Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 43 / 3 / Solution

The logit link makes the probability constant within each group: write $p_i=\{1+\exp [-(\mu +{\alpha }_i)]{\}}^{-1}$, with ${\alpha }_1=0$. Let $S_i={\sum }_{j=1}^k{Y}_{ij}$ and $N=km$. Independence gives $S_i\sim \mathrm{Bin}(N,p_i)$. Up to terms independent of the parameters, the log likelihood isThe likelihood equations areTogether they also imply $S_1-N{\hat{p}}_1=0$. Thus, when all totals satisfy $0<S_i<N$,The group logits are unconstrained apart from their parametrization, and each binomial log likelihood is strictly concave in its logit, so these equations give the unique finite maximum. A total of $0$ or $N$ instead gives a boundary probability and an infinite logit; a finite MLE need not exist in that case.

For $n=2$, keep $0<p_1,p_2<1$ fixed and let $N=km\to \infty$. Since ${\hat{p}}_1=S_1/N$, the central limit theorem gives $\sqrt{N}({\hat{p}}_1-p_1)\stackrel{d}{\to }N(0,p_1(1-p_1))$. The derivative of the logit is $1/[p_1(1-p_1)]$, so the delta method gives the baseline logit estimator in a group-factor binomial model lawThe second group does not improve the baseline estimate because it has its own free offset. Equivalently, the Fisher information matrix for $(\mu ,{\alpha }_2)$ isIts first diagonal entry confirms the same asymptotic variance.