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 Anonymous posted on Friday, March 15, 2002 - 9:27 pm
Dear Linda/Bengt

I have 4 time-point data. The means of the outcome variable are 1.24, 1.29, 1.34, and 1.30. I expected that a quadratic LGM might fit well. However, I obtained a negative variance of the quadratic factor. Thus, I tried equal error variance, error covariance, and/or factor covariance. I still had the problem.

Finally, I decided to free slope factor loadings: 0,*, *, 1, where * means a free parameter. My question is how to interpret the effect of a covariate on the slope factor. For example, if the effect is positive, then how can I interpret the effect of the independent variable on the slope factor? It seems to me that it is not possible to interpret it because the slope is positive and then negative. Am I right?

Then, for this type of free slope models, can't we evaluate the effect of a covariate on the slope factor?

Thank you very much for your time.
 Bengt O. Muthen posted on Saturday, March 16, 2002 - 7:35 am
The relationship between the covariate and the slope growth factor can be interpreted for the time period between 0 and 1, for example, the first and fourth time points. The slope is the rate of change from the first to the fourth time points.

Assuming that your variances are small enough so that the increase followed by a decrease is significant, you could also investigate the quadratic model a bit more by fixing the variance of the quadratic factor to zero (don't forget to fix the corresponding covariances to zero also). The negative quadratic variance may simply reflect that individuals are very similar with respect to the downturn at the fourth time point.
 Anonymous posted on Saturday, March 16, 2002 - 12:23 pm
Dear Bengt

Thanks so much for your prompt answers. May I ask you two questions based on your answers?

1. When we free slope factor loadings, as you said, The slope is the rate of change from the first to the fourth time points.

I still cannot understand how to interpret the effect of a covariate on the slope factor (factor laodings increase and then decrease).

Suppose that we fix slope loadings as 0, 1, 2, and 3. Also suppose that the effect of a covariate on the slope factor is positive. Then, our interpretation would be that the higher on the covariate, the higher on the slope, that is the higher change rate.

However,suppose that we fix slope loadings as 0, *, *, and 1. Also suppose that the effect of a covariate on the slope factor is positive. Then, how can we interpret the effect? Given that the slope is not linear, the interpretation is not quite clear.

2. Suppose that we use a quadratic model where we fix the variance of the quadratic factor to zero. Then, how can we evaluate the effect of a covariate on the quadratic factor when its variance is 0?

Thank you very much again.
 bmuthen posted on Sunday, March 17, 2002 - 10:42 am
When choosing time scores as 0, *, *, 1, the slope is defined as the change from time 1 to time 4. The mean of the slope is the mean difference between time 1 and time 4. So a significant influence of a covariate x on the slope says that x is significantly influencing the change from time 1 to time 4. Note, however, that technically speaking the slope, and therefore the covariate, influences the outcome also at times 2 and 3. The time 2, 3 influence is however multiplied by the estimated time scores for those two time points, which means that the significance of the influence at those points has to be judged using "the delta" method. But you need not concern yourself with that if you don't want to.

Regarding the quadratic factor variance, you want to distinguish between two cases. (1) the variance is zero without covariates. (2) the residual variance is zero with covariates. Having found (1), the residual variance often becomes positive when adding covariates because of using more information - you can try that. In (2), this is ok because if the covariate has an influence it means that the slope actually does have variation (it just was not detected without the covariate). I think this is sometimes referred to as "varying" slopes rather than random slopes in the HLM lit.
 Peter Elliott posted on Friday, October 03, 2003 - 8:03 pm
Thank you very much for Mplus. I am trying to model data collected at four time points; intake, 2 months, 4 months, and 12 months. My problem is being able to specify the terms to include in the model (constant, linear, quadratic, etc). The means for the four time points are 18.4 (intake), 12.7 (2 months), 13.6 (4 months), and 14.4 (12 months) - lower scores are better. That is, people seem to respond to the treatment well initially, then fall back to some extent. This type of data must be common in some treatment fields. Any suggestions as to what sort of curve I should try to fit? Your advice would be greatly appreciated. Thank you. Peter
 Linda K. Muthen posted on Saturday, October 04, 2003 - 7:58 am
You might consider a piecewise growth model with one intercept factor and two slope factors -- one for the initial drop and one for the subsequent rise. The language would be:

i BY y1-y4@1;
s1 BY y1@0 y2@1 y3@1 y4@1;
s2 BY y1@0 Y2@0 y3@1 Y3@5;

You can then have different covariates for the decline and for the subsequent rise.
 Peter Elliott posted on Monday, October 06, 2003 - 10:39 pm
Linda, you are a marvel. Thanks heaps for your helpful advice. I fitted the model as you suggested and came up with a very non-signigicant Chi-sq (p = .35) and a CFI of .99.

With many thanks,

 Linda K. Muthen posted on Saturday, October 11, 2003 - 11:24 am
I'm glad I could help.
 Anonymous posted on Monday, December 01, 2003 - 6:06 am
I ran slp by y1@0, y2@1, y3* y4* (taken measures at baseline, 1 week, 3 months, and 6 months)
and got the following results (Estimates S.E. and Est./S.E.) suggest that the slope is not linear (if linear, expect 0,1,12,24).

Slp BY
y1 0.000 0.000 0.000
y2 1.000 0.000 0.000
y3 36.916 18.477 1.998
y4 4.543 2.010 2.261

How do i determine whether to add a quadratic term or piecewise growth modeling? Could you please let me know the syntax?

Also, say that you are now adding the covariates to the above model. Can I fix the slopes to what I got from the above unconditional model or do i let the program to re-estimate them? Thanks a million for your help.
 Linda K. Muthen posted on Monday, December 01, 2003 - 8:21 am
You can try a quadratic model. I don't think piecewise would work here. Example 22.1B in the Mplus User's Guide shows how to set up a quadratic growth model. Basically, you just add a quadratic growth factor. The time scores for the quadratic factor are the squared linear time scores.

I would not fix the slopes when adding covariates.
 Anonymous posted on Thursday, March 24, 2005 - 3:12 pm
I am running a three time point LGM with the third slope factor pattern coefficient freely estimated (0, 1, *). I have several questions related to the interpretation of the resultant parameters:
1) The mean of the slope was -1.53. I am interpreting this to represent only the change between the first two assessments - is that correct?
2) Following this, does the significance of the slope parameter only indicate significant change between the first two measurements?
3) Similarly, does significant slope variance only refer to significant variability in growth between the first two measurements?
4) Is there a way to estimate significance of change or significance of variability between the second and third measurements?
5) I have two DVs and I estimated this same model for both constructs (0,1,*). The correlation between the slope factors that is estimated via a parallel process growth model was .438. Does this only represent the relationship for the change between the first and second measurements?

