Chapter 2. Saturated array extensions and conditional locality

This chapter follows Definition 3.1, Proposition 3.2, Lemmas 3.3–3.4, and Proposition 3.5 of R (pp. 7–13). It starts with the full family of ordered arrays from Chapter 1. The argument enlarges the local state while preserving conditional product laws, then uses saturation to obtain joint independence of rows and exact conditional kernels for individual cells and grouped rectangles.

Fixed source, from page 7

The complete states at a few positions are added to the parameters. These positions are called parent slots; in a square array, they become parent rows. This additional information is distinct from the original local parameter $u(y)$ at each cell. A slot is simply a position indexed by the finite set $I$.

What remains to be proved

Suppose $k\ge 3$ is the least order at which mixing fails. Mixing below order $k$ gives the ordered array laws $\rho _{n}$, indexed by $I^{n}$, where $I=\{1,\ldots,k\}$. Fixing the last axis index at one slot gives a plane. Under $\rho _{n}$, any family of at most $k-1$ last-axis planes is independent as a family of complete random vectors. Independence here means a joint product law, beyond pairwise independence.

The ordered limits first give this conclusion along the last axis. For $n=2$, this is independence of proper subfamilies of whole columns; the corresponding statement for whole rows is still unavailable. Interchanging rows and columns cannot supply it: the source explicitly assumes no symmetry under permutations of the axes.

Section 3 of R enlarges both the local states and their parameters while keeping the resulting object in the specified array class. Saturation then reduces the new observations in the extension to observations in the original object. This is how row independence is obtained. The difficulty is that independence can be lost after conditioning on additional information. The proof below spells out the local dependence of conditional kernels that prevents this problem.

1. Admissible array objects and three kinds of observation

Definition 3.1 (pp. 7–8) specifies standard Borel probability systems $(Y,m_{0},T_Y)$ and $(U,\nu ,T_U)$, factor maps $\kappa :Y\to X$ and $u:Y\to U$ intertwining time, and, for each $n\ge 0$, an array law $m_{n}$ on $Y^{I^n}$. Here $I^{0}$ is a singleton, so level zero is the local state law $m_{0}$.

The following conditions must hold together:

1. $m_{n}$ is invariant under the simultaneous coordinatewise action of $T_Y$, and its image under coordinatewise $\kappa$ is the fixed law $\rho _{n}$. For $n\ge1$, fixing any axis index at any slot gives $m_{n-1}$. Every proper subfamily of last-axis planes is independent under $m_{n}$.
2. For $n\ge1$, put $\mathcal P_n=\sigma (u(y_v):v\in I^{n})$. For each proper subfamily $J\subsetneq I$, the planes remain independent given the full parameter array $\mathcal P_n$. The conditional law of plane $j$ depends only on its own parameter subarray and equals the conditional law of $m_{n-1}$ given that subarray.
3. Under $m_{0}$, the variables $\kappa (y)$ and $u(y)$ are independent. If $m_0=\int_U\eta_u\,\mathrm d\nu(u)$, this is equivalent to $\kappa _*\eta _u=\mu _X$ for $\nu$-almost every $u$.

Condition 2 has the following precise test-function formulation. For every proper subfamily $J$ and bounded plane tests $F_j$,

Here, $\mathcal P_{n-1,j}$ contains only the parameters of plane $j$. In particular, for a single $j$, the right-hand side still depends only on that plane's parameters. This locality requirement is an essential part of the condition.

For an extension $\theta :Y'\to Y$, the map $\theta$ must preserve all $m_{n}$ and $\kappa$, and the old parameter $u\circ \theta$ must be a factor of the new parameter $u'$. Thus each of the following old $\sigma$-algebras pulls back into the corresponding new $\sigma$-algebra:

Here $\mathcal O_{n,j}$ records all complete states outside the missing plane, including more than their parameters.

2. What saturation actually provides

Proposition 3.2 (pp. 8–9) states that every object has an extension $M$ with the following property. For every further admissible extension of $M$ with map $\theta$, every fixed $n\ge 1$ and $F\in L^{2}(m_{n})$, and $\mathcal D$ equal to $\mathcal P_n$ or one of $\mathcal O_{n,j}$ and $\mathcal G_{n,j}$, the identity below holds. Write $\theta_n$ for the coordinatewise factor map at level $n$, namely $\theta_n((y_v')_{v\in I^n})=(\theta(y_v'))_{v\in I^n}$. It is not an iterate of $\theta$ obtained by composition $n$ times.

