The coin-turning walk and its scaling limit

TL;DR

This paper studies the scaling limits of a coin-turning random walk under various cooling regimes, revealing a phase transition at a critical scaling where a new recurrent process with Beta marginals emerges.

Contribution

It introduces a non-classical invariance principle for coin-turning walks, identifying a critical cooling regime and characterizing the limiting process as Beta-distributed and recurrent.

Findings

Invariance principle holds for certain non-classical scalings.

Critical cooling regime at order const·n^{-1} leads to a Beta-distributed, recurrent process.

Walk exhibits phase transition in scaling behavior at the critical order.

Abstract

Let $S$ be the random walk obtained from "coin turning" with some sequence $\{p_n\}_{n\ge 1}$, as introduced in [6]. In this paper we investigate the scaling limits of $S$ in the spirit of the classical Donsker invariance principle, both for the heating and for the cooling dynamics. We prove that an invariance principle, albeit with a non-classical scaling, holds for "not too small" sequences, the order const$\cdot n^{-1}$ (critical cooling regime) being the threshold. At and below this critical order, the scaling behavior is dramatically different from the one above it. The same order is also the critical one for the Weak Law of Large Numbers to hold. In the critical cooling regime, an interesting process emerges: it is a continuous, piecewise linear, recurrent process, for which the one-dimensional marginals are Beta-distributed. We also investigate the recurrence of the walk and its scaling limit, as well as the ergodicity and mixing of the $n$th step of the walk.

Figures

Equations

$$X_n:=\{\begin{array}{ll}1-X_{n-1},& \text{with probability }{p}_n;\\ X_{n-1},& \text{otherwise},\end{array}$$

$$Y_n:=\{\begin{array}{ll}-Y_{n-1},& \text{with probability }{p}_n;\\ Y_{n-1},& \text{otherwise}.\end{array}$$

$$Y_n:=(-1{)}^{{\sum }_{i=1}^n{W}_i},$$

$$\mathsf{Corr}(Y_i,Y_j)$$

$$=\prod _{i+1}^j\mathbb{E}(-1{)}^{W_k}=\prod _{k=i+1}^j(1-2p_k)=:e_{i,j};$$

$$\mathbb{E}(Y_j\mid Y_i)$$

$${\overline{X}}_N\to \zeta \text{ a.s.},$$

$$\underset{j\to \infty }{\lim }{e}_{ij}=0,\forall i\in \mathbb{N}.$$

$$\underset{j\to \infty }{\lim }\mathsf{Cov}(Y_j,Y_i)=0,$$

$$\sum _n\min (p_n,q_n)=\infty .$$

$$\min (p_i,q_i)=\{\begin{array}{ll}{p}_i,& \text{if }{p}_i\le 1/2;\\ q_i=1-p_i,& \text{if }{p}_i>1/2,\end{array}$$

$$|e_{i,j}|=\left|\prod _{k=i+1}^j(1-2p_k)\right|=\prod _{i<k\le j,p_k\le 1/2}(1-2p_k)\times \prod _{i<k\le j,p_k>1/2}(1-2q_k)=$$

$$\prod _{k=i+1}^j(1-2\min (p_k,q_k)).$$

$$\sum _i\min (p_i,q_i)=\sum _{i\in I}{p}_i+\sum _{i\notin I}{q}_i<\infty ,$$

$$\mathbb{P}\left(\exists n_0\in \mathbb{N}:W_i=1_{I^c}(i)\text{ for all }i\ge n_0\mid {\mathcal{F}}_1\right)=1,$$

$$\mathsf{Law}\left(\frac{S_N}{\sqrt{N}}\right)\to \mathsf{Normal}\left(0,{\sigma }_c^2\right),where{\sigma }_c^2:=1+2\sum _{i=1}^{\infty }\mathsf{Cov}(Y_i,Y_j)=\frac{1-c}{c}.$$

$$p_n=\frac{a}{n},n\ge n_0$$

$$\mathsf{Law}\left(\frac{S_N}{N}\right)\to \mathsf{Beta}(a,a),$$

$$p_n=\frac{a}{n^{\gamma }},n\ge n_0$$

$$\mathsf{Law}\left(\frac{S_N}{\sqrt{N^{1+\gamma }}}\right)\to \mathsf{Normal}\left(0,{\sigma }_{a,\gamma }^2\right),where{\sigma }_{a,\gamma }^2:=\frac{1}{a(1+\gamma )}.$$

$$\underset{n\to \infty }{\lim }(S_{n{t}_1}/n,S_{n{t}_2}/n,...,S_{n{t}_k}/n),$$

$$\hat{X}(t):=\sum _{n=1}^{\infty }\frac{S_n}{n}{1}_{{\tau }_n\le t<{\tau }_{n+1}},t\ge 0.$$

$$\underset{t\to \infty }{\lim }\left(\hat{X}({\tau }_{t{t}_1})t_1,\hat{X}({\tau }_{t{t}_2})t_2,...,\hat{X}({\tau }_{t{t}_k})t_k\right).$$

$$\underset{t\to \infty }{\lim }({\hat{X}}^{({\tau }_{t{t}_1})}(0)t_1,{\hat{X}}^{({\tau }_{t{t}_1})}(\log (t_2/t_1))t_2,...,{\hat{X}}^{({\tau }_{t{t}_k})}(\log (t_k/t_{k-1}))t_k),$$

$$\underset{z\to \infty }{\lim }({\hat{X}}^{(z)}(s_1),{\hat{X}}^{(z)}(s_2),...,{\hat{X}}^{(z)}(s_k))$$

$$Y_n:=(-1{)}^{{\sum }_{i=1}^n{W}_i},$$

$$\mathbb{P}(Y=1\mid Y_1=1)$$

$$\mathbb{P}(Y=1\mid Y_1=-1)$$

$$M=\left({\begin{array}[]{cc}0&1\\ 1&0\\ \end{array}}\right),$$

$$\exp \left(-2\sum _1^n\min \{p_i,1-p_i\}\right).$$

$$S^n(t):=\frac{S_{\lfloor \frac{c}{1-c}nt\rfloor }}{\sqrt{n}},t\ge 0,$$

$$a_n$$

Full text

We prove that an invariance principle, albeit with a non-classical scaling, holds for “not too small” sequences, the order const$\cdot n^{-1}$ (critical cooling regime) being the threshold. At and below this critical order, the scaling behavior is dramatically different from the one above it. The same order is also the critical one for the Weak Law of Large Numbers to hold.

In the critical cooling regime, an interesting process emerges: it is a continuous, piecewise linear, recurrent process, for which the one-dimensional marginals are Beta-distributed.

We also investigate the recurrence of the walk and its scaling limit, as well as the ergodicity and mixing of the $n$th step of the walk.

J.E.’s research was supported in part by Simons Foundation Grant 579110.

S.V.’s research was supported in part by Swedish Research Council grant VR 2014-5157.

1. 1 Introduction

2. 2 Mixing

3. 2.1 Characterization of mixing

4. 2.2 Why is mixing a natural assumption?

5. 3 Review of relevant literature

6. 3.1 Some results from [7]

7. 3.2 Recent results by Benaïm, Bouguet and Cloez

8. 4 Our main results

9. 4.1 The law of the $n$th step for large $n$

10. 4.2 Scaling limits for the walk

11. 4.2.1 The time-homogeneous case

12. 4.2.2 Heating regime

13. 4.2.3 Cooling regime

14. 4.2.4 Neither heating nor cooling regime

15. 4.3 Validity of the WLLN

16. 4.4 Recurrence

17. 5 Examples and open problems

18. 6 Proofs

19. 6.1 Preparation I: The zigzag process

20. 6.2 Preparation II: Approximating the walk with a martingale

21. 6.3 Some specific cases

22. 6.3.1 Cooling, critical

23. 6.3.2 Cooling, subcritical

24. 6.3.3 Cooling, subcritical; the necessity of $\liminf\limits_{n\rightarrow\infty}\frac{p_{n}}{p_{n+1}}>0$

25. 6.3.4 (Heating)

26. 6.4 Proof of Theorem 1

27. 6.5 Proof of Theorem 2

28. 6.6 Proof of Theorem 3

29. 6.6.1 STEP 1

30. 6.6.2 STEP 2

31. 6.7 Proof of Proposition 2

32. 6.8 Proof of Theorem 4 – strongly critical case

33. 6.9 Proof of Theorem 4 – supercritical case

34. 6.10 Proof of Theorem 4 – critical case

35. 6.11 Proof of Theorem 4 – subcritical case

36. 6.12 Proof of Theorem 5

37. 6.13 Proof of Theorem 6

38. 6.14 Proof of Theorem 7

39. 6.15 Proof of Theorem 8

40. 7 Appendix

1. Introduction

We start with reviewing the notion of the coin turning process, which has been introduced recently in [7].

Let $p_{2},p_{3},p_{4}...$ be a given deterministic sequence of numbers such that $p_{n}\in[0,1]$ for all $n$; define also $q_{n}:=1-p_{n}$. We define the following time-dependent “coin turning process” $X_{n}\in\{0,1\}$, $n\geq 1$, as follows. Let $X_{1}=1$ (“heads”) or $=0$ (“tails”) with probability $1/2$. For $n\geq 2$, set recursively

[TABLE]

that is, we turn the coin over with probability $p_{n}$ and do nothing with probability $q_{n}$. (Equivalently, one can define $p_{1}=1/2$ and $X_{1}\equiv 0$.)

Consider $\overline{X}_{N}:=\frac{1}{N}\sum_{n=1}^{N}X_{n}$, that is, the empirical frequency of $1$’s (“heads”) in the sequence of $X_{n}$’s. We are interested in the asymptotic behavior of this random variable when $N\to\infty$. Since we are interested in limit theorems, it is convenient to consider a centered version of the variable $X_{n}$; namely $Y_{n}:=2X_{n}-1\in\{-1,+1\}$. We have then

[TABLE]

Let also $\mathcal{F}_{n}:=\sigma(Y_{1},Y_{2},...,Y_{n}),\ n\geq 1$.

Note that the sequence $\{Y_{n}\}$ can be defined equivalently as follows:

[TABLE]

where $W_{1},W_{2},W_{3},...$ are independent Bernoulli variables with parameters $p_{1},p_{2},p_{3},...$, respectively, and $p_{1}=1/2$.

Remark 1** (Poisson binomial random variable).**

The number of turns that occurred up to $n$, that is $\sum_{i=2}^{n}W_{i}$, is a Poisson binomial random variable.$\hfill\diamond$

For the centered variables $Y_{n}$, we have $Y_{j}=Y_{i}(-1)^{\sum_{i+1}^{j}W_{k}},\ j>i$, and so, using $\mathsf{Corr}$ and $\mathsf{Cov}$ for correlation and covariance, respectively, one has

[TABLE]

The quantity $e_{i,j}$ will play an important role throughout the paper.

Next, we define our basic object of interest.

Definition 1** (Coin-turning walk).**

The random walk $S$ on $\mathbb{Z}$ corresponding to the coin-turning, will be called the coin-turning walk. Formally, $S_{n}:=Y_{1}+...+Y_{n}$ for $n\geq 1$; we can additionally define $S_{0}:=0$, so the first step is to the right or to the left with equal probabilities. As usual, we then can extend $S$ to a continuous time process, by linear interpolation.

Remark 2**.**

Even though $Y$ is Markovian, $S$ is not. However, the 2-dimensional process $U$ defined by $U_{n}:=(S_{n},S_{n+1})$ is Markovian. It lives on a ladder embedded into $\mathbb{Z}^{2}$. See Figure 1.$\hfill\diamond$

In [7], several scaling limits of the form $\lim_{n\to\infty}\mathtt{Law}\left(\frac{S_{n}}{b_{n}}\right)=\mathtt{L}$, have been established, where $\{b_{n}\}_{n\geq 1}$ is an appropriate sequence (depending on the sequence of $p_{n}$’s) tending to infinity and $\mathtt{L}$ is a non-degenerate probability law. In [7] the focus was on the $\lim_{n\to\infty}p_{n}=0$ case.

