Graphing growth curves PreviousNext
Mplus Discussion > Growth Modeling of Longitudinal Data >
 Anonymous posted on Sunday, July 04, 2004 - 8:04 pm
With your graphing function, I have generated the plots of 3 growth curves based on estimated means for 3 separate samples. Each of these curves is shown in a different .gph file because they came from different samples and were estimated with different models.

For presentation's sake, I would like one graph to show all three curves so they may be easily contrasted. I cannot figure out how to do this...essentially how to merge or cut/paste various curves.

Can you help?
 Linda K. Muthen posted on Monday, July 05, 2004 - 9:37 am
This cannot currently be done in Mplus. It should be available in the future.
 Hanno Petras posted on Friday, April 08, 2005 - 6:56 am
Dear Linda & Bengt,

I noticed that when I run a censored or censored inflated growth model that the plots for the estimated growth trajectories are not available. This is not the case in regular or two-part growth models. Is this intended or am I missing something? Thanks.


 Thuy Nguyen posted on Sunday, April 10, 2005 - 12:40 pm
For a growth model with censored or censored inflated outcomes, you should get a plot of estimated means for the censored outcomes and a plot of estimated probabilities for the censored-inflated outcomes. Adding the PLOT option to example 6.3 on the Mplus CD gives us these 2 plots. Note that if there are covariates in the model, then the model estimated means cannot be computed. A note as to why will be printed in the RESIDUAL output section if RESIDUAL is requested.

If you are still not getting these plots without covariates in the model, please send input/output and data to
 Hanno Petras posted on Thursday, April 14, 2005 - 4:36 am
Dear Thuy,

thank you for the information. Could you send me the example for a multiple class model with censored inflated outcomes with the graphics command added? My email is
Also, in addition to defining a growth structure for the censored outcomes, you can run the censored inflated multiple class model allowing the censored inflated variables to be unstructured, i.e., not to define an intercept, slope etc. Both models run, but the first is very computational intensive. I am wondering if the model with unstructured censored variables is a legitimate model to run, or would you always define a growth structure for the censoring part as well? Are there any guidelines or suggestions for testing the correct specifications of the model?


 BMuthen posted on Friday, April 15, 2005 - 1:50 am
Regardng your second question, the inflated parts of the outcomes need to be allowed to correlate in the model. A minimal model for accomplishing this is an intercept only growth model.
 Hanno Petras posted on Friday, April 15, 2005 - 5:54 am
Thank you. Do you have any model building recommedations? For example, when running a two part model, it makes sense to model the u part and the y-part separately before analyzing them in the same model. I am not quite sure how this would be translated into the censored inflated model.


 BMuthen posted on Saturday, April 16, 2005 - 4:31 am
No, for the censored inflated model, the inflation part of the model is latent so you can't do them separately. So model building ideas have to be drawn on from regular growth modeling, that is, start with a simple model and go from there.
 Hanno Petras posted on Monday, April 18, 2005 - 6:07 am
OK, so does that mean that I would assume that for a growth model with intercept, slope and quadratic slope I would also specify the growth for the censored inflated part with an intercept, slope, and quadratic slope? In a second step, I could reduce the number of growth parameters for the censored inflated part if some turn out to be nonsignificant?


 bmuthen posted on Tuesday, April 19, 2005 - 12:13 pm
You could start with the same growth factors in the inflation part but only allow the intercept to be random for the inflation part (so estimating only the means, not the variances, for s and q in the inflation part).
 Hanno Petras posted on Friday, January 05, 2007 - 10:39 am
Dear Linda & Bengt,

I am running a Two-Part model looking at delinquency data over 13 time points. The u-part relates to the probability of engagement in crime and the y-part relates to the count of crimes the person engaged in within that time period. The mean count of crimes range from 1.38 to 2.72. I noticed that treating it as count increases the computation immensely. In the Olson & Schaeffer paper the y-part is logged (if I remember correctly). Given that the outcome is relative rare would it be more appropriate to specify it as count instead of treating it as log continuous. If the latter is advisable, are the estimates accurate or as in the Poisson regression model (treating the outcome as log + an added constant) are they rough approximations. Sorry for the long question. Thanks.

 Bengt O. Muthen posted on Monday, January 08, 2007 - 9:35 am
With two-part modeling, having y as a count variable would call for using a Poisson distribution that is truncated, that is, not including zero, which Mplus does not accomodate. So with counts, Mplus leads to ZIP modeling instead of two-part. It's hard to say which is best.
 Hanno Petras posted on Monday, January 08, 2007 - 11:02 am
Dear Bengt,

I just saw that I filed my question under the wrong heading. My apologies for that.

