Type=RANDOM and standardization
Message/Author
 Daniel E Bontempo posted on Wednesday, November 28, 2007 - 10:37 am
Hi -

I am seeking further insight into why standardized estimates and confidance intervals are not available when type=RANDOM. Is it a computational problem or is it in some way thoretically invalid?

We have a tricotomus item checklist with 5 factor structure and data at 4 waves (intervals vary enough to require inidivdual time basis). Because of the large number of indicators for each factor and other data conditions, we are not able to run a 5-factor multivariate growth model. What we have done is run pairwise growth models, getting the covariance of the slopes in each model. Then, for tabulation purposes,we standardize the covariances using the estimated slope variances. I'm not sure these are true correlations, but they are in a common metric we can compare.

My concern is that the reason standardized values are not available might speak to some gross invalidity to what we are doing. Any further discussion/comment would be appreciated.
 Linda K. Muthen posted on Wednesday, November 28, 2007 - 2:45 pm
The basic idea is that for TYPE=RANDOM, there is not a single variance/covariance matrix but rather one for each value of x. This makes in unclear how one would standardize.
 Daniel E Bontempo posted on Wednesday, November 28, 2007 - 3:32 pm
I see the problem.

Is there some better way to compare associations of slopes if they are not all in the same model?

Setting aside the curves of factors (as describe above). Say I calculate the manifest score of factors A & B and a covariate C at each occasion. I run two multivariate growth models with TYPE=RANDOM. One has trajectory of A with trajectory of C, while the 2nd has trajectory of B with trajectory of C. A and B are not on the same scale because each has a different number of items. IF I want to compare the relative associations A with C and B with C, does our claculation offer any value?
 Bengt O. Muthen posted on Thursday, November 29, 2007 - 8:51 am
Sounds like you are asking about two different things. You can certainly use a standardization of a covariance between two slope growth factors - that correlation is fine. In contrast your last paragraph seems to talk about regressions. But it confuses me. It first talks about factors (I assume DVs) A and B and a covariate C, so it sounds like regressing A or B on C - but then you say "trajectory A with trajectory of C", which I don't understand.
 Daniel E Bontempo posted on Thursday, November 29, 2007 - 9:59 am
Sorry for any confusion. We have done this calculation for a few different growth models and I had them collectively in mind. Let me try to clarify to get us all on the same age. The same underlying data is used and we have 4 waves.

In one case we had factors at each occasion with tricotomous indicators, linear growth model, obtained the covariance of the growth parameters. We did this pairwise because we could not get the 5-factor model to run. I called this the curves of factors model, or 2nd order growth. The growth parameters are not regressed on any covariates. We obtain the covariances of the growth parameters from MPlus and standardize them for tabulation using the estimated variances of the respective growth parameters.

In the 2nd application we used the manifest scores of the factors as well as the manifest score of another instrument. We did five multivariate (two linear growth curves) models using each factor score respectively with the 2nd instrument. (I distinguish this as 1st order growth model). In this model we do regress the growth parameters on covariates such as gender and IQ deficit. We obtain the growth parameter covariances (conditioned on the covariates) and standardize them with the estimated residual variances of the growth parameters.
 Daniel E Bontempo posted on Thursday, November 29, 2007 - 10:00 am
When I refer to "trajectory with" I am only talking about the growth parameter covariances (level-level, level-slope, slope-slope) obtained from our MPlus Model. I guess I, somewhat confusingly, distinguished A & B as factors and C as a covariate because A & B are two dimensions of the instrument with 5 factor structure, and we have some models that include the item level and estimate the factors. C on the other hand is a global health assessment and is always in our models as a summary score. Sorry for any confusion.

If I understand you, you see no problem with our standardization of the covariance.

Is this true in both applications, including the one where the growth parameters are conditioned on a few covariates?

How does this relate to Type=Random and Linda's initial response about individual covariance matrices?
 Bengt O. Muthen posted on Thursday, November 29, 2007 - 8:20 pm
Didn't this thread start with you using Type = Random for which standardizations are not produced? You work with growth modeling so perhaps you used Type=Random to handle individually-varying times of observation. Linda's answer concerned having a random slope in a regression of y on x in which case the y variance is a function of the x values and standardization is therefore not well defined. But it sounds like you have a growth model with regular regression on covariates and standardization would be straightforward and fine. As long as you don't use the standardized coefficients for comparisons across groups - but it sounds like you work with the same group and just different DVs.

