Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 29K / c / Solution

Ignoring the multinomial coefficient, the likelihood is

L (θ) = (1 - θ)^{2 n_{AA}} θ^{2 n_{BB}} [2 θ (1 - θ)]^{n_{A B}} = 2^{n_{A B}} θ^{2 n_{BB} + n_{A B}} (1 - θ)^{2 n_{AA} + n_{A B}} .

The score equation is

\frac{2 n _{BB} + n _{A B}}{θ} - \frac{2 n _{AA} + n _{A B}}{1 - θ} = 0,

and strict concavity of the log-likelihood gives

θ_{MLE} = \frac{2 n _{BB} + n _{A B}}{2 n} = θ_{w^{*}} .

Using the variables

Z_{i}

from part (b), the central limit theorem gives

n (θ_{MLE} - θ) = \frac{1}{n} \sum_{i = 1}^{n} (Z_{i} - θ) d N (0, Var (Z_{1})) .

Therefore the allele-frequency estimator under Hardy-Weinberg equilibrium has limiting law

n (θ_{MLE} - θ) d N (0, \frac{θ ( 1 - θ )}{2}) .

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