1 Introduction

The notion of standardness has been introduced by Vershik [10] in the context of decreasing sequences of measurable partitions indexed by the non-negative integers. Vershik’s definition and characterizations of standardness have been translated from their original ergodic theoretic formulation into a probabilistic language by Émery and Schachermayer [2]. In this framework, the objects of focus are the filtrations indexed by non-positive integers. These are the non-decreasing sequences \((\mathcal{F }_n)_{n \le 0}\) of sub-\(\sigma \)-fields of a probability space \((\Omega ,\mathcal{A },{\mathbb{P }})\).

All the sub-\(\sigma \)-fields of \(\mathcal{A }\) that we will consider are assumed to be complete and essentially separable with respect to \({\mathbb{P }}\). By definition, a sub-\(\sigma \)-field of \((\Omega ,\mathcal{A },{\mathbb{P }})\) is separable if it can be generated as a complete \(\sigma \)-field by a sequence of events, or equivalently, by some real random variable. One can check that a sub-\(\sigma \)-field \(\mathcal{B }\subset \mathcal{A }\) is separable if and only if the Hilbert space \(L^2(\Omega ,\mathcal{B },{\mathbb{P }})\) is separable.

Almost all filtrations that we will consider in this study have the following property: for each \(n, \mathcal{F }_n\) is generated by \(\mathcal{F }_{n-1}\) and by some random variable \(U_n\) which is independent of \(\mathcal{F }_{n-1}\) and uniformly distributed on some finite set with \(r_n\) elements, for some sequence \((r_n)_{n \le 0}\) of positive integers. Such filtrations are called \((r_n)_{n \le 0}\)-adic.

For such filtrations, as shown by Vershik [10], standardness turns out to be tantamount to a simpler, much more intuitive property: an \((r_n)\)-adic filtration \(\mathcal{F }\) is standard if and only if \(\mathcal{F }\) is of product type, that is, \(\mathcal{F }\) is the natural filtration of some process \(V = (V_n)_{n\le 0}\) where the \(V_n\) are independent random variables; in this case, it is easy to see that the process \(V\) can be chosen with the same law as \(U= (U_n)_{n\le 0}\). So, at first reading, ‘standard’ can be replaced with ‘of product type’ in this introduction.

Although intuitive, the notion of product-type filtrations is not as simple as one could believe. For example, the assumption that the tail \(\sigma \)-field \(\mathcal{F }_{-\infty } = \bigcap _{n \le 0} \mathcal{F }_n\) is trivial, and the property \(\mathcal{F }_n = \mathcal{F }_{n-1} \vee \sigma (U_n)\) for every \(n \le 0\) do not ensure that \((\mathcal{F }_n)_{n \le 0}\) is generated by \((U_n)_{n \le 0}\). In the standard case, \((\mathcal{F }_n)_{n \le 0}\) can be generated by some other sequence \((V_n)_{n \le 0}\) of independent random variables which has the same law as \((U_n)_{n \le 0}\). In the non-standard case, no sequence of independent random variables can generate the filtration \((\mathcal{F }_n)_{n \le 0}\).

The first examples of such a situation were given by Vershik [10]. By modifying and generalizing one of these examples, Smorodinsky [8] and Émery and Schachermayer [2] introduced the split-words processes.

The law of a split-words process depends on an alphabet \(A\), endowed with some probability measure, and a decreasing sequence \((\ell _n)_{n \le 0}\) of positive integers (the lengths of the words) such that \(\ell _0=1\) and the ratios \(r_n = \ell _{n-1}/\ell _n\) are integers. For the sake of simplicity, we consider here only finite alphabets endowed with the uniform measure.

A split-words process is an inhomogeneous Markov process \(((X_n,U_n))_{n \le 0}\) such that for every \(n \le 0\):

  • \((X_n,U_n)\) is uniform on \(A^{\ell _n} \times [\![1,r_n ]\!]\).

  • \(U_n\) is independent of \(\mathcal{F }^{(X,U)}_{n-1}\).

  • if one splits the word \(X_{n-1}\) (of length \(\ell _{n-1} = r_n\ell _n\)) into \(r_n\) subwords of lengths \(\ell _n\), then \(X_n\) is the \(U_n\)-th subword of \(X_{n-1}\).

Such a process is well-defined since the sequence of uniform laws on the sets \(A^{\ell _n} \times [\![1,r_n ]\!]\) is an entrance law for the transition probabilities given above. By construction, the natural filtration \(\mathcal{F }^{X,U}\) of \(((X_n,U_n))_{n \le 0}\) is \((r_n)_{n \le 0}\)-adic. One can check that the tail \(\sigma \)-field \(\mathcal{F }^{X,U}_{-\infty }\) is trivial. Thus, it is natural to ask whether \(\mathcal{F }^{X,U}\) is standard or not.

Whether a split-words process with lengths \((\ell _n)_{n \le 0}\) generates a standard filtration or not is completely characterised: the filtration is non-standard if and only if

$$\begin{aligned} \sum _{n}\frac{ \log r_{n}}{ \ell _{n}} < +\infty \quad (\Delta ). \end{aligned}$$

Note that this condition does not depend on the alphabet \(A\).

In this statement, the ‘if’ part and a partial converse have been proved by Vershik [10] (in a very similar framework) and by Laurent [5]. The ‘only if’ part has been proved by Heicklen [4] (in Vershik’s framework) and by Ceillier [1]. The generalization to arbitrary alphabets has been performed by Laurent in [7]: the characterisations and all the results below still hold are when the alphabet is a Polish space endowed with some probability measure.

Although these examples are rather simple to construct, proving the non-standardness requires sharp tools like Vershik’s standardness criterion [2, 10]. One can also use the I-cosiness criterion of Émery and Schachermayer [2] which may be seen as more intuitive by probabilists. Actually, Laurent proved directly that both criteria are equivalent. Moreover, applying these criteria to the examples above leads to rather technical estimations.

Another question concerns what happens to a filtration when time is accelerated by extracting a subsequence. Clearly, every subsequence of a standard filtration is still standard. But Vershik’s lacunary isomorphism theorem [10] states that from any filtration \((\mathcal{F }_n)_{n \le 0}\) such that \(\mathcal{F }_0\) is essentially separable and \(\mathcal{F }_{-\infty }\) is trivial, one can extract a filtration \((\mathcal{F }_{\phi (n)})_{n \le 0}\) which is standard. This striking fact is mind-boggling for anyone who is interested by the boundary between standardness and non-standardness. A natural question arises: when \((\mathcal{F }_n)_{n \le 0}\) is not standard, how close to identity the increasing map \(\phi \) (from \({\mathbb{Z }}_-\) to \({\mathbb{Z }}_-\)) provided by the lacunary isomorphism theorem can be?

Of course, as standardness is an asymptotic property, the extracting map \(\phi \) has to skip an infinity of times integers (equivalently, \(\phi (n)-n \rightarrow -\infty \) as \(n \rightarrow -\infty \)).

In [10], Vershik provides an example of a non-standard dyadic filtration \((\mathcal{F }_n)_{n \le 0}\) such that \((\mathcal{F }_{2n})_{n \le 0}\) is standard. Gorbulsky also provides such an example in [3].

Using the fact that the family of split-words filtrations is stable by extracting subsequences, Ceillier exhibits in [1] an example of a non-standard filtration \((\mathcal{F }_n)_{n \le 0}\) which is as close to standardness as it can be: every subsequence \((\mathcal{F }_{\phi (n)})_{n \le 0}\) is standard as soon as \(\phi \) skips an infinity of integers.

This paper is devoted to the filtrations sharing this property. We call them filtrations at the threshold of standardness.

1.1 Main results and organization of the paper

Some definitions and classical facts used in the paper are recalled in an Appendix, at the end of the paper. In the Sects. 2 and 3 which are the core of the paper, two classes of filtrations are studied, first the filtrations of the split-words processes, second some filtrations inspired by an unpublished example of Tsirelson.

The case of split-words filtrations The first part deals with split-words filtrations.

First, we characterise the filtrations at the threshold of standardness among the split-words filtrations.

Proposition 1

A split-words filtration with lengths \((\ell _n)_{n \le 0}\) is at the threshold of standardness if and only if

$$\begin{aligned} \sum _{n \le 0} \frac{\log r_{n}}{\ell _{n}} < +\infty \quad (\Delta ) \end{aligned}$$

and

$$\begin{aligned} \inf _{n \le 0} \frac{\log r_{n-1}}{\ell _{n}} >0 \quad (\star ). \end{aligned}$$

Next, we give a sufficient condition which ensures that a standard split-words filtration cannot be extracted from any split-words filtration at the threshold of standardness.

Proposition 2

If

$$\begin{aligned} \sum _{n \le 0} \frac{\log r_{n}}{\ell _{n}} = +\infty \quad (\lnot \Delta ) \end{aligned}$$

and

$$\begin{aligned} \lim _{n \rightarrow -\infty } \frac{\log r_{n}}{\ell _{n}} = 0 \quad (\Box ), \end{aligned}$$

then any split-words filtration with lengths \((\ell _n)_{n \le 0}\) is standard but cannot be extracted from a split-words filtration at the threshold of standardness.

One could think that the threshold of standardness is a kind of boundary between standardness and non-standardness. Yet, the situation is not so simple. Indeed, Proposition 2 provides an example (Example 3) of two split-words filtrations, where

  • the first one is non-standard,

  • the second one is standard,

  • the second one is extracted from the first one,

  • yet, no intermediate filtration (for extraction) is at the threshold of standardness.

Furthermore, we provide an example of a non-standard split-words filtration from which no filtration at the threshold of standardness can be extracted (Example ). The proof relies on Theorem A below.

Recall that, given any filtration \((\mathcal{F }_n)_{n \le 0}\) and an infinite subset \(B\) of \({\mathbb{Z }}_-\), the extracted filtration \((\mathcal{F }_n)_{n \in B}\) is standard if and only if the complement \(B^c = {\mathbb{Z }}_- \setminus B\) is large enough in a certain way. Here, the meaning of “large enough” depends on the filtration \(\mathcal{F }\) considered. When \(\mathcal{F }\) is at the threshold of standardness, “large enough” means exactly “infinite”. But various types of transition from non-standardness to standardness are possible, and the next theorem provides some other possible conditions.

