Gibbs sampling and scan schemes

Apply [F1] with the block coordinate as target and the retained coordinates as conditioning variable. Its formula gives probability sections on $D$ by normalization and uses $\rho$ on every exceptional fibre; the supplier proves measurability of evaluations and that the exceptional set has zero marginal mass. Thus $K_B(z,\cdot )$ is a probability measure for every $z$, not only almost every $z$.

Fix any measurable $F\subseteq S_{-B}$. Under the coordinate grouping $S\cong S_B\times S_{-B}$, the inverse image under ${\mathrm{splice}}_B$ of a measurable rectangle $A\times C$ is $C\times A$. Since such rectangles generate the finite-product sigma-algebra (The product sigma-algebra and its finite iterates), the function $(z,u)\mapsto {\mathbf{1}}_E({\mathrm{splice}}_B(z,u))$ is product-measurable for every product-measurable $E\subseteq S$. The kernel integration theorem [F4] therefore makes $z\mapsto K_B(z,E_z)$ measurable. Define ${\nu }_F(E)={\int }_F{K}_B(z,E_z){\pi }_{-B}(dz)$; this integral is defined for every such $E$.

The set function $E\mapsto \pi (E\cap p_{-B}^{-1}(F))$ is a finite measure by restriction of $\pi$. The set function ${\nu }_F$ is also a finite measure: it vanishes at $\mathrm{\varnothing }$, and ${\nu }_F(S)={\int }_{F}1d{\pi }_{-B}={\pi }_{-B}(F)\le 1$. If $(E_n)$ are pairwise disjoint product-measurable sets, their sections $(E_n{)}_z$ are pairwise disjoint, so $K_B(z,({\bigcup }_n{E}_n{)}_z)={\sum }_n{K}_B(z,(E_n{)}_z)$. The partial sums are measurable by step 1.2 and increase; applying [F5] to ${\mathbf{1}}_F(z){\sum }_{n=1}^N{K}_B(z,(E_n{)}_z)$ and [F6] to each finite sum gives ${\nu }_F({\bigcup }_n{E}_n)={\lim }_N{\sum }_{n=1}^N{\nu }_F(E_n)={\sum }_n{\nu }_F(E_n)$.

For a coordinate rectangle $E=A\times C$, the density event identity [F1] gives $\pi (E\cap p_{-B}^{-1}(F))=\pi (A\times (C\cap F))={\int }_{C\cap F}{K}_B(z,A){\pi }_{-B}(dz)={\nu }_F(E)$. The coordinate rectangles form a generating pi-system and include the whole space. The two finite measures agree on that system and have the common finite total mass ${\pi }_{-B}(F)$, so the constant sequence consisting of the whole space is a finite-measure exhaustion in [F3]. Measure uniqueness yields $\pi (E\cap p_{-B}^{-1}(F))={\nu }_F(E)$ for every product-measurable $E$. Since $F$ was arbitrary, [F2] holds for the displayed kernel. The kernel/event construction itself uses no AC; the optional conditional-expectation-class convention is not invoked.

Write $u=(x-m_X)\mathrm{/}{\sigma }_X$ and $v=(y-m_Y)\mathrm{/}{\sigma }_Y$. The quadratic form of $\mathrm{\Sigma }$ is $(a,b)\mathrm{\Sigma }(a,b{)}^T=({\sigma }_{X}a+r{\sigma }_{Y}b{)}^2+(1-r^2){\sigma }_Y^2{b}^2$, so it is positive for every nonzero $(a,b)$; also $\det \mathrm{\Sigma }={\sigma }_X^2{\sigma }_Y^2(1-r^2)>0$. Its inverse is ${\mathrm{\Sigma }}^{-1}=\frac{1}{{\sigma }_X^2{\sigma }_Y^2(1-r^2)}\left(\begin{array}{cc}{\sigma }_Y^2& -r{\sigma }_X{\sigma }_Y\\ -r{\sigma }_X{\sigma }_Y& {\sigma }_X^2\end{array}\right)$. Thus [F2] applies. Substitution into its density formula, using [F3] to identify the product reference measure, gives $p(x,y)=\frac{1}{2\pi {\sigma }_X{\sigma }_Y\sqrt{1-r^2}}\exp [-\frac{u^2-2ruv+v^2}{2(1-r^2)}]$. The covariance convention in [F1] fixes the parameters used here.