Thank you for your answer. So when treating the y part as count Mplus models it as Poisson and not truncated Poisson and that would mean that the estimates are biased. Therefore the log transformation may make more sense. Would you agree?


 Bengt O. Muthen posted on Monday, January 08, 2007 - 11:49 am
 Adam Darnell posted on Monday, February 05, 2007 - 10:28 am
I have an LGM of 11 repeated observations, that includes an intercept, and linear and quadratic growth functions. I have several time invariant predictors of these growth parameters. I would like to display the estimated growth curves for two different levels of one of these predictors, in order to illustrate the effect of the predictor. Is there a way to display multiple estimated growth curves across conditions of a predictor? I have tried specifying covariate sets in the adjusted estimated means window, and just specifying two contrasting values of the predictor I am trying to illustrate. But the same curve is shown for both sets. Thanks in advance for your help,
 Bengt O. Muthen posted on Tuesday, February 06, 2007 - 10:31 am
This should work fine. Take ex6.10 as an example (you have that ex on your CD or find it on our web site), plotting against two values of x1, say -1 and +1. In the Adjusted means menu you enter say "low" in the upper right hand box and then click Name covariate set. Then you click on x1, click on Use and enter -1, plus say ok. Then put the cursor in the covariate set again and type "high" and click on Name covariate set (it is important to click again, otherwise you get 2 overlapping curves) and then enter +1. This gives 2 lines. Note that some models don't imply different curves for different covariate values (such as mixtures where x only influences c).
 Adam Darnell posted on Tuesday, February 06, 2007 - 6:16 pm
Thank you Dr. Muthen for your reply... I used example 6.10 as a guide. I was defining covariate sets in the same way that you described, but it seems that it works fine as long as you don't include multiple observed predictors with unanalyzed associations between them. If the predictors are uncorrelated, it is no problem to specify covariate sets (as is the case with ex6.10 -- x1 and x2 are uncorrelated). But if there are unanalyzed associations between the predictors and then you specify covariate sets the trajectories are always the same for the covariate sets. Could you explain this? In my model I am interested in illustrating the effect of one predictor on linear and quadratic growth, controlling for the effects of 4 observed covariates and the level of the intercept. All of the observed predictors have unanalyzed associations between them, and I am unable to specify covariate sets with any result as long as those correlations are there. Thank you for your assistance.
 Adam Darnell posted on Wednesday, February 07, 2007 - 8:33 am
Another question Dr. Muthen: as I described above, I am interested in the effect of one predictor on the slope controlling for the intercept, and on the quadratic controlling for the intercept and slope. So I have direct effects between the latents where appropriate. I gather that the adjusted means are not adjusted for these effects, correct? That is the effect of my predictor of interest will appear the same in the adjusted means graph regardless of whether the model specifies an unanalyzed association between intercept and slope, or a direct effect. Also, I am guessing that it is not possible to specify covariate sets based on these latent covariates? Thank you very much...
 Linda K. Muthen posted on Wednesday, February 07, 2007 - 9:25 am
Please send your input, data, output, and license number to Give us the sequence you go through to get the graphs and explain where it goes wrong.
 Eva Van de gaer posted on Friday, February 16, 2007 - 5:56 am
Dear Linda and Bengt,

I have estimated a two level (level 1: students, level 2: schools) latent growth curve model and would like to draw the growth curves at level 2. Is this possible in Mplus? I would like to be able to draw growth curves of a number of schools but it seems to me that I only can draw growth curves at the student level.

This is part of my model specication:

iw1 sw1 qw1| aczc1@0 aczc2@1 aczc3@3 aczc4@5;
aczc1-aczc4 (1);

ib1 sb1 qb1| aczc1@0 aczc2@1 aczc3@3 aczc4@5;

OUTPUT: modindices tech4 tech1;
SAVEDATA: results are results.dat;
PLOT: Type = plot1;
type = plot2;
Type = plot3;
Series = aczc1(0) aczc2(1) aczc3(3) aczc4(5);

Thank you!
 Bengt O. Muthen posted on Friday, February 16, 2007 - 7:43 pm
The plot that Mplus provides is for the estimated mean growth curve. There is only one mean per variable. If you want to plot curves for schools, you would have to ask for factor scores - which gives you the estimated school values for each time point. Mplus does not plot that for you, but you have the values in the graph file and can plot them with another program.
 Richard Rivera posted on Wednesday, October 24, 2007 - 12:32 pm
When, I run a two-part growth model with out covariates, can Mplus produce plots? If so, what is the code?
 Linda K. Muthen posted on Wednesday, October 24, 2007 - 5:21 pm
See the PLOT command.
 Julia Dmitrieva posted on Tuesday, October 14, 2008 - 11:29 pm
Dear Linda and Bengt,

I am trying to graph results of a growth curve model for censored inflated data. I am using a covariate, so plot command does not work. Can I manualy compute estimated values for a given value of a covariate (e.g., X=2)?

