CLT for Random Walks of Commuting Endomorphisms on Compact Abelian Groups

Abstract

Let $\mathcal S$ be an abelian group of automorphisms of a probability space $(X, {\mathcal A}, \mu )$ with a finite system of generators $(A_1, \ldots , A_d).$ Let $A^{{\underline{\ell }}}$ denote $A_1^{\ell _1} \ldots A_d^{\ell _d}$, for ${{\underline{\ell }}}= (\ell _1, \ldots , \ell _d).$ If $(Z_k)$ is a random walk on ${\mathbb {Z}}^d$, one can study the asymptotic distribution of the sums $\sum _{k=0}^{n-1} \, f \circ A^{\,{Z_k(\omega )}}$ and $\sum _{{\underline{\ell }}\in {\mathbb {Z}}^d} {\mathbb {P}}(Z_n= {\underline{\ell }}) \, A^{\underline{\ell }}f$, for a function f on X. In particular, given a random walk on commuting matrices in $SL(\rho , {\mathbb {Z}})$ or in ${\mathcal M}^*(\rho , {\mathbb {Z}})$ acting on the torus ${\mathbb {T}}^\rho$, $\rho \ge 1$, what is the asymptotic distribution of the associated ergodic sums along the random walk for a smooth function on ${\mathbb {T}}^\rho$ after normalization? In this paper, we prove a central limit theorem when X is a compact abelian connected group G endowed with its Haar measure (e.g., a torus or a connected extension of a torus), $\mathcal S$ a totally ergodic d-dimensional group of commuting algebraic automorphisms of G and f a regular function on G. The proof is based on the cumulant method and on preliminary results on random walks.

Appendices

Appendix 1: Mixing, Moments and Cumulants

For the sake of completeness, we recall in this appendix some general results on mixing of all orders, moments and cumulants (see [24] and the references given therein). Since our aim is to apply the results to bounded families of trigonometric polynomials, for simplicity we will assume that the families of random variables are uniformly bounded.

For a real random variable Y (or for a probability distribution on ${\mathbb {R}}$), the cumulants (or semi-invariants) can be formally defined as the coefficients $c^{(r)}(Y)$ of the cumulant generating function $t \rightarrow \ln {\mathbb {E}}(e^{tY}) = \sum _{r=0}^\infty \, c^{(r)}(Y) \,{t^r \over r!}$, i.e., $c^{(r)}(Y) = {\partial ^r \over \partial ^r t} \ln {\mathbb {E}}(e^{tY})|_{t = 0}.$ Similarly the joint cumulant of a random vector $(X_1, \ldots , X_r)$ is defined by

$$\begin{aligned} c(X_1, \ldots , X_r) = {\partial ^r \over \partial t_1 \ldots \partial t_r} \ln {\mathbb {E}}(e^{\sum _{j=1}^r t_jX_j})|_{t_1 = \ldots = \, t_r = 0}. \end{aligned}$$

(55)

This definition can be given as well for a finite measure $\nu$ on ${\mathbb {R}}^r$. Its cumulant is noted $c_\nu (x_1, \ldots , x_r)$. The joint cumulant of $(Y, \ldots ,Y)$ (r copies of Y) is $c^{(r)}(Y)$.

For any subset $I = \{i_1, \ldots , i_p\} \subset J_r:= \{1, \ldots , r\}$, we put

$$\begin{aligned} m(I) = m(i_1, \ldots , i_p):= {\mathbb {E}}(X_{i_1} \cdots X_{i_p}), \ s(I)= s(i_1, \ldots , i_p):= c(X_{i_1}, \ldots , X_{i_p}). \end{aligned}$$

The cumulants of a process $(X_j)_{j \in \mathcal J}$, where $\mathcal J$ is a set of indexes, is the family

$$\begin{aligned} \{c(X_{i_1}, \ldots , X_{i_r}), ({i_1}, \ldots , {i_r}) \in \mathcal J^r, r \ge 1\}. \end{aligned}$$

The following formulas link moments and cumulants and vice-versa:

$$\begin{aligned} c(X_1, \ldots , X_r)= & {} s(J_r) = \sum _{\mathcal P} (-1)^{p-1} (p - 1)! \ m(I_1) \cdots m(I_p), \end{aligned}$$

(56)

$$\begin{aligned} {\mathbb {E}}(X_{1} \cdots X_{r})= & {} m(J_r) = \sum _{\mathcal P} s(I_1) \cdots s(I_p), \end{aligned}$$

(57)

where in both formulas, $\mathcal P = \{I_1, I_2, \ldots , I_p\}$ runs through the set of partitions of $J_r = \{1, \ldots , r\}$ into $p \le r$ nonempty intervals, with p varying from 1 to r.

Now let be given a random field of real random variables $(X_{\underline{k}})_{{\underline{k}}\in {\mathbb {Z}}^d}$ and a summable weight R from ${\mathbb {Z}}^d$ to ${\mathbb {R}}$. For $Y := \sum _{{\underline{\ell }}\in {\mathbb {Z}}^d} \, R({\underline{\ell }}) \, X_{\underline{\ell }}$ we obtain using (56):

$$\begin{aligned} c^{(r)}(Y) = c(Y, \ldots , Y) = \sum _{({\underline{\ell }}_1, \ldots , {\underline{\ell }}_r) \, \in ({\mathbb {Z}}^d)^r} \, c(X_{{\underline{\ell }}_1}, \ldots , X_{{\underline{\ell }}_r}) \, R({\underline{\ell }}_1) \cdots R({\underline{\ell }}_r). \quad \end{aligned}$$

(58)

1.1 Limiting Distribution and Cumulants

For our purpose, we state in terms of cumulants a particular case of a theorem of M. Fréchet and J. Shohat, generalizing classical results of A. Markov. Using the formulas linking moments and cumulants, a special case of their “generalized statement of the second limit theorem” can be expressed as follows:

Theorem 6.1

[14] Let $(Z^{(n)}, n \ge 1)$ be a sequence of centered real r.v. such that

$$\begin{aligned} \lim _n c^{(2)}(Z^{(n)}) = \sigma ^2, \ \lim _n c^{(r)}(Z^{(n)}) = 0, \forall r \ge 3, \end{aligned}$$

(59)

then $(Z^{(n)})$ tends in distribution to $\mathcal N(0, \sigma ^2)$. (If $\sigma = 0$, then the limit is $\delta _0$).

Theorem 6.2

(cf. Theorem 7 in [25]) Let $(X_{\underline{k}})_{{\underline{k}}\in {\mathbb {Z}}^d}$ be a random process and $(R_n)_{n \ge 1}$ a summation sequence on ${\mathbb {Z}}^d$. Let $(Y^{(n)})$ be defined by $Y^{(n)} = \sum _{\underline{\ell }}\, R_n({\underline{\ell }}) \, X_{\underline{\ell }}, n \ge 1$. If

$$\begin{aligned} \sum _{({\underline{\ell }}_1, \ldots , {\underline{\ell }}_r) \, \in ({\mathbb {Z}}^d)^r} \, c(X_{{\underline{\ell }}_1}, \ldots , X_{{\underline{\ell }}_r}) \, R_n({\underline{\ell }}_1) \ldots R_n({\underline{\ell }}_r) = o(\Vert Y^{(n)}\Vert _2^r), \forall r \ge 3, \end{aligned}$$

(60)

then ${Y^{(n)} \over \Vert Y^{(n)}\Vert _2}$ tends in distribution to $\mathcal N(0, 1)$ when n tends to $\infty$.

Proof

Let $\beta _n := \Vert Y^{(n)}\Vert _2 = \Vert \sum _{\underline{\ell }}\, R_n({\underline{\ell }}) \, X_{\underline{\ell }}\Vert _2$ and $Z^{(n)} = \beta _n^{-1} Y^{(n)}$.

