Weak tests against likelihood tilts do not ensure almost-sure posterior consistency

Let $c>0$ normalize the density $f(y)=c\mathrm{/}[y^2(\log y{)}^2]$ on $[e,\mathrm{\infty })$. Its normalizing integral is finite and positive, and $\mu :=E_{0}Y=c{\int }_e^{\mathrm{\infty }}[y(\log y{)}^2{]}^{-1}dy=c<\mathrm{\infty }$. For IID $Y_i$ with this density put $X_i=Y_i-\mu$ and $S_n={\sum }_{i=1}^n{X}_i$. These increments are integrable, nonconstant and centered.

For $b\ge e$, direct integration gives $E_0[Y{1}_{\{Y>b\}}]=c\mathrm{/}\log b$ and $P_0(Y>b)\le c\mathrm{/}[b(\log b{)}^2]$. Moreover $E_0[Y^2{1}_{\{Y\le b\}}]=c{\int }_e^b(\log y{)}^{-2}dy=O(b\mathrm{/}(\log b{)}^2)$: split the integral at $\sqrt{b}$, bound the first part by $\sqrt{b}$, and the second by $4b\mathrm{/}(\log b{)}^2$. Put $b_n=n\mathrm{/}(\log n{)}^{3\mathrm{/}2}$ for all sufficiently large $n$. By [F6], the probability that any of the first $n$ observations exceeds $b_n$ is $O((\log n{)}^{-1\mathrm{/}2})\to 0$. The variance of ${\sum }_{i=1}^n{Y}_i{1}_{\{Y_i\le b_n\}}$ is at most $O(n{b}_n\mathrm{/}(\log b_n{)}^2)$ by independence. Dividing this bound by $(n\mathrm{/}\log n{)}^2$ gives $O(b_n\mathrm{/}n)\to 0$. Chebyshev [F4] therefore shows $S_n\mathrm{/}(n\mathrm{/}\log n)=-c\log n\mathrm{/}\log b_n+o_{P_0}(1)\to -c$. Define $a_n=cn\mathrm{/}(2\log n)$ and ${\varphi }_n=1_{\{S_n>-a_n\}}$ for $n\ge 3$, with arbitrary tests for $n<3$. It follows that $P_0^n{\varphi }_n\to 0$.

By [F3], the hitting times $T_j=\inf \{n\ge 1:S_n\ge j\}$ are finite almost surely for every positive integer $j$. Define recursively $N_0=0$ and let $N_j$ be the least integer at least $\max (j,N_{j-1}+1)$ with $P_0(T_j>N_j)\le 2^{-j}$. Such an integer exists since $T_j<\mathrm{\infty }$ almost surely. Put ${ϵ}_j=N_j^{-1}$. Thus ${ϵ}_j>0$ and ${ϵ}_j\to 0$; these constants depend only on the true law, not the realized observations. By [F5], $T_j\le N_j$ eventually almost surely. Also $T_j\to \mathrm{\infty }$ almost surely, since finitely many finite partial sums cannot exceed arbitrarily large levels.

Take sample space $\mathsf{X}=[e,\mathrm{\infty })\bigsqcup \{z\}$ with its disjoint-union Borel sigma-algebra and dominating measure equal to Lebesgue measure plus ${\delta }_z$. Set $p_0(y)=f(y)$ and $p_0(z)=0$. On the continuous part supply $p_{\ast }(y)=f(y)e^{-2y}$ and $p_j(y)=f(y)e^{-y-\mu -{ϵ}_j}$; give each density its remaining mass at $z$. Both continuous masses are strictly between zero and one, so these atom masses are positive, and every density is finite and normalized. Take $\mathrm{\Theta }=\{\ast \}\bigsqcup \{1,2,\dots \}$ with the discrete metric and its Borel sigma-algebra; the supplied density family is jointly measurable. By [F1], $K(\ast )=2\mu <\mathrm{\infty }$ and $K(j)=E_0(Y+\mu +{ϵ}_j)=2\mu +{ϵ}_j$, so ${\mathrm{\Theta }}^{\ast }=\{\ast \}$. All relevant log ratios are integrable because $E_{0}Y<\mathrm{\infty }$.

On the $P_0$-support, the tilt at $j$ has density $r_j(y)=e^{y-\mu -{ϵ}_j}=e^{X-{ϵ}_j}$. Its total mass is infinite, since ${\int }_e^{\mathrm{\infty }}{e}^{y}f(y)dy=\mathrm{\infty }$, but the finite pointwise density makes it sigma-finite by [F2]. Its product density is $e^{S_n-n{ϵ}_j}$. On $\{{\varphi }_n=0\}$ this is at most $e^{-a_n}$, uniformly in $j$, and hence ${\sup }_j{Q}_j^n(1-{\varphi }_n)\le e^{-a_n}\to 0$. For $0<\eta <1$, the set at distance greater than $\eta$ from ${\mathrm{\Theta }}^{\ast }$ is precisely $\{1,2,\dots \}$, so these same tests satisfy both required weak errors. For $\eta \ge 1$ that set is empty and the type-II requirement is vacuous.

Put $r=e^{-1\mathrm{/}4}$ and give the model prior ${\pi }_{\ast }=1\mathrm{/}2$, ${\pi }_j=(1-r)r^{j-1}\mathrm{/}2$. These masses sum to one. Every relative KL neighborhood has mass at least $1\mathrm{/}2$ because it contains $\ast$. On any observed continuous sample, divide the prior-weighted likelihood by ${\prod }_i{p}_{\ast }(Y_i)$ to obtain the alternative numerator $A_n=e^{S_n}{\sum }_{j\ge 1}{\pi }_j{e}^{-n{ϵ}_j}$. It is finite, being at most $e^{S_n}\mathrm{/}2$, and the total relative evidence is $1\mathrm{/}2+A_n>0$. Thus the posterior mass of the alternatives is $A_n\mathrm{/}(1\mathrm{/}2+A_n)$.

Almost surely, for every sufficiently large $j$, step 2.2 gives $T_j{ϵ}_j\le 1$ and $S_{T_j}\ge j$. Consequently $A_{T_j}\ge e^{j-1}{\pi }_j=(1-r)e^{j-1}{r}^{j-1}\mathrm{/}2\to \mathrm{\infty }$. Since $T_j\to \mathrm{\infty }$, the posterior mass outside every discrete neighborhood of radius less than one has limes superior one almost surely. This contradicts the claimed a.s. consistency although both test errors in the refuted statement tend to zero.