Only extensions already shown to belong to the same array class may be compared here. Membership cannot be assumed and then established by (4). Lemma 3.4 avoids this circularity by induction on the number of parent slots.

Closure under inverse limits

Consider a countable tower of extensions. Take the compatible path spaces for the local states and the parameters, with time acting coordinatewise. Countable products of standard Borel spaces, and their Borel subsets of compatible paths, are again standard Borel. The consistent finite-stage laws determine the limit probability laws. Commutativity of the factor diagrams, face consistency along every axis, and time invariance hold first for finite-stage tests and then at the limit by monotone-class approximation.

Unconditional independence passes to the limit in the same way: pull all tests on finitely many planes back from one common finite stage, factor the product expectation there, and then approximate arbitrary bounded tests on the limit. The argument also preserves independence of the local $X$ coordinate $\kappa (y)$ from the parameters.

The conditional product identity needs a separate verification. Fix $J\subsetneq I$, and first take all $F_j$ from one finite stage $r$. For every $s\ge r$, the identity at stage $s$ is

The parameter $\sigma$-algebras on both sides increase with $s$. By forward martingale convergence, the left-hand side converges in $L^{2}$ to conditional expectation given the full limit parameters; every factor on the right also converges in $L^{2}$. If $|F_j|\le 1$, the product error satisfies

Thus the limit on the right is the product of the limit conditional expectations. For a bounded $F_j$ on the limit, let $F_j^{(s)}$ be its conditional expectation onto the information from plane $j$ at stage $s$, using that plane's complete-state $\sigma$-algebra. These approximations are still tests on the same plane. They converge in $L^{2}$ with a uniform supremum bound. Contractivity of conditional expectation and the same product estimate pass the identity to the limit. This preserves all of condition 2, including its full conditional product law.

Scheduling the bounded energy increments

At every stage and for every $n$, the space $L^{2}(m_{n})$ is separable. Choose a countable dense family of bounded tests. At fixed $n$ there are only $1+2k$ kinds of data. List all combinations of introduction stage, level, test, and kind of data in a countable schedule in which every introduced combination is revisited infinitely often.

For a fixed test $F$, use $\|\mathbb E[F|\mathcal D]\|_{2}$ as the energy. It is nondecreasing under admissible extensions and bounded by $\|F\|_{2}$. At each visit, choose an admissible extension whose energy is within $\varepsilon _q$ of the supremum over all currently available admissible extensions, with $\varepsilon _q\downarrow 0$. The supremum need not be attained. The identity extension ensures that the set of candidates is nonempty.

Let $M_\infty$ be the inverse limit of the countable tower and $Z$ any further admissible extension. Composition preserves extensions, so $Z$ is also an admissible candidate over every finite stage. At every sufficiently late visit to an old test, the following comparison holds.

Its projection norm in $Z$ is bounded by the current supremum. That supremum is at most the projection norm at the chosen next stage plus $\varepsilon _q$, and the next-stage norm is bounded by the norm in $M_\infty$. Hence the norm in $Z$ exceeds the norm in $M_\infty$ by at most $\varepsilon _q$. All tests are pulled back to the spaces being compared.

Letting the visit index tend to infinity, $q\to \infty$, shows that $Z$ cannot increase the energy. Finite-stage tests are dense in the inverse-limit $L^{2}$ space. For any $F$, we have the contraction estimate

This estimate extends the conclusion to all $F$. Finally, the pullback of the old $\mathcal D$ is contained in $\mathcal D'$, so orthogonal projections satisfy the Pythagorean identity

The left-hand side vanishes, proving (4). For each given further extension, this is an identity of equivalence classes. It makes no choice of one pointwise conull set common to all possible extensions; such a choice is unnecessary and does not follow by countable intersection.

3. A conditional-independence lemma

Lemma 3.3 (pp. 9–10) concerns standard random elements $E$, $O_{1},\ldots ,O_s$, and $B$, with $E_j=e_j(E)$ and $A_j=a_j(E_j,O_j)$. Assume

The second identity is a separate hypothesis. It says that the conditional law of $B$ depends on each $O_j$ only through the corresponding $A_j$.

