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Counterbalancing steps at random
in a random walk

Jean Bertoin111 Institute of Mathematics, University of Zurich, Switzerland, [email protected]
Abstract

A random walk with counterbalanced steps is a process of partial sums Sˇ(n)=Xˇ1++Xˇn\check{S}(n)=\check{X}_{1}+\cdots+\check{X}_{n} whose steps Xˇn\check{X}_{n} are given recursively as follows. For each n2n\geq 2, with a fixed probability pp, Xˇn\check{X}_{n} is a new independent sample from some fixed law μ\mu, and with complementary probability 1p1-p, Xˇn=Xˇv(n)\check{X}_{n}=-\check{X}_{v(n)} counterbalances a previous step, with v(n)v(n) a uniform random pick from {1,,n1}\{1,\ldots,n-1\}. We determine the asymptotic behavior of Sˇ(n)\check{S}(n) in terms of pp and the first two moments of μ\mu. Our approach relies on a coupling with a reinforcement algorithm due to H.A. Simon, and on properties of random recursive trees and Eulerian numbers, which may be of independent interest. The method can be adapted to the situation where the step distribution μ\mu belongs to the domain of attraction of a stable law.
Keywords: Reinforcement, random walk, random recursive tree, Eulerian numbers, Yule-Simon model.
Mathematics Subject Classification: 60F05; 60G50; 05C05; 05A05.

1 Introduction

In short, the purpose of the present work is to investigate long time effects of an algorithm for counterbalancing steps in a random walk. As we shall first explain, our motivation stems from a nearest neighbor process on the integer lattice, known as the elephant random walk.

The elephant random walk is a stochastic process with memory on \mathbb{Z}, which records the trajectory of an elephant that makes steps with unit length left or right at each positive integer time. It has been introduced by Schütz and Trimper [31] and triggered a growing interest in the recent years; see, for instance, [4, 5, 15, 16, 22, 23], and also [2, 6, 8, 14, 20, 26] for related models. The dynamics depend on a parameter q[0,1]q\in[0,1] and can be described as follows. Let us assume that the first step of the elephant is a Rademacher variable, that is equals +1+1 or 1-1 with probability 1/21/2. For each time n2n\geq 2, the elephant remembers a step picked uniformly at random among those it made previously, and decides either to repeat it with probability qq, or to make the opposite step with complementary probability 1q1-q. Obviously, each step of the elephant has then the Rademacher law, although the sequence of steps is clearly not stationary.

Roughly speaking, it seems natural to generalize these dynamics and allow steps to have an arbitrary distribution on \mathbb{R}, say μ\mu. In this direction, Kürsten [23] pointed at an equivalent way of describing the dynamics of the elephant random walk which makes such generalization non trivial222 Note that merely replacing the Rademacher distribution for the first step of the elephant by μ\mu would not be interesting, as one would then just get the evolution of the elephant random walk multiplied by some random factor with law μ\mu.. Let p[0,1]p\in[0,1], and imagine a walker who makes at each time a step which is either, with probability pp, a new independent random variable with law μ\mu, or, with probability 1p1-p, a repetition of one of his preceding steps picked uniformly at random. It is immediately checked that when μ\mu is the Rademacher distribution, then the walker follows the same dynamics as the elephant random walk with parameter q=1p/2q=1-p/2. When μ\mu is an isotropic stable law, this is the model referred to as the shark random swim by Businger [14], and more generally, when μ\mu is arbitrary, this is the step reinforced random walk that has been studied lately in e.g. [7, 9, 10, 11].

The model of Kürsten yields an elephant random walk only with parameter q[1/2,1]q\in[1/2,1]; nonetheless the remaining range can be obtained by a simple modification. Indeed, let again p[0,1]p\in[0,1] and imagine now an repentant walker who makes at each time a step which is either, with probability pp, a new independent random variable with law μ\mu, or, with probability 1p1-p, the opposite of one of his previous steps picked uniformly at random. When μ\mu is the Rademacher distribution, we simply get the dynamics of the elephant random walk with parameter q=p/2[0,1/2]q=p/2\in[0,1/2].

More formally, we consider a sequence (Xn)(X_{n}) of i.i.d. real random variables with some given law μ\mu and a sequence (εn)n2(\varepsilon_{n})_{n\geq 2} of i.i.d. Bernoulli variables with parameter p[0,1]p\in[0,1], which we assume furthermore independent of (Xn)(X_{n}). We construct a counterbalanced sequence (Xˇn)(\check{X}_{n}) by interpreting each {εn=0}\{\varepsilon_{n}=0\} as a counterbalancing event and each {εn=1}\{\varepsilon_{n}=1\} as an innovation event. Specifically, we agree that ε1=1\varepsilon_{1}=1 for definiteness and denote the number of innovations after nn steps by

i(n)j=1nεjfor n1.{\mathrm{i}}(n)\coloneqq\sum_{j=1}^{n}\varepsilon_{j}\qquad\text{for }n\geq 1.

We introduce a sequence (v(n))n2(v(n))_{n\geq 2} of independent variables, where each v(n)v(n) has the uniform distribution on {1,,n1}\{1,\ldots,n-1\}, and which is also independent of (Xn)(X_{n}) and (εn)(\varepsilon_{n}). We then define recursively

Xˇn{Xˇv(n) if εn=0,Xi(n) if εn=1.\check{X}_{n}\coloneqq\left\{\begin{matrix}-\check{X}_{v(n)}&\text{ if }\varepsilon_{n}=0,\\ X_{{\mathrm{i}}(n)}&\text{ if }\varepsilon_{n}=1.\\ \end{matrix}\right. (1)

Note that the same step can be counterbalanced several times, and also that certain steps counterbalance previous steps which in turn already counterbalanced earlier ones. The process

Sˇ(n)Xˇ1++Xˇn,n0\check{S}(n)\coloneqq\check{X}_{1}+\cdots+\check{X}_{n},\qquad n\geq 0

which records the positions of the repentant walker as a function of time, is called here a random walk with counterbalanced steps. Note that for p=1p=1, i.e. when no counterbalancing events occur, Sˇ\check{S} is just a usual random walk with i.i.d. steps.

In short, we are interested in understanding how counterbalancing steps affects the asymptotic behavior of random walks. We first introduce some notation. Recall that μ\mu denotes the distribution of the first step X1=Xˇ1X_{1}=\check{X}_{1} and write

mk𝔼(X1k)=xkμ(dx)m_{k}\coloneqq\mathbb{E}(X_{1}^{k})=\int_{\mathbb{R}}x^{k}\mu(\mathrm{d}x)

for the moment of order k1k\geq 1 of X1X_{1} whenever X1Lk()X_{1}\in L^{k}(\mathbb{P}). To start with, we point out that if the first moment is finite, then the algorithm (1) yields the recursive equation

𝔼(Sˇ(n+1))=pm1+(1(1p)/n)𝔼(Sˇ(n)),n1,\mathbb{E}(\check{S}(n+1))=pm_{1}+(1-(1-p)/n)\mathbb{E}(\check{S}(n)),\qquad n\geq 1,

with the initial condition 𝔼(Sˇ(1))=m1\mathbb{E}(\check{S}(1))=m_{1}. It follows easily that

𝔼(Sˇ(n))\displaystyle\mathbb{E}(\check{S}(n)) p2pm1nas n;\displaystyle\sim\frac{p}{2-p}\,m_{1}n\qquad\text{as }n\to\infty;

see e.g. Lemma 4.1.2 in [18]. Our first result about the ballistic behavior should therefore not come as a surprise.

Proposition \theproposition.

Let p[0,1]p\in[0,1]. If X1L1()X_{1}\in L^{1}(\mathbb{P}), then there is the convergence in probability

limnSˇ(n)n=p2pm1.\lim_{n\to\infty}\frac{\check{S}(n)}{n}=\frac{p}{2-p}m_{1}.

We see in particular that counterbalancing steps reduces the asymptotic velocity of a random walk by a factor p/(2p)<1p/(2-p)<1. The velocity is smaller when the innovation rate pp is smaller (i.e. when counterbalancing events have a higher frequency), and vanishes as pp approaches 0+0+.

The main purpose of this work is to establish the asymptotic normality when μ\mu has a finite second moment.

Theorem \thetheorem.

Let p(0,1]p\in(0,1]. If X1L2()X_{1}\in L^{2}(\mathbb{P}), then there is the convergence in distribution

limnSˇ(n)p2pm1nn=𝒩(0,m2(p2pm1)232p),\lim_{n\to\infty}\frac{\check{S}(n)-\frac{p}{2-p}m_{1}n}{\sqrt{n}}=\mathcal{N}\left(0,\frac{m_{2}-\left(\frac{p}{2-p}m_{1}\right)^{2}}{3-2p}\right),

where the right-hand side denotes a centered Gaussian variable parametrized by mean and variance.

It is interesting to observe that the variance of the Gaussian limit depends linearly on the square m12m_{1}^{2} of the first moment and the second moment m2m_{2} of μ\mu only, although not just on the variance m2m12m_{2}-m_{1}^{2} (except, of course, for p=1p=1). Furthermore, it is not always a monotonous function333 For instance, in the simplest case when μ\mu is a Dirac point mass, i.e. m2=m12m_{2}=m_{1}^{2}, then the variance is given by 4(1p)m2(32p)(2p)2\frac{4(1-p)m_{2}}{(3-2p)(2-p)^{2}} and reaches its maximum for p=(917)/80.6p=(9-\sqrt{17})/8\approx 0.6. At the opposite, when μ\mu is centered, i.e. m1=0m_{1}=0, the variance is given by m2/(32p)m_{2}/(3-2p) and hence increases with pp. of the innovation rate pp, and does not vanish when pp tends to 0 either.

Actually, our proofs of Proposition 1 and Theorem 1 provide a much finer analysis than what is encapsulated by those general statements. Indeed, we shall identify the main actors for the evolution of Sˇ\check{S} and their respective contributions to its asymptotic behavior. In short, we shall see that the ballistic behavior stems from those of the variables XjX_{j} that have been used just once by the algorithm (1) (in particular, they have not yet been counterbalanced), whereas the impact of variables that occurred twice or more regardless of their signs ±\pm, is asymptotically negligible as far as only velocity is concerned. Asymptotic normality is more delicate to analyze. We shall show that, roughly speaking, it results from the combination of, on the one hand, the central limit theorem for certain centered random walks, and on the other hand, Gaussian fluctuations for the asymptotic frequencies of some pattern induced by (1).

