Two parallel processes growth model PreviousNext
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Message/Author
 Anonymous posted on Wednesday, June 09, 2004 - 7:10 pm
Hello.
I have 2 variables (y1 & y2) measured at 4 time points. At each time point, multiple regression analyses show that the relationship between y1 & y2 is an inverted U shape, i.e., y1 is highest when y2 is at a moderate level, y1 is low when y2 is either low or high.

Growth models show that y1 did not change over time, i.e., there is significant variation among our subjects in the latent intercept factor (i1) but no variation in the latent linear slope factor. On the other hand, both the intercept (i2) and linear slope factors (s2) of y2 have significant variation among subjects. There is no evidence of a quadratic growth factor for either y1 or y2.

Growth modelling results also show that there is no correlation between i1 and i2, and between i1 and s2. This does not contradict my cross-sectional analyses. However, is there any way that I can model the data longitudinally to reflect the cross-sectional relationship?

Thanks a lot for your time.
 Linda K. Muthen posted on Thursday, June 10, 2004 - 8:14 am
You can try to create an interaction variable:

i2sq | i2 XWITH i2;

and then regress

i1 ON i2 i2sq;

I think this will capture the nonlinear relationship between i1 and i2.
 Anonymous posted on Thursday, June 10, 2004 - 8:52 am
Hello Linda:

My growth models has random times of observation.
I got the following error message when I did the above -
*** ERROR in Model command
The model specified with the following set of MODEL statements is not supported for TYPE = RANDOM:
I2 | RESILI1 AT AGE1
I1SQ | I1 XWITH I1
I2 ON I1SQ

My program is listed below -
i1 s1| contol1-contol4 at age1-age4;
i1sq | i1 XWITH i1;
i2| resili1-resili4 at age1-age4;
resili1 resili2 resili3 resili4 (1);
[contol1-contol4@0];
[i1* s1* i2*];
i1 with s1;
i2 ON i1 i1sq;

Any advice is appreciated. Thanks a lot.
 Linda K. Muthen posted on Thursday, June 10, 2004 - 9:04 am
Please send the full output to support@statmodel.com.
 Thomas Olino posted on Wednesday, March 23, 2005 - 7:18 am
I ran a parallel growth curve model that used all categorical indicators. The latent factors were not specified to be categorical factors, so the analysis continued using numerical integration on the 4 dimensions.

After the solution was found, I noticed an input error in my model. After I corrected the error, I tried to re-run the analysis. However, the analysis would not and I received an error message: "THERE IS NOT ENOUGH MEMORY SPACE TO RUN THE PROGRAM ON THE CURRENT INPUT FILE...etc."

First, what would the suggested pc-hardware specifications be for the analysis? Second, any idea how the analysis would run once, but not again? Third, what would the pros and cons be (or reference) for running the analysis using Monte Carlo as the integration method?
 bmuthen posted on Wednesday, March 23, 2005 - 7:48 am
The combination of many dimensions of integration and a large sample can cause the MEMORY SPACE message - although it sounds like you haven't installed the latest update, version 3.12, which I think gives a more explicit message.

I can't say why it worked the first time and not thereafter - would have to have exact details on the two runs.

To bring down the computational demands, you can try working with fewer integration points to give an approximate ML solution, e.g.

integration = 7;

- or, as you say, use MONTECARLO integration, which might be quicker still. They should give similar results.
 Kersi Antia posted on Wednesday, August 17, 2005 - 11:39 am
Hi Bengt and Linda:

I was wondering whether it is possible to specify and estimate a Growth Mixture Model (GMM) for a parallel process model. I have data on 166 salespeoples' monthly sales performance and corresponding usage of a web-based training system and have been able to specify a parallel process growth model thus far (sales performance and website usaage as the growth processes modeled). I was wondering whether it is mathematically possible to have finite classes identified wrt performance and usage with distinct though overlapping covariates, and whether MPLUS can handle this.

