The Naming Game on the complete graph

TL;DR

This paper rigorously analyzes the time to consensus in the naming game on complete graphs, establishing bounds that match previous numerical predictions and developing new probabilistic tools for the analysis.

Contribution

It provides the first rigorous bounds on the consensus time in the naming game on complete graphs, confirming and extending prior numerical findings.

Findings

Consensus time is at least n^{1/2 - o(1)}

Consensus time is at most logarithmic in n when only two words remain

Develops new probabilistic estimates for semimartingales with jumps

Abstract

We consider a model of language development, known as the naming game, in which agents invent, share and then select descriptive words for a single object, in such a way as to promote local consensus. When formulated on a finite and connected graph, a global consensus eventually emerges in which all agents use a common unique word. Previous numerical studies of the model on the complete graph with $n$ agents suggest that when no words initially exist, the time to consensus is of order $n^{1/2}$, assuming each agent speaks at a constant rate. We show rigorously that the time to consensus is at least $n^{1/2-o(1)}$, and that it is at most constant times $\log n$ when only two words remain. In order to do so we develop sample path estimates for quasi-left continuous semimartingales with bounded jumps.

Tables

Table 1. Table 1: The various types of transitions occurring between two individuals

reactants		product	$n\cdot$(rate)
$A+AB$	$\to$	$2AB$	$1/2$
$A+AB$	$\to$	$2A$	$3/2$
$B+AB$	$\to$	$2AB$	$1/2$
$B+AB$	$\to$	$2B$	$3/2$
$AB+AB$	$\to$	$2A$	$1$
$AB+AB$	$\to$	$2B$	$1$
$A+B$	$\to$	$B+AB$	$1$
$A+B$	$\to$	$A+AB$	$1$

Table 2. Table 2: List of transitions with jumps Δ i subscript Δ 𝑖 \Delta_{i} and rates q i subscript 𝑞 𝑖 q_{i}

$n{\Delta }_{i}x$	$n{\Delta }_{i}y$	$n{\Delta }_{i}z$	$n{\Delta }_{i}u$	$n^{-1}{q}_i$
$-1$	$0$	$1$	$-\text{sgn}\left(x-y\right)+\mathrm{𝟏}\left(u=0\right)$	$xz/2$
$1$	$0$	$-1$	$\text{sgn}\left(x-y\right)+\mathrm{𝟏}\left(u=0\right)$	$3xz/2$
$0$	$-1$	$1$	$\text{sgn}\left(x-y\right)+\mathrm{𝟏}\left(u=0\right)$	$yz/2$
$0$	$1$	$-1$	$-\text{sgn}\left(x-y\right)+\mathrm{𝟏}\left(u=0\right)$	$3yz/2$
$2$	$0$	$-2$	$2(\text{sgn}(x-y)+\mathrm{𝟏}(n(x-y)\in \{0,-1\})$	$z\left(z-n^{-1}\right)/2$
$0$	$2$	$-2$	$2(-\text{sgn}(x-y)+\mathrm{𝟏}(n(x-y)\in \{0,1\})$	$z\left(z-n^{-1}\right)/2$
$-1$	$0$	$1$	$-\text{sgn}\left(x-y\right)+\mathrm{𝟏}\left(u=0\right)$	$xy$
$0$	$-1$	$1$	$\text{sgn}\left(x-y\right)+\mathrm{𝟏}\left(u=0\right)$	$xy$

Equations

$$W_t(v)=\{w\in V:vknows\; the\; word\; invented\; byw\}.$$

$$\{W_t(v)=\{w\}for\; allv\in V:w\in V\}$$

$$V_t=\bigcup _v{W}_t(v)$$

$$V=\{1,\dots ,n\}andE=\{\{v,w\}:v,w\in V,v\ne w\},$$

$$U:=\{(t_i,u_i):i\ge 1\}\subset R_{+}\times [0,1]$$

$$(i-1)/k\le u<i/k.$$

$$\mathbb{P}(\underset{a\le t\le b}{\sup }|V_t-\frac{n}{2}|=o(n))=1-o(1).$$

$${\mathcal{C}}_t(w)=\{v:w\in W_t(v)\}$$

$${\Theta }_t(W):V\to {\mathcal{P}}_o(V^{\prime })\setminus \{\varnothing \}for\; allt>0,$$

$$\underset{t\to \infty }{\lim }{\Phi }_t\in \{{\Phi }^A,{\Phi }^B\}.$$

$$\begin{array}[]{l}T_{c}=\inf\,\{t:\Phi_{t}\in\{\Phi^{A},\Phi^{B}\}\}.\end{array}$$

$$\underset{n\to \infty }{\lim }\underset{(X,Y,Z)}{\sup }{\mathbb{P}}_{(X,Y,Z)}(T_c/\log n>\gamma +\alpha )=0$$

$$\underset{n\to \infty }{\lim }\underset{(X,Y,Z)}{\sup }{\mathbb{P}}_{(X,Y,Z)}(T_c/\log n>\gamma -\alpha )=1$$

$$X=X_0+X^m+X^p$$

$$⟨M⟩=⟨M^c⟩+⟨M^d⟩,$$

$$\exp (M-(e^c/2)⟨M⟩)$$

$$E_t=1+(E_{-}\cdot (M-V){)}_t+\frac{1}{2}(E_{-}\cdot ⟨M^c⟩{)}_t+\sum _{s\le t}{E}_{s^{-}}(e^{\Delta M_s}-1-\Delta M_s).$$

$$\frac{e^c}{2}\sum _{s\le t}{E}_{s^{-}}(\Delta M_s{)}^2.$$

$$E^p$$

$$\le E_{-}\cdot (-V+\frac{1}{2}{e}^c⟨M⟩).$$

$${}^p(\Delta X)={}^{p}X-{}^p{X}_{-}=0.$$

$$P(\bullet (X_t-X_0-X_t^p)\ge a+(\lambda e^{\lambda c}/2)⟨X^m{⟩}_t)\le e^{-\lambda a}$$

$$P(\bullet \lambda X_t^m-({\lambda }^2{e}^{\lambda c}/2)⟨X^m{⟩}_t\ge \lambda a)\le e^{-\lambda a}.$$

$$\text{if}0<\lambda c\le 1/2\text{then}P(\bullet (X_t-X_0-X_t^p)\ge a+\lambda ⟨X^m{⟩}_t)\le e^{-\lambda a}.$$

$${\mu }_t(X)=\frac{d}{dt}{X}_t^p,{\sigma }_t^2(X)=\frac{d}{dt}⟨X^m{⟩}_t.$$

$$\mu (x)={\int }_{R}y\alpha (x,dy),{\sigma }^2(x)={\int }_R{y}^2\alpha (x,dy),$$

$$\mu (XY)=\sigma (X,Y)+X_{-}\mu (Y)+Y_{-}\mu (X),$$

$$XY=X_0{Y}_0+[X,Y]+X_{-}\cdot Y+Y_{-}\cdot X.$$

$$(XY{)}^p=⟨X^m,Y^m⟩+X_{-}\cdot Y^p+Y_{-}\cdot X^p.$$

$$|⟨X^m,Y^m{⟩}_t-⟨X^m,Y^m{⟩}_s|\le \sqrt{(⟨X^m{⟩}_t-⟨X^m{⟩}_s)(⟨Y^m{⟩}_t-⟨Y^m{⟩}_s)}.$$

$$\sum _i|⟨X^m,Y^m{⟩}_{t_i}-⟨X^m,Y^m{⟩}_{s_i}|$$

Full text

1 Introduction

The study of social dynamics from the standpoint of statistical physics is an area which has seen increased attention in recent years [4]. Historically, interacting particle system models of opinion dynamics, such as the voter model, have been of interest to mathematicians and studied in detail. However, new models emerging in the physics literature have yet to be given a fully rigorous mathematical treatment. One of these is a model of language development known as the naming game. This is a simple model of invention, sharing, and selection of words that displays eventual consensus towards a common vocabulary. It has been studied, using numerical simulations and heuristic computations, on lattices [1], the complete graph [3] and some random graphs [5]. As a first effort from the standpoint of probability theory, we study the naming game on the complete graph and give rigorous proof of some scaling relations that have been observed numerically.

We first recall the definition of the naming game on a general locally finite undirected graph $G=(V,E)$. Individuals correspond to vertices of the graph, and each individual speaks to its neighbours at a certain rate. The idea is that individuals are attempting to agree on a word to describe a certain object, for which initially, no descriptive words exist. The interaction rules are as follows.

• •

Speaker:

• –

If the speaker does not know a word to describe the object then she invents a word and speaks it to the listener.

• –

On the other hand, if the speaker does know at least one word to describe the object then she selects a word uniformly at random from her vocabulary and speaks it to the listener.

• •

Listener:

• –

If the listener already knows the chosen word, then both speaker and listener delete the remainder of their vocabulary and remember only that word.

• –

Otherwise, the listener adds the chosen word to their vocabulary.

Thus there is a mechanism both for the creation of new words, and for deletion and eventual agreement upon a single word. We now make this description rigorous. The process is denoted $(W_{t})_{t\geq 0}$ with $W_{t}:V\to\mathcal{P}_{o}(V)$ for each $t\geq 0$, where $\mathcal{P}_{o}(V)$ is the collection of finite subsets of $V$. Thus, for each vertex $v\in V$, we have a process $W_{t}(v)$ whose state space consists of all finite subsets of the vertex set $V$ and which is defined as

[TABLE]

The process evolves as follows: For each $v\in V$, at the times of an independent Poisson process with rate one, $v$ chooses a listener $w$ uniformly at random from the set $\{u:uv\in E\}$; say this occurs at time $t$.

• •

If $W_{t^{-}}(v)$ is empty then $v$ speaks word $v$ to $w$, so that $W_{t}(v)=\{v\}$ and $W_{t}(w)=W_{t^{-}}(w)\cup\{v\}$.

• •

If $W_{t^{-}}(v)$ is non-empty then $v$ chooses a uniform random word $u$ from $W_{t^{-}}(v)$ and speaks it to $w$.

• –

If $u\in W_{t^{-}}(w)$ then $W_{t}(v)=W_{t}(w)=\{u\}$.

• –

If $u\notin W_{t^{-}}(w)$ then $W_{t}(v)$ is unchanged and $W_{t}(w)=W_{t^{-}}(w)\cup\{u\}$.

