 Multiplicative penetrance preserves Hardy-Weinberg equilibrium

For ordered alleles $I,J$ independently drawn with probabilities ${\pi }_i$, suppose disease penetrance is $c{\psi }_i{\psi }_j$. Conditional on disease, the allele draws remain independent with probabilities ${\pi }_i^{\ast }={\pi }_i{\psi }_i/{\sum }_u{\pi }_u{\psi }_u$. This follows by normalizing the product $c{\pi }_i{\pi }_j{\psi }_i{\psi }_j$. Unordered homozygote probabilities are $({\pi }_i^{\ast }{)}^2$ and heterozygote probabilities are $2{\pi }_i^{\ast }{\pi }_j^{\ast }$. Unaffected controls do not generally share this factorization: their genotype weights contain $1-c{\psi }_i{\psi }_j$.

// Target: biology.bigb