Thanks for your time,
Kersi
 Linda K. Muthen posted on Wednesday, August 17, 2005 - 2:12 pm
Yes, Mplus can do this.
 fati posted on Friday, November 25, 2005 - 1:06 pm
HI Bengt and Linda;
I ran a growth model for two parallel processes for continuous outcome, I have doing the same thing as example 6.13 in M-plus user's guide.
1-if I want to detect the effect of first processes to the second process (cross-lagged analysis), how can I add to my program the regression between the first and the second processe?
2-it is sufficient to make only the regression between the intercepts and slope in order to mesure this effect?
3- it is logic to make the regression between intercept and slope for the same processes?
4-can I do the same analysis for the dichotomuos outcome?
thank you for your help in advance
 bmuthen posted on Saturday, November 26, 2005 - 6:32 am
1. Do you want the second process to be influenced by the first, or each influencing the other? In the second case you say
s2 on i1; s1 on i2;

2. That depends on your application.

3. Yes, this can make sense.

4. Yes
 FATI posted on Monday, November 28, 2005 - 6:25 am
thank you for your response;

I want just to be sure that I have understand. in basic analysis (FOR EXAMPLE, LINEAR REGRESSION ANALYSIS) if we want to see the effect the processe to the another, I make the regression between two variables (PROCESSES) (for example, detresse and satisfaction), then if I make the regression between slopes and intercepts in Grwowth model analysis for two parallel processes, that is replace the regression between variables in linear regression , if yes, can you tell me the reference to this in order to cite this in my article?
 Linda K. Muthen posted on Monday, November 28, 2005 - 4:17 pm
Growth factors are continuous latent variables. If you regress one on the other, it is a simple linear regression. I don't know of any special citation.
 fati posted on Tuesday, November 29, 2005 - 5:43 am
Excuse me I think that my message wasn't clear enough, when we do two processes growth model, we do a regression between a slope and intercept and not between factors directly, my question is that is sufficient to mesure the effect between the factors (latent variables) or I must add a regression between factors in my model, for example in user's guide example 6.13 (page 89) how can I mesure the effects between the first processe (score of the Y1j) et the second ( score of the Y2j), do I add a regression between the Yij or it is sufficient to use the regression between intercept and slope, I think that is a basic question, I don't have enough information to the growth modeling, thanks you
 Linda K. Muthen posted on Tuesday, November 29, 2005 - 6:47 am
If you want to see the relationship between y1 and y2, it is common to add a residual covariance parameter. y1 WITH y2; Whether you want to do this depends on your data and your model.
 FATI posted on Tuesday, November 29, 2005 - 7:42 am
THANKS YOU FOR YOUR HELP
 Maggie L Liu posted on Thursday, July 10, 2008 - 8:19 pm
Hi,

I am conducting a mediation analysis in which both the mediator and outcome are repeatedly measured (5 waves). I suppose the parallel growth model is what I'm trying to do. The outcome variable is a categorical measure with linear growth over time. I used a multiple indicator growth model for the mediator measure, because that is a multivariate measure with missing values. To assess the mediation effect, we evaluate the path from intervention (0/1 dummy) to the linear trend of the mediator, to the linear trend of the outcome. From analysis results, the mediation path is significant. My question is, I'm not able to obtain fit statistics, such as chi-square, CFI, TLI, and RMSEA. How do I get these values in Mplus?

Thank you very much for your advice!

Best,

Maggie
 Linda K. Muthen posted on Friday, July 11, 2008 - 10:19 am
If you are not obtaining fit statistics, I assume you are using maximum likelihood estimation. You could use WLSMV and obtain fit statistics.
 Bruce A. Cooper posted on Thursday, April 16, 2009 - 5:17 pm
Hi Linda &/or Bengt!

I'm having trouble with estimation, probably due to a small sample (as usual!), but perhaps due to my own ignorance. I have 2 symptoms each measured 7x (e.g., sleep disturbance & morning fatigue), and I want to estimate a parallel process growth model for them with MLR along the lines of ex6.13. The simple growth model for one indicates a significant linear and quadratic slope, the other just a linear slope. When I try to include the quadratic slope in the PP model, I get "NO CONVERGENCE. SERIOUS PROBLEMS IN ITERATIONS" and "THE MODEL MAY NOT BE IDENTIFIED" sorts of warnings. I've tried setting covariances between the uninteresting or nonsignificant latent variables to zero (e.g., i1 with s1, i1 with q1, s1 with q1) with no success. So, here I am "hat in hand" looking for suggestions!

Note: really small sample: 60 paired observations of patients and caregivers; I count 34 or so free parameters without fixing any. Recognizing that this is REALLY a small sample for so many estimates, I'm only looking for some fixed effects to explore/describe these two processes, so I'd be happy fixing some random effects to zero (like random slopes) if it would still allow estimation of the crossed regressions (e.g., caregiver morning fatigue on patient sleep disturbance).

