ACDE model and estimability Why can’t we estimate (co)variances due to A, C, D and E simultaneously in a standard twin design?

Covariances: MZ cov(y i1,y i2 |MZ) = cov(MZ) =  A 2 +  D 2 +  C 2

Covariance: DZ cov(y i1,y i2 |DZ) = cov(DZ) =½  A 2 + ¼  D 2 +  C 2

Functions of covariances 2cov(DZ) – cov(MZ) =  C 2 - ½  D 2 2(cov(MZ) – cov(DZ))=  A / 2  D 2

Linear model y ij =  +b i +w ij  y 2 =  b 2 +  w 2 y, b and w are random variables t =  b 2 /  y 2 –intra-class correlation = fraction of total variance that is attributable to differences among pairs

Data: “Sufficient statistics” (= Sums of Squares / Mean Squares) MZ –variation between pairs (= covariance) –variation within pairs (= residual) DZ –variation between pairs (covariance) –variation within pairs (residual) 4 summary statistics, so why can’t we estimate all four underlying components?

Causal components Between pairsWithin pairs  A 2 +  D 2 +  C 2  E 2 DZ½  A 2 + ¼  D 2 +  C 2 ½  A / 4  D 2 +  E 2 Difference ½  A / 4  D 2 ½  A / 4  D 2 Different combinations of values of  A 2 and  D 2 will give the same observed difference in between and within MZ and DZ (co)variance: confounding (dependency), can only estimate 3 components

In terms of (co)variances “Observed”Expected MZ var  A 2 +  D 2 +  C 2 +  E 2 MZ cov  A 2 +  D 2 +  C 2 DZ var  A 2 +  D 2 +  C 2 +  E 2 DZ cov½  A 2 + ¼  D 2 +  C 2 MZ & DZ variance have the same expectation. Left with two equations and three unknowns

Assumption  D 2 = 0 : the ACE model Between pairsWithin pairs  A 2 +  C 2  E 2 DZ½  A 2 +  C 2 ½  A 2 +  E 2 4 Mean Squares, 3 unknowns –Maximum likelihood estimation (e.g., Mx)