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 Dustin Pardini posted on Sunday, May 11, 2008 - 9:38 am
I am trying to fit a bifactor model of intelligence for the WISC-IV using the following model specifications:

VIQ by sim vocab comp;
PIQ by bd pc matreas;
WM by ds lns;
PS by dsc ss;

g by sim vocab comp bd pc matreas ds lns dsc ss;

g with viq-ps@0;
viq with piq-ps@0;
piq with wm-ps@0;
wm with ps@0;

However, the model gives 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 28.

THE CONDITION NUMBER IS -0.986D-17.

Any suggestions on what is going wrong?
 Bengt O. Muthen posted on Sunday, May 11, 2008 - 10:46 am
When specific factors have only 2 indicators you cannot identify the loading for the second of those indicators. Think of the specific factor as absorbing a residual correlation between those 2 indicators - there is only 1 such correlation and therefore you can only identify 1 parameter, in this case the specific factor variance.
 Li Lin posted on Tuesday, March 02, 2010 - 12:35 pm
Hi,
Can bifactor model be used in EFA? If it can, could you provide an example with Mplus code? Thanks!
 Bengt O. Muthen posted on Tuesday, March 02, 2010 - 12:48 pm
The bi-factor model is a CFA model, not an EFA model - it imposes more than m^2 restrictions. There is a general factor and uncorrelated specific factors. Or am I misunderstanding your question?
 Li Lin posted on Tuesday, March 02, 2010 - 1:19 pm
Thanks, Dr. Muthen. You have got my question correctly. I saw bifactor model was included in "Results of different exploratory factor models" Table in this paper "The role of the bifactor model in resolving dimensionality issues in health outcomes measures" (http://www.springerlink.com/content/jh561175967n4503/), and wondered.
 Kätlin Peets posted on Wednesday, February 23, 2011 - 8:25 am
We are interested in testing the degree of subject-specificity versus subject-generalizability of motivational constructs. We are comparing second-order factor models with bifactor models. We are not sure about the interpretation though. What can we say about subject-specificity of a construct if the second-order factor model does not worsen the fit compared to the bifactor solution?

We also have another question. If we find that a bi-factor model fits the data best but subject-specific factors do not have significant variance, should we still prefer this model over the others? In addition, as our sample size is not very high (less than 200), can the low variance estimates be influenced by sample size?
 Bengt O. Muthen posted on Wednesday, February 23, 2011 - 4:34 pm
I don't know how much power there is to reject a second-order model in favor of a bifactor model under various circumstances. For one thing, it depends on how many items you have per specific factor - at some point the models are not distinguishable. You want to read the literature on this, such as the Yung, Thissen, McLoad (1999) Psychometrika article. Or do a simulation study.

You would want the specific factor variances to be substantial relative to their SEs, but you are right that a small sample may not produce that in which case you can still argue for the model. Again, a simulation study might shed more light.
 Sheila Frankfurt posted on Wednesday, April 20, 2011 - 10:45 am
Hello,

I am running a bifactor CFA with categorical indicators; this is to examine whether PTSD items load onto both a general factor and specific symptom factors. I have fixed the correlations of the general factor with the symptom factors at zero. Do I need to fix the correlations of the symptom factors with eachother at zero? Conceptually, it makes sense that they would be correlated. However, in all the examples the lower-order factors are fixed to be uncorrelated.

MODEL: f1 BY u1-u5;
f2 BY u6-u7;
f3 BY u8-u12;
f4 BY u13-u17;
f5 BY u1-u17;
f5 WITH f1-f4@0;

Could you shed some light on this? Thank you very much for your continued help! Sheila
 Bengt O. Muthen posted on Wednesday, April 20, 2011 - 1:55 pm
In general, I don't think the correlations among the specific factors are identified. If you ask for Modindices when they are fixed at zero, you can see if any MIs are non-zero which would indicate that they could be identified.
 Tracy Waasdorp posted on Wednesday, September 14, 2011 - 10:06 am
We are working on a bifactor CFA we received a warning that the model may not be identified and the parameter involved is the in the PSI matrix between anger and psych. Is there an issue with the input syntax?