Our analysis relies on a natural coupling of the counterbalancing algorithm (1) with a basic linear reinforcement algorithm which has been introduced a long time ago by H.A. Simon [32] to explain the occurrence of certain heavy tailed distributions in a variety of empirical data. Specifically, Simon defined recursively a sequence denoted here by (X^n)(\hat{X}_{n}) (beware of the difference of notation between X^\hat{X} and Xˇ\check{X}) via

X^n{X^v(n) if εn=0,Xi(n) if εn=1.\hat{X}_{n}\coloneqq\left\{\begin{matrix}\hat{X}_{v(n)}&\text{ if }\varepsilon_{n}=0,\\ X_{{\mathrm{i}}(n)}&\text{ if }\varepsilon_{n}=1.\\ \end{matrix}\right. (2)

We stress that the same Bernoulli variables εn\varepsilon_{n} and the same uniform variables v(n)v(n) are used to run both Simon’s algorithm (2) and (1); in particular either Xˇn=X^n\check{X}_{n}=\hat{X}_{n} or Xˇn=X^n\check{X}_{n}=-\hat{X}_{n}. It might then seem natural to refer to (1) and (2) respectively as negative and positive reinforcement algorithms. However, in the literature, negative reinforcement usually refers to a somehow different notion (see e.g. [29]), and we shall thus avoid using this terminology.

A key observation is that (1) can be recovered from (2) as follows. Simon’s algorithm naturally encodes a genealogical forest with set of vertices ={1,2,,}\mathbb{N}=\{1,2,\ldots,\} and edges (v(j),j)(v(j),j) for all j2j\geq 2 with εj=0\varepsilon_{j}=0; see the forthcoming Figure 1. Then Xˇn=X^n\check{X}_{n}=\hat{X}_{n} if the vertex nn belongs to an even generation of its tree component, and Xˇn=X^n\check{X}_{n}=-\hat{X}_{n} if nn belongs to an odd generation. On the other hand, the statistics of Simon’s genealogical forest can be described in terms of independent random recursive trees (see e.g. Chapter 6 in [17] for background) conditionally on their sizes. This leads us to investigate the difference Δ(𝕋k)\Delta(\mathbb{T}_{k}) between the number of vertices at even generations and the number of vertices at odd generations in a random recursive tree 𝕋k\mathbb{T}_{k} of size k1k\geq 1. The law of Δ(𝕋k)\Delta(\mathbb{T}_{k}) can be expressed in terms of Eulerian numbers, and properties of the latter enable us either to compute explicitly or estimate certain quantities which are crucial for the proofs of Proposition 1 and Theorem 1.

It is interesting to compare asymptotic behaviors for counterbalanced steps with those for reinforced steps. If we write S^(n)=X^1++X^n\hat{S}(n)=\hat{X}_{1}+\cdots+\hat{X}_{n} for the random walk with reinforced steps, then it is known that the law of large numbers holds for S^\hat{S}, namely S^(n)/nm1\hat{S}(n)/n\to m_{1} in L1L^{1} when |x|μ(dx)<\int_{\mathbb{R}}|x|\mu(\mathrm{d}x)<\infty, independently of the innovation parameter pp. Further, regarding fluctuations when |x|2μ(dx)<\int_{\mathbb{R}}|x|^{2}\mu(\mathrm{d}x)<\infty, a phase transition occurs for the critical parameter pc=1/2p_{c}=1/2, in the sense that S^\hat{S} is diffusive for p>1/2p>1/2 and superdiffusive for p<1/2p<1/2; see [10, 11]. Despite of the natural coupling between (1) and (2), there are thus major differences444 This should not come as a surprise. In the simplest case when μ=δ1\mu=\delta_{1} is the Dirac mass at 11, one has S^(n)n\hat{S}(n)\equiv n whereas Sˇ\check{S} is a truly stochastic process, even for p=0p=0 when there is no innovation. between the asymptotic behaviors of Sˇ\check{S} and of S^\hat{S}: Proposition 1 shows that the asymptotic speed of Sˇ\check{S} depends on pp, and Theorem 1 that there is no such phase transition for counterbalanced steps and Sˇ\check{S} is always diffusive.

The phase transition for step reinforcement when μ\mu has a finite second moment can be explained informally as follows; for the sake of simplicity, suppose also that μ\mu is centered, i.e. m1=0m_{1}=0. There are i(n)pn\mathrm{i}(n)\sim pn trees in Simon’s genealogical forest, which are overwhelmingly microscopic (i.e. of size O(1)O(1)), whereas only a few trees reach the size O(n1p)O(n^{1-p}). Because μ\mu is centered, the contribution of microscopic trees to S^(n)\hat{S}(n) is of order n\sqrt{n}, and that of the few largest trees of order n1pn^{1-p}. This is the reason why S^(n)\hat{S}(n) grows like nn1p\sqrt{n}\gg n^{1-p} when p>1/2p>1/2, and rather like n1pnn^{1-p}\gg\sqrt{n} when p<1/2p<1/2. For counterbalanced steps, we will see that, due to the counterbalancing mechanism, the contribution of a large tree with size 1\ell\gg 1 is now only of order \sqrt{\ell}. As a consequence, the contribution to Sˇ(n)\check{S}(n) of the largest trees of Simon’s genealogical forest is only of order O(n(1p)/2)O(n^{(1-p)/2}). This is always much smaller than the contribution of microscopic trees which remain of order n\sqrt{n}. We further stress that, even though only the sizes of the trees in Simon’s genealogical forest are relevant for the analysis of the random walk S^\hat{S} with reinforced steps, the study of the random walk Sˇ\check{S} with counterbalanced steps is more complex and requires informations on the fine structure of those trees, not merely their sizes.

The rest of this text is organized as follows. Section 2 focusses on the purely counterbalanced case p=0p=0. In this situation, for each fixed n1n\geq 1, the distribution of Sˇ(n)\check{S}(n) can be expressed explicitly in terms of Eulerian numbers. Section 3 is devoted to the coupling between the counterbalancing algorithm (1) and H.A. Simon’s algorithm (2), and to the interpretation of the former in terms of a forest of random recursive trees induced by the latter. Proposition 1 and Theorem 1 are proved in Section 4, where we analyze more finely the respective contributions of some natural sub-families. Last, in Section 5, we present a stable version of Theorem 1 when μ\mu belongs to the domain of attraction (without centering) of an α\alpha-stable distribution for some α(0,2)\alpha\in(0,2).

2 Warm-up: the purely counterbalanced case

This section is devoted to the simpler situation555 Observe that this case without innovation has been excluded in Theorem 1. when p=0p=0. So εn0\varepsilon_{n}\equiv 0 for all n2n\geq 2, meaning that every step, except of course the first one, counterbalances some preceding step. The law μ\mu then only plays a superficial role as it is merely relevant for the first step. For the sake of simplicity, we further focus on the case when μ=δ1\mu=\delta_{1} is the Dirac mass at 11.

The dynamics are entirely encoded by the sequence (v(n))n2(v(n))_{n\geq 2} of independent uniform variables on {1,,n1}\{1,\ldots,n-1\}; more precisely the purely counterbalanced sequence of bits is given by

Xˇ1=1andXˇn=Xˇv(n)for all n2.\check{X}_{1}=1\quad\text{and}\quad\check{X}_{n}=-\check{X}_{v(n)}\quad\text{for all }n\geq 2. (3)

The random algorithm (3) points at a convenient representation in terms of random recursive trees. Specifically, the sequence (v(n))n2(v(n))_{n\geq 2} encodes a random tree 𝕋{\mathbb{T}}_{\infty} with set of vertices \mathbb{N} and set of edges {(v(n),n):n2}\{(v(n),n):n\geq 2\}. Roughly speaking, 𝕋{\mathbb{T}}_{\infty} is constructed recursively by incorporating vertices one after the other and creating an edge between each new vertex nn and its parent v(n)v(n) which is picked uniformly at random in {1,,n1}\{1,\ldots,n-1\} and independently of the other vertices. If we view 11 as the root of 𝕋{\mathbb{T}}_{\infty} and call a vertex jj odd (respectively, even) when its generation (i.e. its distance to the root in 𝕋{\mathbb{T}}_{\infty}) is an odd (respectively, even) number, then

Xˇn={1 if n is an even vertex in 𝕋,1 if n is an odd vertex in 𝕋.\check{X}_{n}=\left\{\begin{matrix}1&\text{ if $n$ is an even vertex in $\mathbb{T}_{\infty}$,}\\ -1&\text{ if $n$ is an odd vertex in $\mathbb{T}_{\infty}$.}\end{matrix}\right.

Let us now introduce some notation with that respect. For every n1n\geq 1, we write 𝕋n{\mathbb{T}}_{n} for the restriction of 𝕋{\mathbb{T}}_{\infty} to the set of vertices {1,,n}\{1,\ldots,n\} and refer to 𝕋n{\mathbb{T}}_{n} as a random recursive tree with size nn. We also write Odd(𝕋n)\mathrm{Odd}({\mathbb{T}}_{n}) (respectively, Even(𝕋n)\mathrm{Even}({\mathbb{T}}_{n})) for the number of odd (respectively, even) vertices in 𝕋n{\mathbb{T}}_{n} and set

Δ(𝕋n)Even(𝕋n)Odd(𝕋n)=n2Odd(𝕋n).\Delta({\mathbb{T}}_{n})\coloneqq\mathrm{Even}({\mathbb{T}}_{n})-\mathrm{Odd}({\mathbb{T}}_{n})=n-2\mathrm{Odd}({\mathbb{T}}_{n}).

Of course, we can also express

Δ(𝕋n)=Xˇ1++Xˇn,\Delta({\mathbb{T}}_{n})=\check{X}_{1}+\cdots+\check{X}_{n},

which is the trajectory of an elephant full of regrets (i.e. for the parameter q=0q=0).

The main observation of this section is that law of the number of odd vertices is readily expressed in terms of the Eulerian numbers. Recall that kn\langle{}^{n}_{k}\rangle denotes the number of permutations ς\varsigma of {1,,n}\{1,\ldots,n\} with kk descents, i.e. such that #{1j<n:ς(j)>ς(j+1)}=k\#\{1\leq j<n:\varsigma(j)>\varsigma(j+1)\}=k. Obviously kn1\langle{}^{n}_{k}\rangle\geq 1 if and only if 0k<n0\leq k<n, and one has

k=0n1nk=n!.\sum_{k=0}^{n-1}{\displaystyle\left\langle{n\atop k}\right\rangle}=n!.

The linear recurrence equation

nk=(nk)n1k1+(k+1)n1k{\displaystyle\left\langle{n\atop k}\right\rangle}=(n-k){\displaystyle\left\langle{{n-1}\atop{k-1}}\right\rangle}+(k+1){\displaystyle\left\langle{{n-1}\atop k}\right\rangle} (4)

is easily derived from a recursive construction of permutations (see Theorem 1.3 in [30]); we also mention the explicit formula (see Corollary 1.3 in [30])

nk=j=0k(1)j(n+1j)(k+1j)n.{\displaystyle\left\langle{n\atop k}\right\rangle}=\sum_{j=0}^{k}(-1)^{j}{\binom{n+1}{j}}(k+1-j)^{n}.
Lemma \thelemma.

