2.1 Three gap theorem and consequences

Let $(x_n)_{n \ge 0}$ be the sequence defined by $x_n := T_\theta ^n(0) = n\theta - \lfloor n\theta \rfloor $ . The well-known three gap theorem states that, for every $n \ge 1$ , the intervals defined by the subdivision $(x_k)_{0 \le k \le n-1}$ have at most three different lengths. We shall use a very precise statement of the three gap theorem, involving the continued fraction expansion $\theta = [0;a_1,a_2,\ldots ]$ . Let us recall some classical facts. A more complete exposition can be found in [Reference Khintchine5].

Set $a_0=\lfloor \theta \rfloor = 0$ and define two sequences $(p_k)_{k \ge -2}$ and $(q_k)_{k \ge -2}$ of integers by $p_{-2}=0$ , $q_{-2}=1$ , $p_{-1}=1$ , $q_{-1}=0$ , and, for every $k \ge 0$ ,

$$ \begin{align*}p_k = a_kp_{k-1}+p_{k-2},\quad q_k = a_kq_{k-1}+q_{k-2}.\end{align*} $$

Moreover, $p_k$ and $q_k$ are relatively prime since $p_kq_{k-1}-q_kp_{k-1} = (-1)^{k-1}$ .

The fractions $(p_k/q_k)_{k \ge 0}$ are the convergents of $\theta $ . The following inequalities hold.

$$ \begin{align*}\frac{p_0}{q_0} < \frac{p_2}{q_2} < \cdots < \theta < \cdots < \frac{p_3}{q_3} < \frac{p_1}{q_1}.\end{align*} $$

In particular, the difference $\theta - p_k/q_k$ has the same sign as $(-1)^k$ and

$$ \begin{align*}\frac{a_{k+2}}{q_kq_{k+2}} = \bigg|\frac{p_{k+2}}{q_{k+2}} - \frac{p_k}{q_k}\bigg| < \bigg|\theta - \frac{p_k}{q_k}\bigg| < \bigg|\frac{p_{k+1}}{q_{k+1}} - \frac{p_k}{q_k}\bigg| = \frac{1}{q_kq_{k+1}}.\end{align*} $$

Hence, the sequence $(p_k/q_k)_{n \ge 0}$ converges to $\theta $ .

For every $k \ge -1$ , set $\ell _k = (-1)^k(q_k\theta -p_k) = |q_k\theta -p_k|$ . In particular, $\ell _{-1} = 1$ , $\ell _0 = \theta $ and $\ell _1 = 1 - \lfloor 1/\theta \rfloor \theta \le 1-\theta $ . Since, for every $k \ge 0$ ,

$$ \begin{align*}\frac{1}{q_{k+2}} \le \frac{a_{k+2}}{q_{k+2}} < |q_k\theta-p_k| < \frac{1}{q_{k+1}},\end{align*} $$

the sequence $(\ell _k)_{k \ge -1}$ is decreasing. For every $k \ge 1$ , we have $\ell _k = -a_k \ell _{k-1} + \ell _{k-2}$ , so $a_k = \lfloor \ell _{k-2}/\ell _{k-1}\rfloor $ .

We can now give a precise statement of the three gap theorem. The formulation below is close to the formulation given by Alessandri and Berthé [Reference Alessandri and Berthé1].

Under the assumptions of Theorem 3, we have $\ell _{k-1}-(b-1)\ell _k = (\ell _{k-1}-b\ell _k)+\ell _k$ . Moreover, $\ell _{k-1}-b\ell _k<\ell _k$ if and only if $b=a_k$ . From these observations, we deduce the following consequences.

We can now answer precisely the following two questions. Given a semi-open subinterval J of $\mathbf {I}$ , what is the maximal number of iterations of $T_{-\theta }$ to reach J from anywhere in $\mathbf {I}$ ? What is the minimal number of iterations of $T_{-\theta }$ necessary to return in J from anywhere in J?

Proof. We are searching for the least non-negative integer m such that

(1) $$ \begin{align} \bigcup_{k=0}^m T_\theta^k(J) = \mathbf{I}. \end{align} $$

By rotation, one may assume that $J = [0,|J|[$ . In this case, for every $k \ge 0$ ,

$$ \begin{align*}T_\theta^k(J) = \begin{cases} [x_k,x_k+|J|[ & \text{if } x_k+|J| \le 1, \\ [x_k,1[ \cup [0,x_k+|J|-1[ & \text{if } x_k+|J|> 1. \end{cases}\end{align*} $$

Therefore, equality 1 holds if and only if the greatest length of the sub-intervals of $\mathbf {I}$ defined by the subdivision $(x_k)_{0 \le k \le m}$ is at most $|J|$ . Thus, the first item follows from Corollary 5.

Next, given $y \in J$ , we are searching for the least integer $m(y) \ge 1$ such that ${T_{-\theta }^{m(y)}(y) \in J}$ , which implies that

$$ \begin{align*}\min_{0 \le i < j \le m(y)} \|x_j-x_i\| \le \|m(y)\theta\| = \|y-T^{m(y)}(y)\| < |J|.\end{align*} $$

Denoting by $k \ge 0$ the least integer such that $\ell _k \le |J|$ , we derive $m(y) \ge q_k$ by Corollary 5.

2.2 A Markov chain on $\mathbf {I}^2$

In this section, we study the Markov chains on $\mathbf {I}^2$ with kernel K. Our purpose is to give upper bounds on the entrance time in the set $B_\theta = [0,\theta [ \times [\theta ,1[ \cup [\theta ,1[ \times [0,\theta [$ .

The transition probabilities can be also written as

$$ \begin{align*}K((y,y'),\cdot) = \tfrac{1}{2}(\delta_{(T_{-\theta}(y),y')} + \delta_{(y,T_{-\theta}(y'))}) \text{ if } (y,y') \in B_\theta,\end{align*} $$

$$ \begin{align*}K((y,y'),\cdot) = \tfrac{1}{2}(\delta_{(y,y')} + \delta_{(T_{-\theta}(y),T_{-\theta}(y'))}) \text{ if } (y,y') \in B_\theta^c.\end{align*} $$

Note that the transition probabilities from any $(y,y') \in B_\theta ^c$ preserve the difference $y'-y$ modulo $1$ . See Figure 2.

Since D is absorbing and has no intersection with $B_\theta $ , the set $B_\theta $ cannot be reached from D. But the set $B_\theta $ can be reached from any $(y,y') \in D^c$ . We set

$$ \begin{align*}t_{y,y'}:=\inf\{n \ge 0 : (T_{-\theta}^n(y),T_{-\theta}^n(y'))\in B_\theta\},\end{align*} $$

$$ \begin{align*}T_{y,y'} = \inf\{n \ge 0 : (F_n(y),F_n(y')) \in B_\theta\}.\end{align*} $$

Note that we have always $T_{y,y'} \ge t_{y,y'}$ , and that $t_{y,y'}$ and $T_{y,y'}$ are null whenever $(y,y')$ is in $B_\theta $ . Moreover,

$$ \begin{align*}T_{y,y'} +1 = \inf\{n \ge 0 : F_n(y')-F_n(y) \not\equiv y'-y\, \mod\!\!\: 1\}.\end{align*} $$

For every real number $\delta $ , we denote by

$$ \begin{align*}\|\delta\| := \mathrm{dist}(\delta,\mathbb{Z}) = \min(\delta-\lfloor\delta\rfloor,1-\delta+\lfloor\delta\rfloor)\end{align*} $$

the shortest distance from $\delta $ to $\mathbb {Z}y$ .

