Growth modeling with parallel processes PreviousNext
Mplus Discussion > Growth Modeling of Longitudinal Data >
 Jungmeen Kim posted on Friday, November 04, 2005 - 6:42 am
I fitted a latent curve model with parallel processes for self-esteem and depression among children. The model also included covariates of four subtypes of child maltreatment that predict both growth functions of self-esteem and depression.

I found that the intercept of self-esteem is positively related to the slope of depression. The intercept of depression slope is significant and negative, indicating that depression decreased over time for the entire group. So, does the positive value of the regression path from the self-esteem intercept to the depression slope mean that children with higher initial levels of self-esteem showed greater decreases in the depression over time? Or because the depression slope is negative (decreasing), a positive influence indicates slower decreases in the depression?
 bmuthen posted on Friday, November 04, 2005 - 7:55 am
A positive influence means a slower decrease in depression.

Note that the intercept of the depression slope is not the mean - you get the mean in Tech4.
 Jungmeen Kim posted on Friday, November 04, 2005 - 1:18 pm
Dear Bengt,

Your answer was (critically) helpful! Thanks so much for your comment regarding intercept vs. mean of the slope factor. A related question is about testing equality of the means of intercept and slope factors across two groups (boys vs. girls). I put equality contraints giving the identical number in the second 'MODEL: male' statement as in the first MODEL statement like this,
[CDI_L] (3);
[CDI_S] (4);
[SEI_L] (5);
[SEI_S] (6);

In the output, however, the intercepts fo the first group yields all zeros; whereas in the second group, the values of the intercepts of those factors are non-zeros. In the TECH4 output results, the estimated means are not equal between the groups. I know as the default in Mplus, the intercepts of the factors are fixed to zero for the first group and are freed to be estimated in the other groups. Am I still testing the equaility (invariance) of the means of the intercept and slope factors by assiging the same value for the corresponding parameters in the Mplus program? I'm confused because the output shows different values of the parameters across groups. Thanks!
 bmuthen posted on Saturday, November 05, 2005 - 9:15 am
This is hard to answer without seeing your run. Chapter 16 (see the | symbol) lists the default settings for growth models using the old BY language and the new | growth language. If you are not doing this already, I would recommend using the new language. As you see in chapter 16, intercepts being fixed at zero or not for a group depends on the setting, such as having categorical outcomes or not. So I can't answer you specifically. Note also that in your case with the regression of a slope growth factor regressed on an intercept growth factor, testing invariance across groups in growth factor means is not accomplished by holding growth factor intercept parameters equal across groups since the growth factor mean is not just a function of the intercept parameter as I mentioned. If this answer is not sufficient, please send your input, output, data, and license number to
 Michelle Little posted on Sunday, November 26, 2006 - 1:44 pm
I have a question about the interpretation of parallel process growth models using ind. varying time scores.

When regressing a slope factor (construct A) on a slope factor (Construct B), is it best to use the unconditional mean of both slopes to interprete the resulting effects?

So, if process A is declining in the unconditional model, and process B is declining in the unconditional model- and then find that regressing slope B on slope A results in a positive regression estimate (and a positive slope intercept for construct B)- then I would interprete that steeper declines in A is associated with steeper decline in B (thus the remaining intercept of B is now positive).

Is this correct?


 Linda K. Muthen posted on Monday, November 27, 2006 - 10:12 am
If your two processes are declining and you regress b on a, it means that for a one unit decrease in a, b decreases the amount of the regression coefficient.
 Michelle Little posted on Monday, November 27, 2006 - 10:41 am
Ok. thanks much.

But, how about when process A is rising and process b is declining. In that case, would a poistive regression coefficient be interpreted as meaning that a rising A process is associated with less of a decline in B?