Remark 3** (Supercritical cases).**

Note that if $\sum_{n}p_{n}<\infty$ then by the Borel-Cantelli lemma, only finitely many turns will occur a.s.; therefore the $X_{j}$’s will eventually become all ones or all zeros, and hence

[TABLE]

where $\zeta\in\{0,1\}$. By the symmetry of the definition with respect to heads and tails (or, by the bounded convergence theorem), $\zeta$ is a Bernoulli($1/2$) random variable.

Similarly, if $\sum_{n}q_{n}<\infty$, then $S$ will be eventually stuck at two neighboring integers, again, by the Borel-Cantelli lemma. $\hfill\diamond$

These two trivial cases (we call them “lower supercritical” and “upper-supercritical” cases) are not considered, and so we have the following assumption.

Assumption 1** (Divergence).**

In the sequel we are going to assume that $\sum_{n}p_{n}=\infty$ and also $\sum_{n}q_{n}=\infty$.

2. Mixing

Unlike in [7] and in the previous section, we now do not randomize the walk with taking $Y_{1}$ to be a symmetric random variable. Nevertheless, it is still true for the indicators of turns $W_{k}$, that $Y_{j}=Y_{i}(-1)^{\sum_{i+1}^{j}W_{k}},\ j>i$, and that for $e_{i,j}=\prod_{k=i+1}^{j}(1-2p_{k})$ we have ${\mathbb{E}}(Y_{j}\mid Y_{i})=Y_{i}{\mathbb{E}}(-1)^{\sum_{i+1}^{j}W_{k}}=e_{i,j}Y_{i},$ hence ${\mathbb{E}}(Y_{i}Y_{j})=e_{i,j}$.

2.1. Characterization of mixing

We will say that the sequence of random variables $(Y_{n})_{n\geq 1}$ satisfies the mixing condition if

[TABLE]

Under mixing, $\lim_{j\to\infty}{\mathbb{E}}(Y_{j}\mid Y_{i})=0$, so $Y_{j}$ “becomes symmetrized” for $i$ fixed and large $j$. Also, $\lim_{j\to\infty}{\mathbb{E}}(Y_{i}\,Y_{j})=0$ and $\lim_{j\to\infty}{\mathbb{E}}{Y_{j}}=0$, hence

[TABLE]

in accordance with the usual notion of mixing.

Mixing has a very simple characterization in terms of the sequence $\{p_{n}\}_{n\geq 1}$.

Proposition 1** (Condition for mixing).**

Mixing holds if and only if

[TABLE]

Proof of Proposition 1.

Since

[TABLE]

we have

[TABLE]

When $p_{k}\neq 1/2$ for all $k\geq 1$, (3) and (5) are equivalent by a well known result about infinite products; when $p_{k}=1/2$ infinitely often, (3) and (5) are clearly simultaneously satisfied.

In all other cases, define $k_{0}:=\max\{k\in{\mathbb{N}}\mid p_{k}=1/2\}$. For $i<k_{0}$, $e_{i,j}=0$ for all large $j$, while for $i\geq k_{0}$, (3) is tantamount to (5), just like in the first case. ∎

2.2. Why is mixing a natural assumption?

The mixing condition is stronger than Assumption 1 if $p_{k}$ keeps crossing the line $1/2$ (i.e. $\liminf p_{k}<1/2<\limsup p_{k}$), while they are equivalent when $p_{k}$ settles on one side of $1/2$ eventually.

In the first case Assumption 1 is automatically satisfied, as $p_{k}\geq 1/2$ i.o. and also $q_{k}\geq 1/2$ i.o. Defining $I:=\{i\in{\mathbb{N}}:\ p_{i}\leq 1/2\}$, we see that the mixing condition is nevertheless violated if and only if

[TABLE]

that is, when $\sum_{i\in I}p_{i}<\infty$ and $\sum_{i\not\in I}q_{i}<\infty$. In this case, recalling that $W_{i}$ is the indicator of a turn at time $i$, by Borel-Cantelli,

[TABLE]

where $\mathbf{1}_{I^{c}}$ is the characteristic function of the set ${\mathbb{N}}\setminus I$. That is, along $I$, “turning” eventually stops, while along ${\mathbb{N}}\setminus I$, “staying” eventually stops.

Our conclusion is that when mixing does not hold, the random walk is “eventually deterministic”, and thus the setup is less interesting. For example, from the point of view of recurrence, the problem becomes a question about a deterministic process; whether that process takes any integer value infinitely many times depends simply on the set $I$ (as long as $\displaystyle\sum_{i\in I}p_{i}<\infty$ and $\displaystyle\sum_{i\not\in I}q_{i}<\infty$.)

To have a concrete example, let $I=\{2,4,6,\dots\}$ be the set of positive even integers. Then, for large times, the walk will alternate between taking two consecutive steps up and taking two consecutive steps down. This excludes recurrence of course, as the process becomes stuck at some triple of consecutive integers. We summarize the above discussion in Figure 2.

We conclude this Section with some notation.

Throughout the paper, $c_{n}\sim d_{n}$ will mean that $\lim_{n\to\infty}c_{n}/d_{n}=1$; also, $c_{n}=o(d_{n})$ will mean that $\lim_{n\to\infty}c_{n}/d_{n}=0$. Convergence in distribution will also be denoted by $\overset{d}{\rightarrow}$.

3. Review of relevant literature

3.1. Some results from [7]

Some of the basic results of [7] are summarized in the following theorem.

Theorem A**.**

Let $S$ denote the coin-turning walk.

• (i)

(“Time-homogeneous case”.) Let $p_{n}=c$ for all $n\geq 1$, where $0<c<1$. Then

[TABLE]

• (ii)

(“Lower critical case”.) Fix $a>0$ and let

[TABLE]

for some $n_{0}\in\mathbb{N}$. Then111A nice exercise, left to the reader, is to show that when the sequence is precisely $(p_{1}=1/2),p_{2}=1/3,p_{3}=1/4,p_{4}=1/5$, … , $\frac{S_{N}}{N}$ has precisely discrete uniform law for each $N$. This fact, as Márton Balázs pointed out to us, can be related to Pólya urns. The more general connection of our model with Pólya urns will be presented in a forthcoming paper.**

[TABLE]

where ${\sf Beta}(\alpha,\beta)$ denotes the Beta distribution with parameters $\alpha,\beta$.

• (iii)

(“Lower subcritical case”.) Fix $\gamma,a>0$ and let

[TABLE]

for some $n_{0}\in{\mathbb{N}}$. (Since $\gamma>1$ corresponds to the supercritical case, we assume that $0<\gamma<1$.) Then

[TABLE]

3.2. Recent results by Benaïm, Bouguet and Cloez

In a recent follow up paper to [7] by Bouguet and Cloez [2], the setting has been generalized in such a way that instead of two states (heads and tails or $\pm 1$), one considers $D\geq 2$ states, and with probability $p_{n}$ in the $n$th step the state changes according to a given irreducible Markov chain.222E.g. when $D=2$, one can still consider unequal probabilities for switching between the states in different directions. (They also allow a small error term.) They assume that $\{p_{n}\}_{n\geq 1}$ is a decreasing sequence and $p:=\lim_{n}p_{n}$ is not necessarily zero. This excludes the $p=1$ case we consider, except the trivial $p_{n}\equiv 1$ case, and the most interesting case is $p=0$, the one we call cooling dynamics.

Bouguet and Cloez prove several interesting results, generalizing/strengthening those in [7]. For example they show that if $\sum_{n}p_{n}=\infty,\ \lim_{n\to\infty}np_{n}=\infty$ and $\sum_{n}(p_{n}n^{2})^{-1}<\infty$, then the empirical distribution of the states converges almost surely to the unique invariant probability distribution of the Markov chain.

The relationship with [7] is explained in 4.2 in [2].

The paper builds on the authors’ previous results with M. Benaïm in [1], and they point out in [2] that

*“In particular, the results we use provide functional convergence of the rescaled interpolating processes to the auxiliary Markov processes…” *

at which point the authors refer to [1] and another article. Also, after their Theorem 2.8, treating the critical $p_{n}=c/n$ case, they note that

*“It should be noted that our approach for the study of the long-time behavior of … also provides functional convergence for some interpolated process … from which Theorem 2.8 is a straightforward consequence.” *

On the one hand, their Theorem 2.8 is really about the convergence of $S_{n}/n$ only, and the “interpolating process” alluded to is not the random walk $S$, and it is not completely clear if the authors of [2] are trying to say that one in fact can obtain from [1] the functional convergence for $S$ in the critical case as stated in our Theorem 4(3).

On the other hand, it seems that this derivation is after all doable, as we explain this briefly below. The reader may safely skip this part though and return to it only after reading our main results. Indeed, let us suppose we already know tightness and only want to check the convergence of the finite dimensional distributions, that is the existence of the limit (in law)

[TABLE]

for some $0<t_{1}<t_{2}<...<t_{k}$. When $p_{n}=c/n$ for large $n$’s, define $\tau_{t}:=\sum_{1}^{\lfloor t\rfloor}p_{n}$. Define the “pasting process” $\widehat{X}$ by

[TABLE]

After some algebra, the limit in (6) is equivalent to the following one:

[TABLE]

Using the fact that $\lim_{t\to\infty}(\tau_{t}-\log t)$ is a constant, we can rewrite this as

[TABLE]

where $\widehat{X}^{(z)}(t):=\widehat{X}(t+z)$, $z>0$. Now, if the limit (in law)

[TABLE]

is known, we are done, and this latter type of limit of “pseudo-trajectories” is what has been derived in [1] under some suitable assumptions.

In summary, [2] provides a very valuable complement to [7]. Moreover, with some further efforts, our result on the zigzag process in the present paper can apparently be recovered from the results presented in the sequence [1, 2], and vice versa.

4. Our main results

4.1. The law of the $n$th step for large $n$

Recall that

[TABLE]

where $W_{1},W_{2},W_{3},...$ are independent Bernoulli variables with parameters $p_{1},p_{2},p_{3},...$, respectively.

When $p_{k}\leq 1/2$ for all large $k$, $\rho:=\prod_{i=2}^{\infty}(1-2p_{i})$ is well defined as the terms are in $[0,1]$ with finitely many exceptions. In particular, when $\sum p_{i}<\infty$, by Borel-Cantelli, $Y_{i}=Y$ for all large $i$, a.s., and in Proposition 1 in [7] it has been shown that in this case

[TABLE]

This may be generalized is as follows.

Theorem 1**.**

Define $N:=\mathrm{card}\{i:p_{i}>1/2\}\in\mathbb{N}\cup\{\infty\}$.

• (a)

If mixing holds, or if $\exists i:p_{i}=1/2$ then $\lim_{n}{\mathbb{P}}(Y_{n}=1\mid\mathcal{F}_{1})=1/2$.

• (b)

If mixing does not hold, then there are two cases ($k\in\{-1,1\}$):

• (i)

if $N<\infty$, then $\displaystyle\lim_{n\to\infty}{\mathbb{P}}(Y_{n}=1\mid Y_{1}=k)=\frac{1}{2}(1+k\rho)$, and $\rho\neq 0.$

• (ii)

if $N=\infty$ then ${\mathbb{P}}(Y_{n}=1\mid Y_{1}=k)$ has no limit.

