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Let $I,J$ denote the two ordered alleles at the genetic locus. Under Hardy-Weinberg equilibrium they are independent with probabilities ${\pi }_i,{\pi }_j$. Write the multiplicative penetrance as $\mathbb{P}(D\mid I=i,J=j)=k{\psi }_i{\psi }_j$, with parameters chosen so these are valid probabilities, and set $Z={\sum }_u{\pi }_u{\psi }_u$. ThenThe Bayes' theorem givesThus multiplicative penetrance preserves Hardy-Weinberg equilibrium: the affected subjects still have two independent allele draws, with tilted allele frequencies. The displayed product uses ordered allele slots. For the usual unordered genotypes, the corresponding probabilities areThe factor of two is required for heterozygotes and must not be dropped when translating the ordered notation into genotype counts.

For genetic association analysis, the case genotype probabilities factor into allele frequencies, so under this model an allelic comparison with representative population controls is appropriate; separate dominance departures are not required by the penetrance model. The ratio of case allele frequency to population allele frequency is proportional to ${\psi }_i$, and relative risk multipliers satisfy ${\psi }_i/{\psi }_j=({\pi }_i^{\ast }/{\pi }_i)/({\pi }_j^{\ast }/{\pi }_j)$. Population controls are important to this exact statement. Among specifically unaffected controls,which generally does not factor. For a rare disease it is close to the population genotype law, but no rare-disease assumption was needed for the case factorization itself. Population stratification and sampling dependence must also be addressed before using an ordinary allelic chi-squared test.