For every n1n\geq 1, we have

(Odd(𝕋n)=)=1(n1)!n11,\mathbb{P}(\mathrm{Odd}({\mathbb{T}}_{n})=\ell)=\frac{1}{(n-1)!}{\displaystyle\left\langle{{n-1}\atop{\ell-1}}\right\rangle},

with the convention666 Note that this convention is in agreement with the linear recurrence equation (4). that 1 0=1\langle{}^{\ 0}_{-1}\rangle=1 in the right-hand side for n=1n=1 and =0\ell=0.

Proof.

Consider n1n\geq 1 and note from the very construction of random recursive trees that there is the identity

(Odd(𝕋n+1)=)=n(Odd(𝕋n)=)+n+1n(Odd(𝕋n)=1).\mathbb{P}(\mathrm{Odd}({\mathbb{T}}_{n+1})=\ell)=\frac{\ell}{n}\,\mathbb{P}(\mathrm{Odd}({\mathbb{T}}_{n})=\ell)+\frac{n+1-\ell}{n}\,\mathbb{P}(\mathrm{Odd}({\mathbb{T}}_{n})=\ell-1).

Indeed, the first term of the sum in the right-hand side accounts for the event that the parent v(n+1)v(n+1) of the new vertex n+1n+1 is an odd vertex (then n+1n+1 is an even vertex), and the second term for the event that v(n+1)v(n+1) is an even vertex (then n+1n+1 is an odd vertex).

In terms of A(n,k)n!(Odd(𝕋n+1)=k+1)A(n,k)\coloneqq n!\mathbb{P}(\mathrm{Odd}({\mathbb{T}}_{n+1})=k+1), this yields

A(n,k)=(k+1)A(n1,k)+(nk)A(n1,k1),A(n,k)=(k+1)A(n-1,k)+(n-k)A(n-1,k-1),

which is the linear recurrence equation (4) satisfied by the Eulerian numbers. Since plainly A(1,0)=(Odd(2)=1)=1=01A(1,0)=\mathbb{P}(\mathrm{Odd}(2)=1)=1=\langle{}^{1}_{0}\rangle, we conclude by iteration that A(n,k)=knA(n,k)=\langle{}^{n}_{k}\rangle for all n1n\geq 1 and 0k<n0\leq k<n. Last, the formula in the statement also holds for n=1n=1 since Odd(1)=0\mathrm{Odd}(1)=0. ∎

Remark \theremark.

Lemma 2 is implicit in Mahmoud [24]777 Beware however that the definition of Eulerian numbers in [24] slightly differs from ours, namely kn\langle{}^{n}_{k}\rangle there corresponds to k1n\langle{}^{\ n}_{k-1}\rangle here.. Indeed Odd(𝕋n)\mathrm{Odd}({\mathbb{T}}_{n}) can be viewed as the number of blue balls in an analytic Friedman’s urn model started with one white ball and replacement scheme ()1 00 1\left({}^{0\ 1}_{1\ 0}\right); see Section 7.2.2 in [24]. In this setting, Lemma 2 is equivalent to the formula for the number of white balls at the bottom of page 127 in [24]. Mahmoud relied on the analysis of the differential system associated to the replacement scheme via a Riccati differential equation and inversion of generating functions. The present approach based on the linear recurrence equation (4) is more direct. Lemma 2 is also a closed relative to a result due to Najock and Heyde [27] (see also Theorem 8.6 in Mahmoud [24] and Section 6.2.4 in Drmota [17]) which states that the number of leaves in a random recursive tree with size nn has the same distribution as that appearing in Lemma 2.

We next point at a useful identity related to Lemma 2 which goes back to Laplace (see Exercise 51 of Chapter I in Stanley [33]) and is often attributed to Tanny [34]. For every n0n\geq 0, there is the identity in distribution

Odd(𝕋n+1)=(d)U1++Un,\mathrm{Odd}({\mathbb{T}}_{n+1})\,{\overset{\mathrm{(d)}}{=}}\,\lceil U_{1}+\cdots+U_{n}\rceil, (5)

where in the right-hand side, U1,U2,U_{1},U_{2},\ldots is a sequence of i.i.d. uniform variables on [0,1][0,1] and \lceil\cdot\rceil denotes the ceiling function. We now record for future use the following consequences.

Corollary \thecorollary.
  • (i)

    For every n2n\geq 2, the variable Δ(𝕋n)\Delta({\mathbb{T}}_{n}) is symmetric, i.e. Δ(𝕋n)=(d)Δ(𝕋n)\Delta({\mathbb{T}}_{n})\,{\overset{\mathrm{(d)}}{=}}\,-\Delta({\mathbb{T}}_{n}), and in particular 𝔼(Δ(𝕋n))=0\mathbb{E}(\Delta({\mathbb{T}}_{n}))=0.

  • (ii)

    For all n3n\geq 3, one has 𝔼(Δ(𝕋n)2)=n/3\mathbb{E}(\Delta({\mathbb{T}}_{n})^{2})=n/3.

  • (iii)

    For all n1n\geq 1, one has 𝔼(|Δ(𝕋n)|4)6n2\mathbb{E}(|\Delta({\mathbb{T}}_{n})|^{4})\leq 6n^{2}.

Proof.

(i) Equivalently, the assertion claims that in a random recursive tree of size at least 22, the number of odd vertices and the number of even vertices have the same distribution. This is immediate from (5) and can also be checked directly from the construction.

(ii) This has been already observed by Schütz and Trimper [31] in the setting of the elephant random walk; for the sake of completeness we present a short argument. The vertex n+1n+1 is odd (respectively, even) in 𝕋n+1{\mathbb{T}}_{n+1} if and only if its parent is an even (respectively, odd) vertex in 𝕋n{\mathbb{T}}_{n}. Hence one has

𝔼(Δ(𝕋n+1)Δ(𝕋n)𝕋n)=1nΔ(𝕋n),\mathbb{E}(\Delta({\mathbb{T}}_{n+1})-\Delta({\mathbb{T}}_{n})\mid{\mathbb{T}}_{n})=-\frac{1}{n}\Delta({\mathbb{T}}_{n}),

and since Δ(𝕋n+1)Δ(𝕋n)=±1\Delta({\mathbb{T}}_{n+1})-\Delta({\mathbb{T}}_{n})=\pm 1, this yields the recursive equation

𝔼(Δ(𝕋n+1)2)=(12/n)𝔼(Δ(𝕋n)2)+1.\mathbb{E}(\Delta({\mathbb{T}}_{n+1})^{2})=(1-2/n)\mathbb{E}(\Delta({\mathbb{T}}_{n})^{2})+1.

By iteration, we conclude that 𝔼(Δ(𝕋n)2)=n/3\mathbb{E}(\Delta({\mathbb{T}}_{n})^{2})=n/3 for all n3n\geq 3.

(iii) Recall that the process of the fractional parts {U1++Un}\{U_{1}+\cdots+U_{n}\} is a Markov chain on [0,1)[0,1) whose distribution at any fixed time n1n\geq 1 is uniform on [0,1)[0,1). Writing Vn=12UnV_{n}=1-2U_{n} and Wn=2{U1++Un}1W_{n}=2\{U_{1}+\cdots+U_{n}\}-1, we see that V1,V2,V_{1},V_{2},\ldots is a sequence of i.i.d. uniform variables on [1,1][-1,1] and that WnW_{n} has the uniform distribution on [1,1][-1,1] too.

The characteristic function of the uniform variable VjV_{j} is

𝔼(exp(iθVj))=θ1sin(θ)=1θ26+θ4120+O(θ6)as θ0,\mathbb{E}(\exp({\operator@font i}\theta V_{j}))=\theta^{-1}\sin(\theta)=1-\frac{\theta^{2}}{6}+\frac{\theta^{4}}{120}+O(\theta^{6})\qquad\text{as }\theta\to 0,

and therefore for every n1n\geq 1,

𝔼(exp(iθ(V1++Vn)))\displaystyle\mathbb{E}(\exp({\operator@font i}\theta(V_{1}+\cdots+V_{n}))) =(1θ26+θ4120+O(θ6))n\displaystyle=\left(1-\frac{\theta^{2}}{6}+\frac{\theta^{4}}{120}+O(\theta^{6})\right)^{n}
=1n6θ2+(n120+n(n1)72)θ4+O(θ6).\displaystyle=1-\frac{n}{6}\theta^{2}+\left(\frac{n}{120}+\frac{n(n-1)}{72}\right)\theta^{4}+O(\theta^{6}).

It follows that

𝔼((V1++Vn)4)=24(n120+n(n1)72)n2/3.\mathbb{E}((V_{1}+\cdots+V_{n})^{4})=24\left(\frac{n}{120}+\frac{n(n-1)}{72}\right)\leq n^{2}/3.

We can rephrase (5) as the identity in distribution

Δ(𝕋n+1)=(d)V1++Vn+Wn.\Delta({\mathbb{T}}_{n+1})\,{\overset{\mathrm{(d)}}{=}}\,V_{1}+\cdots+V_{n}+W_{n}.

Since 𝔼(Wn4)=1/3\mathbb{E}(W_{n}^{4})=1/3, the proof is completed with the elementary bound (a+b)416(a4+b4)(a+b)^{4}\leq 16(a^{4}+b^{4}). ∎

We now conclude this section with an application of (5) to the asymptotic normality of Δ(𝕋n)\Delta({\mathbb{T}}_{n}). Since 𝔼(U)=1/2\mathbb{E}(U)=1/2 and Var(U)=1/12\mathrm{Var}(U)=1/12, the classical central limit theorem immediately yields the following.

Corollary \thecorollary.

Assume p=0p=0 and μ=δ1\mu=\delta_{1}. One has

limnΔ(𝕋n)n=𝒩(0,1/3)in distribution.\lim_{n\to\infty}\frac{\Delta({\mathbb{T}}_{n})}{\sqrt{n}}={\mathcal{N}}(0,1/3)\qquad\text{in distribution.}

Corollary 2 goes back to [27] in the setting of the number of leaves in random recursive trees; see also [4, 5, 15, 16] for alternative proofs in the framework of the elephant random walk.

3 Genealogical trees in Simon’s algorithm

From now on, μ\mu is an arbitrary probability law on \mathbb{R} and we also suppose that the innovation rate is strictly positive, p(0,1)p\in(0,1). Recall the construction of the sequence (X^n)(\hat{X}_{n}) from Simon’s reinforcement algorithm (2). Simon was interested in the asymptotic frequencies of variables having a given number of occurrences. Specifically, for every n,jn,j\in\mathbb{N}, we write

Nj(n)#{n:X^=Xj}{N}_{j}(n)\coloneqq\#\{\ell\leq n:\hat{X}_{\ell}=X_{j}\}

for the number of occurrences of the variable XjX_{j} until the nn-th step of the algorithm (2), and

νk(n)#{1ji(n):Nj(n)=k},k\nu_{k}(n)\coloneqq\#\{1\leq j\leq\mathrm{i}(n):{N}_{j}(n)=k\},\qquad k\in\mathbb{N} (6)

for the number of such variables that have occurred exactly kk times. Observe also that the number of innovations satisfies the law of large numbers i(n)pn{\mathrm{i}}(n)\sim pn a.s.