Remark 4** (Ergodicity).**

Part (a) in Theorem 1 is interpreted as “mixing implies ergodicity”, since $(1/2,1/2)$ is the invariant distribution for the switching matrix

[TABLE]

and we can consider our model as one where at step $n$ the transition given by $M$ may or may not apply (with probabilities $p_{n}$ and $1-p_{n}$, resp.).$\hfill\diamond$

Remark 5** (Speed of convergence).**

Regarding Theorem 1, one may wonder what the speed of convergence is when the limit exists. A closer look at the proof of Theorem 1 in Section 6 shows that the total variation distance to the limit decays as

[TABLE]

We leave the details to the reader. $\hfill\diamond$

4.2. Scaling limits for the walk

Recently, Sean O’Rourke has asked whether the results of [7] could be extended to convergence in the process sense, in the spirit of the classical Donsker invariance principle (see e.g. [9] for the classical result and its proof). We are now going to answer this question, and moreover, we are also going to consider additional cases, when turns are becoming more and more frequent (i.e. $p_{n}$ is close to one), such as, for example, $p_{n}=1-c/n$ or $p_{n}=1-n^{-\gamma}$, $0<\gamma<1$ for large $n$.

Note: In the rest of the paper, for convenience we assume again that $p_{1}=1/2$, i.e. we symmetrize the setting.

4.2.1. The time-homogeneous case

As a warm up, we start with the time-homogeneous case.

Theorem 2** (Time-homogeneous case:).**

Assume that $p_{n}=c$ for $n\geq n_{0}$. For $n\geq 1$, define the rescaled walk ${S}^{n}$ by

[TABLE]

and let $\mathcal{W}$ denote the Wiener measure. Then $\lim_{n\to\infty}\mathsf{Law}({S}^{n})=\mathcal{W}$ on $C[0,\infty)$.

Remark 6**.**

We will show that Theorem 2 follows trivially from our general martingale approximation method of Subsection 6.2. However, we note that one can also give a direct proof using that the “turning times” are geometrically distributed. Here is a sketch: assuming that e.g. $Y_{1}=1$ we can consider the period consisting of the first run of $1$’s together with the first run of $-1$’s. The second, third etc. periods are defined similarly, and the piece-wise linear “roof-like” processes in these periods are i.i.d. (up to their respective starting values). Since the length of each run is geometrically distributed, and those geometric variables are independent, the Renewal Theorem applies to the lengths of the periods. One then applies the classical invariance principle to the process considered at each second “turning time”, and finally extends the result for all times. We leave the details to the reader.$\hfill\diamond$

Remark 7**.**

Theorem 2 is also covered by those in  [4, 5]. The first one treats the “uniformly strong mixing” condition for Markov chains and weak convergence.$\hfill\diamond$

4.2.2. Heating regime

The following theorem will give an invariance principle for the “heating” case, that is for the case when the $p_{n}$ are getting close to one. But before that we present an important remark.

Remark 8** (Even and odd parts).**

It turns out that in the heating regime, the right approach is to look at the sums of the two sub-series $I=\sum_{\textsl{odd}}:=\displaystyle\sum_{k=1}^{\infty}q_{2k-1}$ and $II=\sum_{\textsl{even}}:=\displaystyle\sum_{k=1}^{\infty}q_{2k}$ separately. If either $I<\infty$ or $II<\infty$, then the invariance principle breaks down.

Indeed, by Borel-Cantelli then, after some finite time, every other step turns the coin a.s., and consequently, $S$ is stuck on a set of size three, which rules out the validity of any invariance principle. We conclude that for an invariance principle to hold, it is not enough to assume merely that $\displaystyle\sum_{k=0}^{\infty}q_{k}=\infty$; one needs to assume that in fact $I=II=\infty$. $\hfill\diamond$

In light of the previous remark, without the loss of generality from now on we will work under the following assumption.

Assumption 2**.**

$I=II=\infty.$**

Before we state the following theorem, we need some more notations, Introduce

[TABLE]

which is well defined as the sum of a Leibniz series,

[TABLE]

and

[TABLE]

so that ${\mathbb{E}}\xi_{i}^{2}=\mathsf{Var}((-1)^{W_{i}})=4p_{i}q_{i}$ and ${\mathbb{E}}\Lambda^{2}_{n}=v_{n}$.

Theorem 3** (Invariance principle; heating regime).**

Assume that $q_{n}\to 0$. Besides Assumption 2, assume that there exists a $C>0$ such that at least one of the following two assumptions is satisfied:

[TABLE]

[TABLE]

• (a)

Define the rescaled walk ${S}^{n}$ by setting

[TABLE]

where $Z(x):=\inf\{n\in\mathbb{N}:\ v_{n}\geq x\}$. Then

[TABLE]

where $\mathcal{W}$ is the Wiener measure.

• (b)

We have $\lim_{n\to\infty}\Lambda_{n}=\infty$ almost surely333Note that Drogin in [6] proves, in fact, two invariance principles. The second one uses the function $s^{2}$ (our $\Lambda^{2}$) for time-change., and $\displaystyle\lim_{n\to\infty}\frac{\Lambda^{2}_{n}}{{\mathbb{E}}\Lambda^{2}_{n}}=1\ \text{in probability}.$

Remark 9** (Equivalent condition).**

One can rewrite (9) in a “backward looking” way:

[TABLE]

as both are equivalent to saying that $q_{n}\geq\mathrm{const}\cdot q_{r}$ for $r>n$ if $n$ is even and $r$ is odd. A similar statement holds for (10).$\hfill\diamond$

4.2.3. Cooling regime

When $\displaystyle\lim_{n\to\infty}p_{n}=0$, one deals with a so-called “cooling dynamics” as the turns become infrequent. In this case, the scaling limit is not necessarily Brownian motion, as the following theorem shows. Loosely speaking, the order const$\cdot n^{-1}$ is the critical one in the sense that for sequences of larger order the invariance principle is in force, however at this order or below it the situation is dramatically different.

Theorem 4** (Cooling regime).**

Let the process $S^{n}$ be defined by $S^{n}(t):=S_{nt}/n,\,t\geq 0$, where for non-integer values of $nt$ we assign $S_{nt}$ using the usual linear interpolation. Let $\mathcal{R}$ be the process (“random ray”) defined by $\mathcal{R}(t):=tR$, where $R$ is a random variable equal to $\pm 1$ with equal probabilities. We have the following limits in the process sense:

1. (1)

Supercritical case:* $\displaystyle\sum_{n=1}^{\infty}p_{n}<\infty$. Then $\lim_{n\to\infty}\|{S}^{n}(\cdot)-\mathcal{R}(\cdot)\|_{\infty}=0$ almost surely.* 2. (2)

Strongly critical case:* $p_{n}=o(1/n)$ but $\displaystyle\sum_{n=1}^{\infty}p_{n}=\infty$. Then $\lim_{n\to\infty}{S}^{n}(\cdot)=\mathcal{R}(\cdot)$ in law.* 3. (3)

Critical case:* $p_{n}=c/n$ for $n\geq n_{0}$. Recalling the notion of the zigzag process (defined in Section 6.1), $\lim_{n\to\infty}S^{(n)}$ is the zigzag process, where the limit is meant in law.* 4. (4)

Subcritical case:* (Cooling but larger order than $1/n$) Let $p_{1}=1/2$. Assume that, as $n\to\infty$,*

• (a)

$A_{n}:=np_{n}\uparrow\infty$;

• (b)

$p_{n}\downarrow 0$.

Then, for the rescaled walk (11) the invariance principle (12) holds.

4.2.4. Neither heating nor cooling regime

The following result generalizes the case when $\lim\limits_{n\rightarrow\infty}p_{n}=a$ with $0<a<1$, as well as the time-homogeneous case of Theorem 2: the invariance principle holds as long as the $p_{n}$ are bounded away from both [math] and $1$.

Theorem 5** (Invariance principle; neither heating nor cooling regime).**

Assume that

[TABLE]

Then for the rescaled walk (11) the invariance principle (12) holds.

4.3. Validity of the WLLN

With regard to the Weak Law of Large Numbers (by which we mean that $S_{n}/n\to 0$ in probability), we know that it breaks down at the critical regime. On the other hand, the following result shows that above that order it is always in force.

Theorem 6** (WLLN).**

Let $p_{n}\leq 1/2$ for all $n\geq 1$ and assume that $\lim_{n\to\infty}np_{n}=\infty$. Then $\displaystyle\lim_{n\to\infty}\frac{S_{n}}{n}=0$ in probability.

4.4. Recurrence

We now turn our attention to the recurrence/transience of the walk and its scaling limit.

Definition 2**.**

We call $S$ recurrent if

[TABLE]

Let us introduce the following mild condition on the walk.

Assumption 3** (Spreading).**

Assume that for all $n,K\in{\mathbb{N}}$,

[TABLE]

Remark 10**.**

Assumption 3 is trivially satisfied when $\sigma_{n}^{2}:=\sf{Var}(S_{n})\to\infty$ and the scaling limit

[TABLE]

holds with $a,b\in{\mathbb{R}}$, $a\leq b$ and $n\in{\mathbb{N}}$, and some probability measure $Q$ such that $Q(\{0\})=0$. These scaling limits we did establish in many cases in [7].

Let us now assume also mixing. Reformulate (15) as

[TABLE]

It is easy to see that the conditioning on $Y_{n}$ could be safely dropped, as the “initial” $n$th step gets forgotten.$\hfill\diamond$

Theorem 7**.**

Besides Assumption 3, assume also mixing. Then $S$ is recurrent.

In the next statement, the part that concerns the walk is a particular case of Theorem 7, provided that one knows that Assumption 3 holds. (For example, this is the case when $p_{n}=c/n$ for $n\geq n_{0}$ with some $n_{0}$ and $c>0$.)

Theorem 8** (Recurrence; lower critical case).**

Suppose that $p_{n}\leq c/n$ for $n\geq n_{0}$ with some $n_{0}$ and $c>0$, and at the same time $\sum_{n}p_{n}=\infty$. Then $S$ is recurrent, and in the $p_{n}=c/n$, $n\geq n_{0}$ case, the scaling limit (zigzag process) is recurrent as well.

Finally, we would like to summarize our scaling results in the diagram on Figure 3.

5. Examples and open problems

In this section, we compute the scaling $Z(\cdot)$ for a few examples in the cooling regime and the heating regime. We first give two concrete examples for the heating regime. Notice that the scaling function $Z(\cdot)$ is the generalized inverse of $v(m):=\sum_{n=1}^{m}4a_{n}^{2}p_{n}q_{n}$. Hence, it suffices to compute $v(m)$ in order to obtain the scaling of $S^{(n)}$.

Example 1** (Heating regime).**

Set $p_{n}=1-\frac{c}{2\,n^{\gamma}}$, for $n\geq n_{0}$, where $0<\gamma<1$. By Proposition 2 in Section 6.2, $\mathsf{Var}(S_{m})=(1+o(1))v(m)$, so we only need to compute $\mathsf{Var}(S_{m})$, and then $Z(\cdot)$ is asymptotically equivalent to the “inverse” of $\mathsf{Var}(S_{m})$. First, note that

[TABLE]

and

[TABLE]

Thus

[TABLE]

Let us now show that

[TABLE]

In the case when $i\leq n-n^{\frac{2\gamma+1}{3}}$ (note that $\gamma<\frac{2\gamma+1}{3}<1$), one has

[TABLE]

yielding

[TABLE]

For $i\geq n-n^{\frac{2\gamma+1}{3}}$, we have

[TABLE]

where

[TABLE]

with $1\leq j\leq n^{\frac{2\gamma+1}{3}}$. Define also

[TABLE]

Note that

[TABLE]

but

[TABLE]

Hence,

[TABLE]

implying

[TABLE]

At the same time,

[TABLE]

so

[TABLE]

Then, combining (17), (18), (19) and (20) we obtain (16). Hence

[TABLE]

*and as a result, $\mathsf{Var}(S_{n})=\frac{c}{2(1-\gamma)}n^{1-\gamma}+o(n^{1-\gamma})$.

