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Multivariate heritability in an independent pathway model

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k.j.kan's picture
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Joined: 01/29/2016 - 12:00
Multivariate heritability in an independent pathway model

Dear Guinea-pig lovers,

Heritability is defined on the basis of phenotypic variance. One may regard a so-called common pathway model as a model that contains one or more latent phenotypes that mediate certain genetic and environmental effects on observed variables that function as indicators of those latent phenotypes. As a result the common pathway model enables us to calculate the heritability of both the mediating latent phenotype as well as the observed phenotypes.

One can also interpret a common pathway model as a means to calculate the heritability of certain common variance among a number of observed variables, thus of phenotypic CO-variance, right?

An independent pathway model only contains observed phenotypic variance, no latent phenotypic variance or phenotypic covariance (though it does contain genetic covariance and/or environmental covariance terms that explanation the existence of covariance among the observed phenotypes). My question is if one can still speak of 'heritabilty of the common variance' in an independent pathway model. If so, what is the appropriate way to calculate this heritability? Intuitively I would calculate communalities (sums of the squared standardized loadings) and divide the 'genetic communality' by the sum of the genetic and environmental communality. However, this procedure also sounds like an old-fashioned, rough proxy method.

Any advise on this?

neale's picture
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Joined: 07/31/2009 - 15:14
Elementary, Dear Watson...

The independent pathway ACE model can be considered a submodel of a 3-factor common pathway model. It is simply a 3-factor common pathway model in which for one factor the heritability is 100% while C and E are zero (1:0:0 A:C:E), another factor has 0:1:0 for A:C:E and the third has 0:0:1 A:C:E. Therefore while it is appropriate to speak of heritability of the factors in this case, it is either 1 or 0 depending on which factor you look at.

I would note in passing that many folks seem to stop at the single factor common pathway model and compare it to a 3 factor independent pathway model. This approach is biased in favor of finding greater support for the independent pathway model. So it is important to test the 3 factor common pathway model as well.

k.j.kan's picture
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Joined: 01/29/2016 - 12:00
identification

Hi Mike, many thanks for your reply.

I believe I can follow your explanation (see attached powerpoint slides, for an example): What you're saying is that a 3-factor (A, C, E) independent pathways model is nested within a 3-factor (ACE, ACE, ACE) model by assuming that C and E (factor 1), A and E (factor 2), and A and C (factor 3) do not explain any of the variance (0), so that the heritability coefficients of those factors (a^2) are 1 (and c^2 =0, e^2 =0), 0 (and c^2=1, e^2=0), and 0 (and c^2=0, e^2=0), right? (Which would imply that certain factors become redundant and the model can be reparameterisized as the one we usually fit).

Next question then: How would such a 3-factor (ACE, ACE, ACE) model be identified? The 3 factors will have the same indicators.

Best,
Kees-Jan

neale's picture
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Joined: 07/31/2009 - 15:14
Identification

So a phenotypic factor model (unrelated persons) is identified when:

  1. The number of measures is greater than the number of factors by an amount which increases with the pattern 2 1 2 11 2 1 1 1 2 etc. This function will test that
isIdentified<-function(nVariables,nFactors) as.logical(1+sign((nVariables*(nVariables-1)/2) -  nVariables*nFactors + nFactors*(nFactors-1)/2))
# if this function returns FALSE then model is not identified, otherwise it is.
isIdentified(nVariables,nFactors)
  1. The rotational indeterminacy is dealt with - the usual approach is to make the factors orthogonal by fixing the first factor loading to zero for the second factor, the first two factor loadings to zero for the third factor, etc.
  2. Sign invariance is also dealt with by, e.g., bounding the first free factor loading of every factor to be non-negative

The same conditions will work when there are genetically informative research designs, such as twins, with a couple of provisos. First, the minimum number of variables m required to identify k factors is fewer, apparently simply k <= m-2. Second, rotational indeterminacy doesn't hold because rotation affects the cross-relative cross-variable covariance structure. I should have written this up like 20 years ago, but perhaps 2016 will be its year :).

k.j.kan's picture
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Joined: 01/29/2016 - 12:00
Escher?

So, this is starting to look like a common pathways Cholesky decomposition, which would be nested in an independent pathways Cholesky. Following the previous rationale, that independent pathway Cholesky would be a special case of a more general common pathway model. Where does this end?

k.j.kan's picture
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Joined: 01/29/2016 - 12:00
purpose

One remark in addition to the comment in the post below (why does this message appear on top of the previous?).

The purpose of my question was the following: We often calculate the heritability of a phenotypic correlation, 'bivariate heritability'. I wondered how to generalize this to the multiviarate case. Is this possible?

neale's picture
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Joined: 07/31/2009 - 15:14
Idk what's up with the ordering

The OpenMx website will transition in the next few months to a different server - perhaps the new version of Drupal to which we are upgrading will unscrew the ordering thing.

I am not a fan of bivariate heritability because it isn't always defined. For example, suppose A:C:E is 33:33:33 for both traits. Also suppose that rG=.8 rC=-.4 and rE=-.4 In this case, there is no phenotypic correlation so bivariate heritability and the C&E proportions of it are all infinite (divide by zero). Or if rE=0 then bivariate "heritability" exceeds 1.0, which for a coefficient that is supposed to be a proportion is not ideal. My preference is to state what the contributions of each component are to the phenotypic correlation, .8.33 and -.4.33 in the above case, without resorting to proportions that aren't.

k.j.kan's picture
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Joined: 01/29/2016 - 12:00
Also not a fan

Hi Mike, yes it is also my preference is to calculate the contributions of each component to the phenotypic correlation (because that is something I can understand). Nevertheless, I was wondering how to generalize bivariate heritability to multivariate heritability (this because others were asking me to calculate such a coefficient).

And thanks for the example!