Set $s_X={\sigma }_X\sqrt{1-r^2}$, $s_Y={\sigma }_Y\sqrt{1-r^2}$, $a(y)=m_X+r{\sigma }_{X}v$, and $b(x)=m_Y+r{\sigma }_{Y}u$. For $s>0$ write $g_{c,s}(t)=s^{-1}\varphi ((t-c)\mathrm{/}s)$. The identities $u^2-2ruv+v^2=(u-rv{)}^2+(1-r^2)v^2=(v-ru{)}^2+(1-r^2)u^2$ therefore give both factorizations $p(x,y)=g_{m_Y,{\sigma }_Y}(y)g_{a(y),s_X}(x)=g_{m_X,{\sigma }_X}(x)g_{b(x),s_Y}(y)$. Expanding each squared term verifies the exponents, and multiplying the factors verifies the prefactors.

For any $c\in \mathbb{R}$, $s>0$, and Borel set $A\subseteq \mathbb{R}$, the diffeomorphism $T(t)=c+st$ in [F4] gives ${\int }_A{g}_{c,s}(z)dz={\int }_{T^{-1}(A)}\varphi (t)dt=N(c,s^2)(A)$ by [F5]. Thus each $g_{c,s}$ is a probability density. Applying [F6] to the first factorization in step 2.1 yields $m_Y(y)=\int p(x,y)dx=g_{m_Y,{\sigma }_Y}(y)$; the second factorization gives $m_X(x)=\int p(x,y)dy=g_{m_X,{\sigma }_X}(x)$. These marginal densities are finite and strictly positive at every real argument.

Apply [F7] to $p$ with Lebesgue reference measures and, for each updated coordinate, the supplied filler $N(0,1)$ from [F5]. Since both marginal densities in step 3.1 are finite and positive everywhere, the density formula is used on every conditioning fibre. For measurable $A\subseteq \mathbb{R}$, the first conditional kernel is $K_X(y,A)=\frac{{\int }_{A}p(x,y)dx}{m_Y(y)}={\int }_A{g}_{a(y),s_X}(x)dx=N(a(y),s_X^2)(A)$, using the factorization in step 2.1 and density identification in step 3.1. Exchanging coordinates gives $K_Y(x,A)=N(b(x),s_Y^2)(A)$. By [F7] these are probability kernels satisfying the conditional event identities; [F8] identifies them as proper full conditionals at every conditioning value. The proof makes no assertion at the excluded singular endpoints.

For each fixed $x$, the map $u\mapsto {\mathrm{splice}}_B(x_{-B},u)$ is measurable by the finite product sigma-algebra. Thus $P_B(x,\mathrm{\varnothing })=0$, $P_B(x,S)=\int 1d{K}_B(x_{-B})=1$, and for pairwise disjoint $C_n$ the indicators of their inverse images sum increasingly to the indicator of the inverse image of ${\bigcup }_n{C}_n$. [F4] gives countable additivity, so $P_B(x,\cdot )$ is a probability measure.

For fixed measurable $C\subseteq S$, the function $(x,u)\mapsto {\mathbf{1}}_C({\mathrm{splice}}_B(x_{-B},u))$ is product-measurable because coordinate projection and splicing are measurable for the finite product sigma-algebra. The lifted kernel $x\mapsto K_B(x_{-B},\cdot )$ is a probability kernel by [F1] and measurability of the projection. Applying [F3] to this kernel gives measurability of $x\mapsto P_B(x,C)$. Thus every evaluation is measurable and every section is a probability measure.

Set $D=S_{-B}$ in [F1]. This identifies the law of $(Z,U)$ with ${\pi }_{-B}(dz)K_B(z,du)$ on every product-measurable event. By [F2], after the current $(Z,U)=(z,u)$ the refreshed block $V$ has law $K_B(z,\cdot )$. Form the product probability kernel $K_B(z,\cdot )\otimes K_B(z,\cdot )$ using [F4] and mix it against ${\pi }_{-B}$ by [F5]. For a rectangle $D\times A\times C$, the first conditional identity gives probability $K_B(z,A)$ for $U\in A$ at retained value $z$, and the update gives probability $K_B(z,C)$ for $V\in C$; iterating yields ${\int }_D{K}_B(z,A)K_B(z,C){\pi }_{-B}(dz)$. This is also the rectangle value of the constructed triple law. Both are probability measures, so [F7] makes them equal on the product sigma-algebra.

