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 Anonymous posted on Friday, October 04, 2002 - 10:21 am
In my 5-time-point longitudinal data with equal time intervals, the outcome measure had a large change (drop) from t1 to t2; then, the change was much smaller over time after t2. So I decided to run a piecewise LGM to capture the non-linear shape of the trajectory. The following is a simple piecewise LGM I tried:

VARIABLE:
NAMES ARE Y1-Y5;
MISSING = ALL (-9);
USEVAR = Y1-Y5;
ANALYSIS: TYPE = MEANSTRUCTURE MISSING H1;
ITERATIONS = 2000;
MODEL:
I BY Y1-Y5@1;
S1 BY Y1@0 Y2@1 Y3@1 Y4@1 Y5@1;
S2 BY Y1@0 Y2@0 Y3@1 Y4@2 Y5@3;
[Y1-Y5@0 I S1 S2];
OUTPUT: SAMPSTAT STANDARDIZED TECH1 TECH3;

Mplus outcome show: "THE MODEL ESTIMATION TERMINATED NORMALLY.""THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL. PROBLEM INVOLVING PARAMETER 11" (Parameter 11 is the variance of S1, according to the TECHNICAL 1 OUTPUT).No fit statistics/indexes were printed in the outcome.
Could you please point out what is wrong in my model specification? Your help will be highly appreciated!
 bmuthen posted on Friday, October 04, 2002 - 6:02 pm
Your first piece consists of only 2 time points which does not allow identification of all the usual growth parameters. To solve this, you can fix the variance of s1 and its covariance with i to zero. This says that individuals do not vary in terms of their piece 1 slopes.
 Anonymous posted on Saturday, October 05, 2002 - 1:53 pm
Thank you very much for your great help! Actually, once Cov(s1,i) was fixed to zero, the model worked and the variances of all the growth factors were estimated. I would like to know if the model still works when only fixing Var(s1) to zero, but freeing Cov(s1,i)? Could you please show me how to fix a variance of growth factor in Mplus (I tried "s1 with s1@0;" it did not work). Many thanks again!
 Linda K. Muthen posted on Monday, October 14, 2002 - 8:07 am
You can fix a variance by mentioning just the name of the variable, for example,

s1@0;
 Anonymous posted on Sunday, June 27, 2004 - 8:26 pm
I have the same problem as the person above. Your solution worked when I did a piecewise with the 2-point piece at the beginning (note my 5 times correspond to ages with the numbers after d showing age and indicating years between each measurement):

MODEL: i s1 | d21@0 d27@6 d43@6 d52@6 d61@6;
i s2 | d21@0 d27@0 d43@16 d52@25 d61@34;
s1 WITH i@0;

But it did not work when I tried to 2-point piece at the end instead:

MODEL: i s1 | d21@0 d27@6 d43@22 d52@31 d61@31;
i s2 | d21@0 d27@0 d43@0 d52@0 d61@9;
s2 WITH i@0;


I get the following error:

THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL. PROBLEM INVOLVING PARAMETER 13.

FACTOR SCORES WILL NOT BE COMPUTED DUE TO NONCONVERGENCE OR NONIDENTIFIED MODEL.

What should I do differently for the second model?
 bmuthen posted on Monday, June 28, 2004 - 7:24 am
In your second alternative you have only one non-zero time score in your "i s2 |" statement, which means that s2 does not get defined as a change factor.
 Daniel Seddig posted on Tuesday, December 23, 2008 - 1:30 pm
hello,

the same situation as above: 5 measurements; two growth pieces.

how about specifying a model with an "overlap" of the two pieces? it could look like this:


i BY v1@1 v2@1 v3@1 v4@1 v5@1;
s1 BY v1@0 v2@1 v3@1.5 v4@1.5 v5@1.5;
s2 BY v1@0 v2@0 v3@1 v4@2 v5@3;

[v1-v5@0];
[i s1 s2];

so the knot is somehow "between" time points 2 and 3 and actually not measured but assumed. this also avoids the identification problems mentioned above. or would you judge this way of modeling as wrong? thanks for comments.
 Linda K. Muthen posted on Friday, December 26, 2008 - 2:26 pm
I think this model would be identified but I think it would be difficult to interpret the timepoints that are included in both pieces.
 S. Jeanne Horst posted on Sunday, March 07, 2010 - 1:55 pm
I am modeling change in self-reported effort across a testing session (7 time-points). A plot of means suggests a linear trend downward for the first three timepoints, followed by a small increase and almost linear trend from there. Both based on the graph and theory, a transition after the 3rd time point makes sense (i.e., switch from hard cognitive test to a noncognitive test). Also, I'm using FIML (not MCAR).