Thanks in advance for your assisstance.
 bmuthen posted on Thursday, March 24, 2005 - 5:27 pm
1. yes

2. I assume you refer to the significance of the slope mean - then, yes.

3. yes

4. yes

5. yes

I guess you want an elaboration of 4 (and 5) :)
An easy way is to do a second run with changed time scores (*, -1, 0), where you need to give a suitable negative starting value for *. A more complicated way is to use the estimates from the run you mention and recognize that the mean change is estimated by the product lambda*[s]. This product can be given a SE since the variance of the product can be computed - but that is more complicated. 5 can also be done via the changed time scores approach.
 Anonymous posted on Friday, May 20, 2005 - 12:20 pm

I am running a growth mixture model and have a question about the time scores.
1) Is there a theory or best practice recommendation concerning fixing the time score to the measurement occasion versus letting it be freely estimated?
2) I am interested in overall growth --looking at an earlier post I am thinking I can code my growth process time measurements as:

i s q| wt1 @0 wt2 @.20 wt3 @.40
wt4 @.60 wt5 @.80 wt6 @1;

3) Do you see any problem with this (given this model has a quadratic term).
4) I would conclude that I can interpret the influence of background variables on the slope as that influence for the process from time one to time 6.
5) I ran both a freely estimated time score model and a fixed time score model. I am wondering why there is such difference in the R squared within class between the two models. In the freely estimated model the R squared of the observed outcomes is around .3 for all three classes. In the model where I fix the time score the R squared is around .17 or .20 within all three classes (except the first time point is .30).
6) Finally, when estimating the freely estimated time score model the graph that is produced is not usable—that is, the axis of time has odd spacing (0, 1 for the first time period—then quite uneven), not what you would like for a publication. What would your recommendation be for graphing a freely estimated time score model?

Thank you so much for your assistance!!
 bmuthen posted on Friday, May 20, 2005 - 5:52 pm
1) With 2 growth factors you have to fix at least 2 time scores - and 0 and 1 are typically used. With 3 factors, fix 3 values. It seems like a good baseline model to start with is with fixed time scores (and I assume you have edquidistant time points here given your scoring). If need be, this can then be followed by freeing some time scores to capture accelerated or decelerated growth. But a quadratic growth curve is already quite flexible so that would be a last resort.

2)-3) That's the way to do it. The quadratic factor will use the square of these time scores automatically.

4) This would be correct for linear growth (2 growth factors), because time 6 has time score 1, but you have to take into consideration that you have both a linear slope s and a quadratic slope q both of which contribute to the mean at time 6.

5) Sounds like your development deviates a bit from quadratic. Take a look at each individual's observed trajectory using the Mplus graphics component. Perhaps you need a cubic, or a piecewise model.

6) The spacing of the time axis for the plot will be that of the time scores (fixed and estimated). You can plot at other time points. But first resolve the growth function; hopefully you don't need estimated time scores in a quadratic.
 QianLi Xue posted on Wednesday, July 18, 2007 - 12:58 pm

I'm trying to fit a GCM for two parallel processes for continuous outcomes using the following syntax:

s1 ON i2;
s2 ON i1;

and suppose I find out through exploratory analysis that the relationship between, for example, s1 and i2 is not linear, but can be approximated by a two-piece linear spline. Can I do this in MPLUS?

 Linda K. Muthen posted on Wednesday, July 18, 2007 - 6:49 pm
No, Mplus does not do splines.
 J.W. posted on Friday, February 15, 2008 - 1:40 pm
My quadratic growth model with 6 time points (specified by “MODEL: I S Q | Y0@0 Y1@1 Y2@2 Y3@3 Y4@4 Y5@5;” in Mplus) works very well. However, orthogonal polynomial time functions did not work. I tried the following orthogonal polynomial coefficients:

1) The integer orthogonal polynomial coefficients from statistics texts:

Intercept Linear Quadratic
1 -3 5
1 -2 0
1 -1 -3
1 0 -4
1 1 -3
1 2 0
1 3 5

2) the general orthogonal polynomial coefficients generated from SAS:

Intercept Linear Quadratic
0.4082 -.5976 0.5455
0.4082 -.3586 -.1091
0.4082 -.1195 -.4364
0.4082 0.1195 -.4364
0.4082 0.3586 -.1091
0.4082 0.5976 0.5455


Are there any examples/studies that apply orthogonal polynomial time functions in growth model using Mplus? Your help will be appreciated!
 Linda K. Muthen posted on Friday, February 15, 2008 - 3:09 pm
Please send the input, data, output, and your license number to
 J.W. posted on Monday, February 18, 2008 - 10:15 am
Linda, the problem was in data format. The Model works fine now. Thanks a lot anyway!
 Tom Hildebrandt posted on Sunday, June 01, 2008 - 7:52 am
I have a question about growth modeling and fitting a particularly difficult growth pattern. I have a data set in which the DV is an oscillating outcome. However, I'm interested in the prediction of specific aspects of this oscillation. In particular, I am interested in predicting the individual variation in amplitude of the "wave" and "frequency" of the wave.

My question is how to specify a growth model with an oscillating or typical wave form.
 Bengt O. Muthen posted on Tuesday, June 03, 2008 - 11:21 am
That falls into the realm of non-linear growth models, that is, models that are non-linear in the random effects. This is in general not possible in Mplus. You may want to search the literature for non-linear growth models. One book is by Marie Davidian, but I don't think it covers oscillation phenomena.
 Tom Hildebrandt posted on Tuesday, June 03, 2008 - 8:52 pm
Thank you Bengt,

I will look up the book by Marie Davidian and continue the search.

If I find something applicable, I will post it for anyone who is interested.
 Bengt O. Muthen posted on Wednesday, June 04, 2008 - 2:53 pm
That would be good.
 Maja Wiest posted on Wednesday, July 29, 2009 - 7:30 am
Dear Linda & Bengt,

it is possible to implement an exponential function of growth in a GMM like it is usually done to model learning curves that level off after a certain time?
I like to model adaptation trajectories after a critical life event and I am mostly interested in then and where people start to level off in ratings of life satisfaction.

Thank you very much.
 Linda K. Muthen posted on Wednesday, July 29, 2009 - 1:47 pm
The relationship between the outcome and the growth factors must be linear. However, a non-linear growth shape can be specified using the values of the time scores. See the Topic 3 course handout.
 anonymous posted on Friday, October 02, 2009 - 2:09 pm
Drs. Muthén and Muthén,

Thank you very much for this site and for the courses you offer. My question concerns modeling non-linear growth using time scores and is related to previous posts. I am using age adjusted standard scores measured at four time points. Based on the spacing of the time points, the appropriate way to set up the time scores would be i s| 0 1 2 2.5. However, examining the means over time, i s|-.25 0 1 2 may be better and the fit is improved. I have not had success in fitting a piecewise model with the four time points. Is it incorrect for the time scores to be set up to reflect non-linearity without using a specific function?
 Linda K. Muthen posted on Friday, October 02, 2009 - 3:33 pm
If you select time scores to fit the data better, I am not sure how you would interpret your results. Time scores should reflect the distance between measurements. If you fit a linear model, what do the modification indices show?
 Bengt O. Muthen posted on Friday, October 02, 2009 - 3:56 pm
Also, you may want to reconsider using age-adjusted standard scores for growth modeling. See an article by Mike Selzer called The Metric Matters which may help explain why. I'm not sure where it is published but you can probably find it by searching Jstor and other venues under Selzer.
 Harald Gerber posted on Tuesday, January 12, 2010 - 2:13 am
I assume that with time scores of: 0 1 3 5
a significant test of slope mean and variance between 0 and 1 also applies to later intervals 1 and 3 or 3 and 5. Is that correct? I assume that because the one unit change is nested within the later intervals.

However, when one has time scores like: 0 1 2.71 4.42 my assumption mentioned above would not hold, because the one unit change is not nested within the later intervals (e.g., from 2.71 to 4.42)!?
 Bengt O. Muthen posted on Tuesday, January 12, 2010 - 1:10 pm
The significance of the slope mean (say alpha) refers to the model-estimated mean change in the outcome from 0 to 1 which is alpha. The change from 0 to 3 is 3*alpha, so the significance of alpha is relevant here too.