For measurable $A,C\subseteq S_B$ and $D\in {\mathcal{S}}_{-B}$, the triple-law formula from step 1.1 gives $\mathbb{P}(Z\in D,U\in A,V\in C)={\int }_D{K}_B(z,A)K_B(z,C){\pi }_{-B}(dz)$. The same formula with one of $A,C$ equal to the whole block space gives the two marginal testing identities. Hence, by testing on every event $\{Z\in D\}$, $K_B(Z,A)$ and $K_B(Z,C)$ are versions of the corresponding conditional expectations, and their product is a version of the conditional expectation of ${\mathbf{1}}_A(U){\mathbf{1}}_C(V)$. The conditional-independence identity [F3] therefore holds for indicator functions.

Finite linear combinations extend the identity in step 2.1 to finite-valued measurable simple functions. For bounded measurable real $f,g$, partition bounded intervals containing their ranges into finitely many subintervals of length below $1\mathrm{/}n$ and choose one value on each inverse-image set; this gives finite-valued measurable simple $f_n,g_n$ with uniform errors tending to zero. By [F6], $h_f(z)=\int fd{K}_B(z,\cdot )$ and $h_g(z)=\int gd{K}_B(z,\cdot )$ are measurable; uniform approximation also gives uniform convergence of $h_{f_n}$ to $h_f$ and $h_{g_n}$ to $h_g$. Since the functions are uniformly bounded, the products $f_n(U)g_n(V)$ and $h_{f_n}(Z)h_{g_n}(Z)$ converge uniformly to their respective limits, so their testing identities over every $\{Z\in D\}$ pass to the limit. Thus $h_f(Z)$ and $h_g(Z)$ are the two conditional-expectation versions and $h_f(Z)h_g(Z)$ is a version for the product, proving [F3] for all bounded $f,g$. AC is used only for this class-valued notation.

The full conditional identity [F1] says that the target law in coordinates $(z,u)$ and the iterated law ${\pi }_{-B}(dz)K_B(z,du)$ agree on coordinate rectangles. Both are probability measures, and rectangles generate the finite product sigma-algebra, so [F8] makes them equal on every product-measurable event. Integrating nonnegative simple functions and then increasing limits gives $\int h(x)\pi (dx)=\int {\pi }_{-B}(dz)\int h({\mathrm{splice}}_B(z,u))K_B(z,du)$ for every nonnegative product-measurable $h$.

Let $M$ be the joint law of the current and updated states under initial law $\pi$ and transition $P_B$, which exists by [F3]–[F4]. For arbitrary measurable $A,C\subseteq S$, the update formula and step 1.1 give $M(A\times C)=\int {\pi }_{-B}(dz)\int K_B(z,du){\mathbf{1}}_A({\mathrm{splice}}_B(z,u))\int K_B(z,dv){\mathbf{1}}_C({\mathrm{splice}}_B(z,v))$. All integrands are bounded measurable functions, so the iterated integrals are defined.

For each $z$, the product measure $K_B(z,\cdot )\otimes K_B(z,\cdot )$ is invariant under swapping its two coordinates: the product measure and its swapped image agree on every rectangle because $K_B(z,A)K_B(z,C)=K_B(z,C)K_B(z,A)$, hence agree on the generated product sigma-algebra by [F8]. Applying this symmetry to the two inner factors in step 2.1 yields $M(A\times C)=M(C\times A)$ for arbitrary $A,C$. By [F6], $P_B$ is reversible with respect to $\pi$.

Set $C=S$ in the detailed-balance identity of step 3.1. Since $P_B(x,S)=1$, its left side is $\pi (A)$ and its right side is $\int P_B(y,A)\pi (dy)=(\pi P_B)(A)$. This is invariance by [F7], for every measurable $A$.

