Function-space pCN and invariant posterior kernels

If $H=\{0\}$, then it has only one Borel probability measure and the conclusion holds. Otherwise, let $\mathcal{C}$ be the family of all finite coordinate cylinder sets ${\pi }_F^{-1}(B)$, where $F\subseteq J$ is finite and $B\subseteq {\mathbb{R}}^F$ is Borel. Each cylinder is Borel by continuity in [F1]. The family contains $H$, and the intersection of two cylinders is the finite-coordinate cylinder obtained on the union of their index sets. Hence $\mathcal{C}$ is a pi-system.

Enumerate the basis if it is infinite, and in the finite case repeat the last partial sum after all basis vectors have appeared. For $x_0\in H$ and $r>0$, put $S_N(x)={\sum }_{j\le N}\mathrm{\mid }⟨x-x_0,e_j⟩{\mathrm{\mid }}^2$. By [F2], $\mathrm{\parallel }x-x_0{\mathrm{\parallel }}^2={\sup }_N{S}_N(x)$, so $B(x_0,r)={\bigcup }_{\begin{array}{c}q\in \mathbb{Q}\\ 0\le q<r^2\end{array}}{\bigcap }_{N\ge 1}\{x:S_N(x)\le q\}\mathrm{.}$ Indeed, if $\mathrm{\parallel }x-x_0{\mathrm{\parallel }}^2<r^2$, choose a rational strictly between these two numbers; conversely, membership in one intersection gives $\mathrm{\parallel }x-x_0{\mathrm{\parallel }}^2\le q<r^2$. By [F1] and [F6], each $S_N$ is Borel, and $\{S_N\le q\}$ is a cylinder over a Borel subset of ${\mathbb{R}}^F$. Thus every open ball belongs to $\sigma (\mathcal{C})$.

Let $D$ be the set of all finite rational linear combinations of basis vectors. If the basis is finite, $D$ is at most countable by [F3]. If it is countably infinite, enumerate it and write $D$ as the union over $N$ of the sets of rational linear combinations of the first $N$ vectors. Each of those sets is at most countable by [F3], so their union is at most countable by the Countable Choice clause of [F3]. The real span of the basis is dense by completeness. Each finite real linear combination is approximated by a rational one: approximate each of its finitely many coefficients by a rational and use the triangle inequality and $\mathrm{\parallel }{e}_j\mathrm{\parallel }=1$. Thus $D$ is dense in $H$. By definition, this makes $H$ separable. Separability: the existence of an at most countable dense subset The balls $B(d,s)$ with $d\in D$ and positive rational $s$ form an at most countable family by [F3], and they are open by [F5]. They are a base: given $x\in U$ with $U$ open, choose $r>0$ with $B(x,r)\subseteq U$, then $d\in D$ with $\mathrm{\parallel }d-x\mathrm{\parallel }<r\mathrm{/}3$, and a rational $s$ with $\mathrm{\parallel }d-x\mathrm{\parallel }<s<r-\mathrm{\parallel }d-x\mathrm{\parallel }$. The triangle inequality gives $x\in B(d,s)\subseteq B(x,r)\subseteq U$. By step 2.1 every such ball lies in $\sigma (\mathcal{C})$, so every open set, as a union of a subfamily of this countable base, lies in $\sigma (\mathcal{C})$.

Every coordinate map is continuous by [F1], so $\sigma (\mathcal{C})\subseteq \mathcal{B}(H)$. Step 3.1 and the definition of the Borel sigma-algebra give the reverse inclusion. Thus $\sigma (\mathcal{C})=\mathcal{B}(H)$.

The assumed equality of finite-coordinate measures makes $\nu$ and ${\nu }^{\mathrm{\prime }}$ agree on every cylinder in $\mathcal{C}$. They are probability measures, so they agree on $H$ as well. The pi-system $\mathcal{C}$ generates $\mathcal{B}(H)$ by step 4.1; [F4] therefore gives $\nu ={\nu }^{\mathrm{\prime }}$.