Theorem A

Let \((\alpha _n)_{n \le 0}\) be any sequence of non-negative real numbers. There exists a split-words filtration \((\mathcal{F }_n)_{n \le 0}\) such that for every infinite subset \(B\) of \({\mathbb{Z }}_-\), the extracted filtration \((\mathcal{F }_n)_{n \in B}\) is standard if and only if

$$\begin{aligned} \sum _{n \in B^c} \alpha _n = +\infty \quad or\quad \sum _{n \le 0} \mathbf{1 }_{[n \notin B,\,n+1\notin B]} = +\infty . \end{aligned}$$

In other words, the extracted filtration \((\mathcal{F }_n)_{n \in B}\) is standard if and only if B skips infinitely many pairs of consecutive integers or if B skips sufficiently many integers to make the series \(\sum _{n \in B^c} \alpha _n\) diverge. The smaller the \(\alpha _n\) are, the most time must be accelerated to get a standard filtration. In contrast, if the sequence \((\alpha _n)_{n \le 0}\) is bounded away from \(0\), the extraction has only to skip infinitely many integers, therefore \((\mathcal{F }_n)_{n \le 0}\) is at the threshold of standardness.

Theorem A immediately provides other interesting examples. For example, it may happen that \((\mathcal{F }_{2n})_{n \le 0}\) is standard while \((\mathcal{F }_{2n-1})_{n \le 0}\) is not, or vice versa. When this phenomenon occurs, we will say that the filtration \((\mathcal{F }_{n})_{n \le 0}\) “interlinks” standardness and non-standardness.

Repeated interlinking is possible. By slowing time suitably in a filtration at the threshold of standardness (Example ), one gets a filtration \((\mathcal{F }_{n})_{n \le 0}\) such that \((\mathcal{F }_{2n})_{n \le 0}, (\mathcal{F }_{4n})_{n \le 0}, (\mathcal{F }_{8n})_{n \le 0},\ldots \) are non-standard, whereas \((\mathcal{F }_{2n-1})_{n \le 0}, (\mathcal{F }_{4n-2})_{n \le 0}, (\mathcal{F }_{8n-4})_{n \le 0},\ldots \) are standard.

Improving on an example of Tsirelson In a second part, we study another type of filtrations inspired by a construction of Tsirelson in unpublished notes [9].

Tsirelson has constructed an inhomogeneous discrete Markov process \((Z_n)_{n \le 0}\) such that the random variables \((Z_{2n})_{n \le 0}\) are independent and such that the natural filtration \((\mathcal{F }^Z_n)_{n \le 0}\) is non-standard although its tail \(\sigma \)-field is trivial. This example is illuminating since “simple” reasons explain why the standardness criteria do not hold and no technical estimates are required. Tsirelson’s construction relies on a particular structure of the triples \((Z_{2n-2},Z_{2n-1},Z_{2n})\) that we explain. We call “bricks” these triples.

In this paper, we give a modified and simpler construction which provides stronger results by requiring more on the bricks: in our construction, for every \(n \le 0, Z_{2n-2}\) is a deterministic function of \(Z_{2n-1}\) and \(Z_{2n-1}\) is a deterministic function of \((Z_{2n-2},Z_{2n})\), hence the filtration \((\mathcal{F }^Z_{2n})_{n \le 0}\) is generated by the sequence \((Z_{2n})_{n \le 0}\) of independent random variables. Yet, \((\mathcal{F }^Z_{2n-1})_{n \le 0}\) is not standard. Thus the filtration \(\mathcal{F }^Z\) “interlinks” standardness and non-standardness. Actually, we have a complete characterisation of the standard filtrations among the filtrations extracted from \(\mathcal{F }^Z\).

Theorem B

There exists a Markov process \((Z_n)_{n \le 0}\) such that

  • for each \(n \le 0, Z_n\) takes its values in some finite set \(F_n\),

  • the random variables \((Z_{2n})_{n \le 0}\) are independent,

  • for each \(n \le 0, Z_{2n-1}\) is a deterministic function of \((Z_{2n-2},Z_{2n})\),

  • the filtration \((Z_n)_{n \le 0}\) is \((r_n)_{n \le 0}\)-adic for some sequence \((r_n)_{n \le 0}\),

  • for any infinite subset \(D\) of \({\mathbb{Z }}_-\), the filtration \((\mathcal{F }^Z_{n})_{n \in D}\) is standard if and only if \(2n-1 \notin D\) for infinitely many \(n \le 0\).

In particular, the filtration \((\mathcal{F }^Z_{2n-1})_{n \le 0}\) is at the threshold of standardness.

In this theorem, the statement that \((\mathcal{F }^Z_{2n-1})_{n \le 0}\) is at the threshold of standardness cannot be deduced from the standardness of \((\mathcal{F }^Z_{2n})_{n \le 0}\) and the non-standardness of \((\mathcal{F }^Z_{2n-1})_{n \le 0}\) only. Indeed, the example of repeated interlinking mentioned above (see Example 11 in Sect. 2) provides a counterexample (modulo a time-translation). The proof that \((\mathcal{F }^Z_{2n-1})_{n \le 0}\) is at the threshold of standardness actually uses the fact that \((Z_n)_{n \le 0}\) is an inhomogeneous Markov process.

2 The case of split-words filtrations

In the whole section, excepted in Sect. 2.5, \(\mathcal{F }= (\mathcal{F }_n)_{n \le 0}\) denotes a split-words filtration associated to a finite alphabet \(A\) (endowed with the uniform measure) and a decreasing sequence \((\ell _n)_{n \le 0}\) of positive integers (the lengths) such that \(\ell _0=1\) and the ratios \(r_n = \ell _{n-1}/\ell _n\) are integers.

First, we prove the characterisation at the threshold of standardness among the split-words filtrations stated in Proposition 1.

2.1 Proof of Proposition 1

Preliminary observations: let \(B\) be an infinite subset of \({\mathbb{Z }}_-\) such that \(B^c\) is infinite. Then \((\mathcal{F }_n)_{n \in B}\) is a split-words filtration with lengths \((\ell _n)_{n \in B}\). The ratios between successive lengths are the integers \((R_n)_{n\in B}\) given by

$$\begin{aligned} R_n = \ell _{m(n)}/\ell _n \quad \mathrm{where}\; m(n) = \sup \{k<n : k \in B\}. \end{aligned}$$

Set \(B_1=B\cap (1+B)\) and \(B_2=B\backslash (1+B)\). Then \(B_2\) is infinite and

  • for \(n \in B_1, R_n = r_n\),

  • for \(n \in B_2, R_n \ge r_nr_{n-1} \ge r_{n-1}\).

Furthermore, if \(B^c\) does not contain two consecutive integers, then for any \(n \in B_2\), one has \(n-2 \in B\) since \(n-1 \notin B\), thus \(m(n)=n-2\) and \(R_n = r_nr_{n-1}\).

Proof of the “if” part: assume that \((\Delta )\) and \((\star )\) hold:

$$\begin{aligned} \sum _{n\le 0} \frac{\log (r_n)}{\ell _n}<+\infty \quad \mathrm{and}\quad \inf _{n\le 0} \frac{\log r_{n-1}}{\ell _n} >0. \end{aligned}$$

Then \(\mathcal{F }\) is not standard, and for every infinite subset \(B\) of \({\mathbb{Z }}_-\) such that \(B^c\) is infinite,

$$\begin{aligned} \sum _{n\in B} \frac{\log (R_n)}{\ell _n} \ge \sum _{n\in B_2} \frac{\log (R_n)}{\ell _n} \ge \sum _{n\in B_2} \frac{\log r_{n-1}}{\ell _n} = +\infty . \end{aligned}$$

since \(B_2=B\backslash (1+B)\) is infinite. Thus, \((\mathcal{F }_n)_{n \in B}\) is standard since the sequence of lengths \((\ell _n)_{n \in B}\) fulfills condition \(\lnot (\Delta )\). Therefore \(\mathcal{F }\) is at the threshold of standardness.

Proof of the “only if” part: condition \((\Delta )\), which is equivalent to the non-standardness of \(\mathcal{F }\), is necessary for \(\mathcal{F }\) to be at the threshold of standardness. Let us show that if \((\Delta )\) and \(\lnot (\star )\) hold, then \(\mathcal{F }\) is not at the threshold of standardness. By \((\Delta ), \log (r_n)/\ell _n \rightarrow 0\) as \(n \rightarrow -\infty \), hence

$$\begin{aligned} \liminf _{n \rightarrow -\infty } \frac{\log (r_nr_{n-1})}{\ell _n} = \liminf _{n \rightarrow -\infty } \frac{\log r_{n-1}}{\ell _n} = 0, \end{aligned}$$

since \(0\) is the greatest lower bound of the positive real numbers \(\log r_{n-1}/\ell _n\). Thus one can find a subsequence such that

$$\begin{aligned} \forall \, n \in {\mathbb{Z }}_-,\; \frac{\log (r_{\phi (n)} r_{\phi (n)-1})}{\ell _{\phi (n)}} \le 2^n \quad \mathrm{and} \quad \phi (n-1)\le \phi (n)-2. \end{aligned}$$

Set \(B=(\phi ({\mathbb{Z }}_-)-1)^c\). Let us show that the filtration \((\mathcal{F }_n)_{n\in B}\) is not standard. By construction, \(\phi ({\mathbb{Z }}_-)\) is infinite and does not contain two consecutive integers. Hence \(B\) and \(B^c\) are both infinite and \(B_2=B\backslash (B+1)=\phi ({\mathbb{Z }}_-)\). Moreover, according to the preliminary observations, \(R_n=r_n\) for every \(n \in B_1\) and \(R_n=r_n r_{n-1}\) for every \(n \in B_2\) since \(B^c\) does not contain two consecutive integers. Thus

$$\begin{aligned} \sum _{n\in B} \frac{\log (R_n)}{\ell _n}&= \sum _{n\in B_1} \frac{\log (r_n)}{\ell _n} + \sum _{n\in \phi ({\mathbb{Z }}_-)} \frac{\log (r_n r_{n-1})}{\ell _n}\\&\le \sum _{n \le 0} \frac{\log (r_n)}{\ell _n} + \sum _{m\le 0} 2^m\\&< +\infty . \end{aligned}$$

Therefore \((\mathcal{F }_n)_{n \in B}\) is not standard. Thus \(\mathcal{F }\) is not at the threshold of standardness.

