Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / O / O4 / b / iii / Solution

Under the null put $p=q=r$. The likelihood becomes the same two binomial constants times $r^{m+n}(1-r{)}^{2M+2N-m-n}$. Maximization gives the pooled estimateUnder an interior null, (ii) has zero mean and variance $r(1-r)(1/(2M)+1/(2N))$. Statistical consistency of $\hat{r}$ and Slutsky's theorem permit replacement of $r$ by $\hat{r}$ in the denominator, giving the stated $Z\Rightarrow N(0,1)$. For a two-sided level-$\alpha$ test, reject when $\boxed{|Z|>{\Phi }^{-1}(1-\alpha /2)}$. The large-sample approximation requires both expected counts of each type to grow; if all observed alleles are identical the stated denominator degenerates and this approximation is not applicable.