The pCN accept-reject kernel preserves the posterior

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.