At this point I have tried many models and have large standardized residuals with each model, except one that may not be identified.

The best-fitting model is a 3-piece piecewise:
i= 1111111
s1=0122222
s2=0001111
s3=0000123

1. Is it correct that s2 is not identified? I tried fixing the errors to 0 for those time points, as a remedy, but fit is worse (i.e., global fit and high standardized residuals).

2. Even though unstandardized residuals are small for all models (0 - 0.085), the standardized residuals are high, particularly for two of the means. Does this sound right?

3. There are 999 in the standardized residuals, but do I correctly understand from the discussion board that this typically indicates that the divisor was 0? Or, should I be concerned that they are indicating something wrong with my models?

Thank you for any help, advice or suggestions that you can offer.
 Linda K. Muthen posted on Monday, March 08, 2010 - 8:26 am
It sounds like you have two different outcomes, a hard cognitive test and a noncognitive test. A single growth model should have the same outcome at each time point.

The growth model for s2 is not identified unless the variance of s2 is fixed at zero. You should not be fixing the residual variances of the outcomes to zero. I think the 999 in standardized comes about because of this.
 S. Jeanne Horst posted on Monday, March 08, 2010 - 9:08 am
Thank you for your help.

Here are several clarifications to my original questions:

1. The outcome is self-reported effort for each of the 7 time points. The cognitive and non-cognitive tests were two different "treatments" within the same testing session. The outcome is effort.

2. Also, I get 999's for most of my models. I only fix the 2 residuals for the one model that has the not-identified 0-1 slope (s2; fix the residuals for the time points related to the 0-1). Often the 999's are the standardized residual for the means for the first & last time point or for the variances.

Thanks for any help you can offer.
 Linda K. Muthen posted on Monday, March 08, 2010 - 10:01 am
Please send your full output and license number to support@statmodel.com.
 Kate Stringer posted on Thursday, April 22, 2010 - 5:02 am
Is there any literature that discusses the use of a piece that only contains two time points and the pros and cons of setting the covariance of the intercept and the slope of that piece to zero?
 Linda K. Muthen posted on Thursday, April 22, 2010 - 9:01 am
I don't know of any specific literature on this. It is a matter of model identification. With two repeated measures, the model is not identified unless you fix the variance of the slope growth factor to zero and the covariance between the intercept and slope growth factors to zero.
 Kate Stringer posted on Friday, April 23, 2010 - 5:48 am
Thanks, Linda. I have one more question-- I have a slope 1 (negative slope, times 1-2 for a piecewise model) and slope 2 (positive slope, times 2-6), and they are positvely correlated. What is the interpretation for this correlation?
And if the slope 2 for the negative slope growth curve and the slope 2 for the positive slope growth curve are negatively associated, what would be the interpretation of the correlation.
 Linda K. Muthen posted on Saturday, April 24, 2010 - 11:58 am
For the piece of the model with only two timepoints, the slope growth factor variance and the covariance between the intercept and slope growth factors should be fixed at zero so I can't understand what you are asking. Please send your full output and license number to support@statmodel.com.
 Sara Anderson posted on Monday, June 21, 2010 - 9:37 am
I am attempting a piecewise growth model with two different, but highly correlated, instruments (measures of reading skills). To complicate matters, there are two time points that overlap for these instruments. Unfortunately, I get a negative slope for one of the growth terms, which definitely shouldn't be the case (both increase). Complicating issues, the growth isn't linear so I'm freeing several of the growth parameters.

I've also tried to run latent growth models by creating latent variables for the common and unique time periods, but the model won't converge.

Is there any hope in using both instruments or should I just treat them as the same instrument or do a sequential model?
 Linda K. Muthen posted on Monday, June 21, 2010 - 11:08 am
I am confused because it sounds like you are treating them as the same instrument in the piecewise model. I would treat them as two separate processes given that there are two different outcomes and see how that goes. If you want help with the model that did not converge, please send the files and your license number to support@statmodel.com.
 Mario Mueller posted on Tuesday, July 13, 2010 - 2:46 am
I have 6 timepoints & 4 indicators. As recommended in topic 4, p. 82 I fitted LGMs for each indicator and sum scores. Mostly I got piecewise models as best fitting models.

Next, I run sequences of longitudinal CFAs to check for measurement invariance. Finally, I added growth factors (2 pieces) to the model that I found initially.

