Growth model with time-varying variable PreviousNext
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 Dan Powers posted on Thursday, October 26, 2000 - 9:46 am
I am considering a growth model over 4 (or more) time points. An important intervening variable is an event time. I have proposed to treat the event as a time-varying covariate, which assumes the value 0 if the event has not occurred by the tth measurement occasion and 1 in all subsequent occasions after the event occurs (assume that no indivuals have this event prior to the 1st measurment occasion).

Question:

Is it reasonable to specify this as a growth model as shown (without arrows) below? y is the oucome and z is the time varying event covariate. Or are there better ways?


y1 y2 y3 y4


ay by
x
az bz


z1 z2 z3 z4
 Linda K. Muthen posted on Thursday, October 26, 2000 - 4:40 pm
An alternative to the model that you propose, which I would refer to as a parallel process growth model, would be a model where the time-varying covariates are directly related to the outcomes. y1 regressed on z1; y2 regressed on z2, etc.
 Dan Powers posted on Monday, October 30, 2000 - 9:38 am
Linda,

Thanks. This actually closer to what had in mind. However, I was thinking of a latent Z that affected Ys growth parameters.

y1 y2 y3
| /\ / |
ya yb
\ /
\/
Z
/ | \
z1 z2 z3


Cheers,
Dan
 Linda K. Muthen posted on Monday, October 30, 2000 - 11:09 am
As we always say, you can estimate any model that is identified. You need to create a model to match your research questions as closely as possible.
 Anonymous posted on Wednesday, October 02, 2002 - 9:25 am
Some questions about random slopes for time-varying covariates:

Example 5 in Mplus 2.12 Addendum (Pp.15-17) deals with random effects of time-varying covariates on outcome measures. To my understanding, each of the time-varying covariates, x1-x4, in this model could have a random effect on the corresponding outcome measure (i.e., x1 on y1, x2 on y2 ..., and x4 on y4). The four different random coefficients could be estimated simultaneously from the LGM, and their interpretation is straightforward. In Example 5 of Mplus 2.12 Addendum, it looks that the four random coefficients were used as indicators of a latent random variable - random slope "st."

My questions are:
1) With my own data, I can estimate different random coefficients (e.g., x1 on y1, x2 on y2, ..., and x4 on y4) only when some of them (at least two of them) were set to equal; otherwise "THE MODEL ESTIMATION DID NOT TERMINATE NORMALLY." Theoretically speaking, is it possible to estimate four different random coefficients for the four time-varying covariates?
2) Using the four random coefficients as indicators of the random slope "st," how to interpret "st"? If some of the random coefficients were statistically significant (e.g., random coefficient of x1 on y1 was significant, but the rest were not), and "st" was not significant, then how to interpret "st"?
3) When RANDOM option was used, Mplus only provided Loglikelihood and Information Criteria for model fit test. Will future version of Mplus provide more fit statistics for GLM with random slope for time-varying covariates?

By the way, where could I download the data set used in Example 5 (i.e., growth.dat) (I did not find it in the Mplus home page)?

Thank you very much for your help.
 bmuthen posted on Thursday, October 03, 2002 - 9:17 am
Here are some answers:

1) Yes, theoretically you can identify and estimate a different random slope for each of your four time-varying covariates. The fact that you need to impose equalities among some of the random slopes probably indicates that not all of them have significant variation in your data - when the random slope variances are (almost) zero, the procedure can experience slow convergence - this is an indication that fixed slopes can be used instead.

2) To work with a second-order factor st behind the 4 random slopes is ok in principle, but you would want each of the 4 slopes to have significant random variation for this to be meaningful. I would not try to interpret st otherwise.

3) In line with typical multilevel modeling, the usual fit indices are not provided for models with random slopes. This is because there is no longer a single covariance matrix structure, but the covariance matrix varies as a function of the covariates that have random slopes. New types of fit indices may be added later.

Example 5 does not correspond to a real data set. Together with Tihomir Asparouhov I am preparing a paper to be put on the Mplus web site, which will have real data corresponding to this example.
 Jennie Jester posted on Tuesday, December 03, 2002 - 10:45 am
I am trying to model the development of attention problems, independent of aggression problems. I attempted to do this by using aggression problems at each time point as a covariate of attention problems at that time point. I believed that this would then produce a model in which the attention problems would be independent of aggression problems and I estimated a 4-class solution in a growth mixture model, using the attention problems to define the growth parameters. However, when I look at the aggression problems in each class, they differ in parallel with the attention problems. I cannot say, then, that the covariates are independently related to attention problems. Can you explain why this is so and what I can do to remedy this?
 bmuthen posted on Tuesday, December 03, 2002 - 7:38 pm
All observed variables contribute to the formation of latent classes in mixture modeling. This is as it should be. Perhaps it is true that the aggression covariates have a similar class profile as the attention problems.

To check this out, you might want to do a parallel process growth mixture model for both aggression and attention, each with its own latent class variable, and see how strongly related those 2 latent class variables are. The solution you found may indicate that these 2 latent class variables are very similar.

Leaving out the aggression covariates when analyzing the attention trajectories in order to get purely attention classes is not recommendable if you think that the aggression covariates are related to attention development. Then classes for attention would be gotten from a misspecified model.
 Kim Henry posted on Friday, February 28, 2003 - 1:17 pm
I am specifying a latent growth curve of "violent media use". I want to examine the effects of a time variying covariate (aggression). Specifically, I want to see if, within each measurement occassion, aggression predicts violent media use above and beyond what is captured by change over time. Also, I want to see if aggression has a lagged effect. That is, if aggression at T1 predicts media use at T2, if aggression at T2 predicts media at T3, etc. I've noticed that many of the published studies examining lagged hypotheses similar to mine have kept the within time effect (T1 aggression on T1 media use, T2 aggression on T2 media use, etc.) in the model when examining the lagged effect. What is the difference in the interpretation of the lagged effects in a model that includes the within time and lagged effects, and one that includes just the lagged effects? Thanks for your advice!
 bmuthen posted on Sunday, March 02, 2003 - 5:54 pm
Perhaps some of the studies you have seen with x1 influencing y1, x2 influencing y2, etc have been done in the multilevel software framework. The SEM-latent variable framework is perhaps more inviting to including lagged effects such as x1 influencing y1 and y2. With lagged effects, an outcome such as y2 is influenced by both x1 and x2 so the x2 effect is the partial regression coefficient controlling for x1 (and the growth factors which always influence y's). So the x2 effect may be a bit smaller here. I think the need to include lagged effects should be investigated because they may make very good substantive sense.
 Kim Henry posted on Sunday, March 02, 2003 - 7:34 pm
Thanks for the advice. I'm still not clear on one thing though. My main hypothesis concerns the lagged effects. If I include just the lagged effects (and not the within time effects...x1 on y1, x2 on y2, etc.) the lagged effects are not significant. However, if I include the within time effects and the lagged effects then the lagged effects are significant. I'm not sure I understand why this would be...is it necessary to adjust for the within time effect in order to properly assess the lagged effect?
 bmuthen posted on Monday, March 03, 2003 - 6:38 am
I am not sure why you are getting these significance results. It would seem reasonable to include within time effect whenever you try lagged effects to cover the possibility of an immediate effect.
 Anonymous posted on Thursday, September 09, 2004 - 11:49 am
What would you say is the main advantage of specifying time varying covariates rather than a simultaneous process model with regression from one random slope to the other? It seems to me that the two approaches would achieve the same basic result (a conditional slope) but the latter tells us nothing about the intra-individual and group trajectory on the covariate.
 Linda K. Muthen posted on Wednesday, September 29, 2004 - 4:12 pm
The first approach is less ambitious in that it does not specify a model for the time-varying covariates. The second approach can allow for direct effects effects between the outcomes in the two processes.
 Anonymous posted on Friday, October 29, 2004 - 12:17 pm
I have a growth model with 3 time points that are unequally spaced. Also I have 1 time-invarying (INVARY) and 2 time-varying variables (VARY11 VARY12 VARY13, VARY21 VARY22 VARY23). My question is: If I suppose one time-varying variable (VARY21 VARY22 VARY23) has the direct effect to another time-varying variable (VARY11 VARY12 VARY13), but it does not have the direct effect to outcome variable (T1 T2 T3). My model look like this:

I BY T1@1 T2@1 T3@1;
S BY T1@0 T2@5 T3@11;
[T1@0 T2@0 T3@0 I S];
I ON INVARY;
S ON INVARY;
T1 ON VARY11;
T2 ON VARY12;
T3 ON VARY13;
VARY11 ON VARY21;
VATY12 ON VARY22;
VARY13 ON VARY23;

1. Is this model correct?
2. I modeled this model, the model fit looks bad. Then I tried to model this time-varying variable has direct effect to outcome and get rid of the effect between time-varying variables. The model fit looks good. Does this mean that the effect between time-varying variables is not significant?

Thank you so much!
 Linda K. Muthen posted on Friday, October 29, 2004 - 2:49 pm
It appears that you have set your model up as intended. Because the model fits better with direct effects from the v2's to the t's, you can say that the v1's don't fully mediate the relationship between the v2's and the t's. To assess the significance of the direct effects of the v2's on the t's, include these paths in the model and look at the ratio of their parameter estimates to their standard errors.
 Anonymous posted on Monday, November 08, 2004 - 7:29 am
Hello,

I have a growth model with 3 time points that are unequally spaced, one time-varying covariate, one time-invarying covariate. whatever I included the time-varying covariate or not, the model fit is good in terms of all those index. But when I included the time-varying covariates, the average intercept (intercepts of slope) is no significant at all. It is not reasonable, right? We cannot have a growth model with 0 slope. Am I right? Thanks!
 bmuthen posted on Sunday, November 14, 2004 - 12:06 pm
You should check Tech4 and see what the slope mean is. The slope intercept is not the same as the slope mean when the slope is regressed on a time-invariant covariate.
 DWANT posted on Friday, February 25, 2005 - 9:04 am
I am considering a growth model over 4 time points for a continuous (depression score) variable. I have dichotamous "other" symptom occurrence data (e.g. GI problems-y/n, fatigue-y/n, etc.)that are time varying covariates that were measured at the same time. From my prior analyses on the depression scores I know that the responders are in known classes of consistently high score, consistently low score, high score with decline and low score with increase. I am attempting to run a GMM with known classes as in example 8.8. (p.166) The modelled example uses a time invariant covariate however, and I am having convergence issues running it with the time varying covariates. It does, however, run with a time invariant covariate. My model is likely wrong. How does one model the time varying variable in such a model where instead of x, there is u1,u2,u3,u4 on y1,y2,y3,y4 with known classes?
 bmuthen posted on Saturday, February 26, 2005 - 5:45 pm
The only difference between time-invariant and time-varying variables would seem to be that the latter would typically not be specified to influence the latent class variable. If this doesn't help, send the output of your failed run to support@statmodel.com.
 MDekker posted on Friday, April 01, 2005 - 7:18 am
We are building age-based growth curve models. So age is the basis for the time structure/accelerated longitudinal design. We have 3 time points with level of psychopathology as outcome continous variable. In addition we have age (6-18 years), and 2 groups (with and without intellectual disability), and sex as covariates). In the fixed part of the full model we have modelled the intercept, and the slopes for: age group sex age*group group*sex age*sex and age*group*sex. In the random part we have modelled the intercept and the slope for age.
With 3 time points we assume we can only model 2 random effects max? Is that correct?
We assume that it doesn't make any sense to have for instance 'group' or sex in the random part of the model, as this is a time-invariant covariate? Is this correct?
How many parameters are estimated is this model? 8 in the fixed part and 3 in te random part? plus the residual? So, 12 in total?
Thank you!
 BMuthen posted on Saturday, April 02, 2005 - 8:51 pm
See the output for the number of parameters estimated. You can also ask for TECH1 to see the free parameters in the model. Yes, it is correct that only two random effects (growth factors) can be used with three time points. Growth factors can have different variances for the levels of group and gender if you do a multiple group analysis with respect to group and sex.
 Anonymous posted on Wednesday, May 11, 2005 - 7:43 am
Hello, If I included the invariant covariates in the LGM. How can I do the test of mean of intercept and slope coming from the TECH4? Thanks!
 bmuthen posted on Wednesday, May 11, 2005 - 8:12 am
That's a bit more involved and would require that you know how to apply "the Delta method" to compute SEs (see stat lit).
 Anonymous posted on Thursday, May 12, 2005 - 7:31 am
continuous about testing the mean of the intercept and slope when including the time-invariate covariates. Thanks about the suggestion DELTA methods, I am not sure bootstrap method is suitable to this kind of testing. Thanks!
 bmuthen posted on Thursday, May 12, 2005 - 12:41 pm
Bootstrap-based testing could work too, but I don't think you can do that in Mplus for this case.
 Anonymous posted on Thursday, May 12, 2005 - 12:54 pm
Thanks for your response. I am trying to bootstap in other program, then feed the data to Mplus, save the results, repeatly.
 Anonymous posted on Friday, May 13, 2005 - 8:12 am
Continue about the DELTA method to calculate the SE of the mean intercept,
Assume one time-invariate covariate(c),
Mean(Intercept)=Intercept(I)+beta*c
D={1 beta}
V={var(I) cov(I,c),
Cov(I,c) var(c)};
SE(mean(Intercept))=sqrt(D*V*D`);
V: variance-covariance matrix coming from the Mplus output
D: derivate matrix
Is it correct?
How about calculate the SE of variance intercept?
Thanks!
 bmuthen posted on Friday, May 13, 2005 - 8:26 am
Looks right - where V comes from Tech3. The variance of the intercept factor just has a bit more complicated derivative.
 Pablo posted on Monday, October 31, 2005 - 11:07 am
I was wondering if you have additional info (e.g., syntax, papers) about multilevel mediation similar to the example from p.92-94 in the m-plus manual. I am interested in a extension of the model in which the predictor variables a1-a4 are, in turn, regressed on or predicted by other time varying covariates (something like
a1 on c1 e1
a2 on c2 e2
a3 on c3 e3, etc).
Any help or suggestion would be greatly appreciated
 Linda K. Muthen posted on Monday, October 31, 2005 - 3:16 pm
I don't have any examples of this but doing what you show is how you would specify such a model.
 Pablo posted on Tuesday, November 01, 2005 - 2:48 pm
Linda,
I have a more clear idea of what I want to do and a syntax that I hope can answer this problem.
To make it more clear I'll try to use equations;
This is the general model to explain the model from p.92
y sub (ti)=b(0i)+b(1i)*a(ti)+b(2i)*w(ti)+e
Then one has a set of three equations predicting each b coefficient with time invariant covariates.
However, for predicting b2i (random slopes) I want to use, in addition to time invariant covariates, a different set of time varying covariates reflecting random slopes.
b(2i)=b0+y*w+y*(time varying cov)+e

Does this make sense?