Lemma \thelemma.

For every k1k\geq 1, we have

limnνk(n)pn=11pB(k,1+1/(1p))in probability,\lim_{n\to\infty}\frac{\nu_{k}(n)}{pn}=\frac{1}{1-p}{\mathrm{B}}(k,1+1/(1-p))\qquad\text{in probability,}

where B{\mathrm{B}} denotes the beta function.

Lemma 3 is essentially due to H.A. Simon [32], who actually only established the convergence of the mean value. The strengthening to convergence in probability can be obtained as in [13] from a concentration argument based on the Azuma-Hoeffding’s inequality; see Section 3.1 in [28]. The right-hand side in the formula is a probability mass on \mathbb{N} known as the Yule-Simon distribution with parameter 1/(1p)1/(1-p). We record for future use a couple of identities which are easily checked from the integral definition of the beta function:

11pk=1B(k,1+1/(1p))=1\frac{1}{1-p}\sum_{k=1}^{\infty}{\mathrm{B}}(k,1+1/(1-p))=1 (7)

and

11pk=1kB(k,1+1/(1p))=1p.\frac{1}{1-p}\sum_{k=1}^{\infty}k{\mathrm{B}}(k,1+1/(1-p))=\frac{1}{p}. (8)

For k=1k=1, Lemma 3 reads

limnn1ν1(n)=p2pin probability.\lim_{n\to\infty}n^{-1}\nu_{1}(n)=\frac{p}{2-p}\qquad\text{in probability.} (9)

We shall also need to estimate the fluctuations, which can be derived by specializing a Gaussian limit theorem for extended Pólya urns due to Bai et al. [1].

Lemma \thelemma.

There is the convergence in distribution

limnν1(n)np/(2p)n=𝒩(0,2p38p2+6p(32p)(2p)2).\lim_{n\to\infty}\frac{\nu_{1}(n)-np/(2-p)}{\sqrt{n}}=\mathcal{N}\left(0,\frac{2p^{3}-8p^{2}+6p}{(3-2p)(2-p)^{2}}\right).
Proof.

The proof relies on the observation that Simon’s algorithm can be coupled with a two-color urn governed by the same sequences of random bits (εn)(\varepsilon_{n}) and of uniform variables (v(n))(v(n)) as follows. Imagine that we observe the outcome of Simon’s algorithm at the nn-step and that for each 1jn1\leq j\leq n, we associate a white ball if the variable X^j\hat{X}_{j} appears exactly once, and a red ball otherwise. At the initial time n=1n=1, the urn contains just one white ball and no red balls. At each step n2n\geq 2, a ball picked uniformly at random in the urn (in terms of Simon’s algorithm, this is given by the uniform variable v(n)v(n)). If εn=1\varepsilon_{n}=1, then the ball picked is returned to the urn and one adds a white ball (in terms of Simon’s algorithm, this corresponds to an innovation and ν1(n)=ν1(n1)+1\nu_{1}(n)=\nu_{1}(n-1)+1). If εn=0\varepsilon_{n}=0, then the ball picked is removed from the urn and one adds two red balls (in terms of Simon’s algorithm, this corresponds to a repetition and either ν1(n)=ν1(n1)1\nu_{1}(n)=\nu_{1}(n-1)-1 if the ball picked is white, or ν1(n)=ν1(n1)\nu_{1}(n)=\nu_{1}(n-1) if the ball picked is red). By construction, the number WnW_{n} of white balls in the urn coincides with the number ν1(n)\nu_{1}(n) of variables that have appeared exactly once in Simon’s algorithm (2).

We shall now check our claim in the setting of [1] by specifying the quantities which appear there. The evolution of number of white balls in the urn is governed by Equation (2.1) in [1], viz.

Wn=Wn1+InAn+(1In)Cn,W_{n}=W_{n-1}+I_{n}A_{n}+(1-I_{n})C_{n},

where In=1I_{n}=1 if a white ball is picked and In=0I_{n}=0 otherwise. In our framework, we further have An=2εn1A_{n}=2\varepsilon_{n}-1 and Cn=εnC_{n}=\varepsilon_{n}. If we write n{\mathcal{F}}_{n} for the natural filtration generated by the variables (Ak,Ck,Ik)kn(A_{k},C_{k},I_{k})_{k\leq n}, then AnA_{n} and CnC_{n} are independent of n1{\mathcal{F}}_{n-1} with

𝔼(An)=2p1,𝔼(Cn)=p,Var(An)=4(pp2),Var(Cn)=pp2.\mathbb{E}(A_{n})=2p-1,\quad\mathbb{E}(C_{n})=p,\quad\mathrm{Var}(A_{n})=4(p-p^{2}),\quad\mathrm{Var}(C_{n})=p-p^{2}.

This gives in the notation (2.2) of [1]:

σM2\displaystyle\sigma^{2}_{M} =p2p4(pp2)+(1p2p)(pp2)+(p1)2p2p(1p2p)\displaystyle=\frac{p}{2-p}4(p-p^{2})+\left(1-\frac{p}{2-p}\right)(p-p^{2})+(p-1)^{2}\frac{p}{2-p}\left(1-\frac{p}{2-p}\right)
=2p38p2+6p(2p)2,\displaystyle=\frac{2p^{3}-8p^{2}+6p}{(2-p)^{2}},

and finally

σ2=2p38p2+6p(32p)(2p)2.\sigma^{2}=\frac{2p^{3}-8p^{2}+6p}{(3-2p)(2-p)^{2}}.

Our claim can now be seen as a special case of Corollary 2.1 in [1]. ∎

We shall also need a refinement of Lemma 3 in which one does not only record the number of occurrences of the variable XjX_{j}, but more generally the genealogical structure of these occurrences. We need to introduce first some notation in that respect.

Fix n1n\geq 1 and let some 1ji(n)1\leq j\leq{\mathrm{i}}(n) (i.e. the variable XjX_{j} has already appeared at the nn-th step of the algorithm). Write 1<<kn\ell_{1}<\ldots<\ell_{k}\leq n for the increasing sequence of steps of the algorithm at which XjX_{j} appears, where k=Nj(n)1k={N}_{j}(n)\geq 1. The genealogy of occurrences of the variable XjX_{j} until the nn-th step is recorded as a tree Tj(n)T_{j}(n) on {1,,k}\{1,\ldots,k\} such that for every 1a<bk1\leq a<b\leq k, (a,b)(a,b) is an edge of Tj(n)T_{j}(n) if and only if v(b)=av(\ell_{b})=\ell_{a}, that is, if and only if the identity X^b=Xj\hat{X}_{\ell_{b}}=X_{j} actually results from the fact that the algorithm repeats the variable X^a\hat{X}_{\ell_{a}} at its b\ell_{b}-th step. Plainly, Tj(n)T_{j}(n) is an increasing tree with size kk, meaning a tree on {1,,k}\{1,\ldots,k\} such that the sequence of vertices along any branch from the root 11 to a leaf is increasing. In this direction, we recall that there are (k1)!(k-1)! increasing trees with size kk and that the uniform distribution of the set increasing trees with size kk coincides with the law of the random recursive tree of size kk, 𝕋k{\mathbb{T}}_{k}. See for instance Section 1.3.1 in Drmota [17].

Refer to caption
Figure 1: Example of a genealogical forest representation of Simon’s algorithm (2) after 18 steps. The dotted edges account for innovation events, i.e. εj=1\varepsilon_{j}=1 and the four genealogical trees are rooted at 1,5,7,141,5,7,14. In each subtree, vertices at even generations are colored in green and vertices at odd generation in white. For instance the genealogical tree T2(18)T_{2}(18) is rooted at 55, it has 33 even vertices and 22 odd vertices.

More generally, the distribution of the entire genealogical forest given the sizes of the genealogical trees can be described as follows.

Lemma \thelemma.

Fix n1n\geq 1, 1kn1\leq k\leq n, and let n1,,nk1n_{1},\ldots,n_{k}\geq 1 with sum n1++nk=nn_{1}+\cdots+n_{k}=n. Then conditionally on Nj(n)=nj{N}_{j}(n)=n_{j} for every j=1,,kj=1,\ldots,k, the genealogical trees T1(n),,Tk(n)T_{1}(n),\ldots,T_{k}(n) are independent random recursive trees with respective sizes n1,,nkn_{1},\ldots,n_{k}.

Proof.

Recall that the set {(v(j),j)\{(v(j),j) for 1jn}1\leq j\leq n\} is that of the edges of the random recursive tree with size nn, 𝕋n{\mathbb{T}}_{n}. The well-known splitting property states that removing a given edge, say (v(j),j)(v(j),j) for some fixed jj, from 𝕋n{\mathbb{T}}_{n} produces two subtrees which in turn, conditionally on their sizes, are two independent random recursive trees. This has been observed first by Meir and Moon [25]; see also [3] and references therein for more about this property.

The genealogical trees T1(n),,Tk(n)T_{1}(n),\ldots,T_{k}(n) result by removing the edges (v(j),j)(v(j),j) in 𝕋n{\mathbb{T}}_{n} for which εj=1\varepsilon_{j}=1 and enumerating in each subtree component their vertices in the increasing order. Our statement is now easily seen by applying iteratively this splitting property. ∎

We shall also need for the proofs of Proposition 1 and Theorem 1 an argument of uniform integrability that relies in turn on the following lemma. Recall that if TT is a rooted tree, Δ(T)\Delta(T) denotes the difference between the number of vertices at even distance from root and that at odd distance.

Lemma \thelemma.

For every 1<β<211p1<\beta<2\wedge\frac{1}{1-p}, one has

supn11nj=1n𝔼(Nj(n)β)<\sup_{n\geq 1}\frac{1}{n}\sum_{j=1}^{n}\mathbb{E}({N}_{j}(n)^{\beta})<\infty

and

supn11n𝔼(j=1i(n)|Δ(Tj(n))|2β)<.\sup_{n\geq 1}\frac{1}{n}\mathbb{E}\left(\sum_{j=1}^{{\mathrm{i}}(n)}|\Delta(T_{j}(n))|^{2\beta}\right)<\infty.
Proof.

The first claim is a consequence of Lemma 3.6 of [9] which states that for β(1,1/(1p))\beta\in(1,1/(1-p)) [beware that the parameter denoted by pp in [9] is actually 1p1-p here], there exists numerical constants c>0c>0 and η(0,1)\eta\in(0,1) such that 𝔼(Nj(n)β)c(n/j)η\mathbb{E}({N}_{j}(n)^{\beta})\leq c(n/j)^{\eta} for all 1jn1\leq j\leq n.