At one point you mention standardizing with respect to the residual variance of a DV, but you want to use the full variance.
 Daniel E Bontempo posted on Friday, November 30, 2007 - 1:33 pm
Bengt, thanks for the followup. I am doing growth modeling, and we do have individual times of observation. We are using Type=Random, and if we ask for STD in the OUTPUT, we get the message that it is not available for Type=Random. Or, at least we do inthe growth model where the growth parameters are regressed on gender and IQ.

Yes. it is the same group and just different DV in the growth model. Glad this is not a problem as I did not think of another way.

Concerning your point about full variances, do you mean that I should run an unconditional model, note the vairances, and then use those to standardize the covariances reported in the conditional model? Residual variances are reported in the conditional model. Or, are the full variances in one of the tech outputs?
 Bengt O. Muthen posted on Friday, November 30, 2007 - 3:13 pm
Yes, Type = Random does not produce standardized values. When the only reason for Type = Random is due to "AT" (individually-varying times of observation) as in your case, it would in principle be possible to provide standardized results (current Mplus just shuts it off whenever RANDOM is specified) - in that case we don't have a random slope for the covariate (only for the time variable), which is the sticking point. To make a long story short, in your case standardization of coefficients for growth factors regressed on covariates is appropriate and you could compute these by hand given the output in Mplus (see standardization formulas in the new Mplus Version 5 User's Guide, chapter 17).

Regarding your last question, you should use the conditional run and just get the full variances from Tech4.
 Daniel E Bontempo posted on Friday, November 30, 2007 - 5:35 pm
Thanks again. This is so helpful.

Unfortunately TECH4 is also not available when Type=RANDOM. But, I do not mund doing the uncnditional run if that is what I need to do.

If I can ask, since the covariance is of the residualized levels and slopes, why is it necessary to use the full variances instead of the residual variances? Is this not conceptually a covariance of residuals which generally would be standardized with residual variances?

That aside, the formulas you mention refer to regression coefficients. No formula is explicitly given for covariances. I am dividing the covariance by the square root of the product of the variances. Correct?
 Bengt O. Muthen posted on Friday, November 30, 2007 - 6:01 pm
Talking about growth factors regressed on covariates - no need to approximate by using the unconditional run - unless you have many covariates - it can be hand calculated.

Talking about WITH parameters - yes, if you are interested in residual correlations you should certainly use the residual variances.

That's right - square root of the product of the variances.
 Anke Schmitz posted on Wednesday, July 16, 2014 - 5:41 am
Dear Drs Muthen,

I’m estimating two interactions in a model.
But with “Type = random and algorithm = integration” the output just offers unstandardized estimates. Do I have to use them or how can I standardized the coefficients? Is it possible to standardize the regression coefficients of interactions between latent factor x dichotomous observed variable and interactions between dichotomous observed variable and metric observed variable?

I have the following variables in my model:

text (1/0) dichotomous observed variable
lg (0-22) metric observed variable
vw= factor estimated by indicators with 0/1/2 (categorical)
tv=factor estimated by indicators 0/1 (categorical)

Interaction 1: text*lg
Interaction2: text*vw

the message goes on in the next post...
 Anke Schmitz posted on Wednesday, July 16, 2014 - 5:41 am
The input is as follows:

title: Prüfung der Moderatoreffekte

data: file = CFA2.dat;
listwise = ON;

define: inter1=lg*text; !interaction between observed metric variable (0/1)and observed dichotomous variable

variable: names = code text textlng alter sex schule testmot1 vw2 vw3
vw5 vw6 vw7 ist inttext tvst testmot2 lgvtwo lgvtge lv1 lv2 lv3
lv4 lv5 lv6 lv7 lv8_1 lv8_2 lv9 lv10 lv11 lv12 lv13 lv14 lv15
lv16 lv17 lv18 lv19 lv20 lv21 lv22 int1 int2 int3 int4 int5 int6 int7 int8
int9 int10 cint lesk1 lesk2 lesk3 lesk4 lesk5 lesk6 lesk7 lesk8 clesk;
missing = all (99999);