Informally, we want to show that the times $t_{y,y'}$ and $T_{y,y'}$ are small when $\|y'-y\|$ is far from $0$ and that they are ‘often’ big when $\|y'-y\|$ is close to $0$ . The restriction ‘often’ cannot be removed since $t_{\theta -\varepsilon _1,\theta +\varepsilon _2}=0$ for every $\varepsilon _1 \in ~]0,\theta ]$ and $\varepsilon _2 \in [0,1-\theta [$ . See Figure 3.

Figure 3 Entrance time in $B_\theta $ from any point under the action of the map $(y,y') \mapsto (T_{-\theta }(y),T_{-\theta }(y'))$ . Points far from the diagonal have low entrance time.

Next, we prove a useful lemma.

Lemma 7. Let $y \in \mathbb {R}$ . There exists a unique sequence $(\zeta ^y_n)_{n \ge 1}$ of independent uniform Bernoulli random variables such that, for every integer $n \ge 0$ ,

$$ \begin{align*}\text{ for all } n \ge 0, \quad F_{n+1}(y) = T_{-\theta}^{\zeta^y_n}(F_n(y)).\end{align*} $$

Proof. The uniqueness holds since $T_{-\theta }(y') \ne y'$ for every $y' \in \mathbf {I}$ . To prove the existence, we construct a sequence $(\zeta ^y_n)_{n \ge 1}$ of random variables by setting, for every $n \ge 1$ ,

$$ \begin{align*}\zeta^y_n := (\eta_n + {\mathbf 1}_{[\theta,1[}(F_{n-1}(y)))\,\mod\!\!\: 2 = {\mathbf 1}_{[\eta_n=0~;~F_{n-1}(y) \ge \theta]} + {\mathbf 1}_{[\eta_n=1~;~F_{n-1}(y) < \theta]}.\end{align*} $$

For every $n \ge 0$ , set $\mathcal {F}_n := \sigma (\eta _1,\ldots ,\eta _n)$ , with the convention $\mathcal {F}_0 = \{\emptyset ,\Omega \}$ .

By construction, $\zeta ^y_n$ is an $\mathcal {F}_n$ -measurable Bernoulli random variable, which is uniform and independent of $\mathcal {F}_{n-1}$ since

$$ \begin{align*}\mathbf{E}[\zeta^y_n|\mathcal{F}_{n-1}] = \tfrac{1}{2} {\mathbf 1}_{[F_{n-1}(y) \ge \theta]} + \tfrac{1}{2} {\mathbf 1}_{[F_{n-1}(y) < \theta]} = \tfrac{1}{2} \text{ almost surely}.\end{align*} $$

Moreover, a computation distinguishing two cases whether $F_{n-1}(y) < \theta $ or $F_{n-1}(y) \ge \theta $ yields

$$ \begin{align*}F_n(y) = f_{\eta_n}(F_{n-1}(y)) = T_{-\theta}^{\zeta^y_n}F_{n-1}(y).\end{align*} $$

The proof is complete.

Proof. Let $\delta = \|y'-y\|$ . Since y and $y'$ play the same role, we may—and we do—assume that $y' = T_\delta (y)$ .

We use the sequence $(\zeta ^y_n)_{n \ge 1}$ introduced in Lemma 7 and abbreviate the notation into $(\zeta _n)_{n \ge 1}$ .

For every integer $n \ge 0$ ,

$$ \begin{align*}\text{ for all } n \ge 0, \quad F_{n+1}(y) = T_{-\theta}^{\zeta_n}(F_n(y)).\end{align*} $$

For every integer $n \in [0,T_{y,y'}-1]$ , the two points $F_n(y')$ and $F_n(y')$ belong to the same interval $[0,\theta [$ or $[\theta ,1[$ , so $F_{n+1}(y') = T_{-\theta }^{\zeta _n}(F_n(y')) \equiv F_n(y')-\zeta _n\theta \,\mod \!\!\: 1$ . By recursion, we get

$$ \begin{align*}F_n(y) = T_{-\theta}^{\zeta_1+\cdots+\zeta_n}(y)\quad\text{ for all } n \ge 0\end{align*} $$

and

$$ \begin{align*} F_n(y') = T_{-\theta}^{\zeta_1+\cdots+\zeta_n}(y') = T_{\delta}(F_n(y))\quad\text{ for all } n \in [0,T_{y,y'}].\end{align*} $$

Hence,

$$ \begin{align*} T_{y,y'} &= \inf\{n \ge 0 : ( F_n(y),T_{\delta}((F_n(y)) ) \in B_\theta\} \\ &= \inf\{n \ge 0 : ( T_{-\theta}^{\zeta_1+\cdots+\zeta_n}(y), T_{\delta}(T_{-\theta}^{\zeta_1+\cdots+\zeta_n}(y)) ) \in B_\theta\} \\ &= \inf\{n \ge 0 : \zeta_1+\cdots+\zeta_n = t_{y,y'}\}. \end{align*} $$

Since $(\zeta _n)_{n \ge 1}$ is a sequence of independent uniform Bernoulli random variables, the random variable $T_{y,y'}$ is negative binomial, with parameters $t_{y,y'}$ and $1/2$ .

If $\delta \le \theta $ , then, for every $x \in \mathbf {I}$ ,

$$ \begin{align*}(x,T_{\delta}(x)) \in B_\theta \!\!\:\iff x \in [\theta-\delta,\theta[ \cup [1-\delta,1[.\end{align*} $$

Thus, $t_{y,y'} = \inf \{n \ge 0 : T_{-\theta }^n(y) \in [\theta -\delta ,\theta [ \cup [1-\delta ,1[\}$ . By Corollary 6, at most ${bq_k+q_{k-1}-1}$ iterations of the map $T_{-\theta }$ are sufficient to reach each one of the intervals $[\theta -\delta ,\theta [$ or $[1-\delta ,1[$ from y. But these two intervals are disjoint since $0<\theta -\delta <\theta < 1/2<1-\delta <1$ . Hence, $t_{y,y'} \le bq_k+q_{k-1}-2$ .