2.2 Proof of Proposition 2 and example

Proof

Assume that \((\lnot \Delta )\) and \((\Box )\) hold and that \(\mathcal{F }\) is extracted from some split-words filtration \(\mathcal{H }\) with lengths \((\ell ^{\prime }_n)_{n \le 0}\), namely \(\mathcal{F }_n = \mathcal{H }_{\phi (n)}\) for every \(n \le 0\), for some increasing map \(\phi \) from \({\mathbb{Z }}_-\) to \({\mathbb{Z }}_-\). Then for every \(n \le 0, \ell _n=\ell ^{\prime }_{\phi (n)}\) and \(r_n = r^{\prime }_{\phi (n-1)+1} \cdots r^{\prime }_{\phi (n)}\) where \(r^{\prime }_k = \ell ^{\prime }_{k-1}/\ell ^{\prime }_{k}\). Let us show that \(\mathcal{H }\) cannot be at the threshold of standardness.

Condition \((\lnot \Delta )\) ensures that \(\mathcal{F }\) is standard. If \(\phi \) skips only finitely many integers, then \(\mathcal{H }\) is standard and the conclusion holds. Otherwise, \(\phi (n-1)\le \phi (n)-2\) for infinitely many \(n\), and for those \(n\),

$$\begin{aligned} \frac{\log \left( r^{\prime }_{\phi (n)-1}\right) }{\ell ^{\prime }_{\phi (n)}} \le \frac{\log \left( r^{\prime }_{\phi (n-1)+1} \cdots r^{\prime }_{\phi (n)}\right) }{\ell ^{\prime }_{\phi (n)}} = \frac{\log r_n}{\ell _n}. \end{aligned}$$

Thus, \((\Box )\) implies that

$$\begin{aligned} \inf _{k\le 0} \frac{\log (r^{\prime }_{k-1})}{\ell ^{\prime }_k}=0. \end{aligned}$$

Since the sequence \((r^{\prime }_n)_{n \le 0}\) does not fulfill condition \((\star ), \mathcal{H }\) is not at the threshold of standardness. \(\square \)

Example 3

Define the sequence of lengths \((\ell _n)_{n \le 0}\) by \(\ell _0=1, \ell _{-1}=2\) and, for every \(n\le -1\),

$$\begin{aligned} \ell _{n-1}=\ell _n 2^{\lfloor \ell _n/|n| \rfloor }, \end{aligned}$$

where \(\lfloor x\rfloor \) denotes the integer part of \(x\).

A recursion shows that for every \(n \le 0, \ell _n\) is a power of \(2\), and that \(\ell _n \ge 2^{|n|} \ge |n|\), hence \(r_n=\ell _{n-1}/\ell _n\ge 2\).

Moreover, for every \(n \le -1\),

$$\begin{aligned} \frac{\log _2(r_n)}{\ell _n} = \frac{{\lfloor \ell _n/|n| \rfloor }}{\ell _n} \in \left[ \frac{1}{2|n|},\frac{1}{|n|}\right] . \end{aligned}$$

Therefore, \((\lnot \Delta )\) and \((\Box )\) hold, hence \(\mathcal{F }\) is standard but cannot be extracted from any split-words filtration at the threshold of standardness.

Yet, since each \(\ell _n\) is a power of \(2, \mathcal{F }\) is extracted from the dyadic split-words filtration \(\mathcal{H }\), which is not standard. Since every filtration extracted from \(\mathcal{H }\) is a split-words filtration, one can deduce that no intermediate filtration (for extraction) between \(\mathcal{H }\) and \(\mathcal{F }\) is at the threshold of standardness.

Remark There are trivial examples of standard split-words filtrations which cannot be extracted from any split-words filtration at the threshold of standardness. For example, consider any split-words filtrations such that \(\lnot (\Delta )\) holds and such that \(r_n\) is a prime number for every \(n \le 0\). The last condition prevents the filtration from being extracted from any other split-words filtration. Yet, it still could be extracted from some filtration at the threshold of standardness which is not a split-words filtration.

2.3 Proof of Theorem A

Replacing \(\alpha _{n}\) by \((\alpha _{n} \vee |n+2|^{-2}) \wedge 1\) for \(n\le -3\) does not change the nature of the series \(\sum _{k \in B^c} \alpha _{k}\), hence we may assume that for \(n\le -3\),

$$\begin{aligned} |n+2|^{-2} \le \alpha _{n} \le 1. \end{aligned}$$

Set \(\ell _0=1, \ell _{-1}=2, \ell _{-2}=8, \ell _{-3}=64, \ell _{-4}=2^{11}=2{,}048\) and \(\ell _{n-2}=2^{\lfloor \alpha _{n-1}\ell _n\rfloor }\) for every \(n\le -3\), where \(\lfloor x\rfloor \) denotes the integer part of \(x\). We begin with two technical lemmas.

Lemma 4

For every \(n \le -1, \ell _n \ge |n|^3\) and \(\ell _{n} \ge 2|n+1|^2 \ell _{n+1}\).

Proof of Lemma 4

The proof of Lemma 4 is done by induction. One checks that the above inequalities hold for \(-4 \le n \le -1\).

Fix some \(n \le -3\). Assume that the inequalities hold for \(n+1, n\) and \(n-1\). Then

$$\begin{aligned} \log _2\ell _{n-2} - \log _2\ell _{n-1}&= \lfloor \alpha _{n-1}\ell _{n}\rfloor - \lfloor \alpha _n\ell _{n+1}\rfloor \\&\ge \alpha _{n-1}\ell _{n}-1-\alpha _n\ell _{n+1}\\&\ge |n+1|^{-2}\ell _{n}-\ell _{n+1}-1\\&\ge \ell _{n+1}-1 \;(\mathrm{since}\; \ell _{n} \ge 2|n+1|^2 \ell _{n+1})\\&\ge |n+1|^3-1\; (\mathrm{since}\; \ell _{n+1} \ge |n+1|^3), \end{aligned}$$

hence

$$\begin{aligned} \ell _{n-2}/\ell _{n-1} \ge 2^{|n+1|^3-1} \ge 2|n-1|^2 \;(\mathrm{since}\; n \le -3). \end{aligned}$$

Since \(\ell _{n-1} \ge |n-1|^3\), one has

$$\begin{aligned} \ell _{n-2} \ge 2|n-1|^2\ell _{n-1} \ge 2|n-1|^5 \ge |n-2|^3 \; (\mathrm{since}\; n \le -3). \end{aligned}$$

Thus the inequalities hold for \(n-2\). The proof is complete. \(\square \)

Lemma 5

For every \(n \le -4\),

$$\begin{aligned} \frac{\log _2 \ell _{n-1}}{\ell _n}&\le \frac{1}{2|n+1|^2},\\ \frac{\alpha _{n-1}}{2} \le \frac{\log _2 \ell _{n-2}}{\ell _n}&\le \alpha _{n-1},\\ \frac{\log _2 \ell _{n-3}}{\ell _n}&\ge 1. \end{aligned}$$

Proof of Lemma 5

For \(n \le -3\), the assumptions made on the sequence \((\alpha _k)_{k \le 0}\), and Lemma 4 entail \(\ell _n\alpha _{n-1}\ge |n|^3/|n+1|^2\ge 1,\) thus \(\alpha _{n-1}\ell _n/2 \le \lfloor \alpha _{n-1}\ell _{n}\rfloor \le \alpha _{n-1}\ell _n\). Thus, the recursion formula \(\ell _{n-2} = 2^{\lfloor \alpha _{n-1}\ell _n\rfloor }\) yields

$$\begin{aligned} \frac{\alpha _{n-1}}{2}\le \frac{\log _2 \ell _{n-2}}{\ell _n} \le \alpha _{n-1}. \end{aligned}$$

When \(n \le -4\), the same inequalities hold for \(n+1\) and \(n-1\), hence by Lemma 4

$$\begin{aligned} \frac{\log _2 \ell _{n-1}}{\ell _n} \le \alpha _n \frac{\ell _{n+1}}{\ell _{n}} \le \frac{\ell _{n+1}}{\ell _{n}} \le \frac{1}{2|n+1|^2}, \end{aligned}$$

and

$$\begin{aligned} \frac{\log _2 \ell _{n-3}}{\ell _n} \ge \frac{\alpha _{n-2}}{2} \frac{\ell _{n-1}}{\ell _{n}} \ge \frac{1}{2|n|^2}2|n|^2 = 1. \end{aligned}$$

The proof is complete. \(\square \)

We now prove Theorem A.

Let us check that the split-words filtration associated to the to the lengths \((\ell _n)_{n \le 0}\) fulfills the properties of the previous proposition.

Let \(B\) be an infinite subset of \({\mathbb{Z }}_-\). Since replacing \(B\) by \(B \setminus \{-2,-1,0\}\) does not change the nature of the filtration \((\mathcal{F }_n)_{n\in B}\), one may assume that \(B \subset ]-\infty ,-3]\).

Set \(m(n) = \sup \{k<n:k \in B\}\) for every \(n \le 0\). Then \((\ell _{m(n)}/\ell _n)_{n \in B}\) is the sequence of ratios associated to the lengths \((\ell _n)_{n \in B}\). Since \((\Delta )\) characterises standardness of split-words filtrations,

$$\begin{aligned} (\mathcal{F }_n)_{n \in B}\; \text{ is } \text{ standard }\; \Longleftrightarrow \sum _{n \in B} \frac{\log _2 (\ell _{m(n)}/\ell _n)}{\ell _n} = +\infty \Longleftrightarrow \sum _{n \in B} \frac{\log _2 \ell _{m(n)}}{\ell _n} = +\infty , \end{aligned}$$

where the last equivalence follows from the convergence of the series \(\sum _n \log _2\ell _n/\ell _n\) since \(\ell _n \ge 2^{|n|}\) for every \(n \le 0\).

Let us split \(B\) into three subsets:

  • \(B_1=\{ n\in B : m(n)=n-1\}\),

  • \(B_2=\{ n\in B : m(n)=n-2\}\),

  • \(B_3=\{ n\in B : m(n)\le n-3\}\).

Then

$$\begin{aligned} \sum _{n \in B} \frac{\log _2 \ell _{m(n)}}{\ell _n} = \sum _{n \in B_1} \frac{\log _2 \ell _{n-1}}{\ell _n} + \sum _{n \in B_2} \frac{\log _2 \ell _{n-2}}{\ell _n} + \sum _{n \in B_3} \frac{\log _2 \ell _{m(n)}}{\ell _n}. \end{aligned}$$

The inequality \(\ell _{m(n)} \ge \ell _{n-3}\) for \(n \in B_3\) and Lemma 5 show that in the right-hand side,

  • the first sum (over \(B_1\)) is always finite,

  • the middle sum (over \(B_2\)) has the same nature as \(\sum _{n \in B_2} \alpha _{n-1}\),

  • the last sum (over \(B_3\)) is finite if and only if \(B_3\) is finite.