Thanks!
Bruce
 Linda K. Muthen posted on Thursday, April 16, 2009 - 6:03 pm
You need to send the problem along with your license number to support@statmodel.com. It is not possible to know why you are having problems without more information.
 Nina L. Alesci posted on Thursday, March 18, 2010 - 8:20 pm
Is it possible to do a twolevel parallel process model? I tried it with a community level variable (n=60) and an individual level variable (n=3636) where individuals have been sampled from those communities. Here is my code:
Analysis:
TYPE = MISSING h1 TWOLEVEL;
ALGORITHM=INTEGRATION;
MITERATIONS=10000;

Model:
%WITHIN%
ienfw senfw | enfsrv1@0 enfsrv3@1 enfsrv5@2;
enfsrv1 (1); enfsrv3 (1); enfsrv5 (1);
%BETWEEN%
ienfb senfb | enfsrv1@0 enfsrv3@1 enfsrv5@2;
enfsrv1@0; enfsrv3@0; enfsrv5@0;

istgw sstgw | stage1@0 stage2@.5 stage3@1 stage4@1.5 stage5@2
stage6@2.5;


The model produced this warning. THE MODEL ESTIMATION DID NOT TERMINATE NORMALLY DUE TO A NON-ZERO
DERIVATIVE OF THE OBSERVED-DATA LOGLIKELIHOOD.
Do you know what I am doing wrong?
 Linda K. Muthen posted on Friday, March 19, 2010 - 7:10 am
This is possible. Did you put the state variables on the BETWEEN list? If not, try that. If that is not it, please send your input, data, output, and license number to support@statmodel.com.
 sigrune posted on Tuesday, May 18, 2010 - 5:10 am
Dear Linda/Bengt,

I ran a parallel growth curve model for two latent variables of authoritative teaching based on four items each measured at three occasions. N= 900. I have missing data. I used a step by step procedure to ensure that the CFA for each variable at each occasion and for the growth models separately yielded good fit. Also the parallel model has good fit. When adding covariates this works well for observed variables (e.g. seniority - 4 categories and type of intervention -2 categories). However, when adding latent variables as covariates there is an error message saying that the model may not be identified:

The standard errors of the model parameter estimates could not be computed. The model may not be identified. Check your model. Problem involving parameter X.

When I include one latent covariate (inovat) based on 7 item, parameter x is the regression of s(lope)2 on s(slope)1. However, if I include the same latent covariate as well as one or two observed variables (intervention and gender), the problem appears in the regression of inovat on s1.

Do you have any suggestions on what the problem might be and how I can solve it?


Best,
Sigrun
 Linda K. Muthen posted on Tuesday, May 18, 2010 - 9:06 am
We would need to see the output or outputs and your license number at support@statmodel.com to be able to comment on this.
 Mine Yildirim posted on Thursday, February 03, 2011 - 6:58 am
Hello,
I have two questions;
1) I want to test mediating effect with parallel process LGM, but i have a clustered data. I have 3-levels, namely 1)time, 2)children, 3)school.

I could not find any previous example, article or any message about multilevel parallel process LGM for mediation analysis (in the Mplus discussion topics also). Did anybody apply this kind of analysis before? I am not sure if it is possible because even without parallel process i had some problems with multilevel LGM part. Mplus did not run without giving any error message for some models.

Is it possible to do multilevel parallel process LGM for mediation and moderation analysis with Mplus?

I appreciate it a lot if you could give some references on that topic?

Thanks beforehand for your response.
Regards.
 Bengt O. Muthen posted on Thursday, February 03, 2011 - 5:47 pm
This should be possible in Mplus. 3-level analysis (2 levels in Mplus) has no extra problem with parallel processes. You should explore each process separately to see if you have enough school variation (and enough schools) to capture the parameters of the school level. I don't know any references. We have 3-level growth model examples in our growth modeling teaching in Topic 8 (coming up this March).
 Nicholas Bishop posted on Friday, February 11, 2011 - 10:12 am
Hello,
I am running a parallel-process model on cognitive and physical health outcomes, and have a question regarding the set-up of interactive effects between latent and observed variables. The observed variable is an interactive term between two categorical variables. When setting up the interaction between the latent variable and the observed interaction, do I need to include the main effects for the observed interaction?

irec - intercept memory
sadl - slope functional limitations

(a)
irecblk | irec XWITH black;
irecm8ed | irec XWITH m8ed;
irecm8e | irec XWITH blkm8e;
sadl ON irecm8e;

Or would I be able to simply include the interaction term?

(b)
irecm8e | irec XWITH blkm8e;
sadl ON irecm8e;

I cannot get (a) to converge but (b) converges without a problem. Thanks!