Here is my model:

i s q| Y1@0 Y2@.1 Y3@.2 Y4@.3
Y5@.4 Y6@.5 Y7@.6 Y8@.7 Y9@.8 Y10@.9 Y11@1.0
Y12@1.1 Y13@1.2;
ii si| Y1#1@0 Y2#1@.1 Y3#1@.2 Y4#1@.3
Y5#1@.4 Y6#1@.5 Y7#1@.6 Y8#1@.7 Y9#1@.8 Y10#1@.9
Y11#1@1.0 Y12#1@1.1 Y13#1@1.2;
ii-si@0; i-q@0;
Y1-Y13 ON X;

Thank you very much,
 Linda K. Muthen posted on Wednesday, October 15, 2008 - 8:07 am
This requires numerical integration so it cannot be calculated by hand.
 Julia Dmitrieva posted on Wednesday, October 15, 2008 - 11:15 am
Hi Linda,
Thank you. What would be your suggestion for reporting results? My ultimate model is a growth mixture model with one static and one time-varying control variable. I would need to somehow show the shape of the developmental trajectories. Should I graph results for trajectories without the covariates? Would that be an adequate representation?
Alternatively, should I run growth mixture models on the residuals (from the regression of the outcome variable on the control variables)? Are there any other options?

Thank you very much,
 Bengt O. Muthen posted on Wednesday, October 15, 2008 - 5:52 pm
I would show the graph without covariates.
 Wei Chun posted on Monday, February 02, 2009 - 6:21 pm
Dear Linda,

In Mplus output, following plots are usually available:
Histograms (sample values, estimated factor scores, estimated values)
Scatterplots (sample values, estimated factor scores, estimated values)
Sample means
Estimated means
Sample and estimated means
Observed individual values
Estimated individual values

But in my output, I can only get:
Histograms (sample values, estimated factor scores, estimated values)
Scatterplots (sample values, estimated factor scores, estimated values)
Sample proportions
Estimated probabilities
Item characteristic curves
Information curves
Estimated individual probability values

Although I have asked for the following:
PLOT: SERIES ARE con1_94 con1_96
con1_98 con1_00 con1_02 con1_04(*);

Could you please advise me on this issue? Many thanks.
 Linda K. Muthen posted on Tuesday, February 03, 2009 - 8:12 am
The plots that you get depend on the model that is estimated. For example, Observed individual values and estimated individual values are given for growth models. It sounds like you have an IRT model. If you have further questions regarding this, please send them along with your outputs and license number to
 Wei Chun posted on Tuesday, February 03, 2009 - 2:24 pm
Thank you, Linda.

It is a 6 wave growth model with quadratic trend. I used WLSMV estimator because of my categorical outcome variables. If I use MLR estimator, I can get plots of
Sample and estimated means
Observed individual values
Estimated individual values
So do I use MLR for getting the plots and use WLSMV for reporting the outcomes?

Many thanks for your advice.
 Linda K. Muthen posted on Tuesday, February 03, 2009 - 3:01 pm
If you are not using Version 5.2, go to Support - Mplus Updates and download it. If you are, please send your input, data, output, and license number to
 Kurt Beron posted on Tuesday, March 10, 2009 - 1:33 pm

I'm trying to plot a censored normal outcome from a growth model. The plot that is produced seems to be the unconditional plot of the latent variables, which includes the zero values (lower bound in my case). However what I'd like is the plot of the more relevant, for me anyway, E(Y|Y>0). Is there a way to generate this in Mplus? Usually this requires computing the inverse Mills Ratio and this isn't obvious to me in Mplus.

 Bengt O. Muthen posted on Tuesday, March 10, 2009 - 4:10 pm
No, I don't see a way to plot this using the Mplus PLOT command. The normal density and distribution function would need to be available. Something to add.
 Dimiter M. Dimitrov posted on Tuesday, March 24, 2009 - 1:50 pm
Dear Linda,
In UG ex.19.6 (Individual development over time), which PLOT command(s) to add in order to plot individual trajectories over time?
 Linda K. Muthen posted on Tuesday, March 24, 2009 - 3:11 pm
You would need PLOT3.
 Dimiter  posted on Tuesday, March 24, 2009 - 4:09 pm
I used PLOT3, but it does not provide individual trajectories. Is there some additional syntax under the command: PLOT: TYPE IS PLOT3; ?
Also, how to make a decision about model fit based on the reported indices (Loglikelihood, AIC, and BIC)?
 Bengt O. Muthen posted on Tuesday, March 24, 2009 - 7:32 pm
There is no UG ex 19.6 so I can't tell which model you are considering. With growth models the PLOT command SERIES = is the one to use to look at estimated mean curves and individual trajectories.

