 Solution

Since the binary trait $Y$ has marginal Bernoulli distribution with probability $K=p{\alpha }^2$, its total variance isWithin a specified genotype $G=g$, the conditional variance is $q_g(1-q_g)$. The within-genotype contribution to total variance is its average over the population:Another Hardy-Weinberg equilibrium average givesHenceBy the law of total variance, the remaining between-genotype component is the genetic variance of penetrance, namelyIt is nonnegative, as a variance must be. It also follows from $\beta -{\alpha }^2=\varphi (1-\varphi )(\theta -1{)}^2\ge 0$. The single-genotype variance and its population average are different quantities; both have been specified.