When \(B_3\) is finite, any pair of consecutive integers excepted a finite number of them contain at least one element of \(B\). Hence, \((B_2 -1)\) only differs from \(B^c\) by a finite set of integers. Thus the sum \(\sum _{n \in B_2} \alpha _{n-1} = \sum _{n \in B_2-1} \alpha _n\) has the same nature as \(\sum _{n \in B^c} \alpha _n\). Theorem A follows.

2.4 Some applications of Theorem A

Choosing particular sequences \((\alpha _n)_{n \le 0}\) in Theorem A provides interesting examples of non-standard filtrations. In what follows, \(\mathcal{F }\) denotes the filtration associated to the sequence \((\alpha _n)_{n \le 0}\) given by Theorem A.

Example 6

If \(\alpha _n\) is bounded away from 0, then \(\mathcal{F }\) is at the threshold of standardness.

Example 7

If \(\alpha _{n}=0\) for every even \(n\) and \(\alpha _{n}=1\) for every odd \(n\), then \((\mathcal{F }_{2n})_{n \le 0}\) is standard whereas \((\mathcal{F }_{2n-1})_{n \le 0}\) is not.

Example 8

If the series \(\sum \alpha _n\) converges, then for every infinite subset \(B\) of \({\mathbb{Z }}_-\), the extracted filtration \((\mathcal{F }_n)_{n\in B}\) is standard if and only if \((B \cup (B-1))^c\) is infinite. In particular, the filtrations \((\mathcal{F }_{2n})_{n \le 0}\) and \((\mathcal{F }_{2n-1})_{n \le 0}\) are at the threshold of standardness.

Example 9

If \(\alpha _n\sim 1/|n|\) as \(n\) goes to \(-\infty \), then \(\mathcal{F }\) is not standard and no filtration at the threshold of standardness can be extracted from \(\mathcal{F }\).

Proof of Example 9

The non-standardness of \(\mathcal{F }\) is immediate by Theorem .

Call \(\mu \) the non-finite positive measure on \({\mathbb{Z }}_-\) defined by

$$\begin{aligned} \mu (B) = \sum _{n \in B} \alpha _n\quad \text{ for }\; B \subset {\mathbb{Z }}_-. \end{aligned}$$

Let \((\mathcal{F }_n)_{n \in B}\) be any non-standard filtration extracted from \(\mathcal{F }\). We show that \((\mathcal{F }_n)_{n \in B}\) cannot be at the threshold of standardness by constructing a subset \(B^{\prime }\) of \(B\) such that \((\mathcal{F }_n)_{n \in B^{\prime }}\) is not standard although \(B \setminus B^{\prime }\) is infinite.

By the Theorem A, we know that \(\mu (B^c) < +\infty \) and

$$\begin{aligned} n \notin B \; \text{ and }\; n+1 \notin B \;\text{ only } \text{ for } \text{ finitely } \text{ many }\; n \in {\mathbb{Z }}_-. \end{aligned}$$

Since \(\mu (B^c)\) is finite, the elements of \(B^c\) get rarer and rarer as \(n \rightarrow -\infty \). In particular, the set \(A = (B-1) \cap B \cap (B+1)\) is infinite.

We get \(B^{\prime }\) from \(B\) by removing a “small” infinite subset of \(A\). Namely, we set \(B^{\prime }= B \backslash A^{\prime }\) where \(A^{\prime }\) is an infinite subset of \(A\) which does not contain two consecutive integers and chosen such that \(\mu (A^{\prime }) < +\infty \). By construction, \(B \backslash B^{\prime } = A^{\prime }\) is infinite and \(\mu ((B^{\prime })^c) < +\infty \) since \((B^{\prime })^c = B^c \cup A^{\prime }\). Thus \(B^{\prime }\) is an infinite subset of \(B\).

Using the definition of \(A\) and the fact that \(A^{\prime }\) does not contain two consecutive integers and by construction of \(A\), one checks that \((B^{\prime }\cup (B^{\prime }-1))=(B\cup (B-1))\), therefore \((B^{\prime }\cup (B^{\prime }-1))^c\) is infinite.

Thus \((\mathcal{F }_n)_{n \in B^{\prime }}\) is not standard, which shows that \((\mathcal{F }_n)_{n \in B}\) is not at the threshold of standardness. \(\square \)

2.5 Interlinking standardness and non standardness

In this subsection, we show that any filtration at the threshold of standardness (not necessarily a split-words filtration) provides a filtration which interlinks repeatedly standardness and non-standardness by a suitable slowing-down of time.

Given any filtration \((\mathcal{F }_n)_{n \le 0}\), a simple way to get a slowed down filtration is to repeat each \(\mathcal{F }_n\) some finite number of times, which may depend of \(n\). We now show that this procedure does not change the nature of the filtration.

Lemma 10

Let \((\mathcal{F }_n)_{n \le 0}\) be any filtration and \(\phi \) an increasing map from \({\mathbb{Z }}_-\) to \({\mathbb{Z }}_-\) such that \(\phi (0)=0\). For every \(n \le 0\), set \(\mathcal{G }_n=\mathcal{F }_k\) if \(\phi (k-1)+1 \le n \le \phi (k)\). Then:

  • \((\mathcal{G }_n)_{n \le 0}\) is a filtration,

  • \((\mathcal{F }_n)_{n \le 0}\) is extracted from \((\mathcal{G }_n)_{n \le 0}\),

  • \((\mathcal{G }_n)_{n \le 0}\) is standard if and only if \((\mathcal{F }_n)_{n \le 0}\) is standard.

Proof of Lemma 10

By construction, \(\mathcal{G }_{\phi (k)} = \mathcal{F }_k\) for every \(k \le 0\) and the sequence \((\mathcal{G }_n)_{n \le 0}\) is constant on every interval \([\![\phi (k-1)+1,\phi (k) ]\!]\). The first two points follow.

The “only if” part of the third point is immediate since \(\mathcal{F }\) is extracted from \(\mathcal{G }\).

Assume that \(\mathcal{F }\) is standard. Then, up to a enlargement of the probability space, one may assume that \(\mathcal{F }\) is immersed in some product-type filtration \(\mathcal{H }\). Define a slowed-down filtration by \(\mathcal{K }_n=\mathcal{H }_k\) if \(\phi (k-1)+1 \le n \le \phi (k)\). Then \(\mathcal{K }\) is still a product-type filtration To prove that \(\mathcal{G }\) is immersed in \(\mathcal{K }\), we have to check that for every \(n \le -1, \mathcal{G }_{n+1}\) and \(\mathcal{K }_n\) are independent conditionally on \(\mathcal{G }_n\). This holds in any case since:

  • when \(\phi (k-1)+1 \le n \le \phi (k)-1, \mathcal{G }_{n+1}=\mathcal{F }_k, \mathcal{K }_n=\mathcal{H }_k\) and \(\mathcal{G }_n=\mathcal{F }_k\);

  • when \(n=\phi (k), \mathcal{G }_{n+1}=\mathcal{F }_{k+1}, \mathcal{K }_n=\mathcal{H }_k\) and \(\mathcal{G }_n=\mathcal{F }_k\).

Hence \(\mathcal{G }\) is standard. \(\square \)

Example 11

Assume that \((\mathcal{F }_n)_{n \le 0}\) is at the threshold of standardness. Set \(\phi (0)=0, \phi (-1)=-1\) and, for every \(k \le -1, \phi (2k) = -2^{|k|}\) and \(\phi (2k-1) = -2^{|k|}-1\). Let \(\mathcal{G }\) be the slowed-down filtration obtained from \(\mathcal{F }\) as above. Then for any \(d \ge 1\), the filtration \((\mathcal{G }_{2^dn})_{n \le 0}\) is not standard, whereas the filtration \((\mathcal{G }_{2^dn-2^{d-1}})_{n \le 0}\) is standard.

Proof of Example 11

Fix \(d \ge 1\). The filtrations \((\mathcal{G }_{2^dn})_{n \le -2}\) and \((\mathcal{G }_{2^dn-2^{d-1}})_{n \le -1}\) can be obtained from \((\mathcal{F }_{n})_{n \le -2d-2}\) and \((\mathcal{F }_{2n-1})_{n \le -d}\) by time-translations and by the slowing-down procedure just introduced. And truncations, time-translations and slowing-down procedure preserve the nature of the filtrations. \(\square \)

3 Improving on an example of Tsirelson

3.1 A construction of Tsirelson

In some non-published notes, Tsirelson gives a method to construct an inhomogeneous Markov process \((Z_n)_{n \le 0}\) whose natural filtration is easily proved to be non-standard, although the tail \(\sigma \)-field \(\mathcal{F }^Z_{-\infty }\) is trivial and the random variables \((Z_{2n})_{n \le 0}\) are independent. This construction relies on a particular structure of each triple \((Z_{2n-2},Z_{2n-1},Z_{2n})\). We will call bricks these triples, since the sequence \((Z_n)_{n \le 0}\) is obtained by gluing the triples \((Z_{2n-2},Z_{2n-1},Z_{2n})\) in a Markovian way. We now give a formal definition which is directly inspired by Tsirelson’s construction.

Definition 12

Fix \(\alpha \in ]0,1[\). Let \(F_{0},F_{1},F_{2}\) be finite sets. We will say that a triple \((Z_{0},Z_{1},Z_{2})\) of uniform random variables with values in \(F_{0},F_{1},F_{2}\) is an \(\alpha \)-brick if

  • the triple \((Z_{0},Z_1,Z_{2})\) is Markov.

  • \(Z_{0}\) and \(Z_{2}\) are independent.

  • for any non-anticipative coupling of two copies \((Z^{\prime }_{0},Z^{\prime }_{1},Z^{\prime }_{2})\) and \((Z^{\prime \prime }_{0},Z^{\prime \prime }_{1},Z^{\prime \prime }_{2})\) of \((Z_{0},Z_{1},Z_{2})\), defined on some probability space \((\bar{\Omega },\bar{\mathcal{A }},\bar{{\mathbb{P }}})\),

    $$\begin{aligned} \bar{{\mathbb{P }}}\left[ Z^{\prime }_{2} \ne Z^{\prime \prime }_{2}\big |\sigma \left( Z^{\prime }_{0},Z^{\prime \prime }_{0}\right) \right] \ge 1-\alpha \quad \text{ on } \text{ the } \text{ event } \left[ Z^{\prime }_{0} \ne Z^{\prime \prime }_{0}\right] . \end{aligned}$$

Here, the expression “non-anticipative” means that the filtrations generated by the three-time processes \(Z^{\prime }\) and \(Z^{\prime \prime }\) are immersed in the natural filtration of \((Z^{\prime },Z^{\prime \prime })\). In particular, \(Z^{\prime }_{1}\) and \(Z^{\prime \prime }_{0}\) are independent conditionally on \(Z^{\prime }_{0}\) (the couple \((Z^{\prime }_{0},Z^{\prime \prime }_{0})\) gives no more information on \(Z^{\prime }_{1}\) than \(Z^{\prime }_{0}\) does). Similarly, \(Z^{\prime }_{2}\) and \((Z^{\prime \prime }_{0},Z^{\prime \prime }_{1})\) are independent conditionally on \((Z^{\prime }_{0},Z^{\prime }_{1})\). And the same holds when the roles of \(Z^{\prime }\) and \(Z^{\prime \prime }\) are exchanged.