Next, combining Jensen’s inequality with Corollary 2(iii), we get that for k2k\geq 2

𝔼(|Δ(𝕋k)|2β)𝔼(|Δ(𝕋k)|4)β/26kβ.\mathbb{E}(|\Delta({\mathbb{T}}_{k})|^{2\beta})\leq\mathbb{E}(|\Delta({\mathbb{T}}_{k})|^{4})^{\beta/2}\leq 6k^{\beta}.

Then recall that conditionally on Nj(n)=k1{N}_{j}(n)=k\geq 1, Tj(n)T_{j}(n) has the law of the random recursive tree with size kk, 𝕋k{\mathbb{T}}_{k}, and hence

𝔼(j=1i(n)|Δ(Tj(n))|2β)\displaystyle\mathbb{E}\left(\sum_{j=1}^{{\mathrm{i}}(n)}|\Delta(T_{j}(n))|^{2\beta}\right) =j=1n(k=1n𝔼(|Δ(𝕋k)|2β)(Nj(n)=k))\displaystyle=\sum_{j=1}^{n}\left(\sum_{k=1}^{n}\mathbb{E}(|\Delta({\mathbb{T}}_{k})|^{2\beta})\mathbb{P}({N}_{j}(n)=k)\right)
6j=1n(k=1nkβ(Nj(n)=k)).\displaystyle\leq 6\sum_{j=1}^{n}\left(\sum_{k=1}^{n}k^{\beta}\mathbb{P}({N}_{j}(n)=k)\right).

We know from the first part that this last quantity is finite, and the proof is complete. ∎

4 Proofs of the main results

As its title indicates, the purpose of this section is to establish Proposition 1 and Theorem 1. The observation that for every n1n\geq 1 and 1ji(n)1\leq j\leq{{\mathrm{i}}(n)}, the variable XjX_{j} appears exactly Even(Tj(n))\mathrm{Even}(T_{j}(n)) times and its opposite Xj-X_{j} exactly Odd(Tj(n))\mathrm{Odd}(T_{j}(n)) times until the nn-step of the algorithm (1), yields the identity

Sˇ(n)i=1nXˇi=j=1i(n)Δ(Tj(n))Xj,\check{S}(n)\coloneqq\sum_{i=1}^{n}\check{X}_{i}=\sum_{j=1}^{{\mathrm{i}}(n)}\Delta(T_{j}(n))X_{j}, (10)

which lies at the heart of our approach. We stress that in (10) as well as in related expressions that we shall use in the sequel, the sequence of i.i.d. variables (Xn)(X_{n}) and the family of genealogical trees (Tj(n))(T_{j}(n)) are independent, because the latter are constructed from the sequences (εn)(\varepsilon_{n}) and (v(n))(v(n)) only.

Actually, our proof analyzes more precisely the effects of the counterbalancing algorithm (1) by estimating specifically the contributions of certain sub-families to the asymptotic behavior of Sˇ\check{S}. Specifically, we set for every k1k\geq 1,

Sˇk(n)j=1i(n)Δ(Tj(n))Xj𝟙Nj(n)=k,\check{S}_{k}(n)\coloneqq\sum_{j=1}^{{\mathrm{i}}(n)}\Delta(T_{j}(n))X_{j}\mathbbm{1}_{N_{j}(n)=k}, (11)

so that

Sˇ(n)=k=1nSˇk(n).\check{S}(n)=\sum_{k=1}^{n}\check{S}_{k}(n).

4.1 Proof of Proposition 1

The case p=1p=1 (no counterbalancing events) of Proposition 1 is just the weak law of large numbers, and the case p=0p=0 (no innovations) is a consequence of Corollary 2. The case p(0,1)p\in(0,1) derives from the next lemma which shows more precisely that the variables XjX_{j} that have appeared in the algorithm (1) but have not yet counterbalanced determine the ballistic behavior of Sˇ\check{S}, whereas those that have appeared twice or more (i.e. such that Nj(n)2N_{j}(n)\geq 2) have a negligible impact.

Lemma \thelemma.

Assume that X1L1()X_{1}\in L^{1}(\mathbb{P}) and recall that m1=𝔼(X1)m_{1}=\mathbb{E}(X_{1}). Then the following limits hold in probability:

  1. (i)

    limnn1Sˇ1(n)=m1p/(2p)\lim_{n\to\infty}n^{-1}\check{S}_{1}(n)=m_{1}p/(2-p),

  2. (ii)

    limnn1k=2n|Sˇk(n)|=0\lim_{n\to\infty}n^{-1}\sum_{k=2}^{n}\left|\check{S}_{k}(n)\right|=0.

Proof.

(i) Recall the notation (6) and that the sequence of i.i.d. variables (Xj)(X_{j}) is independent of the events {Nj(n)=1}\{{N}_{j}(n)=1\}. We see that there is the identity in distribution

Sˇ1(n)=(d)S1(ν1(n)),\check{S}_{1}(n)\,{\overset{\mathrm{(d)}}{=}}\,S_{1}(\nu_{1}(n)),

where S1(n)=X1++XnS_{1}(n)=X_{1}+\cdots+X_{n} is the usual random walk. The claim follows readily from the law of large numbers and (9).

(ii) We first argue that for each fixed k2k\geq 2,

limnn1Sˇk(n)=0almost surely.\lim_{n\to\infty}n^{-1}\check{S}_{k}(n)=0\qquad\text{almost surely.} (12)

Indeed, recall that νk(n)\nu_{k}(n) denotes the number of genealogical trees Tj(n)T_{j}(n) with size kk. It follows from Lemma 3 that conditionally on νk(n)=\nu_{k}(n)=\ell, the sub-family of such Tj(n)T_{j}(n) enumerated in the increasing order of the index jj, is given by \ell i.i.d. copies of the random recursive tree 𝕋k{\mathbb{T}}_{k}. Hence, still conditionally on νk(n)=\nu_{k}(n)=\ell, enumerating the elements of the sub-family {XjΔ(Tj(n)):Nj(n)=k}\{X_{j}\Delta(T_{j}(n)):{N}_{j}(n)=k\} in the increasing order of jj yields \ell independent variables, each being distributed as X1Δ(𝕋k)X_{1}\Delta({\mathbb{T}}_{k}) with X1X_{1} and Δ(𝕋k)\Delta({\mathbb{T}}_{k}) independent. We deduce from Corollary 2(i) that the variable X1Δ(𝕋k)X_{1}\Delta({\mathbb{T}}_{k}) symmetric, and since it is also integrable, it is centered. Since νk(n)n\nu_{k}(n)\leq n, this readily entails (12) by an application of the law of large numbers.

The proof can be completed by an argument of uniform integrability. In this direction, fix an arbitrarily large integer \ell and write by the triangular inequality

1n𝔼(k=n|Sˇk(n)|)\displaystyle\frac{1}{n}\mathbb{E}\left(\sum_{k=\ell}^{n}\left|\check{S}_{k}(n)\right|\right) 1n𝔼(j=1i(n)|Xj|Nj(n)𝟙Nj(n))\displaystyle\leq\frac{1}{n}\mathbb{E}\left(\sum_{j=1}^{{\mathrm{i}}(n)}|X_{j}|{N}_{j}(n)\mathbbm{1}_{{N}_{j}(n)\geq\ell}\right)
=𝔼(|X1|)nj=1n𝔼(Nj(n)𝟙Nj(n))\displaystyle=\frac{\mathbb{E}(|X_{1}|)}{n}\sum_{j=1}^{n}\mathbb{E}\left({N}_{j}(n)\mathbbm{1}_{{N}_{j}(n)\geq\ell}\right)
1β𝔼(|X1|)nj=1n𝔼(Nj(n)β),\displaystyle\leq\ell^{1-\beta}\frac{\mathbb{E}(|X_{1}|)}{n}\sum_{j=1}^{n}\mathbb{E}\left({N}_{j}(n)^{\beta}\right),

where the last inequality holds for any β>1\beta>1. We see from Lemma 3 that the right-hand side converges to 0 as \ell\to\infty uniformly in n1n\geq 1, and the rest of the proof is straightforward. ∎

4.2 Proof of Theorem 1

For p=1p=1 (no counterbalancing events), Theorem 1 just reduces to the classical central limit theorem, so we assume p(0,1)p\in(0,1). The first step of the proof consists in determining jointly the fluctuations of the components Sˇk\check{S}_{k} defined in (11).

Lemma \thelemma.

Assume that m2=𝔼(X12)<m_{2}=\mathbb{E}(X^{2}_{1})<\infty. Then as nn\to\infty, the sequences of random variables

Sˇ1(n)pm1/(2p)n(for k=1)\frac{\check{S}_{1}(n)-pm_{1}/(2-p)}{\sqrt{n}}\qquad\text{(for $k=1$)}

and

Sˇk(n)n(for k2)\frac{\check{S}_{k}(n)}{\sqrt{n}}\qquad\text{(for $k\geq 2$)}

converge jointly in distribution towards a sequence

(𝒩k(0,σk2))k1\left(\mathcal{N}_{k}(0,\sigma^{2}_{k})\right)_{k\geq 1}

of independent centered Gaussian variables, where

σ12pm22pp2m12(32p)(2p)2,\sigma_{1}^{2}\coloneqq\frac{pm_{2}}{2-p}-\frac{p^{2}m_{1}^{2}}{(3-2p)(2-p)^{2}},

σ220\sigma^{2}_{2}\coloneqq 0, and

σk2kpm23(1p)B(k,1+1/(1p))for k3.\sigma^{2}_{k}\coloneqq\frac{kpm_{2}}{3(1-p)}\mathrm{B}(k,1+1/(1-p))\qquad\text{for $k\geq 3$}.
Proof.

For each k1k\geq 1, let (Yk(n))n1(Y_{k}(n))_{n\geq 1} be a sequence of i.i.d. copies of Δ(𝕋k)X\Delta(\mathbb{T}_{k})X, where XX has the law μ\mu and is independent of the random recursive tree 𝕋k\mathbb{T}_{k}. We further assume that these sequences are independent. Taking partial sums yields a sequence indexed by kk of independent random walks

Sk(n)=Yk(1)++Yk(n),n0.S_{k}(n)=Y_{k}(1)+\cdots+Y_{k}(n),\qquad n\geq 0.