Nicholas
 Linda K. Muthen posted on Sunday, February 13, 2011 - 2:57 pm
XWITH is used to specify interactions between latent variables and between an observed and a latent variable. Interactions between observed variables are specified using the DEFINE command.

You ask about including main effects but show two situations one of which has three interactions and one that has one interaction. I see no main effects.
 Nicholas Bishop posted on Monday, February 14, 2011 - 9:14 am
Hi Linda,
Below is the entire model statement that I am using (condensed to fit here), sorry I did not clarify earlier:

Model:

irec srec | totrec98@0 totrec00@.2 totrec02@.4 totrec04@.6 totrec06@.8 totrec08@1;

iadl sadl | r4adla@0 r5adla@.2 r6adla@.4 r7adla@.6 r8adla@.8 r9adla@1;

srec ON iadl;
iadl sadl ON irec;

!irec with srec;
iadl with sadl;

irec srec iadl sadl ON
R4AGEY_B black
FEMALE R4MAR
RBLUCL RHMKR R4NOCC
lthsdeg gthsdeg
ainc1 ainc2 ainc3 aass1 aass2 aass3
m8ed f8ed R4CHOK FWHTCL
;

irecm8e | irec XWITH m8ed;
sadl ON irecm8e;
 Bengt O. Muthen posted on Monday, February 14, 2011 - 6:32 pm
This looks fine. oYou are regressing sadl on an interaction and also on the corresponding two main effects, which is customary.
 Nicholas Bishop posted on Thursday, February 17, 2011 - 9:19 am
Thank you for the input. I have two questions related to the proposed parallel process model:
1) Can you provide a source that explains how to probe interactions between a continuous latent variable and a continuos observed variable?
2) I would like to extend this parallel process model to account for NMAR data on the outcome variables. I have tried to implement Diggle-Kenward and Wu-Carroll selection adjustment and pattern-mixture adjustment to the I and S of both outcomes, but the models would not converge. I am able to successfully implement these NMAR models when only adjusting one of the outcome variables.

What potential biases would be introduced to a parallel-process model when only controlling for data NMAR on one of the outcome trajectories?

Nicholas
 Bengt O. Muthen posted on Thursday, February 17, 2011 - 5:22 pm
1) See the handout for Topic 3, slides 164-171.

2) If the missingness is the same for the 2 processes, then NMAR should be done for both. But perhaps NMAR for one is better than MAR - but hard to argue for in a pub. There shouldn't be a problem doing NMAR for 2 processes.
 Nicholas Bishop posted on Monday, March 07, 2011 - 6:26 am
Regarding 1) and in reference to the relevant slides for Topic 3, in parallel-process output, I am not given the unit variances for the initial status of the outcome variables, only the mean and standard errors for the estimated intercepts. Can these be used like the unit variances to standardize and aid in interpreting the interaction?

Also, in the handout and video you center the intercept for the single growth process at zero using the statement [i@0 s];. In the parallel-process framework, would I want to do this for the intercept of each process? In addition, what does including the slope (s) after i@0 invoke?

Thanks for your time and helpful videos/handouts.
 Bengt O. Muthen posted on Monday, March 07, 2011 - 5:48 pm
Regarding your second paragraph, yes, you would do this for the processes that you want to have an interaction with [i] for. You ask what [s] means inside [i@0 s] - it
means that the mean of s is freely estimated.

Regarding your first paragraph, I don't follow you. The unit variance for the initial status can be obtained from the standardized solution. You cannot standardize by using standard errors.
 Nicholas Bishop posted on Monday, March 07, 2011 - 7:07 pm
When I use the ANALYSIS option TYPE=RANDOM in conjunction with the XWITH statement and request the standardized solution, I am given the following warning:

STANDARDIZED (STD, STDY, STDYX) options are not available for TYPE=RANDOM.
Request for STANDARDIZED (STD, STDY, STDYX) is ignored.

Is there a way to specify the XWITH interaction without using the TYPE=RANDOM option or another way to receive the standardized solution?
 Bengt O. Muthen posted on Tuesday, March 08, 2011 - 8:42 am
This is currently not available. But you can do it by hand - see the FAQ (right margin of the home page, bottom) with title:

"The variance of a dependent variable as a function of latent variables that have an interaction is discussed in Mooijaart and Satorra"
 Nicholas Bishop posted on Wednesday, June 08, 2011 - 10:59 am
Hello again,
I am hoping to clarify the restrictions I am using on the parameters across dropout patterns in my pattern mixture parallel process model. I am using example 4 from Section 5.1.1 from the 2010 manuscript on non-ignorable dropout as a reference.

