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For a binary disease trait $Y$ with genotype $G$, the genetic variance of penetrance is the between-genotype term in the law of total variance. The averaged within-genotype term is $\mathbb{E}[q_G(1-q_G)]$. Under multiplicative diallelic penetrance, the genetic term is $p^2({\beta }^2-{\alpha }^4)$. This decomposition attributes variance to the modeled locus, not automatically to all hereditary causes.

Let $G$ count copies of allele $2$. Under Hardy-Weinberg equilibrium, $G$ has a binomial distribution with parameters $2,\varphi$, and the penetrance is $q_G=p{\theta }^G$. ConsequentlyThus the overall penetrance isThe multiplicative diallelic penetrance model requires $0\le \varphi \le 1$, $p\ge 0$, $\theta \ge 0$, and each occupied genotype's value $p{\theta }^G$ to be at most one. These are probabilities, not unrestricted relative-risk scores.

For binary traits, $Y_1{Y}_2$ is the indicator that both individuals are affected. Hence the pair probability is its expectation, or equivalently the covariance plus the product of individual means. Using the preceding results,These statements retain the same conditional-independence and common-population assumptions as the multiplicative diallelic penetrance covariance calculation.

For full siblings in an outbred pedigree, condition on both being affected under multiplicative diallelic penetrance and on residual disease outcomes being independent given their genotypes. The usual Mendelian IBD sharing of full siblings probabilities $(1/4,1/2,1/4)$ are weighted by $p^2{\alpha }^4,p^2{\alpha }^2\beta ,p^2{\beta }^2$, respectively, giving the displayed conditional sharing probabilities. Since $\beta -{\alpha }^2=\varphi (1-\varphi )(\theta -1{)}^2\ge 0$, $z_1\le 1/2$, strictly for a polymorphic locus with a nontrivial effect.