If $G$ is connected and finite, then with probability one, the system eventually settles into one of the set of absorbing states

[TABLE]

and we would like to know what happens on the way to this consensus. Let

[TABLE]

denote the set of words in existence at time $t$. If $G$ is the complete graph on $n$ vertices, i.e.,

[TABLE]

numerical studies and heuristic computations [2] indicate three distinct phases.

Early phase: $V_{t}$ rises from [math] to about $n/2$ in about $\frac{1}{2}\log n$ time. 2. 2.

Middle phase: $V_{t}$ remains fairly constant up till about $n^{1/2}$ time. 3. 3.

Late phase: $V_{t}$ falls sharply to 1 within about $n^{1/4}$ time.

In this article we consider the early and middle phases, and what we call the final phase, where we assume that $V_{t}$ is initially equal to $2$, and track the dynamics until it goes to $1$. The bulk of the late phase, during which the diversity of language collapses from a large number to a small number of different words, is more difficult to assess, and is not considered here.

In the next section we construct the model as a stochastic process, then describe the main results and give the layout for the rest of the article.

2 Construction and Main Results

We first note a useful “graphical construction” of the process, on a general locally finite graph $G$, from arbitrary initial data. We assume the vertices are totally ordered according to some fixed order. Given $\mu>0$, let $\{(s_{i},u_{i}):i\geq 1\}$ be an independent and identically distributed sequence, with each $s_{i}$ exponentially distributed with mean one and each $u_{i}$ independent of $s_{i}$ and uniform on $[0,1]$, and for $i\geq 1$, let $t_{i}=\mu^{-1}\sum_{j=1}^{i}s_{j}$. Then, the set of points

[TABLE]

defines what we call an augmented Poisson point process with intensity $\mu$, since $(t_{i})$ are the jump times of a Poisson process with intensity $\mu$ and each point $t_{i}$ comes equipped with an independent uniform random variable $u_{i}$ to help with the decision-making process.

Let $F$ denote the set of directed edges $\{(v,w):vw\in E\}$, and associate to each directed edge $(v,w)\in F$ an independent augmented Poisson point process $U(v,w)$ with intensity $(\deg v)^{-1}$. Suppose that $(t,u)\in U(v,w)$ and $|W_{t^{-}}(v)|=k$, with $W_{t^{-}}(v)=\{w_{1},\dots,w_{k}\}$ labelled in increasing order.

• •

If $k=0$ then $v$ speaks word $v$ to $w$ at time $t$.

• •

If $k\geq 1$, then $v$ speaks word $w_{i}$ to $w$ at time $t$ if and only if

[TABLE]

We then follow the rules as described above to determine $W_{t}$. If $G$ is a finite graph, then since the intensity of the union $\bigcup_{(v,w)\in F}U(v,w)$ is finite, its points are well-ordered in time with probability 1, and so $W_{t}$ can be determined from the initial state and the points $U(v,w)$ by updating sequentially in time. If $G$ is an infinite graph, one needs to ensure that for each spacetime point $(v,t)$, a finite number of events suffices to determine $W_{t}(v)$. Although this is not hard to do, we will ignore it since from here on we focus on the case where $G$ is the complete graph on $n$ vertices and thus finite for any $n$.

Recall that $V_{t}=\bigcup_{v}W_{t}(v)$ denotes the set of words in existence at time $t$. The following result gives estimates of $V_{t}$ in the middle phase of the process.

Theorem 1**.**

For any $\epsilon>0$, let $a=(\frac{1}{2}+\epsilon)\log n$ and $b=n^{1/2-\epsilon}$. Then as $n\to\infty$

[TABLE]

The result is proved in two main steps.

First, we show that $n/2+n^{1/2+o(1)}$ are ever created, and within $(\frac{1}{2}+o(1))\log n$ time. 2. 2.

Then, we show that $o(n)$ words are deleted in $n^{1/2-o(1)}$ time.

The proof relies on approximating the size of the cluster $\mathcal{C}_{t}(w)$ corresponding to a given word $w$ by a sort of branching process evolving in a non-stationary random environment. The cluster is defined by

[TABLE]

and is the set of individuals that know word $w$ at time $t$. We also need to control the correlation between distinct clusters $\mathcal{C}_{t}(w_{1},),\mathcal{C}_{t}(w_{2})$. To achieve both tasks we will use a slightly modified graphical construction which is better tailored to tracking the evolution of one or more distinguished clusters.

For the next result we introduce some notation. Let $\Theta_{t}(W)$ denote the configuration at time $t$ when the initial configuration is $W$, and let $V^{\prime}\subset V$. If $W:V\to\mathcal{P}_{o}(V^{\prime})\setminus\{\emptyset\}$ then clearly

[TABLE]

that is, if each vertex has initially a non-empty vocabulary consisting of words in $V^{\prime}$, the same is true at later times. In particular, if $V^{\prime}=\{A,B\}$ for a pair of words $A,B$, then each vertex has one of the three types $A,B$ and $AB$. We note that, starting from $W(v)=\varnothing$ for all $v$, before the process achieves consensus there is a good chance that at some point only two words remain, so we can think of it as the final phase of the process. For the complete graph on $n$ vertices, the rate of change of the number of individuals of each type does not depend on the particular location of the individuals. Therefore, letting $X_{t},Y_{t},Z_{t}$ denote the number of sites at time $t$ with respective types $A,B$ and $AB$, the process $\Phi_{t}=(X_{t},Y_{t},Z_{t})$ is a continuous-time Markov chain. Since the states $\Phi^{A}=(n,0,0)$ and $\Phi^{B}=(0,n,0)$ are the only absorbing states and are both accessible from all other states, it follows that with probability one,

[TABLE]

The following result characterizes how long this takes, for large $n$. Use $\mathbb{P}_{(X,Y,Z)}(\,\cdot\,)$ for the law of the process with initial configuration $(X,Y,Z)$.

Theorem 2**.**

Let $\gamma=1+(-4+2\sqrt{5})^{-1}$, and define the stopping time

[TABLE]

Then, for any $\alpha>0$,

[TABLE]

Notice that, if individuals only remember the last word they heard, then starting from a configuration with two words, we obtain the voter model on the complete graph, for which the time to consensus is of order $n$. The reason it is much faster here is because, once a majority of type $A$ or $B$ develops, it is maintained. To prove this result we use an ODE heuristic to get an idea of what is happening, then carve up the state space into a few pieces and use martingale estimates to control the behavior of sample paths on each piece.

The paper is laid out as follows. In Section 3 we derive a simple and useful sample path estimate for quasi-left continuous semimartingales with bounded jumps, and give some formulas that help with computations later on. This section can be read independently of the rest of the paper, and may be of use in other applications. In Section 4 we prove Theorem 1 in several steps. In Section 4.1 we show that about $n/2$ words are created in about $\frac{1}{2}\log n$ time, using Chebyshev’s inequality and a coupon-collecting argument, respectively. In Section 4.2 we show that $o(n)$ words are deleted in $n^{1/2-\epsilon}$ time, which as noted above is achieved by controlling the number of individuals that know a given word, and which requires the sample path estimates of Section 3. In Section 5 we use an ODE comparison and the estimates of Section 3 to prove Theorem 2. Some additional results are collected in an Appendix, including a general sample path estimate for Poisson processes, and one for semimartingales with sublinear drift.

3 Sample path estimation

Using the semimartingale theory in [6] we derive a useful estimate for quasi-left continuous semimartingales with bounded jumps, which can be found in Lemma 3. It can be thought of as a continuous-time analogue of Azuma’s inequality. In this section, unless otherwise noted, references are to formulas in [6].

Given is a filtered probability space $(\Omega,\operatorname{\mathcal{F}},\mathbf{F},P)$ satisfying the “usual conditions” as described in [6]. Processes are assumed to be optional. Given $X$, $X_{-}$ is the left continuous process $(X_{t^{-}})_{t\geq 0}$ and $\Delta X=X-X^{-}$ is the process of jumps. $X^{p}$ denotes the compensator and $X^{c}$ the continuous martingale part, when they exist.

A semimartingale $X$ is a process (on $\operatorname{\mathbb{R}}$ unless specified otherwise) that can be written as $X=X_{0}+M+A$, where $X_{0}$ is an $\operatorname{\mathcal{F}}_{0}$-measurable random variable, $M$ is a local martingale and $A$ has locally finite variation. Using I.3.17, a semimartingale is special if it can be written as

[TABLE]

where $X^{p}$ is the compensator of $X$ and $X^{m}$ is a uniquely defined local martingale satisfying $X_{0}^{m}=0$. If $X$ is a semimartingale with bounded jumps, that is, $|\Delta X|\leq c$ for some $c>0$ then by I.4.24, $X$ is a special semimartingale and $|\Delta X^{m}|\leq 2c$. If $X$ is also quasi-left continuous, that is, $\Delta X_{T}=0$ a.s. on $\{T<\infty\}$, for any predictable time $T$, then using I.2.35 in the proof of I.4.24, we obtain the slightly stronger estimate $|\Delta X^{m}|\leq c$.

Any (right-continuous) Markov chain with values in $\operatorname{\mathbb{R}}$ is a semimartingale, since it is right-continuous and has locally finite variation, and is also quasi-left continuous, effectively because the jump times of a Poisson process are totally inaccessible; if this explanation is insufficient use Proposition 22.20 in [kallenberg] and note that Markov chains are Feller processes. As shown in I.4.28, a deterministic function $f:\operatorname{\mathbb{R}}_{+}\to\operatorname{\mathbb{R}}$ is a semimartingale iff it is right-continuous with finite variation over each compact interval, and is quasi-left continuous iff it is continuous, since any fixed time is predictable.

We will occasionally assume $X$ is defined only up to some predictable time $\zeta$ that may be finite; in this case, information about $X$ can be recovered from the stopped processes $X^{\tau_{n}}$ defined by $X_{t}^{\tau_{n}}=X_{t\wedge\tau_{n}}$, where $\tau_{n}$ is an announcing sequence for $\zeta$, i.e., an increasing sequence of stopping times with limit $\zeta$.

If $M$ is a local martingale satisfying $M_{0}=0$ and $|\Delta M|\leq c$ for some $c>0$, by I.4.1, $M$ is locally square-integrable, so by I.4.3, $M^{2}$ has a compensator, denoted $\langle M\rangle$ and called the predictable quadratic variation. Relative to the decomposition (I.4.18) $M=M^{c}+M^{d}$ into continuous and discontinuous martingale parts,

[TABLE]

and $\langle M^{d}\rangle$ is the compensator of $[M^{d}]=\sum_{s\leq t}(\Delta M_{s})^{2}$, the quadratic variation of $M^{d}$.