Your second question concerns a big topic. Basically, these indices are meant to be used when comparing two models. For instance, you want the model with the lowest BIC.
 Dimiter  posted on Tuesday, March 24, 2009 - 7:35 pm
it's ex.9.16
 Bengt O. Muthen posted on Tuesday, March 24, 2009 - 7:48 pm
I think the individual plots may not be available for Type = Random.
 Nicole Nugent posted on Tuesday, May 11, 2010 - 10:50 am
Dear Bengt and Linda,

I have conducted a growth model with intercept and slope of count variables over 4 timepoints regressed on treatment (two groups; 0, 1) and covariates. I would like to plot (or use output to manually plot) the curves for each treatment group (0,1) as an illustration of the significant differences in counts over time. Any help with how to do this is much appreciated.

Thank you in advance,
 Linda K. Muthen posted on Wednesday, May 12, 2010 - 8:28 am
You can use TYPE=MIXTURE with the KNOWNCLASS option to obtain a curve for each of the two treatment groups.
 Deborah Bandalos posted on Wednesday, July 07, 2010 - 1:43 pm
I have run a censored-inflated growth model on data in which I know there is some missing data. However, in the censored-inflated output it states that there are 0 missing data patterns and I get no info on the amount of missing data. Why would this be?

My syntax is:


USEVARIABLES = shop2 shop4 shop5-shop7 shop8 shop9;
CENSORED = shop2 shop4 shop5-shop7 shop8 shop9 (bi);
MODEL: i s| shop2@0 shop4@1 shop5@2 shop6@3 shop7@4 shop8@5 shop9@6;
ii si | shop2#1@0 shop4#1@1 shop5#1@2 shop6#1@3 shop7#1@4 shop8#1@5 shop9#1@6;
shop2-shop9 (1);! held = due to convergence problems;
OUTPUT: tech1 tech4;
plot: type is plot3; series = shop2-shop9(*);


 Linda K. Muthen posted on Wednesday, July 07, 2010 - 2:59 pm
Please send the full output and your license number to
 Diane Chen posted on Thursday, December 16, 2010 - 12:36 pm
Is it possible to graph growth curves for a 2nd process (e.g., aggression) within different classes of another process (e.g., peer rejection)? What would the plot/series input statement look like? Thanks!
 Linda K. Muthen posted on Thursday, December 16, 2010 - 2:50 pm
You use a | symbol to separate the two processes in the SERIES option.
 Isaac Petersen posted on Tuesday, February 15, 2011 - 4:55 pm
Is it possible to get confidence intervals or standard error bars for the adjusted estimated means?
 Linda K. Muthen posted on Tuesday, February 15, 2011 - 5:28 pm
No, this is not possible.
 EFried posted on Wednesday, February 22, 2012 - 7:54 am
In the online tutorial video Topic 4, second part, you mention that MPLUS is able to contrast the mean of the dependent variable on measurement point t of (1) people who will be present at t+1 in a longitudinal study vs. (2) people who will drop out at t+1.

I do not find this function, using the newest version of MPLUS, and wonder if you could hint me into the right direction.

Also, is there a paper covering methods to test whether longitudinal data are MAR/MCAR in MPLUS?

Thank you!!
 EFried posted on Wednesday, February 22, 2012 - 8:10 am
(My sample has 5 measurement points, missing data 5% at the first measurement point and 40% at the last. Not atypical for depression studies, but I don't quite know how to handle this)
 Linda K. Muthen posted on Wednesday, February 22, 2012 - 2:11 pm
See the DESCRIPTIVE option of the DATA MISSING command. See also the following paper on the website:

Muthén, B., Asparouhov, T., Hunter, A. & Leuchter, A. (2011). Growth modeling with non-ignorable dropout: Alternative analyses of the STAR*D antidepressant trial. Psychological Methods, 16, 17-33.

MAR cannot be tested. See the following paper on the website for MCAR:

Muthén, B., Kaplan, D. & Hollis, M. (1987). On structural equation modeling with data that are not missing completely at random. Psychometrika, 52:3, 431-462.
 EFried posted on Sunday, February 26, 2012 - 7:25 am
Thank you, Linda.