Thanks in advance.

ANALYSIS:
ESTIMATOR=WLSMV;
processors = 8(starts);

MODEL:
FEAR by P_023 P_022 P_019 P_020 P_018 P_008 P_015 P_017 S_067 S_065 S_068
P_101 P_102 S_066 S_015;
ANGER by P_054 P_052 P_051 P_056 P_047 P_055 P_049 P_046 P_045 P_048 P_050 P_059
P_057 P_058;
OVER by P_007 P_004 P_061 P_012 P_011 P_064 P_118 P_067 P_063 P_112;
COG by P_105 P_090 P_084 P_106 P_095 P_085;
PSYCH by P_023 P_022 P_019 P_020 P_018 P_008 P_015 P_017 S_067 S_065 S_068
P_101 P_102 S_066 S_015
P_054 P_052 P_051 P_056 P_047 P_055 P_049 P_046
P_045 P_048 P_050 P_059
P_057 P_058 P_007 P_004 P_061 P_012 P_011 P_064 P_118 P_067 P_063 P_112
P_105 P_090 P_084 P_106 P_095 P_085;

OUTPUT: standardized res TECH1 TECH2 modindices;
 Bengt O. Muthen posted on Wednesday, September 14, 2011 - 10:30 am
In general, the correlations between the specific factors need to be fixed at zero. Also, the correlations between the specific factors and the general factor need to be fixed at zero. In a model where therefore all factor correlations are fixed at zero, you can specify this conveniently by saying

Model=nocovariances;

See the V6 UG pages 540-541
 Tracy Waasdorp posted on Wednesday, September 14, 2011 - 11:52 am
This worked. Thank you very much.
 Joshua Isen posted on Thursday, October 25, 2012 - 5:39 pm
I'm implementing a bifactor model where the correlations amongst all factors are fixed at zero. (There are no other variables/covariates in the model besides the factor indicators.) The Mplus output indeed confirms that all factor correlations are zero. However, when I save the data as Fscores, and then simply use these factor scores in a follow-up analysis, the correlations amongst factor scores are non-zero. This seems puzzling to me. Why is this happening?
 Bengt O. Muthen posted on Friday, October 26, 2012 - 1:40 pm
Estimated factor scores do not behave like true factors unless they are very well measured. See for example:

Skrondal, A. and Laake, P. (2001). Regression among factor scores. Psychometrika 66, 563-575.
 Linda K. Muthen posted on Friday, October 26, 2012 - 1:42 pm
The correlations among the factors in the model and the correlations using factor scores are not the same unless factor determinacy is one.
 Joshua Isen posted on Saturday, October 27, 2012 - 1:03 pm
Thank you for the reference. I assume the "quality" of my estimated factor scores is based on the factor determinacies. Since there seems to be a rule of thumb that determinacies < .80 are unreliable, this suggests that I shouldn't use my estimated factor scores for further analysis.
 Michael Strambler posted on Monday, December 03, 2012 - 8:06 pm
Drs. Muthen,
We ran a bifactor analysis to specify a new measurement model. The bifactor model fits better than any other plausible model we have compared it to, but we get an error that the residual covariance matrix is not positive definite.

When I check the residual variances for the estimated R^2, there is one small negative residual variance (-.09 for a 17-item model and -.14 for a 15-item model). Also, the item with this problem has a loading on its content factor of .94 and .96 respectively, which seem questionably high. Should we be concerned with this? If so, what are some possible solutions?

Some other models we have tried that eliminate this message include:
1) Using THETA parametrization, which gets rid of this error message, and also reduces the factor loading on the content factor to .87. However, we are unsure about the implications of this switch and whether it is a legitimate "fix."
2) Freeing the item factor loadings and fixing the factor variances to 1 and factor means to 0, as well as allowing the content factors to correlate. This reduces the standardized content factor loading for the item in question to .76. This, however, complicates multiple group analyses.

