= 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> is
$$
L(\theta)=\prod_{i=0}^{n-1}q_i^{mp_i}(1-q_i)^{m(1-p_i)}.
$$
The 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 as
$$
\ell(\eta)=m\sum_{i=0}^{n-1}\left[p_i\eta_i-\log(1+e^{\eta_i})\right]+\text{constant}.
$$
Each derivative is $m(p_i-q_i)$, where $q_i$ is the <logistic function> of $\eta_i$, and each second derivative is $-mq_i(1-q_i)<0$. Because $0<p_i<1$, there is a unique finite global maximum at
$$
\widehat\eta_i=\log\frac{p_i}{1-p_i},\qquad \widehat q_i=p_i.
$$
Converting back gives the <Bradley-Terry maximum-likelihood estimate on a path>:
$$
\boxed{\widehat\theta_n=1,\qquad
\widehat\theta_i=\prod_{k=i}^{n-1}\frac{p_k}{1-p_k},\quad 0\leq i<n.}
$$
Equivalently, recurse backwards using $\widehat\theta_i=\widehat\theta_{i+1}p_i/(1-p_i)$. 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.
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