For each measurable $A$, the finite weighted sum in [F2] is measurable in $x$ by [F1], and it is nonnegative. For pairwise disjoint $A_n$, countable additivity of each $P_j(x,\cdot )$ gives ${\sum }_j{w}_j{P}_j(x,{\bigcup }_n{A}_n)={\sum }_j{w}_j{\sum }_n{P}_j(x,A_n)={\sum }_n{\sum }_j{w}_j{P}_j(x,A_n)$; the interchange is valid because the sum over $j$ is finite and every term is nonnegative. Its total mass is ${\sum }_j{w}_j{P}_j(x,S)={\sum }_j{w}_j=1$. Hence the random-scan kernel is a probability kernel.

Each $P_j$ is a probability kernel by [F1]. Repeated application of the composition result in [F3] shows that $P_1\cdots P_m$ is a probability kernel; associativity makes this finite product independent of parenthesization. This is the full-sweep transition kernel.

By [F2] and nonnegativity of the weights, for every measurable $A,C$, ${\int }_A{P}_{\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{d}}(x,C)\pi (dx)={\sum }_{j=1}^m{w}_j{\int }_A{P}_j(x,C)\pi (dx)$. This follows by applying [F7] to each scalar multiple and [F6] repeatedly to the finite sum.

Reversibility of each $P_j$ in [F1] changes the last expression to ${\sum }_{j=1}^m{w}_j{\int }_C{P}_j(y,A)\pi (dy)={\int }_C{P}_{\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{d}}(y,A)\pi (dy)$. Thus the detailed-balance identity [F3] holds for every $A,C$.

Set $C$ equal to the whole state space in step 2.1 and use $P_{\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{d}}(y,S)=1$ from [F5]. The resulting identity is $\int P_{\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{d}}(y,A)\pi (dy)=\pi (A)$, which is invariance by [F4]. The proof establishes stationarity only; it does not establish irreducibility, uniqueness, ergodicity, or a rate.

For two probability kernels $P,Q$, the iterated-kernel identity [F5] gives $\int (PQ)(x,A)\pi (dx)=\int Q(y,A)(\pi P)(dy)$. This follows by expanding $(PQ)(x,A)=\int Q(y,A)P(x,dy)$ and applying the joint-law iterated-integral identity to the bounded nonnegative function $Q(y,A)$.

To prove that reversibility can fail, give $S=\{0,1{\}}^2$ its full sigma-algebra and target masses $\pi (0,0)=\pi (0,1)=\pi (1,0)=1\mathrm{/}5$ and $\pi (1,1)=2\mathrm{/}5$. Supply the $X$-conditional kernel $K_X(0,\{0\})=K_X(0,\{1\})=1\mathrm{/}2$, $K_X(1,\{0\})=1\mathrm{/}3$, $K_X(1,\{1\})=2\mathrm{/}3$, and the $Y$-conditional kernel with these same respective probabilities when conditioning on $X=0$ or $X=1$. These are probability kernels because their nonnegative rows sum to one and every map between the finite spaces is measurable. For every $x,y$, $\pi (x,y)={\pi }_Y(y)K_X(y,\{x\})={\pi }_X(x)K_Y(x,\{y\})$ by the displayed masses. Summing these identities over arbitrary subsets verifies every conditional event identity in [F6]. They are therefore exact full-conditionals without invoking the downstream Gaussian example.

Define ${\mu }_0=\pi$ and ${\mu }_k={\mu }_{k-1}{P}_k$ for $1\le k\le m$. If ${\mu }_{k-1}=\pi$, then [F1] gives ${\mu }_k=\pi P_k=\pi$. Induction from ${\mu }_0=\pi$ yields ${\mu }_m=\pi$; associativity in [F3] identifies ${\mu }_m=\pi (P_1\cdots P_m)$, and [F4] calls this invariance.

Let $P_X$ refresh $X$ and $P_Y$ refresh $Y$, using the kernels in step 1.2, and set $P=P_X{P}_Y$ in chronological order. By [F3] and [F6], $P((0,0),\{(0,1)\})=K_X(0,\{0\})K_Y(0,\{1\})=1\mathrm{/}4$, whereas $P((0,1),\{(0,0)\})=K_X(1,\{0\})K_Y(0,\{0\})=1\mathrm{/}6$. Thus the forward stationary flow between these singleton states is $1\mathrm{/}20$ and its reverse is $1\mathrm{/}30$. These unequal values violate [F7]. Step 2.1 nevertheless proves invariance for this sweep, since its two updates are exact by step 1.2. This independently supplies the nonreversibility witness and completes both claims.