Using (58), we have $c^{(r)}(Z^{(n)}) = \beta _n^{-r} \sum _{({\underline{\ell }}_1, \ldots , {\underline{\ell }}_r) \, \in ({\mathbb {Z}}^d)^r} \, c(X_{{\underline{\ell }}_1}, \ldots , X_{{\underline{\ell }}_r}) \, R({\underline{\ell }}_1) \ldots R({\underline{\ell }}_r)$. The theorem follows then from the assumption (60) by Theorem 6.1 applied to $(Z^{(n)})$. $\square$

Definition 6.3

A measure preserving ${\mathbb {N}}^d$ (or ${\mathbb {Z}}^d$)-action $T: {{\underline{n}}} \rightarrow T^{\underline{n}}$ on a probability space $(X, \mathcal A, \mu )$ is r-mixing, $r > 1$, if for all sets $B_1, \ldots , B_r \in \mathcal A$

$$\begin{aligned} \lim _{\min _{{1 \le \ell < \ell ' \le r}} \Vert {\underline{n}}_\ell - {\underline{n}}_{\ell '}\Vert \rightarrow \infty } \mu \left( \bigcap _{\ell = 1}^r T^{-{\underline{n}}_\ell } \, B_\ell \right) =\prod _{\ell = 1}^r \mu (B_\ell ). \end{aligned}$$

Notations 6.4

For f in the space $L_0^\infty (X)$ of measurable essentially bounded functions on $(X, \mu )$ with $\int f \, d\mu = 0$, we apply the definition of moments and cumulants to $(T^{{\underline{n}}_1}f,\ldots , T^{{\underline{n}}_r}f)$ and put

$$\begin{aligned} m_f({\underline{n}}_1,\ldots , {\underline{n}}_r) = \int _X \, T^{{\underline{n}}_1} f \cdots T^{{\underline{n}}_r} f \, d\mu , \ \ s_f({\underline{n}}_1,\ldots , {\underline{n}}_r) := c(T^{{\underline{n}}_1} f,\ldots , T^{{\underline{n}}_r} f). \nonumber \\ \end{aligned}$$

(61)

To use the property of mixing of all orders, we need the following lemma.

Lemma 6.5

For every sequence $({\underline{n}}_1^k,\ldots , {\underline{n}}_r^k)_{k \ge 1}$ in $({\mathbb {Z}}^d)^r$, there are a subsequence with possibly a permutation of indices (still written $({\underline{n}}_1^k,\ldots , {\underline{n}}_r^k)$), an integer $\kappa (r) \in [1, r]$, a subdivision $1 = r_{1} < r_{2} < \ldots < r_{{\kappa (r)-1}} < r_{{\kappa (r)}} \le r$ of $\{1, \ldots , r\}$ and integral vectors ${\underline{a}}_j$, for $j \in ]r_{s}, r_{{s+1}}[, s = 1, ..., \kappa (r)$, and $j \in ]r_{\kappa (r)}, r]$, such that

$$\begin{aligned}&\lim _k \min _{1 \le s \not = s' \le \kappa (r)}\Vert {\underline{n}}_{r_{s}}^k - {\underline{n}}_{r_{s'}}^k\Vert = \infty , \end{aligned}$$

(62)

$$\begin{aligned}&{\underline{n}}_j^k = {\underline{n}}_{r_{s}}^k + {\underline{a}}_j, \text { for }r_{s} < j < r_{{s+1}}, \ s = 1, \ldots , \kappa (r)-1, \text { and for } r_{\kappa (r)} < j \le r. \nonumber \\ \end{aligned}$$

(63)

If $({\underline{n}}_1^k,\ldots , {\underline{n}}_r^k)$ satisfies $\lim _k \max _{i \not = j} \Vert {\underline{n}}_i^k - {\underline{n}}_j^k\Vert = \infty$, then $\kappa (r) > 1$.

Proof

Remark that if $\sup _k \max _{i \not = j} \Vert {\underline{n}}_i^k - {\underline{n}}_j^k\Vert < \infty$, then $\kappa (r) = 1$ so that (62) is void and (63) is void for the indexes such that $r_{{s+1}} = r_{s} +1$. The proof of the lemma is by induction. The result is clear for $r= 2$. Suppose that the subsequence for the sequence of $(r-1)$-tuples $({\underline{n}}_1^k,\ldots , {\underline{n}}_{r-1}^k)$ is built.

Let $1 \le r_{1} < r_{2} < \ldots < r_{{\kappa (r-1)}} \le r-1$ be the corresponding subdivision of $\{1, \ldots , r -1\}$, as stated above for the sequence $({\underline{n}}_1^k,\ldots , {\underline{n}}_{r-1}^k)$. If $({\underline{n}}_1^k,\ldots , {\underline{n}}_{r-1}^k)$ satisfies $\lim _k \max _{1 < i < j \le r-1} \Vert {\underline{n}}_i^k - {\underline{n}}_j^k\Vert = \infty$, then $\kappa (r-1) > 1$ by construction in the induction process.

Now consider $({\underline{n}}_1^k,\ldots , {\underline{n}}_r^k)$. If $\lim _k \Vert {\underline{n}}_r^k - {\underline{n}}_i^k\Vert = +\infty$, for $i=1, \ldots , r-1$, then we take $1 \le r_{1} < r_{2} < \ldots < r_{{\kappa (r-1)}} < r_{{\kappa (r)}} = r$ as new subdivision of $\{1, \ldots , r\}$. If $\liminf _k \Vert {\underline{n}}_r^k - {\underline{n}}_{i_s}^k\Vert < +\infty$ for some $s \le \kappa (r-1)$, then along a new subsequence (still denoted ${\underline{n}}_r^k$) we have ${\underline{n}}_r^k = {\underline{n}}_{i_s}^k + {\underline{a}}_r$, where ${\underline{a}}_r$ is a constant integral vector. After changing the labels, we insert $n_r$ in the subdivision of $\{1, \ldots , r-1\}$ and obtain the new subdivision of $\{1, \ldots , r\}$.

For the last condition on $\kappa$, suppose that $\lim _k \max _{1 < i < j \le r} \Vert {\underline{n}}_i^k - {\underline{n}}_j^k\Vert = \infty$. Then, if $\liminf _k \max _{1 < i < j \le r-1} \Vert {\underline{n}}_i^k - {\underline{n}}_j^k\Vert < +\infty$, necessarily, $\kappa (r) > 1$. If, on the contrary, the sequence $({\underline{n}}_1^k,\ldots , {\underline{n}}_{r-1}^k)$ satisfies $\lim _k \max _{1 < i < j \le r-1} \Vert {\underline{n}}_i^k - {\underline{n}}_j^k\Vert = \infty$, then $\kappa (r-1) > 1$, so that $\kappa (r) \ge \kappa (r-1) > 1$. $\square$

Lemma 6.6

If a ${\mathbb {Z}}^d$-dynamical system on $(X, \mathcal A, \mu )$ is mixing of order $r \ge 2$, then, for any $f \in L_0^\infty (X)$, $\underset{\max _{i \not = j} \Vert {\underline{n}}_i - {\underline{n}}_j\Vert \rightarrow \infty }{\lim }s_f({\underline{n}}_1,\ldots , {\underline{n}}_r) = 0$.

Proof

The notation $s_f$ was introduced in (61). Suppose that the above convergence does not hold. Then there is $\varepsilon > 0$ and a sequence of r-tuples $({\underline{n}}_1^k ={\underline{0}},\ldots , {\underline{n}}_r^k)$ such that $|s_f({\underline{n}}_1^k,\ldots , {\underline{n}}_r^k)| \ge \varepsilon$ and $\max _{i \not = j} \Vert {\underline{n}}_i^k - {\underline{n}}_j^k\Vert \rightarrow \infty$ (we use stationarity).

By taking a subsequence (but keeping the same notation), we can assume that, for two fixed indexes i, j, $\lim _k \Vert {\underline{n}}_i^k - {\underline{n}}_j^k\Vert = \infty$. By Lemma 6.5, there is a subdivision $1 = r_{1} < r_{2} < \ldots < r_{{\kappa (r)-1}} < r_{{\kappa (r)}} \le r$ and integer vectors ${\underline{a}}_j$ such that (62) and (63) hold.

