Bradley-Terry maximum-likelihood estimate on a comparison tree

For a connected tree of comparisons with every observed win fraction strictly between zero and one, the Bradley-Terry model can fit every empirical edge probability exactly. Fix one positive strength to remove scaling ambiguity, then propagate strength ratios $p/(1-p)$ along tree edges. Edge log-ratios are independent real coordinates and each binomial log-likelihood term is strictly concave, proving existence and uniqueness of the normalized estimate. Cycles would impose extra compatibility relations.

For observed neighbor win fractions $p_0,\dots ,p_{n-1}\in (0,1)$ on the path graph $0,1,\dots ,n$, the normalized Bradley-Terry model estimate with ${\hat{\theta }}_n=1$ is ${\hat{\theta }}_i={\prod }_{k=i}^{n-1}{p}_k/(1-p_k)$. This is backward propagation of empirical odds along the path. Every edge proportion is fitted exactly, and strict concavity in edge log-ratios proves that this is the unique global maximum.

Orient the edges of a comparison tree and set ${\eta }_{ij}=\log {\theta }_i-\log {\theta }_j$. After anchoring one vertex strength, these edge quantities are unconstrained coordinates. The Bradley-Terry model probability is the logistic function of ${\eta }_{ij}$, and its binomial log-likelihood contribution is $m[p{\eta }_{ij}-\log (1+e^{{\eta }_{ij}})]$. Differentiating gives the fitted odds $p/(1-p)$.