(1) The difference between means of the dependent variable between dropouts and non dropouts after measurement point 1 is highly significant p<.001, I guess that means that people with a higher value on my dependent variable dropout more readily.

(2) Does multiple imputation rely on the MAR assumption?

(3) Bringing covariates into the model with the "x1 x2 x3;" command so that MPLUS uses a different way to handle covariate missing values leads to non-identification very often. Are there any constraints that I would need to implement that MPLUS doesn't use by default and could cause these problems?
 Linda K. Muthen posted on Sunday, February 26, 2012 - 11:06 am
1. That sounds correct.

2. Yes.

3. When you bring a binary covariate into the model, it results in a non-identification message because the mean and variance of a binary variable are not orthogonal. When this is the reason, the message can be ignored.
 EFried posted on Monday, February 27, 2012 - 5:44 am

I'm not sure whether the problems arise from this.

Using a GMM with

type = MIXTURE;
Starts = 400 40;
i s q | phq_t0@0 phq_t1@1 phq_t2@2 phq_t3@3 phq_t4@4;
i s ON sex hisdep;

leads to no problems, 27 free parameters, and a BIC of 14000.

Including "sex hisdep;" into the model (both binary) leads to 38 free parameters, a lot of perturbed starting value runs in both phases, and a BIC of 16000.

Entropy and class solutions are about the same, though.

What could be the reason for this?
 Linda K. Muthen posted on Monday, February 27, 2012 - 6:08 am
It is impossible to answer this question with the information given. Send the relevant outputs and your license number to
 Andreas Stenling posted on Thursday, January 24, 2013 - 5:27 am

We are trying to graphically display individual trajectories in latent growth curves. However, the plots only displays straight lines and no curves , although we request observed individual values. We have tried different syntax to obtain graphs as stated in the Mplus users guide but regardless of the syntax used we only receive straight lines.

So our question is how do we obtain a plot showing individual trajectories? We suspect that there is a simple solution to this problem but we are currently unable to solve this. Below is our model and plot syntax.

i s| Sjalvkansla1@0 Sjalvkansla2@1 Sjalvkansla3@2;

PLOT: SERIES = Sjalvkansla1 (0) Sjalvkansla2 (1) Sjalvkansla3 (2);
 Linda K. Muthen posted on Thursday, January 24, 2013 - 11:55 am
Please send your output, gph file, and license number to
 Laia Becares posted on Thursday, May 30, 2013 - 4:36 am

I have a piecewise latent growth curve which jointly models two outcomes. How can I plot both lines?

I have tried:
Series is czbmi3 (0) czbmi5 (1) czbmi7 (2)czght3 (0) czght5 (1) czhgt7 (2):
Type is plot3;

but don't get the expected graphs.

Thank you
 Bengt O. Muthen posted on Thursday, May 30, 2013 - 1:28 pm
You can try using | in the Series option to separate the two pieces.
 AshleyM posted on Sunday, November 17, 2013 - 1:04 pm

I am trying to get PLOTS for the trajectories of classes for a series of LCGA models. However, I am not getting the expected plots in the output and the option is not available to view the plots.

Here are parts of my syntax that might be helpful:

USEVARIABLES ARE dintk1 dintk2 dintk3 dintk4 dintk5 dintk6 dintk7 dintk8 dintk9 dintk10 dintk11 OHFNP21 OHFNP22 OHFNP23 OHFNP24 OHFNP25 OHFNP26 OHFNP27 OHFNP28 OHFNP29 OHFNP210 OHFNP211;
TSCORES = dintk1-dintk11;
CLASSES = c(3);

STARTS = 100 10;

i s|OHFNP21-OHFNP211 AT dintk1 dintk11;

SERIES = dintk1 - dintk11 (s) ;

As you can see, I am trying to use the TIMESCORES command to account for individually varying times of observation (UG 6.12) and the fact that the data are not time structured.

Is it possible to get plots while using TIMESCORES? Do you know if it is still possible to view the plots using this command? Or is there something else that I am doing incorrectly?

Thank you!
 Linda K. Muthen posted on Sunday, November 17, 2013 - 2:45 pm
Plots are not available for the AT and TIMESCORES options.
 AshleyM posted on Tuesday, November 19, 2013 - 10:27 am
Ok, thank you. If the plots, LMR-LRT, and BLRT options are not available for LCGA models using TIMESCORES, are there other indices of fit I can look to in order to make decisions about the best number of trajectory classes (aside from AIC, BIC, entropy, and posterior probabilities, and theory)? Do you have any suggestions about how to make such decisions?