Also, we did run the model as continuous data and got the same warning. In this model, the same item's residual variance was not estimated, and again the residual variance for the estimated R^2 was negative and small.
 Bengt O. Muthen posted on Monday, December 03, 2012 - 8:34 pm
Perhaps you want to try bi-factor EFA to see if there should be modifications to your bi-factor CFA model.
 Li Lin posted on Wednesday, January 23, 2013 - 12:17 pm
I am trying to fit a bifactor model using the following model specifications:
model: VULD by SFVULb SFVULj SFVULc SFVULk SFVULd SFVULl SFVULh SFVULm SFVULi SFVULn SFVULe SFVULo SFVULf SFVULp SFVULg SFVULq;
LABD by SFVULb SFVULj SFVULc SFVULk SFVULd SFVULl SFVULh SFVULm SFVULi SFVULn;
CLID by SFVULe SFVULo SFVULf SFVULp SFVULg SFVULq;

VULD with LABD-CLID @0;
LABD with CLID @0;

However, error message appeared in output - "NO CONVERGENCE. NUMBER OF ITERATIONS EXCEEDED."

Any suggestion to fix this?
 Bengt O. Muthen posted on Wednesday, January 23, 2013 - 3:20 pm
Please send output to support@statmodel.com
 Richard E. Zinbarg posted on Friday, February 15, 2013 - 10:12 am
I have more like a couple of questions about the bifactor model than a problem per se. I am using the factors in my bifactor model as predictors in a survival analysis. Given that the factors are all constrained to be orthogonal to me, it seems to me that there is no need to run a model with all of them entered simultaneously as predictors given that the results shouldn't be different for a set of orthogonal predictors than the zero-order results for each predictor entered by itself. Does that make sense? I hope so as given that there are several factors (13 actually) in my bifactor model there are too many integration points for memory capacity when I do try to enter all the factors simultaneously as predictors to try to confirm my intuition. When I use Monte Carlo integration with 5000 integration points to run the full model, the results do differ from the zero-order results for each predictor entered by itself. Is the Monte Carlo integration accounting for this pattern going contrary to my intuition? Or is my intuition just wrong in the first place? Thanks!
 Bengt O. Muthen posted on Friday, February 15, 2013 - 1:42 pm
Try it out for a model with only 2 orthogonal factors.

With 13 factors the Monte Carlo integration may not be very precise.
 Sarah Dermody posted on Tuesday, June 25, 2013 - 10:11 am
I have a similar question as Richard E. Zinbarg. I have a bi-factor model with uncorrelated three specific personality factors and a general personality factor. I want to know the unique predictive value of each of the variables on Y. The results do not make sense when I examine the three specific and the general factors simultaneously as predictors of Y (i.e., the sign of the specific factors become reversed but remain significant compared to the model without the general factor). Is it redundant to examine them simultaneously as predictors of they are technically supposed to be orthogonal? Why might my results be changing in this odd way?
 Bengt O. Muthen posted on Tuesday, June 25, 2013 - 2:04 pm
This should work given that all 4 factors are uncorrelated. You may be interested in the article by Gustafsson & Balke in MBR 1993 where they discuss such prediction.
 Alan Taylor posted on Sunday, July 07, 2013 - 5:30 pm
I am trying to fit a bifactor model in which one of the specific factors has only two indicator variables. I note this advice from Dr Muthen (May 11 2008):

When specific factors have only 2 indicators you cannot identify the loading for the second of those indicators. Think of the specific factor as absorbing a residual correlation between those 2 indicators - there is only 1 such correlation and therefore you can only identify 1 parameter, in this case the specific factor variance.

I have been trying various strategies, such as constraining the loadings on the specific factor to be equal or both one, but without success.

Is there a sensible way to proceed in this case?