For each n1n\geq 1, the family of blocks

Bk(n){ji(n):Nj(n)=k}for 1ki(n)B_{k}(n)\coloneqq\{j\leq{\mathrm{i}}(n):N_{j}(n)=k\}\qquad\text{for }1\leq k\leq{\mathrm{i}}(n)

forms a random partition of {1,,i(n)}\{1,\ldots,{\mathrm{i}}(n)\} which is independent of the XjX_{j}’s. Recall that we are using the notation νk(n)=#Bk(n)\nu_{k}(n)=\#B_{k}(n), and also from Lemma 3, that conditionally on the Nj(n)N_{j}(n)’s, the genealogical trees Tj(n)T_{j}(n) are independent random recursive trees. We now see from the very definition (11) that for every fixed n1n\geq 1, there is the identity in distribution

(Sˇk(n))k1=(d)(Sk(νk(n)))k1,\left(\check{S}_{k}(n)\right)_{k\geq 1}\,{\overset{\mathrm{(d)}}{=}}\,\left(S_{k}(\nu_{k}(n))\right)_{k\geq 1},

where in the right-hand side, the random walks (Sk)k1(S_{k})_{k\geq 1} are independent of the sequence of block sizes (νk(n))k1(\nu_{k}(n))_{k\geq 1}.

Next we write, first for k=1k=1,

S1(ν1(n))pn2pm1=S1(pn2p)pn2pm1+j=pn/(2p)ν1(n)Y1(j),S_{1}(\nu_{1}(n))-\frac{pn}{2-p}m_{1}=S_{1}\left(\left\lfloor\frac{pn}{2-p}\right\rfloor\right)-\frac{pn}{2-p}m_{1}+\sum_{j=\lceil pn/(2-p)\rceil}^{\nu_{1}(n)}Y_{1}(j),

second S20S_{2}\equiv 0 (since Δ(𝕋2)0\Delta(\mathbb{T}_{2})\equiv 0) for k=2k=2, and third, for k3k\geq 3,

Sk(νk(n))=Sk(pn1pB(k,1+1/(1p)))+j=pn1pB(k,1+1/(1p))νk(n)Yk(j),S_{k}(\nu_{k}(n))=S_{k}\left(\left\lfloor\frac{pn}{1-p}{\mathrm{B}}(k,1+1/(1-p))\right\rfloor\right)+\sum_{j=\lceil\frac{pn}{1-p}{\mathrm{B}}(k,1+1/(1-p))\rceil}^{\nu_{k}(n)}Y_{k}(j),

with the usual summation convention that j=ab=j=ba\sum_{j=a}^{b}=-\sum_{j=b}^{a} when b<ab<a.

Since the i.i.d. variables Y1()Y_{1}(\cdot) have mean m1m_{1} and variance m2m12m_{2}-m_{1}^{2}, the central limit theorem ensures that there is the convergence in distribution

limnn1/2(S1(pn2p)pn2pm1)=𝒩1(0,p(m2m12)2p).\lim_{n\to\infty}n^{-1/2}\left(S_{1}\left(\left\lfloor\frac{pn}{2-p}\right\rfloor\right)-\frac{pn}{2-p}m_{1}\right)=\mathcal{N}_{1}\left(0,\frac{p(m_{2}-m_{1}^{2})}{2-p}\right). (13)

Similarly, for k3k\geq 3, each Yk(n)Y_{k}(n) is centered with variance km2/3km_{2}/3 (by Corollary 2(i-ii)) and hence, using the notation in the statement, there is the convergence in distribution

limnn1/2Sk(pn1pB(k,1+1/(1p)))\displaystyle\lim_{n\to\infty}n^{-1/2}S_{k}\left(\left\lfloor\frac{pn}{1-p}{\mathrm{B}}(k,1+1/(1-p))\right\rfloor\right) =𝒩k(0,σk2)\displaystyle=\mathcal{N}_{k}\left(0,\sigma_{k}^{2}\right) (14)

Plainly, the weak convergences (13) and (14) hold jointly when we agree that the limits are independent Gaussian variables.

Next, from Lemma 3 and the fact that for k3k\geq 3, the i.i.d. variables Yk(j)Y_{k}(j) are centered with finite variance, we easily get

limnn1/2|j=pn1pB(k,1+1/(1p))νk(n)Yk(j)|=0in L2().\lim_{n\to\infty}n^{-1/2}\left|\sum_{j=\lceil\frac{pn}{1-p}{\mathrm{B}}(k,1+1/(1-p))\rceil}^{\nu_{k}(n)}Y_{k}(j)\right|=0\qquad\text{in }L^{2}(\mathbb{P}).

Finally, for k=1k=1, we write

j=pn/(2p)ν1(n)Y1(j)=m1(ν1(n)pn/(2p))+j=pn/(2p)ν1(n)(Y1(j)m1).\sum_{j=\lceil pn/(2-p)\rceil}^{\nu_{1}(n)}Y_{1}(j)=m_{1}(\nu_{1}(n)-\lfloor pn/(2-p)\rfloor)+\sum_{j=\lceil pn/(2-p)\rceil}^{\nu_{1}(n)}(Y_{1}(j)-m_{1}).

On the one hand, we have from the same argument as above that

limnn1/2|j=pn/(2p)ν1(n)(Y1(j)m1)|=0in L2().\lim_{n\to\infty}n^{-1/2}\left|\sum_{j=\lceil pn/(2-p)\rceil}^{\nu_{1}(n)}(Y_{1}(j)-m_{1})\right|=0\qquad\text{in }L^{2}(\mathbb{P}).

On the other hand, we already know from Lemma 3 that there is the convergence in distribution

limnm1ν1(n)pn/(2p)n=𝒩(0,2p38p2+6p(32p)(2p)2m12).\lim_{n\to\infty}m_{1}\frac{\nu_{1}(n)-\lfloor pn/(2-p)\rfloor}{\sqrt{n}}=\mathcal{N}\left(0,\frac{2p^{3}-8p^{2}+6p}{(3-2p)(2-p)^{2}}m_{1}^{2}\right).

Obviously, this convergence in law hold jointly with (13) and (14), where the limiting Gaussian variables are independent. Putting the pieces together, this completes the proof. ∎

The final step for the proof of Theorem 1 is the following lemma.

Lemma \thelemma.

We have

limKsupn1n1𝔼(|kKSˇk(n)|2)=0.\lim_{K\to\infty}\sup_{n\geq 1}n^{-1}\mathbb{E}\left(\left|\sum_{k\geq K}\check{S}_{k}(n)\right|^{2}\right)=0.
Proof.

We write

kKSˇk(n)=j=1nXjΔ(Tj(n))𝟙Nj(n)K.\sum_{k\geq K}\check{S}_{k}(n)=\sum_{j=1}^{n}X_{j}\Delta(T_{j}(n))\mathbbm{1}_{{N}_{j}(n)\geq K}.

Since the XjX_{j} are independent of the Tj(n)T_{j}(n), we get

𝔼(|kKSˇk(n)|2)\displaystyle\mathbb{E}\left(\left|\sum_{k\geq K}\check{S}_{k}(n)\right|^{2}\right) =𝔼(j,j=1nXjXjΔ(Tj(n))𝟙Nj(n)KΔ(Tj(n))𝟙Nj(n)K)\displaystyle=\mathbb{E}\left(\sum_{j,{j^{\prime}}=1}^{n}X_{j}X_{j^{\prime}}\Delta(T_{j}(n))\mathbbm{1}_{{N}_{j}(n)\geq K}\Delta(T_{j^{\prime}}(n))\mathbbm{1}_{{N}_{j^{\prime}}(n)\geq K}\right)
m2j,j=1n𝔼(Δ(Tj(n))𝟙Nj(n)KΔ(Tj(n))𝟙Nj(n)K).\displaystyle\leq m_{2}\sum_{j,{j^{\prime}}=1}^{n}\mathbb{E}\left(\Delta(T_{j}(n))\mathbbm{1}_{{N}_{j}(n)\geq K}\Delta(T_{j^{\prime}}(n))\mathbbm{1}_{{N}_{j^{\prime}}(n)\geq K}\right).

We evaluate the expectation in the right-hand side by conditioning first on Nj(n)=kN_{j}(n)=k and Nj(n)=kN_{j^{\prime}}(n)=k^{\prime} with k,k3k,k^{\prime}\geq 3. Recall from Lemma 3 that the genealogical trees Tj(n)T_{j}(n) and Tj(n)T_{j^{\prime}}(n) are then two random recursive trees with respective sizes kk and kk^{\prime}, which are further independent when jjj\neq j^{\prime}. Thanks to Corollary 2(i-ii) we get

𝔼(Δ(Tj(n))𝟙Nj(n)KΔ(Tj(n))𝟙Nj(n)K)\displaystyle\mathbb{E}\left(\Delta(T_{j}(n))\mathbbm{1}_{{N}_{j^{\prime}}(n)\geq K}\Delta(T_{j^{\prime}}(n))\mathbbm{1}_{{N}_{j^{\prime}}(n)\geq K}\right)
={13𝔼(Nj(n)𝟙Nj(n)K) if j=j0 if jj.\displaystyle=\left\{\begin{matrix}\frac{1}{3}\mathbb{E}(N_{j}(n)\mathbbm{1}_{{N}_{j}(n)\geq K})&\text{ if }j=j^{\prime}\,\\ 0&\text{ if }j\neq j^{\prime}.\end{matrix}\right.

We have thus shown that

𝔼(|kKSˇk(n)|2)m23j=1n𝔼(Nj(n)𝟙Nj(n)K),\mathbb{E}\left(\left|\sum_{k\geq K}\check{S}_{k}(n)\right|^{2}\right)\leq\frac{m_{2}}{3}\sum_{j=1}^{n}\mathbb{E}(N_{j}(n)\mathbbm{1}_{{N}_{j}(n)\geq K}),

which yields our claim just as in the proof of Lemma 4.1(ii). ∎

The proof of Theorem 1 is now easily completed by combining Lemmas 4.2 and 4.2. Indeed, the identity

pm22p+k=2σk2=m232p\frac{pm_{2}}{2-p}+\sum_{k=2}^{\infty}\sigma^{2}_{k}=\frac{m_{2}}{3-2p}

is easily checked from (8).

5 A stable central limit theorem

The arguments for the proof of Theorem 1 when the step distribution μ\mu has a finite second moment can be adapted to the case when μ\mu belongs to some stable domain of attraction; for the sake of simplicity we focus on the situation without centering. Specifically, let (an)(a_{n}) be a sequence of positive real numbers that is regularly varying with exponent 1/α1/\alpha for some α(0,2)\alpha\in(0,2), in the sense that limnarn/an=r1/α\lim_{n\to\infty}a_{\lfloor rn\rfloor}/a_{n}=r^{1/\alpha} for every r>0r>0, and suppose that

limnX1++Xnan=Zin distribution,\lim_{n\to\infty}\frac{X_{1}+\cdots+X_{n}}{a_{n}}=Z\qquad\text{in distribution}, (15)

where ZZ is some α\alpha-stable random variable. We refer to Theorems 4 and 5 on p. 181-2 in [19] and Section 6 of Chapter 2 in [21] for necessary and sufficient conditions for (15) in terms of μ\mu only. We write φα\varphi_{\alpha} for the characteristic exponent of ZZ, viz.