Let $Q_j$ be the conditional law of $O_j$ given $(E_j,A_j)$. Then

Here is the full verification. The kernel $K_j(E_j,\cdot )$ is also the law of $O_j$ given $E_j$, because $E_j$ is a function of $E$ and the kernel depends only on $E_j$. For a value $x$ of $E_j$, let $M_j(x,\cdot )$ be the image of $K_j(x,\cdot )$ under $o\mapsto a_j(x,o)$. Disintegration with respect to $(E_j,A_j)$ gives

This identity holds for $\operatorname{Law}(E_j)$-almost every $x$. Substituting $x=e_j(E)$ therefore preserves it on a $\operatorname{Law}(E)$-conull set. Writing $\alpha=(\alpha_1,\ldots,\alpha_s)$, the original joint law becomes

The first three factors give the law of $(E,A,B)$, and the last factor is precisely the conditional kernel in (5). The fact that adding $B$ leaves the conditional kernels of the $O_j$ unchanged rests on the locality hypothesis in the second identity. Without it, the conclusion is generally false.

4. Incorporating the first axis into the local state

In Lemma 3.4 (pp. 10–11), the old object is now saturated. Fix any $R\subseteq I$ with $|R|\le k-2$, and choose $a\in I\setminus R$. Set

Here $R$ indexes the parent slots, while $a$ is the projection position retaining the old $X$ factor. The law of $u'$ and the coordinatewise time action define the new parameter system. It retains all old parameters and adds the complete states at positions in $R$; the complete states at other positions are not added.

At level $n$, use the old law $m_{n+1}$ and put the first axis $c\in I$ inside the new local state. The remaining indices $w\in I^{n}$ form the new array. Write the old states as $y_{c,w}$. Admissibility requires every defining condition to be checked at every $n$.

Checks that do not require induction

• The map $\theta _a$ retains $y_{a,w}$ pointwise. Face consistency along every axis of the old array gives its law $m_{n}$, so all old level laws are preserved. The $X$ factor is $\kappa (y_a)$.
• The new parameter contains $u(y_a)$ and thus determines the old parameter. All diagrams commute with the coordinatewise time actions.
• The last axis of the new level is the last axis of the old level. Unconditional independence of proper plane subfamilies, time invariance, and consistency of the other faces are inherited directly.
• The new local $X$ coordinate is independent of the new parameter. Under the old law $m_{1}$, condition on the full old parameter $p=(u(y_c))_c$. The set $R\cup\{a\}$ has at most $k-1$ positions, and its joint kernel is $\bigotimes _{c\in R\cup\{a\}}\eta _{p_c}$. After further conditioning on the states $(y_c)_{c\in R}$ added to the parameter, the conditional law of $y_a$ remains $\eta _{p_a}$. Since $\kappa _*\eta _{p_a}=\mu _X$, the law of $\kappa (y_a)$ given the full new parameter is always $\mu _X$.

Only the last-axis conditional product identity given the new parameters remains. When $R=\varnothing$, the full new parameter is exactly the full old parameter, so the required identity is the old conditional product identity at level $n+1$. This establishes the induction base without using saturation.

Nonempty $R$: first establish locality of the conditional kernel

Assume inductively that the construction gives admissible extensions for every parent set smaller than $|R|$, every permitted projection position, and every level. Fix $n\ge 1$ and a missing last-axis position $j_{0}$.

Use the following notation:

$E=(u(y_{c,w}))_{c,w}$ is the full old parameter; $O_j=(y_{c,w}:w_n=j)$, for $j\ne j_0$, is a complete observed last-axis plane; $O=(O_j)_{j\ne j_0}$; $E_j$ is the parameter array of $O_j$ itself; $A_j=(y_{c,w}:c\in R,w_n=j)$ consists of the states in $O_j$ added to the parameters; and $B=(y_{c,w}:c\in R,w_n=j_0)$ consists of all states in the missing plane added to the parameters.

In the old object, given $E$, the $O_j$ are independent and each conditional kernel depends only on $E_j$. To apply (5), we must prove

Enumerate $R$ as $c_{1},\ldots ,c_r$ and reveal the missing pieces $B_{c_t}=(y_{c_t,w}:w_n=j_{0})$ in that order. At step $t$, set $R'=\{c_1,\ldots,c_{t-1}\}$. Since $|R'|<|R|$ and $c_t\notin R'$, the induction hypothesis already supplies an admissible extension with this smaller parent set and projection position $c_t$.

The observation algebras for this extension can be identified explicitly:

• The pullback of the old $\mathcal G_{n,j_0}$ is $\sigma ((u(y_{c_t,w}))_w,(y_{c_t,w}:w_n\ne j_{0}))$.
• The new $\mathcal P_n'$ is $\sigma (E,(y_{l,w})_{l\in R',w\in I^{n}})$.
• The new $\mathcal O_{n,j_0}'$ is $\sigma (O)$, because the new local state contains the entire internal axis.
• The new $\mathcal G_{n,j_0}'$ therefore equals $\sigma (E,O,B_{R'})$. The part of $R'$ in the observed planes is already included in $O$.