Lemma 1**.**

Let $M$ be a quasi-left continuous local martingale with $M_{0}=0$ and $|\Delta M|\leq c$ for some $c>0$. Then,

[TABLE]

is a local supermartingale with initial value $1$.

Proof.

Let $V_{t}$ be a continuous predictable process with locally finite variation, satisfying $V_{0}=0$, and let $E=\exp(M-V)$. Applying Itô’s formula I.4.57 using the function $x\mapsto e^{x}$,

[TABLE]

Noting that $e^{x}-1-x\leq\frac{1}{2}e^{c}x^{2}$ when $|x|\leq c$, the last term is bounded by

[TABLE]

Using this bound and taking the compensator of both sides in (2),

[TABLE]

The assumption of quasi-left continuity implies that $\langle M\rangle$ can be taken continuous (i.e. it has a continuous version; see I.4.3). Since it is the compensator of $[M,M]$, $\langle M\rangle$ is also predictable and has locally finite variation. Letting $V=(e^{c}/2)\langle M\rangle$, the same is true for $V$. With this choice of $V$, $E^{p}\leq 0$, which implies $E$ is a local supermartingale. ∎

Lemma 2**.**

Let $X$ be a special semimartingale with locally square-integrable martingale part $X^{m}$. Then, $X$ is quasi-left continuous iff $X^{p}$ and $\langle X^{m}\rangle$ are continuous.

Proof.

If $X$ is quasi-left continuous (qlc), then by I.2.35 its predictable projection ${}^{p}X=X_{-}$. Since the ${}^{p}(\,\cdot\,)$ operation is linear (which follows from uniqueness and property (ii) in I.2.28), and since ${}^{p}X_{-}=X_{-}$, it follows that

[TABLE]

Since $X$ is special it has a compensator $X^{p}$, and by I.3.21, $\Delta(X^{p})=\,^{p}(\Delta X)$. By the above, this is [math], i.e., $X^{p}$ is continuous. Using (1), $\Delta X^{m}=\Delta X$ which implies that $X^{m}$ is qlc, by definition of qlc. Since $X^{m}$ is locally square-integrable, by I.4.3, $\langle X^{m}\rangle$ is continuous.

On the other hand, if $\langle X^{m}\rangle$ is continuous then by I.4.3, $X^{m}$ is qlc. If in addition $X^{p}$ is continuous then $\Delta X=\Delta X^{m}$ which implies $X$ is qlc, by definition of qlc. ∎

Lemma 3**.**

Let $X$ be a quasi-left continuous semimartingale such that $|\Delta X|\leq c$ for some $c>0$. Then for $\lambda,a>0$ and $\bullet\in\pm$,

[TABLE]

Proof.

Notice that $X_{t}-X_{0}-X_{t}^{p}=X_{t}^{m}$ and that for $\lambda>0$ and $\bullet\in\pm$, $\langle\bullet\lambda X^{m}\rangle=\lambda^{2}\langle X^{m}\rangle$. As noted just above (1), since $X$ has bounded jumps it is special and by Lemma 2, $X^{m}$ is qlc. Take $M=\bullet\lambda X^{m}$ in Lemma 1, which has $|\Delta M|\leq\lambda c$, and use Doob’s inequality to find

[TABLE]

∎

For practicality’s sake we’ll use a slightly cruder version of (3). Since $1/2\leq\log 2$, if $\lambda c\leq 1/2$ then $e^{\lambda c}\leq 2$, so from (3) it follows that for $a>0$ and $\bullet\in\pm$,

[TABLE]

Using Lemma 2 as inspiration, say that a special semimartingale $X$ with locally square-integrable martingale part $X^{m}$ is quasi-absolutely continuous (qac) if both $X^{p}$ and $\langle X^{m}\rangle$ are absolutely continuous. In this case define the drift $\mu(X)=(\mu_{t}(X))_{t}$ and the diffusivity $\sigma^{2}(X)=(\sigma_{t}(X))_{t}$ for Lebesgue-a.e. $t$ by

[TABLE]

For deterministic processes, qac is equivalent to absolute continuity, since $\mu_{t}(f)=f(t)$, $\sigma^{2}_{t}(f)=0$ and absolute continuity implies locally finite variation. For Markov chains $X$ on $\operatorname{\mathbb{R}}$ with jump measure $\alpha(x,dy)$, if qac holds then $\mu$ and $\sigma$ are given by functions

[TABLE]

i.e., $\mu_{t}(X)=\mu(X_{t})$ and $\sigma^{2}_{t}(X)=\sigma^{2}(X_{t})$. Conversely, if $|\Delta X|\leq c$ and the total intensity $q(x)=\int_{\operatorname{\mathbb{R}}}\alpha(x,dy)$ of the jump measure is bounded on compact subintervals of $\operatorname{\mathbb{R}}$, then $X$ is qac up to the first explosion time $\sup_{r>0}\inf\{t:|X_{t}|\geq r\}$, and $\sigma^{2}(x)\leq c^{2}q(x)$.

Lemma 4**.**

[Product rule] Suppose $X_{t},Y_{t}$ are qac semimartingales on a common filtered probability space. Then $\langle X^{m},Y^{m}\rangle$ exists and is absolutely continuous, and $(XY)^{p}$ exists, is absolutely continuous, and $\mu_{t}(XY)=\frac{d}{dt}(XY)^{p}_{t}$ is given by

[TABLE]

where $\sigma_{t}(X,Y)=\frac{d}{dt}\langle X^{m},Y^{m}\rangle_{t}$.

Proof.

By definition of quadratic variation,

[TABLE]

Since $X=X_{0}+X^{p}+X^{m},\,Y=Y_{0}+Y^{p}+Y^{m}$ and $X_{0}+X^{p},\,Y_{0}+Y^{p}$ have locally finite variation, $[X,Y]=[X^{m},Y^{m}]$. Since $X^{m},\,Y^{m}$ are locally square-int, $[X^{m},Y^{m}]$ has compensator $\langle X^{m},Y^{m}\rangle$, so $XY$ has compensator

[TABLE]

The result will follow if we can show $\langle X^{m},Y^{m}\rangle$ is absolutely continuous. For any $s<t$, applying the Cauchy-Schwarz inequality to the symmetric, bilinear and semidefinite map $(X,y)\mapsto\langle X,Y\rangle_{t}-\langle X,Y\rangle_{s}$ gives

[TABLE]

Absolutely continuity of $t\mapsto\langle X^{m}\rangle_{t},\,\langle Y^{m}\rangle_{t}$ means that for any $\epsilon>0$ there is $\delta>0$ so that if $\sum_{i}|t_{i}-s_{i}|<\delta$ then $\sum_{i}|\langle X^{m}\rangle_{t_{i}}-\langle X^{m}\rangle_{s_{i}}|,\,\sum_{i}|\langle Y^{m}\rangle_{t_{i}}-\langle Y^{m}\rangle_{s_{i}}|<\epsilon$. Using the Cauchy-Schwarz inequality to obtain the second line,

[TABLE]

which shows that $\langle X^{m},Y^{m}\rangle$ is absolutely continuous. ∎

4 Early and middle phases

In this section we consider the behaviour of $|V_{t}|$ for $t\leq n^{1/2-o(1)}$. Define

[TABLE]

respectively the number of words created up to time $t$, and the number of words created and then deleted by time $t$. Theorem 1 is implied by the following two propositions, whose proof is the objective of this section.

Proposition 1**.**

For each $\epsilon>0$, $\lim_{n\to\infty}\mathbb{P}(\sup_{t\geq(\frac{1}{2}+\epsilon)\log n}|\ |V_{t}^{o}|-\frac{n}{2}|\geq n^{1/2+\epsilon})=0.$

Proposition 2**.**

For each $\epsilon>0$, $\lim_{n\to\infty}\mathbb{P}(\sup_{t\leq n^{1/2-\epsilon}}|V_{t}^{\times}|=o(n))=1.$

In words, in order to estimate $|V_{t}|$ we obtain good control on $|V_{t}^{o}|$, then show that $|V_{t}^{\times}|$ is not too big. We begin with $V_{t}^{o}$.

4.1 Creation of vocabulary

Our first task is to prove Proposition 1, and to do so we show that $|V_{t}^{o}|$ rises from [math] to $n/2+O(n^{1/2+o(1)})$ within $\frac{1}{2}\log n$ time, then remains constant. For a vertex $v$ let $N_{t}(v)=|W_{t}(v)|$ denote the size of the vocabulary of individual $v$, and let

[TABLE]

be the first time that every individual knows at least one word. Clearly $V_{t}^{o}$ is non-decreasing as a set, so $V_{\infty}^{o}=\lim_{t\to\infty}V_{t}^{o}$ exists and $|V_{\infty}^{o}|\leq n$. Once everyone knows a word, no new words are created, so $V_{t}=V_{T_{o}}^{o}=V_{\infty}^{o}$ for $t\geq T_{o}$. Proposition 1 is implied by the following two lemmas, in which we estimate $T_{o}$ and $V_{T_{o}}^{o}$.

Lemma 5**.**

For $c\geq 0$,

[TABLE]

Lemma 6**.**

Let $X=|V_{T_{o}}^{o}|$ be the number of words ever created. Then,

[TABLE]

Proof of Lemma 5.

Let $M_{t}=\{v:N_{t}(v)=0\}$ denote mute vertices, those not yet knowing a word, and observe that $T_{o}\leq t$ is equivalent to $|M_{t}|=0$. For each distinct ordered pair of vertices $(v,w)$, at rate $(n-1)^{-1}$, the directed edge $(v,w)$ has an event, and both $v$ and $w$ are removed from $M_{t}$, if either or both still belongs. If we let $Z_{t}=|M_{t}|$ denote the number of mute vertices at time $t$, it follows that $Z_{t}$ is a Markov chain with $Z_{0}=n$ and transitions

[TABLE]

We find that

[TABLE]

Letting $m(t)=\operatorname{\mathbb{E}}[Z_{t}]$, $m(0)=n$ and taking expectations in the above, $m^{\prime}(t)=-2m(t)$, which has the unique solution $m(t)=ne^{-2t}$. Fix $c\in\operatorname{\mathbb{R}}$ and let $t_{c}=\frac{1}{2}\log n+c$. Using Markov’s inequality,

[TABLE]

To get a lower bound we turn to $Z_{t}^{2}$, which has transitions

[TABLE]

so

[TABLE]

Letting $\nu(t)=\operatorname{\mathbb{E}}[Z_{t}^{2}]$, $\nu(0)=n^{2}$ and taking expectations above,

[TABLE]

so letting $\gamma=4-2/(n-1)$, using $m(t)=ne^{-2t}$ and solving the above DE, we find

[TABLE]

As above let $t_{c}=\frac{1}{2}\log n+c$, then $m(t_{c})=e^{-2c}$ and for fixed $c$,

[TABLE]

so $\operatorname{Var}(Z_{t_{c}})=\nu(t_{c})-m(t_{c})^{2}=e^{-2c}+o(1)$. Using Chebyshev’s inequality,

[TABLE]

The result follows by taking a union bound of both estimates. ∎

We note in passing that $|V_{0}^{o}|=0$ and $|V_{t}^{o}|$ increases by $1$ at rate $Z_{t}$. Heuristically, $Z_{t}\approx ne^{-2t}$, so $|V_{t}^{o}|\approx(n/2)(1-e^{-2t})$, for $t\leq\frac{1}{2}\log n$. This can be made precise using stochastic calculus, although we do not pursue it here.

