Coverage level is not the posterior probability of the realized confidence set

By [L1] each $\theta \in \mathrm{\Theta }$ determines a probability measure ${\mathbb{P}}_{\theta }$ on $(S,\mathcal{S})$ with ${\mathbb{P}}_{\theta }(\{s\})=1\mathrm{/}4$ for $s\in D_{\theta }$ and ${\mathbb{P}}_{\theta }(\{s\})=0$ otherwise. Define $C(s):=\{\begin{array}{ll}\{s_1-1\}\cap \mathrm{\Theta },& s_1=s_2,\\ \left\{{\displaystyle \frac{s_1+s_2}{2}}\right\}\cap \mathrm{\Theta },& s_1\mathrm{\ne }{s}_2\mathrm{.}\end{array}$ Each $C(s)$ is a subset of $\mathrm{\Theta }$, and each set $\{\theta \in C\}$ is a subset of $S$ and therefore an element of the full power set $\mathcal{S}=2^S$, so $C$ is a confidence set by [L2].

For fixed $\theta$ the four points of $D_{\theta }$ evaluate under $C$ as follows: $(\theta +1,\theta +1)$ gives $C=\{(\theta +1)-1\}=\{\theta \}$, which contains $\theta$; $(\theta -1,\theta -1)$ gives $C=\{\theta -2\}\cap \mathrm{\Theta }$, which is empty for $\theta =14$ and equals $\{\theta -2\}$ for $\theta =16,18$; in every case it does not contain $\theta$; and the two points with distinct coordinates, $(\theta +1,\theta -1)$ and $(\theta -1,\theta +1)$, both give $C=\{\theta \}$. Hence ${\mathbb{P}}_{\theta }(\theta \in C)=3\mathrm{/}4$ for every $\theta \in \mathrm{\Theta }$, and the coverage level of $C$ is $3\mathrm{/}4$ by [L2].

Let $s^{\ast }:=(15,17)$. Since $s^{\ast }$ has distinct coordinates, step 1.1 gives $C(s^{\ast })=\{(15+17)\mathrm{/}2\}=\{16\}$. Moreover, $s^{\ast }\in D_{\theta }$ holds exactly for $\theta =16$, so ${\mathbb{P}}_{\theta }(\{s^{\ast }\})$ equals $1\mathrm{/}4$ for $\theta =16$ and $0$ for $\theta \in \{14,18\}$.

Fix weights $\pi (14),\pi (16),\pi (18)\ge 0$ summing to $1$ with $\pi (16)>0$, and put $W:=\{(\theta ,s):\theta \in \mathrm{\Theta },s\in D_{\theta }\}$, a twelve-element set, and give $(\theta ,s)\in W$ weight $\pi (\theta )\mathrm{/}4$. These nonnegative weights sum to $1$, so [L3] defines a probability measure $Q$ on the finite space $W$. In events below, $\theta$ and $s$ denote the two coordinate maps on $W$. The event $\{s=s^{\ast }\}$ has $Q$-probability $\pi (16)\cdot \frac{1}{4}>0$, so the conditional probability given $s^{\ast }$ is defined by [L3].

The event $\{\theta \in C(s),s=s^{\ast }\}$ equals $\{\theta =16,s=s^{\ast }\}$, because $C(s^{\ast })=\{16\}$ by step 2.2; its $Q$-probability is $\pi (16)\cdot \frac{1}{4}$, the same as $Q(\{s=s^{\ast }\})$ by step 2.2. Hence $Q(\theta \in C(s)\mid s=s^{\ast })=\frac{\pi (16)\mathrm{/}4}{\pi (16)\mathrm{/}4}=1.$

The confidence set $C$ has level $3\mathrm{/}4$ by step 2.1, while for the prior $\pi$ and the datum $s^{\ast }$ the conditional probability that $\theta \in C(s)$ is $1$ by step 4.1; since $1\mathrm{\ne }3\mathrm{/}4$, the refuted statement is false.