The posterior excludes large scales and mixing mass far from compact data

The latent representation is and $X_i\mid {\theta }_i,\sigma \sim N({\theta }_i,{\sigma }^2)$ independently. Use the test ${\phi }_n=\mathbf{1}\{\text{some }{X}_i\mathrm{\notin }I\}$. Its type-I error is zero. Write $\mathrm{\Phi }(t)={\int }_{-\mathrm{\infty }}^t\varphi (z)dz$; [F9] gives ${\mathrm{\Phi }}^{\mathrm{\prime }}(t)=\varphi (t)$. The normal formula [F5] is even and satisfies ${\varphi }^{\mathrm{\prime }}(t)=-t\varphi (t)$, so it decreases on $[0,\mathrm{\infty })$. For fixed $\theta$ and $\sigma$, set $g(\theta )=P(\theta +\sigma Z\in I)=\mathrm{\Phi }((U-\theta )\mathrm{/}\sigma )-\mathrm{\Phi }((L-\theta )\mathrm{/}\sigma )$. Then $g^{\mathrm{\prime }}(\theta )={\sigma }^{-1}[\varphi ((L-\theta )\mathrm{/}\sigma )-\varphi ((U-\theta )\mathrm{/}\sigma )]$. For $\theta \le (L+U)\mathrm{/}2$, $\mathrm{\mid }L-\theta \mathrm{\mid }\le \mathrm{\mid }U-\theta \mathrm{\mid }$, so $g^{\mathrm{\prime }}(\theta )\ge 0$; for $\theta \ge (L+U)\mathrm{/}2$ the inequality reverses and $g^{\mathrm{\prime }}(\theta )\le 0$. Thus $g$ is maximized at the midpoint, where the component leakage is $2\mathrm{\Phi }(-R\mathrm{/}\sigma )$. Every component with $\sigma >s_n$ therefore puts probability at least $q_n=2\mathrm{\Phi }(-R\mathrm{/}{s}_n)$ outside $I$, and $P_{F,\sigma }^n({\phi }_n=0)\le (1-q_n{)}^n\le e^{-n{q}_n}$.

By [F8] the latent DP model is a measurable prior law. Introduce the latent locations ${\theta }_{1:n}$. Given these locations, $X_{1:n}$ and $\sigma$, the observation likelihood is independent of $F$, so [F1] gives the conditional law $DP(\alpha +{\sum }_i{\delta }_{{\theta }_i})$. Given this updated $F$, the next latent location has conditional law $F$; applying [F3] to that predictive draw gives $E[F(A)\mid {\theta }_{1:n},X_{1:n},\sigma ]=\frac{\alpha (A)+{\sum }_i{\mathbf{1}}_A({\theta }_i)}{\alpha (\mathbb{R})+n}$ for every Borel set $A$. With $A=I_d^c$, conditional Markov inequality yields ${\mathrm{\Pi }}_n(F(A)>{ϵ}_n^2,\sigma \le s_n\mid X_{1:n})\le \frac{\alpha (A)+{\sum }_{i}P({\theta }_i\in A,\sigma \le s_n\mid X_{1:n})}{{ϵ}_n^2(\alpha (\mathbb{R})+n)}\mathrm{.}$

Fix an index $i$ and condition on ${\theta }_{-i},X_{1:n},\sigma$. The latent vector is exchangeable because its conditional law given $F$ is IID. Reordering ${\theta }_i$ last and applying [F2] shows its prior predictive law given ${\theta }_{-i}$ is $H_i=(\alpha +{\sum }_{k\mathrm{\ne }i}{\delta }_{{\theta }_k})\mathrm{/}(\alpha (\mathbb{R})+n-1)$. Bayes' formula multiplies this measure by the likelihood ${\varphi }_{\sigma }(X_i-t)$. Almost surely $X_i\in I$. On $A=I_d^c$, $\mathrm{\mid }{X}_i-t\mathrm{\mid }\ge d$, so the likelihood numerator is at most $C{\sigma }^{-1}{e}^{-d^2\mathrm{/}(2{\sigma }^2)}$. For $\sigma \le s_n$, the denominator is at least the base-measure contribution $\frac{1}{\alpha (\mathbb{R})+n-1}{\int }_{X_i-\sigma }^{X_i+\sigma }{\varphi }_{\sigma }(X_i-t)\alpha (dt)\ge \frac{c}{\alpha (\mathbb{R})+n-1},$ because the base density is bounded below near $I$ and $[X_i-\sigma ,X_i+\sigma ]$ lies there for all sufficiently large $n$. Consequently $P({\theta }_i\in A\mid {\theta }_{-i},X_{1:n},\sigma )\le Cn{\sigma }^{-1}{e}^{-d^2\mathrm{/}(2{\sigma }^2)}\le Cn{s}_n^{-1}{e}^{-d^2\mathrm{/}(2s_n^2)},$ where $s^{-1}{e}^{-d^2\mathrm{/}(2s^2)}$ is increasing on $0<s<d$.

Let $N_n^H$ be the likelihood-ratio numerator integrated over $\{\sigma >s_n\}$. By change of density and [F6], $E_0[(1-{\phi }_n)N_n^H]\le e^{-n{q}_n}$. On the denominator event, [F7] therefore gives, for every $\zeta >0$, $P_0^n\{{\mathrm{\Pi }}_n(\sigma >s_n\mid X_{1:n})>\zeta \}\le o(1)+{\zeta }^{-1}{e}^{(C_0+2)n{\delta }_n^2-n{q}_n}⟶0.$

Substitute the step 1.3 bound into step 1.2 and use $\alpha (A)\le \alpha (\mathbb{R})$. The resulting upper bound is $\frac{\alpha (\mathbb{R})}{{ϵ}_n^2(\alpha (\mathbb{R})+n)}+Cn{ϵ}_n^{-2}{s}_n^{-1}{e}^{-d^2\mathrm{/}(2s_n^2)}+{\mathrm{\Pi }}_n(\sigma >s_n\mid X_{1:n}),$ which tends to zero by $n{ϵ}_n^2\to \mathrm{\infty }$, the second premise, and step 2.1. Thus the posterior probability of $F(I_d^c)>{ϵ}_n^2$ tends to zero. Combining with step 2.1 proves the sieve-complement claim.