When specifying classes = c(1); and type = mixture; and %overall% in the model statement, are the means for the un-identified parameters held equal to those of the pattern corresponding to dropping out one time point later for all un-identified parameters (for both trajectories in the parallel process model)?

I would be happy to post more of my syntax if that would help. Thanks.

Nick Bishop
 Bengt O. Muthen posted on Wednesday, June 08, 2011 - 6:05 pm
Yes, each process would have to have its equalities of this kind.
 Nicholas Bishop posted on Wednesday, June 08, 2011 - 7:31 pm
Thanks Bengt.

Is the default assumption for the pattern mixture model provided in example 5.1.1 that the means for the un-identified parameters are held equal to the pattern corresponding to dropping out one time point later? In the Enders piece that accompanies your work on NMAR in Psychological Methods, he places a number of identifying restrictions on the pattern mixture model to produce three latent sub-groups based on patterns of missing data.

I want to be clear that without these identifying restrictions, the default in mplus is that the means of the un-identified parameters are held equal to the pattern corresponding to dropout at t + 1.

Sorry if this is redundant.
 Bengt O. Muthen posted on Thursday, June 09, 2011 - 7:36 am
No, it is not the default. The analyst has to apply the restrictions he/she deems appropriate since there are many possible choices as seen in the Demitras et al article I refer to. I just happen to like the approach I used in my Psych Meth article.
 Nicholas Bishop posted on Thursday, June 09, 2011 - 10:34 am
Using the Psychological Methods article and corresponding syntax (5.1.1), please tell me if this is correct:

The identifying restrictions constrain the effect of the slope on d1-d2 to be equal, and the effect of d1-d3 on q to be equal. For the slope constraint, the mean of patterns corresponding to d1-d2 are held equal to the mean of the pattern corresponding to dropout at d3. For the quadratic constraint, the mean of patterns corresponding to d1-d3 are held equal to the mean of the pattern corresponding to dropout at d4.


i-q on d4-d5;
i on d1 d2 d3;
s on d3;
s on d1 (1);
s on d2 (1);
q on d1 (2);
q on d2 (2);
q on d3 (2);

Thanks for your help.
 Linda K. Muthen posted on Thursday, June 09, 2011 - 2:08 pm
Compare your input to the pattern mixture input that is on the website with the Psych Methods article.
 Dana Garbarski posted on Wednesday, July 06, 2011 - 6:09 am
Hello,
I am running a parallel process latent growth model with two dichotomous sets of repeated measures and wlsmv estimation. The univariate latent growth models show that each of these curves have intercepts, linear slopes, and quadratic slopes (one curve has no variance in the quadratic). I was hoping to be able to regress the linear slope of one curve on the other to say something about how the two processes are related over time (I expect that growth in one of the processes is driving growth in the other), but I'm not quite sure how the quadratic slopes should be modeled to fit into this interpretation, especially since one of the quadratic slopes has zero variance and is thus not correlated with the other intercepts and slopes.
Any thoughts or advice would be most appreciated!
 Bengt O. Muthen posted on Wednesday, July 06, 2011 - 12:59 pm
It is hard to work with a model that's predicting from a linear and a quadratic slope because those 2 growth factors are not separately interpretable. In contrast, I can imagine that the intercept of one process influences the 2 slope factors of the other.

It is ok to have predictors of a growth factor with zero residual variance - you may still find a signficant predictor. This happens quite frequently.
 Caroline Vancraeyveldt posted on Wednesday, December 07, 2011 - 5:19 am
Dear Linda and Bengt,

I am currently struggling with the fit of a parallel proces growth model. Both seperate growth models have a good fit, but when we put them together, the fit of the "global" model is not good. Do you have suggestions to solve this problem, or the reasons for the occurence of this problem? Is it possible that the two outcomes are too highly correlated? (correlation of the two slopes was .70).

Thank you in advance!
Kind regards,
Caroline
 Linda K. Muthen posted on Wednesday, December 07, 2011 - 11:51 am
It may be that you need residual covariances across processes. I would ask for modification indices to explore this.
 Caroline Vancraeyveldt posted on Thursday, December 08, 2011 - 12:26 am
Thank you for your response.
We have tried to improve the model fit by following the modification indices. However, the correlation between slopes remains extremely high (.90). Can we report this, or is this comparable to multicollinearity?