For the missing piece $B_{c_t}$, take any bounded test $F$ and regard it as a test on the projected old level $n$. Applying saturation (4) to this already admissible extension gives equation (8) of the source:

The right-hand side is a fixed conditional kernel from the old law $m_{n}$. It depends only on $E$ and the part of internal row $c_t$ contained in $O$. Since $c_t\in R$, that part is included in the $A_j$. In particular, the kernel no longer depends on the previously revealed parent-state pieces $B_{R'}$ in the missing plane.

Write $A=(A_j)_{j\ne j_0}$, and denote this kernel by $H_{c_t}(E,A;\cdot)$. The chain rule for conditional distributions and repeated use of the preceding identity give

For example, for the test $\prod _{t\le r}b_t(B_{c_t})$, first condition on $E,O,B_{c_1},\ldots ,B_{c_{r-1}}$ and replace the last factor by $H_{c_r}b_r$. This depends only on $E,A$ and can be taken outside the next conditional expectation. Repetition gives the product of the $H_{c_t}b_t$. Product tests determine the joint kernel, establishing (7).

Every use of saturation has involved an extension with fewer parent slots. Only after the extension for the current set $R$ has been proved admissible may it be used for larger parent sets. Thus the extension being constructed is never used to justify its own admissibility.

Recovering the conditional product identity

Apply Lemma 3.3 with $E$, $O_j$, $A_j$, and $B$. Given $E$, all the $A_j$, and $B$, the $O_j$ remain independent. The conditional kernel of plane $j$ is the kernel of the old law $m_{n}$ given $(E_j,A_j)$.

Now $\sigma (E,(A_j)_{j\ne j_{0}},B)$ is exactly the full parameter algebra at new level $n$, and $(E_j,A_j)$ is exactly the parameter of new plane $j$. This plane has the old level-$n$ law $m_{n}$, which is also the new level-$n-1$ law. The kernel is therefore precisely the one required by Definition 3.1.

This proves the assertion first for the $k-1$ planes obtained by omitting one position. Every smaller proper subfamily lies in such a family, and its kernel follows by taking a marginal of the product kernel. Since $n$ was arbitrary, the conditional identity holds at every level, completing the induction.

5. Deriving row independence

For Proposition 3.5 (pp. 11–13), write the square as $y_{a,j}$, with row index $a$ and column index $j$. Fix the full parameter $E=(u_{a,j})$, the row parameters $p_a=(u_{a,j})_j$, and the column parameters $q_j=(u_{a,j})_a$. Denote the conditional law of the old $m_{1}$ given the parameter tuple $p$ by $\lambda _p$.

The definition already gives independence of every proper subfamily of whole columns given $E$, with column $j$ having kernel $\lambda _{q_j}$. The row assertion follows next.

Fix $R\subseteq I$, $|R|\le k-2$, and $a\notin R$, and use the admissible extension just constructed by incorporating the first axis. At its first level, take a bounded test of the whole row $Y_{a,I}$. For this test $F$, saturation with the $\mathcal P_1$ data gives:

Choose any $r\le k-1$ distinct rows $a_{1},\ldots ,a_r$ and reveal them in order. Before row $t$ is revealed, there are only $t-1\le k-2$ parent rows, so (14) applies. Successive conditional expectations give

This proves a conditional product law for every proper subfamily of whole rows. Independence of all $k$ rows is neither established nor needed.

The conditional product condition in the same extension gives Proposition 3.5(2): given $E$ and the parent rows $Y_{R,I}$, every proper subfamily of columns is independent, and each column is further conditioned only on its own parent entries $Y_{R,j}$.

Finally, saturation for $\mathcal G_{1,j_0}$ gives equation (12) of the source:

The left-hand side observes other rows, whereas the right-hand side uses only the remaining entries and parameters of the same row. This local cell kernel is needed for the operator rectangle identity in the next chapter.