Our conclusion is that $Z(x)\sim\lfloor(2x(1-\gamma)/c)^{\frac{1}{1-\gamma}}\rfloor$, that is, for the rescaled walk (11) the limit in (12) holds.*

Example 2** (Heating regime).**

Let $p_{n}=1-\frac{c}{n}$, $n\geq n_{0}$, for some $n_{0}\geq 1$. From Lemma 1 in Section 6, $\lim_{n\to\infty}a_{n}=1/2$, hence

[TABLE]

Thus, for the rescaled walk (11), the limit (12) holds, but now with $Z(x)\sim\lfloor e^{x/c}\rfloor$.

Next is an example for the cooling regime.

Example 3** (Subcritical case; cooling regime).**

If $p_{n}=\frac{c}{n^{\gamma}}$ for some $c>0$, $\gamma\in(0,1)$ and all $n\geq n_{0}$, then for the rescaled walk (11) the invariance principle (12) holds. Indeed, similarly to the previous examples, one only needs to know the order of $\mathsf{Var}(S_{m})$. By Theorem 2 of [7], $\mathsf{Var}(S_{n})=(1+o(1))\frac{n^{1+\gamma}}{c(1+\gamma)}$, so $Z(x)\sim\left\lfloor[c(1+\gamma)]^{\frac{1}{1+\gamma}}(x)^{\frac{1}{1+\gamma}}\right\rfloor$.

We finally present a few open problems.

Problem 1** (When $p_{n}$ is not comparable to $1/n$; different PPP’s).**

What can be said about the case when $\liminf_{n}np_{n}=0$ and $\limsup_{n}np_{n}=\infty$? A somewhat related question is whether the following is possible for some situations: the scaling limit is a piecewise deterministic process and the turning points form a PPP but the intensity is different from const$/x\,\mathrm{d}x$.

Problem 2** (Random temporal environment).**

One can also consider a random walk in a random temporal environment (as opposed to the more usual random spatial environment) as follows. Assume now that the $p_{n}$ are i.i.d. random and follow the same distribution (supported on $[a,b]$, for $0<a<b<1$) or a family of distributions on $[a,b]$. What can one say about the walk in the quenched or in the annealed case?

6. Proofs

The rest of the paper is organized as follows. After presenting two preparations sections on martingale approximation and on a piecewise deterministic process, we give the proofs of the main results.

6.1. Preparation I: The zigzag process

We now define a stochastic process, which we will relate to the critical case in the cooling regime.

Definition 3** (Zigzag process).**

Consider a Poisson point process (PPP) on $[0,\infty)$ with intensity measure $\frac{a}{x}\,\mathrm{d}x$ with $a>0$. Once the realization is fixed, the value of the process at $t\geq 0$ is obtained as follows. Starting with the segment containing $t$ and going backwards towards the origin, color the first, third, fifth, etc. segments between the points blue. The second, fourth, etc. will be colored red. Given this Poisson intensity, we will have infinitely many segments towards zero (and also towards infinity) almost surely.

Let $\lambda_{b}(t)$ and $\lambda_{r}(t)$ denote the Lebesgue measure of the union of blue, resp. red segments between [math] and $t$. Then we define the zigzag process $X$ by

[TABLE]

where $W$ is a random sign, that is $W=-1$ or $W=1$ with equal probabilities. See Figure 4.

It is easy to check directly that the law of the process is invariant under scaling both axes by the same number.

Remark 11** (One-dimensional marginals).**

It is more challenging to check directly that the one-dimensional marginals of the zigzag process are $\mathsf{Beta}(a,a)$, although this follows immediately from Theorem 4 along with the scaling limit result for the one-dimensional distributions in [7]. Edward Crane has shown us a nice direct proof for this fact though. The interested reader may enjoy trying to find such a proof him/herself.$\hfill\diamond$

6.2. Preparation II: Approximating the walk with a martingale

We are interested in the scaling limit of the random walk $S$, and in particular, whether we have a Donsker-style invariance principle, leading eventually to Brownian motion. Following the general principle that “it always helps to find a martingale”, in this section we investigate the following important, though still somewhat vague, question.

Question 1** (M).**

For a given sequence $\{p_{n}\}_{n\geq 1}$ is the walk $S$ “sufficiently close” to some martingale $M$?

After Question (M), the next question is of course:

Question 2** (INV.M).**

Is there an invariance principle for $M$?

Focusing now on Question (M) only, we recall from (1) and (2) the identity $e_{i,j}={\mathbb{E}}(Y_{j}\mid Y_{i})/Y_{i},$ and that for $1\leq i<j<k$, $e_{i,j}e_{j,k}=e_{i,k}.$ With the convention $e_{i,i}:={\mathbb{E}}(Y^{2}_{i})=1$, recall the definition of $a_{n}=\sum_{i=0}^{\infty}e_{n,n+i}$ from (7), assuming that the series is convergent (if $p_{n}\geq 1/2$ for large $n$, then it always is; see below). Then

[TABLE]

is a martingale. Indeed,

[TABLE]

which is identically zero, since $a_{n+1}e_{n,n+1}=a_{n}-1$, as

[TABLE]

Observe also that

[TABLE]

since $\mathsf{Var}(Y_{n})={\mathbb{E}}(Y_{n}^{2})=1$ for each $n$.

To understand what we mean by being sufficiently close to a martingale, recall that the rescaled walk ${S}^{n}$ is defined by

[TABLE]

where

[TABLE]

Since $|Y_{k}|=1$, if the $a_{n}$ are not too large, then it suffices to analyze the sequence of the rescaled martingales $M^{n}(t):=\frac{M_{Z(nt)}}{\sqrt{n}}$ instead of the sequence of the rescaled random walks. Thus, we have the answer in the affirmative to Question (M), provided that

1. (a)

$a_{n}$ is well-defined; 2. (b)

$a_{Z(n)}=o(\sqrt{n})$ (e.g. when $a_{n}$ remains bounded) as $n\to\infty$. (We dropped $t$ as it is just a constant.)

Proposition 2** (Equivalent conditions for (b)).**

Set

[TABLE]

Since the martingale differences $M_{i}-M_{i-1}$ are uncorrelated and centered, one has

[TABLE]

where $v_{n}$ is defined by (8) and $\mathsf{Var}(M_{i}-M_{i-1})$ is given by (6.2). Then the conditions

• (b.1)

$v_{n}\to\infty,a_{n}=o\left(\sqrt{v_{n}}\right)$;

• (b.2)

$\sigma_{n}\to\infty,a_{n}=o(\sigma_{n})$**

are equivalent; and when they are satisfied, $\sqrt{v_{n}}\sim\sigma_{n}$.

Of course, ${\sf(b.1)}\Leftrightarrow{\sf(b.2)}\Rightarrow{\sf(b)}$. Moreover, if $v_{n}\to\infty$, then the condition $a_{n}=o\left(\sqrt{v_{n}}\right)$ is in fact equivalent to (b). The proofs of these statements are provided later.$\hfill\diamond$

To answer Question (INV.M), we refer the invariance principle of Drogin.

Proposition D1972** (Part of Theorem 1 in [6]).**

Let $(X_{i})_{i\geq 1}$ be a sequence of square integrable random variables adapted to the filtration $(\mathcal{F}_{i})_{i\geq 1}$. Assume that they are martingale differences: ${\mathbb{E}}(X_{i}\mid\mathcal{F}_{i-1})=0$, and that $v_{m}:=\sum_{i=1}^{m}{\mathbb{E}}(X^{2}_{i}\mid\mathcal{F}_{i-1})\to\infty$ a.s. The processes $S$ and $S^{n},\ n\geq 1$ by $S(v_{m})=\sum_{i=1}^{m}X_{i},\ S(0):=0,$ and by $S^{n}(t):=S(nt)/\sqrt{n},\ t\geq 0$, using linear interpolation between integer times. Then the following are equivalent:

• (i)

If $\epsilon>0$, then

[TABLE]

• (ii)

As $n\to\infty$, the law of $S^{n}$ converges to the Wiener measure and

[TABLE]

Note that, in our setting, both $v_{m}$ and $Z(n)$ are deterministic. To summarize the discussion of Question(M) and (INV.M) above, in our setting, once the limit process is Brownian motion, we need to check the following conditions,

1. (a)

$a_{n}$ is well-defined; 2. (b)

$a_{Z(n)}=o(\sqrt{n})$, or equivalently, $a_{n}=o(\sigma_{n})$ (given $v_{n}\rightarrow\infty$), as $n\to\infty$. 3. (c)

$v_{n}\rightarrow\infty$ and (22) holds.

Here (a), (b) guarantee (M) affirmative , and (c) guarantees (IN.V) affirmative.

6.3. Some specific cases

The first two cases we are looking at are in the cooling regime, the last one is in the heating regime. We will use the conditions discussed in the last paragraph in Proposition 2.

6.3.1. Cooling, critical

Let $p_{n}=c/n$ for large $n$. If $c\geq 1/2$, then (a) fails to hold, because then $a_{n}=\infty$. Otherwise $a_{n}$ is of order $n^{1-2c}$, and $\sqrt{v_{n}}$ is of the same order, and thus (b) fails to hold. In both cases, the answer to Question (M) is negative.

6.3.2. Cooling, subcritical

Let $p_{n}\leq 1/2$ for all444We may assume this without the loss of generality, as the validity of the invariance principle does not depend on a finite number of terms. $n\geq 1$ and $p_{n}=c/n^{\gamma}$ for $n$ large, where $0<\gamma<1$. In this case the answers to (M) and to (INV.M) are both in the affirmative, and one can compute that $a_{n}=\frac{n^{\gamma}}{2c}(1+o(1))$.

6.3.3. Cooling, subcritical; the necessity of $\liminf\limits_{n\rightarrow\infty}\frac{p_{n}}{p_{n+1}}>0$

One can see that assumption (a) in Theorem 4(4) guarantees that

[TABLE]

The following example shows indeed the necessity of this bound, that is, by showing that the property that $a_{n}=o(v_{n})$ can break down if this $liminf$ is zero. Indeed, let

[TABLE]

Then

[TABLE]

and $\sum_{k}\frac{(k+1)!-k!}{k^{k}}<\infty$, so $a_{n}$ is well-defined. Moreover,

[TABLE]

At the same time,

[TABLE]

since for $k!<i\leq(k+1)!$,

[TABLE]

At the same time it is worth noting that with these $p_{i}$’s, the assumption (4)(a) in Theorem 4 is violated too, since for $i=(m+1)!$, one has

[TABLE]

while

[TABLE]

6.3.4. (Heating)

Let $p_{n}=1-q_{n}$ and $q_{n}\to 0$ but $\sum q_{n}=\infty$. We have

[TABLE]

and, since $1-2p_{n}=2q_{n}-1<0$ for large $n$, using the Leibniz criterion, along with the assumption that $\sum q_{n}=\infty$, it follows that $a_{n}$ is well defined. The validity of the martingale approximation follows from the fact that $a_{n}\leq 1$ but $v_{n}\to\infty$; see the proof of Theorem 3.

6.4. Proof of Theorem 1

Clearly, if $p_{i}=1/2$ for some $i\in{\mathbb{N}}$ then the process “gets symmetrized” from time $i$ on (and $\rho=0$), and the statement is trivial. We will thus assume in the rest of the proof that $p_{i}\neq 1/2,\ \forall i\in{\mathbb{N}}$.