I have the following syntax:


model:
isah ssah | csah1-isah1 at t0-t3;

stvc |csah1 on cadls;
stvc |esah1 on eadls ;
stvc |gsah1 on gadls;
stvc |isah1 on iadls ;

stvc2 |cadls on cototsx cphysact ;
stvc2 |eadls on eototsx ephysact ;
stvc2 |gadls on gototsx gphysact ;
stvc2 |iadls on iototsx iphysact ;

isah ssah stvc stvc2 on sex age ctposaff ctngaffe ;
isah with ssah stvc stvc2;
ssah with stvc stvc2;

Is this the right syntax for doing what I want to do?
If so, when I tried to run this model I was asked to add algorithm=integration, is that ok? Why do I need to add this?
Thank you vary much for your help.
 bmuthen posted on Tuesday, November 01, 2005 - 3:45 pm
You specify the model

(1) ysub(ti)=b(0i)+b(1i)*a(ti)+b(2i)*w(ti)+e

(2) b(2i)=b0+y*w+y*(time varying cov)+e

I assume that in (1) a(ti) is time-related and w(ti) is a time-varying covariate. I assume that in (2) y is a slope - is that right, or not? I am confused because you have y in eqn (1).

Calling the eqn (2) "time varying cov" z, and inserting (2) in (1), it looks like you want the interaction z*w to be a time-varying covariate influencing y_(ti) - is that right?
 Pablo posted on Tuesday, November 01, 2005 - 6:44 pm
Oops,it was a typo. The answer to the first two questions is yes, a(ti) is the time-related var and w is the time-varying covariate. Y is the slope (sorry again, I was trying to find a letter in the keyboad similar to gamma and didn't realize that I used y as the outcome). At any rate, the answer to the las question is yes too. I want to insert 2 in 1.
Thanks, a lot.
 bmuthen posted on Wednesday, November 02, 2005 - 5:12 am
Ok, then we are on the same page. I think your model is doable, but wouldn't it be simpler to create the interaction variable as the product z*w in DEFINE and use that in (1) as an extra term?
 Pablo posted on Wednesday, November 02, 2005 - 9:07 am
Actually, that sounds better and simpler. Now, if I am interested in two predictors and create a three-way interaction term, how would you suggest I interpret significant results? I can handle linear regression following cohen et al procedure but I am not sure in this case how to address the issue.
Just one more question, when I tried to run the model I pasted, I was asked to add algorithm=integration, why is that?
Thank you very much for your help.
 BMuthen posted on Wednesday, November 02, 2005 - 6:34 pm
You would interpret a three-way interaction just like in linear regression.

The model you tried to run needs numerical integration because your random slope was a dependent variable.
 Pablo posted on Friday, November 04, 2005 - 7:33 am
Thanks you very much. Sorry to bombard you with so many questions but I've seen very little published on models like the ones we're discussing.
I have one more basic question, in a randon slope model with time varying and time invariant covariates I get a nonsignifican slope for y. However, when I run the intercept and slope model (repeated measures) for y with no covariates the slope is significant. I can't make sense of this. Any thoughts would be greatly appreciated

By the way, mplus support as well as the software is by far the best.
 bmuthen posted on Friday, November 04, 2005 - 7:50 am
When you say "nonsignificant slope for y", perhaps you mean that the intercept estimate for the slope growth factor is insignificant? With covariates, the mean of the slope factor is not estimated but is obtained as a function of the slope intercept and the regression coefficient for the covariate times the covariate mean. This mean is given in Tech4. Without covariates, the mean (not the intercept) is reported for the slope growth factor.
 Clinton Anderson posted on Saturday, March 18, 2006 - 7:54 am
When I use the following syntax

MODEL: i s | M1T1-M1T4 AT V19@0 V31 V43 V55;
PLOT: TYPE IS PLOT1;
SERIES IS M1T1 (s) M1T2 (s) M1T3 (s) M1T4 (s);

All the data points are plotted on the y-axis.

How do I get the data to spread out over the x-axis using the time scores?
 Linda K. Muthen posted on Saturday, March 18, 2006 - 2:06 pm
I don't think this work with individually-varying times of observation. Send your question to support@statmodel.com along with your license number for a more definitive answer.
 uclaalice86 posted on Thursday, January 18, 2007 - 1:15 am
I am running a growth model with time-varying covariates as an alternative to a parallel process model. How do you interpret the effects of time-varying covariates? Specifically, I am looking at perceived disrimination over four time-pts with two time-invarying covariates (school composition and gender) and am interested in the role of group identity (my time-varying covariate) across the time-pts as well.
Here is the model I am testing:
i s | y3@0 y4@1 y5@2 y6@3;
i s on sch gender;
x3 on y3;
x4 on y4;
x5 on y5;
x6 on y6;

There is significant change in slope, significant sch on s, and the time-varying covariate is significant at all time points. How do you interpret the significant slope change and the effect of the time-varying covariates together? Does effect of the group identity at each time point take into account how percieved discrimination is changing over time in subsequent time points?
 uclaalice86 posted on Thursday, January 18, 2007 - 8:29 am
Sorry.. typo in previous post. I flipped the x and y in my post above. The model is:
i s | y3@0 y4@1 y5@2 y6@3;
i s on sch gender;
y3 on x3;
y4 on x4;
y5 on x5;
y6 on x6;
Sorry about that!
 Linda K. Muthen posted on Thursday, January 18, 2007 - 9:42 am
When you have a time-varying covariate, the outcome is regressed not only on the growth factors but also on the time-varying covariate, so the effect is a partial effect.
 Andrea Dalton posted on Friday, October 26, 2007 - 4:47 pm
I am a new user of Mplus and am trying to get the hang of it by replicating analyses that I previously conducted using HLM.

In this case, I'm a bit stuck. HLM allows you to model the effects of time-varying covariates without necessarily including a factor accounting for the effect of linear time. I have not yet been able to do this in Mplus, and I'm sure it's because I'm not using the correct syntax.

Here's the best guess I have so far:
MODEL: i | pa1-pa6@1;
stvc |pa1 ON pen1;
stvc |pa2 ON pen2;
stvc |pa3 ON pen3;
stvc |pa4 ON pen4;
stvc |pa5 ON pen5;
stvc |pa6 ON pen6;
stvc WITH i;

Also, the time-varying covariate is a random effect, but can I model it as fixed or non-randomly varying and still retain the latent variable "stvc" so that I can regress it on some person-level predictors?

