Exponential tests and KL mass force posterior exclusion

Choose $\alpha$ with $ϵ<\alpha <\beta$. By [A1] and [F1], the denominator $D_n$ is at least $e^{-n\alpha }$ eventually almost surely. By the type-I premise, ${\varphi }_n(X_1,\dots ,X_n)\to 0$ almost surely under the same infinite IID law. Intersect these two full-probability events.

Define $N_n:={\int }_V{\prod }_{i=1}^n\frac{p_{\theta }(X_i)}{p_{{\theta }^{\ast }}(X_i)}(1-{\varphi }_n)\mathrm{\Pi }(d\theta ),Y_n:=e^{n\alpha }{N}_n\mathrm{.}$ By [F2], [F4], [F8], and the type-II bound, ${\mathbb{E}}_{P_0^n}{Y}_n\le e^{n\alpha }\mathrm{\Pi }(V)e^{-n\beta }\le e^{-n(\beta -\alpha )}\mathrm{.}$

Set $a_n=e^{-n(\beta -\alpha )\mathrm{/}2}$. For every sufficiently large $n$, Markov's inequality [F5] gives $P_0^n\{Y_n>a_n\}\le e^{-n(\beta -\alpha )\mathrm{/}2}$. The right-hand side is summable, and the finite initial terms do not matter. Thus [F6] and [F7] imply $Y_n\to 0$ almost surely.

On the full-probability event from the denominator lemma, the ordinary evidence is $Z_n:={\int }_{\mathrm{\Theta }}{\prod }_i{p}_{\theta }(X_i)\mathrm{\Pi }(d\theta )=({\prod }_i{p}_{{\theta }^{\ast }}(X_i))D_n,$ which is finite and positive. Bayes' formula [F3] therefore gives the posterior as the normalized likelihood integral, and division by this positive common factor expresses it using $D_n$. Decompose the tested likelihood integral over $V$ into its ${\varphi }_n$ and $1-{\varphi }_n$ parts. Since $0\le {\varphi }_n\le 1$, ${\int }_V{R}_n(\theta )\mathrm{\Pi }(d\theta )={\int }_V{\varphi }_n{R}_n(\theta )\mathrm{\Pi }(d\theta )+N_n\le {\varphi }_n{\int }_{\mathrm{\Theta }}{R}_n(\theta )\mathrm{\Pi }(d\theta )+N_n={\varphi }_n{D}_n+N_n,$ where $R_n(\theta )={\prod }_i{p}_{\theta }(X_i)\mathrm{/}{p}_{{\theta }^{\ast }}(X_i)$. Therefore $\mathrm{\Pi }(V\mid X_1,\dots ,X_n)\le {\varphi }_n+\frac{N_n}{D_n}\le {\varphi }_n+e^{n\alpha }{N}_n={\varphi }_n+Y_n\mathrm{.}$ Steps 1.1 and 2.1 make the right-hand side tend to zero.