The last condition helps to negate the \(I\)-cosiness criterion recalled in Sect. 3.3 of the present paper. Actually, Tsirelson used another formulation involving iterated Kantorovitch–Rubinstein metrics to negate Vershik’s criterion.

The next example is a slight simplification of the example provided by Tsirelson, which is enlightening.

Example 13

Let \(p\) be a prime number, and \({\mathbb{Z }}_p\) be the finite field with \(p\) elements. Note \(F_{0}\) the set of all two-dimensional linear subspaces of \(({\mathbb{Z }}_p)^5, F_{1}\) the set of all two-dimensional affine subspaces of \(({\mathbb{Z }}_p)^5\) and \(F_{2} = ({\mathbb{Z }}_p)^5\). Construct a Markovian triple \((Z_{0},Z_{1},Z_{2})\) as follows:

  • choose uniformly \(Z_{0}\) in \(F_{0}\);

  • given \(Z_{0}\), choose uniformly \(Z_{1}\) among the affine planes with direction \(Z_{0}\);

  • given \(Z_{0}\) and \(Z_{1}\), choose uniformly \(Z_{2}\) on the affine plane \(Z_{1}\).

Then \((Z_{0},Z_{1},Z_{2})\) is an \(\alpha \)-brick with \(\alpha = 1/p\).

Indeed, one checks that \(Z_{1}\) is uniform on \(F_{1}, Z_{2}\) is uniform on \(F_{2}\) and that \(Z_{0}\) and \(Z_{2}\) are independent. Now, let \((Z^{\prime }_{0},Z^{\prime }_{1},Z^{\prime }_{2})\) and \((Z^{\prime \prime }_{0},Z^{\prime \prime }_{1},Z^{\prime \prime }_{2})\) be any non-anticipative coupling of two copies of \((Z_{0},Z_{1},Z_{2})\), defined on some probability space \((\bar{\Omega },\bar{\mathcal{A }},\bar{{\mathbb{P }}})\).

Conditionally on \((Z^{\prime }_{0},Z^{\prime \prime }_{0},Z^{\prime }_{1},Z^{\prime \prime }_{1}), Z^{\prime }_{2}\) is uniform on the affine plane \(Z^{\prime }_{1}\) and \(Z^{\prime \prime }_{2}\) is uniform on the affine plane \(Z^{\prime \prime }_{1}\). Since two affine planes with different directions are distinct and since two distinct planes have at most \(p\) common points, one has

$$\begin{aligned} \bar{{\mathbb{P }}}\left[ Z^{\prime }_2 \ne Z^{\prime \prime }_2\big |\sigma \left( Z^{\prime }_{0},Z^{\prime \prime }_{0},Z^{\prime }_{1},Z^{\prime \prime }_{1}\right) \right] \ge \frac{p^2-p}{p^2} \mathbf{1 }_{\left[ Z^{\prime }_{1} \ne Z^{\prime \prime }_{1}\right] } \ge \frac{p-1}{p}\mathbf{1 }_{\left[ Z^{\prime }_{0} \ne Z^{\prime \prime }_{0}\right] }. \end{aligned}$$

Thus

$$\begin{aligned} \bar{{\mathbb{P }}}\left[ Z^{\prime }_2 \ne Z^{\prime \prime }_2\big |\sigma \left( Z^{\prime }_{0},Z^{\prime \prime }_{0}\right) \right] \ge \left( 1-1/p \right) \mathbf{1 }_{\left[ Z^{\prime }_{0} \ne Z^{\prime \prime }_{0}\right] }. \end{aligned}$$

Remark In Tsirelson’s original example, \(Z_1\) was an affine line uniformly chosen among the lines whose direction are included in the linear plane \(Z_{0}\), and \((Z_{0},Z_{1},Z_{2})\) was an \(\alpha \)-brick with \(\alpha = 2/(p+1)\).

Note that the size of \(F_2\), namely \(|F_{2}| = p^5\), can be as large as one wants, and that

$$\begin{aligned} |F_{0}| = \frac{\left( p^5-1\right) \left( p^5-p\right) }{\left( p^2-1\right) \left( p^2-p\right) } = \left( p^4+p^3+p^2+p+1\right) \left( p^2+1\right) \end{aligned}$$

is much larger. But \(|F_{0}|\) has at least two prime divisors since the greatest common divisor of \(p^4+p^3+p^2+p+1\) and \(p^2+1\) is \(1\), hence \(|F_{0}|\) cannot be a power of a prime number. Thus, such bricks cannot be glued together. Fortunately, a slight modification solves this problem, and we shall see examples of this later on.

Up to the detail just mentioned, the next theorem achieves Tsirelson’s construction.

Theorem C

Let \((Z_n)_{n \le 0}\) be a sequence of uniform random variables with values in finite sets \((F_n)_{n \le 0}\) and \((\alpha _n)_{n \le 0}\) be an \(]0,1[\)-valued sequence such that the series \(\sum _n \alpha _n\) converges. Assume that

  • the sets \(F_{2n}\) are not singles,

  • \((Z_n)_{n \le 0}\) is a non-homogeneous Markov process,

  • for each \(n \le 0\), the subprocess \((Z_{2n-2},Z_{2n-1},Z_{2n})\) is an \(\alpha _n\)-brick.

Then the natural filtration \(\mathcal{F }^X\) is not standard. Moreover, if the tail \(\sigma \)-field \(\mathcal{F }^Z_{-\infty }\) is trivial, then \(|F_{2n}| \rightarrow +\infty \) as \(n \rightarrow -\infty \).

In one excepts the necessary condition \(|F_{2n}| \rightarrow +\infty \) as \(n \rightarrow -\infty \) for \(\mathcal{F }^Z_{-\infty }\) to be trivial, which is due to the authors of the present paper, the proof is adapted from Tsirelson’s notes.

Proof of Theorem C

First, we show that \(Z_0\) does not fulfills the I-cosiness criterion (see Sect. 3.3). Indeed, set

$$\begin{aligned} c = \prod _{k \le 0}(1-\alpha _k) > 0 \end{aligned}$$

and consider any non-anticipative coupling \((Z^{\prime }_n)_{n \le 0}\) and \((Z^{\prime \prime }_n)_{n \le 0}\) of the process \((Z_n)_{n \le 0}\), defined on some probability space \((\bar{\Omega },\bar{\mathcal{A }},\bar{{\mathbb{P }}})\). By assumption, for every \(n \le 0\),

$$\begin{aligned} \bar{{\mathbb{P }}}\left[ Z^{\prime }_{2n} \ne Z^{\prime \prime }_{2n}\big |\sigma \left( Z^{\prime }_{2n-2},Z^{\prime \prime }_{2n-2}\right) \right] \ge \left( 1-\alpha _n\right) \mathbf{1 }_{\left[ Z^{\prime }_{2n-2} \ne Z^{\prime \prime }_{2n-2}\right] }. \end{aligned}$$

By induction, for every \(n \le 0\),

$$\begin{aligned} \bar{{\mathbb{P }}}\left[ Z^{\prime }_0 \ne Z^{\prime \prime }_0\big |\sigma \left( Z^{\prime }_{2n},Z^{\prime \prime }_{2n}\right) \right] \ge \left( \prod _{k=n+1}^0 \left( 1-\alpha _k\right) \right) \mathbf{1 }_{\left[ Z^{\prime }_{2n} \ne Z^{\prime \prime }_{2n}\right] } \ge c \mathbf{1 }_{\left[ Z^{\prime }_{2n} \ne Z^{\prime \prime }_{2n}\right] } \end{aligned}$$

If, for some \(N \le 0\), the \(\sigma \)-fields \(\mathcal{F }^{Z^{\prime }}_{2N}\) and \(\mathcal{F }^{Z^{\prime \prime }}_{2N}\) are independent, then

$$\begin{aligned} \bar{{\mathbb{P }}}\left[ Z^{\prime }_0 \ne Z^{\prime \prime }_0\right] \ge c \bar{{\mathbb{P }}}\left[ Z^{\prime }_{2N} \ne Z^{\prime \prime }_{2N}\right] = c\left( 1-|F_{2N}|^{-1}\right) \ge c/2. \end{aligned}$$

Hence \(\bar{{\mathbb{P }}}\left[ Z^{\prime }_0 \ne Z^{\prime \prime }_0 \right] \) is bounded away from \(0\), which negates the I-cosiness criterion. The non-standardness of \(\mathcal{F }^X\) follows.

The second part of the theorem directly follows from the next proposition, applied to the sequence \((Y_n)_{n \le 0} = (Z_{2n})_{n \le 0}\). \(\square \)

Proposition 14

Let \((\gamma _n)_{n \le 0}\) be a sequence of real numbers in \([0,1]\) such that

$$\begin{aligned} \prod _{n \le 0}\gamma _n > 0. \end{aligned}$$

Let \((Y_n)_{n \le 0}\) be a family of random variables which are uniformly distributed on finite sets \((E_n)_{n \le 0}\). Let \((Y^{\prime }_n)_{n \le 0}\) and \((Y^{\prime \prime }_n)_{n \le 0}\) be independent copies of the process \((Y_n)_{n \le 0}\), defined on some probability space \((\bar{\Omega },\bar{\mathcal{A }},\bar{{\mathbb{P }}})\). Assume that \(\mathcal{F }^Y_{-\infty }\) is trivial and that for every \(n \le 0\),

$$\begin{aligned} \bar{{\mathbb{P }}}\left[ Y^{\prime }_{n} \ne Y^{\prime \prime }_{n}|\sigma \left( Y^{\prime }_{n-1},Y^{\prime \prime }_{n-1}\right) \right] \ge \gamma _n \mathbf{1 }_{\left[ Y^{\prime }_{n-1} \ne Y^{\prime \prime }_{n-1}\right] }. \end{aligned}$$

Then \(|E_n| \rightarrow 1\) or \(|E_n| \rightarrow +\infty \) as \(n \rightarrow -\infty \).

