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Are any pairs of these four structural models nested in another? I think the second-order single factor model is nested in the second-order two factor model (second order single factor model is when covariance between the two second order factors in second-order two factor model is 1) but I can't determine whether other pairs are nested or not.

  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ – gung - Reinstate Monica Jun 22 '16 at 4:01
  • $\begingroup$ It is not from a course or textbook. It is from my own research project. $\endgroup$ – J. S. Jun 23 '16 at 2:25

I have taken the liberty of editing the figure to add labels a)-d) and re-uploading it within the question.

Three definitions of nesting, in SEM are:

"whenever one model has all the same free parameters as does a second model but also has other free parameters not shared by the other model. In other words, the two models are equivalent except for a subset of free parameters in one model that are fixed or constrained in the other (Maruyama, 1998, p. 235). ).

Loehlin (1992) defines nested as being when "the model with the smaller number of free variables can be obtained from the model with the larger number of free variables by fixing one or more of the latter" (p. 67).

Kline (2011) defines nested as "if one is a proper subset of the other. For example, if a free parameter is dropped from model A ... to form B ... model B is nested in model A" (p. 214).

You are correct that model b) is nested in model c). The one-factor model is a restricted version of the two-factor model. To see this, consider what happens if we fix the correlation between the two factors in model c) to 1. In this case the 2 factors will be identical which is the same as replacing them both with a single factor.

By similar reasoning, we can also see that d) is nested in c) if we fixed both paths from Fatique in model d) to 1.

Loehlin, J.C., (1992), Latent Variable Models: An introduction to factor, path and structural analysis, 2nd Ed, Hillsdale, New Jersey

Kline, R.B. (2011), Principles and Practice of Structural Equation Modeling, 3rd Ed, Guildford Press

Maruyama, G. (1998), Basics of Structural Equation Modeling. Thousand Oaks CA: Sage.


a) is the saturated model. All models are nested within the saturated model. That's the basis of the chi-square test. You're testing against the saturated model, which has 0 df and 0 chi-square.

c) and d) are equivalent models. d) isn't identified without a constraint on the loadings, so there is only one parameter to estimate.

All that we are left with is the relationship between b), and c)/d). These are nested, but not (quite) in the usual way. Constraing the correlation to 1 in model b) gives you nested models, but you can't test these nested models using the chi-square test, because the parameter is on the boundary of the permitted space. In a sense it's a one-tailed test - when you relax the correlation in model c) we know that the parameter will change, but it must get smaller, not larger. The nested model chi-square test assumes that the parameter can change in any direction. It's surprisingly tricky. Here's a paper: http://psycnet.apa.org/?&fa=main.doiLanding&doi=10.1037/1082-989X.13.2.150

  • $\begingroup$ Thank you for your feedback! What I am particularly interested is whether or not I can do a chi-square test between a and each of b, c, d. If a is the saturated model and all three other models are nested within a, I can do chi-square tests between a & b, a & c, and a & d, correct? I would greatly appreciate your feedback on this. Thank you very much! $\endgroup$ – J. S. Jun 23 '16 at 1:46
  • $\begingroup$ Yes, you can do those tests. $\endgroup$ – Jeremy Miles Jun 23 '16 at 1:47

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