 Linda K. Muthen posted on Monday, November 27, 2006 - 11:13 am
 Belinda Needham posted on Tuesday, August 19, 2008 - 12:25 pm
I would like to run a parallel process growth model for BMI and symptoms of depression (where I regress the slope of BMI on the intercept of depression and vice versa). In the unconditional latent growth models for BMI and depression, there is a significant quadratic term. How do I account for nonlinearity in the parallel process growth model? Is it necessary to do so? Below is the model command I have so far:

MODEL:ibmi sbmi qbmi| bmi05@0 bmi10@1 bmi15@2 bmi20@3;
idep sdep qdep| depsum05@0 depsum10@1 depsum15@2 depsum20@3;
ibmi on ex3_age female black nocoll05 unmard05 exsum05;
sbmi on ex3_age female black nocoll05 unmard05 exsum05;
qbmi on ex3_age female black nocoll05 unmard05 exsum05;
idep on ex3_age female black nocoll05 unmard05 exsum05;
sdep on ex3_age female black nocoll05 unmard05 exsum05;
qdep on ex3_age female black nocoll05 unmard05 exsum05;
***sbmi on idep;
***sdep on ibmi;
idep with ibmi;
idep with sdep;
ibmi with sbmi;
sdep with sbmi;

Should I change the astericked lines to the following:

sbmi qbmi on idep;
sdep qdep on ibmi;
 Linda K. Muthen posted on Tuesday, August 19, 2008 - 3:13 pm
You should fit each growth process separately as a first step. They do not need to have the same shape. Then you can fit a model with both processes. I would not add covariates until after that. You do not want to have ON and WITH statements for the same variables.
 dan berry posted on Wednesday, January 07, 2009 - 1:57 pm
Dear Drs. Muthen,
In a parallel process model with covariates and a distal outcome Id like to leave the intercept/mean of the outcome variable (WJG5) in the scale of the of the scaling indicator. But when I fit the model (v4.2)tech4 gives a mean of -8; the indicator scales are in the 100s. When I use [wj5*] it says the model may not be identified. Im sure Im missing something basic, but not sure what.


Analysis:type= missing meanstructure;
Estimator= mlr;
MODEL: wjg5 BY wjbmwcg5 wjbrwcg5;

att54 slatt | lattg54@0 lgmattk@1 lgmatt1@2 lgmatt3@4 lattg4@5;
ktcon sltcon | cnfl_tkf@0 cnfl_t1s@1 cnfl_t2s@2 cnfl_tg3@3 cnfl_tg4@4;
ktcon ON att54;
att54 WITH totagr54;
slatt on ktcon;
sltcon on slatt;
slatt on totagr54;
WJg5 ON ktcon;
wjg5 ON att54;
WJg5 ON SLatt;

OUTPUT: SAMPSTAT modindices;
tech4 tech1;
 Linda K. Muthen posted on Thursday, January 08, 2009 - 8:07 am
You need to add [wjbmwcg5@0] to the MODEL command for identification purposes.
 dan berry posted on Thursday, January 08, 2009 - 12:25 pm
Thank you for your quick response. But could it be something else? When I add [wjbmwcg5@0] along with [wjg5*], the model never converges (even when I set it for huge numbers of iterations). If I cannot get WJG5 in the metric of the scaling indicator, Im unclear about what the -8 tech4 mean for WJG5 represents. Is it that the default mean is zero, and that -8 is the conditional mean based on the average values of all the predictors in which the WJG5 is regressed upon?
 Linda K. Muthen posted on Thursday, January 08, 2009 - 1:39 pm
You will need to send your input, data, output, and license number to for further help.
 Tim Stump posted on Tuesday, November 03, 2009 - 12:09 pm
I have two piecewise parallel processes characterized by the following statements:

i1 s1 | ueia0@0 ueia2@2 ueia4@2 ueia8@2 ueia12@2;
i1 s2 | ueia0@0 ueia2@0 ueia4@2 ueia8@6 ueia12@10;

i2 s3 | seia0@0 seia2@2 seia4@2 seia8@2 seia12@2;
i2 s4 | seia0@0 seia2@0 seia4@2 seia8@6 seia12@10;