If $\delta>\theta $ , then, for every $x \in \mathbf {I}$ ,

$$ \begin{align*}(x,T_{\delta}(x)) \in B_\theta \!\!\:\iff x \in [0,\theta[ \cup [1-\delta,1-\delta+\theta[.\end{align*} $$

Thus, $T = \inf \{n \ge 0 : F_n(y) \in [0,\theta [ \cup [1-\delta ,1-\delta +\theta [\}$ . By Corollary 6 applied with $k=1$ and $b=1$ , at most $q_1$ iterations of the map $T_{-\theta }$ are sufficient to reach each one of the intervals $[0,\theta [$ or $[1-\delta ,1-\delta +\theta [$ from y. But these two intervals are disjoint since $0<\theta <1/2<1-\delta <1-\delta +\theta <1$ . Hence, $t_{y,y'} \le q_1-1$ .

The proof is complete.

Proof. By Lemma 8, if $\theta \le \|y'-y\|$ , then $t_{y,y'} \le q_1-1 \le q_k+q_{k-1}-2$ .

From now on, we focus on the case where $\|y'-y\| \le \theta $ . Observe that

$$ \begin{align*}\ell_k+\ell_{k+1} \le a_{k+1}\ell_k+\ell_{k+1} = \ell_{k-1} \le \|y'-y\|.\end{align*} $$

If $\|y'-y\| < \ell _{k-1}+\ell _k$ , then Lemma 8 applied to the integers k and $b=1$ yields $t_{y,y'} \le q_k+q_{k-1}-2$ .

Otherwise, $k \ge 2$ since $\ell _{k-1} < \ell _{k-1}+\ell _k \le \|y'-y\| \le \theta = \ell _0$ . Furthermore, the least integer $k' \ge 1$ such that $\ell _{k'}+\ell _{k'+1} \le \|y'-y\|$ is at most $k-1$ and Lemma 8 applies to the integers $k' \ge 1$ and some $b \in [1,a_{k'+1}]$ , so

$$ \begin{align*}t_{y,y'} \le bq_{k'}+q_{k'-1}-2 \le a_{k'+1}q_{k'}+q_{k'-1}-2 = q_{k'+1}-2 \le q_k-2.\end{align*} $$

Hence, in all cases, we get $t_{y,y'} \le q_k+q_{k-1}-2$ .

2.3 A time-changed symmetric random walk

For every real number x, denote by ${\overline {x} = x+\mathbb {Z}}$ its equivalence class in $\mathbb {R}/\mathbb {Z}$ . Then Lemma 8 and the strong Markov property show that the process $(\overline {F_n(y')-F_n(y)})_{n \ge 0}$ is a time-changed symmetric random walk on $\mathbb {R}/\mathbb {Z}$ with steps $\pm \overline {\theta }$ .

Lemma 10. For every $n \ge 0$ , set $\mathcal {F}_n = \sigma (\eta _1,\ldots ,\eta _n)$ , with the convention $\mathcal {F}_0 = \{\emptyset ,\Omega \}$ . Consider the non-decreasing process $(A_n)_{n \ge 0}$ defined by

$$ \begin{align*}A_n := \sum_{k=0}^{n-1} {\mathbf 1}_{B_\theta} (F_k(y),F_k(y')) \quad\textrm{with the convention } A_0=0,\end{align*} $$

and set $A_\infty :=\lim _n A_n$ . Consider its inverse $(N_a)_{a \ge 0}$ , defined by

$$ \begin{align*}N_a := \inf\{n \ge 0 : A_n \ge a\}\quad \textrm{with the convention } \inf \emptyset = +\infty.\end{align*} $$

On the event $[N_a<+\infty ] = [a \le A_\infty ]$ , set

$$ \begin{align*}\Delta N_a := N_{a+1}-N_a \quad\text{and}\quad W_a := \sum_{b=1}^a (2\zeta^y_{N_b}-1),\end{align*} $$

with the notation of Lemma 7.

1. (1) For every $n \ge 0$ ,

   $$ \begin{align*}F_n(y')-F_n(y) \equiv y'-y + \theta W_{A_n} \,\mod\!\!\: 1.\end{align*} $$

2. (2) The random times $(N_a-1)_{a \ge 1}$ are stopping times in the filtration $(\mathcal {F}_n)_{n \ge 0}$ .

3. (3) The process $(W_a)_{a \ge 0}$ is a simple symmetric random walk on $\mathbb {Z}$ , with possibly finite lifetime $A_\infty $ . More precisely, for every positive integer $a \ge 0$ , conditionally on the event $[A_\infty \ge a]$ , the increments $W_1-W_0,\ldots ,W_a-W_{a-1}$ are independent and uniform on $\{-1,1\}$ .

4. (4) One has $A_\infty = A_{T^D_{y,y'}}$ , where $T^D_{y,y'} := \inf \{n \ge 0 : F_n(y) = F_n(y')\}$ . If $y'-y \in \mathbb {Z}+\theta \mathbb {Z}$ , then $A_\infty $ is almost surely finite. Otherwise, $A_\infty $ is infinite.

5. (5) Assume that $y'-y \notin \mathbb {Z}y+\theta \mathbb {Z}$ . Conditionally on the process $(((F_{N_a}(y), F_{N_a}(y')))_{a \ge 0}$ , the random variables $(\Delta N_a)_{a \ge 0}$ are independent and each $\Delta N_a - 1$ is negative binomial with parameters $t_{F_{N_a}(y),F_{N_a}(y')}$ and $1/2$ .

Proof. We use the sequence $(\zeta _n)_{n \ge 1}:=(\zeta ^y_n)_{n \ge 1}$ introduced in Lemma 7 and define $(\zeta ^{\prime }_n)_{n \ge 1}:=(\zeta ^{y'}_n)_{n \ge 1}$ in the same way. By construction,

$$ \begin{align*}F_n(y) = f_{\eta_n}(F_{n-1}(y)) \equiv F_{n-1}(y)-\zeta_n\theta \,\mod\!\!\: 1.\end{align*} $$

Then

$$ \begin{align*}F_n(y')-F_n(y) \equiv F_{n-1}(y')-F_{n-1}(y) +(\zeta_n-\zeta^{\prime}_n)\theta \,\mod\!\!\: 1.\end{align*} $$

A recursion yields, for every $n \ge 0$ ,

$$ \begin{align*}F_n(y')-F_n(y) \equiv y'-y+M_n \theta \,\mod\!\!\: 1 \quad\text{where } M_n := \sum_{k=1}^n(\zeta_k-\zeta^{\prime}_k),\end{align*} $$

with the convention $M_0=0$ . For every $k \ge 1$ , $\zeta ^{\prime }_k = 1-\zeta _k$ if $(F_{k-1}(y),F_{k-1}(y')) \in B_\theta $ , whereas $\zeta ^{\prime }_k = \zeta _k$ if $(F_{k-1}(y),F_{k-1}(y')) \in B_\theta ^c$ , so

$$ \begin{align*} \zeta_k-\zeta^{\prime}_k &= {\mathbf 1}_{B_\theta} (F_{k-1}(y),F_{k-1}(y')) \times (2\zeta_k-1) \\ &= (A_k-A_{k-1}) \times (2\zeta_k-1). \end{align*} $$

By definition, for every $a \ge 0$ , on the event $[N_a<+\infty ]$ , the process $(A_n)_{n \ge 0}$ remains constant and equal to a during the time interval $[N_a,N_{a+1}-1]$ . Hence,

$$ \begin{align*} M_n = \sum_{k=1}^n (M_k-M_{k-1}) = \sum_{b=1}^{A_n} (M_{N_b}-M_{N_b-1}) = \sum_{b=1}^{A_n} (2\zeta_{N_b}-1) = W_{A_n}. \end{align*} $$

Item (1) follows.