There would be an issue if the time scores were estimated parameters, but that doesn't seem to be the case here.
 Hugo Hesser posted on Thursday, January 28, 2010 - 4:18 am
Thank you for Mplus! I’m trying to model data from an RCT with a short and long-term follow-up, collected at four time points: pre intervention, 2 month, 6 month and 42 month. We are interested in comparing the effects of two treatments and overall improvement for the entire sample over time. Sample means and individual slopes seem to suggest that individuals decrease rapidly in symptoms between pre and two month (intervention phase), and continue to decrease between 2 and 6 month but at a slower rate (short-term follow-up), and then slightly increase between 6 and 42 month (long term follow-up). This development over time seems to be similar in both treatments (ie., linear piecewise)

Is it possible to run a piecewise growth model with only four time points? And more importantly, do you believe that this might be the right way to go, as there seem to be three unique phases (i.e., initial large drop, continued improvement, and a slight deterioration at the final time point) in our data. If not, could you please guide us in another direction. We have tried a quadratic model but it fits very poor. We would greatly appreciate any advice that you can give us regarding a shape that might fit our data.
 Linda K. Muthen posted on Thursday, January 28, 2010 - 9:47 am
Four time points is not enough for a piecewise model without some restrictions. You could consider the following model or some variation of it:

MODEL: i s1 | y11@0 y12@1 y13@1 y14@1;
i s2 q | y11@0 y12@0 y13@1 y14@2;
s1@0; q@0;
 Hugo Hesser posted on Saturday, January 30, 2010 - 3:56 am
Dear Linda,

Thank very much for your useful suggestion of a model, which resulted in a very nice fit (although maybe over fitted). This is definitely in the right direction we want to go. Unfortunately, we ran into some negative residual variance at the last time point. This is probably due to that the variable at that time point is not normally distributed (evidence of a floor effect, many people are doing good). Hopefully we can figure this out on our own (estimator = MLR). But if you have some ideas we would love to hear them. We have some additional questions though:

1. Just to be sure, our time scores for our piecewise model were:1 piece = 0111, 2 piece = 0017, to specify unequal distance between measurement occasion in the 2nd piece (0, 2, 6, 42 month). Is this correct?
2. We would like to center at the last time point (42 month), to be able to interpret the effect of a time-invariant covariate on endpoint. Is this doable in our model. If so, how would you set it up?

We are so glad that you are willing to share you extensive knowledge. With many thanks,

 Linda K. Muthen posted on Saturday, January 30, 2010 - 10:45 am
To obtain the time scores with centering at the last time point, try:

-1 0 0 0
-7 -7 -6 0

Be sure you get the same chi-square.
 Kate Stringer posted on Wednesday, February 17, 2010 - 5:55 am
I have a very basic question: What is the difference between the piecewise approach and freeing slope loading factors to account for an initial decline and then an increase? For example, I have 6 times of data equally spaced by 1 year. From time 1 to time 2, there is a decrease; then from times 2 through 6, there is an increase. If I free the slope loading factors would I do the following:

i s | cp1 cp2 cp3@0 cp4@1 cp5@2 cp6@3;

If I do the piecewise approach, would it look like the following:

i BY cp1-cp6@1;
s1 BY cp1@0 cp2@1 cp3@1 cp4@1 cp5@1 cp6@1;
s2 BY cp1@0 cp2@0 cp3@1 cp4@2 cp5@3 cp6@4;
 Linda K. Muthen posted on Wednesday, February 17, 2010 - 11:15 am
The two models cannot really be compared. They have different parameters and model the data differently. I would probably use two sequential models where each piece has an intercept growth factor. You cannot identify a linear model for two measurement occasions so for that part you would need to fix the variance of the slope growth factor to zero.
 Kate Stringer posted on Wednesday, February 17, 2010 - 11:19 am
Thanks, Linda. Is my code correct? When I use the piecewise code listed above, I get the error


Parameter 15 is S1
 Linda K. Muthen posted on Wednesday, February 17, 2010 - 11:31 am
Please send the full output and your license number to
 Gregory Kirkner posted on Thursday, May 13, 2010 - 9:15 am
I want to fit latent growth models for some continuous observed variables measured at 3 equally spaced time points. Because the observation times were equally spaced, I began my analyses using fixed time scores of 0, 1 and 2. However, plots, descriptives, model fit output, etc. indicated that some of the observed data were non-linear. Also, some of the observed data have a pattern of growth followed by a marked reversal of growth (e.g., increasing means from time 0 to 1, followed by decreasing means from time 1 to 2).

For the data indicating non-linear growth, I can use free, logarithmic or exponential time scores, but can I also manually fix self-determined non-linear time scores (e.g., 0, 1, 1.3) to improve fit?

For the data having a reversal in growth, can I use a negative time score (e.g., 0, 1, -.75) to improve fit?

Finally, when evaluating the chi-square test of model fit, higher p-values are desired, correct?

 Linda K. Muthen posted on Thursday, May 13, 2010 - 1:41 pm
I would free the last time score.
 Gregory Kirkner posted on Friday, May 14, 2010 - 12:03 pm
Dr. Muthen, Thank you for your prompt reply. I free the last time scores, but on some models I get a message stating "NO CONVERGENCE. NUMBER OF ITERATIONS EXCEEDED. FACTOR SCORES WILL NOT BE COMPUTED DUE TO NONCONVERGENCE OR NONIDENTIFIED MODEL". can you tell me why this happens?

Also, when comparing model fit between free time scores, fixed time scores, logarithmic time scores, is there a fit statistic that is preferred for use in this regard?

 Linda K. Muthen posted on Saturday, May 15, 2010 - 8:08 am
Please send the full output and your license number to
 Wayne deRuiter posted on Friday, May 21, 2010 - 5:25 am
Dr. Muthen,

I am current trying to develop unconditional models for three health behaviours using linear and quadratic LCM. I have run both linear and quadratic LCM's for all three health behaviours that I am evaluating. I noticed that the RMSEA demonstrates slightly better fit for each quadratic model over its linear counterpart; difference of 0.004 for physical activity, difference of 0.006 for alcohol, and difference of 0.02 for smoking. Are these differences substantial enough to warrent using a quadratic LCM over a linear LCM? Is there another kind of test to determine the shape of the curve rather than just comparing RMSEA's between linear and quadratic models?

I also noticed that when comparing the linear and quadratic LCM, the slope means change substantially in magnitude and sign (ie: positive to negative). Is this typical when comparing linear and quadratic LCM?

 Linda K. Muthen posted on Friday, May 21, 2010 - 8:41 am
If the mean of the quadratic growth factor is not significant, I would not include the quadratic growth factor. For difference testing of nested models, you should use chi-square or the loglikelihood.
 Arina Gertseva posted on Tuesday, June 15, 2010 - 9:58 am
Dear Linda/Bengt,
I am trying to identify the best fitting growth function for offending (measured with the use of a cumulative scale) measured repeatedly over the course of 6 years (five measures were done on consecutive years and the last one was made 3 years after the fifth measure). The observed means of offending are 17.46; 15.06; 21.23; 16.79; 20.53; and 49.02.
At first, I tried to free slope loadings: 0, 1, *, 3, 4, *, but I got the warning about “no convergence.”
Next, I tried to fit piecewise regression with four different slopes, but I got the warning concerning “non-positive latent variable covariance matrix.” When I fixed variance to zero, the warning disappeared, but I would rather not do that because I want to estimate the effect of a covariate on the slope(s).

