Poisson ACE model
Message/Author
 Bill Howells posted on Wednesday, July 26, 2006 - 10:03 am
I have twin data with a count variable that I want to fit with the traditional ACE model to estimate heritability of the phenotype, and model the count variable as Poisson. Therefore, due to the count variable, I specified this multiple groups model as a TYPE=MIXTURE with KNOWNCLASS = the two zygosity groups, mz and dz. After attempting to specify a residual variance or "specific environment" (E term in the ACE model), I learned that Poisson variables do not have the residual variance defined in Mplus, presumably because the mean and variance of a Poisson rv are equal.

Without the E term, it looks like the model is not identified. Examining the path model with tracing rules, it looks like the covariance for MZ twins is identical to the total variance of the phenotype, ie. both are 1*a^2 + 1*c^2, where a = the path coefficient for the additive genetic effect and c = path coefficient for common environment. And without the E term, Mplus gives me a "cov matrix not invertible" error, definitely a bad sign.

So, the Poisson ACE model does not exist and the Poisson AC model is not identified? And I should just "forgetahboutit"?

Bill H, MS, Dept Psychiatry, Wash U, St Louis

ps--analyzed as a continuous var with GROUP= specification as in Prescott (2004), the traditional ACE model runs fine
 Bengt O. Muthen posted on Wednesday, July 26, 2006 - 4:32 pm
You are right that Poisson regression does not have a residual variance parameter. So the E variance component in the ACE model is not a free parameter but is obtained as a remainder,

E = 1 - A + C;

where A and C are variances from those components. This is in line with the case of a categorical outcome which Prescott has written about in the WLSMV context. This is also doable in the ML context.

If it is helpful to you, I can send you an input for ML of a categorical outcome - just email me.