𝔼(exp(iθZ))=exp(φα(θ))for all θ;\mathbb{E}\left(\exp({\operator@font i}\theta Z)\right)=\exp(-\varphi_{\alpha}(\theta))\qquad\text{for all }\theta\in\mathbb{R};

recall that φα\varphi_{\alpha} is homogeneous with exponent α\alpha, i.e.

φα(θ)=|θ|αφα(sgn(θ))for all θ0.\varphi_{\alpha}(\theta)=|\theta|^{\alpha}\varphi_{\alpha}(\textrm{sgn}(\theta))\qquad\text{for all }\theta\neq 0.

Recall the definition and properties of the Eulerian numbers kn\langle{}^{n}_{k}\rangle from Section 2, and also the Pochhammer notation

(x)(k)Γ(x+k)Γ(x)=j=0k1(x+j),x>0,k,(x)^{(k)}\coloneqq\frac{\Gamma(x+k)}{\Gamma(x)}=\prod_{j=0}^{k-1}(x+j),\qquad x>0,k\in\mathbb{N},

for the rising factorial, where Γ\Gamma stands for the gamma function. We can now claim:

Theorem \thetheorem.

Assume (15). For each p(0,1)p\in(0,1), we have

limnSˇ(n)an=Zˇin distribution,\lim_{n\to\infty}\frac{\check{S}(n)}{a_{n}}=\check{Z}\qquad\text{in distribution},

where Zˇ\check{Z} is an α\alpha-stable random variable with characteristic exponent φˇα\check{\varphi}_{\alpha} given by

φˇα(θ)=p(1p)k=1=0k1φα((k2)θ)(1+1/(1p))(k)k11,θ.\check{\varphi}_{\alpha}(\theta)=\frac{p}{(1-p)}\sum_{k=1}^{\infty}\sum_{\ell=0}^{k-1}\frac{\varphi_{\alpha}((k-2\ell)\theta)}{(1+1/(1-p))^{(k)}}{\displaystyle\left\langle{{k-1}\atop{\ell-1}}\right\rangle},\qquad\theta\in\mathbb{R}.

The proof of Theorem 5 relies on a refinement of Simon’s result (Lemma 3) to the asymptotic frequencies of genealogical trees induced by the reinforcement algorithm (2). We denote by 𝒯{{\mathcal{T}}^{\uparrow}} the set of increasing trees (of arbitrary finite size), and for any τ𝒯\tau\in{\mathcal{T}}^{\uparrow}, we write |τ||\tau| for its the size (number of vertices) and Δ(τ)\Delta(\tau) for the difference between its numbers of even vertices and of odd vertices. Refining (6), we also define

ντ(n)j=1i(n)𝟙{Tj(n)=τ},τ𝒯.\nu_{\tau}(n)\coloneqq\sum_{j=1}^{{\mathrm{i}}(n)}\mathbbm{1}_{\{T_{j}(n)=\tau\}},\qquad\tau\in{\mathcal{T}}^{\uparrow}.
Lemma \thelemma.

We have

τ𝒯|τ|+|Δ(τ)|2(1+1/(1p))(|τ|)=4p3(1p),\sum_{\tau\in{{\mathcal{T}}^{\uparrow}}}\frac{|\tau|+|\Delta(\tau)|^{2}}{(1+1/(1-p))^{(|\tau|)}}=\frac{4p}{3(1-p)},

and there is the convergence in probability

limnτ𝒯(|τ|+|Δ(τ)|2)|ντ(n)np(1p)(1+1/(1p))(|τ|)|=0.\lim_{n\to\infty}\sum_{\tau\in{{\mathcal{T}}^{\uparrow}}}(|\tau|+|\Delta(\tau)|^{2})\left|\frac{\nu_{\tau}(n)}{n}-\frac{p}{(1-p)(1+1/(1-p))^{(|\tau|)}}\right|=0.
Proof.

We start by claiming that for every k1k\geq 1, and every tree τ𝒯\tau\in{{\mathcal{T}}^{\uparrow}} with size |τ|=k|\tau|=k, we have

limnντ(n)n=p(1p)(1+1/(1p))(k)in probability.\lim_{n\to\infty}\frac{\nu_{\tau}(n)}{n}=\frac{p}{(1-p)(1+1/(1-p))^{(k)}}\qquad\text{in probability.} (16)

Indeed, the distribution of the random recursive tree 𝕋k\mathbb{T}_{k} of size kk is the uniform probability measure on the set of increasing trees with size kk, which has (k1)!(k-1)! elements. We deduce from Lemma 3 and the law of large numbers that

ντ(n)νk(n)/(k1)!.\nu_{\tau}(n)\sim\nu_{k}(n)/(k-1)!.

The claim (16) now follows from Lemma 3 and the identity

B(k,1+1/(1p))=(k1)!(1+1/(1p))(k).{\mathrm{B}}(k,1+1/(1-p))=\frac{(k-1)!}{(1+1/(1-p))^{(k)}}.

We now have to prove that (16) holds in L1(|τ|+|Δ(τ)|2,𝒯)L^{1}(|\tau|+|\Delta(\tau)|^{2},{\mathcal{T}}^{\uparrow}). On the one hand, one has obviously for every n1n\geq 1

τ𝒯|τ|ντ(n)=n.\sum_{\tau\in{{\mathcal{T}}^{\uparrow}}}|\tau|\nu_{\tau}(n)=n.

On the other hand, there are (k1)!(k-1)! increasing trees with size kk and hence

p1pτ𝒯|τ|(1+1/(1p))(|τ|)=p1pk=1kB(k,1+1/(1p))=1,\frac{p}{1-p}\sum_{\tau\in{{\mathcal{T}}^{\uparrow}}}\frac{|\tau|}{(1+1/(1-p))^{(|\tau|)}}=\frac{p}{1-p}\sum_{k=1}^{\infty}k{\mathrm{B}}(k,1+1/(1-p))=1,

where the second equality is (8). We deduce from Scheffé’s Lemma and (16) that there is the convergence in probability

limnτ𝒯|τ||ντ(n)np(1p)(1+1/(1p))(|τ|)|=0.\lim_{n\to\infty}\sum_{\tau\in{{\mathcal{T}}^{\uparrow}}}|\tau|\left|\frac{\nu_{\tau}(n)}{n}-\frac{p}{(1-p)(1+1/(1-p))^{(|\tau|)}}\right|=0.

Similarly, we deduce from Corollary 2(ii) and Lemma 3 that, for every n0n\geq 0

𝔼(τ𝒯Δ(τ)2ντ(n))\displaystyle\mathbb{E}\left(\sum_{\tau\in{{\mathcal{T}}^{\uparrow}}}\Delta(\tau)^{2}\nu_{\tau}(n)\right) =𝔼(j=1i(n)Δ(Tj(n))2)=13𝔼(j=1i(n)|Tj(n)|)=n/3,\displaystyle=\mathbb{E}\left(\sum_{j=1}^{{\mathrm{i}}(n)}\Delta(T_{j}(n))^{2}\right)=\frac{1}{3}\mathbb{E}\left(\sum_{j=1}^{{\mathrm{i}}(n)}|T_{j}(n)|\right)=n/3,

and further, since there are (k1)!(k-1)! increasing trees with size kk and 𝕋k\mathbb{T}_{k} has the uniform distribution on the set of such trees,

p1pτ𝒯Δ(τ)2(1+1/(1p))(|τ|)\displaystyle\frac{p}{1-p}\sum_{\tau\in{{\mathcal{T}}^{\uparrow}}}\frac{\Delta(\tau)^{2}}{(1+1/(1-p))^{(|\tau|)}} =p1pk=1𝔼(Δ(𝕋k)2)B(k,1+1/(1p))\displaystyle=\frac{p}{1-p}\sum_{k=1}^{\infty}\mathbb{E}(\Delta(\mathbb{T}_{k})^{2}){\mathrm{B}}(k,1+1/(1-p))
=p1pk=1k3B(k,1+1/(1p))=13.\displaystyle=\frac{p}{1-p}\sum_{k=1}^{\infty}\frac{k}{3}{\mathrm{B}}(k,1+1/(1-p))=\frac{1}{3}.

We conclude again from Scheffé’s Lemma that

limnτ𝒯Δ(τ)2|ντ(n)pnp(1p)(1+1/(1p))(|τ|)|=0,\lim_{n\to\infty}\sum_{\tau\in{{\mathcal{T}}^{\uparrow}}}\Delta(\tau)^{2}\left|\frac{\nu_{\tau}(n)}{pn}-\frac{p}{(1-p)(1+1/(1-p))^{(|\tau|)}}\right|=0,

and the proof is complete. ∎

We now establish Theorem 5.

Proof of Theorem 5.

We denote the characteristic function of μ\mu by

Φ(θ)=eiθxμ(dx)for θ.\Phi(\theta)=\int_{\mathbb{R}}{\mathrm{e}}^{\mathrm{i}\theta x}\mu(\mathrm{d}x)\qquad\text{for }\theta\in\mathbb{R}.

Fix r>0r>0 small enough so that |1Φ(θ)|<1|1-\Phi(\theta)|<1 whenever |θ|r|\theta|\leq r, and then define the characteristic exponent φ:[r,r]\varphi:[-r,r]\to\mathbb{C} as the continuous determination of the logarithm of Φ\Phi on [r,r][-r,r]. In words, φ\varphi is the unique continuous function on [r,r][-r,r] with φ(0)=0\varphi(0)=0 and such that Φ(θ)=exp(φ(θ))\Phi(\theta)=\exp(-\varphi(\theta)) for all θ[r,r]\theta\in[-r,r]. For definitiveness, we further set φ(θ)=0\varphi(\theta)=0 whenever |θ|>r|\theta|>r.

Next, observe from the Markov’s inequality that for any 1<β<2(1p)11<\beta<2\wedge(1-p)^{-1} and any a>0a>0

(ji(n):|Δ(Tj(n))|an)a2βnβ𝔼(j=1i(n)|Δ(Tj(n))|2β),\mathbb{P}\left(\exists j\leq{\mathrm{i}}(n):|\Delta(T_{j}(n))|\geq a\sqrt{n}\right)\leq a^{-2\beta}n^{-\beta}\mathbb{E}\left(\sum_{j=1}^{{\mathrm{i}}(n)}|\Delta(T_{j}(n))|^{2\beta}\right),

so that, thanks to Lemma 3,

limn1nmax1ji(n)|Δ(Tj(n))|=0 in probability.\lim_{n\to\infty}\frac{1}{\sqrt{n}}\max_{1\leq j\leq{\mathrm{i}}(n)}|\Delta(T_{j}(n))|=0\qquad\text{ in probability.}

In particular, since the sequence (an)(a_{n}) is regularly varying with exponent 1/α>1/21/\alpha>1/2, for every θ\theta\in\mathbb{R}, the events

Λ(n,θ){|θΔ(Tj(n))/an|<r for all j=1,i(n)},n1\Lambda(n,\theta)\coloneqq\{|\theta\Delta(T_{j}(n))/a_{n}|<r\text{ for all }j=1,\ldots{\mathrm{i}}(n)\},\qquad n\geq 1

occur with high probability as nn\to\infty, in the sense that limn(Λ(n,θ))=1\lim_{n\to\infty}\mathbb{P}(\Lambda(n,\theta))=1.