For each Borel set $B$ in the extended potential range, the preimage ${\bar{\mathrm{\Phi }}}^{-1}(B)$ is the union of $D\cap {\mathrm{\Phi }}_D^{-1}(B)$ and $H\setminus D$ when $+\mathrm{\infty }\in B$, and is just $D\cap {\mathrm{\Phi }}_D^{-1}(B)$ otherwise. Since $D$ is Borel and ${\mathrm{\Phi }}_D$ is Borel on the trace Borel structure of $D$, every such preimage is Borel in $H$. Thus the extended potential used by the algorithm is Borel.

Write $\rho =\sqrt{1-{\beta }^2}$ and define $Q_{\beta }(u,B)={\int }_H{\mathbf{1}}_B(\rho u+\beta z){\mu }_0(dz)$ for Borel $B\subseteq H$. For fixed $u$, this is the pushforward of the probability ${\mu }_0$ under the continuous map $z\mapsto \rho u+\beta z$, so it is a probability measure in $B$. The joint map $(u,z)\mapsto \rho u+\beta z$ is continuous; by [F6], its Borel preimages are product-measurable for $\mathcal{B}(H)\otimes \mathcal{B}(H)$. Hence $(u,z)\mapsto {\mathbf{1}}_B(\rho u+\beta z)$ is product-measurable. The constant kernel $u\mapsto {\mu }_0$ is a probability kernel, so [F3] makes $u\mapsto Q_{\beta }(u,B)$ measurable. Therefore $Q_{\beta }$ is a probability kernel by [F2].

The finite-potential set $E=\{u:\bar{\mathrm{\Phi }}(u)<+\mathrm{\infty }\}$ is Borel by step 1.1. The coordinate projections on $H\times H$ are continuous and hence Borel by [F5]; [F6] identifies their source Borel sigma-algebra with the product sigma-algebra. Composing them with the Borel $\bar{\mathrm{\Phi }}$ makes $(u,v)\mapsto \bar{\mathrm{\Phi }}(u)$ and $(u,v)\mapsto \bar{\mathrm{\Phi }}(v)$ Borel. On $E\times E$ their difference is real- valued and Borel, and [F7] makes its exponential Borel. Extending $\min \{1,\exp (\bar{\mathrm{\Phi }}(u)-\bar{\mathrm{\Phi }}(v))\}$ by $0$ off $E\times E$ therefore makes the algorithm's $\alpha$ a Borel function with values in $[0,1]$. Put $a(u)={\int }_H\alpha (u,v)Q_{\beta }(u,dv)$. By [F3], $a$ is measurable; the probability rows of $Q_{\beta }$ give $0\le a(u)\le 1$.

For each $u$ and Borel $A\subseteq H$, define $P_{\beta }(u,A)={\int }_A\alpha (u,v)Q_{\beta }(u,dv)+{\mathbf{1}}_A(u)(1-a(u))\mathrm{.}$ The first term is a measure in $A$ by [F4] and step 2.1. The second term is also a measure: among any pairwise-disjoint sequence of sets, at most one contains $u$, so the indicator of their union is the sum of their indicators at $u$. Their sum has total mass $a(u)+1-a(u)=1$. For fixed $A$, the first term is measurable in $u$ by [F3], and the second is measurable because both ${\mathbf{1}}_A(u)$ and $a(u)$ are measurable. Thus $P_{\beta }$ is a probability kernel by [F2].

Steps 1.1, 1.2, 2.1, and 3.1 establish that the potential extension, the proposal, the acceptance function, and the accept-reject transition have the claimed Borel and probability properties.