Proof of Lemma 6.

Letting $X_{v}$ for each vertex $v\in V$ be the Bernoulli random variable equal to one if and only if $v$ speaks before listening, by construction and obvious symmetry, we have

[TABLE]

It follows that the expected number of words is given by

[TABLE]

To also compute the variance, fix $v,w\in V$ and let $B$ be the event that the first edge becoming active starting from $v$ or $w$ is edge $vw$. Since there are $n-1$ edges starting from each vertex,

[TABLE]

In addition, the two vertices cannot both speak before listening when $B$ occurs whereas the two events are independent on the event $B^{c}$ therefore

[TABLE]

Combining (7)–(8), we deduce that

[TABLE]

which, together with some basic algebra, gives the variance

[TABLE]

From (6) and (9) and Chebyshev’s inequality, we conclude that

[TABLE]

for all $\alpha>1/2$. This completes the proof. ∎

4.2 Maintenance of vocabulary

Next, we prove Proposition 2, that says that with probability tending to $1$ as $n\to\infty$,

[TABLE]

Clearly $V_{t}^{\times}$, like $V_{t}^{o}$, is non-decreasing, since once a word vanishes from the population, it does not come back. We first bound $|V_{t}^{\times}|$ by a simpler quantity. Say that agreement upon word $y$ occurs at $(v,w,t)$ if

[TABLE]

If word $w$ is created at some time $s\leq t$, then $w\in W_{s}(w)$, and remains in individual $w$’s vocabulary at least until the first time $t>s$ that agreement occurs at $(\cdot,w,t)\ \hbox{or}\ (w,\cdot,t)\}$. This implies

[TABLE]

In words, in order to delete a word $w$ from the population, it must at least be deleted from its source. Since each agreement contributes at most 2 to $H_{t}$, it follows that

[TABLE]

In order to control $A_{t}$ we first define some useful observable quantities. For $w\in V$ we recall the cluster $\mathcal{C}_{t}(w)$ of $w$, that is, the set of individuals that know word $w$ at time $t$:

[TABLE]

Recall that $N_{t}(v)=|W_{t}(v)|$ denotes the size of the vocabulary of individual $v$, and let

[TABLE]

denote the rate at which word $w$ is spoken. Let $J(w,v)$ denote the times at which $w$ speaks to $v$, and let

[TABLE]

noting that $N_{t}(v)\leq N_{t}^{\ell}(v)$ and $\{(N_{t}^{\ell}(v)):v\in V\}$ is a collection of independent Poisson processes with intensity 1. If we let

[TABLE]

then $N_{t}(v)=N_{t}^{\ell}(v)-N_{\tau_{a}(v,t)^{-}}^{\ell}(v)$, and in particular,

[TABLE]

Let $S_{t}(w)=|\mathcal{C}_{t}(w)|$ and $P_{t}(w)=(S_{t}(w)-1)/(n-1)$, and let $S_{t}=\max_{w}S_{t}(w)$. Each site $v$ that knows word $w$ speaks it at rate $N_{t}(v)^{-1}/(n-1)$ to each of the other $S_{t}(w)-1$ sites in $\mathcal{C}_{t}(w)$. Letting

[TABLE]

so that $A_{t}=\sum_{w}A_{t}(w)$, it follows that $A_{t}(w)$ increases by $1$ at rate

[TABLE]

Since $\sum_{w\in V}R_{t}(w)=n$ is the total speaking rate and $P_{t}(w)\leq(S_{t}-1)/(n-1)\leq S_{t}/n$, summing the above display over $w\in V$ we find

[TABLE]

We have reduced the problem of controlling $|V_{t}^{\times}|$ to that of controlling $S_{t}$. The following becomes the goal of this subsection. Since its proof has a few parts, we call it a theorem.

Theorem 3**.**

For small $\epsilon>0$,

[TABLE]

Before moving onto the proof of Theorem 3 we first use it to obtain Proposition 2.

Proof of Proposition 2.

From (11), for any $T>0$, $\sup_{t\leq T}A_{t}\leq\textrm{Poisson}(\int_{0}^{T}S_{u}du)$. Using Theorem 3, with probability $1-o(1)$ as $n\to\infty$

[TABLE]

Since $\mathbb{P}(\textrm{Poisson}(\lambda)\leq 2\lambda)\to 1$ as $\lambda\to\infty$ it follows that $\sup_{t\leq n^{1/2-\epsilon}}A_{t}=o(n)$ with probability $1-o(1)$, and since $|V_{t}^{\times}|\leq 2A_{t}$, the same is true for $|V_{t}^{\times}|$. ∎

To begin the proof of Theorem 3 we introduce a modified construction to help us make a coupling. First, for each ordered triple $(y,z,v)$ let $R_{t}(y,z,v)$ be the rate at which word $y$ is spoken by site $z$ to $v$, let $R_{t}(y,v)=\sum_{z}R_{t}(y,z,v)$ be the rate at which site $v$ hears word $y$, and as above let $R_{t}(y)=\sum_{v}R_{t}(y,v)$ be the rate at which word $y$ is spoken. We calculate

[TABLE]

Clearly $\sum_{y}R_{t}(y,v)=1$ for each $v,w$ and $t\geq 0$. Fix an ordering $v_{1}<\dots<v_{n}$ of $V$ and define an independent family $\{U_{v}:v\in V\}$ of augmented Poisson point processes with intensity 1, that will correspond to listening events. For $v\in V$, $1\leq i,j\leq n$ and $t\geq 0$ let

[TABLE]

noting that $\{I_{t}(v,i,j):1\leq i,j\leq n\}$ partitions $[0,1)$. Then, if $(t,u)\in U_{v}$ and $u\in I_{t^{-}}(v,i,j)$, word $v_{i}$ is spoken by $v_{j}$ to $v$, which defines the process. Using this construction and given $C,R>0$ we obtain upper bounds $\mathbf{C}_{t}(w),\mathbf{R}_{t}(w)$ on $\mathcal{C}_{t}(w),R_{t}(w)$ for all $w\in V$, valid up to the time

[TABLE]

That is, we obtain for each $w\in V$ a pair of processes $\mathbf{C}_{t}(w),\mathbf{R}_{t}(w)$ with nice properties, such that $\mathcal{C}_{t}(w)\subseteq\mathbf{C}_{t}(w)$ and $R_{t}(w)\leq\mathbf{R}_{t}(w)$ for $t\leq T_{C,R}$ pointwise on realizations of the process. Given $w\in V$, $\mathbf{C}_{t}(w),\mathbf{R}_{t}(w)$ are non-decreasing and defined as follows. For $i\in\{1,\dots,n\}$ let

[TABLE]

Define

[TABLE]

Let $i$ be such that $w=v_{i}$. Initially, $\mathbf{C}_{0}(w)=\{w\}$ and $\mathbf{R}_{0}(w)=1+C$. $\mathbf{R}_{t}(w)$ is defined as follows.

[TABLE]

Then, $\mathbf{C}_{t}(w)$ is defined as follows.

[TABLE]

We demonstrate the claimed comparison.

Lemma 7**.**

For each $w\in V$, $\mathcal{C}_{t}(w)\subseteq\mathbf{C}_{t}(w)$ and $R_{t}(w)\leq\mathbf{R}_{t}(w)$ for $t<T_{C,R}$.

Proof.

Let

[TABLE]

then $\mathcal{C}_{t}(w)\subset\{w\}\subseteq\mathbf{C}_{t}(w)$ and $R_{t}(w)\leq 1\leq\mathbf{R}_{t}$ for $t<\tau_{c}(w)$. For the remainder, assume $t\geq\tau_{c}(w)$ and let $i$ be such that $w=v_{i}$. By construction, $v\in V$ is added to $\mathcal{C}_{t}(w)$ if

[TABLE]

and otherwise, $\mathcal{C}_{t}(w)$ does not increase. If $t\geq\tau_{c}(w)$ then $N_{t}(w)\geq 1$, and if $z\notin H_{t}$ then $N_{t}^{\ell}(z)=N_{t}(z)$. So, from the second line of (12),

[TABLE]

If $w\in\mathcal{C}_{t}(w)$ then $N_{t}(w)^{-1}\leq 1$. By definition of $T_{C,R}$, if $\mathcal{C}_{t}(w)\subseteq\mathbf{C}_{t}(w)$ and $t<T_{C,R}$ then

[TABLE]

If $v\in\mathbf{C}_{t}(w)$ and $t<T_{C,R}$ then $N_{t}^{\ell}(v)\geq N_{t}^{\ell}(v,R)+1$, since this implies existence of a point in

[TABLE]

which is counted in $N_{t}^{\ell}(v)$ but not in $N_{t}^{\ell}(v,R)$. If $\mathcal{C}_{t}(w)\subseteq\mathbf{C}_{t}(w)$ it follows that $R_{t}(w,v)\leq\mathbf{R}_{t}(w)/(n-1)$ for each $v$ which implies the containment $\mathcal{C}_{t}(w)\subseteq\mathbf{C}_{t}(w)$ is preserved across transitions (13) that cause $\mathcal{C}_{t}(w)$ to increase. Since $\mathbf{C}_{t}(w)$ is non-decreasing and transitions are well-ordered this implies $\mathcal{C}_{t}(w)\subseteq\mathbf{C}_{t}(w)$ for $t<T_{C,R}$. It remains to check $R_{t}(w)\leq\mathbf{R}_{t}(w)$ for $\tau_{c}(w)\leq t<T_{C,R}$. But in this case, (10) and the previous argument give