Thanks,
Andrea Dalton
 Bengt O. Muthen posted on Friday, October 26, 2007 - 5:05 pm
It sounds like you want neither a fixed nor a random growth slope in which case your i | statement is correct. If you want a fixed slope you say

i s | ....
s@0;

You can define stvc as you have and still regress it on say x:

stvc on x;

where you can say stvc@0 to specify a zero residual in this regression to indicate that x is the only reason stvc varies.
 Andrea Dalton posted on Friday, October 26, 2007 - 5:18 pm
Okay - but now I see that if I use the statement stvc@0, I get a coefficient of 0 under the heading "Means", and HLM gives me a nonzero coefficient when the slope of the time-varying covariate is fixed. Have I missed something?
 Linda K. Muthen posted on Friday, October 26, 2007 - 6:28 pm
It sounds like you need to send your HLM output, your Mplus output, and your license number to support@statmodel.com so we can see exactly what you are doing in both programs. If set up correctly, you will obtain the same results.
 Bengt O. Muthen posted on Friday, October 26, 2007 - 6:30 pm
I don't see how that would happen - if you go by the User's Guide example 6.18 and fix st@0 with or without x, you do get a coefficient for st. So please send your input, output, data, and license number to support@statmodel.com.
 Heather See posted on Monday, December 10, 2007 - 4:59 pm
Hi-

I am conducting a latent growth model very similar to the one in example 6.12 where I am estimating the slope of a time-varying covariate. I want to examine the moderating effect of early engagement in home learning activities (time invariant) on the impact of family income (time-varying covariate) on school readiness skills (continuous outcome). I wanted to clarify if both the time invariant and time-varying covariate, as in this example, can be continuous or if one (or both) has to be categorical to model an interaction. I'm trying to determine whether to use a total sum of activities as my variable or to categorize the variable and create "high"/ "low" engagement groups (by way of a cluster analysis). Any help would be greatly appreciated!
 Bengt O. Muthen posted on Monday, December 10, 2007 - 6:35 pm
They can both be continuous. Since the time-invariant covariate predicts the random slope for the time-varying covariate, the time-invariant covariate ends up moderating (interacting with; multiplying) the time-varying covariate.
 Leah Rohlfsen posted on Tuesday, March 18, 2008 - 11:43 am
I'm running parallel processes two-part growth models. My outcome is functional limitations (measured in 94, 96, 98, 00, and 02). The time-varying covariate is income (to see if/how the i and s of inc affects funct) and is measured in 92, 94, 96, 98, and 00. Because the initial income wave is prior to the initial wave of the outcome, which would be better/correct?
iu su |bin2@0 bin3@2 bin4@4 bin5@6 bin6@8;
iy sy |cont2@0 cont3@2 etc.;
Inci Incs |Inc1@0 Inc2@2 Inc3@4 Inc4@6 Inc5@8;
OR:
iu su |bin2@2 bin3@4 bin4@6 bin5@8 bin6@10;
iy sy |cont2@2 cont3@4 etc.;
Inci Incs |Inc1@0 Inc2@2 etc.;
Thanks
 Linda K. Muthen posted on Friday, March 21, 2008 - 7:47 am
You need a zero time score in each growth process to define the intercept growth factor. The time scores 0 1 2 3 4 would be appropriate for the outcomes measured in 94, 96, 98, 00, and 02 and 92, 94, 96, 98, and 00. You say that income is a time-varying covariate. Your specification shows it as a parallel process growth model. Example 6.13 shows a parallel process growth model. Example 6.10 shows a growth model with time-varying covariates. See these examples to see the difference.
 Miranda Witvliet posted on Wednesday, April 16, 2008 - 3:01 pm
I`m trying to run a latent growth model with several categorical time-varying covariates. The outcome in the model is continuous. Is it a problem to have missings (at random) on the time-varying covariates?
Thanks
 Linda K. Muthen posted on Wednesday, April 16, 2008 - 5:09 pm
If an observation is missing on the time-varying covariate, is it also missing on the outcome?
 Miranda Witvliet posted on Thursday, April 17, 2008 - 6:41 am
Most of the missings are only on the (categorical) time-varying covariates and not on the outcome. There ar though a few missings on both the covariate and the outcome,
Thanks for your respons
 Linda K. Muthen posted on Thursday, April 17, 2008 - 12:59 pm
Any observation with missing on one or more covariates is not included in an analysis because missing data theory applies to dependent variables. In your case, if you don't want these observation eliminated, you can bring the covariates into the model by mentioning their means, variances, or covariances in the MODEL command. When you do this the same distributional assumptions are made about these variables as for the dependent variables in the model.
 Kaigang Li posted on Sunday, June 15, 2008 - 10:49 pm
Hello Professor Muthen,

Can one or both outcome variables in the Example 13 (GROWTH MODEL FOR TWO PARALLEL
PROCESSES FOR CONTINUOUS OUTCOMES WITH
REGRESSIONS AMONG THE RANDOM EFFECTS) be categorical variable(s)?

I tried it. The program went but couldn't converge.

Thanks,

Kaigang
 Linda K. Muthen posted on Monday, June 16, 2008 - 6:05 am
Yes, they can. You need to send your input, data, output, and license number to support@statmodel.com if you can't figure the problem out on your own.
 Michael Spaeth posted on Monday, March 16, 2009 - 10:24 am
Regarding time invariant vs. time variant.

Imagine I have a growth model of problem behavior. Risk factors (covariates) are measured across all waves as the outcome. Most authors would include risk factors as time invariant covariates in the model, and predict intercept and slope of problem behavir, which seems to me like a "programming-effect" of the specific risk factor. However, I also thougt of time variant effects of my risk factors.

Because of:

theory: I assume immediate effects (concurrent and lagged) besides "programming-effects" on the slope.

However, are there any "hard" criteria to decide on that issue? What came into my mind is --> 1.) to compute growth curves of each risk factor to judge on stability over time (instability supports time variant definition) and, in addition, 2.) to have a look on the residualvariances of the outcomes of the growth curve pertaining problem behavior (high residualvariances would suggest that there is much to predict over and beyond the general growth shape).
Lastly, 3.) would it be reasonable to test all risk factors as invariant covariates first and then retest the insignificant ones as time variant (plus lagged effects) afterwards?

Do you know any paper discussing that distinction or comparing both kinds of effects?
 Bengt O. Muthen posted on Monday, March 16, 2009 - 10:54 am
I don't know of any papers on this, but I would include some of the risk factors (the early ones) as time-invariant covariates and some as time-varying and then see if the time-varying covariates add any significant effects.
 Brewery Lin posted on Tuesday, November 09, 2010 - 6:49 pm
I am considering a growth model over 3 time points. It's so wonderful that model fit so perfect while using Mplus (:D). However, when I try the same data in LISREL, the result is inverse. I feel confused about that. Are there any possible reasons to explain the situation? Truly appreciate your help in resolving the problem.

Syntaxes used in the two different softwares are described below.

TITLE: MI growth model with time-varying covariate
DATA: FILE IS D:\20101108.dat;
VARIABLE: NAMES ARE gender y1-y3 s1-s3 f1-f3;
USEVARIABLES ARE y1-y3 s1-s3;
MODEL: i s | y1@0 y2@1 y3@2;
y1 ON s1;
y2 ON s2;
y3 ON s3;


MI with time-varying predictors
RAW DATA FROM FILE 20101108.psf
Observed Variables: y1 y2 y3 s1 s2 s3
Latent variables: INT SLP SC1 SC2 SC3
Sample Size: 2494
Relationships
s1 = 1.0*SC1
s2 = 1.0*SC2
s3 = 1.0*SC3
set variance of s1 equal to 0
set variance of s2 equal to 0
set variance of s3 equal to 0
y1 = 1*INT+0*SLP+SC1
y2 = 1*INT+1*SLP+SC2
y3 = 1*INT+2*SLP+SC3
INT = Const
SLP = Const
path diagram
End of Problem
 Linda K. Muthen posted on Tuesday, November 09, 2010 - 6:54 pm
If you are specifying the same model in Lisrel and Mplus using the same data and estimator, you will obtain the same results. If you want to know what the differences in your results are due to, send the Mplus and Lisrel outputs and your license number to support@statmodel.com.
 Jing Zhang posted on Thursday, August 25, 2011 - 2:59 pm
Dear Dr. Muthen,

I am testing a growth model with time varying predictors and mediators.
X – M – Y
X: predictor
M: mediator
Y: outcome variable

X, M, and Y have data for 4 time points.

