General-state Markov kernels, invariance, and reversibility — Examples

Every matrix entry is nonnegative and each row sums to one, so [F4] makes $P$ a probability kernel. Also $a+b>0$ and $\pi (0)+\pi (1)=(a+b)\mathrm{/}(a+b)=1$, with both entries positive, so $\pi$ is a probability measure by [F3].

Each diagonal flow atom is unchanged by swap, and the off-diagonal atoms agree since ${\mathrm{\Lambda }}_{\pi ,P}(\{0\}\times \{1\})=\pi (0)a=ab\mathrm{/}(a+b)=\pi (1)b={\mathrm{\Lambda }}_{\pi ,P}(\{1\}\times \{0\})$. The four atoms generate the finite product sigma-algebra, so ${\mathrm{\Lambda }}_{\pi ,P}={\mathrm{\Lambda }}_{\pi ,P}^{\mathrm{\top }}$ and $P$ is reversible by [F1].

The general-state implication [F2] now gives $\pi P=\pi$, so $\pi$ is invariant.

For fixed $A\in \mathcal{B}(\mathbb{T})$, $x\mapsto {\mathbf{1}}_A({\tau }_{\pm h}x)$ is measurable by [F2]; the two translated points are distinct since $0<2h<1$, each row has total mass one, and disjoint-event indicators add pointwise, so [F4] makes $P$ a probability kernel.

For product-measurable $C$, the graph maps ${\mathrm{\Gamma }}_{\pm h}(x)=(x,{\tau }_{\pm h}x)$ are measurable because rectangle preimages are measurable by [F5]; with $g_{\pm }(x)={\mathbf{1}}_C(x,{\tau }_{\pm h}x)$, the joint-law formula [F6] gives ${\mathrm{\Lambda }}_{m_{\mathbb{T}},P}(C)=\frac{1}{2}\int g_{+}d{m}_{\mathbb{T}}+\frac{1}{2}\int g_{-}d{m}_{\mathbb{T}}$.

The swap map is measurable by [F5]. The identities $\tau \circ {\mathrm{\Gamma }}_h={\mathrm{\Gamma }}_{-h}\circ {\tau }_h$ and $\tau \circ {\mathrm{\Gamma }}_{-h}={\mathrm{\Gamma }}_h\circ {\tau }_{-h}$ and [F6] give ${\mathrm{\Lambda }}_{m_{\mathbb{T}},P}^{\mathrm{\top }}(C)=\frac{1}{2}\int g_{-}({\tau }_{h}x)d{m}_{\mathbb{T}}+\frac{1}{2}\int g_{+}({\tau }_{-h}x)d{m}_{\mathbb{T}}$. By [F3] and [F7] this equals the expression for ${\mathrm{\Lambda }}_{m_{\mathbb{T}},P}(C)$ in step 2.1, for every product-measurable $C$.

By [F8], symmetry makes $P$ reversible; [F9] then makes $m_{\mathbb{T}}$ invariant. Countable choice is inherited only through the supplied circle measure; the kernel calculation uses no further choice.

The uniform values sum to one, so [F6] makes $\pi$ a probability; for every set $A$, $K_j(i,A)=1$ exactly when ${\sigma }_j(i)\in A$ and is otherwise zero, so each row is a probability measure and its evaluations are measurable on the finite space by [F7]. The flow has mass $1\mathrm{/}3$ at $(i,{\sigma }_j(i))$; since ${\sigma }_j^2$ is the identity, its support and masses are symmetric, so both kernels are reversible by [F1].

Chronological composition [F3] gives $Q=K_1{K}_2$ with updates $0\mapsto 2$, $1\mapsto 0$, and $2\mapsto 1$; [F4] makes it a probability kernel. Each state has exactly one predecessor, so $\pi Q(\{j\})=1\mathrm{/}3=\pi (j)$ for every j and [F5] gives $\pi Q=\pi$.

By [F2] and step 1.1, each component kernel preserves the same uniform probability $\pi$.

The composite flow has ${\mathrm{\Lambda }}_{\pi ,Q}(\{0\}\times \{2\})=1\mathrm{/}3$ but ${\mathrm{\Lambda }}_{\pi ,Q}(\{2\}\times \{0\})=0$, so it is not symmetric by [F1]. Thus the composition is invariant but not reversible, although both updates are reversible and invariant by steps 1.1 and 2.1.