Could you please advise me on what sort of curve I should try to fit? Your advice would be greatly appreciated. Thank you. Arina.
 Bengt O. Muthen posted on Tuesday, June 15, 2010 - 11:01 am
This sequence of means seem difficult to fit a curve to. I would investigate why the means jump up and down so much. Perhaps that is a function of using a cumulative scale.
 Wayne deRuiter posted on Tuesday, July 06, 2010 - 8:00 pm
Dear Dr Muthen

If I wanted to test between using a linear or quadratic growth factor I should be using difference testing of nested models using chi-square or the loglikelihood. How would I do this? Is there an example somewhere?

 Bengt O. Muthen posted on Wednesday, July 07, 2010 - 10:02 am
It's a little tricky to do this testing with chi-square because the more restricted model has not only the quadratic growth factor mean but also its variance fixed at zero - a variance fixed at zero is on the border of the admissible parameter space so chi-square is not formally valid.

You can certainly test if a fixed, i.e. zero variance, growth factor is needed by simply looking at the significance of the quadratic growth factor mean.
 Wayne deRuiter posted on Wednesday, July 07, 2010 - 7:59 pm
Dear Dr Muthen

Could I use a loglikelihood for difference testing or does that have the same problems as the chi-square?

If the quadratic growth factor mean is significant does that mean that I should choose and quadratic growth curve over a linear one?

If the quadratic growth factor mean is not significant, does that mean I should use a linear model?

Sorry for all the questions, but I am not familiar with difference testing.

Thank you
 Linda K. Muthen posted on Thursday, July 08, 2010 - 10:15 am
Loglikelihood difference testing has the same problems as chi-square difference testing.


 Wayne deRuiter posted on Monday, July 12, 2010 - 7:57 pm

I know that "q" refers to a quadratic growth curve model. What is the syntax for a cubic growth curve model?

Thank you
 Linda K. Muthen posted on Tuesday, July 13, 2010 - 7:59 am
Growth factors do not need to have a specific name. They are named as any other variable. See pages 601-608 of the Version 6 User's Guide for a description of how to specify various types of growth models.
 Wayne deRuiter posted on Thursday, July 15, 2010 - 8:11 pm

I recently ran linear and quadratic growth curve models. Both models fit the data fairly well. Is there an alternative to the linear and quadratic models that I could test to see if it fits the data better?

Would allowing some of the factor loadings to be free be one alternative in which the data may fit better?

 Linda K. Muthen posted on Friday, July 16, 2010 - 9:31 am
You should look at the mean of the quadratic growth factor. If it is not significant, you should use the linear model.
 Rod Bond posted on Thursday, November 25, 2010 - 4:23 am
Hi, I am modeling change in quality of life at 4 times – 0, 3, 6, 12 months. Sample means indicate a significant increase from T1 to T2 but thereafter no mean change.
I have tried:
1. Freely estimating time scores for T3 and T4. The slope variance is negative and the residual variance at T1 is large. If s@0, fit is worse and still large T1 residual. If residuals constrained to be equal, fit better and the slope variance significant.
2. Fitting a quadratic function. Gives poor fit and a negative q variance. If q@0, fit poor and large T1 residual.
3. Transforming the time scale. Best is a root transformation where time is raised to the power of 0.1. This provides a good fit, although T1 residual now smaller than at other 3 times. Constraining residuals to be equal does not significantly worsen the fit and the slope variance is now significant.
4. Piecewise model. Time for s1 are 0 1 1 1, and for s2 are 0 0 1 3. s1@0 for identification. The model fits well, but gives large T1 residual.
My inclination is to go with the root transformation of time because it fits the data well and gives me a significant slope variance that can be related to covariates. But do you agree?
Also, I am unclear about whether to constrain the residual variances to be equal. In effect, should lack of fit be pushed into slope factor variance?
 Bengt O. Muthen posted on Thursday, November 25, 2010 - 9:44 am
It's a common situation. Alt. 3 might be good in terms of fitting well, but I am concerned that 1-3 all make if harder to interpret covariate effects. Alt 4 is better that way, although you say it has a T1 problem. What about Alt 5, using sequential processes - the first one for T1 and T2 and the second one for T3 and T4. Adding a second intercept makes the fitting easier. True, you have to give up a slope variance for each piece, but you gain the advantage of having covariates be able to influence 4 different growth factors, each with a clear interpretation.
 Hemant Kher posted on Thursday, November 25, 2010 - 2:32 pm
Dear Professors,

I have a question regarding my exponential growth model that involves 6 waves of data. I am able to determine the exponential rate (alpha), but not its variance. In other words the model implies that all worhers learn at the same rate. The code I have used for my model is given below:

i BY y1-y6@1;
s BY y1* (L0)
y2 (L1)
y3 (L2)
y4 (L3)
y5 (L4)
y6 (L5);

y1-y6 (1);

[i*30 s*20];
i*1 s*1;
i with s*0;

rate BY y1@0;
rate with i@0 s@0;

model constraint:
New (rate);
L0 = 1-(exp((0)*(-1*rate)));
L1 = 1-(exp((1)*(-1*rate)));
L2 = 1-(exp((2)*(-1*rate)));
L3 = 1-(exp((3)*(-1*rate)));
L4 = 1-(exp((4)*(-1*rate)));
L5 = 1-(exp((5)*(-1*rate)));

Is it possible to estimate the variance for the exponential parameter "rate" in this model?
 Rod Bond posted on Friday, November 26, 2010 - 7:04 am
Thanks for your advice. I know this has come up before in this forum, but I am confused as to how you can relate a slope with the variance fixed at 0 to a covariate. I can see that the model runs but what does it mean if, say, I find a significant effect of gender on the s1 slope?
 Bengt O. Muthen posted on Friday, November 26, 2010 - 8:08 am
It simply implies that the slope mean is different for males and females.
 Rod Bond posted on Friday, November 26, 2010 - 8:29 am
If the slope mean differs according to values on a covariate, then the slope has a variance? Would the interpretation of the gender effect be different if, with more time points, the slope variance was free? I'm still confused!
 Bengt O. Muthen posted on Friday, November 26, 2010 - 9:13 am
If you regress s1 on gender you have 2 potential sources of variance in s1, gender and the residual in the regression. I am saying that if the residual variance is 0 your model says that s1 has different means for males/females. And yes, s1 has a variance due to gender. But the residual variance is 0 - so 100% R-square.

Note that when s1 is regressed on a covariate, the Mplus statement


refers to its residual variance.

Or, maybe you are referring to an uncondional analysis (no covariates) finding that s1 has no variance. In that case, it is still of interest to regress s1 on gender for 2 reasons. Adding the covariate changes the model and an s1 variance may be detected - due to both covariate and residual.
 Rod Bond posted on Friday, November 26, 2010 - 10:51 am

Thanks. I now understand!
 Ting Dai posted on Tuesday, March 20, 2012 - 7:30 pm
Dear Drs. Muthen,

I have 5 waves of data, and descriptives show it might be a cubic growth pattern, however, the complex PSI matrix had a non-positive definite. My questions are:

1) Should I try to fit a reduced cubic model?

2) In Mplus, how can I fit a reduced cubic model that reduces the slope and the quadratic term? Is it by setting
[q@0]; , or should I only set the factor variances to be zero?

3) Would a linear model with intercept and slope but freeing the Times 2-4 slope paths be a better option than a cubic or a reduced cubic?

4) Any sources I could cite to support the use of a model described in 3)?

Thanks very much!
 Linda K. Muthen posted on Wednesday, March 21, 2012 - 9:20 am
1-2. You would fix only the variances to zero not the means.