My question is how I would specify such a model. Is there an example in the user’s guide?
 Bengt O. Muthen posted on Thursday, August 25, 2011 - 5:32 pm
In modeling longitudinal data, a first decision you have to make is if you want to specify some sort of growth model for the outcomes, or simply take into account that observations are correlated over time, assuming no change over time. That is, is time an important factor to study or a nuisance to just control for?
 Jing Zhang posted on Thursday, August 25, 2011 - 7:31 pm
Dear Dr. Muthen,

Thanks for your response. In my study, time is an important factor as I am looking at the cognitive development and emotion regulation of young children over years. In this situation, what model would you suggest?

Thanks,
Jing
 Bengt O. Muthen posted on Saturday, August 27, 2011 - 12:17 pm
Then you have to ask yourself if X, M, and Y all should have a growth model or only say Y, or only M and Y. X could perhaps be treated as a time-varying covariate. More than one outcome in growth modeling is sometimes called parallel processes. For these kinds of growth models, see the handout and video for Topic 3 on our web site.
 Vanessa posted on Tuesday, January 24, 2012 - 8:23 pm
If a time-invariant covariate is centred and the slope is regressed on it, in the output does the intercept of the slope now represent the slope mean (conditional on the covariate)?
 Linda K. Muthen posted on Wednesday, January 25, 2012 - 11:33 am
When a continuous covariate is centered, the intercept in the regression of the slope growth factor on the covariate is the mean.
 Vanessa posted on Monday, January 30, 2012 - 9:47 pm
Thank you for the response.

What if there is a combination of centered continuous variables and dichotomous variables as predictors of the slope - if the dichotomous variables are coded as -1 and 1 will the intercept in the regression of the slope growth factor on the covariates still be the slope mean?
 Linda K. Muthen posted on Wednesday, February 01, 2012 - 11:22 am
I don't believe in this case the intercept is the mean. You can check this out by asking for RESIDUAL in the OUTPUT command and comparing the model estimated mean to the intercept.
 Youngoh Jo posted on Wednesday, February 15, 2012 - 3:30 pm
Dear Dr. Muthen,

Below is the syntax I used, but there was no p value on the output.

What's wrong?

Thanks in advance,


data: file is "E:\group 3.dat";
VARIABLE:
NAMES ARE sc2-sc6 pa1-pa5 mo1-mo5 ab1-ab5 ta1-ta5 dp1-dp5 nne2-nne5;
USEVARIABLES ARE sc2-sc6 pa1-pa5 mo1-mo5 ab1-ab5 ta1-ta5 dp1-dp5 nne2-nne5;
MISSING ARE ALL (999);
CLASSES = c (1);
ANALYSIS: TYPE = MIXTURE;
START = 100 10;
MODEL:
%OVERALL%
i s | sc2@0 sc3@1 sc4@2 sc5@3 sc6@4;
sc2 ON pa1 mo1 ab1 ta1 dp1;
sc3 ON sc2 pa2 mo2 ab2 ta2 dp2 nne2;
sc4 ON sc3 pa3 mo3 ab3 ta3 dp3 nne3;
sc5 ON sc4 pa4 mo4 ab4 ta4 dp4 nne4;
sc6 ON sc5 pa5 mo5 ab5 ta5 dp5 nne5;

OUTPUT: tech1 tech8
 Linda K. Muthen posted on Wednesday, February 15, 2012 - 4:40 pm
You must be using a version of Mplus from before the p-values were added.
 Youngoh Jo posted on Thursday, February 16, 2012 - 8:30 am
Dear Dr. Muthen,

I am using the Version 4.2.

Do I need a higher version to get p values?

Thanks,
 Linda K. Muthen posted on Thursday, February 16, 2012 - 2:51 pm
Yes.
 EFried posted on Tuesday, February 21, 2012 - 6:56 am
In a Growth Mixture Model with both time-varying and time-invariant growth factors on a continuous outcome variable, is there a way in MPLUS to assess how much percentage of variance both groups of variables taken together explain on the outcome (similar to what latent-state-trait models do)?

I am interested whether my outcome variable is affected mostly by baseline factors (e.g. genes, personality) or life events.

Thank you
EFried
 Linda K. Muthen posted on Tuesday, February 21, 2012 - 7:32 am
Mplus gives an overall R-square for each dependent variable.
 EFried posted on Tuesday, February 21, 2012 - 9:26 am
In which output can that value be found? I can't find it in my current output (tech1 4 7 10 11 13 14 sampstat residuals modindices), and neither in the Mplus manual.
Thank you!
 Linda K. Muthen posted on Tuesday, February 21, 2012 - 10:03 am
Ask for STANDARDIZED in the OUTPUT command. R-square comes at the end of the standardized output.
 Youngoh Jo posted on Saturday, February 25, 2012 - 1:17 pm
I am using version 4.2 which does not show p values when I run a growth modeling according to your comments above. How can I get p values by calculating manually?

Thanks,
 Linda K. Muthen posted on Saturday, February 25, 2012 - 1:39 pm
Look up the ratio of the parameter estimate to its standard error in a z-table.
 Edith Visser posted on Tuesday, March 27, 2012 - 6:59 am
Dear Dr. Muthen,

I use growth models with age as individually-varying times of observations (Tscores). Therefore I use analysis TYPE = RANDOM.

In another discussion it was stated:

"Chi-square and related fit measures are not available when means, variances, and covariances are not sufficient statistics for model estimation as with TYPE=RANDOM."

How can I evaluate the modelfit. I am not interested in comparing models. Is it possible to compute RSMEA and TLI by hand?

Do you have a reference in which age was used as time-varying times of observations?

Thanks,

Edith Visser
 Linda K. Muthen posted on Tuesday, March 27, 2012 - 6:58 pm
Chi-square and related fit statistics cannot be computed for these models. There are no absolute fit statistics. Nested models can be compared using -2 times the difference in the loglikelihoods for two nested model.

See Papers on the website.
 Artemis Koukounari posted on Sunday, July 22, 2012 - 10:11 am
I am trying to fit a growth model with time varying covariates (these are SH in the code below) and the data are arranged by age so they are partially missing. Part of my code follows:

ANALYSIS: TYPE=RANDOM missing;
ESTIMATOR =ML;
MCONVERGENCE=.001;
MODEL: i s| HE7@0 HE8@1 HE9@2 HE10@3 HE11@4 HE12@5 HE13@6 HE14@7 HE15@8;
HE7 ON SH7;
HE8 ON SH8;
HE9 ON SH9;
HE10 ON SH10;
HE11 ON SH11;
HE12 ON SH12;
HE13 ON SH13;
HE14 ON SH14;
HE15 ON SH15;

I get the following msg:

*** ERROR One or more variables in the data set have no non-missing values. Check your data and format statement.

Do I need a FORMAT statement and what that could be?