I'd like to determine if the two slopes of one process (s1 and s3) are the same as the slopes from the other process (s2 and s4), i.e., s1-s3=0 and s2-s4=0. What syntax would I add to carry out these tests?
 Linda K. Muthen posted on Tuesday, November 03, 2009 - 12:36 pm
You can do this using difference testing of nested models where one model allows the parameters to be free and the other constrains them to be equal which is described in Chapter 13 of the user's guide or you can use MODEL TEST. See the user's guide for more information.
 Tim Stump posted on Wednesday, November 04, 2009 - 5:10 am
Linda, thanks for your reply to my post on november 3. I have a followup question. How do I use the parameter labels (eg, p1, p2, etc from chapter 16 of manual) and MODEL TEST statement when specifying the growth model with the "|" symbol?
 Linda K. Muthen posted on Wednesday, November 04, 2009 - 6:04 am
When a bar symbol is involved, it is the means and variances of the random effects that are the parameters in the model. So you would label those parameters. If I am not understanding the question, you need to send the full output and your license number to so I can see the exact situation.
 Wayne deRuiter posted on Monday, July 12, 2010 - 7:55 pm

How can I test for the assumptions of a parallel process model?

 Linda K. Muthen posted on Tuesday, July 13, 2010 - 8:01 am
Are you asking how to test the fit of a parallel process model? If so, fit is assessed as for any model.
 Wayne deRuiter posted on Tuesday, July 13, 2010 - 7:43 pm
Sorry, I was thinking about a statistic for multivariate normality for the outcomes, random effects, and residuals.

Also, I would like to test for the independence between random effects and residuals as well as the independence of the residuals.

Please correct me if I am wrong, are these not the assumptions for growth curve models.
 Linda K. Muthen posted on Wednesday, July 14, 2010 - 3:38 pm
We don't have any specific tests for normality of the outcomes. The MLR estimator is robust to non-normality of the outcomes. I am not aware of tests of multivariate normality of random effects and residuals.

Regarding independence, perhaps you are referring to Hausman-type tests of uncorrelatedness between residuals and exogenous variables. I am not aware of such tests when the exogenous variable is a random effect.

Uncorrelatedness of residuals can be explored using WITH statements when those parameters are identified.
 Mine Yildirim posted on Thursday, February 17, 2011 - 5:07 am
Dear Linda,
I have a related question about your post on Tuesday, August 19, 2008 - 3:13 pm. It is still not clear to me how to add or if there is need to add quadratic term into parallel process LGM for mediation test. Without quadratic term my model is;
ib sb| bmi0@0 bmi1@1 bmi2@1.5 bmi3@2.5;
is ss| ssb0@0 ssb1@1 ssb2@1.5 ssb3@2.5;
sb ON is ss;
ss ON ib;
ib sb ON group;
is ss ON group;

When I add a quadratic term to one of these indivudial growth models (for example to BMI model), should i regress quadratic growth factor on linear growth factor of ssb to find a mediating effect? Like this;
ib sb qb| bmi0@0 bmi1@1 bmi2@1.5 bmi3@2.5;
is ss| ssb0@0 ssb1@1 ssb2@1.5 ssb3@2.5;
qb ON is ss;
ss ON ib;
ib sb qb ON group;
is ss ON group;

 Linda K. Muthen posted on Thursday, February 17, 2011 - 10:45 am
You could do this.
 Mine Yildirim posted on Thursday, April 07, 2011 - 8:25 am
Thank you for your answer. I have a following question on the same topic.

I have a quadratic growth both for the mediator and the outcome variable. I have a difficulty to calculate the mediated effect in this situation. Since I have a quadratic growth in both variable, should I still use the classic method as;

The action theory test (a-path) Group -> Slope1
The conceptual theory test (b-path) - Slope 1 -> Slope 2

Should I include the quadratic slope into the calculation of mediating effect?
For instance as;

a-path- Group -> Slope 1
b-path- Slope 1 -> Quadratic 2

OR something like combination of linear and quadratic slope factors;

a-path- Group -> Slope 1 + Group -> Slope 2
b- path - Slope 1 -> Slope 2 + Quadratic 1 -> Quadratic 2

OR something else?

