Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 39 / 5 / Solution

The Bradley-Terry model assigns comparison probability $q_i={\theta }_i/({\theta }_i+{\theta }_{i+1})$ to each observed edge. Up to factors independent of the parameters, its likelihood function isThe comparison graph is a path graph, hence a tree. After fixing ${\theta }_n=1$, its edge ratios are unconstrained positive coordinates: every collection $r_i={\theta }_i/{\theta }_{i+1}>0$ determines uniquely ${\theta }_i={\prod }_{k=i}^{n-1}{r}_k$.

For a convenient strict-concavity calculation, use the edge log-ratios in a Bradley-Terry comparison tree ${\eta }_i=\log r_i$. The log-likelihood separates asEach derivative is $m(p_i-q_i)$, where $q_i$ is the logistic function of ${\eta }_i$, and each second derivative is $-m{q}_i(1-q_i)<0$. Because $0<p_i<1$, there is a unique finite global maximum atConverting back gives the Bradley-Terry maximum-likelihood estimate on a path:Equivalently, recurse backwards using ${\hat{\theta }}_i={\hat{\theta }}_{i+1}{p}_i/(1-p_i)$. All estimates are finite and positive. The sample size $m$ changes the curvature of the likelihood but not this maximizer; absence of comparison cycles is what permits every empirical edge proportion to be fitted simultaneously.