1. Is this an adequate method?
2. When I fit the model

t1 by a07 a16 a35 a62 ;
t2 by b07 b16 b35 b62 ;
t3 by c07 c16 c35 c62 ;
t4 by d07 d16 d35 d62 ;
t5 by e07 e16 e35 e62 ;
t6 by f07 f16 f35 f62 ;
i s1 | t1@0 t2@1 t3@2 t4@2 t5@2 t6@2;
i s2 | t1@0 t2@0 t3@0 t4@1 t5@2 t6@2;
s1 with i@0;
s2 with i@0;

...I get the error message:

THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL. PROBLEM INVOLVING PARAMETER 127.

P. 127 refers to Alpha at S1.
 Linda K. Muthen posted on Tuesday, July 13, 2010 - 6:14 am
Please send the full output and your license number to support@statmodel.com.
 Annie Desrosiers posted on Thursday, July 28, 2011 - 9:59 am
Hi Dr. Muthen, a run a piecewise model with 5 time points:
I BY Y1-Y5@1;
S1 BY Y1@0 Y2@1 Y3@1 Y4@1 Y5@1;
S2 BY Y1@0 Y2@0 Y3@1 Y4@2 Y5@3;

Because I have only 2 time points for my first piece, I know I have to fix variance and covariance with I of s1 @0.
I have 2 questions:
-1-
Can I have a covariable on s1, even if the variance is fixed at 0 ??
-2-
In the output (tech4), all the mean values of I, s1 and s2 are 0. Is that normal ? I would need the mean values for those latent variables like a linear model.
Thank you
 Bengt O. Muthen posted on Thursday, July 28, 2011 - 1:24 pm
1. Yes. Take for example a covariate male. This means that s1 has different means for the two genders.

2. Using the BY approach to growth modeling implies a CFA approach where factor means are zero by default. You get what you want if instead you use the bar ( | ) approach. See the User's Guide Version 6 pp. 602-607 for the relationship between BY and |.
 Kate Stringer posted on Wednesday, August 10, 2011 - 10:16 am
I have a model in which I specify the following piecewise model:

i s1 | ind1@0 ind2@1 ind3@1 ind4@1 ind5@1 ind6@1;
i s2 | ind1@0 ind2@0 ind3@1 ind4@2 ind5@3 ind6@4;

How do I specify time in order to center it at ind6 (time 6)?
 Bengt O. Muthen posted on Wednesday, August 10, 2011 - 4:16 pm
You say

i s1 | ind1@-1 ind2@0 ind3@0 ind4@0 ind5@0 ind6@0;

i s2 | ind1@-4 ind2@-4 ind3@-3 ind4@-2 ind5@-1 ind6@0;
 Stephanie Fitzpatrick posted on Saturday, February 11, 2012 - 12:17 pm
Hello,

I am trying to do piecewise growth modeling within growth mixture modeling. I have weight at baseline, 3 months, 6 months, 12 months, and 18 months. I would like to look at change in weight from baseline to 6 months and then from 6 months to 18 months. When I tried to overlap the piecewise models my intercepts were way off. The only way I get parameter estimates that make sense is if I model baseline to 6 months and then 12 to 18 months. But that leaves a gap between 6 to 12 months so I can only interpret the slopes for the first 6 months and the last 6 months as opposed to the last year. See my syntax below. Thank you in advance for your help.

MODEL: %Overall%
I1 S1 | KB@0 K3M@3 K6M@6;
I2 S2 | K12M@-6 K18M@0;

S1@0;
S2@0;
 Linda K. Muthen posted on Saturday, February 11, 2012 - 4:23 pm
The time scores for the piecewise model you describe should be:

MODEL: %Overall%
I1 S1 | KB@0 K3M@1 K6M@2 K12M@2 K18M@2;
I2 S2 | KB@0 K3M@0 K6M@0 K12M@1 K18M@2;

S1@0;
S2@0;
 Stephanie Fitzpatrick posted on Sunday, February 12, 2012 - 9:47 am
Hello Linda,

Thank you for your quick response. I tried the syntax above, but the intercepts I get in my output, especially I1 does not match what comes out in the graph. The I1 and I2 are very low to me in some classes to be considered intercepts. Am I interpreting these intercepts wrong? I thought they were the average starting points (I1 for Baseline and I2 for 6 months). Do these intercepts represent something different?

