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ACDE model and estimability

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

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ACDE model and estimability

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

  2. Covariances: MZ cov(yi1,yi2|MZ) = cov(MZ) = sA2 + sD2 + sC2

  3. Covariance: DZ cov(yi1,yi2|DZ) = cov(DZ) = ½sA2 + ¼sD2 + sC2

  4. Functions of covariances 2cov(DZ) – cov(MZ) = sC2 - ½sD2 2(cov(MZ) – cov(DZ)) = sA2 + 3/2sD2

  5. Linear model yij = m + bi + wij sy2 = sb2 + sw2 • y, b and w are random variables • t = sb2/sy2 • intra-class correlation = fraction of total variance that is attributable to differences among pairs

  6. 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?

  7. Causal components Between pairs Within pairs MZ sA2 + sD2 + sC2sE2 DZ ½sA2 + ¼sD2 + sC2 ½sA2 + 3/4sD2 + sE2 Difference ½sA2 + 3/4sD2 ½sA2 + 3/4sD2 Different combinations of values of sA2 and sD2 will give the same observed difference in between and within MZ and DZ (co)variance: confounding (dependency), can only estimate 3 components

  8. In terms of (co)variances “Observed” Expected MZ var sA2 + sD2 + sC2 + sE2 MZ cov sA2 + sD2 + sC2 DZ var sA2 + sD2 + sC2 + sE2 DZ cov ½sA2 + ¼sD2 + sC2 MZ & DZ variance have the same expectation. Left with two equations and three unknowns

  9. Assumption sD2 = 0 : the ACE model Between pairs Within pairs MZ sA2 + sC2sE2 DZ ½sA2 + sC2 ½sA2 + sE2 • 4 Mean Squares, 3 unknowns • Maximum likelihood estimation (e.g., Mx)

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