6. From the single-cell identity (12) to the grouped identity (13)

The last step on pp. 12–13 concerns grouping. Partition the rows into three nonempty sets $P,A,B$ and the columns into three nonempty sets $r,s,t$. Fix $E$ and write $\Lambda _A=\bigotimes _{a\in A}\lambda _{p_a}$. The required identity is

First fix one missing column $j_{0}$. Given $E$, all whole rows in $P$, and every other column, reveal the entries for $a\in A$ in column $j_{0}$ one at a time. When entry $a$ is revealed, the parent rows comprise $P$ and the previously revealed rows in $A$. Their other entries are already known from the observed columns. There are at most

parent rows. Thus (12) applies at every step, and its right-hand side does not depend on previously revealed missing entries. Successive multiplication gives equation (15) of the source:

Now take $j_{0}\in t$, put $J=t\setminus\{j_0\}$ and $C=r\cup s$, and retain $W=Y_{A,J}$. The set $C\cup J$ consists of exactly $k-1$ columns. Proposition 3.5(2) makes them independent given $E$ and all whole rows in $P$. Within any column of $J$, the row set $P\cup A$ is proper because $B$ is nonempty. Hence, given its $P$ entries, the marginal of its $A$ entries is still $\bigotimes _{a\in A}\eta _{u_{a,j}}$. Consequently,

In (15), integrate out the other states in the $J$ columns, retaining only $W$. The right-hand side depends on those $J$ columns only through $W$, so its kernel is unchanged. The joint conditional kernel of $(W,Y_{A,j_{0}})$ is therefore the product marginal in (16) followed by the row kernels in (15).

Under $\Lambda _A$, the rows are independent. Within each row, $C\cup J$ is a proper coordinate set and has the same product marginal. Composing these two kernels therefore gives exactly the conditional law of $\Lambda _A$ given $Y_{A,C}$, proving (13). If $J=\varnothing$, the first kernel is the empty product, and the calculation still applies.

The other grouped assertions also have explicit sources. A group of rows is a proper subfamily of whole rows and therefore has the product of their $\lambda$ laws. Within one row, the union of any two column groups remains a proper column set, so the three rectangular cells of the grouped row are pairwise independent. The column argument is the same. Given all whole rows in $P$, the union of any two column groups is a proper column set, so the required conditional independence of grouped columns follows from part (2). Each argument starts from a joint product law for a proper subfamily and then groups coordinates. It does not infer joint independence of three groups from pairwise independence.

7. Conull sets and quantifiers

All state spaces are standard Borel, so regular conditional distributions exist, and a countable generating algebra can determine each probability kernel. Verify a kernel identity first on this determining class, then obtain equality of probability measures on one conull set. This supplies the common set needed to use the identity for bounded tests, beyond separate almost-everywhere statements for individual tests.

There are countably many levels $n$. For fixed $k$, there are only finitely many parent subsets, projection positions, missing positions, and square groupings; integer time shifts are countable. The required kernel identities can therefore hold simultaneously on a common conull set at each relevant level. When time covariance is needed, intersect over all integer shifts as well.

Random variables at different levels live in different spaces, so common conull sets are understood level by level. At $n=2$, disintegrating over the law of the full parameter $E$ gives a conull parameter set. For every $E$ in that set, the assertions hold almost everywhere under its conditional square law. They need not hold at every state point over that parameter.

The proof does not give any of the following stronger assertions:

• “Saturation holds for all extensions on one common pointwise conull set.”
• “Conditional independence automatically survives the addition of parent slots.”
• “Column independence gives row independence by interchanging the axes.”
• “The extended $T_Y$ is still mixing.”
• “Product laws for every proper subfamily imply a product law for the complete $k$-tuple.”

The source retains mixing only in the $X$ coordinate. The enlarged system $Y$ is not required to be mixing.

Aside: why column independence alone is insufficient

The following example shows why a shortcut omitting saturation can fail. It is not a counterexample to R, and no claim is made that this finite square extends to the full array tower required there.

Take $k=3$. Let $U_a$ and $V_a$, for $a=1,2,3$, together with $H$, be seven independent fair bits. With $\oplus$ denoting addition modulo two, define

The three entries in each row are independent: an output triple in a fixed row uniquely determines $U_a$, $V_a$, and $H$, so it has probability 1/8. The three entries in each column are also independent. Any two whole columns are independent. This is immediate for the first two. Given the first column, the third is the independent uniform vector $V$ translated by a vector, and its uniform law is independent of the first column. The second and third columns are treated in the same way. The parameters may be constant.

Yet the parity $y_{a,1}\oplus y_{a,2}\oplus y_{a,3}$ of every row is the same variable $H$. Observing one complete parent row $P$ therefore determines $H$. After the first two whole columns are observed as well, the missing entry of another row $A$ is

Its conditional law is a point mass. Under the marginal law of row $A$ alone, given $y_{A,1},y_{A,2}$, the third entry is still a fair bit. Thus the two sides of (12) can differ in this finite example. The original column product laws do not justify (12). R first derives locality of the conditional kernels from saturation and extensions with smaller parent sets, and then obtains (12). That part of the local argument is necessary.