We then deduce from (10) and the fact that the variables XjX_{j} are i.i.d. with law μ\mu that for every θ\theta\in\mathbb{R},

𝔼(exp(iθSˇ(n)/an)𝟙Λ(n,θ))=𝔼(exp(1nj=1i(n)nφ(θan1Δ(Tj(n))))𝟙Λ(n,θ)).\mathbb{E}(\exp(\mathrm{i}\theta\check{S}(n)/a_{n})\mathbbm{1}_{\Lambda(n,\theta)})=\mathbb{E}\left(\exp\left(-\frac{1}{n}\sum_{j=1}^{{\mathrm{i}}(n)}n\varphi\left(\theta a_{n}^{-1}\Delta(T_{j}(n))\right)\right)\mathbbm{1}_{\Lambda(n,\theta)}\right).

We then write, in the notation of Lemma 5,

1nj=1i(n)nφ(θan1Δ(Tj(n)))=τ𝒯nφ(θan1Δ(τ))ντ(n)n.\frac{1}{n}\sum_{j=1}^{{\mathrm{i}}(n)}n\varphi\left(\theta a_{n}^{-1}\Delta(T_{j}(n))\right)=\sum_{\tau\in{\mathcal{T}}^{\uparrow}}n\varphi\left(\theta a_{n}^{-1}\Delta(\tau)\right)\frac{\nu_{\tau}(n)}{n}.

Recall that we assume (15). According to Theorem 2.6.5 in Ibragimov and Linnik [21], φ\varphi is regularly varying at 0 with exponent α\alpha, and since α<2\alpha<2, the Potter bounds (see Theorem 1.5.6 in [12]) show that for some constant CC:

n|φ(θan1Δ(τ))|C|θΔ(τ)|2.n|\varphi\left(\theta a_{n}^{-1}\Delta(\tau)\right)|\leq C|\theta\Delta(\tau)|^{2}. (17)

We deduce from Lemma 5 that for every fixed θ\theta\in\mathbb{R}, there is the convergence in probability

limnτ𝒯n|φ(θan1Δ(τ))||ντ(n)np(1p)(1+1/(1p))(|τ|)|=0.\lim_{n\to\infty}\sum_{\tau\in{\mathcal{T}}^{\uparrow}}n|\varphi\left(\theta a_{n}^{-1}\Delta(\tau)\right)|\left|\frac{\nu_{\tau}(n)}{n}-\frac{p}{(1-p)(1+1/(1-p))^{(|\tau|)}}\right|=0.

Furthermore, still from Theorem 2.6.5 in Ibragimov and Linnik [21], we have

limnnφ(θ/an)=φα(θ),for every θ,\lim_{n\to\infty}n\varphi(\theta/a_{n})=\varphi_{\alpha}(\theta),\qquad\text{for every }\theta\in\mathbb{R},

and we deduce by dominated convergence, using Lemma 5 and (17), that

limnτ𝒯|nφ(θan1Δ(τ))φα(θΔ(τ))|p(1p)(1+1/(1p))(|τ|)=0.\lim_{n\to\infty}\sum_{\tau\in{\mathcal{T}}^{\uparrow}}|n\varphi\left(\theta a_{n}^{-1}\Delta(\tau)\right)-\varphi_{\alpha}(\theta\Delta(\tau))|\frac{p}{(1-p)(1+1/(1-p))^{(|\tau|)}}=0.

Putting the pieces together, we have shown that

limn𝔼(exp(iθSˇ(n)/an))=exp(τ𝒯φα(θΔ(τ))p(1p)(1+1/(1p))(|τ|)).\lim_{n\to\infty}\mathbb{E}(\exp(\mathrm{i}\theta\check{S}(n)/a_{n}))=\exp\left(-\sum_{\tau\in{{\mathcal{T}}^{\uparrow}}}\varphi_{\alpha}(\theta\Delta(\tau))\frac{p}{(1-p)(1+1/(1-p))^{(|\tau|)}}\right).

It only remains to check that the right-hand side above agrees with the formula of the statement. This follows from Lemma 2 and the fact that for every k1k\geq 1, 𝕋k\mathbb{T}_{k} has the uniform distribution on {τ𝒯:|τ|=k}\{\tau\in{{\mathcal{T}}^{\uparrow}}:|\tau|=k\}. ∎

References

  • [1] Bai, Z. D., Hu, F., and Zhang, L.-X. Gaussian approximation theorems for urn models and their applications. Ann. Appl. Probab. 12, 4 (2002), 1149–1173.
  • [2] Baur, E. On a class of random walks with reinforced memory. Journal of Statistical Physics 181 (2020), 772–802.
  • [3] Baur, E., and Bertoin, J. Cutting edges at random in large recursive trees. In Stochastic analysis and applications 2014, vol. 100 of Springer Proc. Math. Stat. Springer, Cham, 2014, pp. 51–76.
  • [4] Baur, E., and Bertoin, J. Elephant random walks and their connection to Pólya-type urns. Phys. Rev. E 94 (Nov 2016), 052134.
  • [5] Bercu, B. A martingale approach for the elephant random walk. J. Phys. A 51, 1 (2018), 015201, 16.
  • [6] Bercu, B., and Laulin, L. On the center of mass of the elephant random walk. Stochastic Processes and their Applications 133 (2021), 111 – 128.
  • [7] Bertenghi, M. Asymptotic normality of superdiffusive step-reinforced random walks. arXiv:2101.00906.
  • [8] Bertenghi, M. Functional limit theorems for the multi-dimensional elephant random walk. Stoch. Models 38, 1 (2022), 37–50.
  • [9] Bertoin, J. Noise reinforcement for Lévy processes. Ann. Inst. H. Poincaré Probab. Statist. 56, 3 (2020), 2236–2252.
  • [10] Bertoin, J. Scaling exponents of step-reinforced random walks. Probab. Theory Relat. Fields 179 (2021), 295–315.
  • [11] Bertoin, J. Universality of noise reinforced Brownian motions. In In and out of equilibrium 3. Celebrating Vladas Sidoravicius, vol. 77 of Progr. Probab. Birkhäuser/Springer, Cham, [2021] ©2021, pp. 147–161.
  • [12] Bingham, N. H., Goldie, C. M., and Teugels, J. L. Regular variation, vol. 27 of Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 1987.
  • [13] Bollobás, B., Riordan, O., Spencer, J., and Tusnády, G. The degree sequence of a scale-free random graph process. Random Structures Algorithms 18, 3 (2001), 279–290.
  • [14] Businger, S. The shark random swim (Lévy flight with memory). J. Stat. Phys. 172, 3 (2018), 701–717.
  • [15] Coletti, C. F., Gava, R., and Schütz, G. M. Central limit theorem and related results for the elephant random walk. J. Math. Phys. 58, 5 (2017), 053303, 8.
  • [16] Coletti, C. F., Gava, R., and Schütz, G. M. A strong invariance principle for the elephant random walk. J. Stat. Mech. Theory Exp., 12 (2017), 123207, 8.
  • [17] Drmota, M. Random trees. Springer Wien, NewYork, Vienna, 2009. An interplay between combinatorics and probability.
  • [18] Durrett, R. Random graph dynamics, vol. 20 of Cambridge Series in Statistical and Probabilistic Mathematics. Cambridge University Press, Cambridge, 2007.
  • [19] Gnedenko, B. V., and Kolmogorov, A. N. Limit distributions for sums of independent random variables. Translated from the Russian, annotated, and revised by K. L. Chung. With appendices by J. L. Doob and P. L. Hsu. Revised edition. Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills., Ont., 1968.
  • [20] Gut, A., and Stadtmüller, U. The number of zeros in elephant random walks with delays. Statist. Probab. Lett. 174 (2021), Paper No. 109112, 9.
  • [21] Ibragimov, I. A., and Linnik, Y. V. Independent and stationary sequences of random variables. Wolters-Noordhoff Publishing, Groningen, 1971. With a supplementary chapter by I. A. Ibragimov and V. V. Petrov, Translation from the Russian edited by J. F. C. Kingman.
  • [22] Kubota, N., and Takei, M. Gaussian fluctuation for superdiffusive elephant random walks. J. Stat. Phys. 177, 6 (2019), 1157–1171.
  • [23] Kürsten, R. Random recursive trees and the elephant random walk. Phys. Rev. E 93, 3 (2016), 032111, 11.
  • [24] Mahmoud, H. M. Pólya urn models. Texts in Statistical Science Series. CRC Press, Boca Raton, FL, 2009.
  • [25] Meir, A., and Moon, J. Cutting down recursive trees. Mathematical Biosciences 21, 3 (1974), 173 – 181.
  • [26] Miyazaki, T., and Takei, M. Limit theorems for the ‘laziest’ minimal random walk model of elephant type. J. Stat. Phys. 181, 2 (2020), 587–602.
  • [27] Najock, D., and Heyde, C. C. On the number of terminal vertices in certain random trees with an application to stemma construction in philology. J. Appl. Probab. 19, 3 (1982), 675–680.
  • [28] Pachon, A., Polito, F., and Sacerdote, L. Random graphs associated to some discrete and continuous time preferential attachment models. J. Stat. Phys. 162, 6 (2016), 1608–1638.
  • [29] Pemantle, R. A survey of random processes with reinforcement. Probab. Surveys 4 (2007), 1–79.
  • [30] Petersen, T. K. Eulerian numbers. Birkhäuser Advanced Texts: Basler Lehrbücher. [Birkhäuser Advanced Texts: Basel Textbooks]. Birkhäuser/Springer, New York, 2015. With a foreword by Richard Stanley.
  • [31] Schütz, G. M., and Trimper, S. Elephants can always remember: Exact long-range memory effects in a non-Markovian random walk. Phys. Rev. E 70 (Oct 2004), 045101.
  • [32] Simon, H. A. On a class of skew distribution functions. Biometrika 42, 3/4 (1955), 425–440.
  • [33] Stanley, R. P. Enumerative combinatorics. Vol. 1, vol. 49 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1997. With a foreword by Gian-Carlo Rota, Corrected reprint of the 1986 original.
  • [34] Tanny, S. A probabilistic interpretation of Eulerian numbers. Duke Math. J. 40, 4 (12 1973), 717–722.