Kind regards,
Caroline
 Linda K. Muthen posted on Thursday, December 08, 2011 - 9:47 am
Multicollinearity refers to high correlations among predictors not outcomes.
 Paraskevas Petrou posted on Friday, April 19, 2013 - 1:59 am
Dear Mplus users,

I'm working on a growth model of moderated mediation with a 3-wave dataset and I have some questions:

1. My 2 interaction terms are between the intercept of one variable and the slope of another one. Should I specify this as:
Interaction | s1 XWITH i6 ?

2. Many of my slopes should not follow necessarily an 1-2-3 fashion, but most likely 1-2-1 or 1-2-2. If for 1-2-3 I use 0-1-2 as factor loadings, what should the factor loadings be for 1-2-1 and 1-2-2?

3. Some of my variables should not have slope but only intercept. Does it simply suffice to enter s@0 in the model?

Any help would be very much appreciated, thanx in advance!
Paris
 Linda K. Muthen posted on Friday, April 19, 2013 - 10:08 am
1. Yes.
2. For 1 2 2, use 0 1 1. For 1 2 1, use 0 1 *.
3. Yes.
 Paraskevas Petrou posted on Friday, April 26, 2013 - 6:17 am
Thank you Linda.

A couple of more questions regarding your advice:

1. Using 0 1 * as factor loadings results in the following error:
Error in parsing line:
"X_T1@0 X_T2@1 X_T3@*"

2. If next to i and s I also specify a "q" (quadratic term), don't I actually test the above question?

3. One term of my interactions is a variable measured three times which I assume to be stable over time. There are two ways to model that: i) Specify its i and s and constrain s to be equal to 0 or ii) Make a latent factor with the three measurments loading onto the factor. What method is the most preferable?

Best,
Paris
 Bengt O. Muthen posted on Friday, April 26, 2013 - 2:13 pm
1. Don't say x_T3@*. Instead say x_T3*.

2. A general quadratic model won't be identified with only 3 time points, but needs restrictions like zero var-covs among growth factors.

3. You can use i | ...., which is the same as a factor with unit fixed loadings. Intercept only growth models are perfectly fine.
 Ginnie posted on Saturday, April 27, 2013 - 4:35 pm
Hi Dr. Muthen,

I would like to run a parallel processes LGM, and I have some questions:

1: do I have to test each growth process separately before doing the parallel processes?

2: if one of the individual growth model has no sig. mean slope but has sign. variance, can it still be used in the parallel processes model?

Thanks!
Ginnie
 Linda K. Muthen posted on Sunday, April 28, 2013 - 10:01 am
It is a good idea to test each model separately before estimating a model with both processes. If the mean of the slope growth factor is not signficant, this means that there is no development over time for the process.
 Ginnie posted on Thursday, May 02, 2013 - 3:17 pm
Hi Dr. Muthen,

I have a follow-up question on the parallel processes LGCM. One of the model does not show a sig. mean growth factor, and the other one does. When analyzing them in the parallel process, both slopes are sig. correlated as well as their intercepts. I am just wondering whether the results are valid for interpretation.

Thanks.
Ginnie
 Bengt O. Muthen posted on Thursday, May 02, 2013 - 9:05 pm
I think that is ok.
 Dayuma Vargas posted on Sunday, December 22, 2013 - 9:35 pm
Hello,

I ran a parallel process LGM with both processes having positive linear slope, a negative quadratic slope, and a positive cubic term. I found that the linear terms covary significantly and positively with one another, which means that individuals who show faster gains over time in x show faster gains over time in y as well. The quadratic terms covary significantly and negatively with one another, but I am having some difficulty trying to interpret this. Would you be able to advice me on this? Thank you.
 Bengt O. Muthen posted on Monday, December 23, 2013 - 5:02 pm
It is hard to give a separate interpretation of linear and quadratic growth factors since they interact.
 Neha Goyal posted on Monday, July 14, 2014 - 4:19 pm
I have two outcomes which I seperately modeled. The best unconditional model for one variable was a piecewise model with one linear piece from 0-8 months and second linear piece from 8-60 months. Furthermore, I had to set the slope variance of the first linear piece to zero since I encountered a non-positive definite covariance matrix. The best unconditional model for the second variable was a piecewise model with one linear piece from 0-18 months and the second linear piece from 18-60 months. In other words, they have different growth trajectories. My questions are

1) If the variance of one piece for a variable is zero, does it even make sense to do a parallel process model.
2) How do you do a parallel process model with two variables with different piecewise growth trajectories, if at all?
 Bengt O. Muthen posted on Monday, July 14, 2014 - 5:36 pm
1) You still have variance in the intercept and therefore you want to correlate them over processes.