Many thanks for any help you can provide
 Linda K. Muthen posted on Sunday, July 22, 2012 - 10:53 am
If your data are in fixed format, you need a format statement. I would simply check if you are reading your data correctly, for example, do you have blanks in your data set, do the number of names in the NAMES statement match the number of columns in the data set?
 Artemis Koukounari posted on Monday, July 23, 2012 - 11:23 am
Thank you very much Prof Muthen for your helpful input but I have checked what you proposed and no my data are in free format. I am sorry I did not express myself correctly before.
I have repeated measures at unequal time points. The time points are childrens' ages and the age at which children entered in the study varied. Data are in wide format so I guess I need to use tscores? Ages are 7-15. Each child has data at 3 occassions. So I guess if a child had data on ages 7, 8 and 10 timescores for these would be 0, 1 and 3, right?And for all other ages timescores would be missing, right?

And in the code do I put?

ANALYSIS: TYPE =RANDOM MISSING;
ESTIMATOR=ML;

Can y recommend any reference for such a problem? I have gone through MPLUS Short Courses Topic 3, but if there is also a paper/study dealing with this similar problem, that would help me understand better too.
Thanks so much!
Kind Regards,

Artemis
 Bengt O. Muthen posted on Tuesday, July 24, 2012 - 8:06 am
Given your data structure, you should have a look at UG ex6.18.
 Leah J Welty posted on Thursday, July 26, 2012 - 8:07 am
I am estimating a two class growth mixture model for a categorical outcome with 10 time points and a time-varying covariate. The model also includes time-invariant covariates. By design, my time-varying covariate (and my outcome) are missing at certain time points because random subsamples of the cohort were given certain interviews (e.g. only half the sample got a 3.5 and 4.0 year interview, a different selection got an additional interview at 10 years).

When I estimate the model without the time-varying covariate, it uses my ~ 1800 participants. When I include the time-varying covariate, it uses only 244 participants -- those who were in all the random subsamples.

Do you have suggestions for how to handle this? I understand the default is to drop observations that are missing time-varying (or time invariant) covariates. I see a post above that touches on this issue, but it wasn't clear to me how this might apply to my situation.

Many thanks,
Leah
 Linda K. Muthen posted on Thursday, July 26, 2012 - 12:01 pm
You can recode the missing value to any non-missing value for those individuals who are missing on the outcome and the time-varying covariate. These individuals do not contribute to the results but you avoid the person being deleted.
 Emily Midouhas posted on Friday, August 10, 2012 - 9:22 am
Hello. I have three timepoints of data on children's behavioural difficulties (dependent var) and family poverty (independent var). Then we have a child psychological construct variable we believe moderates this relationship but it is measured at the last timepoint, and we can't be sure the construct is stable across the study period (prior to the third timepoint, so change in difficulties could not be attributed to an interaction between poverty and the construct). However, rather than not include our three timepoints of dependent variable data and only include the third timepoint of data we want to account for the clustering of difficulties within children over time. Can we do this without having to make the assumption that our construct is stable throughout childhood/estimate the change in difficulties?

Can you have a model with many timepoints for your time-varying independent variables and only one timepoint for your dependent variable?
 Linda K. Muthen posted on Saturday, August 11, 2012 - 10:30 am
A time-varying covariate needs an outcome at that time point to be used.
 Emily Midouhas posted on Sunday, August 12, 2012 - 1:54 am
Thank you. Can one model the individual trajectories to simply control for clustering of outcome within the child but not to determine whether the time-varying predictor influences average change between time points? Would this still be a growth model but one where there is no interaction between the predictor and time/age? Also, can the intercept be centred on the last time point?
 Linda K. Muthen posted on Sunday, August 12, 2012 - 9:21 am
The answer to each of your questions is yes.
 lam chen posted on Wednesday, November 14, 2012 - 8:32 am
I am modeling a GMM with covariates to explore the influenced factor on the latent class variable.
x is a time-varying variable. Can i use the syntax in mplus as follow?
c on x1 x2 x3 x4;(four times measurement of x)
As x1-x4 are repeated measurement,i don't know the answer.
 Linda K. Muthen posted on Thursday, November 15, 2012 - 9:12 am
There is no technical reason not to do this. But the question of at which time point c resides may make it substantively incorrect.
 lam chen posted on Monday, November 19, 2012 - 7:22 pm
As x1-x4 were repeated measurement,the correlation coefficients among them were 0.4 to 0.5, a moderate level.Would the multi-collinearity among this variables lead to bias in parameters estimating? Or Mplus has considered this issue?
Thanks a lot for your help, Dr. Muthen.
 Linda K. Muthen posted on Tuesday, November 20, 2012 - 12:01 pm
This does not seem too high.
 rongqin posted on Friday, December 07, 2012 - 4:34 am
I am doing a latent growth model with time-varying and random slope. I could not get chi-square statistics. I understand why. however, how do i report the results for my growth model? i know i should compare the linear and quadratic model to see which one is better. However, how can i know whether the model itself is good enough or not?
 Linda K. Muthen posted on Friday, December 07, 2012 - 9:44 am
There are no absolute fit statistics unless you have categorical outcomes in which case you can look at TECH10.
 rongqin posted on Saturday, December 08, 2012 - 4:50 am
All my variables are continuous.
I did latent growth model for two age cohorts (one from 12-16, the other 16-20. they were measured each year). I want to know the growth of one variable (e.g., friendship quality)across whole period of adolescence with these two group of adolescent measured at the same time points but different ages. I used TSCORE to define different time points for these two groups. namely young group time1@0 time2@2...for old group time1@4 time2@5...
I used type = random (since i assume slope would be different for different person especially for adolescents in different group). I do not have a time-varying covariate. is my following model correct?
usevariables are
vnriabma wnriabma xnriabma ynriabma znriabma T04 T15 T26 T37 T48;
MISSING ARE ALL (999.00);
TSCORES are T04-T48;
ANALYSIS:
Type IS MISSING RANDOM;
ESTIMATOR IS MLR;
MODEL:
iabma sabma | vnriabma-znriabma AT T04-T48;

Is there any other alternative analysis (latent growth models) i can do to answer my question? Thank you!
 Linda K. Muthen posted on Sunday, December 09, 2012 - 5:42 pm
You might try the multiple group multiple cohort approach shown in Example 6.18.
 rongqin posted on Monday, December 10, 2012 - 10:15 am
Thank you. However, in 6.18, the slope is not assumed as random, can i specify the model in 6.18 type = random? my interest is not to compare these two groups, but to use the intercept and slope of the whole group to predict another outcome. in that case, maybe the 6.12 model is better?

what i also want to ask is that: in the model I specified in my last post, if i specify TSCORE and type = random without specific time-varying covariant, is it still a RANDOM slope calculated?

or is there a anyway to calculate the chi-square with the model i specified above? if not, do you have any reference articles reporting results without model fit?
 Linda K. Muthen posted on Monday, December 10, 2012 - 2:30 pm
Example 6.18 is an example of how to deal with multiple cohorts. The growth model shown has a random slope growth factor.