Furthermore, we will handle the conditional probability ${\mathbb{P}}(\cdot\mid Y_{1}=1)$ only, the argument for ${\mathbb{P}}(\cdot\mid Y_{1}=-1)$ is similar. In terms of the $W_{i}$, one has $Y_{n}:=(-1)^{\sum_{i=1}^{n}W_{i}},$ where $W_{1},W_{2},W_{3},...$ are independent Bernoulli variables with parameters $p_{1},p_{2},p_{3},...$, respectively and we will handle the $p_{1}=0$ (i.e. $W_{1}\equiv 0$) case. In particular, $\prod_{i=1}^{n}(1-2p_{i})=\prod_{i=2}^{n}(1-2p_{i}).$

Let $x_{n}:={\mathbb{P}}(Y_{n}=1)$. We have the recursion

[TABLE]

and the substitution $y_{n}:=x_{n}-1/2$ yields $y_{n+1}=(1-2p_{n})y_{n}$ with $y_{1}=1/2$. Hence,

[TABLE]

Case 1: $N=\infty$. We have to prove that $x_{n}$ converges to 1/2 or has no limit, according to whether $\min(p_{i},q_{i})$ is summable or not.

Let $N_{n}:=\mathrm{card}\{i\leq n:\ p_{i}>1/2\};$ then $\lim_{n\to\infty}N_{n}=\infty.$ Since

[TABLE]

we have

[TABLE]

Given that $\lim_{n}(-1)^{N_{n}}$ does not exist, there are two cases:

(i) the right-hand side converges because the product (without the $(-1)^{N_{n}}$ factor) converges to zero and mixing holds ($\sum_{i=1}^{\infty}\min(p_{i},q_{i})=\infty)$, in which case $\lim_{n}y_{n}=0$ and $\lim_{n}x_{n}=1/2$.

(ii) the right-hand side has no limit and mixing does not hold in which case $y_{n}$ (hence $x_{n}$) has no limit.

Case 2: $N<\infty$. Let us assume first that $N=0$, that is, $p_{i}<1/2,i\geq 1$. If $\sum_{i=1}^{\infty}p_{i}=\infty$ then in (23) we have $\prod_{i=1}^{n}(1-2p_{i})\downarrow 0$, implying $\lim_{n}y_{n}=0$ and $\lim_{n}x_{n}=1/2$. If $\sum_{i=1}^{\infty}p_{i}<\infty$, then $\rho=\prod_{i=1}^{\infty}(1-2p_{i})>0$ and $\lim_{n}y_{n}=\frac{1}{2}\rho$, that is, $\lim_{n}x_{n}=\frac{1}{2}(1+\rho).$

In the general case, for large $i$, $\min(p_{i},q_{i})=p_{i}<1/2$, and mixing is tantamount to $\sum_{i=1}^{\infty}p_{i}=\infty$. The proof is very similar as before, using the fact that the product has positive terms for large enough indices.

6.5. Proof of Theorem 2

The martingale method is applicable in this case too. Indeed, direct computation gives $a_{n}=\frac{1}{2p}$, $\forall n$ and $v_{n}=\overset{n}{\sum}4a_{i}^{2}p_{i}q_{i}=\frac{1-p}{p}n.$ Hence $a_{n}=o(\sqrt{v_{n}})$, $a^{2}_{i}\xi^{2}_{i}$ are bounded, so (22) holds, and thus we can apply Proposition D1972, yielding the answer to (INV.M) in the affirmative.∎

6.6. Proof of Theorem 3

First we will prove the statement under the more restrictive assumption that

[TABLE]

and then we upgrade it for showing the statement under the condition appearing in the theorem.

6.6.1. STEP 1

We start with a simple lemma.

Lemma 1**.**

Assume that for the non-negative sequence $(q_{n})$, such that

• •

$q_{n}\to 0$,

• •

$\sum_{n}q_{n}=\infty$,

• •

the condition (24) holds.

Then $\liminf_{n\to\infty}a_{n}>0$, where the $a_{n}$ are defined by (7). Moreover, if $\displaystyle\lim_{n\to\infty}\frac{q_{n+1}}{q_{n}}=1,$ then $\displaystyle\lim_{n\to\infty}a_{n}=1/2$.

Remark 12**.**

The condition that $q_{n}\rightarrow 0$ is really necessary in Lemma 1. Indeed, fix $c_{1},c_{2}>0$, $c_{1}\neq c_{2}$ and let $q_{n}=c_{1}/n$ if $n$ is odd and $q_{n}=c_{2}/n$ if $n$ is even. Then $q_{n+1}/q_{n}\not\to 1$, though (24) still holds. In this case $a_{n}\not\to 1/2$, rather (as it is not hard to show) $\displaystyle\lim_{k\to\infty}a_{2k}=\frac{c_{1}}{c_{1}+c_{2}}\neq\frac{c_{2}}{c_{1}+c_{2}}=\lim_{k\to\infty}a_{2k+1}$. $\hfill\diamond$

Proof.

Fix some $n$, and for $m\geq n$ let

[TABLE]

and note that $w_{m}\downarrow 0$ as $m\to\infty$ due to $\sum q_{i}=\infty$. Then

[TABLE]

Now take any finite $c>\limsup_{n}q_{n+1}/q_{n}$, and assume that $n$ is so large that the quantity $q_{\ell+1}/q_{\ell}<c$ for all $\ell\geq n$. Then

[TABLE]

As a result,

[TABLE]

where the telescopic sum converges due to the fact that $w_{m}\to 0$. Since $c>\limsup_{n}q_{n+1}/q_{n}$ is arbitrary, we can even conclude that

[TABLE]

To prove the second part of the claim, observe that from (26) we already have $\liminf_{n}a_{n}\geq 1/2$. To show the counterpart of this, fix an arbitrary $\varepsilon>0$ and let $n$ be so large that $\frac{q_{n+2k+2}}{q_{n+2k+1}}\geq 1-\varepsilon/2$ and $q_{n+2k+2}\leq\varepsilon/4$ for all $k\geq 0$. Then

[TABLE]

given that $\displaystyle\frac{q_{n+2k+2}}{q_{n+2k+1}}\to 1$ and $q_{n+2k+2}\to 0$. Hence (see (25))

[TABLE]

and

[TABLE]

Since $\varepsilon>0$ is arbitrary, we conclude that $\limsup_{n}a_{n}\leq 1/2$, which completes the proof. ∎

We now continue the proof of the theorem under the assumption that (24) holds.

Proof of Theorem 3 (a):

Noting that all conditions at the end of Section 6.2 to Question(M),(INV.M) are in the affirmative (as $a_{n}$ is well-defined and stays bounded), except (22). Since in our case $X_{i}=a_{i}\xi_{i}Y_{i-1}$ and $|Y_{i}|=1$, what we need is to show that

[TABLE]

(Note that $Z(n)$ in our case it is deterministic, and so is $v_{m}$.) Since

[TABLE]

as $a_{i}$ is a Leibniz series, all but finitely many terms in the sum in (27) are zero, proving (27). We conclude that (22) holds.

Next, a direct computation shows that $v_{m}=4\sum_{i=1}^{m}a_{i}^{2}p_{i}q_{i}.$ Then

[TABLE]

follows from Lemma 1 and from the assumptions $p_{n}\to 1$ and $\sum q_{n}=\infty$. The proof of (a) is thus complete.

Proof of Theorem 3 (b): First, we prove that $\Lambda_{n}^{2}=\sum_{i=1}^{n}a_{i}^{2}\xi_{i}^{2}\to\infty$. Recall that $\xi_{i}=(-1)^{W_{i}}-E(-1)^{W_{i}}=2p_{i}-1+(-1)^{W_{i}}$ satisfies ${\mathbb{E}}\xi_{i}^{2}=\mathsf{Var}((-1)^{W_{i}})=4p_{i}q_{i}$. Let also $U_{i}:=a_{i}^{2}\xi_{i}^{2}\in[0,4]$.

Since the $W_{i}$ are independent, so are the $\xi_{i}$, and hence, for $\Lambda_{n}^{2}$, the Three Series Theorem applies: the non-negative series $\sum_{i}U_{i}$ diverges if for some $A>0$, $\sum_{i}{\mathbb{E}}[U_{i};|U_{i}|\leq A]$ diverges. But for $A>4$,

[TABLE]

as $a_{i}$ is bounded away from zero, $p_{i}\to 1$ and $\sum q_{i}=\infty.$

Alternatively, let $\epsilon>0$. Then $p_{i}\to 1$ and $\sum q_{i}=\infty$ along with the second Borel-Cantelli lemma guarantee that $\xi_{i}=2p_{i}-1+(-1)^{W_{i}}\geq 2-\epsilon$ for infinitely many $i$’s almost surely. We are done because the $a_{i}$ are bounded away from zero.

For the second statement, by using Chebyshev’s inequality, it is enough to show that

[TABLE]

Since $a_{n},p_{n},q_{n}\in[0,1]$,

[TABLE]

Moreover, for large $n$’s,

[TABLE]

for some $c>0$, since $\displaystyle\liminf_{i\to\infty}a_{i}>0$ by Lemma 1 and $p_{i}\to 1$. Given that $\sum_{i=1}^{n}q_{i}\to\infty$, (29) and (30) together yield (28), thus completing the proof of the statement. $\hfill\qed$

6.6.2. STEP 2

We now upgrade the result obtained in STEP 1, by dropping the restriction that (24) holds. We need the following

Lemma 2** (Comparison with “regular” sequences).**

Let $0\leq q_{n}\leq 1/2,\ n\geq 1$.

(i) Assume that there exists a sequence $q^{*}_{k}\to 0$ such that $q^{*}_{n}$ is not summable, regular, in the sense that (24) holds, and $q_{n}\leq q^{*}_{n}$ for even $n$, while $q_{n}\geq q^{*}_{n}$ for odd $n$. Then $\displaystyle\liminf_{k\to\infty}a_{2k}>0$.

(ii) Assume that there exists a sequence $\tilde{q}_{k}\to 0$ such that $\tilde{q}_{n}$ is not summable, regular in the sense that (24) holds, and $q_{n}\leq\tilde{q}_{n}$ for odd $n$, while $q_{n}\geq\tilde{q}_{n}$ for even $n$. Then $\displaystyle\liminf_{k\to\infty}a_{2k+1}>0$.

Proof of Lemma 2.

Since $0\leq q_{n}\leq 1/2$ for $n\geq 1$, it is easy to check the following (for example by observing that for $k>n$, the coefficients of $q_{k}$ in $a_{n}$ form a Leibniz series as well):

• •

Let $n=2k$. Then $a_{n}$ is decreasing555The terms increasing and decreasing are not used in the strict sense. in all $q_{i}$ for which $i$ is even and increasing in all $q_{i}$ for which $i$ is odd.

• •

Let $n=2k+1$. Then $a_{n}$ is increasing in all $q_{i}$ for which $i$ is even and decreasing in all $q_{i}$ for which $i$ is odd.

Turning to the proof of (i) (a similar proof works for (ii), which we omit), note that, because of its monotonicity and non-summability (use $I=\infty$ and $q^{*}_{2k}\geq q_{2k}$), STEP 1 yields that $(q^{*}_{n})$ is such that $\liminf a_{n}>0$, and in particular, $\liminf_{k}a_{2k}>0$. Hence, by the first bullet point above, $\liminf_{k}a_{2k}>0$ also for $(q_{n})$, proving (i). ∎

Proof of Theorem 3.

First, without the loss of generality, we assume that $m_{0}=1$ (changing a finite number of terms does not change the validity of the invariance principle). Similarly, we may and will assume that $q_{n}\leq 1/2$ for all $n\geq 1$, as we assume anyway that $q_{n}\to 0$.

