 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-221/2/i/solution

Take all components to be centered and let their marginal variances be

V_{A}, V_{C}, V_{E}

C_{i 1} = C_{i 2} = C_{i}, Y_{ij} = A_{ij} + C_{i} + E_{ij},

where

C_{i}, E_{i 1}, E_{i 2}

are mutually independent and the

E_{ij}

are identically distributed. The causal directed acyclic graph has arrows

A_{i 1} \to Y_{i 1} \leftarrow C_{i} \to Y_{i 2} \leftarrow A_{i 2}, E_{i 1} \to Y_{i 1}, E_{i 2} \to Y_{i 2} .

A_{i 1} = A_{i 2} = G_{i}

with

Var (G_{i}) = V_{A}

; the genetic cause is completely shared. For dizygotic twins, one explicit construction is

A_{i 1} = G_{i} + G_{i 1}, A_{i 2} = G_{i} + G_{i 2},

where

G_{i}, G_{i 1}, G_{i 2}

are independent with variance

V_{A} /2

. Then both additive genetic terms have variance

V_{A}

, while their covariance is

V_{A} /2

1/2

. In a fuller graph, the shared

G_{i}

points to both genetic components and the unique

G_{ij}

points only to its own component.

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