3. I would use a cubic model over a model with free time scores. I would use a free time score model when one value is not in line with the others. I would not free several time scores.

4. No.
 Ting posted on Wednesday, March 21, 2012 - 9:37 am
Thanks, Dr. Muthen, for a quick response.

Just so I am sure I get what you said: A cubic or reduced cubic model is better than a linear model with slope paths set as 0,*,*,*,1?

And you wouldn't worry that 5 time points are too few to fit a cubic model?

 Linda K. Muthen posted on Wednesday, March 21, 2012 - 10:58 am
Yes, I think if cubic fits it is better than a model with free time scores.

With so many fixed parameters, five time points may be okay. Try it out and see.
 Isaura Olivares posted on Sunday, April 15, 2012 - 9:56 pm
Dear Drs. Muthen,
I am fairly new to Mplus and I am trying to run a growth mixture model using data with four time points. I have used the following syntax to run the model:
i s q| YDEser@0 DE3ser@1.5 DE4ser@3 DE5ser@4.5;
i s q@0; I conducted a series of analyses and determined based on the BIC and on LMR that a four-class model fits the data best. Unfortunately, I have not been able to replicate the best log-likelihood, regardless of the starting values I use. I am not sure what starting values I should use, but I have used up to 2000 random starts. I am afraid that the solution I have is only a local one. (1) How can I over come this problem? Is my model mispecified? Thanks in advance!
 Isaura Olivares posted on Sunday, April 15, 2012 - 9:57 pm
I also have a set of 5 time-invariant covariates that I would like to include in the model. I ran the model with the four classes and included the covariates even though I wasn’t sure whether the 4-class model described above was a good one. After running the model, I found that the classifications/proportions for each class changed, which I attributed to the covariates because some have missing data. However, after conducting a search on this discussion board, I found that this may be due to a misspecification in my 4-class model and not to missing data. (2) What can I do to address this issue?
Thank you, I appreciated any help you can give me!
 Linda K. Muthen posted on Monday, April 16, 2012 - 8:57 am
You don't say how many final starts you give. It should be about 1/4 of the initial starts, for example, 2000 500. You must replicate the best loglikelihood before you add covariates.

When classes change when covariates are added, it points to the need to direct effects from the covariates to the outcomes. See the following paper on the website for further information:

Muthén, B. (2004). Latent variable analysis: Growth mixture modeling and related techniques for longitudinal data.
 Harald Gerber posted on Tuesday, May 01, 2012 - 8:41 am
Related to the post on july, 07, 2010 to july, 16, 2010...
In my growth model the quadratic mean is not significant (the quadratic variance, however, is significant). Based on the assumption that a quadratic model is only needed when the quadratic growth factor mean is significant I wanted to report the necessity of a quadratic model very briefly (and elegant!?) as a chi-square test with 1df. More specifically, I transformed the t-test of the quadratic mean to a chi-square test by squaring the z-value of the quadratic-mean test and stated that including a quadratic factor does not provide a better model fit as compared to the linear model. Is that correct? Do I have to take the quadratic growth factor variance into account in any way?
Btw., BIC of the quadratic model was worse as compared to the linear model.
 Linda K. Muthen posted on Wednesday, May 02, 2012 - 12:04 pm
I would report that the quadratic growth factor mean is not significantly different from zero using the z-test and that the linear model has a lower BIC. I don't think more needs to be done.
 Harald Gerber posted on Thursday, May 03, 2012 - 7:29 am
Thank you. Recently I've heard or read (may be it was in this forum but I can't remember) that the Q-variance (besides Q-mean) also can be used as a criterion to accept or reject a quadratic growth model (so, following this, you need a non-significant Q-mean AND a non-significant Q-variance to reject a quadratic growth model). What do you or others think about this argument (or does it depend on the specific research question whether Q-variance is of interest to decide about the growth function?).
 Linda K. Muthen posted on Friday, May 04, 2012 - 10:07 am
I would not focus on the variance when the mean is zero. I think this may be over interpreting the details of the variance structure.
 Mihyun Park posted on Saturday, May 05, 2012 - 11:02 am
Dear. Linda

I studied and analyzed using LGM. My data was non linear and the model was be with free time scores(T1@0 T2* T3* T4* T5@1). How can be the numerical formula of growth factor loading and estimates of the slope mean expressed ? I can find an formula of either linear growth modeling or piecewise growth modeling but an non-linear with time free scores.
 Linda K. Muthen posted on Sunday, May 06, 2012 - 5:22 pm
You would use the formulas in the Topic3 course handout on the website. Substitute the estimated time scores for the asterisks.
 Mihyun Park posted on Thursday, May 10, 2012 - 9:28 am
I¡¯d like to thank you for your help.

In the formula,

yti = ¥ç0i + ¥ç1ixt + ¥åti

I can't understand how 'xt(estimated time scores)' is determined?
What's formular of the estimated time scores of non-linear model or spline model?
 Linda K. Muthen posted on Thursday, May 10, 2012 - 11:26 am
It would come from the value that is estimated in the analysis.
 Mihyun Park posted on Thursday, May 10, 2012 - 6:12 pm
I am most grateful to you for such a prompt reply. I know that the value is estimated in the analysis. But I would like to know how can the value estimated ..and what is the formula for estimating the value? I tried to search for the formula in Technical Appendices. But I can't find how can the
'x parameters' estimated?
 Linda K. Muthen posted on Friday, May 11, 2012 - 10:25 am
It is estimated as a factor loading in a CFA.
 Meg posted on Friday, May 18, 2012 - 4:33 pm
Hi Linda,

I am running a LGM with 4 waves of data. The data is not equally spaced (1 year, 5 years, and 7 years). I have set the slope factor loadings to 0 * * 1. The estimates for factors 2 and 3 are -.014 and .271. I'm not sure how to interpret the negative estimate for the second loading.
 Linda K. Muthen posted on Friday, May 18, 2012 - 5:26 pm
Please send your output and license number to
 Charu Mathur posted on Saturday, May 19, 2012 - 5:30 pm
Dear Drs.Muthen,

I am conducting a cohort-sequential latent growth curve analysis with 3 time points. Is it acceptable to model a curvilinear model (with quadratic term) with 3 just time points ? I ask because a naive analysis suggests the growth to be non linear and my understanding was that one would need a minimum of 4 time points to examine curvilinear growth.

Any suggestions will be very helpful


Charu Mathur
 Linda K. Muthen posted on Saturday, May 19, 2012 - 6:40 pm
You need at least four time points to identify a quadratic growth model.
 Charu Mathur posted on Saturday, May 19, 2012 - 7:27 pm
Thanks a lot, Dr.Muthen. Will this requirement of a minimum of 4 data points hold true in the following scenario as well?
we are linking 8 cohorts from 1-18 years, each cohort being measured every 5
years, for 3 times each.


Charu Mathur
 Linda K. Muthen posted on Sunday, May 20, 2012 - 6:04 am
If you are using 1 to 18 as time, then you have 18 timepoints so you are fine.
 Charu Mathur posted on Friday, May 25, 2012 - 8:10 am
Thanks Much,Dr.Muthen. I have a couple of follow-up questions regarding the cohort-sequential latent growth analysis, I am conducting. We have 8 cohorts, and each cohort has 3 waves of data with 5 year interval. EDA shows a quadratic growth

I examined a CSLGC with Q term with BMI as an outcome. The variances were negative and also, there were several warning messages therefore, I constrained the variances to zero and the model converged. Did this happen because the growth factors had zero variance and I was estimating a random effects model? Notably, Q term was significant in both random and fixed effects models.
Subsequently, I ran a dual domain or parallel process model to examine simultaneous growth between BMI and screen time. A univariate analysis showed BMI to have Q growth with fixed effects, and screen time exhibited linear growth with random effects. However, the dual domain model did not converge. I changed the screen time growth factor variances to zero, lowered convergence criteria etc but nothing helped. Interestingly, when I modeled growth as linear for both BMI and screen time with random effects, the dual domain model converged.
Do you think the problem is over identification or is it something else?