Let $d\mu _k(x_1, \ldots , x_r)$ denote the probability measure on ${\mathbb {R}}^r$ defined by the distribution of the random vector $(T^{{\underline{n}}_1^k} f(.), \ldots , T^{{\underline{n}}_r^k}f(.))$. We can extract a converging subsequence from the sequence $(\mu _k)$, as well as for the moments of order $\le r$. Let $\nu (x_1, \ldots , x_r)$ (resp. $\nu (x_{i_1}, \ldots , x_{i_p})$) be the limit of $\mu _k(x_1, \ldots , x_r)$ (resp. of its marginal measures $\mu _k(x_{i_1}, \ldots , x_{i_p})$ for $\{i_1, \ldots , i_p\} \subset \{1, \ldots , r\}$). Let $\varphi _i, i=1, \ldots , r$, be continuous functions with compact support on ${\mathbb {R}}$. Mixing of order r and condition (62) imply

$$\begin{aligned}&\nu (\varphi _1 \otimes \varphi _2 \otimes \ldots \otimes \varphi _r) =\lim _k \int _{{\mathbb {R}}^d} \varphi _1 \otimes \varphi _2 \otimes \ldots \otimes \varphi _r \, {\text {d}}\mu _k\\&\quad = \lim _k \int \prod _{i=1}^r \varphi _i(f(T^{{\underline{n}}_i^k}x))\ {\text {d}}\mu (x) \\&=\lim _k \int \ \left[ \prod _{s=1}^{\kappa (r)-1} \ \prod _{r_{s} \le j < r_{{s+1}}} \varphi _j(f(T^{n_{r_s}^k + {\underline{a}}_j} x))\right] \, \prod _{r_{\kappa (r)} \le j \le r} \varphi _j(f(T^{{\underline{n}}_{r_{\kappa (r)}}^k + {\underline{a}}_j} x)) \ d\mu (x) \\&= \left[ \prod _{s=1}^{\kappa (r)-1} \, \left( \int \prod _{r_{s} \le j < r_{{s+1}}} \varphi _j(f(T^{{\underline{a}}_j}x)) \ {\text {d}}\mu (x) \right) \right] \, \left[ \int \prod _{\kappa (r) \le j \le r} \varphi _j(f(T^{{\underline{a}}_j} x)) \ {\text {d}}\mu (x)\right] . \end{aligned}$$

Therefore, $\nu$ is the product of marginal measures corresponding to disjoint subsets: there are $I_1 = \{i_1, \ldots , i_p\}, I_2 = \{i_1', \ldots , i_{p'}'\}$ two nonempty subsets of $J_r = \{1, \ldots , r\}$, such that $(I_1, I_2)$ is a partition of $J_r$ and $d\nu (x_1, \ldots ,x_r) = d\nu (x_{i_1}, \ldots , x_{i_p}) \times d\nu (x_{i_1'}, \ldots , x_{i_{p'}'})$.

With $\Phi (t_1, \ldots , t_r) = \ln \int e^{\sum t_j x_j} \ d\nu (x_1, \ldots ,x_r)$ and the analogous formulas for $\nu (x_{i_1}, \ldots , x_{i_p})$ and $\nu (x_{i_1'}, \ldots , x_{i_p'})$, we get $\Phi (t_1, \ldots , t_r) = \Phi (t_{i_1}, \ldots , t_{i_p}) + \Phi (t_{i_1'}, \ldots , t_{i_p'})$; hence $c_\nu (x_1, \ldots , x_r) = {\partial ^r \over \partial t_1 \ldots \partial t_r}\Phi (t_1, \ldots , t_r)|_{t_1 = \ldots = \, t_r = 0} = 0$, which contradicts $\liminf _k |s_f({\underline{n}}_1^k,\ldots , {\underline{n}}_r^k)| > 0$. $\square$

Proof of Lemma 3.10

Let $(\chi _{\underline{k}}, {\underline{k}}\in \Lambda )$ be a finite set of characters on G, $\chi _{0}$ the trivial character. If $f = \sum _{{\underline{k}}\in \Lambda } c_{\underline{k}}(f) \, \chi _{\underline{k}}$, the moments of the process $(f(A^{\underline{n}}.))_{{\underline{n}}\in {\mathbb {Z}}^d}$ are

$$\begin{aligned}&m_f({\underline{n}}_1, \ldots , {\underline{n}}_r) = \int f(A^{{\underline{n}}_1} x) \ldots f(A^{{\underline{n}}_r} x) \ dx\\&\quad = \sum _{{\underline{k}}_1, \ldots , {\underline{k}}_r \in \Lambda } c_{k_1} \ldots c_{k_r} 1_{A^{{\underline{n}}_1} \chi _{{\underline{k}}_1} \ldots A^{{\underline{n}}_r} \chi _{{\underline{k}}_r} = \, \chi _{0}}. \end{aligned}$$

We apply Theorem 6.2. Let us check (60), i.e., in view of (58) and (61),

$$\begin{aligned} \left| \sum _{({\underline{\ell }}_1, \ldots , {\underline{\ell }}_r) \, \in \, ({\mathbb {Z}}^d)^r} s_f({\underline{\ell }}_1, \ldots , {\underline{\ell }}_r) \, R_n({\underline{\ell }}_1)\ldots R_n({\underline{\ell }}_r)\right| = o(\Vert Y^{(n)}\Vert _2^r), \forall r \ge 3. \end{aligned}$$

(64)

For r fixed, the function $({\underline{n}}_1, \ldots , {\underline{n}}_r) \rightarrow m_f({\underline{n}}_1, \ldots , {\underline{n}}_r)$ takes a finite number of values, since $m_f$ is a sum with coefficients 0 or 1 of the products $c_{k_1} \ldots c_{k_r}$ with $k_j$ in a finite set. The cumulants of a given order take also a finite number of values according to (56).

Therefore, since mixing of all orders implies $\underset{\max _{i,j} \Vert {\underline{\ell }}_i - {\underline{\ell }}_j\Vert \rightarrow \infty }{\lim }\, s_f({\underline{\ell }}_1,\ldots , {\underline{\ell }}_r) = 0$ by Lemma 6.6, there is $M_r$ such that $s_f({\underline{\ell }}_1, \ldots , {\underline{\ell }}_r) = 0$ if $\max _{i,j} \Vert {\underline{\ell }}_i - {\underline{\ell }}_j\Vert > M_r$. The absolute value of sum in (64) reads

$$\begin{aligned}&\left| \sum _{\max _{i,j} \Vert {\underline{\ell }}_i - {\underline{\ell }}_j\Vert \le M_r} s_f({\underline{\ell }}_1, {\underline{\ell }}_2, \ldots , {\underline{\ell }}_r) \, R_n({\underline{\ell }}_1)\ldots R_n({\underline{\ell }}_r)\right| \\&\ \ \le \sum _{{\underline{\ell }}} \sum _{\Vert {\underline{j}}_2\Vert , \ldots , \Vert {\underline{j}}_r\Vert \le M_r, \, {\underline{j}}_1 ={\underline{0}}} |s_f({\underline{\ell }}, {\underline{\ell }}+ {\underline{j}}_2, \ldots , {\underline{\ell }}+{\underline{j}}_r)| \, \prod _{k=1}^r |R_n({\underline{\ell }}+{\underline{j}}_k)|\\&\ \ = \sum _{\underline{\ell }}\sum _{\Vert {\underline{j}}_2\Vert , \ldots , \Vert {\underline{j}}_r\Vert \le M_r, \, {\underline{j}}_1 ={\underline{0}}} |s_f({\underline{j}}_1,{\underline{j}}_2, \ldots , {\underline{j}}_r)| \ \prod _{k=1}^r |R_n({\underline{\ell }}+j_k)|. \end{aligned}$$

The right-hand side is $<C \sum _{\underline{\ell }}\sum _{\Vert {\underline{j}}_2\Vert , \ldots , \Vert {\underline{j}}_r\Vert \le M_r, \, j_1 ={\underline{0}}} \ \prod _{k=1}^r |R_n({\underline{\ell }}+j_k)|$. Therefore, it is less than a finite sum of sums of the form $\sum _{{\underline{\ell }}\in {\mathbb {Z}}^d} \prod _{k= 1}^r |R_n({\underline{\ell }}+ {{\underline{j}}}_k)|$ of order $o(\sigma _n^{r}(f)), \forall \{{\underline{j}}_1, \ldots , {\underline{j}}_r\} \in {\mathbb {Z}}^d, \forall r \ge 3$, by hypothesis (14) of the lemma. Hence (60) is satisfied. $\square$

Appendix 2: Self-Intersections of a Centered r.w.

In this appendix, we prove Theorem 4.13. As already noticed, we can assume aperiodicity in the proofs.

1.1 d=1: a.s. Convergence of $V_{n, p} / V_{n, 0}$

Lemma 7.1

If W is a 1-dimensional r.w. with finite variance and centered, then, for a.e. $\omega$, there is $C(\omega ) > 0$ such that

$$\begin{aligned} V_n(\omega ) \ge C(\omega ) \, n^{\frac{3}{2}} \, (\mathrm{Log}\mathrm{Log}\, n)^{-\frac{1}{2}}. \end{aligned}$$

(65)

Proof

By the law of iterated logarithm, there is a constant $c > 0$ such that, for a.e. $\omega$, the inequality $|Z_n(\omega )| > c \, (n \ \mathrm{Log}\mathrm{Log}\, n)^\frac{1}{2}$ is satisfied only for finitely many values of n. This implies that, for a.e. $\omega$, there is $N(\omega )$ such that $|Z_n(\omega )| \le (c \, n \ \mathrm{Log}\mathrm{Log}\, n)^\frac{1}{2}$, for $n \ge N(\omega )$. It follows, for $N(\omega ) \le k < n$, $|Z_k(\omega ) \, | \le (c \, k \ \mathrm{Log}\mathrm{Log}k)^\frac{1}{2} \le (c \, n \ \mathrm{Log}\mathrm{Log}\, n)^\frac{1}{2}$.