By [F2], take a finite or countable complete orthonormal basis $(e_j)$ of $H$. Give $H\times H$ the inner product $⟨(u,v),(x,y){⟩}_{\times }=⟨u,x⟩+⟨v,y⟩$. Its induced norm is $\sqrt{\mathrm{\parallel }u{\mathrm{\parallel }}^2+\mathrm{\parallel }v{\mathrm{\parallel }}^2}$. It is complete because a Cauchy sequence has Cauchy coordinate sequences in $H$, whose limits in $H$ give its product limit. Thus this is a Hilbert space. The family $(e_j,0),(0,e_j)$ is orthonormal, and any vector orthogonal to every listed vector has both components orthogonal to every $e_j$, hence both components are zero by completeness; it is a complete orthonormal basis. Let $D\subseteq H$ be countable and dense. The balls $B(d,s)$ with $d\in D$ and positive rational $s$ form a countable base for $H$: for $x\in U$ open, choose $r>0$ with $B(x,r)\subseteq U$, then $d\in D$ with $\mathrm{\parallel }d-x\mathrm{\parallel }<r\mathrm{/}3$ and rational $s$ with $\mathrm{\parallel }d-x\mathrm{\parallel }<s<r-\mathrm{\parallel }d-x\mathrm{\parallel }$. Thus $H$ is second-countable. Then $D\times D$ is countable by [F2] and dense, since each coordinate of a pair can be approximated within $\epsilon \mathrm{/}2$ to make the product norm error less than $\epsilon$. Balls with centers in $D\times D$ and positive rational radii form a countable base: for $z$ in an open set choose $r>0$ with $B(z,r)$ contained there, then a center $d$ with $\mathrm{\parallel }d-z\mathrm{\parallel }<r\mathrm{/}3$ and rational $s$ with $\mathrm{\parallel }d-z\mathrm{\parallel }<s<r-\mathrm{\parallel }d-z\mathrm{\parallel }$ give $z\in B(d,s)\subseteq B(z,r)$. Thus $H\times H$ is second-countable, and [F10] identifies its Borel sigma-algebra with the product Borel sigma-algebra. Since $U$ and $\xi$ are Borel random elements, the pair $(U,\xi )$ is measurable into the product sigma-algebra: the preimage of each measurable rectangle is the intersection of the two coordinate preimages, and rectangles generate the product sigma-algebra. Continuous addition and scalar multiplication then make $(U,V)$ and $(V,U)$ Borel random elements of $H\times H$. Their Borel laws are therefore defined.

Fix a finite list of continuous linear functionals on $H\times H$. Each has the form $(u,v)\mapsto \mathrm{\ell }(u)+m(v)$, because its restrictions to the two coordinate subspaces sum to the functional. Any real linear combination of the listed projections is another such functional. Under $(U,V)$ that combination is $(\mathrm{\ell }+\rho m)(U)+\beta m(\xi )$; under $(V,U)$ it is $(\rho \mathrm{\ell }+m)(U)+\beta \mathrm{\ell }(\xi )$. By [F3], the two summands in each display are centered normal, and [F5] makes them independent; [F6] makes their sum normal, including a zero coefficient. Hence every finite projection vector under either pair is multivariate normal by the projection definition in [F7], with zero mean.

Define the symmetric covariance form $\mathrm{\Gamma }(\mathrm{\ell },m)=\mathbb{E}[\mathrm{\ell }(U)m(U)]$. These products are integrable: [F4] gives finite second moments of $\mathrm{\parallel }U\mathrm{\parallel }$, boundedness of $\mathrm{\ell },m$ gives finite second moments of their evaluations, and $\mathrm{\mid }ab\mathrm{\mid }\le (a^2+b^2)\mathrm{/}2$. The same argument applies to evaluations of $\xi$. For functionals $(\mathrm{\ell },m)$ and $({\mathrm{\ell }}^{\mathrm{\prime }},m^{\mathrm{\prime }})$, the projection under $(U,V)$ is $(\mathrm{\ell }+\rho m)(U)+\beta m(\xi )$. Independence and centering make its mixed covariance terms zero by [F9], so its covariance with the corresponding projection for $({\mathrm{\ell }}^{\mathrm{\prime }},m^{\mathrm{\prime }})$ is $\mathrm{\Gamma }(\mathrm{\ell },{\mathrm{\ell }}^{\mathrm{\prime }})+\rho \mathrm{\Gamma }(\mathrm{\ell },m^{\mathrm{\prime }})+\rho \mathrm{\Gamma }(m,{\mathrm{\ell }}^{\mathrm{\prime }})+\mathrm{\Gamma }(m,m^{\mathrm{\prime }}),$ using ${\rho }^2+{\beta }^2=1$. Under $(V,U)$ the projections are $(\rho \mathrm{\ell }+m)(U)+\beta \mathrm{\ell }(\xi )$ and $(\rho {\mathrm{\ell }}^{\mathrm{\prime }}+m^{\mathrm{\prime }})(U)+\beta {\mathrm{\ell }}^{\mathrm{\prime }}(\xi )$; expansion gives the same expression because $\mathrm{\Gamma }$ is symmetric. Therefore the finite projection vectors from step 2.1 have equal covariance matrices and zero means.