Thanks
 Stephanie Fitzpatrick posted on Sunday, February 12, 2012 - 10:19 am
Hello again Linda,

Also, when I try the syntax you gave me up above I am now getting this error message:

ONE OR MORE PARAMETERS WERE FIXED TO AVOID SINGULARITY OF THE INFORMATION MATRIX. THE SINGULARITY IS MOST LIKELY BECAUSE THE MODEL IS NOT IDENTIFIED, OR BECAUSE OF EMPTY CELLS IN THE JOINT DISTRIBUTION OF THE CATEGORICAL VARIABLES IN THE MODEL.

I tried fixing two of the residual variances equal, but the message still comes up, which I believe means my model is way underidentified. My slope variances are already fixed at 0. Anything else to try to fix this error message?

Thanks
 Linda K. Muthen posted on Sunday, February 12, 2012 - 10:32 am
Please send the relevant outputs and your license number to support@statmodel.com.
 Stephanie Fitzpatrick posted on Sunday, February 12, 2012 - 10:45 am
Hi Linda,

I seemed to have fixed the intercept problem just specifying one intercept; so instead of I1 and I2, I just have I. This also seemed to resolve the error message issue. Now my new question is how to interpret the slopes. Originally, I was interpreting as # of pounds loss on average per month. But with the new coding that you gave me, the slopes seem way too large to interpret that way (e.g., class lost 26 lbs on average per month). How should I interpret the slopes now with the 01222 and 00012 coding scheme now? Is it still change per month or something else?

P.S. not sure how to find my license number.

Thanks
 Linda K. Muthen posted on Sunday, February 12, 2012 - 11:20 am
I need to see an output. I will give you your license number when I respond to your support question.
 pauline posted on Tuesday, June 12, 2012 - 4:59 pm
hello

I would like to model two-piece growth with an upward slope 1 followed by a flat slope 2 between the last two time periods.

should I specify:

i s1| y1@0 y2@1 y3@2 y4@3 y5@3 y6@3;
i s2| y1@0 y2@0 y3@0 y4@0 y5@1 y6@1;

OR

i s1| y1@0 y2@1 y3@2 y4@3 y5@4 y6@4;
i s2| y1@0 y2@0 y3@0 y4@0 y5@0 y6@1;
s2@0;

Thank you
 Linda K. Muthen posted on Tuesday, June 12, 2012 - 5:42 pm
I think the first one with s2@0 gets what you want. You can't identify all of the parameters in a piece with only two time points.
 matteo giletta posted on Tuesday, January 15, 2013 - 2:17 pm
Hi,

Iím trying to model some data using a pre-post-post design in a SEM framework but Iím having problems to define my model. I would like to obtain three latent constructs, one representing baseline levels of the variable, one the reactivity (discrepancy between pre and post assessment), and one the recovery (discrepancy between post1 and post2 assessment).
It seems like with these assessments Iím unable to define a piecewise growth model, so I wonder how I can model these data to obtain an intercept and two slopes or latent constructs.

Is this correct:

f1 by ind1@1 ind2@1 ind3@1;
f2 by ind1@1 ind2@2 ind3@0;
f3 by ind1@0 ind1@1 ind3@2;

Thanks!
 Bengt O. Muthen posted on Tuesday, January 15, 2013 - 3:35 pm
There is not enough information from 3 time points to define piecewise growth modeling or 3 latent variables. For growth modeling you want to estimate the growth factor means and variances and covariances, plus at least one outcome residual variance. You have 3 means and 6 var-covs as sample information, so total 9, but 3 growth factors already uses 9 parameters. You would have to assume zero slope variances and/or zero covariances among the growth factors, which is not very satisfactory.
 Justin Benzer posted on Thursday, April 18, 2013 - 12:57 pm
Hi - I have data measured quarterly for five years and I am estimating a piecewise LGM. I would like to test the hypothesis that the slope for year 3 (s3) is different from the slope for year 4 (g1). I thought the easiest way to do this would be to constrain the means of the two slopes to be equal and calculate a chi-square difference. I tried to do this but it didn't work (see excerpt below).

Thanks!

i s3 | y11-y24@0 y31@1 y32@2 y33@3 y34@4 y41-y54@4;

i g1 | y11-y34@0 y41@1 y42@2 y43@3 y44@4 y51-y54@4;

[s1 g1] (1);
 Bengt O. Muthen posted on Thursday, April 18, 2013 - 3:57 pm
Shouldn't be problematic. Please send the output so we can see exactly what you are doing.
 xiaoyu bi posted on Thursday, January 02, 2014 - 5:53 pm
Hello,
I am running a piecewise growth curve model with 5 waves data. There is a negative decreased slope from T1 to T4, and an increased slope from T4 to T5. Does my codes look right (I fixed the variance of the second slope as 0).
Thank you!