Therefore, if $\mathcal R_n(\omega )$ is the set of points visited by the random walk up to time n, we have $\mathrm{Card}(\mathcal R_n(\omega )) \le 2 (c_1(\omega ) \, n \ \mathrm{Log}\mathrm{Log}\, n)^\frac{1}{2}$, with an a.e. finite constant $c_1(\omega )$.

We have $n = \sum _{{\underline{\ell }}\in {\mathbb {Z}}} \sum _{k=0}^{n-1} 1_{Z_k(\omega ) \, = {\underline{\ell }}} \le (\sum _{{\underline{\ell }}\in \mathcal R_n(\omega )} (\sum _{k=0}^{n-1} 1_{Z_k(\omega ) \, = {\underline{\ell }}})^2)^\frac{1}{2} \, \mathrm{Card}(\mathcal R_n(\omega ))^\frac{1}{2}$ by Cauchy-Schwarz inequality; hence: $V_n(\omega ) \ge {n^2 / \mathrm{Card}(\mathcal R_n(\omega ))}$, which implies (65). $\square$

Lemma 7.2

For an aperiodic r.w. in dimension 1 with finite variance and centered, we have:

$$\begin{aligned} \sup _n \, \left| \sum _{j=1}^n [2\,{\mathbb {P}}(Z_j = 0) - {\mathbb {P}}(Z_j = {\underline{p}}) - {\mathbb {P}}(Z_j = -{\underline{p}})]\right| < + \infty , \ \forall {\underline{p}}\in L(W). \nonumber \\ \end{aligned}$$

(66)

Proof

Since $1 - \Psi (t)$ vanishes on the torus only at $t = {\underline{0}}$ with an order 2, we have:

$$\begin{aligned}&\left| \sum _{j=1}^{N-1} [2{\mathbb {P}}(Z_j = 0) - {\mathbb {P}}(Z_j = {\underline{p}}) - {\mathbb {P}}(Z_j = -{\underline{p}})]\right| \\&\quad = 4 |\int _{{\mathbb {T}}} \sin ^2 \pi \langle {\underline{p}}, t\rangle \, \mathfrak {R}e \left( {1 - \Psi ^N(t) \over 1 - \Psi (t)}\right) \, {\text {d}}t|\\&\quad \le 4 \int _{{\mathbb {T}}} \sin ^2 \pi \langle {\underline{p}}, t\rangle \, |{1 - \Psi ^N(t) \over 1 - \Psi (t)}| \, {\text {d}}t \le 8 \int _{{\mathbb {T}}} { \sin ^2 \pi \langle {\underline{p}}, t\rangle \, \over |1 - \Psi (t)|} \, {\text {d}}t < + \infty . \end{aligned}$$

$\square$

Proof of Theorem 4.13

(for $d=1$: $\displaystyle \lim _n {V_{n, p}(\omega ) \over V_{n}(\omega )} = 1$, a.e. for every $p \in L(W)$.)

Recall that $V_{n}(\omega ) - V_{n, p}(\omega ) \ge 0$ (cf. Ex. 2.4). By (26) we can bound $n^{-1} \, {\mathbb {E}}[V_{n} - V_{n, p}]$ by

$$\begin{aligned} 1 + n^{-1} \sum _{k = 1}^{n-1} \, \left| \sum _{j = 0}^{n-k-1} ({\mathbb {P}}(Z_j = {\underline{p}}) - {\mathbb {P}}(Z_j = 0))+\sum _{j = 0}^{n-k-1} ({\mathbb {P}}(Z_j = -{\underline{p}}) - {\mathbb {P}}(Z_j = 0))\right| \end{aligned}$$

which is bounded according to (66).

Let $\delta > 0$. The bound $\displaystyle {\mathbb {E}}\, \left( {V_{n} - V_{n, p} \over n^{\frac{3}{2}}}\right) = O\left( n^{-\frac{1}{2}}\right)$ implies by the Borel-Cantelli lemma that $\displaystyle \left( {V_{n} - V_{n, p} \over n^{\frac{3}{2}} \, (\mathrm{Log}\mathrm{Log}\, n)^{-\frac{1}{2}}} \right)$ tends to 0 a.e. along the sequence $k_n = [n^{2+\delta }], \, n \ge 1$.

Since $V_{n}(\omega ) \ge C(\omega ) \, n^\frac{3}{2} \, (\mathrm{Log}\mathrm{Log}\, n)^{-\frac{1}{2}}$ with $C(\omega ) > 0$ by (65), we obtain

$$\begin{aligned} 0 \le 1 - {V_{n, {\underline{p}}}(\omega ) \over V_{n}(\omega )} \le {V_{n} (\omega ) - V_{n, p}(\omega ) \over V_{n}(\omega )} < {V_{n}(\omega ) - V_{n, p}(\omega ) \over C(\omega ) \, n^\frac{3}{2} \, (\mathrm{Log}\mathrm{Log}\, n)^{-\frac{1}{2}}}. \end{aligned}$$

Therefore $({V_{n, {\underline{p}}}(\omega ) \over V_{n}(\omega )})_{n \ge 1}$ converges to 1 a.e. along the sequence $(k_n)$.

To complete the proof, it suffices to prove that a.s. $\lim _n \max _{k_n \le j<k_{n+1}} |V_{j,p} / V_j - V_{k_n ,p} / V_{k_n}| = 0$. By monotonicity of $V_{j,p}$ and $V_j$ with respect to j, we have

$$\begin{aligned} \frac{ V_{k_{n},p }}{V_{k_{n}}} \frac{V_{k_{n}}}{ V_{k_{n+1}}} \le \frac{V_{j,p} }{ V_{j}}\le \frac{ V_{k_{n+1},p }}{V_{k_{n+1}}} \frac{V_{k_{n+1}}}{ V_{k_n}}, \ k_n \le j <k_{n+1}. \end{aligned}$$

Therefore, since the first factors in the left and right terms of the inequality tend to 1, it is enough to prove that $\frac{V_{k_{n+1}}-V_{k_n}}{V_{k_n}}\rightarrow 0$ a.s.

By (42), for $d= 1$, for each $\varepsilon > 0$, there is an a.e. finite constant $c_\varepsilon (\omega )$ such that $\sup _{{\underline{\ell }}\in {\mathbb {Z}}} R_n(\omega ,{\underline{\ell }}) = \sup _{{\underline{\ell }}\in {\mathbb {Z}}} \sum _{k=0}^{n-1} 1_{Z_k(\omega ) \, = {\underline{\ell }}} \le c_\varepsilon (\omega ) \, n^{\frac{1}{2} + \varepsilon }$. This implies

$$\begin{aligned}&V_{k_{n+1}}-V_{k_n}=\sum _{\ell , j=0}^{k_{n+1}} 1_{\{Z_\ell =Z_j\}}- \sum _{\ell , j=0}^{k_{n}} 1_{\{Z_\ell =Z_j\}}\\&\quad \le \sum _{\ell =k_n+1}^{k_{n+1}}\sum _{j=1}^{k_{n+1}} 1_{\{Z_\ell =Z_j\}}+ \sum _{j=k_n+1}^{k_{n+1}} \sum _{\ell =1}^{k_{n+1}} 1_{\{Z_\ell =Z_j\}}\\&\le 2 (k_{n+1}- k_n) \, \sup _{p \in {\mathbb {Z}}} \sum _{j=1}^{k_{n+1}} 1_{\{Z_j = p\}} \le 2 c_\varepsilon \, (k_{n+1}- k_n) \, k_{n+1}^{\frac{1}{2} + \varepsilon } \le K \, n^{1+\delta + (2 + \delta )(\frac{1}{2} + \varepsilon )}, \end{aligned}$$

where K is an a.e. finite random variable.