The process $(A_n)_{n \ge 0}$ defined by

$$ \begin{align*}A_n = \sum_{k=0}^{n-1} {\mathbf 1}_{B_\theta} (F_k(y),F_k(y'))\end{align*} $$

is $(\mathcal {F}_n)_{n \ge 0}$ -predictable: the random variable $A_0=0$ is constant and, for every $n \ge 1$ , the random variable $A_n$ is $\mathcal {F}_{n-1}$ -measurable.

For every $a \ge 0$ , the random variable $N_a$ is a stopping time in the filtration $(\mathcal {F}_n)_{n \ge 0}$ . If $a \ge 1$ , the random variable $N_a-1$ is still a stopping time since, for every $n \ge 0$ ,

$$ \begin{align*}[N_a-1 \le n] = [N_a \le n+1] = [A_{n+1} \ge a] \in \mathcal{F}_n.\end{align*} $$

Item (2) follows.

Let $a \ge 0$ . The event $[N_a<+\infty ] = [N_a-1<+\infty ]$ belongs to $\mathcal {F}_{N_a-1}$ . On this event, the sequence $(\tilde {\eta }_{n})_{n \ge 0}:=(\eta _{N_a+n})_{n \ge 0}$ is an independent and identically distributed sequence of uniform Bernoulli random variables and is independent of $\mathcal {F}_{N_a-1}$ . Therefore, we know the following.

• • The Bernoulli random variable $\zeta _{N_a} = (\eta _{N_a}+{\mathbf 1}_{[0,\theta [}(F_{N_a-1}(y)))\,\mod \!\!\: 2$ is uniform, independent of $\mathcal {F}_{N_a-1}$ and $\mathcal {F}_{N_a}$ -measurable. Thus the random variable $W_a-W_{a-1} = 2\zeta _{N_a}-1$ is uniform on $\{-1,1\}$ , independent of $\mathcal {F}_{N_a-1}$ and $\mathcal {F}_{N_a}$ -measurable. Item (3) follows.

• • The random variable $(Y,Y') :=(F_{N_a}(y),F_{N_a}(y'))$ is $\mathcal {F}_{N_a}$ -measurable.

• • The process $(\tilde {F}_{n})_{n \ge 0}$ , defined by $\tilde {F}_n := f_{\tilde {\eta }_n} \circ \cdots \circ f_{\tilde {\eta }_1}$ , is independent of $\mathcal {F}_{N_a}$ and has the same law as $(F_{n})_{n \ge 0}$ .

Observe that, with obvious notation,

$$ \begin{align*}\Delta N_a - 1 = (N_{a+1}-1)-N_a = \tilde{T}_{Y,Y'} := \inf\{n \ge 0 : (\tilde{F}_n(Y),\tilde{F}_n(Y')) \in B_\theta\}.\end{align*} $$

Hence, conditionally on $\mathcal {F}_{N_a}$ and on the event $[N_a<+\infty ] \cap [Y \ne Y']$ , the random variable $\Delta {N_a}-1$ is almost surely finite and negative binomial with parameters $t_{Y,Y'}$ and $1/2$ . Item (5) follows. Moreover, for every $a \ge 0$ , we have almost surely

$$ \begin{align*}N_{a+1}<+\infty \!\!\:\iff (N_a<+\infty \text{ and } F_{N_a}(y) \ne F_{N_a}(y')).\end{align*} $$

Hence, almost surely,

$$ \begin{align*}A_\infty = \inf\{a \ge 0 : N_{a+1}=+\infty\} = \inf\{a \ge 0 : N_a<+\infty~;~F_{N_a}(y) = F_{N_a}(y')\}.\end{align*} $$

During each time interval $[N_a,N_{a+1}-1]$ , the process $(F_n(y')-F_n(y))$ remains constant modulo 1 and the process $(A_n)_{n \ge 0}$ remains constant and equal to a. Thus,

$$ \begin{align*}A_\infty = A_{T^D_{y,y'}} \quad\text{where } T^D_{y,y'} := \inf\{n \ge 0 : F_n(y) = F_n(y')\}.\end{align*} $$

Since, for every $n \ge 0$ , $F_n(y') - F_n(y) \equiv y'-y + \theta W_{A_n} \,\mod \!\!\: 1$ , the difference ${F_n(y') - F_n(y)}$ remains in the coset $y'-y + \theta \mathbb {Z} + \mathbb {Z}$ . Therefore, the time $T^D_{y,y'}$ cannot be finite unless $y'-y \in \theta \mathbb {Z} + \mathbb {Z}$ . Conversely, if $y'-y \in \theta \mathbb {Z} + \mathbb {Z}$ , then $T^D_{y,y'}$ is almost surely finite by recurrence of the symmetric simple random walk on $\mathbb {Z}$ . Item (4) follows.

By density of the subgroup $\theta \mathbb {Z} + \mathbb {Z}$ in $\mathbb {R}$ and by recurrence of the symmetric simple random walk on $\mathbb {Z}$ , we derive the following consequence.

Corollary 11. Let y and $y'$ be two points in $\mathbf {I}$ such that $y'-y \notin \mathbb {Z}+\theta \mathbb {Z}$ . Then, almost surely,

$$ \begin{align*}\liminf_{n \to +\infty} \|F_n(y')-F_n(y)\| = 0 \quad\text{and}\quad \limsup_{n \to +\infty} \|F_n(y')-F_n(y)\| = 1/2.\end{align*} $$

Proof. We use the notation and the results of Lemma 10. The recurrence of the symmetric simple random walk on $\mathbb {Z}$ and the density of the subgroup $\theta \mathbb {Z} + \mathbb {Z}$ in $\mathbb {R}$ yields

$$ \begin{align*}\inf_{n \ge 0} \|F_n(y')-F_n(y)\| = \inf_{a \ge 0} \|y'-y + \theta W_a\| = \inf_{w \in \mathbb{Z}} \|y'-y + \theta w\| = 0 \text{ almost surely}.\end{align*} $$

Since the quantities $\|F_n(y')-F_n(y)\|$ are all positive, the greatest lower bound of the sequence $(\|F_n(y')-F_n(y)\|)_{n \ge 0}$ is not achieved and is therefore a lower limit. We argue in the same way with the least upper bound.

To prove that the Cesàro averages of the sequence $(\|F_n(y')-F_n(y)\|)_{n \ge 0}$ tend to zero, we observe that the time change in the process $(W_{A_n})_{n \ge 0}$ slows down the random walk much more when the difference is $\|y'-y\|$ is small. Indeed, Corollary 9 shows that if $\|y'-y\|$ is not close to zero, then the process $((F_n(y),F_n(y'))_{n \ge 0}$ reaches $B_\theta $ quickly, and the difference $F_n(y')-F_n(y)$ changes immediately after.

