Growth modeling with parallel processes
Message/Author
 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!
Jungmeen
 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 support@statmodel.com.
 Michelle Little posted on Sunday, November 26, 2006 - 1:44 pm
Hello:
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?

thanks

ML
 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?

Thanks

Michelle
 Linda K. Muthen posted on Monday, November 27, 2006 - 11:13 am
Yes.
 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 I’d 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. I’m sure I’m missing something basic, but not sure what.

Thanks!
-dan

Analysis:type= missing meanstructure;
ITERATIONS = 5000;
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;
[wjg5*];
ktcon ON att54;
att54 WITH totagr54;
slatt on ktcon;
sltcon on slatt;
slatt on totagr54;
WJg5 ON ktcon;
wjg5 ON att54;
WJg5 ON SLatt;
SLTCON WITH ATT54@0;
SLTCON WITH TOTAGR54@0;

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, I’m 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 support@statmodel.com 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 support@statmodel.com so I can see the exact situation.
 Wayne deRuiter posted on Monday, July 12, 2010 - 7:55 pm
Hi

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

Thanks
Wayne
 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;

Thanks.
 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

OR
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?

I’d 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.

Regards
 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!
Luna
 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
Yes.
 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
 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.

Thanks.
 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.
 anonymous Z posted on Thursday, February 26, 2015 - 12:21 pm
I fitted a latent curve model with parallel processes for mothers’ and children’s self-esteem. I have treatment and control condition. In the treatment condition, the slopes of mothers’ and children’s self-esteem were significantly correlated; in contrast, in the control condition, the slopes of mothers’ and children’s self-esteem wasn’t. What does this result mean? Usually Mplus can do multiple group comparison in terms of the mean of intercept and slope. Does it make sense to do a comparison on the variance?
 Bengt O. Muthen posted on Friday, February 27, 2015 - 10:48 am
Yes, any parameter can be affected by treatment, so this could be an interesting finding. As long as you can interpret what it means substantively.
 anonymous Z posted on Friday, February 27, 2015 - 11:04 am
Dr. Muthen,

Thank you very much for your response. My followup question is that if Mplus can do multiple group comparison on variance. I know it can do multiple group comparison on the mean of intercept and slope. Does comparison on variance make sense?

Another question is:
In the parallel modelling, is "S1 WITH S2" the correlation of residual variance or the correlation between S1 and S2?

Thanks,
 Bengt O. Muthen posted on Friday, February 27, 2015 - 11:52 am
Q1. Yes.

Q2. That depends on whether S1 and S2 are regressed on something (if they are on the left-hand side of ON). If yes, then it is a residual covariance, if no then it is a covariance. It is not a correlation unless you consider the StdYX output.
 anonymous Z posted on Friday, February 27, 2015 - 11:59 am
How to do multiple group comparison on variance then? I have no idea how to constrain the variance to be equal. What the syntax should be like?

Thank you very much.
 Bengt O. Muthen posted on Friday, February 27, 2015 - 1:08 pm
Variance equality is like other equalities, using say

y (1);

in each group.

This assumes that y is not regressed on anything because if it is, then

y(1);

refers to the residual variance.
 Amber Fahey posted on Monday, March 27, 2017 - 11:11 am
Hello,

I am new to growth models and very novice in multivariate techniques; however, I am looking to conduct a parallel process model to examine the relationship between a measure of orientation and one of cognition, over time. The population is traumatic brain injury patients in an acute rehabilitation hospital. These two measures are highly correlated. Theoretically, this makes sense. Can I still conduct a parallel process model if they are significantly correlated?
 Bengt O. Muthen posted on Thursday, March 30, 2017 - 9:21 am
This can work unless the correlation approaches 1 in which case they are not really different processes.
 Julia Hammett posted on Thursday, June 08, 2017 - 10:36 am
I am interested in predicting the slope of a factor from the intercept of the same factor – My general question is whether it is okay to do that? If so, are there any particular steps I need to take? Maybe you could refer me to any references with guidelines? If this is not something that can be done generally, what could I do to examine whether baseline levels of a certain construct influence changes in this construct?

When I run a model that does this, I receive the following warning message

WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE
DEFINITE. THIS COULD INDICATE A NEGATIVE VARIANCE/RESIDUAL VARIANCE FOR A
LATENT VARIABLE, A CORRELATION GREATER OR EQUAL TO ONE BETWEEN TWO LATENT
VARIABLES, OR A LINEAR DEPENDENCY AMONG MORE THAN TWO LATENT VARIABLES.

(And the residual variances of my two outcome variables are in fact negative.)

Any hep would be much appreciated!
 Bengt O. Muthen posted on Saturday, June 10, 2017 - 12:13 pm
We talk about predicting the growth slope from the growth intercept in our handouts and videos from our short courses on our website. See also the paper on our website:

Muthén, B., Khoo, S.T., Francis, D. & Kim Boscardin, C. (2003). Analysis of reading skills development from Kindergarten through first grade: An application of growth mixture modeling to sequential processes. Multilevel Modeling: Methodological Advances, Issues, and Applications. S.R. Reise & N. Duan (Eds). Mahaw, NJ: Lawrence Erlbaum Associates, pp.71-89.