Thank you!
 Sebastian Teran Hidalgo posted on Monday, December 09, 2013 - 10:59 am

I am running a ZIP growth model. I have successfully run an unconditional version of the model. I requested the plots of the estimated and sample means and they look fine. When I do the same model but now conditional on some dummies related with race and request the sample means I only get a straight line. Why am I getting a straight line?

Thank you so much


i s q| DRINKS13@0
DRINKS14@0.1 ... DRINKS32@1.9;

ii si qi| DRINKS13#1@0
DRINKS14#1@0.1 ... DRINKS32#1@1.9;

DRINKS14@0 ... DRINKS32@0 i s q];

DRINKS14#1@0 ... DRINKS32#1@0 ii si qi];

i on race_h race_w race_b;
s on race_h race_w race_b;
q on race_h race_w race_b;
ii on race_h race_w race_b;
si on race_h race_w race_b;
qi on race_h race_w race_b;



Plot: type=plot2;
series = DRINKS13(13)
DRINKS14(14) ... DRINKS32(32);
 Linda K. Muthen posted on Monday, December 09, 2013 - 11:06 am
Please send the output and your license number to
 xiaoyu bi posted on Monday, January 27, 2014 - 11:43 pm
Dear Dr. Muthen,
How to plot a variable using a long-format data? For example, for wide-format variables, x1 x2 x3 x4 x5, I can plot it by series = x1 x2 x3 x4 x5 (*);
But, how about in long-format data to plot variable X?
Thank you!
 Linda K. Muthen posted on Tuesday, January 28, 2014 - 11:53 am
The PLOT option is not available for long format.
 ellen liao posted on Monday, June 02, 2014 - 9:58 am
Hi, when covariates are included in the growth curve model,what does the 'estimated means'in the PLOT function present? Holding all covariates at their average level (how about sex??)?

Thank you.
 Bengt O. Muthen posted on Tuesday, June 03, 2014 - 9:26 am
Yes, holding all covariates at their average value (which you can change). For a variable such as Female (1=female, 0=male), the average represents the proportion of females in the sample, so you probably want to change that to either 0 or 1.
 Stine Hoj posted on Monday, September 29, 2014 - 7:23 pm
I am running an unconditional GMM with time as a continuous variable (using TYPE=RANDOM and TSCORES) and accounting for clustering using TYPE=COMPLEX.

I am generally able to obtain a plot of sample means for each class using PLOT3. However, if I identify the observed variable as being censored from below then 'sample means' disappears from the plot options. Is this unavailable with censored data, or have I omitted something else important from my code?

Thanks in advance.

TSCORES ARE TimeT1 TimeT2 TimeT3;
CLASSES = c(3);

STARTS = 100 20;

i s | DETA1 DETA2 DETA3 AT TimeT1 TimeT2 TimeT3;

TYPE = plot1 plot2 plot3;
SERIES = DETA1(0.924) DETA2(2.715) DETA3(5.095);
 Linda K. Muthen posted on Tuesday, September 30, 2014 - 6:12 am
The plot is not available when the AT option is used.
 Stine Hoj posted on Tuesday, September 30, 2014 - 7:30 pm
Hi Linda, thanks for your reply.

I had read this on the message board previously but I am able to get the plot using AT so long as I don't also include the CENSORED option. This seems peculiar.

Is there a reason the plot should not be available with AT - e.g. is it inappropriate when there is significant variability in the timing of observations between subjects?
 Linda K. Muthen posted on Wednesday, October 01, 2014 - 9:31 am
Please send the output where you got a growth plot with AT. To the best of the team's knowledge, this is not possible. It is not inappropriate. We have simply not implemented it.
 Terry Ng-Knight posted on Tuesday, January 13, 2015 - 5:01 am
I have run a latent growth model (3 time points) which has been regressed onto 10 covariates.
I have used the regression weights/parameter estimates to calculate model-estimated means at varying levels of my covariates and used these to create graphs in excel. Is there any way I can get standard error or SD estimates for these model-estimated means from Mplus to add to my graphs?

Thank you for your help.
 Bengt O. Muthen posted on Tuesday, January 13, 2015 - 8:04 am
You can use Model Test to get this. Just express the model-estimated means as NEW parameters.
 Philipp Jugert posted on Tuesday, February 10, 2015 - 11:21 am
I have run a conditional latent growth model (4 time points) with one covariate. For illustration, I would like to plot growth trajectories as a function of the continuous covariate (0-15 years) in one graph. The trajectories should include both intercept and slope information. Could you please tell me whether and how this is possible using the PLOT function?