Conversely, if $\|y'-y\|$ is small, we hope that it will take a long time for the process $((F_n(y),F_n(y')))_{n \ge n_0}$ to reach $B_\theta $ . During this time, the difference $F_n(y')-F_n(y)$ remains unchanged. Yet, we may be unlucky, so the times $t_{y,y'}$ and $T_{y,y'}$ may be small. Fortunately, the distance $\|F_n(y')-F_n(y)\|$ is close to $\theta $ at time $N_1 = T_{y,y'}+1$ , so it will change soon. With probability $1/2$ , we will get $W_2=0$ , so $\|F_{N_2}(y')-F_{N_2}(y)\| = \|y'-y\|$ will be small, and this time the distance $\|F_n(y')-F_n(y)\|$ will remain unchanged during a long time. The role of the next Lemma is to provide a lower bound for this time.

Lemma 12. Let y and $y'$ be two distinct points in $\mathbf {I}$ such that $(y,y') \in [0,1-\theta [^2 \cup [\theta ,1[^2$ and $\|y'-y\| < \ell _2$ . Let $k \ge 3$ be the integer such that $\ell _k \le \|y'-y\| < \ell _{k-1}$ . Keep the notation of Lemma 10. Then

$$ \begin{align*}\textrm{ on the event } [W_2=0], \quad \Delta N_2-1 \ge t_{F_{N_2}(y),F_{N_2}(y')} \ge q_k+1-q_2-q_1.\end{align*} $$

Before proving Lemma 12, it is important to note that all probability transitions $K((y,y'),\cdot )$ give full measure to the set $[0,1-\theta [^2 \cup [\theta ,1[^2$ (see Figure 4). Hence, the additional hypothesis $(y,y') \in [0,1-\theta [^2 \cup [\theta ,1[^2$ will not be a true restriction. Actually, this assumption could be removed by changing the lower bound into $q_k-q_2-q_1$ .

Figure 4 The squares $[0,\theta [ \times [1-\theta ,1[$ and $[1-\theta ,1[ \times [0,\theta [$ cannot be reached from anywhere via the kernel K. If $0<\delta <\theta $ , then $B_\theta \cap \{(y,T_\delta (y)); y \in \mathbf {I}\}$ is the union of two ‘semi-open segments’, defined by $\theta -\delta \le y <\theta $ and $1-\delta \le y <1$ . The second semi-open segment cannot be reached from anywhere via the kernel K and is the image of the first one by the map $(y,y') \mapsto (T_{-\theta }(y),T_{-\theta }(y'))$ .

Proof. The inequality $\Delta N_2-1 \ge t_{F_{N_2}(y),F_{N_2}(y')} \ge 0$ follows from item (5) of Lemma 10.

The assumption $(y,y') \in [0,1-\theta [^2 \cup [\theta ,1[^2$ ensures that $(F_n(y),F_n(y'))$ belongs to $[0,1-\theta [^2 \cup [\theta ,1[^2$ not only for every $n \ge 1$ , but also when $n=0$ .

Set $\delta = \|y'-y\|$ . By symmetry, we may assume that $y'-y \equiv \delta \,\mod \!\!\: 1$ , that is, ${y'=T_\delta (y)}$ . Hence, for every $a \ge 0$ , $F_{N_{a+1}-1}(y')-F_{N_{a+1}-1}(y) \equiv \delta + \theta W_a \,\mod \!\!\: 1$ .

On the event $[W_2=0]$ , we have

$$ \begin{align*}F_{N_3-1}(y')-F_{N_3-1}(y) \equiv F_{N_1-1}(y')-F_{N_1-1}(y) \equiv \delta \,\mod\!\!\: 1.\end{align*} $$

Since the points $(F_{N_1-1}(y),F_{N_1-1}(y'))$ and $(F_{N_3-1}(y),F_{N_3-1}(y'))$ belong at the same time to $B_\theta $ and to $[0,1-\theta [^2 \cup [\theta ,1[^2$ , we deduce that $F_{N_1-1}(y)$ and $F_{N_3-1}(y)$ belong to $[\theta -\delta ,\theta [$ . But $F_{N_3-1}(y)$ is obtained from $F_{N_1-1}(y)$ by applying the transformation $T_{-\theta }$ exactly $\zeta ^y_{N_1}+t_{F_{N_1}(y),F_{N_1}(y')}+\zeta ^y_{N_2}+t_{F_{N_2}(y),F_{N_2}(y')}$ times.

On the event $[W_2=0]$ , we have $\zeta ^y_{N_1}+\zeta ^y_{N_2}=1$ , so Corollary 6 yields

$$ \begin{align*}t_{F_{N_1}(y),F_{N_1}(y')}+t_{F_{N_2}(y),F_{N_2}(y')}+1 \ge q_k.\end{align*} $$

But $F_{N_1}(y')-F_{N_1}(y) \equiv \delta \pm \theta \,\mod \!\!\: 1$ , so

$$ \begin{align*}\|F_{N_1}(y')-F_{N_1}(y)\| \ge \theta - \delta> \ell_0 - \ell_k \ge \ell_1.\end{align*} $$

Corollary 9 yields $t_{F_{N_1}(y),F_{N_1}(y')} \le q_2+q_1-2$ .Thus,

$$ \begin{align*}t_{F_{N_2}(y),F_{N_2}(y')}+q_2+q_1-1 \ge q_k.\end{align*} $$

The desired inequality follows.

2.4 Martingales and symmetric simple random walk on $\mathbb {Z}$

In this subsection, we establish three more results that which help us in the proof of Proposition 2. The first of them is a kind of law of large numbers.

Lemma 13. Let $(\mathcal {G}_n)_{n \ge 0}$ be a filtration of $(\Omega ,\mathcal {A},\mathbf {P})$ and let $(Z_n)_{n \ge 1}$ be a sequence of real random variables, bounded in $L^2(\mathbf {P})$ , and adapted to the filtration $(\mathcal {G}_n)_{n \ge 1}$ . Then

$$ \begin{align*}\frac{1}{n} \sum_{k=1}^n (Z_k-\mathbf{E}[Z_k|\mathcal{G}_{n-1}] ) \to 0 \quad\textrm{almost surely as } n \to +\infty.\end{align*} $$

Proof. For every $n \ge 0$ , let

$$ \begin{align*}M_n = \sum_{k=1}^n \frac{1}{k} (Z_k-\mathbf{E}[Z_k|\mathcal{G}_{n-1}] ).\end{align*} $$

By construction, the sequence $(M_n)_{n \ge 0}$ is a square-integrable martingale in $(\mathcal {G})_{n \ge 0}$ . The increments $M_k-M_{k-1}$ are pairwise orthogonal in $L^2(\mathbf {P})$ and

$$ \begin{align*} \sum_{k \ge 1} \|M_k-M_{k-1}\|_2^2 = \sum_{k \ge 1}\frac{1}{k^2} \|Z_k-\mathbf{E}[Z_k|\mathcal{G}_{n-1}] \|_2^2 \le \sum_{k \ge 1}\frac{1}{k^2} \|Z_k \|_2^2 < +\infty. \end{align*} $$

