= Solution
Use the intended <hierarchical Bayesian model>: the breed parameters are conditionally independent given $\alpha,\beta$, and observations are independent given their breed parameters. Put $m_i=\max(\beta,M_i)$. The <uniform-Pareto model evidence> factorizes over breeds:
$$
\boxed{p(\mathbf y_1,\ldots,\mathbf y_I\mid\alpha,\beta)
=\prod_{i=1}^I\frac{\alpha\beta^\alpha}
{(\alpha+n_i)m_i^{\alpha+n_i}}.}
$$
Identical marginal <prior distributions> alone would not determine this product; <conditional independence> is the additional assumption.
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