Thank you.
 Bengt O. Muthen posted on Tuesday, February 10, 2015 - 12:06 pm
Yes, you can do that - I think the plot menu option is called something like Adjusted estimated means...
 Philipp Jugert posted on Tuesday, February 10, 2015 - 1:30 pm
Thank you for your answer. But from my understanding this option will still plot the covariate adjusted trajectories as a function of the time points. However, I would like to have the covariate on the x-axis, not the time points. How is this possible with the plot menu options?
 Bengt O. Muthen posted on Tuesday, February 10, 2015 - 4:29 pm
Ok; I don't think you can do that via the PLOT command. How about using Model Constraint and the PLOT and LOOP options (see UG ex 3.18)? You can use model parameter labels to express the estimated means as a function of a covariate. You get a 95% CI also.
 Nicholas Bishop posted on Wednesday, January 20, 2016 - 12:31 pm
I'd like to plot a quadratic growth model using individually varying time scores as the time axis (individual age centered on mean age of sample at baseline, divided by 10). As plots are not available when using the TIMESCORES option, what method can I use to plot the estimated growth trajectory for each year of age in Excel? I've attempted to use the formula below but the estimates start to fall outside the range of the outcome measure quickly. I've also divided the mean slope and quadratic estimates by 10 to reflect change over 1 year rather than 10.


Thanks for your input.
 Bengt O. Muthen posted on Wednesday, January 20, 2016 - 7:33 pm
Instead of the 0, 1, etc values you should use each individual's value on the TSCORE variable.
 Nicholas Bishop posted on Thursday, January 21, 2016 - 7:59 am
To clarify, I'm interested in plotting the estimated trajectory for the complete sample. The data is collected biennially and individual time scores are centered on mean age at time 0, divided by 10. If I am following you, I would need to adjust my equations as shown below to properly identify the estimated outcome for the average time score at each observation wave.

 Bengt O. Muthen posted on Thursday, January 21, 2016 - 4:17 pm
I was thinking about individuals' estimated means at their particular time points. Sounds like you want to work with average time points, which seems fine.
 saravanelst posted on Thursday, September 22, 2016 - 6:48 am
I've conducted a GMM with 5 time points. The best model is a 4-class model. I would like to add a 95% CI to my plot. Is there a way to do this in Mplus? I tried the loop option, but this doesn't work.

Is it possible to calculate or receive the average of the outcome at each time point and the SE? Then I could create the plot in Excel.
 Bengt O. Muthen posted on Thursday, September 22, 2016 - 9:41 am
Using Model Constraint with our without Loop you can express the estimated mean values at each time point for each class. The formulas follow those in the handout for Topic 3 of our short courses on our website - see, e.g., slide 98.
 saravanelst posted on Monday, November 28, 2016 - 2:16 pm
Thank you.

I used your formulas and it worked. One question though: when I go to 'sample means' in the 'view plot' option the means at the last time points are lower than the means I calculated using the formula: intercept + (slope*t) + (quadratic slope*tsquared). The rest of the means are the same.
Could this be due to a lower N at the last time points? Or due to taking into account the posterior probabilities?
 Bengt O. Muthen posted on Monday, November 28, 2016 - 2:20 pm
The sample means at the end may be biased due to attrition if they are computed for one variable (time point) at a time.
 sed48 posted on Tuesday, April 25, 2017 - 4:38 am
MPlus is not producing the plot for estimated and sample means for my LGCM. Is it not available with some of the other options I have?

Estimator is MLR;

i s | mfq0@0 mfq6@1.2 mfq12@1.8
mfq36@4.3 mfq52@6 mfq86@9.5;

i WITH s;


SERIES = mfq0(0) mfq6(1.2) mfq12(1.8)
mfq36(4.3) mfq52(6) mfq86(9.5);

Any help is appreciated.
 Bengt O. Muthen posted on Tuesday, April 25, 2017 - 5:30 pm
Seems like that should work - send your input, output, and data to Support along with your license number.
 Huiru posted on Wednesday, May 30, 2018 - 5:02 pm
Dear Bengt and Linda,

I would like to show the impact of a time-varying covariate using a plot. Based on the previous posts, I understand that I can use "Adjusted mean manu".

Using ex6.10 data as an example, I used the following code to obtain constant effect of a31-a34 over time, but allow for random slop.

TITLE: this is an example of a linear growth model for a continuous outcome with time-invariant and time-varying covariates

DATA: FILE IS ex6.10.txt;

VARIABLE: NAMES ARE y11-y14 x1 x2 a31-a34;


MODEL: i s | y11@0 y12@1 y13@2 y14@3;
i s ON x1 x2;
tvcov | y11 ON a31;
tvcov | y12 ON a32;
tvcov | y13 ON a33;
tvcov | y14 ON a34;

plot: type=plot3; series=y11-y14(*);

The model works. Now I want to show two lines indicating the effect of tvcov. With the above code, "Adjusted mean manu" is no longer provided.