Thanks.
 Bengt O. Muthen posted on Monday, July 08, 2013 - 6:29 am
Equality of loadings should work. Please send input, output, data, and license number to support.
 Benjamin Miller posted on Wednesday, August 07, 2013 - 3:16 pm
I estimated a bifactor model with categorical indicators. There are two specific factors and one general factor, and all factors are uncorrelated. This model fit the data well.

I am using the latent factors as outcome variables and I have found significant interactions in predicting the latent factors.

However since the latent factors have categorical indicators, I read that I won't get an intercept for the analysis nor will specifying [factor] produce the intercept coefficient. I would like to obtain the intercept so I can interpret the interaction effect.

I tried a "trick" specified in a previous post of putting a factor behind each observed variable (e.g. f1 by item1@1; item1@0;) and then used these factors as the indicators. I used the theta parameterization.

However, the model would not run and I get this error message:
THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL.

Any suggestions or ideas of what I may be failing to do in order to get the model to run? Or why the model won't run? I have 34 indicators. All 34 load onto the general factor and 8 and 24 items on the specific factors, respectively.
 Bengt O. Muthen posted on Wednesday, August 07, 2013 - 3:56 pm
You don't need to estimate an intercept to interpret an interaction effect. The factor intercept is zero.
 wong hua posted on Friday, September 13, 2013 - 6:37 am
I'm trying to fit a bifactor model. I was trying this command:

TITLE: bifactor analysis
DATA: FILE IS listening.dat;
VARIABLE: NAMES ARE y1-y5;
ANALYSIS: ESTIMATOR = MLM;
MODEL: g BY y1-y5;
f1 BY y1-y3;
f2 BY y4-y5;
g with f1 f2@0;
f1 with f2@0;

Mplus gives the follwong error:
THE DEGREES OF FREEDOM FOR THIS MODEL ARE NEGATIVE. THE MODEL IS NOT
IDENTIFIED. NO CHI-SQUARE TEST IS AVAILABLE. CHECK YOUR MODEL.

THE MODEL ESTIMATION TERMINATED NORMALLY

WARNING: THE LATENT VARIABLE COVARIANCE MATRIX (PSI) IS NOT POSITIVE
DEFINITE. THIS COULD INDICATE A NEGATIVE VARIANCE/RESIDUAL VARIANCE FOR A
LATENT VARIABLE, A CORRELATION GREATER OR EQUAL TO ONE BETWEEN TWO LATENT
VARIABLES, OR A LINEAR DEPENDENCY AMONG MORE THAN TWO LATENT VARIABLES.
CHECK THE TECH4 OUTPUT FOR MORE INFORMATION.
PROBLEM INVOLVING VARIABLE G.


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

I've seen that you answered a same question, suggesting that equality of loading should work.
But I don't know how to set equality of loading. Can you give an example? Thank you.
 Linda K. Muthen posted on Friday, September 13, 2013 - 10:24 am
The problem is that

g with f1 f2@0;

does not fix the covariance of g with f1 at 0.

Try

g with f1@0 f2@0;
 wong hua posted on Sunday, September 15, 2013 - 7:15 pm
I tried the following command, as you suggested,
TITLE: bifactor analysis
DATA: FILE IS listening.dat;
VARIABLE: NAMES ARE y1-y5;
ANALYSIS: ESTIMATOR = ML;
MODEL: g BY y1-y5;
f1 BY y1-y3;
f2 BY y4-y5;
g with f1@0 f2@0;
f1 with f2@0;
but it still didn't work,Mplus gives the follwong error:
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 19.
 Linda K. Muthen posted on Sunday, September 15, 2013 - 8:38 pm
Please send the output and your license number to support@statmodel.com.
 Tyler Moore posted on Friday, November 22, 2013 - 9:58 am
I have a bifactor measurement model within an SEM, in which a single predictor is "causing" all of the latent factors. Because the measurement model changes when I change predictors, I've fixed all paths within the measurement model (based on the coefficients estimated with no predictor at all). So, I believe the only things being estimated in my SEM are the paths from the predictor to the latent variables (and the residual variances of said latent variables).