We only need that $v_{n}=4\sum_{i=1}^{n}a_{i}^{2}\,p_{i}\,q_{i}\to\infty,$ what is left is very similar to STEP 1. This will follow from $p_{n}\to 1$ and Assumption 2, provided that either $\liminf_{k}a_{2k}>0$ or $\liminf_{k}a_{2k+1}>0$. By Lemma 2, it is sufficient to construct either a sequence $(q_{n}^{*})$ or a sequence $(\tilde{q}_{n})$ satisfying the properties in the lemma. These sequences will be automatically divergent, given Assumption 2 and that $(q^{*}_{n})$ resp. $(\tilde{q}_{n})$ dominate $(q_{n})$ for even resp. odd $n$’s. Now, assume for example (9) (assuming (10) leads to a similar argument). Define

[TABLE]

Then $(\tilde{q}_{n})$ is regular because $\frac{\tilde{q}_{n+1}}{\tilde{q}_{n}}\leq\max\{C^{-1},C\}$ for all $n\geq 1$, and trivially $q_{2m}\geq\tilde{q}_{2m}$ and $\tilde{q}_{2m+1}\geq q_{2m+1}$. Hence, $\liminf_{k}a_{2k+1}>0$ by Lemma 2(ii). ∎

Remark 13** (One of the two subsequences can be arbitrary).**

Chose an arbitrary “odd” subsequence, satisfying the conditions that it tends to zero and yet not summable. Then take a sufficiently large “even” subsequence that dominates it in the sense of (9), but still tends to zero (for example, let $q_{2n}:=1/\sqrt{2n}$ and $q_{2n+1}:=1/(2n+1)$). Then (9) holds, while the condition $\limsup_{n}q_{n+1}/q_{n}<\infty$ (cf. (24) in the proof) fails to hold, as $\lim_{n}q_{2n}/q_{2n+1}=\infty$.

By the same token, one can first chose an arbitrary non-summable “even” sequence, with the terms tending to zero and then a dominating “odd” one.$\hfill\diamond$

6.7. Proof of Proposition 2

Recall that $S_{n}=M_{n}+(1-a_{n})Y_{n}$, hence

[TABLE]

where, by Cauchy-Schwarz, $|\mathsf{Cov}(M_{n},Y_{n})|\leq\sqrt{{\mathbb{E}}(M_{n}^{2})}=\sqrt{v_{n}}$, so

[TABLE]

where $|A_{n}|\leq 1$. Then

[TABLE]

if $v_{n}\to\infty$ and $a_{n}=o(\sqrt{v_{n}})$ as $n\to\infty$, hence $\sqrt{v_{n}}\sim\sigma_{n}$ follows.

Similarly, we have

[TABLE]

Using the shorthands $\displaystyle w_{n}:=\frac{\sqrt{v_{n}}}{\sigma}_{n}$ and $\displaystyle b_{n}:=\frac{1-a_{n}}{\sigma_{n}}$, one obtains the quadratic equation $w_{n}^{2}+b_{n}A_{n}w_{n}+b_{n}^{2}-1=0,$ where $b_{n}\to 0$. Hence

[TABLE]

but of course $w_{n}\geq 0$. Therefore, $b_{n}\to 0$ implies that $w_{n}\to 1$, that is, $\sqrt{v_{n}}\sim\sigma_{n}$. This is clearly the case when $\sigma_{n}\to\infty$ and $a_{n}=o(\sigma_{n})$ as $n\to\infty.\hfill\qed$

6.8. Proof of Theorem 4 – strongly critical case

First, it is easy to see that if $X$ is a symmetric random variable, concentrated on $[-t,t]$, then $\mathsf{Var}(X)\leq t^{2}$, with equality if and only if the law of $X$ is $\frac{1}{2}(\delta_{-t}+\delta_{t})$.

Now assume that $\lim_{n\to\infty}np_{n}=0$. By a well-known criterion for tightness (see Theorem 4.10 in [9]), the laws of the $S^{(n)}$ are tight on $C([0,T])$ if besides the condition $\lim_{\eta\to+\infty}\sup_{n\geq 1}{\mathbb{P}}(S^{(n)}(0)>\eta)=0$, one also has

[TABLE]

Since $S^{(n)}(0)=0,\ n\geq 1$, the first condition clearly holds. The second one is satisfied by the uniform Lipschitz-ness: $|S^{(n)}(t)-S^{(n)}(s)|\leq|t-s|,\ n\geq 1$.

Given tightness on $C([0,T])$, it is sufficient to show that the limit at time $t>0$ is $\frac{1}{2}(\delta_{-t}+\delta_{t})$, that is, it satisfies $\mathsf{Var}(X)\geq t^{2}$. Indeed, the only continuous functions $f$ on $(0,T)$ satisfying $|f(t)|=t$ are $f(t)=t$ and $f(t)=-t$. For simplicity we will work with $t=1$ (otherwise use a simple scaling), that is we will show that every partial limit at time $t=1$ is such that its variance is at least one.

To achieve this, fix $N\geq 1$ and recall from [7] (see the two displayed formulae right before Theorem 3 there) that

[TABLE]

This quantity is monotone decreasing in all $p_{n}$’s as long as they are all less or equal than $1/2$, because the same holds for each fixed $e_{i,j}$. Fix $\epsilon>0$ and let $N=N(\epsilon)$ be such that $\epsilon/N\leq 1/2$ and that also $\epsilon/n>p_{n}$ holds for all $n>N$. Define $\hat{p}_{n}$ so that it coincides with $p_{n}$ for $n\leq N$ and $\hat{p}_{n}=a/n$ for $n>N$. By monotonicity,

[TABLE]

where $\hat{S}$ is the walk for the sequence $(\hat{p}_{n})$.

In [7] it was shown that

[TABLE]

Since $\epsilon>0$ was arbitrary,

[TABLE]

Now, if $S_{n_{j}}/n_{j}\to X$ in law, then

[TABLE]

because ${\mathbb{E}}(S_{n})=0$ and the variables are all supported in $[-1,1]$ (and so the test function $f(x)=x^{2}$ is admissible). From the last two displayed formula, we have that $\mathsf{Var}(X)\geq 1$ and we are done. ∎

6.9. Proof of Theorem 4 – supercritical case

By the Borel-Cantelli Lemma, for almost every $\omega$, either $S_{n}(\omega)=1$ for all large $n$ or $S_{n}(\omega)=-1$ for all large $n$. As $n\to\infty$, in the first case the path converges uniformly to a straight line with slope $1$; in the second case it converges uniformly to a straight line with slope $-1$. ∎

6.10. Proof of Theorem 4 – critical case

Fix $T>0$, denote by ${{\mathcal{M}}}_{T}$ the set of all locally finite point measures on the interval $(0,T]$, and denote by $N^{(n)}=N^{(n,T)}$ the laws of the point processes induced by the turns of the walk $S^{(n)}$ on the time interval $(0,T]$.

Let $t\in(0,T)$; we assign a continuous (zigzagged) path that increases at666I.e. it increases on $[t,t+\epsilon]$ for some small $\epsilon>0$. $t$ to each point measure.

Definition 4** (Assigning paths).**

Define the map $\Phi_{t}:{{\mathcal{M}}}_{T}\rightarrow C[0,T]$ as follows.

• •

First, label the (countably many) atoms on $(0,t]$ from right to left as $a_{1},a_{2},...,$ i.e., the closest one on the left to $t$ as $a_{1}$, the second closest as $a_{2}$, etc., and note that $t=a_{1}$ is possible; also label the atoms on $(t,T]$, from the closest to the furthest as $b_{1},b_{2}$,…;

• •

assign “$+$” sign to the intervals (the union of which is denoted by $S^{+}_{t}$)

[TABLE]

• •

assign “$-$” sign to the intervals (the union of which is denoted by $S^{-}_{t}$)

[TABLE]

Let $\mu\in{{\mathcal{M}}}_{T}$. For $0<r\leq T$, define

[TABLE]

where $L$ is the Lebesgue measure on the real line. Then $\Phi_{t}(\mu)(\cdot)$ is well-defined and continuous on $[0,T]$. Intuitively, it describes the difference between the total length of increasing parts and the total length of decreasing parts, assuming increase at $t$. Clearly,

[TABLE]

Remark 14**.**

The case $t=0$ is excluded, i.e. one cannot set the path $\Phi_{t}(\mu)(\cdot)$ to first increase at $t=0$, as our point measures may not be locally finite around [math]. For instance, we will show that $N_{n}$ converges to a limiting Poisson Point Process (PPP) $N$, and this $N$ explodes at [math]. However, for $t>0$, $\Phi_{t}(r)\rightarrow 0$, as $r\rightarrow 0$.

We now turn to the case of a PPP with intensity $\frac{c}{x}$ (we replaced the constant $a$ of Theorem 4 by $c$ in the proof to avoid confusion).

Proposition 3**.**

(Turning points $\to$ PPP with intensity $\frac{c}{x}$) Given $0<a<b<\infty$, $c>0$, set $p_{n}=\frac{c}{n}\wedge 1,$ and denote the number of turns from step $\lceil an\rceil+1$ to step $\lceil bn\rceil$ by $N^{(n)}((a,b])$. Denoting $\mu_{c;a,b}:=c\ln(b/a)=\int_{a}^{b}\frac{c}{x}\,\mathrm{d}x,$ one has

• (i)

for $k\geq 0,\ 0<a<b$, as $n\to\infty$,

[TABLE]

• (ii)

given $0<t_{1}<t_{2}<...<t_{l}<\infty$, the random variables

[TABLE]

are independent (independent increments), and

[TABLE]

where ${\sf Poiss}(c)={\sf Poiss}((0,\infty),\frac{c}{x}\,\mathrm{d}x)$ is the law of the PPP with intensity $\frac{c}{x}\,\mathrm{d}x$ on $(0,\infty)$.

Proof.

(of Proposition 3:)

STRATEGY OF THE PROOF: We first prove part (i). Once that is done, since the turns from step $\lceil t_{i}n\rceil+1$ to step $\lceil t_{j}n\rceil$ and from $\lceil t_{l}n\rceil+1$ to $\lceil t_{j}n\rceil$ are independent for any $0<t_{i}<t_{j}\leq t_{l}<t_{r}<\infty$, part (ii) will immediately follow.

Regarding part (i), we only need to prove equation (33), and then (34) will easily follow. In fact we only give here the proof (in three steps) of (33) for $a,b$ integers, i.e., $\lceil an\rceil=an$, $\lceil bn\rceil=bn$, for $n$ large enough; the proof for general $0<a<b$ can then be easily adjusted.

STEP 1: Given $c>0$, and $n$ large enough, define

[TABLE]

We now provide an estimate for $\Pi_{c,n}$, namely

[TABLE]

Indeed,

[TABLE]

The exponent tends to $(1-c)\ln\frac{b}{a}$, and so $\lim_{n\rightarrow\infty}\Pi_{c,n}=\exp(-c\ln(b/a))=\exp(-\mu_{c;a,b})$. Indeed,

[TABLE]

hence

[TABLE]

where $\displaystyle\lim_{n\rightarrow\infty}\overset{bn-an}{\sum_{i=1}}\frac{1}{an+i-c}=\lim_{n\rightarrow\infty}\overset{bn-an}{\sum_{i=1}}\frac{1}{an+i-1}=\ln(b/a)$. leading to (35).