Therefore, $V_{k_{n+1}}-V_{k_n} \le K \, n^{2+ \frac{3}{2}\delta + 2\varepsilon + \varepsilon \delta }$ and $\frac{V_{k_{n+1}}-V_{k_n}}{V_{k_n}}$ is a.s. bounded by

$$\begin{aligned}&{V_{k_{n+1}}-V_{k_n} \over k_n^{3/2} \, (\mathrm{Log}\mathrm{Log}\, k_n)^{-\frac{1}{2}}} \le 2 K \, {n^{2+ \frac{3}{2}\delta + 2\varepsilon + \varepsilon \delta } \over (n^{2+\delta })^{3/2} \, (\mathrm{Log}\mathrm{Log}\, (n^{2+\delta }))^{-\frac{1}{2}}} \\&\le 2 K \, {n^{2+ \frac{3}{2}\delta + 2\varepsilon + \varepsilon \delta } \, n^{-(3+ \frac{3}{2} \delta )} \, (\mathrm{Log}\mathrm{Log}\, (n^{2+\delta }))^{\frac{1}{2}}} \!=\! 2 K \, {n^{-1 + 2\varepsilon + \varepsilon \delta } \, (\mathrm{Log}\mathrm{Log}\, (n^{2+\delta }))^{\frac{1}{2}}}, \end{aligned}$$

which tends to 0, if $2\varepsilon + \varepsilon \delta < 1$.

1.2 d=2: Variance and SLLN for $V_{n,p}$

For $d = 2$, in the centered case with finite variance, the a.s. convergence $V_{n, {\underline{p}}} / V_{n, {\underline{0}}} \rightarrow 1$ for $p \in L(W)$ follows from the strong law of large numbers (SLLN): $\lim _n V_{n, {\underline{p}}}/ {\mathbb {E}}V_{n, {\underline{p}}} = 1$, a.s. We adapt the method of [27] to the case ${\underline{p}}\not = {\underline{0}}$ in the estimation of $\mathrm{Var}(V_{n, {\underline{p}}})$. See also [11] for the computation of the variance of $V_{n, {\underline{0}}}$. We need two auxiliary results.

Lemma 7.3

There is C such that, if ${\underline{p}}, {\underline{q}}$ are in L and n, k are such that $n {\underline{\ell }}_1 = {\underline{p}}\mathrm{\ mod \,} D$ and $(n+k) {\underline{\ell }}_1 = {\underline{q}}\mathrm{\ mod \,} D$, then

$$\begin{aligned} |{\mathbb {P}}(Z_{n+k} = {\underline{q}}) - {\mathbb {P}}(Z_n = {\underline{p}})| \le C \left( {1 \over (n+k)^{\frac{3}{2}}} + { k\over n (n+k)}\right) , \, \forall n, k \ge 1. \end{aligned}$$

(67)

Proof

We have ${\mathbb {P}}(Z_{n+r} = {\underline{q}}) - {\mathbb {P}}(Z_n = {\underline{p}}) = \int _{{\mathbb {T}}^2} G_{n,r}({\underline{t}}) \, d{\underline{t}}$, with

$$\begin{aligned} G_{n,r}({\underline{t}}) := \mathfrak {R}e \, [e^{-2\pi i \langle {\underline{q}}, {\underline{t}}\rangle } \, \Psi ({\underline{t}})^{n+r} -e^{-2\pi i \langle {\underline{p}}, {\underline{t}}\rangle } \, \Psi ({\underline{t}})^{n}]. \end{aligned}$$

The functions $e^{- 2\pi i \langle {\underline{q}}, {\underline{t}}\rangle } \, \Psi ({\underline{t}})^{n+r}$ and $e^{- 2\pi i \langle {\underline{p}}, {\underline{t}}\rangle } \, \Psi ({\underline{t}})^{n}$ are invariant by translation by the elements of $\Gamma _1$ and have a modulus $< 1$, except for ${\underline{t}}\in \Gamma _1$ (cf. Lemma 4.7). To bound the integral of $G_{n,r}$, it suffices to bound its integral $I_n^0 := \int _{U_0} G_{n,r}({\underline{t}}) _, d{\underline{t}}$ restricted to a fundamental domain $U_0$ of $\Gamma _1$ acting on ${\mathbb {T}}^2$.

Denote by $B(\eta )$ the ball with a small radius $\eta$ and center ${\underline{0}}$ in ${\mathbb {T}}^2.$ If $\eta >0$ is small enough, on $U_0 \setminus B(\eta )$, we have $|\Psi ({\underline{t}})| \le \lambda (\eta )$ with $\lambda (\eta ) < 1$, which implies:

$$\begin{aligned} |I_n^0| \le 2 C \lambda ^n(\eta ) + \int _{B(\eta )} |G_{n,r}({\underline{t}})| \, {\text {d}}{\underline{t}}. \end{aligned}$$

and we have

$$\begin{aligned}&|G_{n,r}({\underline{t}})| := |\mathfrak {R}e \, [e^{-2\pi i \langle {\underline{q}}, {\underline{t}}\rangle } \, \Psi ({\underline{t}})^{n+r} -e^{-2\pi i \langle {\underline{p}}, {\underline{t}}\rangle } \, \Psi ({\underline{t}})^{n}]|\\&\quad \le |\Psi ({\underline{t}})|^{n} \, |\Psi ({\underline{t}})^{r} - e^{2\pi i \langle {\underline{q}}- {\underline{p}}, {\underline{t}}\rangle } |. \end{aligned}$$

Since the distribution $\nu$ of the centered r.w. W is assumed to have a moment of order 2, by Lemma 4.8, for $\eta$ sufficiently small, there are constants $a, b>0$ such that

$$\begin{aligned} |\Psi (t)| < 1-a \Vert t\Vert ^2, \ |1-\Psi (t)| <b \Vert t\Vert ^2, \forall t \in B(\eta ). \end{aligned}$$

We distinguish two cases: if ${\underline{p}}= {\underline{q}}$, then $|G_{n,r}({\underline{t}})| \le C(r) (1-a \Vert t\Vert ^2)^n \Vert {\underline{t}}\Vert ^2$; if ${\underline{p}}\not = {\underline{q}}$, $|G_{n,r}({\underline{t}})| \le C(r) |\Psi ({\underline{t}})|^{n} \Vert {\underline{t}}\Vert \le C(r) (1-a \Vert t\Vert ^2)^n \, \Vert {\underline{t}}\Vert$.

Now we bound the integral $\int _{B(\eta )}\ (1 - a {\Vert {\underline{t}}\Vert ^2})^{n} \, \Vert {\underline{t}}\Vert ^2\, {{\text {d}}t_1 \, {\text {d}}t_2}$. By the change of variable $(t_1, t_2) \rightarrow ({s_1, \over \sqrt{n}}, {s_2, \over \sqrt{n}})$, then $(s_1, s_2) \rightarrow (\rho \cos \theta , \rho \sin \theta )$, it becomes successively:

$$\begin{aligned} {1 \over n^2} \, \int _{B(\eta )}\ \left( 1 - a{\Vert {\underline{s}}\Vert ^2 \over n}\right) ^n \, \Vert {\underline{s}}\Vert ^2 \, ds_1 \, ds_2 \le {C \over n^2} \, \int _{\mathbb {R}}e^{- a \rho ^2} \, \rho ^2 \, d\rho = O\left( {1 \over n^2}\right) . \end{aligned}$$

So for ${\underline{p}}= {\underline{q}}$, we get the bound $O(n^2)$. Likewise, if ${\underline{p}}\not = {\underline{q}}$, we get the bound $O(n^{3/2})$.

Recall the L / D is a cyclic group (Lemma 4.4). If n, k satisfy the condition of the lemma, we write: $k= r+ uv$, where r and v ($= |L/D|$) are the smallest positive integers such that respectively: $r {\underline{\ell }}_1 = {\underline{q}}- {\underline{p}}\mathrm{\ mod \,} D$, $v {\underline{\ell }}_1 = {\underline{0}} \mathrm{\ mod \,} D$. Then, using the previous bounds and writing the difference ${\mathbb {P}}(Z_{n+k} = {\underline{q}}) - {\mathbb {P}}(Z_n = {\underline{p}})$ as

$$\begin{aligned}&{\mathbb {P}}(Z_{n+k} \!= {\underline{q}}) \!- {\mathbb {P}}(Z_{n+r+(u-1)v} \!= {\underline{q}}) \!+\! {\mathbb {P}}(Z_{n+r+(u-1)v} \!= {\underline{q}}) \!- {\mathbb {P}}(Z_{n+r+(u-2)v} = {\underline{q}}) \\&+ \ldots + {\mathbb {P}}(Z_{n+r+v} = {\underline{q}}) - {\mathbb {P}}(Z_{n+r} = {\underline{q}}) + {\mathbb {P}}(Z_{n+r} = {\underline{q}}) - {\mathbb {P}}(Z_n = {\underline{p}}), \end{aligned}$$

the telescoping argument gives (67). $\square$

Lemma 7.4

For $d \ge 1$, let $M_n^{(d)} := \{(m_1, \ldots , m_d) \in {\mathbb {N}}^d: \, \sum _i m_i = n\}$. Let $\tilde{M}_n^{(5)} := \{(m_1, \ldots , m_5) \, \in M_n^{(5)}, m_3, \, m_4 > 0 \}$. If $|\lambda |, |\alpha |, |\beta |, |\gamma | < 1$, then

$$\begin{aligned} \sum _{n= 0}^\infty \lambda ^{n} \sum _{(m_1, \ldots , m_5) \, \in \tilde{M}_n^{(5)}} \alpha ^{m_2} \, \beta ^{m_3} \, \gamma ^{m_4} = {1 \over (1 - \lambda )^2} {\lambda ^2 \beta \gamma \over (1 - \lambda \alpha ) (1 - \lambda \beta ) (1 - \lambda \gamma )}. \qquad \end{aligned}$$