By [F7], the finite projection laws in step 3.1 are equal, in particular the finite coordinate laws for the product basis from step 1.1. That step also shows that $H\times H$ is a separable Hilbert space. By [A1], Countable Choice follows from AC; the coordinate-law result [F8] therefore identifies the two Borel probability laws. Hence $\mathrm{Law}(U,V)=\mathrm{Law}(V,U)$.

Step 4.1 is the claimed invariance of the joint law under coordinate swap, for every $\beta \in [0,1]$, including the identity proposal at $\beta =0$ and the independent-reference proposal at $\beta =1$.

Apply [F2] with parameter space $H$, domain $D=H$, one-point data space $\{\ast \}$, potential $\tilde{\mathrm{\Phi }}(u,\ast )=\mathrm{\Phi }(u)$, and envelope $M_1=0$. The domain has full ${\mu }_0$-mass. For any $u\in H$, choose $R>\mathrm{\parallel }u\mathrm{\parallel }$; all sufficiently close $v$ satisfy $\max \{\mathrm{\parallel }u\mathrm{\parallel },\mathrm{\parallel }v\mathrm{\parallel }\}<R$, so the stated Lipschitz bound proves continuity at $u$. Also ${\int }_H{e}^{M_1(\mathrm{\parallel }u\mathrm{\parallel })}{\mu }_0(du)=1$. Hence $0<Z<\mathrm{\infty }$ and $\mu$ is a probability. The density $e^{-\mathrm{\Phi }}$ is strictly positive everywhere. If $\mu (A)=0$, then ${\int }_H{e}^{-\mathrm{\Phi }}{\mathbf{1}}_{A}d{\mu }_0=0$, so [F4] gives $e^{-\mathrm{\Phi }}{\mathbf{1}}_A=0$ ${\mu }_0$-almost everywhere; strict positivity yields ${\mu }_0(A)=0$. Conversely, ${\mu }_0(A)=0$ implies $\mu (A)=0$ directly from the integral formula. Thus $\mu$ and ${\mu }_0$ are equivalent.

Fix any sequence ${\beta }_n\in [0,1]$ with ${\beta }_n\to 0$. For every realization with $U,\xi \in H$, ${\rho }_n:=\sqrt{1-{\beta }_n^2}\to 1$ and $\mathrm{\parallel }{V}_{{\beta }_n}-U\mathrm{\parallel }\le \mathrm{\mid }1-{\rho }_n\mathrm{\mid }\mathrm{\parallel }U\mathrm{\parallel }+{\beta }_n\mathrm{\parallel }\xi \mathrm{\parallel }⟶0.$ Choose $R>\mathrm{\parallel }U\mathrm{\parallel }+1$ for that realization. Eventually $\max \{\mathrm{\parallel }U\mathrm{\parallel },\mathrm{\parallel }{V}_{{\beta }_n}\mathrm{\parallel }\}<R$, and the assumed ball-wise Lipschitz bound gives $\mathrm{\mid }\mathrm{\Phi }(V_{{\beta }_n})-\mathrm{\Phi }(U)\mathrm{\mid }\le L_R\mathrm{\parallel }{V}_{{\beta }_n}-U\mathrm{\parallel }⟶0.$ Therefore ${\mathrm{\Delta }}_n:=\mathrm{\Phi }(V_{{\beta }_n})-\mathrm{\Phi }(U)\to 0$ almost surely. Its measurability, and that of ${\alpha }_{{\beta }_n}$, was established in [F6].