Proof of Proposition 14

By the independence of \((Y^{\prime }_n)_{n \le 0}\) and \((Y^{\prime \prime }_n)_{n \le 0}\), the following exchange properties apply (see [11])

$$\begin{aligned} \bigcap _{m \le 0} \bigcap _{n \le 0} \left( \mathcal{F }^{Y^{\prime }}_m \vee \mathcal{F }^{Y^{\prime \prime }}_n \right)&= \bigcap _{m \le 0} \left( \mathcal{F }^{Y^{\prime }}_m \vee \left( \bigcap _{n \le 0} \mathcal{F }^{Y^{\prime \prime }}_n \right) \right) \\&= \bigcap _{m \le 0} \left( \mathcal{F }^{Y^{\prime }}_m \vee \mathcal{F }^{Y^{\prime \prime }}_{-\infty } \right) \\&= \left( \bigcap _{m \le 0} \mathcal{F }^{Y^{\prime }}_m \right) \vee \mathcal{F }^{Y^{\prime \prime }}_{-\infty } \\&= \mathcal{F }^{Y^{\prime }}_{-\infty } \vee \mathcal{F }^{Y^{\prime \prime }}_{-\infty }. \end{aligned}$$

But, using that \(\mathcal{F }^{Y^{\prime }}_m \vee \mathcal{F }^{Y^{\prime \prime }}_n\) is non-decreasing with respect to \(m\) and \(n\), one gets

$$\begin{aligned} (\mathcal{F }^{Y^{\prime }} \vee \mathcal{F }^{Y^{\prime \prime }})_{-\infty } = \bigcap _{n \le 0} \left( \mathcal{F }^{Y^{\prime }}_n \vee \mathcal{F }^{Y^{\prime \prime }}_n \right) = \bigcap _{m \le 0} \bigcap _{n \le 0} \left( \mathcal{F }^{Y^{\prime }}_m \vee \mathcal{F }^{Y^{\prime \prime }}_n \right) . \end{aligned}$$

Hence the tail \(\sigma \)-field \((\mathcal{F }^{Y^{\prime }} \vee \mathcal{F }^{Y^{\prime \prime }})_{-\infty }\) is trivial. Thus the asymptotic event

$$\begin{aligned} \liminf _{n \rightarrow -\infty } \left[ Y^{\prime }_n \ne Y^{\prime \prime }_n \right] \end{aligned}$$

has probability \(0\) or \(1\).

But a recursion shows that for every \(n \le 0\)

$$\begin{aligned} \bar{{\mathbb{P }}} \left( \bigcap _{n \le k \le 0} \left[ Y^{\prime }_{k} \ne Y^{\prime \prime }_{k} \right] \Big | \sigma \left( Y^{\prime }_{n},Y^{\prime \prime }_{n}\right) \right) \ge \left( \prod _{n+1 \le k \le 0} \gamma _k \right) \mathbf{1 }_{\left[ Y^{\prime }_{n} \ne Y^{\prime \prime }_{n}\right] }. \end{aligned}$$

By taking expectations,

$$\begin{aligned} \bar{{\mathbb{P }}} \left( \bigcap _{n \le k \le 0} \left[ Y^{\prime }_{k} \ne Y^{\prime \prime }_{k}\right] \right) \ge \left( 1-|E_n|^{-1} \right) \prod _{n+1 \le k \le 0} \gamma _k. \end{aligned}$$

If \(|E_n| \ge 2\) for infinitely many \(n \le 0\), then

$$\begin{aligned} \bar{{\mathbb{P }}} \left( \bigcap _{k \le 0} \left[ Y^{\prime }_{k} \ne Y^{\prime \prime }_{k}\right] \right) \ge \frac{1}{2} \prod _{k \le 0} \gamma _k > 0. \end{aligned}$$

Thus \(|E_n| \ge 2\) for every \(n \le 0\) and

$$\begin{aligned} \bar{{\mathbb{P }}}\left( \liminf _{n \rightarrow -\infty } \left[ Y^{\prime }_n \ne Y^{\prime \prime }_n\right] \right) =1. \end{aligned}$$

But by Fatou’s lemma,

$$\begin{aligned} \bar{{\mathbb{P }}}\left( \liminf _{n \rightarrow -\infty } \left[ Y^{\prime }_n \ne Y^{\prime \prime }_n\right] \right) \le \liminf _{n \rightarrow -\infty } \bar{{\mathbb{P }}}\left[ Y^{\prime }_n \ne Y^{\prime \prime }_n\right] . \end{aligned}$$

Hence \(1-|E_n|^{-1} = \bar{{\mathbb{P }}}\left[ Y^{\prime }_n \ne Y^{\prime \prime }_n\right] \rightarrow 1\) thus \(|E_n| \rightarrow +\infty \) as \(n \rightarrow -\infty \). \(\square \)

3.2 Strong bricks

The slight variation we made on Tsirelson’s example of a brick when replacing the affine lines with the affine planes provides two additional properties: \(Z_{0}\) becomes a deterministic function of \(Z_{1}\) and \(Z_{1}\) becomes a deterministic function of \((Z_{0},Z_{2})\). As we shall see, these properties allow to simplify the construction and the proofs, and enable us to get stronger results. The next definition exploits this idea and requires large family of partitions, each with a fixed number of blocks, each block having a fixed size, such that any two blocks chosen in any two different partitions have a small intersection.

Definition 15

(Strong bricks) Fix \(\alpha \in ]0,1[\) and two positive integers \(r_{1},r_{2}\). Let \(F_{0},F_{1},F_{2}\) be finite sets such that \(|F_{1}|=|F_{0}|r_1\) and \(|F_{2}|=r_{1}r_{2}\). Let \(h\) be a bijection from \(F_{0} \times [\![1,r_{1} ]\!]\) to \(F_{1}\). Let \((\Pi _z)_{z \in F_{0}}\) be a family of partitions of \(F_{2}\) indexed by \(F_{0}\) such that

  • each partition \(\Pi _z\) has \(r_{1}\) blocks \(S_{z,1},\ldots ,S_{z,r_{1}}\);

  • each block has \(r_{2}\) elements.

  • for any distinct \((z^{\prime },i^{\prime })\) and \((z^{\prime \prime },i^{\prime \prime })\) in \(F_{0} \times [\![1,r_{1} ]\!], |S_{z^{\prime },i^{\prime }} \cap S_{z^{\prime \prime },i^{\prime \prime }}| \le \alpha r_{2}\).

We will say that a triple \((Z_{0},Z_{1},Z_{2})\) of random variables with values in \(F_{0},F_{1},F_{2}\) is a strong \((r_{1},r_{2})\)-adic \(\alpha \)-brick, associated to the family \((\Pi _z)_{z \in F_{0}}\), if

  • \(Z_{0}\) and \(Z_{2}\) are independent random variables, uniformly distributed in \(F_{0}\) and \(F_{2}\).

  • \(Z_{1} = h(Z_{0},J)\), where \(J\) is the index of the only block of \(\Pi _{Z_{0}}\) which contains \(Z_{2}\) (that is to say \(Z_{2} \in S_{Z_{0},J})\).

The next lemma shows that strong bricks are bricks with some additional properties.

Lemma 16

Let \((Z_{0},Z_{1},Z_{2})\) be strong \((r_{1},r_{2})\)-adic \(\alpha \)-brick. Then

  • \(Z_{0}\) is a deterministic function of \(Z_{1}\);

  • \(Z_{1}\) is a deterministic function of \((Z_{0},Z_{2})\);

  • the conditional law of \(Z_{1}\) given \(Z_{0}\) is uniform on some finite random set of size \(r_{1}\);

  • the conditional law of \(Z_{2}\) given \(Z_{1}\) is uniform on some finite random set of size \(r_{2}\);

  • for all different \(z^{\prime }_{1}\) and \(z^{\prime \prime }_{1}\) of \(Z_{1}\),

    $$\begin{aligned} \sum _{z \in F_{2}} {\mathbb{P }}\left[ Z_{2}=z\big |Z_{1}=z^{\prime }_{1}\right] \wedge {\mathbb{P }}\left[ Z_{2}=z\big |Z_{1}=z^{\prime \prime }_{1}\right] \le \alpha ; \end{aligned}$$
    (1)

  • if \((Z^{\prime }_{0},Z^{\prime }_{1},Z^{\prime }_{2})\) and \((Z^{\prime \prime }_{0},Z^{\prime \prime }_{1},Z^{\prime \prime }_{2})\) is any non-anticipative coupling of \((Z_{0},Z_{1},Z_{2})\), defined on some probability space \((\bar{\Omega },\bar{\mathcal{A }},\bar{{\mathbb{P }}})\), then

    $$\begin{aligned} {\mathbb{P }}\left[ Z^{\prime }_{2} \ne Z^{\prime \prime }_{2}|\mathcal{G }\right] \ge (1-\alpha ) \mathbf{1 }_{\left[ Z^{\prime }_{1} \ne Z^{\prime \prime }_{1}\right] } \ge (1-\alpha ) \mathbf{1 }_{\left[ Z^{\prime }_{0} \ne Z^{\prime \prime }_{0}\right] }; \end{aligned}$$

  • \((Z_{0},Z_{1},Z_{2})\) is an \(\alpha \)-brick.

Proof of Lemma 16

Keep the notations of Definition 15.

The first two statements directly follow from the definition, and the triple \((Z_{0},Z_{1},Z_{2})\) is Markov since \(Z_{0}\) is a function of \(Z_{1}\).

For every \(z_{0} \in F_{0}, j \in [\![1,r_{1} ]\!]\) and \(z_2 \in F_{2}\),

$$\begin{aligned} {\mathbb{P }}\left[ Z_{0}=z_{0} ; J=j ; Z_{2}=z_2\right]&= \mathbf{1 }_{\left[ z_2 \in S_{z_{0},j}\right] }\ {\mathbb{P }}\left[ Z_{0}=z_{0}; Z_{2}=z_2\right] \\&= \mathbf{1 }_{\left[ z_2 \in S_{z_{0},j}\right] }\ \times \frac{1}{|F_{0}|}\ \times \frac{1}{r_{1}r_{2}}. \end{aligned}$$

Summing over \(z_2\) yields

$$\begin{aligned} {\mathbb{P }}\left[ Z_{0}=z_{0};J=j \right] = \frac{1}{|F_{0}|} \times \frac{1}{r_{1}}, \end{aligned}$$

and, by division,

$$\begin{aligned} {\mathbb{P }}\left[ Z_{2}=z_2 \big | Z_{0}=z_{0}; J=j\right] = \mathbf{1 }_{\left[ z_2 \in S_{z_{0},j}\right] } \times \frac{1}{r_{2}}. \end{aligned}$$

Thus \((Z_{0},J)\) is uniform on \(F_0 \times [\![1,r_{1} ]\!]\), and given \((Z_{0},J), Z_{2}\) is uniform on the block \(S_{Z_{0},J}\). Using the equality \(Z_{1} = h(Z_{0},J)\), one gets the third and the fourth statements and the fact that \(Z_1\) is uniform on \(F_1\).