TYPE=RANDOM with the AT option also estimates a model with a random slope growth factor.
 Cameron McIntosh posted on Tuesday, December 11, 2012 - 1:41 pm
Hello,

I am estimating a two-level growth curve for an outcome y with a continuous time-varying covariate x. The cluster variable is city of residence. The code for the model is below:

ANALYSIS: TYPE=TWOLEVEL RANDOM;
ESTIMATOR=ML;
MODEL:
%WITHIN%
iw sw | y1@0 y2@1 y3@2 y4@3;
y1 on x1;
y2 on x2;
y3 on x3;
y4 on x4;
%BETWEEN%
ib sb | y1@0 y2@1 y3@2 y4@3;
y1 on x1;
y2 on x2;
y3 on x3;
y4 on x4;

As shown in the above TVC regressions, I am not estimating random slopes, but I am trying to make sure I understand exactly what is fixed vs. random with regard to the intercepts. Specifically,

1. At the within level, is it correct to say that the level 1 intercepts in the TVC regressions are random across individuals (i.e., random across persons within each occasion) AND random across the four occasions (although this cross-time variability in the intercepts would not be separately computed and reported)?
2. Does this same interpretation hold at the city level?

Thanks,

Cam
 Linda K. Muthen posted on Wednesday, December 12, 2012 - 8:48 am
The random intercepts y1-y4 vary over cities. The random growth factors iw and sw vary over individuals.
 rongqin posted on Friday, December 14, 2012 - 8:19 am
Dear Professors,

If i don't have a specific time varying covariant for my latent growth model, is it recommended not to use example 6.12 where random slope is estimated? is it then wise to just use multiple group analyses to deal with different age cohorts?

what is the advantages and when it is really necessary to use type = random in latent growth curve models?

what mplus do differently for type=random and type=general in latent growth model?

many thanks for your answer!
 Linda K. Muthen posted on Friday, December 14, 2012 - 2:31 pm
You can have a growth model without time-varying covariates. See Example 1.

You need RANDOM when you are estimating random effects except for growth factor means and variances. It is just a different algorithm
 Wen-Hsu Lin posted on Monday, January 21, 2013 - 11:25 pm
Dear Professors,
I am estimating the a LGM with two time varying and two time invariant variables. In addition, I want to use the latent growth variable (I & S) to predict another outcome variable. My code as follow:
model:
i s|swb1@0 swb2@1 swb3@2;
i s on sex ses;
crime4 on i s sex money;

swb1 on sup1
friend1;
swb2 on sup2
friend2;
swb3 on sup3
friend3;

my question is:
(1)In a HLM sense, there will only be one estimated regression parameter for level one predictor. In LGM case above, I have different regression parameters for each time point. Do I need to make it equal?
 Linda K. Muthen posted on Tuesday, January 22, 2013 - 7:05 am
You can constrain them to be equal if you want to, but one of the advantages of the wide format you use here and the long format of most multilevel programs is that you can estimate these parameters and see if they are actually equal. You don't need to force them to be equal.
 Celia Matte-Gagné posted on Tuesday, May 28, 2013 - 12:25 pm
Hi, I'm doing a multilevel growth model and I want to know if PSI (a time-varying predictor) have a fixed or a random effect on hostilit (measured at four time points). When I'm running the following syntax (see below) I received a warning message: THE ESTIMATED BETWEEN COVARIANCE MATRIX IS NOT POSITIVE DEFINITE AS IT SHOULD BE. COMPUTATION COULD NOT BE COMPLETED. PROBLEM INVOLVING VARIABLE S1. THE PROBLEM MAY BE RESOLVED BY SETTING ALGORITHM=EM AND MCONVERGENCe TO A LARGE VALUE.

Syntax:

Define:
ageBYPSI = age * PSI;

variable : names = idno Age PSI Home Sensitiv Structur Hostilit Responsi Involvem ;

usevar = hostilit psi ageBYPSI ;

missing = all (999);

cluster = idno;
within= psi ageBYPSI ;

analysis: type = random twolevel;
estimator = ml;

model:

%within%
s|hostilit on psi;
s1|hostilit on agebypsi;

%between%
hostilit with s;
hostilit with s1;
s with s1;

output: tech4;

When I try to keep only "s| hostilit on psi", I have the warning message of a ill-conditonned fisher matrix. But, when I try only the fixed effect of PSI (hostilit on PSI) I did'nt get an error message. How can I understand that? Can I only test the fixed effect and forget the random one and still have a correct statistical procedure ? or do I need to remove this variable in the analysis.
 Linda K. Muthen posted on Tuesday, May 28, 2013 - 12:56 pm
Please send the output with the random effect and your license number to support@statmodel.com.
 Celia Matte-Gagné posted on Tuesday, May 28, 2013 - 2:10 pm
I am using a public computer with the software on it, how can I find the license number?

Thank you for your help.
 Linda K. Muthen posted on Tuesday, May 28, 2013 - 2:49 pm
You would need to check with your IT department.
 Jennie Jester posted on Thursday, August 01, 2013 - 8:37 am
I am modeling the effect of a time-varying covariate (drinking) on growth of social expectancy. Modification indices showed that drinking at time4 was highly related to expectancy at time5. I tried this syntax:
i by social2 - social5@1;
s by social2@0 social3*3 social4@6 social5*9;
[social2 - social5@0 i s];
social5@0;
social3 on drinkt3;
social4 on drinkt4;
social5 on drinkt5 drinkt4;
This gives a decent-fitting model, chisquare p=.026; RMSEA = .039, CFI = .94

The variables drinkt3-drinkt5 are binary, whether or not the person has onset drinking at that time.

I would like to argue that the effect of drinkt4 on socialt5 is the cumulative effect of drinking over time4 and time5.

Do you think this makes sense? Or is there a better way to set up the model?
 Bengt O. Muthen posted on Thursday, August 01, 2013 - 1:56 pm
One alternative approach is discussed in our Topic 3 course on slides 157-159. See also

Curran, P.J., Muthén, B., & Harford, T.C. (1998). The influence of changes in
marital status on developmental trajectories of alcohol use in young adults.
Journal of Studies on Alcohol, 59, 647-658.
 Jennie Jester posted on Thursday, August 15, 2013 - 1:39 pm
Thank you, that is helpful. I looked at the paper and I believe this syntax would reproduce that model:
deldrink3 by drink3;
deldrink4 by drink4;
deldrink5 by drink5;
deldrink3 with deldrink4;
deldrink4 with deldrink5;
social5 on deldrink5 deldrink4 deldrink3;
social4 on deldrink4 deldrink3;
social3 on deldrink3;

Where drink3-drink5 are binary indicators and deldrink3-5 are the latent intercept factors (incremental status change in the model you referred me to).
My understanding is that drink3-drink5 are indicators of change in status, so if the person started drinking at time4, they would have drink3=0, drink4=1, drink5=0.
Does this syntax reproduce the model ?
 Bengt O. Muthen posted on Thursday, August 15, 2013 - 6:13 pm
You see the input on slide 159 of Topic 3.
 Jennie Jester posted on Friday, August 16, 2013 - 10:13 am
Thank you, that is helpful. I hope that you might look at the syntax that I believe will reproduce that model:
deldrink3 by drink3;
deldrink4 by drink4;
deldrink5 by drink5;
deldrink3 with deldrink4;
deldrink4 with deldrink5;
social5 on deldrink5 deldrink4 deldrink3;
social4 on deldrink4 deldrink3;
social3 on deldrink3;

Where drink3-drink5 are binary indicators and deldrink3-5 are the latent intercept factors (incremental status change in the model you referred me to).
My understanding is that drink3-drink5 are indicators of change in status, so if the person started drinking at time4, they would have drink3=0, drink4=1, drink5=0.
Does this syntax reproduce the model ?
Thanks for looking this over!
 Jennie Jester posted on Friday, August 16, 2013 - 12:27 pm
I don't know why I missed that before. Is the status 1 only for the time of change or does it change to 1 and remain at one following the change?
 Bengt O. Muthen posted on Sunday, August 18, 2013 - 2:05 pm
I don't see the similarity between your input and the one I gave on slide 159. In line with the figure on slide 158, my factors are defined as:

f1 BY alcuse1-alcuse4@1;
f2 BY alcuse2-alcuse4@1;
f3 BY alcuse3-alcuse4@1;
f4 BY alcuse4@1;
f1-f4@0;

where the alcuse variables are the outcomes of the growth model. Your deldrink factors are very different.