[TABLE]

∎

Next we fix $w$ and examine $\mathbf{C}_{t}(w),\mathbf{R}_{t}(w)$ assuming $t<T_{C,R}$, and dropping the $(w)$ for neatness. Notice that $|\mathbf{C}_{t}|$ is non-decreasing and increases by $1$ at rate at least $(1+C)(n-|\mathbf{C}_{t}|)/(n-1)$, which implies $\lim_{t\to\infty}|\mathbf{C}_{t}|=n$. Since $|\mathbf{C}_{t}|$ increases by one at a time, let $y_{1},\dots,y_{n}$ be the order in which vertices are added to $\mathbf{C}_{t}$, with $w=y_{1}$, and condition on $(y_{1},\dots,y_{n})$. We track $Z_{t}=|\mathbf{C}_{t}|$ and $N_{t}^{i}=N_{t}^{\ell}(y_{i},R),\ i=1,\dots,n$ which suffices to determine $\mathbf{C}_{t},\mathbf{R}_{t}$. Let $t_{i}=\inf\{Z_{t}=i\}$ denote the time at which $y_{i}$ is added to $\mathbf{C}_{t}$, and let $k$ be such that $w=v_{k}$. For $i\in\{2,\dots,n\}$, $t_{i}$ is the least value of $t$ such that there is a point

[TABLE]

and in addition, this point belongs to $U_{y_{i}}$. Using this and basic properties of exponential random variables, together with the thinning property of the Poisson process, we find that conditioned on $(y_{1},\dots,y_{n})$,

[TABLE]

is a Markov chain with the following transitions:

[TABLE]

In particular, $\{(N_{t}^{i})_{t<T_{C,R}}:i=1,\dots,n\}$ is an i.i.d. collection of Poisson processes with intensity $1-R/(n-1)$. Since the above does not depend on the choice of values for $(y_{1},\dots,y_{n})$ the same holds unconditionally. Thus $Z_{t}$ can be viewed as follows: initially $Z_{0}=1$, then subject to the random environment determined by the $\{(N_{t}^{i})\}_{i=2}^{n}$, $Z_{t}$ increases by $1$ at rate $\mathbf{R}_{t}(n-Z_{t})/(n-1)$. Let

[TABLE]

and let $(X_{t})$ denote the process with $X_{0}=1$ that increases by $1$ at rate $\Lambda_{t}(X_{t})$. Since $(n-Z_{t})/(n-1)\leq 1$ and $\Lambda_{t}$ is non-decreasing in $z$, it follows that

[TABLE]

We can think of $(X_{t})$ as a branching process with immigration rate $1+C$, in which individual $i$ produces offspring at the time-decreasing rate $1/(1+N_{t}^{i})$. Two tasks lie ahead. The first is to estimate $(X_{t})$. The second is to estimate $T_{C,R}$. We then combine the results to obtain Theorem 3. This is outlined as follows.

Proposition 3**.**

Let $b=1+C$. For small $\epsilon>0$, $b\leq(8\log n)^{4}$ and $R=o(n)$,

[TABLE]

Proposition 4**.**

For small $\epsilon>0$, $b=(8\log n)^{4}$ and $R=b+(\log n)^{11}$,

[TABLE]

Proof of Theorem 3.

Use Propositions 3 and 4 with $b=(8\log n)^{4}$ and $R=b+(\log n)^{11}$. ∎

4.2.1 Estimation of $(X_{t})$

Since $n$ does not appear in the definition of $(X_{t})$ we may as well define it using an infinite sequence $\{(N_{t}^{i})_{t\geq 0}:i=1,2,\dots\}$ of Poisson processes with intensity $r=1-R/(n-1)$. Clearly $r\leq 1$. Since $R$ will be chosen $o(n)$, we will have $r\to 1$ as $n\to\infty$, so throughout we assume $r\geq 1/2$.

We begin with a useful heuristic. Let $b=1+C$. Replacing $N_{t}^{i}$ with its expectation $rt$, $X_{t}$ increases by $1$ at rate $b+X_{t}(1+rt)^{-1}$, which we approximate with the differential equation

[TABLE]

Let $m(t)=\exp(\int_{0}^{t}(1+rs)^{-1}ds)=(1+rt)^{1/r}$. The above equation is linear and has solution

[TABLE]

If $r$ is close to $1$ then $x(t)$ grows just a bit faster than linearly in time. In order to analyze $(X_{t})$ we break it up into two steps:

Up to a fixed time $T$, when the $N_{t}^{i}$ are fairly small. 2. 2.

From time $T$ to $\infty$, when the $N_{t}^{i}$ are fairly large.

The reason to do this is because the estimates that say $|N_{t}^{i}-rt|=o(rt)$ are only effective once $rt$ has had time to increase. The following is the main result of this subsection.

Proposition 5**.**

Let $b=1+C$. There exist $M,x_{0}\in[1,\infty)$ so that for $r\geq 1/2$ and $x\geq b\vee x_{0}$,

[TABLE]

Recall $S_{t}=\max_{w}|\mathcal{C}_{t}(w)|$. Using this result we can prove Proposition 3.

Proof of Proposition 3.

For each $w\in V$, using Lemma 7 and (14),

[TABLE]

Applying the result of Proposition 5 and taking a union bound over $w$, if $r\geq 1/2$ and $x\geq x_{0}$ then

[TABLE]

If $R=o(n)$ then recalling that $r=1-R/(n-1)$, $1/r\leq 1+\epsilon/3$ for large $n$. Since $r\leq 1$, $1+rt\leq 1+t$, and if $\epsilon>0$ is small then $r\geq 1/2$. Letting $x=(8\log n)^{4}$, the probability is $o(1)$ and since $\log(1+t)=O((1+t)^{\epsilon/2})$, it follows that

[TABLE]

uniformly in $t$, as $n\to\infty$. ∎

We tackle the proof of Proposition 5 in a couple of steps.

Step 1. We obtain a somewhat crude upper bound on $(X_{t})$ that has the virtue of being effective starting at time [math]. For $i\geq 1$ let $t_{i}=\inf\{t:X_{t}=i\}$, define $N_{i}=N_{t_{i}}^{i}$ then define $Y_{t},\ Q_{t}$ by

[TABLE]

In words, at the moment $t_{i}$ an individual $i$ is added to the process, the corresponding counting process $N_{t}^{i}$ is stopped, so that $i$ always contributes $(1+N_{t_{i}}^{i})^{-1}$ to $Q_{t}$. Since $(1+N_{t_{i}}^{i})^{-1}\geq(1+N_{t}^{i})^{-1}$ for $t\geq t_{i}$, $(X_{t})$ is dominated by $(Y_{t})$. The next result controls $(Y_{t})$.

Lemma 8**.**

There is $M_{1}\in[1,\infty)$ so that for $a\geq 2,\ b\geq 1$ and $r\geq 1/2$,

[TABLE]

Proof.

Begin by observing that $(Q_{t})$ has the concise description

[TABLE]

where the increment $\Delta_{t}\stackrel{{\scriptstyle d}}{{=}}(1+\textrm{Poisson}(rt))^{-1}$ is independently sampled every time there is a jump. Our first task is to control the size of $Q_{t}$. We compute the drift:

[TABLE]

Let $g(t)=\exp(\int_{0}^{t}\operatorname{\mathbb{E}}[(1+\textrm{Poisson}(rs))^{-1}]ds)$. Using Lemma 24 with $b(t)=0$ and $c=1$, for $a\geq 2$ we find

[TABLE]

This translates to a bound on $(Y_{t})_{t\geq 0}$ as follows. Since $\mu_{t}(Y)=Q_{t}$,

[TABLE]

Since $(Y_{t})_{t\geq 0}$ has transition rate $Q_{t}$ and jump size exactly 1, $\sigma^{2}(Y_{t})=Q_{t}$. Taking $\lambda=1/2$ in (4) (which satisfies $c\lambda\leq 1/2$) while noting $Y_{0}=1$,

[TABLE]

Combining with (16) and taking a union bound,

[TABLE]

Intuitively, $g(t)$ grows roughly like $m(t)$. Let $\xi=\textrm{Poisson}(\lambda)$. Since $x\mapsto(1+x)^{-1}$ is convex, the inequality $\operatorname{\mathbb{E}}[(1+\xi)^{-1}]\geq(1+\operatorname{\mathbb{E}}[\xi])^{-1}$ goes in the wrong direction for an upper bound on $g(t)$. Anticipating our needs, we let $x=\lambda^{\alpha}/2$ in (35) to find

[TABLE]

Using the fact that $(1+\xi)^{-1}\leq 1$ and that probabilities are at most $1$, then using (34),

[TABLE]

Also, if $0<\alpha<1/2$ and $0<r\leq 1$ then

[TABLE]

Let $c(r)=\inf\{c(r,\alpha):\alpha\in(0,1/2)\}$ and let $c=c(1/2)$. Since $c(r,\alpha)$ decreases with $r$, it follows that $c(r)\leq c$ for $r\geq 1/2$. Recalling $m(t)=\exp(\int_{0}^{t}ds/(1+rs))$ defined earlier, it follows that

[TABLE]

Since $m(t)=(1+rt)^{1/r}$ and $1/(r(1+1/r))=1/(r+1)$, and since $r>0$, it follows that

[TABLE]

If $a\geq 2$ and $b\geq 1$ then since $c>0$, $1+a(1-3be^{c}/2)\leq 0$ and

[TABLE]

To conclude, take $M_{1}=\frac{3}{2}e^{c}$, use (17) and note $e^{-a/2}+e^{-(a-2)b/4}\leq 2e^{-(a-2)/2}$ for $b\geq 2$. ∎

Step 2. Next, we do two things.

Lemma 9. We control the environment $\{(N_{t}^{i})\}_{i\geq 1}$ for $t\in[T,\infty)$. 2. 2.

Lemma 10. We use this to get an upper bound on $(X_{t})$ for $t\in[T,\infty)$.

Fix $\alpha\in(0,1/2)$, then let

[TABLE]

denote the last passage time of $N_{t}^{i}$ below the curve $v(t)=rt-(rt)^{1/2+\alpha}/2$, and for $t\geq 0$ let

[TABLE]

Note that $\Lambda_{t}(x)\leq b+x/(1+v(t))$ for $x\leq I_{t}$.

Lemma 9**.**

There is $T_{0}>0$ so that for $r\geq 1/2$ and $\alpha\in(0,1/2)$,

[TABLE]

Proof.