Will appreciate your help,

Charu Mathur
 Linda K. Muthen posted on Friday, May 25, 2012 - 11:31 am
I would need to see the outputs to understand your situation. Please send them and your license number to
 Sami Yli-Piipari posted on Saturday, June 02, 2012 - 6:35 am
Dear Linda and Bengt,
I have been working on the non-linear growth models (4 time points) and I would like to have some help interpreting the results. Best fit model is the model
i s | y1@0 y24@1 y3* y4* with free loading time scores(also the theory supports this conclusion).

1. Slope in the first psycholocial functioning is -.29 and the estimated time scores are .91 (T2) and .93 (T3)
2. Slope of the second functioning is .13 with estimated time scores -.09 (T2) -.18 (T3).

I need help to intepret these results regarding the free loads. I know the slope represents the growth between time points 0 and 1. But what does the estimated time scores indicate?
Thank you very much.
 Linda K. Muthen posted on Saturday, June 02, 2012 - 11:23 am
Their deviations from the linear time scores 2 and 3 tell you about their deviation from linearity.
 Sami Yli-Piipari posted on Sunday, June 03, 2012 - 5:31 am
Thank you very much. Have a great weekend.
 Greg Hildebrandt posted on Friday, June 29, 2012 - 4:22 pm
I recall that there was a previous message discussing Mplus' inability to estimate BLUPs. In a quick search of the discussion board, I was unable to find the message.

When random intercepts and/or coefficients are used, might there be a way of labeling a plot of the randomly determined curves by relevant observation? This strikes me as an important capabilty. In fact, because it may be difficult to test whether the use of random coefficients meets the necessary assumptions, such labeling increases the interest in using such random coefficients.

Also, I haven't been able to analyze all the complications associated with obtaining BLUPs in the SEM setting. However, I believe these would be quite important, when feasible.

In longititudinal regression analysis, one of BLUPs interesting properties is significantly attenuating the multicollinearity that is inevitably present when numerous categorical dummy varaibles are used, say, to analyze the effect of, individual schools on some dependent variable. So, obtaining BLUPs (and the associated test statistics), when they make sense in SEM, would also be a very nice addition to Mplus.
 Mihyun Park posted on Friday, August 24, 2012 - 9:06 am
Dear Linda and Bengt,

I am running a sequential latent growth model analysis with multiple group analysis across gender.

The syntax is following;

i1 s1 | su11@0 su22* su33* su44* su55@1;
i2 s2 | ax11@-1 ax22* ax33* ax44* ax55@0;
ax11 with su11;
ax22 with su22;
ax33 with su33;
ax44 with su44;
ax55 with su55;
i1 with s1;
i2 with s2;

s2 on i1 s1;
i2 on i1 s1;

It was worked ... but there are only the means estimate of i1 and s1 without i2 and s2 in the result output. Can I know the mean estimate of i2 and s2?
 Linda K. Muthen posted on Friday, August 24, 2012 - 9:42 am
s2 and is are dependent variables so intercepts rather than means are estimated. See TECH4 in the OUTPUT command to see the model estimated means.
 Mihyun Park posted on Friday, August 24, 2012 - 10:09 am
Dear Linda and Bengt,

Thank you for your time and prompt response. Your answer really help me out.

I already tried to do TECH 4. The TECH 4 output have ESTIMATED MEANS FOR THE LATENT VARIABLES without S.E. I would like to analysis whether the s2 difference of gender is significantly different or not. If it's different across sex... what is the means and S.E. estimate of boys and girls.
And then... Is it impossible?
 Linda K. Muthen posted on Friday, August 24, 2012 - 10:31 am
If gender is your only covariate and if male is zero and female is one, the regression of s2 on gender will tell you if the means are different.
 Mihyun Park posted on Friday, August 24, 2012 - 11:36 am
I am so grateful for their help.

(Gender is only covariate)
You mean...

i1 s1 | su11@0 su22* su33* su44* su55@1;
i2 s2 | ax11@-1 ax22* ax33* ax44* ax55@0;
ax11 with su11;
ax22 with su22;
ax33 with su33;
ax44 with su44;
ax55 with su55;
i1 with s1;
i2 with s2;

s2 on i1 s1;
i2 on i1 s1;

s2 on gender; !Is that right?

I'm trying to do that.
But I still can't find the means and S.E. estimate of s2 for boys and girls. I only can know whether the sex difference of s2 is significantly different or not. And then...Multiple group analysis with sequential latent growth model having only gender covariate is useful method for comparing the difference of path across gender but it's not valid method for finding the difference of latent variables(e.g.slope).
Is it correct?
 Linda K. Muthen posted on Saturday, August 25, 2012 - 12:23 pm
Your model has more than one covariate for s2. You have

s2 ON i1 s1 gender;

Your model does not estimate a mean for s2. It estimates an intercept for s2. It is a conditional model.

One way to test if the means are different is to define the means for boys and girls in MODEL CONSTRAINT and test if these are different.
 Kara Thompson posted on Thursday, October 25, 2012 - 9:34 am
Good Morning Linda and Bengt,
I just finished watch your Topic 3 seminar on latent growth modeling. I learned a great deal, so thank you. I am working on trying to model alcohol use across 5 waves of data. As this is an adolescent-emerging adulthood sample, the trends are clearly nonlinear. I have been running quadratic and unspecified models. My main problem however, is that after identifying my functional form, I want to move to a multivariate model, as well as be able to use the slopes as predictors of distal harm outcomes. This makes my decision about quadratic/unspecified models much more difficult as both have limits in terms of interpretability. I have several questions about unspecified models:
1. Based on your mplus discussions, it seems as though you recommend only freeing one time score, rather than 3 for example (T1@0 T2* T3* T4* T5@1)? Why is this?
2. I understand that the slope value indicates a 1-unit linear change, so it is not actually capturing the nonlinear trend – therefore, does it make sense to use the slope as a predictor?
3. Once the time scores are estimated, could you re-run the model with those specific time scores? Would the slope value than represent the nonlinear trend over time?
4. Can you conduct multivariate LGM’s using unspecified models? And if so, do all your processes need to be on the same time metric?

Thank you
 Linda K. Muthen posted on Friday, October 26, 2012 - 1:51 pm
When you have free time scores, you should not predict from the slope growth factor. You can predict from the intercept growth factor changing the centering point.