Let \(z^{\prime }_1\) and \(z^{\prime \prime }_1\) be distinct elements in \(F_{1}\). Since the conditional laws \(\mathcal{L }(Z_{2}|Z_{1}=z^{\prime }_1)\) and \(\mathcal{L }(Z_{2}|Z_{1}=z^{\prime \prime }_1)\) are uniform on the blocks \(S_{h^{-1}(z^{\prime }_1)}\) and \(S_{h^{-1}(z^{\prime \prime }_1)}\), one has

$$\begin{aligned} \sum _{z \in F_{2}} {\mathbb{P }}\left[ Z_{2}=z\big |Z_{1}=z^{\prime }_1\right] \wedge {\mathbb{P }}\left[ Z_{2}=z\big |Z_{1}=z^{\prime \prime }_1\right] = \sum _{z \in S_{h^{-1}(z^{\prime }_1)} \cap S_{h^{-1}(z^{\prime \prime }_1)}} \frac{1}{r_2} \le \alpha , \end{aligned}$$

which is the the fifth statement.

Let \((Z^{\prime }_{0},Z^{\prime }_{1},Z^{\prime }_{2})\) and \((Z^{\prime \prime }_{0},Z^{\prime \prime }_{1},Z^{\prime \prime }_{2})\) be any non-anticipative coupling of \((Z_{0},Z_{1},\) \(Z_{2})\), defined on some probability space \((\bar{\Omega },\bar{\mathcal{A }},\bar{{\mathbb{P }}})\). Set \(\mathcal{G }= \sigma (Z^{\prime }_{0},Z^{\prime }_{1},Z^{\prime \prime }_{0},Z^{\prime \prime }_{1})\). By the non-anticipative and the Markov properties,

$$\begin{aligned} \mathcal{L }\left( Z^{\prime }_{2}\big |\mathcal{G }\right) = \mathcal{L }\left( Z^{\prime }_{2}\big |\sigma \left( Z^{\prime }_{0},Z^{\prime }_{1}\right) \right) = \mathcal{L }\left( Z^{\prime }_{2}\big |\sigma \left( Z^{\prime }_{1}\right) \right) \end{aligned}$$

and the same holds with \(Z^{\prime \prime }\).

Thus for all different \(z^{\prime },z^{\prime \prime }\) in \(F_{1}\), one has, on the event \([Z^{\prime }_{1}=z^{\prime }; Z^{\prime \prime }_{1}=z^{\prime \prime }]\),

$$\begin{aligned} {\mathbb{P }}\left[ Z^{\prime }_{2} = Z^{\prime \prime }_{2}\big |\mathcal{G }\right]&= \sum _{z \in F_{2}} {\mathbb{P }}\left[ Z^{\prime }_{2} = z; Z^{\prime \prime }_{2}=z\big |\mathcal{G }\right] \\&\le \sum _{z \in F_{2}} {\mathbb{P }}\left[ Z^{\prime }_{2} = z\big |\mathcal{G }\right] \wedge {\mathbb{P }}\left[ Z^{\prime \prime }_{2}=z\big |\mathcal{G }\right] \\&= \sum _{z \in F_{2}} {\mathbb{P }}\left[ Z^{\prime }_{2} = z\big |Z^{\prime }_{1}=z^{\prime }\right] \wedge {\mathbb{P }}\left[ Z^{\prime \prime }_{2}=z\big |Z^{\prime \prime }_{1}=z^{\prime \prime }\right] \\&= \sum _{z \in F_{2}} {\mathbb{P }}\left[ Z_{2} = z\big |Z_{1}=z^{\prime }\right] \wedge {\mathbb{P }}\left[ Z_{2}=z\big |Z_{1}=z^{\prime \prime }\right] \\&\le \alpha . \end{aligned}$$

Hence

$$\begin{aligned} {\mathbb{P }}\left[ Z^{\prime }_{2} \ne Z^{\prime \prime }_{2}\big |\mathcal{G }\right] \ge (1-\alpha ) \mathbf{1 }_{\left[ Z^{\prime }_{1} \ne Z^{\prime \prime }_{1}\right] } \ge (1-\alpha ) \mathbf{1 }_{\left[ Z^{\prime }_{0} \ne Z^{\prime \prime }_{0}\right] }, \end{aligned}$$

since \([Z^{\prime }_{0} \ne Z^{\prime \prime }_{0}] \subset [Z^{\prime }_{1} \ne Z^{\prime \prime }_{1}]\), which shows the sixth statement.

Conditioning by \(\sigma (Z^{\prime }_{0},Z^{\prime \prime }_{0})\) completes the proof of the last statement. \(\square \)

Examples of strong bricks

One checks that Example 13 provides a \((p^3,p^2)\)-adic \(1/p\)-brick.

More generally, any finite family of partitions \((\Pi _z)_{z \in F_{0}}\) on a finite set \(F_2\) satisfying the conditions of Definition 15 provides a strong brick: simply take two independent random variables \(Z_{0}\) and \(Z_{2}\), uniformly distributed in \(F_{0}\) and \(F_{2}\), set \(Z_{1} = (Z_{0},J)\) and \(F_{1} = F_{0} \times [\![1,r_{1} ]\!]\), where \(J\) is the index of the only block of \(\Pi _{Z_{0}}\) which contains \(Z_{2}\).

Algebra helps us to construct large such families of partitions. Here are two examples using a finite field \(K\) with \(q\) elements, where \(q\) can be any power of a prime number.

Example 17

We set \(r_{1}=r_{2}=q^4, F_{0} = K^{16}\) identified with the set \(\mathcal{M }_4(K)\) of all \(4 \times 4\) matrices with entries in \(K\) and \(F_{2} = K^8\) identified with \(K^4 \times K^4\).

To each matrix \(A \in \mathcal{M }_4(K)\), one can associate the partition of \(K^8\) given by all four-dimensional affine subspaces of \(K^8\) with equations \(y=Ax+b\) where \(b\) ranges over \(K^4\). Each of these subspaces has size \(q^4\). But two subspaces of equations \(y=A^{\prime }x+b^{\prime }\) and \(y=A^{\prime \prime }x+b^{\prime \prime }\) intersect in at most \(q^3\) points (a three dimensional affine subspace) when \(A^{\prime } \ne A^{\prime \prime }\). Hence these partitions provide a \((q^4,q^4)\)-adic \(1/q\)-brick.

Example 18

We set \(r_{1}=r_{2}=q, F_{0} = K^4\) and \(F_{2} = K^2\).

To each quadruple \((a,b,c,d) \in K^4\), one can associate the partition of \(K^2\) given by the \(q\) graphs of equations \(y=ax^4+bx^3+cx^3+dx+e\) where \(e\) ranges over \(K\). Each of these graphs has size \(q\). But two such graphs with different \((a,b,c,d) \in K^4\) intersect in at most \(4\) points. Hence, if \(p \ge 5\) these partitions provide a \((q,q)\)-adic \(4/q\)-brick.

The bricks provided by these two examples can be glued together since the sets \(K^{16}\) and \(K^4\) can be identified by bijective maps with \(L^8\) and \(L^2\), where \(L\) denotes the field with \(q^2\) elements. For each \(n \le 0\), call \(K_n\) the field with \(q_n=q^{2^{|n|}}\) elements and set

$$\begin{aligned} \forall \,n \le 0,\quad F_{2n} = K_n^8,\; r_{2n-1} = r_{2n} = q_n^4 \quad \text{ and }\quad \alpha _n = 1/q_n, \end{aligned}$$

or

$$\begin{aligned} \forall \, n \le 0,\quad F_{2n} = K_n^2,\; r_{2n-1} = r_{2n} = q_n\quad \text{ and }\quad \alpha _n = 4/q_n. \end{aligned}$$

Start with a sequence \((Z_{2n})_{n \le 0}\) of independent random variables uniformly distributed in the \((F_{2n})_{n \le 0}\). For each \(n \le 0\), one can construct a random variable \(Z_{2n-1}\) with values in \(F_{2n-1} = F_{2n-2} \times [\![1,r_{2n-1} ]\!]\), using \(Z_{2n-2}\) and \(Z_{2n}\) and the family of partitions as above, to get an \((r_{2n-1},r_{2n})\)-adic \(\alpha _n\)-brick \((Z_{2n-2},Z_{2n-1},Z_{2n})\). The process \((Z_n)_{n \le 0}\) thus defined provides an example which proves the existence stated in Theorem B.

3.3 Proof of Theorem B

Indeed, Theorem B directly follows from the construction above and the theorem below.

Theorem D

Let \((\alpha _n)_{n \le 0}\) be a sequence of reals in \(]0,1[\) such that the series \(\sum _n \alpha _n\) converges. Let \((Z_n)_{n \le 0}\) be any sequence of random variables taking values in some finite sets \((F_n)_{n \le 0}\) of size \(\ge 2\). Assume that

  • the random variables \((Z_{2n})_{n \le 0}\) are independent;

  • for each \(n \le 0, (Z_{2n-2},Z_{2n-1},Z_{2n})\) is a strong \((r_{2n-1},r_{2n})\)-adic \(\alpha _n\)-brick.

Then

  • \((Z_n)_{n \le 0}\) is a Markov process which generates a \((r_n)\)-adic filtration;

  • for every infinite subset \(D\) of \({\mathbb{Z }}_-, (\mathcal{F }^Z_{n})_{n \in D}\) is standard if and only if \(2n-1 \notin D\) for infinitely many \(n \le 0\).

In particular, the filtration \((\mathcal{F }^Z_{2n-1})_{n \le 0}\) is at the threshold of standardness.

Proof of Theorem D

We now prove the statements.

