Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 29K (Principles of Statistics)

The two parental chromosomes are independent, with probabilities $1-\theta$ and $\theta$ for alleles $A$ and $B$. Therefore the Hardy-Weinberg principle gives

The expectation isFor this polynomial to equal $\theta$ for every $\theta \in (0,1)$, comparison of its constant, linear, and quadratic coefficients givesThus the unique choice is

For fly $i$, let $Z_i$ be half its number of $B$ alleles. Then $Z_i\in \{0,1/2,1\}$, the $Z_i$ are independent and identically distributed,and ${\hat{\theta }}_{w^{\ast }}=n^{-1}{\sum }_i{Z}_i$. The weak law of large numbers gives ${\hat{\theta }}_{w^{\ast }}\to \theta$ in probability. Hence the estimator is unbiased and consistent.

Ignoring the multinomial coefficient, the likelihood isThe score equation isand strict concavity of the log-likelihood gives

Using the variables $Z_i$ from part (b), the central limit theorem givesTherefore the allele-frequency estimator under Hardy-Weinberg equilibrium has limiting law