Hence, the martingale $(M_n)_{n \ge 0}$ is bounded in $L^2(\mathbf {P})$ and it converges almost surely and in $L^2(\mathbf {P})$ to some real random variable $M_\infty $ .

$$ \begin{align*} \frac{1}{n} \sum_{k=1}^n (Z_k-\mathbf{E}[Z_k|\mathcal{G}_{n-1}] ) &= \frac{1}{n} \sum_{k=1}^n k(M_k-M_{k-1}) \\ &= \frac{1}{n} \bigg(nM_n + \sum_{k=1}^{n-1} kM_k - M_0 - \sum_{k=1}^{n-1}(k+1)M_k \bigg) \\ &= M_n - \frac{1}{n} \sum_{k=1}^{n-1} M_k. \end{align*} $$

The conclusion follows from Cesàró’s lemma.

The second result is a refinement of the gambler’s ruin theorem [Reference Resnick10].

Lemma 14. Let $(W_n)_{n \ge 0}$ be a symmetric simple random walk on $\mathbb {Z}$ , that is, $W_0=0$ and $(W_n-W_{n-1})_{n \ge 1}$ is an independent and identically distributed sequence of Rademacher random variables. Fix two non-negative integers a and b and set

$$ \begin{align*}\tau_{-a,b} = \inf\{n \ge 0 : W_n = \{-a,b\}\}.\end{align*} $$

Then $\tau _{-a,b}<+\infty $ almost surely. Moreover:

• • $\mathbf {P}[W_{\tau _{-a,b}} = -a] = b/(a+b)$ and $\mathbf {P}[W_{\tau _{-a,b}} = b] = a/(a+b)$ ;

• • $\mathbf {E}[\tau _{-a,b}] = ab$ ; and

• • for every $\unicode{x3bb} \ge 0$ , $\mathbf {E}[(\cosh \unicode{x3bb} )^{-\tau _{-a,b}}] = \cosh ((({-a+b})/{2})\unicode{x3bb} ) / \cosh ((({a+b})/{2})\unicode{x3bb} ).$

Proof. One checks that the processes $(W_n)_{n \ge 0}$ , $(W_n^2-n)_{n \ge 0}$ and $(e^{\pm \unicode{x3bb} W_n}(\cosh \unicode{x3bb} )^{-n})_{n \ge 0}$ are martingales, and that the martingales

$$ \begin{align*}(W_n)_{n \ge 0}, \quad \bigg( \bigg(W_n-\frac{b-a}{2}\bigg)^2-n \bigg)_{n \ge 0}, \quad \bigg(\cosh\bigg(\unicode{x3bb}\bigg(W_n-\frac{b-a}{2}\bigg)\bigg)(\cosh \unicode{x3bb})^{-n}\bigg)_{n \ge 0}\end{align*} $$

are uniformly bounded on the time interval $[0,\tau _{-a,b}]$ . Hence, the optional stopping time theorem applies. Writing that the expectations at time $\tau _{-a,b}$ are equal to the expectations at time $0$ yields items (1), (2) and (3).

The third result states an equidistribution modulo 1.

Lemma 15. Let $(W_n)_{n \ge 0}$ be a symmetric simple random walk on $\mathbb {Z}y$ and let $\theta $ be any irrational number. Then, for every continuous $1$ -periodic function $f : \mathbb {R} \to \mathbf {C}$ ,

$$ \begin{align*}\frac{1}{n}\sum_{k=0}^{n-1} f(\theta W_k) \to \int_0^1 f(x)\, {\mathrm{d}} x \quad\textrm{almost surely as } n \to +\infty.\end{align*} $$

Proof. Since the trigonometric polynomials form a dense subspace of the vector space of all continuous $1$ -periodic functions, we need only to prove the result when f is the function $e_m : x \mapsto e^{i2\pi mx}$ for some $m \in \mathbb {Z}$ . The case where $m=0$ is trivial, so we fix $m \ne 0$ .

For every $n \ge 1$ , let

$$ \begin{align*}S_n := \sum_{k=0}^{n-1} e^{i2\pi m\theta W_k}.\end{align*} $$

Then

$$ \begin{align*} |S_n|^2 = S_n \overline{S_n} = \sum_{k=0}^{n-1} \sum_{\ell=0}^{n-1}e^{i2\pi m\theta(W_k-W_\ell)}. \end{align*} $$

Let $z = \mathbf {E}[e^{i2\pi m\theta (W_1-W_0)}] = \cos (2\pi m\theta )$ . Then $z \in ~]-1,1[$ since $m \ne 0$ and $\theta $ is irrational. By independent equidistribution of the increments of the random walk W, we derive

$$ \begin{align*} \mathbf{E}[|S_n|^2] &= \sum_{k=0}^{n-1} \sum_{\ell=0}^{n-1} z^{|k-\ell|} = n + 2\sum_{d=1}^{n-1} (n-d)z^d \\ &\le n \bigg(1+2\sum_{d=1}^{n-1}|z|^d \bigg) \le n \frac{1+|z|}{1-|z|}. \end{align*} $$

Thus,

$$ \begin{align*}\sum_{n=1}^{+\infty} \bigg\Vert \frac{1}{n^3}S_{n^3} \bigg\Vert_2 \le \sum_{n=1}^{+\infty} \bigg(\frac{1+|z|}{1-|z|}\bigg)^{-1/2} n^{-3/2}< +\infty.\end{align*} $$

The series $\sum _n S_{n^3}/n^3$ is absolutely convergent in $L^2(\mathbf {P})$ , so it converges almost surely and in $L^2(\mathbf {P})$ . Therefore, $S_{n^3}/n^3 \to 0$ almost surely as $n \to +\infty $ . To deduce that $S_n/n \to 0$ almost surely as $n \to +\infty $ , it suffices to set $r_n = \lfloor n^{1/3} \rfloor $ and to note that

$$ \begin{align*}|S_n - S_{r_n^3}| \le n-r_n^3 \le 3n^{2/3}(n^{1/3}-r) \le 3n^{2/3}.\\[-38pt]\end{align*} $$

5.1 Almost sure continuity of the isomorphism and its inverse

We come back to the transformation $[T_\theta ,\mathrm {Id}]$ defined on $X \times \mathbf {I}$ , where $X = \{0,1\}^{\mathbb {Z}y_+}$ and $\mathbf {I} = [0,1[$ . We endow X with the product topology, $\mathbf {I}$ with the topology deriving from the metric defined by $d(y,y') := \|y'-y\|$ and $X \times \mathbf {I}$ with the product topology. Since the map $T_\theta $ is continuous on $\mathbf {I}$ , the map $[T_\theta ,\mathrm {Id}]$ is continuous on $X \times \mathbf {I}$ .