STEP 2: we now estimate

[TABLE]

Note that the turning step can happen at step $an+i$, for $i=1,2,...,bn-an$, with corresponding probabilities $(\frac{an+1-c}{an+1}\cdot\frac{an+2-c}{an+2}\cdot...\cdot\frac{bn-c}{bn})\cdot\frac{c}{an+i}=\Pi_{c,n}\cdot\frac{c}{an+i},$ $i=0,1,...,bn-an-1$. Thus,

[TABLE]

where

[TABLE]

Since

[TABLE]

one has

[TABLE]

and then (35), (36) give

[TABLE]

STEP 3: we verify (33) using induction, and so we assume that

[TABLE]

and show that $k$ can be replaced by $k+1$ as well. On the the event $\{N^{(n)}((a,b])=k\}$, there should be $k$ turns from step $an+1$ to step $bn+1$, say the turns happen at $an+i_{1},an+i_{2},...,an+i_{k}$, where $i_{1},...,i_{k}$ is an increasing sequence taking values in $\{0,1,...,bn-an-1\}$. Similarly to the $k=1$ case, the probability for this to happen is

[TABLE]

Then ${\mathbb{P}}(N^{(n)}((a,b])=k)$ is the sum of all such terms, i.e.,

[TABLE]

By assumption (37) and the estimate (35), we have

[TABLE]

Similarly,

[TABLE]

where the sequence $i_{1}<i_{2}<...<i_{k}<i_{k+1}$ takes values in $\{0,1,...,bn-an-1\}$. Now

[TABLE]

for $j=0,1,...,bn-an-1$. Now consider the sum

[TABLE]

In each sum on the right-hand side, there are two different kinds of terms: terms of the type

[TABLE]

where $i_{m},m=1,2,...,k+1$ are all different (no repetitions), and terms of the type

[TABLE]

where $i_{m},m=1,2,...,k$ are all different (one repetition). We then rearrange the right-hand side: sum the “non-repeating” terms as one group, denoted by $I$; sum the “once repeating” ones where the term $\frac{c}{an+j}$ is the one repeated by $I_{j}$, $j=0,1,...,bn-an-1$, and we estimate $I$, $I_{j}$ separately.

[TABLE]

since each product $\frac{c}{an+i_{1}}\frac{c}{an+i_{2}}...\frac{c}{an+i_{k}}\frac{c}{an+i_{k+1}}$ appears $k+1$ times in sum $I$. Further,

[TABLE]

hence

[TABLE]

Now, by estimations of $I,I_{j}$s, (39) is written as

[TABLE]

and we conclude that (37) holds with $k$ replaced by $k+1$. This completes the proof of Step 3, and of the proposition altogether. ∎

Note: We use the endpoints $\lceil an\rceil+1,\lceil bn\rceil$ because $\frac{\lceil an\rceil+1,}{n}\rightarrow a^{+},\frac{\lceil bn\rceil}{n}\rightarrow b$, so the above limit represents the number of turns in (a,b] in the scaling limit.

In the sequel we will consider measures equipped with both the weak and the vague topologies. When we consider laws on $C([0,T],\|.\|_{[0,T]})$ where $\|.\|_{[0,T]}$ denotes supremum norm, weak convergence is denoted by $\overset{w}{\rightarrow}$. When one uses vague topology for measures and random measures are considered, $\mathcal{X}_{n}\overset{vd}{\rightarrow}\mathcal{X}$ will be used for convergence in distribution.

Proposition 4** (Convergence for point measures and paths).**

Let $0<t<T$. Then

• (i)

As $n\to\infty$, $N^{(n)}\overset{vd}{\rightarrow}{\sf Poiss}(c)$ on $\mathcal{M}_{T}$ equipped with the vague topology, where ${\sf Poiss}(c)$ is the PPP on $(0,T]$ with intensity $\frac{c}{x}\,\mathrm{d}x$.

• (ii)

$\Phi_{t}:{{\mathcal{M}}}_{T}\to C[0,T]$* is a continuous and uniformly bounded functional, when the former space is equipped with the vague topology, and the latter with the supremum norm $\|.\|_{[0,T]}$.*

• (iii)

As $n\to\infty$, $\Phi_{t}(N^{(n)})\overset{w}{\rightarrow}\Phi_{t}({\sf Poiss}(c))$ on $C([0,T],\|\cdot\|)$.

Proof.

(of Proposition 4:) (i) In order to use Lemma 4 of the Appendix, one needs to define a new metric on $(0,T]$ by $\rho(x,y):=|1/x-1/y|$. Then $\Delta:=((0,T],\rho)$ is a complete separable metric space; notice that $(0,\epsilon]$ is not bounded under $\rho$. Setting ${\mathcal{I}}:=\{(a,b],0<a<b\leq T\}$, it is obvious that ${\mathcal{I}}$ is a semi-ring of bounded Borel sets in $\Delta$, and $\mu(\partial(a,b])=\mu(\{a\}\cup\{b\})=0$, hence ${\mathcal{I}}\subset\widehat{\Delta}_{E{\sf Poiss}(c)}$, where $\widehat{\Delta}_{E{\sf Poiss}(c)}$ is the class of all bounded sets $A\subset\Delta$ with $E\sf Poiss(c)(\partial A)=0$. Then by Lemma 4 of the Appendix, we only need to prove, $N^{(n)}(f)\overset{d}{\rightarrow}{\sf Poiss}(c)(f)$, for any $f\in\hat{{\mathcal{I}}}_{+}$, i.e., any $f$ with $f=\overset{k}{\sum_{i=1}}c_{i}1_{(a_{i},b_{i}]}$, where $(a_{i},b_{i}]\in{\mathcal{I}}$ and $a_{i}>0$. Note that $f$ is undefined on $(0,\min a_{i}]$. Then $N^{(n)}\overset{vd}{\rightarrow}{\sf Poiss}(c)$ on $(0,T]$ follows from $N^{(n)}(1_{(a,b]})\overset{d}{\rightarrow}{\sf Poiss}(c)(1_{(a,b]})$ for $0<a<b\leq T,$ which in turn, follows from Proposition 3.

(ii) Assume that $\mu_{n},\mu\in{{\mathcal{M}}}_{T}$, and $\mu_{n}\overset{v}{\rightarrow}\mu$. Then for any $\varepsilon>0$ small enough, $\mu_{n}\overset{v}{\rightarrow}\mu\ \text{on}\ [\varepsilon,T].$ Since $\mu$ is locally finite, it has finitely many atoms on $[\varepsilon,T]$, say $\varepsilon\leq x_{1}\leq...\leq x_{l}\leq T$. It easy to see that $\exists\ n_{0}$ such that for any $n\geq n_{0}$, $\mu_{n}$ also has $l$ atoms there. Moreover, $\exists\ K=K(\varepsilon,l)\geq n_{0}$, such that, for any $n\geq K$,

[TABLE]

By (32), $|\Phi_{t}(\mu_{n})(\varepsilon)-\Phi_{t}(\mu)(\varepsilon)|\leq 2\varepsilon,$ and by definition (31), we have

[TABLE]

Hence, $\Phi_{t}$ is continuous. Moreover, $\|\Phi_{t}(\mu)\|_{[0,T]}\leq T$, so $\Phi_{t}$ is also uniformly bounded.

Finally, (iii) immediately follows from (i), (ii) and Lemma 4, completing the proof of Proposition 4. ∎

Having Proposition 4 at our disposal, it is now easy to prove that the processes $S^{(n)}$ in the statement of the theorem converge in law to the zigzag process, by checking the convergence of the finite dimensional distributions, and then tightness.

Convergence of fidi’s: Given $0<t_{1}<t_{2}<...<t_{k}$, to check that the law of $(S^{(n)}_{t_{1}},...,S^{(n)}_{t_{k}})$ converges as $n\to\infty$, let $A_{1},A_{2},...,A_{k}\subset{\mathbb{R}}$ be Borel sets, and denote

[TABLE]

[TABLE]

[TABLE]

Moreover, $\{S^{(n)}_{s}=+\}$ ($\{S^{(n)}_{s}=-\}$) will denote the event that the zigzag path is increasing (decreasing) at $s^{+}$, by which we mean that there exists a small interval $[s,s+\epsilon]$ such that $S^{(n)}$ has slope $1$ ($-1$) on $(s,s+\epsilon)$. Then

[TABLE]

where, by symmetry, ${\mathbb{P}}\left(S^{(n)}(t_{1})=+\right)={\mathbb{P}}\left(S^{(n)}(t_{1})=-\right)=\frac{1}{2},$ and

[TABLE]

By Proposition 4, $\Phi_{t_{1}}(N^{(n)})\overset{w}{\to}\Phi_{t_{1}}({\sf Poiss}(c))$ on $C[0,t_{k}]$; composing with projections yields

[TABLE]

Similarly,

[TABLE]

tends to ${\mathbb{P}}\left(\Phi_{t_{1}}({\sf Poiss}(c))_{\vec{t}}\in-\vec{A}\right)$ as $n\rightarrow\infty$, hence

[TABLE]

Tightness: We repeat the argument in the proof of Theorem 4 here. Use that $\lim_{\eta\to+\infty}\sup_{n\geq 1}{\mathbb{P}}(S^{(n)}(0)>\eta)=0$ together with

[TABLE]

are sufficient for tightness on $C([0,T])$. These are indeed satisfied because $S^{(n)}(0)=0,\ n\geq 1$ and because of the uniform Lipschitz-ness. This completes the proof of the theorem in the critical case. $\hfill\qed$

Note: One can use any $\Phi_{s},\,s>0$, instead of $\Phi_{t_{1}}$ (again, $s=0$ is excluded), without causing too much change; then

[TABLE]

Remark 15**.**

*We can also generalize the condition $A_{n}:=np_{n}=c$ a bit, namely, one can mimic the proof in Proposition 3 to show the following.

If the $A_{n}$ are stable in the sense that*

[TABLE]

then the turns $N^{(n)}$ tend to a PPP with intensity $\lambda(x)=\frac{c}{x}\,\mathrm{d}x$. Hence the law of $S^{(n)}$ tends to that of the same zigzag process, i.e., we have the same scaling limit. This includes, for example, the following cases:

• •

$A_{n}\equiv c$* for all large $n$;*

• •

$\lim\limits_{n\rightarrow\infty}A_{n}=c$;

• •

$A_{n}$* is periodic with average period $c$,*

where $c$ is a positive constant.$\hfill\diamond$

6.11. Proof of Theorem 4 – subcritical case

Following the martingale approximation approach and again to prove all conditions at the end of Section 6.2, we will prove the result in the following steps:

• (i)

The $a_{n}\geq 1$ are well-defined; furthermore $a_{n}=o(n)$;

• (ii)

$\lim_{m\to\infty}v_{m}=\infty$;

• (iii)

$a^{2}_{n}=o(v_{n})$;

• (iv)

As $n\to\infty$,

[TABLE]

Step (i). Since $1-x\leq e^{-x}$, $x>0$, and $A_{n}$ is a monotone increasing sequence, we have

[TABLE]

since $\sum_{j=a}^{b}\frac{1}{j}\geq\int_{a}^{b}\frac{{\rm d}x}{x}$ for all integers $a,b$ with $b>a\geq 2$. So

[TABLE]

for large $n$. Since $A_{n+1}\to\infty$, we have $a_{n}=o(n)$.

Step (ii). There exists an $N\geq 1$ such that for all $n\geq N$ we have $p_{n}\geq\frac{1}{n}$ and $q_{n}\geq\frac{1}{4}$. Also, $a_{n}\geq 1$. Hence, for $m$ large enough, $v_{m}=\sum_{n=1}^{m}4a_{n}^{2}p_{n}q_{n}\geq\overset{m}{\underset{n=N}{\sum}}4p_{n}q_{n}\geq\sum_{n=N}^{m}\frac{1}{n}\to\infty$ as $m\rightarrow\infty$.