usevar = text lg vw2 vw3 vw5 vw6 vw7
lv4 lv8_2 lv9 lv11 lv13 lv7 lv14
lv18 lv19 lv21 inter1;

categorical = vw2 vw3 vw5 vw6 vw7 !categorical means 0/1/2
lv4 lv8_2 lv9 lv11 lv13 lv7 lv14
lv18 lv19 lv21;

analysis: type= random;
algorithm = integration;

model:

TV by lv4 lv8_2 lv9 lv11 lv13 lv7 lv14
lv18 lv19 lv21;

VW by vw2 vw3 vw5 vw6 vw7;

inter2 | vw XWITH text; !interaction between factor and dichotomous observed variable

tv on text lg vw inter1 inter2;
 Bengt O. Muthen posted on Wednesday, July 16, 2014 - 11:08 am
We ask that posts be limited to one window.

Type = Random does not offer standardized coefficients because the variance of the DVs vary over observations.

With an observed dichotomous variable being part of the interaction you may be better off treating it as a multiple-group (or Knowclass) analysis, in which case XWITH and Type=Random is not needed.
 Anke Schmitz posted on Thursday, July 17, 2014 - 5:39 am
Dear Bengt,
sorry, that I have to ask once more, but for me its a bit tricky.

As the observed dichotomous variable interacts with the latent variable AND with the metric observed variable I do think multipl-group does not work.

That means I test two parallel interactions:
x (dichotomous observed) with factor
AND
x (the same dichotomous observed) with observed metric variable.

(Besides, I test a three-way interaction).

Is it possible to standardize the
regression coefficients of the interaction between the atent factor and the dichotomous observed variable (x) and interaction between the dichotomous
observed variable (x) and the metric observed variable?
 Bengt O. Muthen posted on Thursday, July 17, 2014 - 11:59 am
If you also have an interaction between the factor and a metric observed variable, what I suggested isn't relevant.

Note that you can standardize your metric observed variable before the analysis is done - if the model is "scale-free" as for instance typically is the case with covariates.

You can also study the latent variable interaction note discussing standardization - see our FAQ Latent variable interactions.

I would still recommend multiple-group (or Knownclass) analysis with the dichotomous variable.
 Anke Schmitz posted on Friday, July 18, 2014 - 2:40 am
Dear Bengt,
thanks for your kind reply.

"You can also study the latent variable interaction note discussing
standardization - see our FAQ Latent variable interactions."

I read the FAQ and how to standardize the regression coefficients with the SDs already. This fomula is very helpful.
But me and my colleagues are unsure how to estimate the variance of the latent dependend variable in the interaction model? All other relevant values I have and I standardized the metric observed variable already. Just the variance of the latent dependend variable is missing.
 Bengt O. Muthen posted on Friday, July 18, 2014 - 4:08 pm
The FAQ can also be used to derive the variance of the latent DV: it is a sum of several components.
 Jordan posted on Wednesday, June 03, 2015 - 3:09 pm
Hello, I am analyzing a first-stage moderated mediation model, using the XWITH command to generate the latent interaction factor (MLR estimator, type=random).

I understand that standardized estimates are not given with the type=random estimation. Is there any way to calculate them manually (despite having random? Or is that a hopeless endeavor...? All variables in the model are latent.
 Linda K. Muthen posted on Wednesday, June 03, 2015 - 3:59 pm
See the FAQ on the website called Latent variable interactions.
 Jordan posted on Thursday, June 11, 2015 - 4:56 pm
Thanks. This FAQ addresses how to find the variance of n3 (the mediator latent variable), and thus how to standardize the corresponding coefficients.

However, I also need to find the variance for n4 (the dependent variable). Is the method analogous to what's in the FAQ? Or is it changed due to there only being on independent variable (n3)?
 Bengt O. Muthen posted on Friday, June 12, 2015 - 8:38 am
Please send your output and data if possible to support so I can see exactly what your model is.
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