For each $i$, using Lemma 23 with $\lambda=r$ and $\tau_{lp}(i)=\tau_{2}$,

[TABLE]

Let $f(t)=2t^{1/2-\alpha}e^{-(rt)^{2\alpha}/4}$, so the right-hand side above is $1/f(t)^{2}$. Then, a union bound and the fact that $f(t)^{-1}\leq 1$ gives

[TABLE]

For $T>0$ let

[TABLE]

Note that $c_{1},c_{2}\in(0,\infty)$ and $\lim_{T\to\infty}c_{1},c_{2}=1$ uniformly for $r\in[1/2,1]$. Since $I_{t}$ is non-decreasing, if $I_{t}\geq f(t)$ and $t>T$ then

[TABLE]

Taking a union bound over the estimate at times $T+k$, $k\geq 0$ gives

[TABLE]

The right-hand side is at most

[TABLE]

and using (33), this is at most

[TABLE]

Then note that $c_{1}\geq 1/2$ and $16r^{-2\alpha}c_{1}c_{2}\leq 17$ for $T$ large enough, uniformly for $r\in[1/2,1]$. ∎

Lemma 10**.**

Given $\alpha\in(0,1/2)$ let $\tau=\inf\{t>T:X_{t}>I_{t}\}$. There is $M_{2}\in[1,\infty)$ so that for $r\in[1/2,1)$ and $a\geq 2$,

[TABLE]

Proof.

Since, as noted before, $\Lambda_{t}(y)\leq b+y/(1+v(t))$ for $y\leq I_{t}$, it follows that for $T\leq t<\tau$ and conditioned on $X_{T}\leq x$, $(X_{t})$ is dominated by the process $\tilde{X}_{t}$ with $\tilde{X}_{T}=x$ that increases by $1$ at rate $b+\tilde{X}_{t}/(1+v(t))$. We proceed as in the proof of Lemma 8. We have

[TABLE]

For $a>0$ let $E_{a}=\{\sup_{t\geq T}\tilde{X}_{t}/g(t)\geq ax+b\int_{T}^{t}ds/g(s)\}$, where $g(t)=\exp(\int_{T}^{t}\ell(s)ds)$, and let $\beta=b\int_{T}^{\infty}ds/g(s)^{2}$. Using Lemma 24 with $c=1$, for $a\geq 2$ we find

[TABLE]

Recall $v(t)=rt-(rt)^{1/2+\alpha}/2$. Using (34) with $\lambda=rt$,

[TABLE]

Let $c(r)=\int_{0}^{\infty}(1+rs)^{-3/2+\alpha}ds$, which is finite for $\alpha\in(0,1/2)$ and $r\in(0,1]$ and decreases with $r$. Let $c=c(1/2)$, so that $c(r)\leq c$ for $r\in[1/2,1]$. Combining and noting $1+rt\leq 1+t$,

[TABLE]

Using that $\ell(t)\geq 1/(1+rt)$, we obtain the complementary bound $g(t)\geq((1+rt)/(1+rT))^{1/r}$. In this way

[TABLE]

Using the more generous lower bound $g(t)\geq(1+rt)/(1+rT)^{1/r}$ and noting $\log(1+rT)\geq 0$ and $1/2\leq r\leq 1$, we find

[TABLE]

Let $M_{2}=e^{c}$ and rearrange terms in the formula for $E_{a}$ to complete the proof. ∎

Proof of Proposition 5.

We note the result of Lemma 8 applies to $(X_{t})$ since it is dominated by $(Y_{t})$. Fix $\alpha=1/4$ in Lemma 9 and 10. Fix $T>0$ and let $L(t)=\frac{1}{2}e^{(rt)^{1/2}/8}$ and $x_{1}=abM_{1}(1+rT)^{1+1/r}$. Let

[TABLE]

be the complement of the event from, respectively, Lemma 8, 9 and 10. On $E$,

[TABLE]

In particular, $X_{T}\leq x_{1}$, so using Lemma 10, for $b\geq 1$ and $T$ large enough,

[TABLE]

Using our choice of $x_{1}$, on $G$ we find

[TABLE]

Since $a,M_{2}\geq 1$, on $E\cap G$ the inequality holds for all $t<\tau$. Taking $a=T^{1/2}$, since $r\geq 1/2$,

[TABLE]

Using Lemmas 8 and 9, for $T>T_{0}$ and some $T_{0}>0$, we find $x_{1}\geq 1$ and

[TABLE]

If $T$ is large enough uniformly for $r\in[1/2,1]$ the above and (19) show

$\mathbb{P}(E^{c}),\mathbb{P}(G^{c}\cap E)\leq\frac{1}{2}T^{3/4}e^{-(rT)^{1/2}/4}$, so a union bound gives

[TABLE]

Choose $T$ so that for $x$ from the statement of the Proposition, $x=(1+rT)^{2}$. It suffices to check that $\tau=\infty$ on $E\cap F\cap G$. But on $E\cap F\cap G$, noting that $2M_{2}\leq M$ and $r\leq 1$ on the second line,

[TABLE]

Since $\tau=\inf\{t>T:X_{t}>I_{t}\}$ and by assumption, $b\leq x=(1+rT)^{2}$, it suffices that

[TABLE]

for $t\geq T$, which is true for $T$ large enough, uniformly for $r\in[1/2,1]$. ∎

4.2.2 Estimation of $T_{C,R}$

Write $T_{C,R}=T_{C}\wedge T_{R}$, where

[TABLE]

Proposition 6**.**

Let $b=1+C$. If $R\geq b+(\log n)^{11}$ and $b\leq(8\log n)^{4}$ then

[TABLE]

Proposition 7**.**

For each $\epsilon>0$ and $b=(8\log n)^{4}$, if $R\leq(\log n)^{12}$ then

[TABLE]

Proof of Proposition 4.

Notice that

[TABLE]

then use Propositions 6 and 7 and take a union bound. ∎

Next we prove Proposition 6, which is the simpler of the two.

Proof of Proposition 6.

Since for $t<T_{C,R}$, each $\mathbf{R}_{t}(w)$ is dominated by $\Lambda_{t}(X_{t})$, taking a union bound we find

[TABLE]

For a given function $\Phi(t)$,

[TABLE]

Taking $x=(8\log n)^{4}$ in Proposition 5,

[TABLE]

We have the trivial bound $\Lambda_{s}(x)\leq b+x$, and it follows from the bound on $\ell(t)$ given in the proof of Lemma 10 that

[TABLE]

Let $L(s)=s^{1/4}e^{(rs)^{1/2}/4}$. Taking $T=2(8\log n)^{2}$, if $n$ is large enough uniformly for $r\in[1/2,1]$ then $\Phi(s)\leq L(s)$ for $s\geq T$ so using Lemma 9 and the above bounds, for $1/2\leq r\leq 1$ and large $n$ we find

[TABLE]

if $T$ is large enough. If $R=o(n/\log n)$, then noting $r=1-R/(n-1)$ and $r\geq 1/2$ for large $n$, if $s\leq n^{1/2}$ then for large $n$,

[TABLE]

which approaches $2$ as $n\to\infty$. It follows that

[TABLE]

and a similar estimate shows that $\Phi(2(8\log n)^{2})=O((\log n)^{10})$. The result follows. ∎

It remains to prove Proposition 7. Define the non-decreasing spacetime set of points

[TABLE]

To get a more workable quantity we will use the fact that

[TABLE]

This way,

[TABLE]

So, to estimate $T_{C}$ we control contributions to $C_{t}(w)$. Let $Q_{t}(w)$ denote the rate at which $\mathcal{A}_{t}(w)$ increases. Let $\{N_{t}^{i}:t\geq 0,\ i\geq 1\}$ be an independent collection of Poisson processes with intensity 1, let $Q,T>0$ and let $N(t)$ be an independent Poisson process with intensity $Q$. Let $t_{i}=\inf\{t:N(t)=i\}$ and let

[TABLE]

Let $T_{Q}(w)=\inf\{t:Q_{t}(w)>Q\}$ and $T_{Q}=\min_{w}T_{Q}(w)$. Then for any $w$,

[TABLE]

In the next lemma we control $B_{t}$.

Lemma 11**.**

For $T>0$, $T_{0}\geq 1$ and $Q\geq 1$,

[TABLE]

Proof.

We first control the value of $B_{T}$, then of $B_{t}$ for $t\in[T-1,T]$, then take a union bound to control the value over the interval $[0,T]$. Let $\tilde{N}(t)=N(T)-N(T-t)$ and fix $T_{0}<T$. Using (4) with $X_{t}=\tilde{N}(t)$, $X_{t}^{p}=\langle X^{m}\rangle_{t}=Qt$, $c=1$, $a=QT_{0}/2$, $\lambda=1/2$ and $\bullet=+$ and noting that $c\lambda\leq 1/2$,

[TABLE]

since $Q\geq 1$. Using the same result except with $X_{t}=N_{t}^{i}$, $X_{t}^{p}=\langle X^{m}\rangle_{t}=t$, $\bullet=-$, $a=T_{0}/4$ and $\lambda=1/4$, and taking a union bound,

[TABLE]

Let $S=\{t\leq T:\tilde{N}(t)>\tilde{N}(t^{-})\}$ be the jump times of $\tilde{N}(t)$, and label them in increasing order as $\tilde{t}_{1},\tilde{t}_{2},\dots,\tilde{t}_{\tilde{N}_{T}}$. On the complement of both events above, $\tilde{N}(T)<2QT$ and $\tilde{t}_{i}>i/2Q$ for $i\geq 2QT_{0}$, and so $N_{\tilde{t}_{i}}^{i}\geq i/4Q$, and this gives

[TABLE]

To see that this also bounds $B_{t}$ for $t\in[T-1,T]$, replace $N_{\tilde{t}_{i}}$ with $N_{\tilde{t}_{i}-1}\geq(i/2Q-1)/2$ and use in the above to obtain the bound

[TABLE]

which has the same upper bound, assuming $T_{0}\geq 1$ so that $\log(1/2+T_{0}/2)\geq 0$. The same works for $T_{1}<T$ and $T_{0}\geq 1$ to give

[TABLE]

since $2QT_{0}e^{-T_{0}/16}\geq e^{-T_{0}/4}$. Taking a union bound over $T_{1}=1,2,\dots,\lfloor T\rfloor,\lfloor T\rfloor+1$ and noting $\lfloor T\rfloor\leq T$ gives the result. ∎

It remains to prove the following result.