I am wondering how should I achieve such plot.

A similar example is on page 157 of the slides for topic 3. I really wish to generate plot like this in order to more effectively communicate with other knowledge users.

Thank you very much!
 Bengt O. Muthen posted on Friday, June 01, 2018 - 3:15 pm
Slide 157 of Topic 3 is for a binary time-varying covariate with fixed slope. I don't know how this relates to your random slope situation and how you envision "two lines".
 Yu-Chih Chen posted on Wednesday, June 06, 2018 - 10:56 am
Dear Bengt and Linda,

I am fitting a parallel process LGCM for two outcomes (depression and satisfaction) with 3 time points, and I have two categorical variable, female (2 categories) and race (white, black, others). Here is my syntax:

I would like to see an interaction effect between gender and race on the trajectory for my 2 outcomes. I have reviewed comments from the forum and the ex.3.18 from the UG, but I still do not figure out. I am wondering how can I get these plots?

Here I attached my syntax (I remove data and variable paty):

fw =female*white;
fb =female*black;

id sd | d1@0 d2@1 d3@2;
is ss | s1@0 s2@1 s3@2;

id sd ON female white black fw fb;
is ss ON female white black fw fb;

SERIES = d1-d3(sd) | s1-s3(ss);

Thank you so much for your help!
 Bengt O. Muthen posted on Wednesday, June 06, 2018 - 4:29 pm
Use the plot option "Adjusted estimated means" where you give values for your covariates for which you want to plot the estimated growth curve.
 James V Ray posted on Wednesday, June 27, 2018 - 8:45 am
Dear Bengt and Linda,

I ran a Multiple Group Multiple Cohort Growth Model with time-varying covariates. A reviewer asked that I present the effects of the tcovs on the growth curve in graphical format. Do you have any guidance on how this might be done? Is it possible to do this using the plot option in Mplus?

Thanks in advance!
 Bengt O. Muthen posted on Wednesday, June 27, 2018 - 4:12 pm
No such plot option; you have to plot it with some other software using the estimated outcome means.
 LS posted on Thursday, August 15, 2019 - 8:59 am
Dear Drs. Muthen,

I have run a conditional quadratic GMM using the manual 3 step approach.

In the first step, I selected as the final unconditional model the 3-class one.

In the last step, I am predicting life satisfaction class membership using a count covariate (0 to 7 reported critical events).

Is it possible to differentially plot the effect of the count variable on the three classes obtained? I would like to have something visual for the conditional model as the plot I obtained for the unconditional one.

Thank you,
 Bengt O. Muthen posted on Thursday, August 15, 2019 - 5:25 pm
Yes, mean curves adjusted for covariate values is one of the plot menu options.
 Marina Epstein posted on Tuesday, August 27, 2019 - 4:14 pm
I am trying to graph the estimated probabilities in a linear growth model with categorical outcomes. Since the intercept is set to zero, how do I obtain the estimate?

That is, I am graphing the following equation using Excel:

Time 1: exp(intercept + timescore1*slope)
Time 2: exp(intercept + timescore2*slope)

What is the value of intercept that I should be using?

Thank you!
 Bengt O. Muthen posted on Wednesday, August 28, 2019 - 3:02 pm
Do either i or s have variances? If so, you need to do numerical integration to get the probabilities. Mplus computes these in the PLOT command.

I don't know why you use exp.
 Marina Epstein posted on Thursday, August 29, 2019 - 10:48 am
Thanks for responding, Bengt. I am able to plot the predicted probabilities in Mplus. I wanted to be able to plot in Excel. My understanding is that in order to obtain the predicted probability of a binary variable, you exponentiate the i + s*x equation for each time point. Am I incorrect?
 Bengt O. Muthen posted on Saturday, August 31, 2019 - 5:12 pm
Yes, you are incorrect. Have a look at the end of Chapter 14 of our UG. See pages 552-557. Also, these formulas are not appropriate if you have variances for i or s.
 Marina Epstein posted on Tuesday, September 03, 2019 - 9:37 am
Thank you very much, this is super helpful. I have one more question. I am doing a sequential piecewise growth model with two intercepts and two slopes. Is there a way to plot both parts of the model? I have tried to specify only slope2 or only slope1 in the SERIES command, but I only get a single plot with one slope.
 Bengt O. Muthen posted on Tuesday, September 03, 2019 - 9:59 am
Check the bar ( | ) feature of the Series option on page 850 of the Version 8 UG.
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