warning indicating that the PSI is not positivedefinite and inspection of Tech4 shows that my latent variables are consistently correlating at greater than one with earlier measures of the same variable. That is, latent variable measured at time 1 correlates with the same latent variable measured at time 2 and time 3 > 1. Collapsing them is obviously not an option  do you have any advice as to how I can go about addressing this problem? Or is the only alternative to conduct comparable analyses using LGCM instead? Thank you for providing such great support, Emma 


You will probably get a fuller answer to this on SEMNET. 


Hi, I am running a growth curve with 10 items at 3 time points and 7 timeinvariant predictors. The model estimation terminated normally, but I received the following warnings: WARNING: THE SAMPLE CORRELATION OF IAEIQ55R AND HAEIQ56R IS 1.000 DUE TO ONE OR MORE ZERO CELLS IN THEIR BIVARIATE TABLE. INFORMATION FROM THESE VARIABLES CAN BE USED TO CREATE ONE NEW VARIABLE. THE MODEL ESTIMATION TERMINATED NORMALLY THE MODEL COVARIANCE MATRIX IS NOT POSITIVE DEFINITE. FACTOR SCORES WILL NOT BE COMPUTED. CHECK YOUR MODEL. Should I trust the results? If not, can you offer any suggestions? Thanks! Hillary 


Re the first problem, you can follow the advice or delete on of those variables that create the 1 correlation (which is used with WLSMV). Alternatively, you can switch to ML although that may lead to heavy computations due to numerical integration. Re the second problem, you need to find where and then why this happens and modify your model. 


Hi Dr. Muthen, Thank you for your response! I ran the growth curve using ML estimation and both of these warnings disappeared. However, I didn't get fit statistics (as displayed below). Do you have any suggestions on how to obtain fit statistics? THE MODEL ESTIMATION TERMINATED NORMALLY MODEL FIT INFORMATION Number of Free Parameters 38 Loglikelihood H0 Value 5431.386 Information Criteria Akaike (AIC) 10938.772 Bayesian (BIC) 11130.545 SampleSize Adjusted BIC 11009.845 (n* = (n + 2) / 24) Thanks! Hillary 


You can use TECH10 to get bivariate fit info. Overall model fit is not available for ML with categorical outcomes. 


Thanks for your suggestions Dr. Muthen! I tried using TECH 10 and am getting the following warning: TECH10 OUTPUT FOR CATEGORICAL VARIABLES IS NOT AVAILABLE FOR MODELS WITH COVARIATES. What are your suggestions? I had fit statistics in my original model with the warnings. I am wondering what are the risks of ignoring the warnings and using those results? Note that I only need the gamma estimates, not the fit statistics to answer my research questions. Thanks! Hillary 


You don't want to report results from a run with the message: THE MODEL ESTIMATION TERMINATED NORMALLY THE MODEL COVARIANCE MATRIX IS NOT POSITIVE DEFINITE. FACTOR SCORES WILL NOT BE COMPUTED. CHECK YOUR MODEL. Even if you have to give up on fit measures, I'd rather use the ML estimates  making sure that the TECH4 growth factor correlations aren't too close to 1 (or that you have negative variances). 


This model was successful! Thank you! Are you aware of any citations for using ML over WLSMV or eliminating fit statistics for accurate gamma estimates for growth curve models? For a similar model with the same error originally (a growth curve with 10 items at 3 time points and 11 timeinvariant predictors), I tried using ML estimation again and obtained the following: THE MODEL ESTIMATION DID NOT TERMINATE NORMALLY DUE TO A NONZERO DERIVATIVE OF THE OBSERVEDDATA LOGLIKELIHOOD. THE MCONVERGENCE CRITERION OF THE EM ALGORITHM IS NOT FULFILLED. CHECK YOUR STARTING VALUES OR INCREASE THE NUMBER OF MITERATIONS. ESTIMATES CANNOT BE TRUSTED.THE LOGLIKELIHOOD DERIVATIVE FOR PARAMETER 29 IS 0.31577533D+00. Do you have any suggestions? Thank you!!! Hillary 


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

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