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Hi OpenMX Community!

I have 3 traits, and I would like to assess whether maybe 2 come from the same genes, while the third has different genes.

That is, I would like to run a IP model with 2 Ac factors, one loading onto 2 traits, and the other one loading onto the last trait. I understand that this would not be identified.

-Is it possible to simplify this by having only 1 Ac factor that loads onto 2 traits, while the last trait only has As factors?

-Would I have to constrain the factor loadings of Ac to be equal since Ac only loads onto two traits?

-I noticed, that this artificially inflates the Cc factor loading of the trait that does not have a Ac component. Can this be somehow corrected?

-Should this be solved with a CP model, where LP only loads onto 2 traits? Would I need to equate the loadings of LP also in this case since there are only 2 factor loadings?

A attached a script assessing these problems using the MCV Cardiovascular Twin Study data.

Thank you for your help!

More recognized experts could answer your questions more precisely for sure, but I have also come across similar problems and got to the conclusions:

Simplifying your model to include only 1 common A factor that only loads onto 2 phenotypes does not model your question correctly. Since there is “similarity” between all three phenotypes, but since Ac only loads onto 2 phenotypes, variance common between all variables would be decomposed into the common C factor which loads onto all 3, thus artificially inflating Cc. Probably you are better off with the Common pathway model, where your latent phenotype would only load onto 2 phenotypes leaving the third only with specific estimates.

In an Independent pathway model, I believe you must constrain the factor loadings of the common factor with only 2 indicators, or you could estimate a correlation between the residual factors. In case of a Common pathway model, if the latent phenotype only loads onto 2 phenotypes I don’t think you need to constrain the factor loadings to be equal.

All in all, it is a really interesting question, how to model: one phenotype is independent of the other two (or more). I would also be very interested what others might think about this problem.

If I understand correctly you have:

There's nothing necessitating such a model being under-identified. You could implement as a reduced ACE model, or a 2-factor Common pathway model, or, as you wish, as a 2-factor IP model.

An easy way to play with this is to install the umx library, and experiment with umxCP and umxIP.

In particular the nFac parameter, i.e.,

Then just remove loadings from the first common factor onto the third trait, and from the second common factor on the first two traits.

umx has a uniform labelling system so they're coded like "cp_loadings_r1c1"

Set the appropriate paths to zero (the umx helper umxModify can make this more brief once you get used to it.

Sing out if you need more help.

PS: It's a common misunderstanding (mostly it seems because people are using 1-factor example scripts) that IP and CP models can have only 1 latent common factor.

Interesting fact: The basic IP model is a reduced three-factor CP model, where factor 1 has C& E set to zero, factor 2 has A& E set to zero, and factor 3 has A& C set to zero. Hence IP nested inside CP (hat-tip @Mike_Neale)

Hopefully, T. Bates' post answers most of your questions. I have a few remarks myself.

Yes, that should work. The point is that a trait can't simultaneously have a specific A factor as well as a "common" A that has one-headed paths going only to it, since that WOULD be unidentified.

A common factor with freely estimated loadings from only two traits is not identified without some further constraint. The simplest thing is to equate the two loadings. As martonandko says, you could alternately drop the two traits' common A factor, and allow a nonzero correlation between their specific A factors.

Attempting to correct this would be even more "artificial." Like martonandko said, you're seeing this happen because familial common variance in the third trait will get "soaked up" by the common C factor.