Sat, 10/30/2010 - 04:56

Hi,

I've tried setting up an ACDE model as I have siblings (up to 3) data for my twins.

Sorry, basic question but I've managed to confuse myself with how to interpret best fit. I've embedded all submodels within the ACDE model. As per below, the AE model has the lowest AIC, so that is the best fit right? So the p value is lower than the ACE model because of the extra df compared to the main model?

> # Compare all models

> ACDENested <- list(multiCholACEFit,multiCholADEFit, multiCholAEFit, multiCholCEFit, multiCholEFit)

> tableFitStatistics(multiCholACDEFit,ACDENested)

Name ep -2LL df AIC diffLL diffdf p

Model 1 : multiCholACDE 18 10103.08 4076 1951.08 - - -

Model 2 : multiCholACE 15 10103.8 4079 1945.8 0.72 3 0.87

Model 3 : multiCholADE 15 10106.97 4079 1948.97 3.89 3 0.27

Model 4 : multiCholAE 12 10106.97 4082 1942.97 3.89 6 0.69

Model 5 : multiCholCE 12 10165.75 4082 2001.75 62.68 6 0

Model 6 : multiCholE 9 10459.53 4085 2289.53 356.45 9 0

Hi

In the univariate case, siblings don't identify the ACDE model because they are assumed to have the same A, C & D predicted covariances as DZ twins.

E(cov(DZ)) = E(cov(Sib)) = .5a^2 + c^2 + .25d^2

In the event that we imagine c to be different for sibs than twins, there would be a different expectation for sib pairs, say .5a^2 + s^2 +.25d^2 this doesn't help (the additional statistic provided by the correlation between sibs serves to identify this different c for sibs). The standard errors probably look wacky for the ACDE model. If you had a different type of relative, for whom the coefficients of resemblance (e.g., the .5, 1, and .25) are not a linear combination of other expectations (the MZ & DZ), then you could identify the model.

I note, however, that you get slightly better fit (lower -2lnL) of the ACDE model than of the ACE model. Judging from the additional 3df it appears that you are using a bivariate model. In this situation, a common D factor may be estimated, but not the entire 3 parameter Cholesky.

Thanks Mike,

I misunderstood the identification concept. It's the number of 'unique' covariances - not just extra covariances generated by the addition of family members with the same covariance as DZ's (ie siblings). So while they might help with power, their addition doesn't assist with identifying a model with 3 parameters (if I now have this clear in my mind).

Right, but note:

1. The number of statistics N with *unique expectations under the model* has to be greater than or equal to the number of free parameters. This is a necessary but not sufficient condition.

2. The N different expected covariances under the model must not be linear combinations of each other. Adding a group of relatives with, say .25a^2 + .25c^2 + .25d^2 (if such a group existed), would not help the classical twin study where MZ's covary a^2 + c^2 + d^2 and DZ's covary .5a^2 + c^2 + .25d^2, because the additional fictitious group has expectations which are simply .25 times those of MZ's.

3. If there are observed statistics with "duplicate" expectations (like the variances of twin 1 and twin 2, which are equal in the ACE model), these contribute to the degrees of freedom of the model, one for each duplicate statistic.