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I'm testing a model with five latent constructs of which two are measured by only two indicators each. How do I take this into account? David Kenny writes (http://davidakenny.net/cm/identify_formal.htm#B3b): "The construct has at least two indicators whose errors are uncorrelated and either a. both the indicators of the construct correlate with a third indicator of another construct but neither of the two indicators' errors is correlated with the error of that third indicator, or b. the two indicators' loadings are set equal to each other." Can I check these error correlations in the Mplus output (e.g., RESIDUAL) or do I have to add something to the syntax in order to force the errors of the two indicatiors to be uncorrelated? Thanks for the help! 


Your model is ok as is using the Mplus defaults. The residuals of the indicators are uncorrelated by default. Because you have other variables in the model than the two twoindicator sets the model is identified. 


Hi, I'm trying to fit a model with 2 latent factors. One factor has three indicators, and the other has two. Indicators are determined by previous theory. Typically, I have seen people deal with the 2 indicator problem by setting the indicators' loadings equal to each other. e.g., A BY F1 F2 F3; B BY F4* F5 (1); However, when I do that in mplus, I get a nonpositive definite error and a warning that the model may not be identified (though, I do get model fit stats, and they look good). When I run the model without setting the indicators equal to one another, it runs fine, with no errors and basically the same model fit statistics (which was surprising because my understanding was that this would be required to identify the model). E.g., A BY F1 F2 F3; B BY F4 F5; The errors also go away if I continue to set the indicators equal to one another and then also fix the variance of the latent factor to 1 (e.g., B@1). I've also seen this in the literature. I'm just curious what the best way to handle this is? Thanks for your thoughts. 


You don't need to hold the loadings equal when there are only 2 indicators. There are special concerns with 2indicator cases for bifactor models but you don't have that. The right way is simply A BY F1 F2 F3; B BY F4 F5; 

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