2) Just correlate all the growth factor that have variances.
 Myung Lee posted on Monday, December 22, 2014 - 3:44 pm
I estimated a parallel processes latent growth curve model with two continuous outcomes measured at three time points. I would like to study the correlation between the two slope factors. In the model, slope loadings on the first factor were specified as linear (0, 1, 2) and those on the second factor were specified as nonlinear using a free time score at T2 (0, *, 1).

Given that the two slope factors have different shapes of the growth trajectories, would I still be able to interpret the slope factor correlation (e.g., r = .70) as: the rate of change in the first outcome from T1 to T3 traveled together with the rate of change in the second outcome from T1 to T3?

Thanks in advance for your inputs.
 Bengt O. Muthen posted on Monday, December 22, 2014 - 5:32 pm
I think so.
 Stefanie Schmidt posted on Thursday, March 19, 2015 - 10:07 am
Hi, when I tried to run a growth model with an ordinal variable (scaled 1-6) using the categorical option I got the following error message: The categorical variables in the growth model do not have the same number of categories. Use the CATEGORICAL option to allow the number of categories to
differ for maximum likelihood estimation. Problem with: I S Do you know what went wrong?

Another question is if it is possible to test 4 potential mediators (2 continous, 2 ordinal) between an intervention (binary, randomized) and a continous outcome variable within one model using parallel process latent growth curve modeling (do you know any example)?

Thank you, Stefanie
 Bengt O. Muthen posted on Thursday, March 19, 2015 - 1:22 pm
Q1. See page 544 in the V7 UG

Q2. Yes, but I can't think of an example right now. Check our papers under Growth Modeling.
 Danielle Kingdon posted on Saturday, August 20, 2016 - 11:02 am
I am working with a parallel process latent growth curve model, with two continuous outcomes measured at four time points. Reviewers have requested that I provide an analysis of how the common variance in the two outcomes is changing across time and how variables predict this common variance. Are you aware of an analysis in Mplus that could answer this question?

Thank you in advance for your response.
 Bengt O. Muthen posted on Saturday, August 20, 2016 - 6:19 pm
You see that in the Estimated Model part of the output when requesting the Residual and standardized options of the Output command.
 Danielle Kingdon posted on Sunday, August 21, 2016 - 11:16 am
Thank you for your reply. From the Estimated Model part of the output I can see the covariances of outcomes 1 and 2 at each time point. I notice that the covariances are decreasing over time. Is there a way to model the predictors of the covariance at each time point? Also, is it possible to model the predictors of the non-shared or error variance in order to understand why the trajectories become increasingly differentiated over time?
 Bengt O. Muthen posted on Sunday, August 21, 2016 - 4:49 pm
It is not easy, but you could try the Constraint=z; option where z is a predictor that can be used in Model Constraint. See the QTL example in the UG.
 Danielle Kingdon posted on Sunday, August 21, 2016 - 5:06 pm
Thank you!
 Liu BAI posted on Friday, November 03, 2017 - 8:02 am
Dear Linda and Bengt,
I'm running a parallel growth curve model to test if two people change together. My model statements are as follows and I have 2 questions.
MODEL:
ii is| x1976@0 x1977@1 x1978@5.44 x1979@5.63;

mi ms| x1966@0 x1967@1.86 x1968@5.29 x1969@5.50;

x1976-x1979 PWITH x1966-x1969;

is with mi;
ms with ii;

First Question: When I test if one's intercept could impact the other one's slope using:
is on mi;
ms on ii;

I get a warning saying that: "THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE DEFINITE."
But if I change it to WITH statement as above, the model fits good with no warning. I wonder why this happened. And, do I have to use ON statement if I'm not interested in the effects from one to the other?

Second Question: I wonder if it is possible to get the covariances between the two slope for each dyad, because I'm interested how this co-change cause influence or influenced by other variables? I tried SAVE = FSCORES, but it only gave the variance for each variable no the covariances.

Thank you!
 Bengt O. Muthen posted on Friday, November 03, 2017 - 5:17 pm
Q1: Send the output with the warning to Support along with your license number.

Q2: You don't have to use ON.

Q3: Use TECH4.
 Liu BAI posted on Friday, November 03, 2017 - 6:16 pm
Thank you for your reply!
About my third question. I think I didn't make it clear enough.

I have two people in the same family and their cortisol variables. I want to see if these two family members have similar cortisol patterns across the day, therefore I used the parallel growth curve model.

