Compact-uniform LAN approximates compact-prior posteriors

By [F4], integration against the prior makes $m_n$ measurable; [F6] gives $E_{P_{n,0}}{m}_n=1$. Also $\mathrm{\mid }{m}_n(x)-m({\mathrm{\Delta }}_n(x))\mathrm{\mid }\le a_n(x),$ so $E_{P_{n,0}}m({\mathrm{\Delta }}_n)\to 1$. The function $m$ is continuous: if $z_j\to z$, then $g_h(z_j)\to g_h(z)$ for each $h$, and compactness of $H$ gives a common finite bound on $g_h(z_j)$; [F8] passes the limit through the prior integral. Thus [F1] and [F7] give $m({\mathrm{\Delta }}_n)\Rightarrow m(Z)$. By [F3], $Em(Z)=1$. Write $U_n=m({\mathrm{\Delta }}_n)$ and $U=m(Z)$, and discard finitely many indices so $E{U}_n<\mathrm{\infty }$. For $Q>0$, weak convergence of the bounded continuous cutoff and convergence of the means give $E(U_n-Q{)}_{+}=E{U}_n-E(U_n\wedge Q)\to EU-E(U\wedge Q)=E(U-Q{)}_{+}$. Since $u{\mathbf{1}}_{\{u>2Q\}}\le 2(u-Q{)}_{+}$, the limsup of the tail expectations is at most $2E(U-Q{)}_{+}$, which tends to zero by [F2]. The finitely many initial integrable terms also have vanishing tails; thus $(U_n)$ is uniformly integrable on this eventual index range.

Fix $R>0$. On the compact set $\{z:\mathrm{\parallel }z{\mathrm{\parallel }}_2\le R\}$, the continuous function $m(z)$ has a positive minimum $c_R$, since each $g_h(z)>0$ and $\pi$ is a probability. On the event $\mathrm{\parallel }{\mathrm{\Delta }}_n{\mathrm{\parallel }}_2\le R$ and $a_n\le c_R\mathrm{/}2$, one has $m_n\ge c_R\mathrm{/}2$. For the two posterior densities with respect to $\pi$, ${\int }_H\mid \frac{L_n(h,x)}{m_n(x)}-\frac{g_h({\mathrm{\Delta }}_n(x))}{m({\mathrm{\Delta }}_n(x))}\mid \pi (dh)\le \frac{2a_n(x)}{m_n(x)}\le \frac{4a_n(x)}{c_R}\mathrm{.}$ Thus the posterior total variation distance is at most $4a_n\mathrm{/}{c}_R$ on this event.

The sequence ${\mathrm{\Delta }}_n$ is tight by [F10] because it converges in distribution. For any $Q>0$, $E[m({\mathrm{\Delta }}_n){\mathbf{1}}_{\{\mathrm{\parallel }{\mathrm{\Delta }}_n{\mathrm{\parallel }}_2>R\}}]\le E[m({\mathrm{\Delta }}_n){\mathbf{1}}_{\{m({\mathrm{\Delta }}_n)>Q\}}]+QP(\mathrm{\parallel }{\mathrm{\Delta }}_n{\mathrm{\parallel }}_2>R)\mathrm{.}$ First choose $Q$ large using uniform integrability from step 1.1, then $R$ large using tightness. Since $m_n\le m({\mathrm{\Delta }}_n)+a_n$ and $E{a}_n\to 0$, the predictive probability of $\mathrm{\parallel }{\mathrm{\Delta }}_n{\mathrm{\parallel }}_2>R$ is correspondingly small.

On $\{\mathrm{\parallel }{\mathrm{\Delta }}_n{\mathrm{\parallel }}_2\le R\}$, $m({\mathrm{\Delta }}_n)\le M_R$ for some finite $M_R$. By [F5], $M_n(\mathrm{\parallel }{\mathrm{\Delta }}_n{\mathrm{\parallel }}_2\le R,a_n>c_R\mathrm{/}2)\le M_R\frac{2E{a}_n}{c_R}+E{a}_n\to 0,$ using $m_n\le m+a_n$. On the complement of the exceptional events established in the preceding two steps, step 1.2 bounds the posterior $L^1(\pi )$ distance by $4a_n\mathrm{/}{c}_R$, and $m_n\le M_R+c_R\mathrm{/}2$. Its predictive expectation is therefore at most $4(M_R+c_R\mathrm{/}2)E{a}_n\mathrm{/}{c}_R\to 0$. The distance is at most two on the exceptional events, whose predictive probabilities vanish as first $n\to \mathrm{\infty }$ and then $R\to \mathrm{\infty }$. This proves the first conclusion.

For bounded continuous $f$, the predictive law satisfies $E_{M_n}f({\mathrm{\Delta }}_n)=E_{P_{n,0}}[m_{n}f({\mathrm{\Delta }}_n)]=E_{P_{n,0}}[m({\mathrm{\Delta }}_n)f({\mathrm{\Delta }}_n)]+o(1)\mathrm{.}$ For a cutoff $Q$, $(m\wedge Q)f$ is bounded and continuous, so its expectation converges by [F1]. The tail error is at most $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}E[m({\mathrm{\Delta }}_n){\mathbf{1}}_{\{m({\mathrm{\Delta }}_n)>Q\}}]$, which tends uniformly to zero as $Q\to \mathrm{\infty }$ by step 1.1. Thus the limit is $E[m(Z)f(Z)]$. This is the Gaussian-shift mixture with base law $N_d(0,K)$ and density $m(z)$.