Proposition 8**.**

For small $\epsilon>0$ and any $k>0$, $Q=\log n$ and $R\leq(\log n)^{k}$,

[TABLE]

Before proving it, we show how it implies Proposition 7. Use whp (with high probability) to denote an event whose probability tends to 1 as $n\to\infty$. Note that if $E_{1},E_{2}$ whp then $E_{1}\cap E_{2}$ whp.

Proof of Proposition 7.

We want to show that $T_{C}>n^{1/2-\epsilon}\wedge T_{R}$ whp. Since $T_{C}\geq T_{C}\wedge T_{Q}$, if $T_{C}\wedge T_{Q}>n^{1/2-\epsilon}\wedge T_{R}$ then $T_{C}>n^{1/2-\epsilon}\wedge T_{R}$. Moreover

[TABLE]

Proposition 8 says that $T_{Q}>n^{1/2-\epsilon}\wedge T_{R}\wedge T_{C}$ whp, so it is enough to show that if $b=(8\log n)^{4}$ and $Q=\log n$ then $T_{C}>n^{1/2-\epsilon}\wedge T_{R}\wedge T_{Q}$ whp, or equivalently that

[TABLE]

In Lemma 11 take $T=n^{1/2}$, $T_{0}=48\log n$ and $Q=\log n$ to find that

[TABLE]

Then, using (21) and Proposition 8 and taking a union bound over the $n$ possible values of $w$,

[TABLE]

The result then follows from (20) and the fact that $98(\log n)^{2}<(8\log n)^{4}$ for large $n$. ∎

By taking a union bound over $w$ and noting the probability does not depend on $w$, to obtain Proposition 8 it is sufficient to show that for any $w$ and small $\epsilon>0$,

[TABLE]

noting that the probability is the same for any $w$. There are three ways that $\mathcal{A}_{t}(w)$ increases:

a site already in $H_{t}$ is added to $\mathbf{C}_{t}$, 2. 2.

agreement occurs at a site already in $\mathbf{C}_{t}$, or 3. 3.

a site is added simultaneously to $\mathbf{C}_{t}$ and $H_{t}$.

Let $Q_{t}^{i}(w),\ i=1,2,3$ denote the rate of each event, so that $Q_{t}(w)=\sum_{i=1}^{3}Q_{t}^{i}(w)$. Since each site in $V\setminus\mathbf{C}_{t}(w)$ is added to $\mathbf{C}_{t}(w)$ at rate $\mathbf{R}_{t}(w)/(n-1)\leq R/(n-1)$,

[TABLE]

Since there are $|\mathbf{C}_{t}(w)\cap\mathcal{C}_{t}(v)|$ sites in $\mathbf{C}_{t}(w)$ that can agree on word $v$, and each word is spoken at rate at most $R/(n-1)$ to each site,

[TABLE]

recalling that $S_{t}=\max_{w}|\mathcal{C}_{t}(w)|$ is the size of the largest cluster. Each time a person speaks, the probability that agreement occurs is at most $S_{t}/(n-1)$. Since $\mathbf{C}_{t}(w)$ increases at rate $\leq R$, it follows that

[TABLE]

The reader may think that $Q_{t}^{3}(w)$ should be [math], since a new addition to a cluster does not yet know the word. However, the upper bound cluster $\mathbf{C}_{t}(w)$ can grow when in the process itself, a word other than $w$ is being spoken. Using Proposition 5 we control $Q_{t}^{1}(w)$ and $Q_{t}^{3}(w)$, which is two thirds of Proposition 8.

Lemma 12**.**

For each $w$, small $\epsilon>0$, $R\leq n^{\epsilon}$ and $i=1,3$,

[TABLE]

Proof.

From (23) and (25) and the choice of $R$, it suffices to show that

[TABLE]

The result of Proposition 3 holds with the probability being $o(1/n)$ – to see this, take $x=(12\log n)^{4}$ in the proof. This gives

[TABLE]

while $\Phi(t,x)$ is still $o((\log n)^{9}(1+t)^{1+\epsilon})$, uniformly in $t$ as $n\to\infty$. The desired result for $i=3$ then follows from (25), since $\sup_{t\leq n^{1/2-\epsilon}}(\log n)^{9}(1+t)^{1+\epsilon}=(\log n)^{9}(1+n^{1/2-\epsilon})^{1+\epsilon}=o(n^{1-\epsilon})$. To get the result for $i=1$ recall from the beginning of this section that $|H_{t}|\leq 2A_{t}$, the number of agreements up to time $t$, and from (11) that $A_{t}\leq\textrm{Poisson}(\int_{0}^{u}S_{u}du)$. Using the above bound on $S_{t}$, with probability $1-o(1/n)$,

[TABLE]

for large $n$. From Lemma 22, $\mathbb{P}(\textrm{Poisson}(\lambda)>2\lambda)\leq e^{-\lambda/3}$, so it follows that

[TABLE]

∎

It remains to control $|\mathbf{C}_{t}(w)\cap\mathcal{C}_{t}(v)|$. We’ll make use of the estimates from Lemma 11, namely that for a Poisson process $N(t)$ with intensity 1,

[TABLE]

First we modify slightly the construction from the beginning of Section 4.2, using a randomization trick. The reason it needs modifying is to ensure the growth of $\mathbf{C}_{t}(w)$ and any $\mathcal{C}_{t}(v)$ are not strongly correlated. Since we only randomize the location of “excess” events that expand $\mathbf{C}_{t}(w)$, the reader may verify that up to a random permutation of certain vertices, the marginal distribution of each $\mathbf{C}_{t}(w)$, and its domination of $\mathcal{C}_{t}(w)$, are unchanged.

To carry out the modification, make the $\{U_{v}\}$ doubly-augmented, that is, each $U_{v}$ is again a Poisson point process with intensity 1, but on $[0,\infty)\times[0,1]^{2}$ instead of $[0,\infty)\times[0,1]$. $\mathbf{R}_{t}(w)$ is defined in the same way as before, and $\mathbf{C}_{t}(w)$ is defined as follows.

[TABLE]

In other words,

• •

if $\mathcal{C}_{t}(w)$ was about to include $v$, then $\mathbf{C}_{t}(w)$ will too, and

• •

if $\mathbf{C}_{t}(w)$ increases when $\mathcal{C}_{t}(w)$ does not, then with respect to

what other clusters are doing, it does so as randomly as possible.

We now control the size of $\mathbf{C}_{t}(w)\cap\mathcal{C}_{t}(v)$, for any $v\neq w$. “wp” is shorthand for “with probability”.

Lemma 13**.**

For any $\epsilon,k>0$, if $R\leq n^{\epsilon/4}$ then

[TABLE]

Proof.

Let $K_{t}=\sum_{v\neq w}|\mathbf{C}_{t}(w)\cap\mathcal{C}_{t}(v)|$. There are three ways $K_{t}$ can increase.

$\mathbf{C}_{t}(w)$ acquires a site that belongs to some (possibly many) $\mathcal{C}_{t}(v)$, $v\neq w$, 2. 2.

some $\mathcal{C}_{t}(v)$, $v\neq w$ acquires a site that belongs to $\mathbf{C}_{t}(w)$, and 3. 3.

$\mathbf{C}_{t}(w)$ and some $\mathcal{C}_{t}(v)$, $v\neq w$ simultaneously acquire the same site.

It suffices to show the contribution to $\sup_{t\leq n^{1/2-\epsilon}\wedge T_{C,R}}K_{t}$ from each item is $o(n/(\log n)^{k})$ wp $1-o(1/n)$. For item 1, the increase is at most $\max_{v}N_{t}^{\ell}(v)$, while for items 2,3 the increase is at most $1$, since at each transition, $C_{t}(v)$ increases for at most one $v$, and by at most $1$. Let $R_{1}(t),R_{2}(t),R_{3}(t)$ denote the rate of each event. Then,

[TABLE]

Since $N_{t}^{\ell}(v)\leq N_{n^{1/2}}^{\ell}(v)$ for $t\leq n^{1/2}$ and each $v$, and since each $N_{n^{1/2}}^{\ell}(v)\sim\textrm{Poisson}(n^{1/2})$, using (26) with $t=T=n^{1/2}$ and a union bound,

[TABLE]

For $t<T_{C,R}$, $\mathbf{R}_{t}(w)\leq R$, so wp $1-o(1/n)$, the contribution from item 1 is at most $\textrm{Poisson}(n^{1/2+\epsilon/2}Rn^{1/2-\epsilon})=\textrm{Poisson}(Rn^{1-\epsilon/4-\epsilon^{2}/2})$ which if $R\leq n^{\epsilon/4}$ is at most $2n^{1-\epsilon^{2}/2}=o(n/(\log n)^{k})$ for any fixed $k$ wp $1-o(1/n)$. Similarly, but more simply since the increase per transition is 1, the contribution from item 3 is at most $\textrm{Poisson}(Rn^{1/2-\epsilon})$ which is at most $n^{1/2}=o(n/(\log n)^{k})$ wp $1-o(1/n)$.

For item 2, note that $\sum_{v\neq w}R_{t}(v)\leq\sum_{v}R_{t}(v)=n$ and that for $t<T_{C,R}$, $\mathbf{C}_{t}(w)$ is dominated by $X_{t}$. Applying Proposition 5, bounding $\log(1+t)$ by $\log(1+n^{1/2})$ for $t\leq n^{1/2}$ and using the trivial but convenient $n/(n-1)\leq 2$ for $n\geq 2$, we find that for $x\geq 1+C$ large enough,

[TABLE]

Taking $x=(12\log n)^{4}$ the probability above is $o(1/n^{2})$. Thus the contribution from item 2 is at most $\textrm{Poisson}(f(n^{1/2-\epsilon}))$, where

[TABLE]

using $\log(1+n^{1/2})\leq(12\log n)^{4}$ and $1+1/r\geq 1$. If $R=o(n)$ then for any $\epsilon>0$, $1/r\leq 1+\epsilon$ for large $n$. Therefore

[TABLE]

It follows as before that $\textrm{Poisson}(f(n^{1/2-\epsilon}))=o(n/(\log n)^{k})$ wp $1-o(1/n)$, and the proof is complete. ∎

Combining this with the other term in (24) we control $Q_{t}^{2}(w)$.

Lemma 14**.**

For any $k>0$ and small $\epsilon>0$, each $w$ and $R\leq(\log n)^{k}$,

[TABLE]

Proof.