View the identity map $x\mapsto x$ on $(H,\mathcal{B}(H),{\mu }_0)$ as a centered Borel Gaussian random element. By [F3], there is $a>0$ such that ${\int }_H{e}^{a\mathrm{\parallel }x{\mathrm{\parallel }}^2}{\mu }_0(dx)<\mathrm{\infty }$. For every $q>0$, the function $r\mapsto r^q{e}^{-a{r}^2}$ is bounded on $[0,\mathrm{\infty })$, so ${\int }_H\mathrm{\parallel }x{\mathrm{\parallel }}^q{\mu }_0(dx)<\mathrm{\infty }$. Since $e^{-\mathrm{\Phi }}\le 1$ and $Z>0$, $\mathbb{E}\mathrm{\parallel }U{\mathrm{\parallel }}^q=Z^{-1}{\int }_H\mathrm{\parallel }x{\mathrm{\parallel }}^q{e}^{-\mathrm{\Phi }(x)}{\mu }_0(dx)\le Z^{-1}{\int }_H\mathrm{\parallel }x{\mathrm{\parallel }}^q{\mu }_0(dx)<\mathrm{\infty }\mathrm{.}$ The same Gaussian moment bound gives $\mathbb{E}\mathrm{\parallel }\xi {\mathrm{\parallel }}^q<\mathrm{\infty }$ for every $q>0$.

The growth bound and $\mathrm{\parallel }{V}_{{\beta }_n}\mathrm{\parallel }\le \mathrm{\parallel }U\mathrm{\parallel }+\mathrm{\parallel }\xi \mathrm{\parallel }$ imply $\mathrm{\mid }{\mathrm{\Delta }}_n\mathrm{\mid }\le 2K+K\mathrm{\parallel }U{\mathrm{\parallel }}^p+K(\mathrm{\parallel }U\mathrm{\parallel }+\mathrm{\parallel }\xi \mathrm{\parallel }{)}^p\le C_{p,K}(1+\mathrm{\parallel }U{\mathrm{\parallel }}^p+\mathrm{\parallel }\xi {\mathrm{\parallel }}^p),$ where the last inequality follows from $(a+b{)}^p\le 2^p(a^p+b^p)$ for $a,b\ge 0$ (use $a+b\le 2\max (a,b)$ and $\max (a,b{)}^p\le a^p+b^p$), with $C_{p,K}:=\max \{2K,K(1+2^p),K{2}^p\}$. By step 2.1 the right side is integrable and does not depend on $n$. Applying [F5] to the measurable sequence $\mathrm{\mid }{\mathrm{\Delta }}_n\mathrm{\mid }$ from [F6] and step 1.2 gives $\mathbb{E}\mathrm{\mid }{\mathrm{\Delta }}_n\mathrm{\mid }\to 0$.

The scalar function $g(t)=\min \{1,e^{-t}\}$ is continuous and $g(0)=1$. Step 1.2 therefore implies ${\alpha }_{{\beta }_n}=g({\mathrm{\Delta }}_n)\to 1$ almost surely. Since $0\le {\alpha }_{{\beta }_n}\le 1$, [F5] with the integrable dominating constant $1$ yields $\mathbb{E}{\alpha }_{{\beta }_n}\to 1$. The sequence ${\beta }_n\to 0$ was arbitrary, so the sequential criterion for real limits proves ${\lim }_{\beta \downarrow 0}\mathbb{E}[{\alpha }_{\beta }]=1$. This is the stationary mean acceptance limit; it asserts no convergence or mixing rate for the resulting chain.

AC is assumed as recorded in [A1]. By [F9], $H$ is second-countable and the Borel sigma-algebra of $H\times H$ is the product sigma-algebra. Put $\rho =\sqrt{1-{\beta }^2}$ and let ${\mathrm{\Lambda }}_0(du,dv)={\mu }_0(du)Q_{\beta }(u,dv)$ be the joint flow of the reference probability and proposal kernel. It exists by [F4]. Take the product probability space $(H\times H,{\mu }_0\otimes {\mu }_0)$ and its coordinate random elements $U_0,\xi$; [F6] gives their independence and common law ${\mu }_0$. For every nonnegative product-Borel $h$, the iterated-integral identity in [F4] and [F1] give $\int hd{\mathrm{\Lambda }}_0={\int }_H{\int }_{H}h(u,\rho u+\beta z){\mu }_0(dz){\mu }_0(du)=\mathbb{E}[h(U_0,\rho U_0+\beta \xi )],$ where the proposal-row pushforward and the change-of-variables identity [F5] identify the flow integral with expectation under $U_0\sim {\mu }_0$, and the product joint law [F6] together with the measurable proposal map identifies that expectation with the displayed one. Thus ${\mathrm{\Lambda }}_0$ is the law in [F3], so ${\mathrm{\Lambda }}_0$ is symmetric.