1. You can free more than one time score.
2. No.
3. No.
4. Yes. No.
 Caroline Vancraeyveldt posted on Friday, November 09, 2012 - 1:09 am
Dear Dr. Muthén,
In your post of March 24, you mention "product lambda". How can I compute this, or where can I find this in the output?
Thank you very much!
 Linda K. Muthen posted on Friday, November 09, 2012 - 11:24 am
I can't find that post. Can you copy the entire post so I can see it? Lambda is the matrix of factor loadings.
 Caroline Vancraeyveldt posted on Monday, November 12, 2012 - 12:57 am
Dear Linda,
This was the post.
Kind regards,

bmuthen posted on Thursday, March 24, 2005 - 5:27 pm
1. yes

2. I assume you refer to the significance of the slope mean - then, yes.

3. yes

4. yes

5. yes

I guess you want an elaboration of 4 (and 5) :-)
An easy way is to do a second run with changed time scores (*, -1, 0), where you need to give a suitable negative starting value for *. A more complicated way is to use the estimates from the run you mention and recognize that the mean change is estimated by the product lambda*[s]. This product can be given a SE since the variance of the product can be computed - but that is more complicated. 5 can also be done via the changed time scores approach.
 Caroline Vancraeyveldt posted on Monday, November 12, 2012 - 5:52 am
Dear Dr.Muthén,

Here you find the post I was referring to:

bmuthen posted on Thursday, March 24, 2005 - 5:27 pm
1. yes

2. I assume you refer to the significance of the slope mean - then, yes.

3. yes

4. yes

5. yes

I guess you want an elaboration of 4 (and 5) :-)
An easy way is to do a second run with changed time scores (*, -1, 0), where you need to give a suitable negative starting value for *. A more complicated way is to use the estimates from the run you mention and recognize that the mean change is estimated by the product lambda*[s]. This product can be given a SE since the variance of the product can be computed - but that is more complicated. 5 can also be done via the changed time scores approach. "
 Bengt O. Muthen posted on Monday, November 12, 2012 - 1:25 pm
My message mentioned "the product lambda*[s]". I was referring to the product of the time score at a certain time point (lambda) multiplied by the mean of the slope growth factor ([s]). If you look at the Topic 3 handout, you see how estimated means for the outcomes are computed using such a product.
 C. Lechner posted on Tuesday, June 04, 2013 - 11:46 am
Dear Linda and Bengt,

I have a growth model with 4 time points. A quadratic model fits well but, as is typical for quadratic models, I find it somewhat hard to interpret the parameters, and even more so the effects of a set of covariates on the growth factors.

I now fit a logarithmic growth model with transformed time scores (0 0.69 1.10 1.39) that has almost identical fit. However, I wonder whether that really facilitates interpretation. For instance, what would the effect of a covariate of b=1 imply? Does it mean its effect depends on the time point?
Moreover, how can I compute estimated values for each time point in this case (i.e., what exactly is the model equation)?

In an empirical application, I have also seen an alternative logarithmic transformation of the time score, which is ln(t)/ln(tmax), resulting in the following vector of timescores:
(0 0.50 0.79 1)
Does this affect interpretation in any way?

Thanks in advance!
 C. Lechner posted on Tuesday, June 04, 2013 - 1:49 pm
Ok, sorry – I think I already solved part of the problem. Means of intercept and slope were estimated at zero, for whatever reason. When using the short language (i s | … ) , I get values that make more sense.

So, whatever transformation of the timescores I use, I guess the slope still represents the linear change in outcome Y for a one-unit change in time, right?
In other words, in the second transformation above the slope would indicate linear change between t1 and t4. Am I correct?
 Linda K. Muthen posted on Wednesday, June 05, 2013 - 12:30 pm
The mean of the slope growth factor is the change from time score zero to time score one.
 C. Lechner posted on Thursday, June 13, 2013 - 6:03 am
That's what I thought, thanks! Allow me one follow-up question:

I want to test how a time-invariant moderator affects intercept and slope, using a multigroup framework.
With freely estimated time scores (e.g., 0 * * 1), the default is that the estimated timescores are held equal across groups.
However, if one assumes that the shape of the curves may differ across groups, one could estimate different time scores in each group, resulting in differentially shaped group trajectories. So, for example:

i s | y1@0 y2* y3* y4@1;

model group2:
i s | y1@0 y2* y3* y4@1;
[i s];

What I was now wondering is whether in such a model one could still compare the means of the latent growth variables across groups. I guess it would work fine with the intercept, but what about the slope?
Given that it represents change from time score 0 to time score 1 in each group, I would assume I can compare its means across groups, even though the estimated time scores in between 0 and 1 would be allowed to differ across groups (e.g., 0 0.5 0.7 1 vs. 0 0.33 0.66 1). Is that correct?
 Linda K. Muthen posted on Thursday, June 13, 2013 - 8:45 am
You need to have the same growth model in all groups to compare growth factor means across groups. The same growth model represents measurement invariance which is required to test means across groups.
 EunJee Lee posted on Thursday, August 14, 2014 - 7:52 pm
I have a few questions of LGM linear spline.

1. Is it okay to constrain residual variance because of negative covariance warning in unconditional model? I got good model fit after constrained residual variance but I'm not sure that my data is okay to fit on lgm model if it needs constrain at the first step.

2. When I run linear spine with four time points, I gave time scores like 0 * * 1. However, mplus output showed time scores at Time 2 and 3 over 1 like 0 1.04 1.08 1.
Why this happened and is it okay? How I can interprete or deal with?

3. If slope variance is not significant, is it not meaninful to go with growth mixture because no heterogeneous in trajectory is found in lgm?

4. If spline and quadratic model shows similar good model fit, which one is recommended to choose? I know I cannot do Chi different test to compare two models so I wonder how I can judge the better one.

Thank you.
 Bengt O. Muthen posted on Friday, August 15, 2014 - 7:58 am
1. Equal residual variances for the outcomes if often a good approximation.

2. That is a complex model. You could try a quadratic instead.

3. Yes, because you can still have variance in the intercept growth factor.

4. Seems like a quadratic is simpler.
 EunJee Lee posted on Friday, August 15, 2014 - 11:31 pm
Thank you very much.

I have two more questions based on your comments.

As follow your recommendation on question number 2, I tried quadratic but it showed not very good model fit like below.

Chi-Square Test of Model Fit

Value 48.339
Degrees of Freedom 7
P-Value 0.0000

CFI = .918; TLI = .930; RMSEA = 0.113

The means of Time 1 to Time 4 are:
T1: 62.283; T2: 59.674; T3:59.630; T4:59.565.

So I thought linear is fitted as the shape of trajectory. However the model fit of linear was similar with that of quadratic so I tried linear spline and got excellent model fit but strange factor loading. How can I deal with factor loading problem? Or could you suggest another option please?

And I have another question.
When I compare the models, linear and linear spline, I usually use chi different test. If I give only spline model modification like variance constraint, should I give same modification on linear too for model comparison, or is it okay to compare with each model's best fit with different modification options?

Thank you for your help.
 Bengt O. Muthen posted on Saturday, August 16, 2014 - 8:22 am
You may want to take a look at the Mplus-related paper on our website under Papers, Growth Modeling:

Kevin J. Grimm , Joel S. Steele , Nilam Ram & John R. Nesselroade (2013) Exploratory latent growth models in the structural equation modeling framework. Structural Equation Modeling: A Multidisciplinary Journal, 20:4, 568-591, DOI: 10.1080/10705511.2013.824775
 EunJee Lee posted on Saturday, August 16, 2014 - 6:14 pm
Thank you for your all kindness.
 Allison Tackman posted on Wednesday, August 05, 2015 - 6:31 pm

In my data, I have participants self-report their positive affect and negative affect each hour they are awake. Since positive affect shows a diurnal pattern (increases after waking, peaks in late afteroon, and then decreases), most people have modeled this change using a model that includes both a sine and cosine wave (so in SAS NLMIXED, the equation is: c0 + c1*sin((2*_pi*t/p)) + c2*cos((2*_pi*t/p)), where t = time and p = period (period is 24 hours in this type of analysis)). I have only seen this done in an MLM framework using the HLM software or SAS NLMIXED, and I'm wondering if a non-linear growth model that models change in terms of a sine and cosine function can be done in an SEM framework, and if so, can this be done in MPLUS? Thank you!
 Jon Heron posted on Thursday, August 06, 2015 - 1:28 am
I think you can fit an SEM-based latent growth model for anything you can describe a mean function for.