This model runs fine, and the path coefficients are reasonable. However, I noticed that when I remove one of the paths from the predictor to a latent variable, the remaining paths change (sometimes substantially). Do you know why this is happening?

Thanks!
 Bengt O. Muthen posted on Sunday, November 24, 2013 - 3:11 pm
I would not fix the measurement parameters because they change when adding predictors. That is a sign that you have direct effects from some predictors to some indicators, that is, measurement non-invariance. Instead, explore the need for direct effects.

With a bi-factor model I can imagine that paths change as you describe because the factors are defined relative to each other - for instance if the general factor changes due to changing which predictors point to it, the specific factors change meaning and therefore their their relationships to predictors change.
 jml posted on Wednesday, January 29, 2014 - 10:40 am
Dear Drs. Muthen,

I am attempting to fit a bifactor model with two specific factors, but where some items only load on the general factor (as opposed to the general factor and one specific factor). I hadn't heard of this being done before, but saw it recently in Cai, Yang, and Hansen (2011) Generalized full-information item bifactor analysis, and I think it would make theoretical sense in my case. However, I can't get the model to converge. My (abbreviated) model and analysis statements are:

MODEL:
G BY Quant1 Quant2 Quant3 Quant5 Qual1 Qual2 Qual3 Qual4 Ctrl1 Ctrl5 Ctrl6 Dis2 Dis3;
DISTRESS BY Ctrl1 Ctrl5 Ctrl6 Dis2 Dis3;
ENGAGED BY Quant3 Quant5 Qual1 Qual2;
G WITH DISTRESS@0 ENGAGED@0;
DISTRESS WITH ENGAGED@0;

ANALYSIS:
TYPE IS GENERAL;
ESTIMATOR IS ML;
ALGORITHM IS EM;
ITERATIONS = 1000;
CONVERGENCE = 0.00005;

Can you see any obvious mistakes? The items that should not load on any specific factor are Quant1, Quant2, Qual3, and Qual4. I use Mplus 6.12. Thank you!
 Bengt O. Muthen posted on Wednesday, January 29, 2014 - 11:24 am
That should work fine and is generally speaking a good idea given that it more clearly defines what the general factor is.

Please send output and data to Support.
 Arshiya Ashkan posted on Sunday, March 02, 2014 - 2:45 am
Hello Dear Prof. Muthen

I'm trying to fit a bifactor model. How I can to obtain, percentage of variance accounted for by general and group factors.

Thanks!
 Linda K. Muthen posted on Sunday, March 02, 2014 - 5:04 pm
You can divide the sum of the squared factor loadings by the sum of the variances of the factor indicators. This works because all factors are uncorrelated.
 Linda K. Muthen posted on Sunday, March 02, 2014 - 5:05 pm
You can do this using MODEL CONSTRAINT.
 Marloes Muijselaar posted on Monday, March 10, 2014 - 6:54 am
Dear Prof. Muthen,

I'm trying to run a bifactor model with dichotomous indicators, which, unfortunately, does not converge.

However, a second order (hierarchical) factor model (with the generic factor of the bifactor model as second order factor) runs without problems.

Could you please help me?

Thanks in advance.

Marloes
 Linda K. Muthen posted on Monday, March 10, 2014 - 8:24 am
Please send the output for the non-converged model and your license number to support@statmodel.com.
 Ted Fong posted on Tuesday, April 01, 2014 - 3:10 am
Dear Dr. Muthén,

For bifactor modeling, Mplus 7.11 allows bifactor EFA (ESEM), CFA, and BSEM. I would like to ask if one can perform a Bayes bifactor EFA? I have tried running a bifactor EFA using Bayes estimator but the output file is blank despite running for nearly 20,000 iterations.

Is Bayes estimator available for bifactor EFA yet? If not, is WLSMV currently the only feasible choice for 20 categorical indicators with 4 latent factors?

Thanks a lot,
Ted
 Bengt O. Muthen posted on Tuesday, April 01, 2014 - 8:33 am
Please send the data, input and license number to Support@statmodel.com.
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