Just-identified model PreviousNext
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 Ting posted on Monday, July 16, 2012 - 8:58 am
Dear Dr. Muthen,

I understand that it is not meaningful to consider the model fit of a just-identified model (say, a path model), because the df is zero. But I haven't read a lot of discussion about the coefficients of just-identified models.
1) Can the path coefficients of this just-identified model be trusted?
2) Is achieving over-identification the top priority, i.e.,if this just-identified path model is what I am interested in, should I sacrifice some parameters(e.g., constrain some path coefficients to be equal or set a nonsig. one to be zero), in order to obtain an over-identified model?

 Linda K. Muthen posted on Tuesday, July 17, 2012 - 11:02 am
1. Yes.
2. The left-out paths should be based on the theory you are testing.
 K Wilkins posted on Sunday, January 27, 2013 - 3:44 pm
Hi Dr. Muthen,

I had a related question. I know that model fit indices such as chi-square are not calculated when analyzing a just-identified model. It seems AIC and BIC are calculated when I run my model, I was curious if I am able to report these?

 Linda K. Muthen posted on Monday, January 28, 2013 - 10:28 am
AIC and BIC are not absolute fit statistics. They are used to compare models. I can't see that they would be useful to report for a single model.
 Orpha de Lenne  posted on Monday, September 17, 2018 - 5:43 am
Dear Dr. Muthen,

I am trying to test a measurement model. The model is just-identified and as such I do not get model fit statistics. Is there another way to still test the fit of this factorial model?

Thank you.
 Bengt O. Muthen posted on Monday, September 17, 2018 - 2:16 pm
Perhaps you have one factor with only 3 indicators. Then there is no test of fit. If that's not the case, please send output to Support along with your license number.
 Timothy Stahl posted on Wednesday, October 23, 2019 - 6:25 pm
I want to follow-up on the comment re: one factor with only 3 indicators.

I understand there is no test of fit (the model is just identified). Does this mean that one should not use this latent factor in further analyses?
 Bengt O. Muthen posted on Thursday, October 24, 2019 - 11:21 am
Q1: No.
In fact, adding other model parts, that factor model will have testable implications because the correlations with variables in other model parts have to be explainable via that factor. So for instance, if the factor predicts an observed Y, you will have a testable model.
 Timothy Stahl posted on Thursday, October 24, 2019 - 11:32 am
Thank you for your quick response!

To follow-up and assure my understanding - does this then mean that I can move past the measurement model (of this latent factor, it is the only one in the model) and run the structural piece with the latent construct in it, and see if the latent factor fits, in addition to fit of overall model (with predictors and outcomes)?
 Bengt O. Muthen posted on Thursday, October 24, 2019 - 4:00 pm
If I understand you correctly, yes.
 Timothy Stahl posted on Thursday, October 24, 2019 - 4:07 pm
That's very helpful, thank you!
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