= Solution
{c}
Let $y_{(n)}=\max_i y_i$ and $c=\max(\beta,y_{(n)})$. For the observed positive data, the <uniform distribution> <likelihood function> is
$$
L(\theta;y)=\theta^{-n}\mathbf1_{\{\theta>y_{(n)}\}}.
$$
Multiplying by the <Pareto distribution> <prior density> gives the <posterior density> kernel $\theta^{-(\alpha+n+1)}\mathbf1_{\{\theta>c\}}$. Since
$$
\int_c^\infty\theta^{-(\alpha+n+1)}\,d\theta
=\frac{c^{-(\alpha+n)}}{\alpha+n},
$$
the normalized <posterior distribution> is
$$
\boxed{\theta\mid y\sim\operatorname{Pareto}(\alpha+n,c),\qquad
\pi(\theta\mid y)=\frac{(\alpha+n)c^{\alpha+n}}{\theta^{\alpha+n+1}}\mathbf1_{\{\theta>c\}}.}
$$
This is <uniform-Pareto conjugacy>. The <posterior density> is strictly decreasing over its support, so the <highest posterior density interval> starts at $c$, where the admissible density is largest in the limiting sense. Its upper endpoint $u$ satisfies
$$
\mathbb P(\theta>u\mid y)=\left(\frac cu\right)^{\alpha+n}=0.05.
$$
Hence a 95% <highest posterior density interval> is
$$
\boxed{\left(c,\;c\,20^{1/(\alpha+n)}\right),\qquad c=\max\!\left(\beta,\max_iY_i\right).}
$$
Inclusion of endpoints does not change its <posterior probability>. Every point inside has higher <posterior density> than every admissible point to its right, and points below $c$ have density zero. This is the <Pareto highest density interval>; equal endpoint densities are unnecessary because one endpoint is the support boundary.
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