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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-45/1/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 45 / 1 / a / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Treat

G_{i}

as the entire phase-known genotype of individual

i

at the genetic loci being studied, and let

\overline{F}

denote the nonfounders. Assume unrelated pedigree founders have independent genotype priors, each nonfounder's genotype depends only on its parents' genotypes, and the observed phenotypes are conditionally independent given the genotypes. These are the standard assumptions behind the pedigree likelihood; shared environmental effects or related pedigree founders would require additional factors.

The directed ancestry graph is acyclic, even when there is inbreeding. Order individuals with parents preceding offspring and apply the chain rule for probabilities. The parental conditional independence assumption gives

P (G_{1}, \dots, G_{n}) = \prod_{i \in F} P (G_{i}) \prod_{i \in \overline{F}} P (G_{i} ∣ G_{m (i)}, G_{f (i)}) .

(1)

Conditional phenotype independence gives

P (X_{1}, \dots, X_{n} ∣ G_{1}, \dots, G_{n}) = \prod_{i} P (X_{i} ∣ G_{i})

. Multiplying these two expressions and summing over every possible latent genotype configuration proves

P (X_{1}, \dots, X_{n}) = G_{1} \sum \dots G_{n} \sum i \prod P (X_{i} ∣ G_{i}) i \in F \prod P (G_{i}) i \in \overline{F} \prod P (G_{i} ∣ G_{m (i)}, G_{f (i)}) .

(2)

This is the law of total probability, not a claim that relatives have independent unconditional genotypes.

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