The map $\alpha _\theta : X \times \mathbf {I} \to \{0,1\}$ associated to Parry’s partition $\alpha _\theta := \{A_0^\theta ,A_1^\theta \}$ is continuous $\mu \otimes \nu $ -almost everywhere since it is given by

$$ \begin{align*} \alpha_\theta(x,y) = {\mathbf 1}_{A_1^\theta}(x,y) &= ( {\mathbf 1}_{[(1-x_0)\theta + x_0(1-\theta),1[}(y) ) \,\mod\!\!\: 2 \\ &= ( x_0+{\mathbf 1}_{[\theta,1[}(T_\theta^{x(0)}(y)) ) \,\mod\!\!\: 2. \end{align*} $$

The ‘ $\alpha _\theta $ -name’ map $\Phi _\theta : X \times \mathbf {I} \to \mathbf {I}$ is given by

$$ \begin{align*}\Phi_\theta(x,y) := ((\alpha_\theta \circ [T_\theta,\mathrm{Id}]^n)(x,y))_{n \ge 0}.\end{align*} $$

Since $[T_\theta ,\mathrm {Id}]$ is continuous and preserves $\mu \otimes \nu $ , we deduce that $\Phi _\theta $ is continuous $\mu \otimes \nu $ -almost everywhere.

We have proved that the partition $\alpha _\theta $ is an independent generator, so $\Phi _\theta $ is an isomorphism transforming the dynamical system $(X \times [0,1[,\mathcal {B}(X) \otimes \mathcal {P}([0,1[),\mu \otimes \nu ,[T_\theta ,\mathrm {Id}])$ into $(X,\mathcal {B}(X),\mu ,S)$ . Thouvenot asked me whether or not $\Phi _\theta ^{-1}$ is continuous $\mu $ -almost everywhere. Answering this question is difficult since we do not have simple formulas to recover $(x,y) \in X \times \mathbf {I}$ from $\Phi _\theta (x,y)$ .

Once again, we reformulate our problem in probabilistic terms. Remember that

$$ \begin{align*}\Phi_\theta((\xi_{-i})_{i \ge 0},Y_0) = (\eta_{-i})_{i \ge 0}.\end{align*} $$

We know that $Y_0$ is almost surely a function of $(\eta _{-i})_{i \ge 0}$ . To prove that $\Phi _\theta ^{-1}$ is continuous $\mu $ -almost everywhere, we just have to prove that this function is almost surely continuous, thanks to the recursion relations

$$ \begin{align*} \xi_{-i} &= ( \eta_{-i} + {\mathbf 1}_{[\theta,1[}(Y_{{-i}-1}) ) \,\mod\!\!\: 2 \\ &= ( \eta_{-i} + {\mathbf 1}_{[\theta,1[}(T_\theta^{\xi_{-i}+\cdots+\xi_0}(Y_0)) ) \,\mod\!\!\: 2. \end{align*} $$

For every $n \ge 0$ , we know that

$$ \begin{align*}Y_0 = f_{\eta_0} \circ \cdots \circ f_{\eta_{-n+1}}(Y_{n+1}),\end{align*} $$

that $(\eta _0,\ldots ,\eta _{-n+1})$ is independent of $\mathcal {F}^{\xi ,Y}_{-n}$ and therefore of $Y_{-n}$ , and $Y_{-n}$ is uniform on  $\mathbf {I}$ .

Given $n \ge 0$ , the random map $P_n = f_{\eta _0} \circ \cdots \circ f_{\eta _{-n+1}}$ has the same law as ${F_n = f_{\eta _n} \circ \cdots \circ f_{\eta _1}}$ . Yet, there is a great difference between the processes $(P_n)_{n \ge 0}$ and $(F_n)_{n \ge 0}$ . When we couple in the past, the range $P_n(\mathbf {I})$ can only decrease or stay equal as n increases. But when we couple in the future, no such inclusion holds. We only know that the Lebesgue measures $|F_n(\mathbf {I})|$ can only decrease or stay unchanged, since, for every $n \ge 0$ , the set $F_{n+1}(\mathbf {I})$ is the image of $F_n(\mathbf {I})$ by the random piecewise translation $f_{\eta _{n+1}}$ .

Moreover, we have the following result.

This convergence is quite slow. In an online version of the paper, we establish that $\mathbf {E}[|F_n(\mathbf {I})|] \le \mathbf {E}[|E_n|] \le 2/q_k$ whenever $n \ge 2(q_k+q_{k-1})q_k^2$ . Since, for each $n \ge 0$ , the random variables $|P_n(\mathbf {I})|$ and $|F_n(\mathbf {I})|$ have the same distribution, and since the sequence $(P_n(\mathbf {I}))_{n \ge 0}$ is non-decreasing, we deduce that $|P_n(\mathbf {I})|$ also goes to zero almost surely as n goes to infinity. These facts and simulations (see Figure 8) lead us to formulate the following conjecture.

Figure 8 Five simulations showing the evolution of the length of the smallest interval of $\mathbf {I}$ (viewed as the circle $\mathbb {R}/\mathbb {Z}$ ) containing $P_n(\mathbf {I})$ , for $1 \le n \le 3000$ , when $\theta = (\sqrt {5}-1)/2$ .

Actually, we expect a still slower convergence. Observe that the conclusion will be false if we replace $P_n(\mathbf {I})$ by $F_n(\mathbf {I})$ .

If Conjecture 20 is true, this gives a positive answer to Thouvenot’s question. Indeed, given $\delta \in ~]0,1/2[$ , we can find almost surely an integer $N \ge 1$ such that the diameter of $P_n(\mathbf {I})$ is at most $\delta $ . Then, the knowledge of $\eta _0,\ldots ,\eta _{-N+1}$ only is sufficient to provide a random variable $\tilde {Y}_0$ such that $\|Y_0 - \tilde {Y}_0\| \le \delta $ . Hence, $Y_0$ can be recovered with probability $1$ as an almost everywhere continuous function of $(\eta _{-i})_{i \ge 0}$ .

Yet, proving this conjecture requires new ideas and looks difficult.

5.2 Changing the partition

To define the random sequence $(\eta _n)_{n \in \mathbb {Z}}$ , we split the interval $\mathbf {I} = [0,1[$ at $\theta $ . An advantage of the (non-trivial) partition $\iota = \{[0,\theta [,[\theta ,1[\}$ is that, for every $n \ge 0$ , the partition

$$ \begin{align*}\bigvee_{k=0}^n T_\theta^k\iota\end{align*} $$

comprises only $n+1$ intervals given by the subdivision $((k\theta ) \,\mod \!\!\: 1)_{0 \le k \le n}$ , which is the least number possible. This facilitates the study of the random maps $F_n = f_{\eta _n} \circ \cdots \circ f_{\eta _1}$ .

Yet, if the definition of the sequence $(\eta _n)_{n \in \mathbb {Z}}$ is modified only by a modification of the splitting point— $\alpha $ instead of $\theta $ , say—a large part of our preceding analysis remains true, namely, Lemma 10 and its variant for $[T_\theta ,T_\theta ^{-1}]$ . Yet, the set $B_\theta $ must be replaced by $B_\alpha $ , and estimating the reaching time of $B_\alpha $ from any point in $\mathbf {I}^2$ is less simple.