Define $w(u)=e^{-\bar{\mathrm{\Phi }}(u)}$ and $m(u,v)=\min \{w(u),w(v)\}$. For every $(u,v)$, if both potential values are finite, then $w(u)\alpha (u,v)=e^{-\bar{\mathrm{\Phi }}(u)}\min \{1,e^{\bar{\mathrm{\Phi }}(u)-\bar{\mathrm{\Phi }}(v)}\}=\min \{e^{-\bar{\mathrm{\Phi }}(u)},e^{-\bar{\mathrm{\Phi }}(v)}\}=m(u,v)\mathrm{.}$ If either value is $+\mathrm{\infty }$, the algorithm sets $\alpha (u,v)=0$ and $m(u,v)=0$, so the identity still holds. In particular, $m$ is a symmetric nonnegative Borel function.

Set $r(u)=1-a(u)\in [0,1]$. The diagonal map $d(u)=(u,u)$ is continuous from $H$ into the product topology on $H\times H$, so by [F10] it is Borel. Thus $u\mapsto {\mathbf{1}}_C(u,u)={\mathbf{1}}_C(d(u))$ is measurable for every Borel $C\subseteq H\times H$. By the diagonal term in [F1] and the joint-flow formula [F4], the rejection part of the target joint flow has mass on $C$ ${\mathrm{\Lambda }}_{\mathrm{r}\mathrm{e}\mathrm{j}}(C)={\int }_H{\mathbf{1}}_C(u,u)r(u)\mu (du)\mathrm{.}$ It is symmetric because $\tau (d(u))=d(u)$ for every $u$: for every Borel $C$, ${\mathbf{1}}_{{\tau }^{-1}C}(u,u)={\mathbf{1}}_C(u,u)$.

For Borel $C\subseteq H\times H$, the accepted part of the target joint flow is ${\mathrm{\Lambda }}_{\mathrm{a}\mathrm{c}\mathrm{c}}(C):={\int }_H{\int }_H{\mathbf{1}}_C(u,v)\alpha (u,v)Q_{\beta }(u,dv)\mu (du)=Z^{-1}{\int }_{H\times H}{\mathbf{1}}_C(u,v)m(u,v){\mathrm{\Lambda }}_0(du,dv)\mathrm{.}$ The equality follows from the target density in [F2], identity 1.2, and the iterated-integral formula [F4]. Since ${\mathrm{\Lambda }}_0$ is symmetric by step 1.1 and $m(u,v)=m(v,u)$, this accepted measure is symmetric. Indeed, for $\tau (u,v)=(v,u)$, the change-of-variables identity [F5] gives $\int {\mathbf{1}}_C(u,v)m(u,v)d{\mathrm{\Lambda }}_0=\int {\mathbf{1}}_C(v,u)m(v,u)d{\mathrm{\Lambda }}_0=\int {\mathbf{1}}_{{\tau }^{-1}C}(u,v)m(u,v)d{\mathrm{\Lambda }}_0\mathrm{.}$ Equality for every Borel $C$ is symmetry under $\tau$.

By the decomposition of $P_{\beta }$ in [F1] and the joint-flow identity [F4], ${\mathrm{\Lambda }}_{\beta }={\mathrm{\Lambda }}_{\mathrm{a}\mathrm{c}\mathrm{c}}+{\mathrm{\Lambda }}_{\mathrm{r}\mathrm{e}\mathrm{j}}$. The accepted and rejection pieces are symmetric by steps 2.1 and 1.3, hence ${\mathrm{\Lambda }}_{\beta }$ is symmetric. By [F7], $P_{\beta }$ is reversible with respect to $\mu$.

The hypotheses of [F8] hold: [F1] gives a probability kernel and the algorithm gives the probability measure $\mu$. Therefore [F8] yields ${\int }_H{P}_{\beta }(u,A)\mu (du)=\mu (A)$ for every Borel $A$. The proof uses only the symmetric-flow identity; it establishes no uniqueness or convergence claim.