Now I wonder if I can get a score measuring the synchrony between the two family members ( (i.e., each family has their only score)? I'm think the the slope covariances for each family. Maybe something like Empirical Bayes covariance. I wonder if it is possible?
 Bengt O. Muthen posted on Saturday, November 04, 2017 - 4:10 pm
In your model, the covariance between the slopes is not a random parameter that varies across units (families). It is assumed to be a regular, fixed parameter. Therefore, you cannot get an estimate of a covariance score for each family.
 Amber posted on Tuesday, July 24, 2018 - 2:58 am
Hello,

I have a quick question about interpreting the output of a parallel process model in MPlus. I have two continuous variables (Variable A and Variable B) both measured at 3 time points over the life course. If I had a model like this:

MODEL:

i_varA s_varA | varA@0 varA@1 varA@2;
i_varB s_varB | varB@0 varB@1 varB@2;

s_varA ON i_varB ;
s_varB ON i_varA ;
i_varA WITH i_varB ;
s_varA WITH s_varB ;

And the output showed a significant negative relationship between the slope of variable A and the slope of variable B, how could I interpret this. Does this mean that as the slope of variable A increases over time, the slope of variable B decreases?

Thanks in advance for your help with this!
 Bengt O. Muthen posted on Tuesday, July 24, 2018 - 2:43 pm
The estimate says that the residual covariance between the A and B slopes is negative. This means that the variation in the slopes that is not explained by the intercepts is negatively correlated (due to left-out influences that end up in the residuals).
 Hannah Schacter posted on Thursday, August 02, 2018 - 9:23 am
I am running a parallel process model w/ 2 continuous variables measured 3 times. I 1st specified a non-directional model, including cov b/t intercepts and slopes. Var A decreases over time (-.072, p<.001), & Var B increases over time (.056, p<.001).

I_varA S_varA|varA1@0 varA2@1 varA3@2;
I_varB S_varB|varB1@0 varB2@1 varB3@2;

I_varA I_varB WITH S_varA S_varB;
I_varA S_varB WITH I_varA S_varB;

I then re-ran the model w/ directional effects.

I_varA S_varA|varA1@0 varA2@1 varA3@2;
I_varB S_varB|varB1@0 varB2@1 varB3@2;

I_varA S_varA pwith I_varB S_varB;

S_varA on I_varA I_varB;
S_varB on I_varB I_varA;

But, now the slope for varA changes sign-it's positive (.121, p=.001), & the positive slope for varB increased (.114, p=.006). Is this b/c the slopes are now conditional effects (i.e., when intercept is 0) given the inclusion of timeXintercept effects? Or is there another reason for the change in size/direction of the slopes?

Thank you.
 Bengt O. Muthen posted on Thursday, August 02, 2018 - 2:08 pm
Perhaps you are looking at the intercept of the slope - which is printed in the regular output - instead of the mean of the slope - which is printed in TECH4.
 Yaqiong Wang posted on Sunday, September 30, 2018 - 5:08 pm
I'm running LGM on two processes: academic achievement and self-efficacy. The linear trajectory fits well with self-efficacy, but there's a drop in academic achievement at the last time point (time 4). I'm thinking of fitting a quadratic term to academic achievement and use the code below:

MODEL:
i1 s1 | T1AM@0 T2AM@1 T3AM@2 T4AM@3;
i2 s2 q | T1GPA@0 T2GPA@1 T3GPA@2 T4GPA@3;
s1 on i2;
s2 on i1;

I have two questions regarding this model:
1. Is this the correct way to add a quadratic term? Do I need to add relationships between q and i1, or other things?
2. If I would like to examine the relationship between the slopes of AM and GPA, should I add a statement such as "s1 on s2"? How would adding a "q" change the relationship between the two linear slopes?

Thank you very much for your help!
 Bengt O. Muthen posted on Sunday, September 30, 2018 - 6:00 pm
The specifications of the growth models are fine. It is probably easiest to simply correlate all 5 factors. It is hard to argue that one slope predicts another.
 Yaqiong Wang posted on Sunday, September 30, 2018 - 7:40 pm
Thank you for you reply! Based on what you just said, if I would like to see if the intercept and slope of process 1 predict the intercept and slope of process 2, can I specify my model like this:

i1 s1 | T1AM@0 T2AM@1 T3AM@2 T4AM@3;
i2 s2 q | T1GPA@0 T2GPA@1 T3GPA@2 T4GPA@3;
s2 on s1;
i2 on i1;
q on s1;

I felt kind of confused about how to build my model after adding "q"
 Bengt O. Muthen posted on Tuesday, October 02, 2018 - 5:49 am
This question is suitable for SEMNET.
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