(68)

Proof

We have for $\lambda , \alpha _1, \ldots , \alpha _d$ such that $|\lambda \alpha _1|, \ldots , |\lambda \alpha _d| \, \in [0, 1[$:

$$\begin{aligned} \sum _{n= 0}^\infty \lambda ^{n} \sum _{(m_1, \ldots , m_d) \, \in M_n^{(d)}} \alpha _1^{m_1} \, \alpha _2^{m_2} \, \ldots \, \alpha _d^{m_d} = \prod _{i=1}^d {1 \over (1 - \lambda \alpha _i)}. \end{aligned}$$

(69)

Indeed, the left hand of (69) is the sum over ${\mathbb {Z}}$ of the discrete convolution product of the functions $G_i$ defined on ${\mathbb {Z}}$ by $G_i(k) = 1_{[0, \infty [}(k) (\lambda \alpha _i)^k$, hence is equal to

$$\begin{aligned}&\sum _{k \, \in {\mathbb {Z}}} \, (G_1 * \ldots * G_d)(k) = \prod _{i=1}^d \, \bigr (\sum _{k \in {\mathbb {Z}}} \, G_i(k)\bigr ) = \prod _{i=1}^d {1 \over (1 - \lambda \alpha _i)}. \end{aligned}$$

For $d=5$ we find

$$\begin{aligned} \sum _{n= 0}^\infty \lambda ^{n} \sum _{(m_1, \ldots , m_5) \in M_n^{(5)}} \alpha ^{m_2} \, \beta ^{m_3} \gamma ^{m_4} = {1 \over (1 - \lambda )^2} {1 \over (1 - \lambda \alpha ) (1 - \lambda \beta ) (1 - \lambda \gamma )}. \end{aligned}$$

The left hand of (68) reads

$$\begin{aligned}&\sum _{n= 0}^\infty \lambda ^{n} \sum _{(m_1, \ldots , m_5) \, \in M_n^{(5)}} \alpha ^{m_2} \, \beta ^{m_3} \, \gamma ^{m_4} - \sum _{n= 0}^\infty \lambda ^{n} \sum _{(m_1, m_2, m_4, m_5) \, \in M_n^{(4)}} \alpha ^{m_2} \, \gamma ^{m_4}\\&- \sum _{n= 0}^\infty \lambda ^{n} \sum _{(m_1, m_2, m_3, m_5) \, \in M_n^{(4)}} \alpha ^{m_2} \, \beta ^{m_3} + \sum _{n= 0}^\infty \lambda ^{n} \sum _{(m_1, m_2, m_5) \, \in M_n^{(3)}} \alpha ^{m_2}\\&= {1 \over (1 - \lambda )^2} [{1 \over (1 - \lambda \alpha ) (1 - \lambda \beta ) (1 - \lambda \gamma )} - {1 \over (1 - \lambda \alpha ) (1 - \lambda \gamma )}\\&-{1 \over (1 - \lambda \alpha ) (1 - \lambda \beta )} + {1 \over (1 - \lambda \alpha )}] = {1 \over (1 - \lambda )^2} {\lambda ^2 \beta \gamma \over (1 - \lambda \alpha ) (1 - \lambda \beta ) (1 - \lambda \gamma )}. \end{aligned}$$

$\square$

Proof of Theorem 4.13

(Case $d=2$) A) Below we consider:

$$\begin{aligned} \sum _{0 \le i_1 < j_1 < n; \ 0 \le i_2 < j_2 < n} \, {\mathbb {P}}(Z_{j_1} - Z_{i_1} = {\underline{r}}, \, Z_{j_2} - Z_{i_2} = {\underline{s}}). \end{aligned}$$

(70)

The (disjoint) sets of possible configurations for $(i_1, j_1, i_2, j_2)$ are the following: (1a) $i_1 \le i_2 < j_1 < j_2$, (1b) $i_2 \le i_1 < j_2 < j_1$, (2a) $i_1 < i_2 < j_2 \le j_1$, (2b) $i_2 < i_1 < j_1 \le j_2$, (3a) $i_1 < j_1 \le i_2 < j_2$, (3b) $i_2 < j_2 \le i_1 < j_1$, (4) $i_2 = i_1 < j_1 = j_2$.

For the case 4), by the LLT the sum (70) restricted to the configurations (4) is less than $\sum _{0 \le i_1 < j_1 < n} 1_{{\underline{r}}= {\underline{s}}} \, {\mathbb {P}}(Z_{j_1} - Z_{i_1} = {\underline{r}}) \le \sum {C \over j_1 - i_1} \le C n \, \mathrm{Log}\, n$.

(1a) and (1b), (2a) and (2b), (3a) and (3b) are, respectively, the same up to the exchange of indices 1 and 2. To bound (70), it suffices to consider the subsums corresponding to 1a), 2a), 3a).

1a) ($i_1 \le i_2 < j_1 < j_2$) Setting $m_1 = i_1, \ m_2 = i_2 - i_1, \ m_3 = j_1 - i_2, \ m_4 = j_2 - j_1, \ m_5 = n - j_2$, we have: $Z_{j_1} - Z_{i_1} = \tau ^{m_1} Z_{m_2 + m_3}$, $Z_{j_2} - Z_{i_2} = \tau ^{m_1+m_2} Z_{m_3 + m_4}$ with $m_1, m_2 \ge 0, m_3, m_4 > 0$; hence:

$$\begin{aligned}&{\mathbb {P}}(Z_{j_1} - Z_{i_1} = {\underline{r}}, \, Z_{j_2} - Z_{i_2} = {\underline{s}}) = {\mathbb {P}}(Z_{m_2+ m_3} = {\underline{r}}, \, \tau ^{m_2} Z_{m_3 + m_4} = {\underline{s}}) \nonumber \\&= \sum _{\underline{\ell }}[{\mathbb {P}}(Z_{m_2} = {\underline{\ell }}) \, {\mathbb {P}}(Z_{m_3} = {\underline{r}}- {\underline{\ell }}) \, {\mathbb {P}}(Z_{m_4} = {\underline{s}}- {\underline{r}}+ {\underline{\ell }})] \nonumber \\&= \int e^{-2\pi i(\langle {\underline{r}}, {\underline{u}}\rangle + \langle {\underline{s}}- {\underline{r}}, {\underline{v}}\rangle )} \sum _{\underline{\ell }}e^{-2\pi i \langle {\underline{\ell }}, \, {\underline{t}}- {\underline{u}}+ {\underline{v}}\rangle } \ \Psi ({\underline{t}})^{m_2} \, \Psi ({\underline{u}})^{m_3} \, \Psi ({\underline{v}})^{m_4} \, d{\underline{t}}\, d{\underline{u}}\, {\text {d}}{\underline{v}}. \nonumber \\ \end{aligned}$$

(71)

The last equation can be shown by approximating the probability vector $(p_j)_{j \in J}$ by probability vectors with finite support and using the continuity of $\Psi$.

2a) ($i_1 < i_2 < j_2 \le j_1$) Setting $m_1 = i_1, \ m_2 = i_2 - i_1, \ m_3 = j_2 - i_2, \ m_4 = j_1 - j_2, \ m_5 = n - j_1$, we have: $Z_{j_1} - Z_{i_1} = \tau ^{m_1} Z_{m_2 + m_3 + m_4}$, $Z_{j_2} - Z_{i_2} = \tau ^{m_1+m_2} Z_{m_3}$, with $m_1, m_4 \ge 0, m_2, m_3 > 0$.