My factors are then regressed on the status change indicators.
 Jennie Jester posted on Tuesday, August 27, 2013 - 11:40 am
Thank you.
I tried this syntax:
i s |arouse2@0 arouse3@3 arouse4@6 arouse5@9;

dldrink3 by arouse3-arouse5@1;
dldrink4 by arouse4-arouse5@1;
dldrink5 by arouse5@1;
dldrink3-dldrink5@0;
dldrink3 on drkdelt3;
dldrink4 on drkdelt4;
dldrink5 on drkdelt5;

My arouse variables are the outcomes of the growth model and the drinking onset is akin to marital status.
However, when I tried this I got the dreaded "THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE
COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL.
PROBLEM INVOLVING PARAMETER 15." Parameter 15 is in the PSI matrix, variance of the slope.
Can you help me figure out what is the problem?

Thanks,

Jennie
 Linda K. Muthen posted on Tuesday, August 27, 2013 - 11:58 am
Please send the full output and your license number to support@statmodel.com.
 Fong Chun Tat Ted posted on Thursday, September 05, 2013 - 3:26 am
Dear Dr. Muthen,

I have done a latent growth model for continuous outcomes with time-invariant and time-varying covariates, like the Example 6.10 in the UG. I would like to ask a question relevant to that example.

For my data, the LGM with time-varying covariates did not provide adequate model fit. When I specify a path from y11 to a32, the model fit significantly improved and became very good. Do you think adding this path would invalidate the LGM results as a32 now becomes a dependent variable rather than a covariate? Do you think adding this path is correct?

Thanks a lot
Ted Fong
 Linda K. Muthen posted on Thursday, September 05, 2013 - 10:32 am
This is fine if it makes substantive sense and if a32 happens after y11.
 Laura Alexandra posted on Thursday, March 27, 2014 - 1:12 pm
Dear Mr. or Ms. Muthén

I am running a latent growth curve analysis in mplus. (Actually I am fitting two latent curves simultaneously but I don`t think that this makes a difference concerning my question)

With regard to my conditional growth curve models I am including several dichotomous time-invariant predictors (Dummy variables coded 1 or 0). Among these predictors I include an interaction between two observed Dummy-variables.

Model:

Latent Intercepts and Latent slopes
on x1 + x2 + (x1 * x2)

As the interaction term is based on two observed dichotomous covariates, may I interpret the results similar to as I would interpret them in standard linear regression analysis?

Thank you very much for your support!
 Linda K. Muthen posted on Thursday, March 27, 2014 - 2:10 pm
Yes.
 Laura Alexandra posted on Thursday, March 27, 2014 - 11:43 pm
Dear Ms. Muthén

Thank you so much for your prompt reply.

May I just add one more question to make sure that I have interpreted things correctly:

Regression latent intercept and latent slope on b1* x1 + b2*x2 + b3*x3

With regard to the above model including two dichotomous covariates (coding is 1 and 0) and an interaction term based on these I would interpret my findings the following:

My results are: there is significant mediation with regard to the latent intercept as the interaction term is significant. Thus

1. Persons who are coded 1 on x1 and x2 start with an initial Intercept value of: constant (Intercept) + b1 + b2 + b3

2. For persons coded as 1 on x2 and 0 on x1 the effect on the intercept is b2, i.e. constant(intercept) + b2, for these persons I also find a significant effect on their rate of change - is it that b2 with regard to the effect on the slope represents the rate of change this persons experience when moving on the (linear) time scale by one unit (so as I look at five years 4*b2 is their whole change over this period)?

Thank you very much for your valuable time and help!

Best wishes
 Bengt O. Muthen posted on Friday, March 28, 2014 - 8:20 am
This is just like standard regression so we don't want to take up much time with it (e.g. "don't interpret main effects when there are interactions"). So with 2 dummies you should simply think in terms of 4 means of your DV, which in your case is your slope. So you have 4 slope means, which in turn implies that you have 4 growth rates over time. What growth rate is has been explained in our Topic 3 course handout and video on our website and relates to growth from the time point with time score 0 to the time point with time score 1.
 Theo Niyonsenga posted on Tuesday, August 12, 2014 - 11:13 pm
I am running a two-level growth model like the one in example 9.12 with the difference that observations were taken at different time points. When I use the key word "AT" with Tscores to define within and between intercepts and slopes, I get the following error message:
--------------------------------------------
*** ERROR in MODEL command
Random effect variables declared with the AT keyword for TYPE = TWOLEVEL
must be declared on the WITHIN level.
--------------------------------------------
I have no problem when using t=0, 1, 2,.... Does it mean that "AT T1,T2,T3,.." is not supported in version 7?
 Bengt O. Muthen posted on Wednesday, August 13, 2014 - 2:16 pm
You give the AT statement only in the Within part of the model.
 Theo Niyonsenga posted on Monday, August 25, 2014 - 6:37 pm
Prof Muthen,

Thank you for your answer.
What about the Between part, do you keep it at fixed time points t=0, 1, 2,....?
Do you sue the intercept & slope define in the within part as random variables at the between level?
The complications as well in my case is that I do have 2 growth curves and I would like to regress one on the other. Some covariates to control for are at the cluster level (between).

I am also wondering how I can get access to the Mplus scrpits from your 2009 paper with T Asparouhov published in JR Stat Soc A.

Thank you in advance for your help.
Theo.
 Linda K. Muthen posted on Tuesday, August 26, 2014 - 11:35 am
With AT, you specify the growth model only on within. You can use the growth factors on both levels.

The paper is on the website under Papers:

Muthén, B. & Asparouhov, T. (2009). Multilevel regression mixture analysis. Journal of the Royal Statistical Society, Series A, 172, 639-657.
 B Chenoworth posted on Monday, November 24, 2014 - 3:05 am
I have estimated a growth curve model with three time points. The unstandardised estimate of my slope in the unconditional model was -0.182. I added a time-varying covariate to the model. It was significant with the outcome variable at Time 1 only (and not at Time 2 or Time 3). When I added the time-varying covariate to the model the TECH4 mean of the slope became -0.06. What does this mean?
 Bengt O. Muthen posted on Monday, November 24, 2014 - 8:11 pm
The slope changes meaning with time-varying covariates. For instance, if the means of the tvc's increase over time with positive influence on the outcomes, your slope mean gets lower because the increase in the outcome means is taken care of by the tvc influence. Try centering the tvc's.
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