Step (iii). Since $p_{n}\downarrow 0$, one has

[TABLE]

From its definition it follows that $v_{n}$ is monotone; we also know that $v_{n}\to\infty$. Hence, by the Stolz–Cesàro Theorem777This is the discrete version of L’Hospital’s rule — see e.g. Problem 70 in [10]., we have

[TABLE]

since $4p_{n}q_{n}a_{n}^{2}=(2p_{n}a_{n})\cdot q_{n}\cdot 2a_{n}$, and $q_{n}\to 1$, $p_{n}a_{n}\geq 1/2$, $a_{n-1}\leq a_{n}$. Next,

[TABLE]

We have (e.g. by integrating by parts)

[TABLE]

From the monotonicity of $p_{n}$ and $np_{n}$, we get $p_{n}\geq p_{n+i}$ and $\frac{p_{n+i}}{p_{n}}\geq\frac{n}{n+i}.$ Then, from (6.11) and (43), it follows that888The last equality is elementary: $\left(1+\frac{1}{n-1}\right)^{2A_{n+1}}=(1+\frac{1}{n-1})^{(n-1)2p_{n+1}}\cdot(1+\frac{1}{n-1})^{4p_{n+1}}=O(e^{2p_{n+1}})=O(1+2p_{n+1})=O(1+p_{n})$.

[TABLE]

Hence

[TABLE]

so the righthand side of (42) tends to zero.

Step (iv). We show how, in our case, (iii) implies (iv). Since $a_{i}$ increases in $i$, and $|\xi_{i}|\leq 2$ gives $a_{i}^{2}\xi_{i}^{2}\leq 4a_{i}^{2}$, we have

[TABLE]

where $f(\cdot)$ is the linear interpolation such that $f(i)=a_{i}$, and here $v$ can be treated also as a positive strictly increasing function on $[0,\infty)$ with $v(m)=v_{m}$, so both $(f^{2})^{-1},v^{-1}$ are well-defined, positive and strictly increasing. Using that $Z(n)=v^{-1}(n)$, Drogin’s condition (40) will be verified if we show that

[TABLE]

for $n$ large enough, because then, for $n$ large enough, $a_{i}^{2}\xi_{i}^{2}<\varepsilon n$ for $i\leq Z(n)$, that is,

[TABLE]

Since $a^{2}_{m}=o(v_{m})$, i.e. $f^{2}(x)=o(v(x))$, for this $\varepsilon$, there is an $M$ such that for $l\geq M$, $\displaystyle{f^{2}(l)}/{v(l)}<{\varepsilon}/4$, and for such an $M$, there is an $N$ such that for $x\geq N$ we have $v^{-1}(x)\geq M$. Hence,

[TABLE]

that is, (44) holds for $n\geq N$. This completes the proof of (iv) and that of the theorem altogether.

6.12. Proof of Theorem 5

We again use the martingale approximation approach of section 6.2. Notice that

[TABLE]

Without the loss of generality, we may assume that $0<a<p_{n}<b<1$. Then $r:=\max\{|2a-1|,|2b-1|\}<1,$ and

[TABLE]

which is why the sum in (45) is well-defined, that is, the $a_{n}$ are well-defined, for all $n\geq 1$. Furthermore,

[TABLE]

which gives $|a_{n}|\leq\frac{1}{1-r}$ for all $n$.

Next, we prove that $v(m)\overset{m\rightarrow\infty}{\longrightarrow}\infty$, or equivalently, that $\sigma_{n}\overset{n\rightarrow\infty}{\longrightarrow}\infty$:

(i) If $p_{n}\leq 1/2$, $\forall n$, then $a_{n}>1$, $\forall n$, and we immediately have $v(m)\overset{m\rightarrow\infty}{\longrightarrow}\infty$.

(ii) Otherwise we have a subsequence $\{p_{n_{k}}\}_{n_{k}}$ such that $n_{k+1}-n_{k}>1$ and $p_{n_{k}}>1/2$, for all $n_{k}$. Notice that, by (45) and a direct computation, we have

[TABLE]

and thus for the subsequence one has

[TABLE]

So the two subsequences $\{a_{n_{k}-1}-1\}_{k\geq 1},\{a_{n_{k}}-1\}_{k\geq 1}$ have opposite signs, hence we have a subsequence of $\{a_{n}\}_{n\geq 1}$ such that its terms are larger than $1$. Consequently, $v(m)\overset{m\rightarrow\infty}{\longrightarrow}\infty$.

Moreover, the condition that $\displaystyle\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{Z(n)}a_{i}^{2}\xi_{i}^{2}\mathbf{1}_{\{a_{i}^{2}\xi_{i}^{2}>n\epsilon\}}=0$ is easy to verify, since our $a_{n}$ are bounded.

In conclusion, the answers to (M) and to (INV.M) are both in the affirmative, yielding the invariance principle (12).

6.13. Proof of Theorem 6

Fix $a>0$ and let $N=N(a)$ be such that $a/N\leq 1/2$ and that also $a/n<p_{n}$ holds for all $n>N$. Define $\hat{p}_{n}$ so that it coincides with $p_{n}$ for $n\leq N$ and $\hat{p}_{n}=a/n$ for $n>N$. Let $\hat{S}$ denote the walk for the sequence $(\hat{p}_{n})$, and note that this walk depends on the parameter $a>0$. By the monotonicity established in the proof of Theorem 4,

[TABLE]

In [7] it was shown that

[TABLE]

Since $a>0$ was arbitrary,

[TABLE]

implying WLLN. ∎

6.14. Proof of Theorem 7

We first need a lemma.

Lemma 3**.**

For every $m$, $n$ and $\ell$ such that $\ell>n\geq m\geq 1$ we have that ${\mathbb{P}}(S_{\ell}\leq S_{n}\mid Y_{m})\geq\frac{1}{2}(1-|e_{m,n+1}|)$.

Proof of Lemma.

We do the proof for $Y_{m}=1$, for $Y_{m}=-1$ the proof is essentially the same. Writing out $e_{m,n+1}=E(Y_{n+1}\mid Y_{m}=1),$ one obtains

[TABLE]

Next, we claim that

[TABLE]

Indeed, let us start our walk at time $n$ instead of time zero at the location $S_{n}$, such that its first step is random and equals $1$ or $-1$ with equal probabilities. Then the LHS of (47) is the probability that $n-\ell$ times later this walk ends up at a position which is not larger than its initial position. By symmetry, this value is at least $1/2$. By (46) and (47) and Markov property,

[TABLE]

as claimed. ∎

We now turn to the proof of Theorem 7 and show e.g. that ${\mathbb{P}}(S_{n}<0\ \mathrm{i.o.}\mid\mathcal{F}_{1})=1$; one can similarly show that ${\mathbb{P}}(S_{n}>0\ \mathrm{i.o.}\mid\mathcal{F}_{1})=1$. It turns out that is enough to construct a sequence $(\ell_{k})_{k\geq 0}$ such that ${\mathbb{P}}(S_{\ell_{i+1}}<0\mid\mathcal{F}_{\ell_{i}})\geq r$ holds with some $r>0$, and the statement then follows from the extended Borel-Cantelli Lemma. Below we define such a sequence recursively, for $r=1/6$.

Let $\ell_{0}:=1$. Once $\{\ell_{i},\ 0\leq i\leq k\}$ have been constructed, we construct $\ell_{k+1}$ as follows. By mixing, we can pick an $N_{k}$ (depending on $\ell_{k}$ only) such that $|e_{\ell_{k},\ell}|<1/3$ for all $\ell\geq N_{k}$. By Lemma 3 then, for all $\ell\geq N_{k}$,

[TABLE]

Using that $|S_{\ell_{k}}|\leq\ell_{k}$ along with Assumption 3,

[TABLE]

Hence, $\exists\ \ell_{k+1}>\max\{\ell_{k},N_{k}\}$ that depends only on $\ell_{k}$ such that

[TABLE]

By combining (48) and (49) we conclude that

[TABLE]

The sought sequence $(\ell_{k})_{k\geq 0}$ has thus been constructed.$\hfill\qed$

6.15. Proof of Theorem 8

Let $\tau_{0}:=0$ and

[TABLE]

Since $\sum p_{n}=\infty$, by the Borel-Cantelli Lemma, there are infinitely many turns. As a result, with probability $1$, all $\tau_{n}$ are well-defined and finite. Clearly, $\tau_{n}\to\infty$,  as $n\to\infty$.

Let

[TABLE]

and note that $A_{n}\subseteq\{S_{2\tau_{n}}\leq 0\}=:B_{n}.$ If we show that $\sum_{n}{\mathbb{P}}(A_{n}\mid\mathcal{G}_{n-1})=\infty$ then by the extended Borel-Cantelli lemma (see Corollary 5.29 in [3]), it follows that ${\mathbb{P}}(A_{n}\text{ i.o.})=1$; hence ${\mathbb{P}}(B_{n}\text{ i.o.})=1$, and so ${\mathbb{P}}(S_{n}\leq 0\text{ i.o.})=1$.

Now, for $n\geq n_{0}$,

[TABLE]

when $k$ is admissible (i.e. the condition has positive probability). Since obviously $\tau_{n}\geq n$, we know that this is never the case for $k<n$.

Since the product on the right hand side tends999A more detailed calculation shows that the RHS equals $2^{-c}\left[1-\frac{c(c-1)}{4k}+O(k^{-2})\right]$ but we do not need it here. to $2^{-c}$, as $k\to\infty$, and only $k$’s for which $k\geq n$ are admissible,

[TABLE]

holds for all large enough $n$ and admissible $k$’s. Thus ${\mathbb{P}}(A_{n}\mid\mathcal{G}_{n-1})\geq 2^{-c-1}$ holds for all large enough $n$, and we are done. A completely symmetric argument shows that also ${\mathbb{P}}(S_{n}\geq 0\text{ i.o.})=1$, thus proving the recurrence of the walk $S$.

A similar proof, left to the reader, establishes that the scaling limit (zigzag process) is recurrent as well.

7. Appendix

Here we invoke some background on random measures that we utilized in the proof of Proposition 4. Much more material on random measures can be found in [8].

Assume that we are given a complete separable metric space $S$.

Definition 5** (Dissecting subsets).**

Denote by $\widehat{S}$ the set of all bounded Borel sets of $S$. A subset ${\mathcal{I}}\subset\widehat{S}$ is called dissecting if

• (a)

every open set $G\subset S$ is a countable union of sets in ${\mathcal{I}}$;

• (b)

every set $B\in\widehat{S}$ is covered by finitely many sets in ${\mathcal{I}}$.

The following lemma is a useful result concerning the weak convergence of random measures. (The measures are equipped with the vague topology, recall Notation 1.)

Lemma 4** (Theorem 4.11 in [8]).**

Let $\xi,(\xi_{n})_{n}$ be random measures on $S$ and let $E$ denote the expectation for $\xi$. Furthermore, let

1. (1)

$\widehat{C}_{s}$* be the set of all continuous compactly supported functions on $S$;* 2. (2)

$\widehat{S}_{E\xi}$* be the class of all bounded sets $A\subset S$ with $E\xi(\partial A)=0$;* 3. (3)

$\widehat{{\mathcal{I}}}_{+}$* be the set of all non-negative simple ${\mathcal{I}}-measurable$ functions for a fix dissecting semi-ring ${\mathcal{I}}\subset\widehat{S}_{E\xi}$.*

Then, as $n\to\infty$, $\xi_{n}\overset{vd}{\longrightarrow}\xi$ if and only if $\xi_{n}(f)\overset{d}{\longrightarrow}\xi(f)$ holds either for all $f\in\widehat{C}_{s}$ or for all $f\in\widehat{{\mathcal{I}}}_{+}$.

Acknowledgements: We are grateful to Márton Balázs, Bertrand Cloez, Edward Crane, Laurent Miclo and Sean O’Rourke for helpful discussions, and also to an anonymous referee for his/her close reading of the manuscript. S.V. thanks the University of Colorado for the hospitality during his visit in May 2018.