Proof that \((Z_n)_{n \le 0}\) is a Markov process and generates a \((r_n)\) -adic filtration

First, note that the filtration \((\mathcal{F }^Z_{2n})_{n \le 0}\) is generated by the independent random variables \((Z_{2n})_{n \le 0}\) since for every \(n \le 0, Z_{2n-1}\) is a deterministic function of \((Z_{2n-2},Z_{2n})\). Hence, for every \(n \le 0\),

$$\begin{aligned} \mathcal{F }^Z_{2n-2} = \sigma \left( Z_{2n-2}\right) \vee \mathcal{F }^Z_{2n-4}, \end{aligned}$$

Moreover, since \(Z_{2n-2}\) is a deterministic function of \(Z_{2n-1}\),

$$\begin{aligned} \mathcal{F }^Z_{2n-1} = \sigma (Z_{2n-1}) \vee \mathcal{F }^Z_{2n-2}= \sigma (Z_{2n-1}) \vee \mathcal{F }^Z_{2n-4}. \end{aligned}$$

By independence of \((Z_{2n-2},Z_{2n-1},Z_{2n})\) and \(\mathcal{F }^Z_{2n-4}\), we get

$$\begin{aligned} \mathcal{L }\left( Z_{2n-1}\big |\mathcal{F }^Z_{2n-2}\right)&= \mathcal{L }\left( Z_{2n-1}\big |\sigma \left( Z_{2n-2}\right) \vee \mathcal{F }^Z_{2n-4}\right) = \mathcal{L }\left( Z_{2n-1}\big |\sigma \left( Z_{2n-2}\right) \right) ,\\ \mathcal{L }\left( Z_{2n}\big |\mathcal{F }^Z_{2n-1}\right)&= \mathcal{L }\left( Z_{2n}\big |\sigma \left( Z_{2n-1}\right) \vee \mathcal{F }^Z_{2n-4}\right) = \mathcal{L }\left( Z_{2n}\big |\sigma \left( Z_{2n-1}\right) \right) . \end{aligned}$$

The Markov property follows. But for every \(n \le 0, (Z_{2n-2},Z_{2n-1},Z_{2n})\) is an \((r_{2n-1},r_{2n})\)-adic \(\alpha _n\)-brick. The \((r_n)\)-adic character of \(\mathcal{F }^Z\) follows.

Proof that \((\mathcal{F }^Z_n)_{n \in D}\) is not standard when \(D\) contains all but finitely many odd negative integers

First, we show that \((\mathcal{F }^Z_{2n-1})_{n \le 0}\) is not standard. To do this, we check that the random variable \(Z_{-1}\) does not satisfy the I-cosiness criterion. Note that \((\mathcal{F }^Z_{2n-1})_{n \le 0}\) is the natural filtration of \((Z_{2n-1})_{n \le 0}\) only since for every \(n \le 0, Z_{2n-2}\) is some deterministic function \(f_n\) of \(Z_{2n-1}\).

Let \((Z^{\prime }_{2n-1})_{n \le 0}\) and \((Z^{\prime \prime }_{2n-1})_{n \le 0}\) be two copies of the process \((Z_{2n-1})_{n \le 0}\), defined on some probability space \((\bar{\Omega },\bar{\mathcal{A }},\bar{{\mathbb{P }})}\). Set \(Z^{\prime }_{2n-2} = f_n(Z^{\prime }_{2n-1})\) and \(Z^{\prime \prime }_{2n-2} = f_n(Z^{\prime \prime }_{2n-1})\) for every \(n \le 0\). Then \((Z^{\prime }_{n})_{n \le 0}\) and \((Z^{\prime \prime }_{n})_{n \le 0}\) are copies of the process \((Z_n)_{n \le 0}\). Moreover, \((\mathcal{F }^{Z^{\prime }}_{2n-1})_{n \le 0}\) and \((\mathcal{F }^{Z^{\prime \prime }}_{2n-1})_{n \le 0}\) are the natural filtrations of \((Z^{\prime }_{2n-1})_{n \le 0}\) and \((Z^{\prime \prime }_{2n-1})_{n \le 0}\).

Assume that these filtrations are immersed in some filtration \((\mathcal{G }_{2n-1})_{n \le 0}\).

Then, for every \(n \le -1\),

$$\begin{aligned} \mathcal{L }\left( Z^{\prime }_{2n+1}\big |\mathcal{G }_{2n-1}\right) = \mathcal{L }\left( Z^{\prime }_{2n+1}\big |\mathcal{F }^{Z^{\prime }}_{2n-1}\right) = \mathcal{L }\left( Z^{\prime }_{2n+1}\big |\sigma \left( Z^{\prime }_{2n-1}\right) \right) , \end{aligned}$$

and since \(Z^{\prime }_{2n}\) is a deterministic function of \(Z^{\prime }_{2n+1}\),

$$\begin{aligned} \mathcal{L }\left( Z^{\prime }_{2n}\big |\mathcal{G }_{2n-1}\right) = \mathcal{L }\left( Z^{\prime }_{2n}\big |\sigma \left( Z^{\prime }_{2n-1}\right) \right) . \end{aligned}$$

The same holds with the process \(Z^{\prime \prime }\).

Hence, since \([Z^{\prime }_{2n+1} = Z^{\prime \prime }_{2n+1}] \subset [Z^{\prime }_{2n} = Z^{\prime \prime }_{2n}]\), the same proof as in Lemma  yields

$$\begin{aligned} \bar{{\mathbb{P }}}\left[ Z^{\prime }_{2n+1} \ne Z^{\prime \prime }_{2n+1}\big |\mathcal{G }_{2n-1}\right] \ge \bar{{\mathbb{P }}}\left[ Z^{\prime }_{2n} \ne Z^{\prime \prime }_{2n}\big |\mathcal{G }_{2n-1}\right] \ge (1-\alpha _n) \mathbf{1 }_{\left[ Z^{\prime }_{2n-1} \ne Z^{\prime \prime }_{2n-1}\right] }. \end{aligned}$$

By recursion,

$$\begin{aligned} {\mathbb{P }}\left[ Z^{\prime }_{-1} \ne Z^{\prime \prime }_{-1}\big |\mathcal{G }_{2n-1}\right] \ge \prod _{n \le k \le -1} (1-\alpha _k) \mathbf{1 }_{\left[ Z^{\prime }_{2n-1} \ne Z^{\prime \prime }_{2n-1}\right] }. \end{aligned}$$

Taking the expectations, one gets

$$\begin{aligned} {\mathbb{P }}\left[ Z^{\prime }_{-1} \ne Z^{\prime \prime }_{-1}\right] \ge \prod _{n \le k \le -1} (1-\alpha _k) {\mathbb{P }}\left[ Z^{\prime }_{2n-1} \ne Z^{\prime \prime }_{2n-1}\right] . \end{aligned}$$

If, for some \(N > -\infty \), the \(\sigma \)-fields \(\mathcal{F }^{\prime }_{2N-1}\) and \(\mathcal{F }^{\prime \prime }_{2N-1}\) are independent, then

$$\begin{aligned} {\mathbb{P }}\left[ Z^{\prime }_{2N-1} \ne Z^{\prime \prime }_{2N-1}\right] = 1-\frac{1}{|F_{2N-1}|} \ge \frac{1}{2}, \end{aligned}$$

since \(Z^{\prime }_{2N-1}\) and \(Z^{\prime \prime }_{2N-1}\) are independent and uniform on \(F_{2N-1}\), and

$$\begin{aligned} {\mathbb{P }}\left[ Z^{\prime }_{-1} \ne Z^{\prime \prime }_{-1}\right] \ge \frac{1}{2} \prod _{N \le k \le -1} (1-\alpha _k) \ge \frac{1}{2} \prod _{k \le -1} (1-\alpha _k) > 0, \end{aligned}$$

which shows that \(Z_{-1}\) does not satisfy the I-cosiness criterion.

Thus \((\mathcal{F }^Z_{2n-1})_{n \le 0}\) is not standard. Thus, if \(D\) is any subset of \({\mathbb{Z }}_-\) which contains all odd negative integers, the filtration \((\mathcal{F }^Z_n)_{n \in D}\) is not standard (since standardness is preserved by extraction). This conclusion still holds when \(D\) contains all but finitely many odd negative integers (since standardness is an asymptotic property).

Proof that \((\mathcal{F }^Z_n)_{n \in D}\) is standard when \(D\) skips infinitely many odd negative integers

Since standardness is preserved by extraction, one only needs to consider the case where \(D\) not only skips infinitely many odd negative integers, but also contains all even ones. In this case, the filtration \((\mathcal{F }^Z_n)_{n \in D}\) is generated by \((Z_n)_{n \in D}\) only. Indeed, if \(n\) is any integer in \({\mathbb{Z }}_- \setminus D\), then \(n\) is odd, hence \(n-1 \in D, n+1 \in D\) and \(Z_n\) is a function of \((Z_{n-1},Z_{n+1})\).

For each \(n \le 0\), the conditional law \(\mathcal{L }(Z_n|\mathcal{F }^Z_{n-1})=\mathcal{L }(Z_n|Z_{n-1})\) is (almost surely) uniform on some random subset of \(F_n\) with \(r_n\) elements. By fixing a total order on the set \(F_n\), one can construct a uniform random variable \(U_n\) on \([\![1,r_n ]\!]\), independent of \(\mathcal{F }^Z_{n-1}\), such that \(Z_n\) is a function of \(Z_{n-1}\) and \(U_n\). Set \(Y_n = Z_n\) if \(n-1 \notin D\) (which may happen only for even \(n\)) and \(Y_n = U_n\) otherwise. Then \(Y_n\) is \(\mathcal{F }^Z_n\)-measurable. This shows that \(\mathcal{F }^Y_n \subset \mathcal{F }^Z_n\) for every \(n \in D\).

Let us prove the reverse inclusion. Fix \(n \in D\), and call \(m \le n\) the integer such that \(m-1 \notin D\) but \(k \in D\) for all \(k \in [\![m,n ]\!]\). Then \(Z_n\) is \(\mathcal{F }^Y_n\)-measurable as a function of \(Y_m=Z_m,Y_{m+1}=U_{m+1},\ldots ,Y_n=U_n\).

Last, for every \(n \in D, Y_n\) is independent of \(\mathcal{F }^Z_{n-1}\) if \(n-1 \in D\) and \(Y_n\) is independent of \(\mathcal{F }^Z_{n-2}\) otherwise. This shows the independence of the random variables \((Y_n)_{n \in D}\). Hence the filtration \((\mathcal{F }^Z_n)_{n \in D}\) is of product type, which completes the proof. \(\square \)