Since ${\mathbb {P}}(Z_{m_2} + \tau ^{m_2+m_3} Z_{m_4} = {\underline{q}}) = \sum _{\underline{\ell }}{\mathbb {P}}(Z_{m_2} = {\underline{\ell }}, \tau ^{m_2+m_3} Z_{m_4} = {\underline{q}}- {\underline{\ell }}) = \sum _{\underline{\ell }}{\mathbb {P}}(Z_{m_2} = {\underline{\ell }}) \, {\mathbb {P}}(\tau ^{m_2+m_3} Z_{m_4} = {\underline{q}}- {\underline{\ell }}) = \sum _{\underline{\ell }}{\mathbb {P}}(Z_{m_2} = {\underline{\ell }}) \, {\mathbb {P}}(\tau ^{m_2} Z_{m_4} = {\underline{q}}- {\underline{\ell }}) = {\mathbb {P}}(Z_{m_2+m_4} = {\underline{q}})$, we have:

$$\begin{aligned}&{\mathbb {P}}(Z_{j_1} - Z_{i_1} = {\underline{r}}, \, Z_{j_2} - Z_{i_2} = {\underline{s}})= {\mathbb {P}}(Z_{m_2+ m_3+m_4} = {\underline{r}}, \, \tau ^{m_2}Z_{m_3} = {\underline{s}}) \nonumber \\&= {\mathbb {P}}(Z_{m_2} + \tau ^{m_2+m_3} Z_{m_4} = {\underline{r}}- {\underline{s}}, \, \tau ^{m_2} Z_{m_3} = {\underline{s}}) \nonumber \\&= {\mathbb {P}}(Z_{m_2} + \tau ^{m_2+m_3} Z_{m_4} = {\underline{r}}- {\underline{s}}) \, {\mathbb {P}}(Z_{m_3} = {\underline{s}}) = {\mathbb {P}}(Z_{m_2 + m_4} = {\underline{r}}- {\underline{s}}) \, {\mathbb {P}}(Z_{m_3} = {\underline{s}}). \end{aligned}$$

Hence, in case 2a), we get:

$$\begin{aligned} {\mathbb {P}}(Z_{j_1} - Z_{i_1} = {\underline{r}}, \, Z_{j_2} - Z_{i_2} = {\underline{s}}) = {\mathbb {P}}(Z_{m_2 + m_4} = {\underline{r}}- {\underline{s}}) \, {\mathbb {P}}(Z_{m_3} = {\underline{s}}). \quad \end{aligned}$$

(72)

3a) ($i_1 < j_1 \le i_2 < j_2$) The events $Z_{j_1} - Z_{i_1} = {\underline{r}}$ and $Z_{j_2} - Z_{i_2} = {\underline{s}}$ are independent.

Following the method of [27], now we estimate $\mathrm{Var}(V_{n, {\underline{p}}}) = {\mathbb {E}}(V_{n, {\underline{p}}}^2) - ({\mathbb {E}}V_{n, {\underline{p}}})^2$. Recall that $V_{n, {\underline{p}}} = \sum _{0 \le i, j < n} \, 1_{Z_{j} - Z_{i} = {\underline{p}}} = \sum _{0 \le i < j < n} \, (1_{Z_{j} - Z_{i} = {\underline{p}}} + 1_{Z_{j} - Z_{i} = -{\underline{p}}}) + n \, 1_{{\underline{p}}= {\underline{0}}}$. We have

$$\begin{aligned} V_{n, {\underline{p}}}^2= & {} \sum _{0 \le i_1 < j_1 < n; \ 0 \le i_2 < j_2 < n} \, (1_{Z_{j_1} - Z_{i_1} = {\underline{p}}} \!+\!\! 1_{Z_{j_1} - Z_{i_1} = - {\underline{p}}}) \ (1_{Z_{j_2} - Z_{i_2} = {\underline{p}}} \!+\!\! 1_{Z_{j_2} - Z_{i_2} = - {\underline{p}}})\\&+\,2n \, 1_{p={\underline{0}}} \sum _{0 \le i < j < n} \, (1_{Z_{j} - Z_{i} = {\underline{p}}} + 1_{Z_{j} - Z_{i} = - {\underline{p}}}) + n^2 \, 1_{p={\underline{0}}}. \end{aligned}$$

The last term gives 0 in the computation of the variance, so that it suffices to bound

$$\begin{aligned}&\sum _{0 \le i_1 < j_1 < n; \ 0 \le i_2 < j_2 < n} \, {\mathbb {E}}[ (1_{Z_{j_1} - Z_{i_1} = {\underline{p}}} + 1_{Z_{j_1} - Z_{i_1} = - {\underline{p}}}) \ (1_{Z_{j_2} - Z_{i_2} = {\underline{p}}} + 1_{Z_{j_2} - Z_{i_2} = - {\underline{p}}})]\\&\quad -\sum _{0 \le i_1 < j_1 < n; \ 0 \le i_2 < j_2 < n} \, [{\mathbb {P}}(Z_{j_1} - Z_{i_1} = {\underline{p}}) + {\mathbb {P}}(Z_{j_1} - Z_{i_1} = - {\underline{p}})]\nonumber \\&\quad [{\mathbb {P}}(Z_{j_2} - Z_{i_2} = {\underline{p}}) + {\mathbb {P}}(Z_{j_2} - Z_{i_2} = - {\underline{p}})], \end{aligned}$$

i.e., the sum over the finite set $({\underline{r}},{\underline{s}}) \in \{\pm {\underline{p}}\}$ of the sums

$$\begin{aligned}&\sum _{0 \le i_1 < j_1 < n; \ 0 \le i_2 < j_2 < n} \, [{\mathbb {P}}(Z_{j_1} - Z_{i_1} = {\underline{r}}, \ Z_{j_2} - Z_{i_2} = {\underline{s}}) - {\mathbb {P}}(Z_{j_1} - Z_{i_1} = {\underline{r}}) \\&\quad {\mathbb {P}}(Z_{j_2} - Z_{i_2} = {\underline{s}})]. \end{aligned}$$

B) By the previous analysis, we are reduced to cases (1a), (2a), (3a). For (3a), by independence, we get 0. As ${\mathbb {E}}(V_{n, {\underline{p}}}^2) - ({\mathbb {E}}V_{n, {\underline{p}}})^2 \ge 0$, for the bound of (1a), we can neglect the corresponding centering terms, since they are subtracted and nonnegative.

(1a) For ${\underline{r}}, {\underline{s}}\in {\mathbb {Z}}^2$, let $a_{{\underline{r}}, {\underline{s}}}(n):= \sum _{0 \le i_1 \le i_2 < j_1 < j_2 < n} \, {\mathbb {P}}(Z_{j_1} - Z_{i_1} = {\underline{r}}, \, Z_{j_2} - Z_{i_2} = {\underline{s}}),$ $a({\underline{p}}, n):= \sum _{{\underline{r}}, {\underline{s}}\in \{\pm {\underline{p}}\}} a_{{\underline{r}}, {\underline{s}}}(n)$.

Setting $m_1 = i_1, m_2= i_2 - i_1, m_3 = j_1 - i_2, m_4 = j_2 - j_1$, $m_5 = n- j_2$ (so that $(m_1, \ldots , m_5)$ runs in the set $\tilde{M}_n^{(5)}$ of Lemma 7.4), by (71) we have for the generating function $A_{{\underline{r}}, {\underline{s}}}(\lambda ) := \sum _{n \ge 0} \lambda ^n a_{{\underline{r}}, {\underline{s}}}(n)$, $0 \le \lambda < 1$:

$$\begin{aligned} A_{{\underline{r}}, {\underline{s}}}(\lambda )= & {} \int _{{\mathbb {T}}^2}\int _{{\mathbb {T}}^2} \, e^{-2\pi i (\langle {\underline{r}}, {\underline{t}}\rangle + \langle {\underline{s}}, {\underline{u}}\rangle )} \, \sum _n \lambda ^n\\&\sum _{(m_1, \ldots , m_5) \, \in \, \tilde{M}_n^{(5)}} \, \Psi ({\underline{t}})^{m_2} \, \Psi ({\underline{t}}+ {\underline{u}})^{m_3} \, \Psi ({\underline{u}})^{m_4} \, d{\underline{t}}\, d{\underline{u}}. \end{aligned}$$