I S1 | phsym1@0 phsym2@1 phsym3@2 phsym4@3 phsym5@3;
I S2 | phsym1@0 phsym2@0 phsym3@0 phsym4@0 phsym5@1;
S2@0;
 Linda K. Muthen posted on Friday, January 03, 2014 - 9:16 am
This looks correct.
 xiaoyu bi posted on Wednesday, January 08, 2014 - 4:57 pm
Hello, Linda,
Thanks for your response.
Another question I have is because the second slope only has 2 timepoints, I fix the variance of the second slope as 0. So, the slope growth factor variance and the covariance between the intercept and slope growth factors are fixed at zero too. What is the appropriate way to explain this statistical issue in the paper? Any advice?
Thank you so much!
 Linda K. Muthen posted on Wednesday, January 08, 2014 - 6:42 pm
With two time points, you could not identify those parameters.
 xiaoyu bi posted on Friday, January 24, 2014 - 5:36 pm
Dear Dr. Muthen,
How to run a piecewise LGM using long format data? I know how to do it using wide format data, but have no idea about how to do it using long format data. The reason that I need to use long format is all persons in my dataset have data in all 12 waves. I need to use "Age" as my time variable, but these persons did not join this study at the same age, and the age has a wide range from 18 to 100. So, if I want to test the piecewise LGM (age 18-44 as the first piece, age 45-64 as the second piece, and age 65 or older as the third piece), then how to do that using long format data? The two variables will be used: age, PC.

Thank you so much!
 Bengt O. Muthen posted on Sunday, January 26, 2014 - 12:28 pm
I assume that you know how to do regular growth modeling in long format using Mplus, so that time is a variable occupying a column in your data, the different time points corresponding to different rows in the data (see UG ex 9.16. So piecewise is no different, you just have a different time variable for each of the different pieces. This way you can get a random slope for each time variable, i.e. each piece. I haven't tried it but seems like that would work. This may also be described in the Raudenbush-Bryk book.
 fred posted on Wednesday, February 05, 2014 - 2:21 am
Dear Dr Muthen,
I am trying to run a single class picewise LGM as a first step to a mixture model.
My first slope only contains 2 time pointes, and the second slope consists of 3 time points. And I have constrained varainces and covariance to zero.

this is my input


MODEL:
i s1| T1@0 T4@1 T5@1 T6@1;
i s| T1@0 T4@0 T5@1.5 T6@2;
s@0;
i-s@0;

s1@0;
i-s1@0;

The output produces a CFI equaling 0.
How can I acquire fit indices for this model (and possibly compare it to a model with only one slope)?


Thanks
 Linda K. Muthen posted on Wednesday, February 05, 2014 - 12:23 pm
You need to fix only the variance of s1 to zero.

Please send the output with a CFI of zero and your license number to support@statmodel.com.
 fred posted on Friday, February 07, 2014 - 12:36 am
Thank you Dr Muthen!
Constraining s1 only seems to have fixed the problem.
However when I try to correlate the error terms of T1-T4 with:


i s1| T1@0 T4@1 T5@1 T6@1;
i s| T1@0 T4@0 T5@1.5 T6@2;
s1@0;
T1 T4 T5 PWITH T4 T5 T6;

I get the error message:

" THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL. PROBLEM INVOLVING PARAMETER 13.THE CONDITION NUMBER IS -0.146D-16.FACTOR SCORES WILL NOT BE COMPUTED DUE TO NONCONVERGENCE OR NONIDENTIFIED MODEL."


Do I have to make additional constrains to make the model with correlated error terms identified? Which constraints are appropriate?

Thanks

Fred
 Linda K. Muthen posted on Friday, February 07, 2014 - 5:39 am
You cannot include all of the residual covariances. They cannot all be identified.
 fred posted on Monday, February 10, 2014 - 12:23 am
Thanks Linda,
the problem of identification remains even if I correlate only the last 3 residual cov. i.e. T4 T5 PWITH T5 T6.

In a case like this, is it at all possible to have any pair of correlated residuals?

And does it make sense to compare a piecewise model which only can be identified with errors freely estimated with a one slope model with correlated errors?
Am I correct to assume that the piecewise model is nested within the single slope model (or vice versa)?


Thanks

Fred
 Linda K. Muthen posted on Monday, February 10, 2014 - 3:33 pm
You should be able to specify residual covariances. Please send the outputs with and without residual covariances and your license number to support@statmodel.com.
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