Id really appreciate any help. Thanks beforehand.
 Mine Yildirim posted on Monday, May 02, 2011 - 12:57 am
Dear Linda and Bengt Muthen,

I would appreciate it a lot if you could give your opinion and guidance on the question that I asked at the previous message at April 7. Thank you in advance.

 Bengt O. Muthen posted on Monday, May 02, 2011 - 8:50 am
I don't think there is any concensus on how to approach mediation in a parallel growth setting. And, with a quadratic function you have the added difficulty of not being able to separate effects of linear and quadratic growth.

One possibility is to focus on mediation at a specific time point and only consider mediation via the intercept growth factor defined for that time point. That is, setting the time score at zero for that time point for both processes.
 Luna Munoz Centifanti posted on Monday, September 03, 2012 - 8:56 am
Dear Linda and Bengt Muthen,
I have a question about doing a parallel process growth model. I see in Topic 3 (slide 181) that the slope of one of the growth factors is restricted to the first two time points with the last 2 time points free. However, the manual shows both processes estimated similarly (for example, i1 s1 | y11@0 y12@1 y13@2 y14@3). Does this difference have anything to do with the inclusion of covariates?
I am running a model where I've run the growth models separately, then together, then added then covariates, and finally included an interaction term between i2 and one of the covariates (following slide 165). I would like to make sure my model is identified properly.
Thanks very much, as always!
 Linda K. Muthen posted on Monday, September 03, 2012 - 9:16 am
No, this has nothing to do with the inclusion of covariates. These are two different growth models. The first has two free time scores. The second is a linear growth model.
 Luna Munoz Centifanti posted on Thursday, September 06, 2012 - 2:22 am
Thanks, Linda, for your quick response.
I was wondering about the example models for parallel process latent growth models in the Topic 3 and in the guide, since one shows covariates and the other does not. I would like to test if i1 predicts s2 and i2 on s1, but controlling for possible covariates. Here is the model statement I have:
i1 s1| x11@0 x12@1 x13@2;
i2 s2| x21@0 x22@1 x23@2;
i1-s2 on gender ses;
s2 on i1;
s1 on i2;

Does that seem a correct merging of Topic 3 and the guide? Is it correct not to have the intercepts correlate or is that dependent on theory?
 Linda K. Muthen posted on Thursday, September 06, 2012 - 10:07 am
I would think the intercept growth factors would typically be correlated but your theory would be the last word.
 xiaoyu bi posted on Thursday, December 26, 2013 - 9:46 am
Hi, Linda,
I fitted a parallel process LGM for arguments and somatic symptoms. Both the slopes are negative, indicating both arguments and somatic symptoms decrease over time.
The intercept between arguments is significantly and negatively related to slope of somatic symptoms. Does that mean people with higher initial level of argument have a greater decrease in somatic symptoms over time?
Thank you!
 Bengt O. Muthen posted on Friday, December 27, 2013 - 11:52 am
 christine meng posted on Monday, March 24, 2014 - 11:18 pm
Hi Drs. Muthen,

I modeled joint book reading and receptive vocabulary using the parallel process model. The intercepts of the slopes of joint book reading and receptive vocabulary are negative. The slopes of joint book reading and receptive vocabulary are significantly and negatively related. How should I interpret the significant association between the slopes?
 Linda K. Muthen posted on Tuesday, March 25, 2014 - 10:06 am
Please send the output and your license number to
 christine meng posted on Monday, March 31, 2014 - 8:52 am
Hi Dr. Muthen,

I am sorry that I don't have the license number, as I am using the university license. But I do have another question. Are there ways to test the moderating effect on the association between the two slopes? For instance, child gender moderates the association between the two slopes.

 Bengt O. Muthen posted on Monday, March 31, 2014 - 6:34 pm
You can do that in a multiple-group analysis based on child gender. Then you can test equality of that association.
Back to top
Add Your Message Here
Username: Posting Information:
This is a private posting area. Only registered users and moderators may post messages here.
Options: Enable HTML code in message
Automatically activate URLs in message