Using $\sum _{{\underline{r}}, {\underline{s}}\in \{\pm {\underline{p}}\}} e^{-2\pi i (\langle {\underline{r}}, {\underline{t}}\rangle + \langle {\underline{s}}, {\underline{u}}\rangle )} = \cos 2\pi \langle {\underline{p}}, {\underline{t}}\rangle \, \cos 2 \pi \langle {\underline{p}}, {\underline{u}}\rangle$, for the generating function $A_{\underline{p}}(\lambda ) := \sum _{n \ge 0} \lambda ^n a({\underline{p}}, n)$, we have by (69) with $\alpha = \Psi ({\underline{t}}), \beta = \Psi ({\underline{u}}), \gamma = \Psi ({\underline{t}}+{\underline{u}})$:

$$\begin{aligned} A_{\underline{p}}(\lambda ) = 4 {\lambda ^2 \over (1 - \lambda )^2} \int \int {\cos 2\pi \langle {\underline{p}}, {\underline{t}}\rangle \, \cos 2 \pi \langle {\underline{p}}, {\underline{u}}\rangle \, \Psi ({\underline{u}}) \Psi ({\underline{t}}+{\underline{u}}) \over (1 - \lambda \Psi ({\underline{t}})) \, (1- \lambda \Psi ({\underline{u}})) \, (1 - \lambda \Psi ({\underline{t}}+ {\underline{u}}))} d{\underline{t}}d{\underline{u}}. \end{aligned}$$

For an aperiodic r.w., the bound obtained for $A_{\underline{p}}(\lambda )$ in [27] for ${\underline{p}}= {\underline{0}}$ is valid. Indeed, the bounds for the domains $T_{i, \delta }$ and $E_\delta$ hold: for $T_{i, \delta }$ this is clear; for $E_\delta$ this is because $(1 - \Psi ({\underline{t}})) \, (1- \Psi ({\underline{u}})) \, (1 - \Psi ({\underline{t}}+ {\underline{u}}))$ does not vanish on $E_\delta$. (See [27] p. 227 for the notation for the domains.) The main contribution comes from a small neighborhood $U_\delta$ of diameter $\delta > 0$ of ${\underline{0}}$ and the factor $\cos 2\pi \langle {\underline{p}}, {\underline{t}}\rangle \, \cos 2 \pi \langle {\underline{p}}, {\underline{u}}\rangle$ plays no role.

It follows that $\displaystyle A_{\underline{p}}(\lambda ) \le {C \over (1 - \lambda )^3}$. Since $a({\underline{p}}, n)$ increases with n, we obtain $a({\underline{p}}, n) = O(n^2)$ by using the following elementary “Tauberian” argument:

Let $(u_k)$ be a non-decreasing sequence of nonnegative numbers. If there is C such that $\displaystyle \sum _{k=0}^{+\infty } \lambda ^k u_k \le {C \over (1 - \lambda )^3}, \ \forall \lambda \in [0, 1[$, then $u_n \le C' n^2, \ \forall n \ge 1$. Indeed, we have:

$$\begin{aligned} \lambda ^n u_n = (1 - \lambda )\left( \sum _{k=n}^{+\infty } \lambda ^k\right) \, u_n \le (1 - \lambda ) \sum _{k=n}^{+\infty } \lambda ^k u_k \le (1 - \lambda ) \sum _{k=0}^{+\infty } \lambda ^k u_k \le {C \over (1 - \lambda )^2}. \end{aligned}$$

For $\lambda = 1 - {1\over n}$ in the previous inequality, we get: $u_n \le C n^2 \, (1 - {1\over n})^{-n} \le C' n^2$.

(2a) Let $b_{{\underline{r}}, {\underline{s}}}(n)$ be the sum

$$\begin{aligned}&\sum _{0 \le i_1 < i_2 < j_2 \le j_1 < n} [{\mathbb {P}}(Z_{j_1} - Z_{i_1} = {\underline{r}}, \, Z_{j_2} - Z_{i_2} = {\underline{s}})\\&\quad - {\mathbb {P}}(Z_{j_1} - Z_{i_1} = {\underline{r}}) \, {\mathbb {P}}( Z_{j_2} - Z_{i_2} = {\underline{s}})] \end{aligned}$$

and $b({\underline{p}}, n) := \sum _{{\underline{r}}, {\underline{s}}\in \{\pm {\underline{p}}\}} b_{{\underline{r}}, {\underline{s}}}(n)$.

Putting $m_1 = i_1, m_2 = i_2 - i_1, m_3 = j_2 - i_2, m_4 = j_1 - j_2$, we have by (72):

$$\begin{aligned} b({\underline{p}}, n)= & {} \sum _{{\underline{r}}, {\underline{s}}\in \{\pm {\underline{p}}\}} \sum _{{\underset{m_1, m_4 \ge 0, \, m_2, m_3 \ge 1}{m_1 + m_2 + m_3 + m_4 < n}}} {\mathbb {P}}(Z_{m_3} = {\underline{s}}) [{\mathbb {P}}(Z_{m_2 + m_4}\\&\quad = {\underline{r}}- {\underline{s}}) - {\mathbb {P}}(Z_{m_2 + m_3+ m_4} = {\underline{r}})]. \end{aligned}$$

We will use Lemma 7.3 to bound $b({\underline{p}}, n)$.

If $m_3 {\underline{\ell }}_1 = {\underline{s}}\mathrm{\ mod \,} D$, then either $(m_2 + m_4) {\underline{\ell }}_1 = {\underline{r}}- {\underline{s}}\mathrm{\ mod \,} D$ and $(m_2+m_4 + m_3) {\underline{\ell }}_1 = {\underline{r}}\mathrm{\ mod \,} D$, or $(m_2 + m_4) {\underline{\ell }}_1 \not = {\underline{r}}- {\underline{s}}\mathrm{\ mod \,} D$ and $(m_2+m_4 + m_3) {\underline{\ell }}_1 \not = {\underline{r}}\mathrm{\ mod \,} D$. In the latter case, the corresponding probabilities are 0. Moreover, ${\mathbb {P}}(Z_{m_3} = {\underline{s}}) =0$ if $m_3 {\underline{\ell }}_1 \not = {\underline{s}}\mathrm{\ mod \,} D$. Therefore, the sum reduces to those indices such that $m_3 {\underline{\ell }}_1 = {\underline{s}}\mathrm{\ mod \,} D$, $(m_2 + m_4) \, {\underline{\ell }}_1 = {\underline{r}}- {\underline{s}}\mathrm{\ mod \,} D$ and $(m_2+m_4 + m_3) \, {\underline{\ell }}_1 = {\underline{r}}\mathrm{\ mod \,} D$ and the bound (67) applies for the nonzero terms. We obtain:

$$\begin{aligned} b({\underline{p}}, n)\le & {} 4 \sum _{{\underset{m_1, m_4 \ge 0, \, m_2, m_3 \ge 1}{m_1 + m_2 + m_3 + m_4 < n}}} \left[ {1 \over m_3} {m_3 \over (m_2 + m_4) (m_2 + m_3 + m_4)}\right. \\&\left. +\,{1 \over m_3 \, (m_2 + m_4 + m_3)^\frac{3}{2}}\right] . \end{aligned}$$

The sum of the first term restricted to $\{m_i\ge 1, m_1+m_2+m_3+m_4\le n\}$ reads:

$$\begin{aligned}&\sum _{k=1}^n\sum _{m_1+m_2+m_3+m_4=k}\frac{1}{(m_2+m_4)(m_2+m_4+m_3)}\\&\quad = \sum _{k=1}^n\sum _{\ell =1}^{k}\sum _{m_2+m_4=\ell , m_1+m_3=k-\ell }\frac{1}{\ell (\ell +m_3)} \\&\le \sum _{k=1}^n\sum _{\ell =1}^{k}\sum _{m_1+m_3=k-\ell }\ell \cdot \frac{1}{\ell (\ell +m_3)} \le \sum _{k=1}^n\sum _{\ell =1}^{k}\left( \frac{1}{\ell }+\cdots +\frac{1}{k}\right) = \frac{1}{2} n(n+1). \end{aligned}$$

Likewise, we have:

$$\begin{aligned}&\sum _{k=1}^n\sum _{m_1+m_2+m_3+m_4=k}\frac{1}{m_3(m_2+m_4+m_3)^\frac{3}{2}} \le n \sum _{k=1}^n\sum _{m_2+m_3+m_4=k}\frac{1}{m_3 \, k^\frac{3}{2}} \\&\le n\sum _{k=1}^n\frac{1}{k^{3/2}}\sum _{m_3=1}^k\frac{k}{m_3} \le n\sum _{k=1}^n\frac{\mathrm{Log}k}{\sqrt{k}}\le C\, n^{3/2}\mathrm{Log}n. \end{aligned}$$

Therefore, we obtain $b(n, {\underline{p}}) = O(n^2)$.

The previous estimations imply $\mathrm{Var}(V_{n, {\underline{p}}}) = O(n^2)$. By (48), for ${\underline{p}}\in L(W)$, $\lim _n V_{n, {\underline{p}}} / {\mathbb {E}}V_{n, {\underline{p}}} = 1$ a.e. follows as in [27] and therefore $\lim _n V_{n, {\underline{p}}} / V_{n, {\underline{0}}} = 1$ a.e.