This might be a good start (more for the references than the content itself)

Psychol Methods. 2015 Mar;20(1):84-101. doi: 10.1037/met0000028.
Meaningful aspects of change as novel random coefficients: a general method for reparameterizing longitudinal models.
Preacher KJ, Hancock GR.
 Bengt O. Muthen posted on Thursday, August 06, 2015 - 7:29 am
See also the 2 Grimm et al articles on our web site under Papers, Growth Modeling. They show SAS and Mplus approaches to nonlinear growth modeling.
 YH, Son posted on Tuesday, November 24, 2015 - 6:14 pm
piecewise LGM with 5 time.
i s1| g1@0 g2@1 g3@1 g4@1 g5@1;
i s2| g1@0 g2@0 g3@1 g4@1 g5@1;
i s2| g1@0 g2@0 g3@0 g4@1 g5@2;

some problems
1.Is it possible 3 piecewise LGM with only 5 time?

2.I tried piecewise LGM with 5 time, with the var with 2 time fixed to 0.
But I try above model, with just s1 fixed to 0.
which is right? just fix s1 or all var to 0?

3.multivariate LGM.
ig s1g| g1@0 g2@1 g3@1 g4@1 g5@1;
ig s2g| g1@0 g2@0 g3@1 g4@2 g5@3;
ia sa | a1@0 a2@1 a3@2 a4@3 a5@4;

because of the var with 0, Can't I estimate the cov(sa,s1g) in multivariate LGM?
 Bengt O. Muthen posted on Wednesday, November 25, 2015 - 4:02 pm
1-2. with only 2 timepoints of increasing time scores you have to fix the variance of the slope ot zero. so add s2@0.

3. You cannot estimate a covariance when a variable has zero variance.
 Albert Reijntjes posted on Thursday, December 15, 2016 - 2:16 am
Dear Dr. Muthen,

Recently, our group conducted a study among young adolescents (4 waves during 1 one schoolyear). A battery of measures was administered, including children’s (peer-nominated) use of so-called coercive and prosocial strategies, aimed to acquire scarce, but valuable resources. A main focus of the research was to examine whether among those high in power/social dominance a) the use of coercive strategies would decrease after initially being high and increasing, and b) the use of prosocial strategies would increase subsequent to, or at the same time, that the use of coercive strategies decreases.
To address this issue properly, it appears to us that detecting a subgroup (latent class) of children who show this profile of changes in strategy use over time, requires specifying group trajectories in which the linear and/or quadratic slope alternates between being positive and negative.
We could find no example in the manual or published where this kind of changes in slopes are presented. Maybe it is just not (yet) possible, but maybe you can advise us as to how we can use Mplus in another way to determine whether this hypothesized subgroup exists in the real world? Thank very much in advance.

Albert Reijntjes, Utrecht University, the Netherlands
 Bengt O. Muthen posted on Friday, December 16, 2016 - 3:04 pm
I assume that you don't know the timing of the slope sign switches - or it isn't the same for everyone. The new book Growth Modeling by Grimm and others has a chapter 11 that deals with spline modeling with unknown knot points which might be useful for you. They give the Mplus script for this. I don't know how well this works with only 4 time points, however.

In the statistical literature they talk about this in terms of "change-point analysis".

I think growth mixture modeling might have a difficult time finding these different patterns because the classification is also influenced by other discerning features.
 George Youssef posted on Monday, January 22, 2018 - 5:19 pm
I am wanting to estimate nonlinear growth in hormone data using a logistic growth as outlined in this paper I am using this as a guide for estimating the logistic growth, which uses MODEL CONSTRAINT to define the lambda and alpha parameters for the logistic model (and thus the factor loadings for the slope). Estimating this is fine, however I wish to save subject specific effects for these lambda and alpha parameters as done in the above mentioned paper by Marceau et al using empirical bayes (see middle paragraph page 10) to estimate subject specific alpha and lambda parameters. I can save FSCORES for the intercept and slope (i.e., representing individual differences in logistic growth) but my understanding is that this slope is simply broad individual differences in this change over time (but not more specific information about individual difference in their lambda and alpha parameters). Could you advise on whether this is possible to do? Otherwise, would this be possible to do if I run the analysis using the multilevel framework in mplus? Thanks for your time.
 Kevin Grimm posted on Tuesday, January 23, 2018 - 8:32 pm
The model in allows has a fixed lower asymptote and change to the upper asymptote with random timing and rate. For timing and rate to be random, they need to be latent variables, which is not shown in To allow the timing and rate to be random, you need to rewrite the model using a first-order Taylor Series Expansion (see Browne & du Toit, 1991). This requires a bit of programming, so I uploaded an example script at In this script, the variables were collected from age 9 through age 15 and the outcome is tanner stage (lower asymptote=1, upper asymptote=5). Depending on when your data was collected, you'll need to update the constraints - inserting the age of participants for each variable (replacing the 9, 10, 11, 12, 13, 14, 15) in those statements. Hope this helps.
 George Youssef posted on Sunday, January 28, 2018 - 4:00 am
Thanks so much for your reply Kevin! I just want to thank you for all your code and papers that provide instruction on all this (that are all so freely available). I am going to spend some time this week trying to follow the math for the linearisation. Your code is extremely helpful for my learning. Thanks so much!
 Magdalena Formanowicz posted on Wednesday, February 14, 2018 - 4:36 am
Hi Mplus Team,

I am modeling change in arousal at 5 times. Sample means indicate a significant decrease from T1 to T2 and a very slight decrease after that

Sample means:

Neither linear nor quadratic curves fit the data. I have also tried a piecewise model. Time for s1 are 0 1 1 1 1, and for s2 are 0 0 1 2 3. s1@0 for identification. The model does not fit well.

When I freely estimate time scores for T3, T4 and T5 the fit is good but I was wondering whether you may have a different suggestion on how to model the time scores after the sharp decrease from T1 to T2.

Thank you
 Bengt O. Muthen posted on Wednesday, February 14, 2018 - 4:22 pm
I wonder if some variation of exp(-t) would fit.
 Magdalena Formanowicz posted on Thursday, February 15, 2018 - 6:01 am
Thank you for your answer. I tried variations of the exponential function and it results in a poor fit.

Again, the means fot measurement are:


I ran a number of analyses in which adjacent time points were contrasted against each other @0 and @1, and all other time points freed. This approach indicated that there is a significant drop between T1 and T2 and T3 and T4.

The differences between T2 and T3 and also T4 and T5 - give insiginificant means and variances of the slope.

The two separate processes T1-T3 and T3-T5 make sense in terms of the theory, and both could have different covariates.

So my question is, can I model it as two separate processes in this way:

i1 s1| T1@0 T2@1 T3@1;

i2 s2| T3@0 T4@1 T5@1;

If not, could you please recommend how could I solve the problem of lack of change between T2 and T3 and also T4 and T5?

Kind regards
 Bengt O. Muthen posted on Thursday, February 15, 2018 - 4:53 pm
Yes, that shape can be hard to capture. I don't know if Pareto distribution might be possible. But, I think it is perfectly fine to use what you showed (with a slight correction):

i1 s1| T1@0 T2@1 T3@1;

i2 s2| T3@0 T4@0 T5@1;
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