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ACDE model and estimability

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# ACDE model and estimability - PowerPoint PPT Presentation

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

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

Covariances: MZ

cov(yi1,yi2|MZ) = cov(MZ)

= sA2 + sD2 + sC2

Covariance: DZ

cov(yi1,yi2|DZ) = cov(DZ)

= ½sA2 + ¼sD2 + sC2

Functions of covariances

2cov(DZ) – cov(MZ) = sC2 - ½sD2

2(cov(MZ) – cov(DZ)) = sA2 + 3/2sD2

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
• 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 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

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

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)