As far as the error message goes, send output and license number to Support.
 Julia Hammett posted on Monday, June 12, 2017 - 11:29 am
Thank you very much for the quick response, Dr. Muthen. Would you be willing to share the input specifications for the analyses of the growth mixture model as referenced in the 2003 article on page 16?
Thanks again!
 Bengt O. Muthen posted on Monday, June 12, 2017 - 6:08 pm
These are no longer available, but you have examples of this type of analysis in the User's Guide.
 Yoon Oh posted on Friday, June 16, 2017 - 10:46 am
I wanted to estimate a cross-domain growth model for constructs A and B. Before estimating this, I ran separete latent growth models for each domain, and found that slopes for both constructs are positive. But the slope of construct B turned to be negative for a cross-domain growth model, and this does not substantively make sense to me. My models are as below. why would the slope coefficient change so much? Would you have any suggestion for the model? Any comments would be greatly appreciated. Thanks.

1. Latent Growth Model for A

MODEL:
im sm | w1-w7 AT age1-age7;
im with sm;

2. Latent Growth Model for B

MODEL:
iy sy | m1-m7 AT age1-age7;
iy with sy;

3. Cross-Domain Model for A and B

MODEL:
im sm | w1-w7 AT age1-age7;
iy sy | m1-m7 AT age1-age7;
im with sm;
iy with sy;
im with iy;
sm on iy;
sy on im sm;
 Bengt O. Muthen posted on Friday, June 16, 2017 - 6:04 pm
Perhaps you are confusing intercepts with means in the output - you get an intercept estimate, not a mean estimate when you regress on something. When you regress a slope on something, you find the mean in TECH4.
 Yoon Oh posted on Friday, June 16, 2017 - 8:25 pm
Thank you for your response, Dr. Muthen.

My model uses TSCORE, and thus ANALYSIS: TYPE=RANDOM.

The output has a warning message that "TECH4 option is not available for TYPE=RANDOM with time score variables."

What should I do to get a mean slope?

Your help is greatly appreciated. Thanks so much.
 Julia Hammett posted on Saturday, June 17, 2017 - 8:38 am
Dr. Muthen,

As I am going through the resources you referenced, I am having trouble finding an example that exactly reflects my situation. So, just to clarify: Can one predict the slope of a factor from the intercept of the same factor? This is the code that I am using to estimate my single latent growth model with the intercept predicting the slope:

MODEL:
inter slope | a1@0 a2@1 a3@2 a4@3;
slope ON inter;

Thank you again.
 Claudia Kruzik posted on Tuesday, July 03, 2018 - 10:02 am
Hello, I have longitudinal data on over 6 waves, and I am trying to disentangle the relative influence of time in school and time in summer break using multilevel growth models with individual-varying times of observations. I have been trying to run syntax similar to that you would use for parallel processes, except that instead of modeling two separate outcomes, I am trying to model two separate time schemes on one outcome.

Syntax:

Variable:
tscores = MSCH1 MSCH2 MSCH3 MSCH4 MSCH5 MSCH6 MSUM1 MSUM2 MSUM3 MSUM4 MSUM5 MSUM6;
cluster = S1_ID;

Analysis:
type = twolevel random;
algorithm = integration;
integration = montecarlo;

Model:
%within%

This syntax has not been running. I get the error, "THE MODEL ESTIMATION DID NOT TERMINATE NORMALLY DUE TO A NON-ZERO
DERIVATIVE OF THE OBSERVED-DATA LOGLIKELIHOOD."

Is it possible to run these sorts of models or is there a better modeling strategy to employ in this case?

My apologies if this is in the wrong subsection; I wasn't entirely sure of the most appropriate place to post this.

 Bengt O. Muthen posted on Tuesday, July 03, 2018 - 3:08 pm
I think it seems more natural to have in school as a binary time-varying covariate.
 Claudia Kruzik posted on Thursday, July 05, 2018 - 8:21 am
Thank you for your reply and insight. To follow-up with your suggestion, I am concerned that including a binary time-varying covariate will not work as - in addition to trying to disentangle these two slopes - I am trying to examine mediating factors associated with both times (i.e. the mediating role of school process variables in the relationship between early status and growth over time in school and summer process variables mediating relationship between early status and growth over time in summer).

So after the model section of the syntax posted previously, my syntax looks somethings like this:

i on covs;
s1 on covs SCHPROC1 SCHPROC2;
s2 on covs SUMPROC1 SUMPROC2;

SCHPROC1 on SES;
SCHPROC2 on SES;
SUMPROC1 on SES;
SUMPROC2 on SES;

Now these models have not been running whether I include just one slope term (i.e. look at summer growth alone) or include both slope terms at the same time. As I mentioned before, I keep getting errors that the model estimation did not terminate normally.
 Bengt O. Muthen posted on Thursday, July 05, 2018 - 5:46 pm
 Yaqiong Wang posted on Saturday, October 06, 2018 - 3:50 pm
If my focus of interest is to examine how slope1 and intercept1 of process A interact with slope2 and intercept2 of process B, should I model the relationship between the slope and intercept of the same process (e.g., slope 1 and intercept 1) as a regression or correlation?

I found that when I included (s1 on i1) instead of (s1 with i1), the relationships between i1 & s2 or i2 & s1 are no longer significant.
 Bengt O. Muthen posted on Sunday, October 07, 2018 - 12:38 pm
SEMNET is suitable for this kind of analysis choice question.