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 Anonymous posted on Thursday, July 01, 2004 - 6:31 am
I have been playing around with the demo version of Mplus 3 to better understand the nonlinear factor analysis capabilities -- I think this is a great addition. I've run a multiple observed groups factor analysis with a quadratic effect using the knownclass option (2 groups). The factor is f1.

The output reports an estimated mean for the quadratic term f1f1 in the first class but the mean of f1f1 is fixed to zero in the second class. This confuses me. The mean of f1f1 will generally not be zero. Additionally, if f1 is normally distributed (which I thought was an assumption of the Klein ML approach implemented in Mplus 3) shouldn't the mean of f1f1 be a direct function of the mean and variance of f1 (vis a vis equations in Joreskog & Yang) and thereby fixed?

So my questions are (1) Should the mean of f1f1 be estimated at all (is this to relax the normality assumption on f1)?, and (2) Should it really be fixed to 0 in the second class?

Thanks in advance.
 bmuthen posted on Thursday, July 01, 2004 - 9:18 am
Please send the output and also preferrably the input and data to support@statmodel.com so we can have a closer look at this.
 Michael J. Zyphur posted on Tuesday, March 07, 2006 - 12:26 pm
Hi Bengt/Linda!
I was wondering about non-linear CFA, where each additional latent variable is an exponential function of an initial latent variable. As shown in the manual, such a model is made simple in Mplus with the XWITH command (e.g., f2 | f1 XWITH f1). However, with this command, the models seem to be limited to specifying the observed variables as being a function of only a linear and quadratic latent variable, as there seems to be no way to compute a cubic latent factor (although a quartic latent factor seems possible - f4 | f2 XWITH f2). Any thoughts on how to do construct a cubic factor, or, for that matter, any odd-numbered exponentiation?

Thanks for your time and Mplus 4!

mike
 Bengt O. Muthen posted on Tuesday, March 07, 2006 - 12:54 pm
I would think

f2 | f1 XWITH f1;
f3 | f2 with f1;

would give 3rd order poly and similarly to get 5th order.

General nonlinear functions of latent variables is more complex, but can perhaps be tricked out of the program.
 Michael J. Zyphur posted on Tuesday, March 07, 2006 - 1:32 pm
Of course, the exponents are additive with multiplication... whoops.

Two questions now (1): Does anything to do with the XWITH command, and the associated numerical integration, invalidate the assumptions (e.g., regularity) associated with testing -2LL differences for models with different numbers of exponentiated factors? Testing such differences is likely a better way to go about investigating the usefulness of additional non-linear factors than looking for significant amounts of variance for the new factors, yes?

(2) The exponentiated factors will always be orthogonal by default. Will this change the substantive interpretation of the results from one where the factors were correlated? In more traditional regression it wouldn't have to when, for example, constructing an orthogonal "power polynomial" equation, yes?
 Bengt O. Muthen posted on Thursday, March 09, 2006 - 6:34 am
1. -2LL is still fine.

2. With exogenous factors f1 and f2,

f1x | f1 xwith f1;
f2x | f2 xwith f2;

f1 and f2 are correlated and so f1x and f2x will be implicitly correlated.
 Edward Barker posted on Wednesday, October 18, 2006 - 5:19 am
Hi:

Is it possible to do a multiple group analysis with non-linear factor analysis (page 54 of the user guide)?

I am trying to establish measurement invariance accross time prior to estimating GGMM models. I am using measurement occassion as the GROUP.

I get this error when applying the user guide example:

ALGORITHM = INTEGRATION is not available for multiple group analysis.
 Linda K. Muthen posted on Wednesday, October 18, 2006 - 5:27 am
When the GROUPING option cannot be used, the KNOWNCLASS option is available. However, if you are testing measurement invariance across time, it is not correct to do it using multiple group analysis. In multiple group analysis, the subjects in the groups must be independent. Instead, set the equalities of intercepts and factor loadings across time and test nested models. See Example 6.14. Just take away the growth part of the model.
 Edward Barker posted on Wednesday, October 18, 2006 - 5:37 am
Please disregard previous message -- I have found Bauer (2005):

VARIABLE:
CLASSES= C(2);
KNOWNCLASS=C (time1=1 time2=2);

ANALYSIS:
TYPE=MIXTURE RANDOM;
ALGORITHM=INTEGRATION;
 Edward Barker posted on Wednesday, October 18, 2006 - 5:38 am
Ok - Thanks!
 Ringo Ho posted on Friday, December 08, 2006 - 9:41 pm
hi Prof Muthen

I have two questions regarding the estimation of nonlinear CFA procedure in mplus:

1) the estimation method is based on
Klein-Moosbrugger (2000) Psychometrika paper, right?

2) if i fit a quadratic factor analysis model with multiple items:
y_j=b1_j*f + b2_j*fsq + e_ij
(fsq is created by : fsq | f xwith f)

I choose to identify f is by setting var(f) = 1 (and mean(f)=0) but free all factor loadings.

Under standard normal assumption on f, mean(fsq)=1 and var(fsq)=2. Should I impose these constraints in the mplus prg as well [your example 5.17 does not have these constraints]? I wonder internally if Mplus set the mean(fsq) and var(fsq) to some specific value. If so, what are these values?

Thanks a lot for your help.
 Linda K. Muthen posted on Saturday, December 09, 2006 - 9:44 am
Mplus uses the same estimator, FIML, but a slightly different algorithm.

The interaction does not have mean and variance parameters so you don't need to do anything.
 luke fryer posted on Monday, October 11, 2010 - 5:07 am
Dr.s Muthen,


Under what conditions would you suggest I employ Non-linear CFA (EXAMPLE 5.7: NON-LINEAR CFA) instead of normal CFA (EXAMPLE 5.2: CFA WITH CATEGORICAL FACTOR
INDICATORS). I am working with dichotomous data (10 indicators per latent variable;3 variables; Monte carlo suggests sufficient sample for Example 5.2).


Luke
 Linda K. Muthen posted on Monday, October 11, 2010 - 11:18 am
I would only use this if I had a specific hyothesis to test.
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