A remarkable phenomenon occurs when we study $[T_\theta ,T_\theta ^{-1}]$ and split $\mathbf {I}$ at $1/2$ : namely, when we set

$$ \begin{align*}\eta_n := ( {\mathbf 1}_{[0,1/2[}(Y_{n-1}) - {\mathbf 1}_{[1/2,1[}(Y_{n-1}) )\xi_n.\end{align*} $$

Indeed, replacing the Markov chain $((\xi _n,Y_n))_{n \in \mathbb {Z}}$ by $((-\xi _n,(1-Y_n) \,\mod \!\!\: 1)_{n \in \mathbb {Z}}$ preserves its law and modifies the sequence $(\eta _n)_{n \in \mathbb {Z}}$ only on a null set, namely, on the negligible event $[\text {there exists } n \in \mathbb {N} : Y_n \in \{0,1/2\}]$ . Hence, the knowledge of $(\eta _n)_{n \in \mathbb {Z}}$ is not sufficient to recover almost surely the Markov chain $((\xi _n,Y_n))_{n \in \mathbb {Z}}$ . By symmetry, at least one bit of information is lost.

We complete this observation with a coupling argument to prove that only one bit of information is lost. Given y and $y'$ in $\mathbf {I}$ , look at the Markov chain $((F_n(y),F_n(y'))_{n \ge 0}$ , where $F_n = f_{\eta _n} \circ \cdots \circ f_{\eta _1}$ .

• • On the event $[(F_n(y),F_n(y')) \in B_{1/2}^c]$ , we have

  $$ \begin{align*}\|F_{n+1}(y')-F_{n+1}(y)\| = \|F_n(y')-F_n(y)\| \le \|F_n(y')+F_n(y)\|.\end{align*} $$

• • On the event $[(F_n(y),F_n(y')) \in B_{1/2}]$ , we have

  $$ \begin{align*}\|F_{n+1}(y')+F_{n+1}(y)\| = \|F_n(y')+F_n(y)\| \le \|F_n(y')-F_n(y)\|.\end{align*} $$

As a result, the quantity $\min (\|F_n(y')+F_n(y)\|,\|F_n(y')-F_n(y)\|)$ can only decrease or stay unchanged, so it tends to zero since $\liminf _{n \to +\infty }\|F_n(y')-F_n(y)\| = 0$ .

Therefore, if $Y' := (Y^{\prime }_n)_{n \in \mathbb {Z}}$ is a copy of $Y := (Y_n)_{n \in \mathbb {Z}y}$ such that Y and $Y'$ are independent and identically distributed conditionally on $\eta := (\eta _n)_{n \in \mathbb {Z}}$ , the process $((Y_n,Y^{\prime }_n))_{n \in \mathbb {Z}}$ thus obtained is a stationary Markov chain with transition kernel K, so it lives almost surely on $D \cup D'$ , where $D'$ is the anti-diagonal

$$ \begin{align*}D' = \{(y,y') \in \mathbf{I}^2 : y+y' \in \mathbb{Z}\} = \{(0,0)\} \cup \{(y,1-y) : y \in~]0,1[\}.\end{align*} $$

Moreover, $D'$ is an absorbing set, like D. Since

$$ \begin{align*}\mathcal{L}((Y,Y')|\eta) = \mathcal{L}(Y|\eta) \otimes \mathcal{L}(Y|\eta),\end{align*} $$

we derive that the conditional law $\mathcal {L}(Y|\eta )$ is almost surely carried by two points, which gives the desired conclusion.

This symmetry is an exceptional case. Intuitively, we may expect that no information is lost in the other cases, namely, when the splitting point is different from $1/2$ , or when we work with $[T_\theta ,\mathrm {Id}]$ .

5.3 Working with more general skew products

So far, we have worked only with $[T_\theta ,\mathrm {Id}]$ and $[T_\theta ,T_\theta ^{-1}]$ . A natural generalization of these two transformations is $[T_\alpha ,T_\beta ]$ , where $\alpha $ and $\beta $ are real numbers such that $\beta -\alpha $ is irrational. The map $[T_\alpha ,T_\beta ]$ can be defined on the product $\{\alpha ,\beta \}^{\mathbb {Z}_+} \times \mathbf {I}$ by

$$ \begin{align*}[T_\alpha,T_\beta]((x_n)_{n \ge 0},y) := ((x_{n+1})_{n \ge 0},T_{x_0}(y)).\end{align*} $$

Let $(Y_n)_{n \in \mathbb {Z}}$ be a Markov chain $(Y_n)_{n \in \mathbb {Z}}$ governed by the recursion relation

$$ \begin{align*}Y_n = T_{-\xi_n}(Y_{n-1}), \text{ where } \xi_n \textrm{ is independent of } \mathcal{F}^{\xi,Y}_{n-1} \textrm{ and uniform on } \{\alpha,\beta\}.\end{align*} $$

Let $\theta = \beta -\alpha $ . Define $f_0$ and $f_1$ from $\mathbf {I}$ to $\mathbf {I}$ by

$$ \begin{align*}f_0(y) := {\mathbf 1}_{[0,\theta[}(y) T_{-\alpha(y)} - {\mathbf 1}_{[\theta,1[}(y)T_{-\beta}(y),\end{align*} $$

$$ \begin{align*}f_1(y) := {\mathbf 1}_{[0,\theta[}(y) T_{-\beta(y)} - {\mathbf 1}_{[\theta,1[}(y)T_{-\alpha}(y).\end{align*} $$

The graphs of $f_0$ and $f_1$ are drawn in Figure 9. The recursion governing $(Y_n)_{n \in \mathbb{Z}}$ can be written $Y_n = f_{\eta_n}(Y_{n-1})$ by adapting the definition of $(\eta_n)_{n \in \mathbb{Z}}$ accordingly.

Figure 9 The modified maps $f_0$ and $f_1$ .

Once again, we study the Markov chain $((F_n(y),F_n(y')))_{n \ge 0}$ , where $(F_n)_{n \ge 0}$ is the sequence of random maps defined by $F_n = f_{\eta _n} \circ \cdots \circ f_{\eta _1}$ . Given y and $y'$ in $\mathbf {I}$ , the difference $F_n(y')-F_n(y)$ is still (modulo 1) $\theta $ times a time-changed symmetric simple random walk on $\mathbb {Z}y$ , where the time change is given by the past sojourn time in $B_\theta $ of the process $((F_n(y),F_n(y')))_{n \ge 0}$ .

But this time, when the Markov chain $((F_n(y),F_n(y')))_{n \ge 0}$ is in $B_\theta ^c$ , the steps it makes are uniform on $\{(-\alpha ,-\alpha ),(-\beta ,-\beta )\}$ . In this general context, estimating the reaching time of $B_\theta $ from any point in $\mathbf {I}^2$ is much harder.