From the proof of Lemma 12 we know that $\mathbb{P}(\sup_{t\leq n^{1/2-\epsilon}\wedge T_{C,R}}S_{t}>n^{1-\epsilon})=o(1/n)$. Using this, (24), $R\leq(\log n)^{k}$ and Lemma 13,

[TABLE]

If $n$ is large then $(n/(\log n)^{k}+n^{1-\epsilon})(\log n)^{k}/(n-1)\leq 2$ and the result follows. ∎

Proof of Proposition 8.

This follows from (22), and Lemmas 12 and 14. ∎

5 Final phase

5.1 Markov chain and ODE heuristic

Using the notation of chemical reactions, we describe the eight types of interactions between any pair of individuals in Table 1. Using this as a reference, we write down the eight transitions for the three coordinates of our Markov chain as well as for $u=|x-y|$ which, as we will see later, is a key quantity in our analysis in Table 2. Note that we have rescaled $(X_{t},Y_{t},Z_{t})$ to $(x_{t},y_{t},z_{t})=n^{-1}(X_{t},Y_{t},Z_{t})$. Also note that $\Delta_{i}(\dots)$ and $q_{i}$ are respectively the change in quantity $\dots$ and the transition rate at the $i^{th}$ transition.

Note that $x_{t}+y_{t}+z_{t}=1$ for $t\geq 0$. To get an idea of what to expect, notice that as $n\to\infty$, sample paths approach solutions to the ODE system with $z=1-(x+y)$ and

[TABLE]

that has the invariant set

[TABLE]

The subset $\ell=\{(x,y)\in\Lambda:x=y\}$ is also invariant, since if $x=y$ then $(x-y)^{\prime}=(x-y)z=0$. Adding the $x^{\prime}$ and $y^{\prime}$ equations, the dynamics on $\ell$ is described by

[TABLE]

that has the stable fixed point $z^{*}=-2+\sqrt{5}$. Thus, (27) has the equilibrium point

[TABLE]

whose stable manifold contains $\ell$. For the dynamics off $\ell$, let $u=|x-y|$ as defined above, taking values in $[0,1]$. From (27), we derive

[TABLE]

We see that $u$ is non-decreasing, so $u(\infty):=\lim_{t\to\infty}u(t)$ is well-defined. If $u(0)>0$ then $u(\infty)>0$, and if in addition $u(\infty)<1$ then according to (28), $z(t)$ has a positive limit, which contradicts $\lim_{t\to\infty}u^{\prime}(t)=0$, therefore we must have $u(\infty)=1$.

To see the connection to Theorem 2, note that if $u(0)\geq n^{-1}$, then since the eigenvalues of the linearization near $u=0$ and $u=1$ are both non-zero, it should take about constant times $\log n$ amount of time for $u$ to exceed $1-n^{-1}$. Next, we delve into the land of martingales to make this intuition precise.

5.2 Controlling sample paths

Defining the process $u$ by $u_{t}=|x_{t}-y_{t}|$, we are interested in the time to consensus, that we can express as

[TABLE]

Using the notation from just above, the drift and diffusivity take the form

[TABLE]

We’ll write for now with $u$ but the same holds for $x,y,z$ and other functions of the state variables. For efficiency of notation, we’ll allow the function to change depending on the variable, so $\mu(u)$ is different from $\mu(x)$ and $\mu(y)$. Also, instead of the compensator $u^{p}$ we’ll use the predictor $\bar{u}=u_{0}+u^{p}$ which includes the initial value, and we’ll denote $u^{m}$ by $M(u)$, and $\langle u^{m}\rangle$ simply by $\langle u\rangle$. Then, $M(u)=u-\bar{u}$, and $\bar{u}$ and $\langle u\rangle$ can be written

[TABLE]

Define the jump size $c_{\Delta}(u)=\sup_{u,i}|\Delta_{i}(u)|$. From (4), if $a>0$, $0<c_{\Delta}(u)\lambda\leq 1/2$ and $\bullet\in\pm$ then

[TABLE]

Defining the maximum transition rate $c_{q}(u)=\sup_{u}\sum_{i}q_{i}(u)$, we have the basic inequality $\sigma^{2}(u)\leq c(u):=(c_{q}c_{\Delta}^{2})(u)$ and we obtain the corollary

[TABLE]

For any quantity $\cdot$, we always have $c_{q}(\cdot)\leq n$, since there are $n(n-1)$ directed edges each ringing at rate $1/(n-1)$, and for most quantities of interest, $c_{\Delta}(\cdot)\leq jn^{-1}$ for a smallish integer $j$, giving $c_{q}c_{\Delta}^{2}\leq j^{2}n^{-1}$, allowing us to take $\lambda$ equal to a small multiple of $n$ while still keeping $\lambda c_{q}c_{\Delta}^{2}t=O(1)$. When the context is clear, we omit the variable and simply write $c,c_{q},c_{\Delta}$.

The workflow of estimates is as follows. For any $\alpha>0$, we find $\epsilon>0$ so that the following holds with probability $1-o(1)$ as $n\to\infty$. Item numbers correspond to the Lemmas where they are proved.

So long as $u_{t}\leq 2\epsilon$, get $|z_{t}-z^{*}|<2\epsilon$ within constant time and keep $|z_{t}-z^{*}|<3\epsilon$ for $n$ time. 2. 2.

So long as $|z_{t}-z^{*}|\leq 3\epsilon$, get $u_{t}>2\epsilon$ within $((2z^{*})^{-1}+\alpha)\log n$ time, and find initial conditions so that $u_{t}\leq 2\epsilon$ for at least $((2z^{*})^{-1}-\alpha)\log n$ time. 3. 3.

Once $u_{t}>2\epsilon$, keep $u_{t}\geq\epsilon$ for $n^{1/2}$ amount of time. 4. 4.

So long as $u_{t}<1-\epsilon$, get $z_{t}>\epsilon/4$ within constant time and keep $z_{t}\geq\epsilon/12$ for $n$ time. 5. 5.

So long as $u_{t}\geq\epsilon$ and $z_{t}\geq\epsilon/12$, get $u_{t}>1-\epsilon$ within constant times $\epsilon^{-2}$ time. 6. 6.

Once $u_{t}>1-\epsilon$, keep $u_{t}\geq 1-2\epsilon$ for $n^{1/2}$ time, and show that if $u_{t}\leq 1-\epsilon+2n^{-1}$ then so long as $u_{t}\geq 1-2\epsilon$, $u_{t}<1$ for at least $(1-\alpha)\log n$ time. 7. 7.

So long as $u_{t}\geq 1-2\epsilon$, get $u_{t}=1$ within $(1+\alpha)\log n$ time.

Propositions 9–11 stitch together Lemmas 15–16, Lemmas 17–19, and Lemmas 20–21, respectively. The combination of these propositions into the proof of Theorem 2 is given at the end of this section.

Lemma 15**.**

Let $b_{t}=z_{t}-z^{*}$ and let

[TABLE]

If $\epsilon\leq 1/4$ and $n\geq 8/\epsilon^{2}$ then

[TABLE]

and for integer $N>0$,

[TABLE]

Proof.

Using Table 2, we find that

[TABLE]

Since $c_{\Delta}(b)\leq 2n^{-1}$ and $c_{q}(b)\leq n$, $\sigma^{2}(b)\leq(c_{q}c_{\Delta}^{2})(b)\leq 4n^{-1}$, and using the product rule from Lemma 4 on $b\cdot b$,

[TABLE]

If $u<2\epsilon$ and $|b|>\epsilon$ then, since $\sqrt{5}\geq 2$ and $z\geq 0$,

[TABLE]

so if in addition $\epsilon,2n^{-1}/\epsilon^{2}\leq 1/4$, then since $zb\leq 1$,

[TABLE]

and since $b_{t}^{2}-b_{0}^{2}\geq-1$, we find

[TABLE]

Moreover, if $b\in[0,1]$ then for $\delta\in\operatorname{\mathbb{R}}$, $|(b+\delta)^{2}-b^{2}|=|2\delta b+\delta^{2}|\leq 2|\delta|+\delta^{2}$, which implies that $c_{\Delta}(b^{2})\leq 2(2n^{-1})+(2n^{-1})^{2}\leq 8n^{-1}$. So, we can take $\lambda\in(0,n/16)$ and $c=c_{q}c_{\Delta}^{2}=64n^{-1}$. Choosing $\lambda=\epsilon^{2}n/64$ gives $c\lambda t=\epsilon^{2}t$. If $\tau_{0}>T$ we find

[TABLE]

Taking $T=2/\epsilon^{2}$ gives a lower bound of $1$. So, taking $a=1$ and $\bullet=+$ gives the first statement. Next, let $\tau_{2}=\inf\{t:|b_{t}-2\epsilon|\geq\epsilon\,\,\hbox{or}\,\,u_{t}\geq 2\epsilon\}$. Using (29), $\mu(b_{t}^{2})\leq-2\epsilon^{2}$ for $t<\tau_{2}$. Thus, if $|b_{0}-2\epsilon|\leq n^{-1}$, $b_{\tau_{2}}\geq 3\epsilon$ then

[TABLE]

Taking $\lambda=\epsilon^{2}n/32$, $c\lambda t=2\epsilon^{2}t$. If $n\geq 5/\epsilon$ then $b_{\tau_{2}}^{2}-\bar{b}_{\tau_{2}}^{2}-c\lambda\tau_{2}\geq 4\epsilon^{2}$. Noting that $\mathbb{P}(\tau_{2}<\infty)=1$, then taking $a=4\epsilon^{2}$ and $\bullet=+$,

[TABLE]

On the other hand, since $z+2+\sqrt{5}\leq 6$ and $|b_{t}|\leq 1$, $\mu(b_{t})\geq-7/2-6n^{-1}$. If $|b_{0}-2\epsilon|\leq n^{-1}$, $\tau_{2}\leq T$ and $b_{\tau_{2}}\leq\epsilon$ then

[TABLE]

If $n\geq 4/\epsilon$ then $4\epsilon n^{-1}\leq\epsilon^{2}$. Taking $\lambda=n/64$, $c\lambda=1$. If $n\geq 6$ then $(7/2)+6n^{-1}\leq 5$. Letting $T=\epsilon^{2}/6$,

[TABLE]

and taking $a=\epsilon^{2}$ and $\bullet=-$, it follows that

[TABLE]

The result follows by stopping the process each time $|b-2\epsilon|\leq n^{-1}$, using (30) and (31), then using the Markov property and taking a union bound while noting that $e^{-\epsilon^{2}n/64}\leq e^{-\epsilon^{4}n/8}$ for $\epsilon\leq 1/2\sqrt{2}\leq 1/4$. ∎

Lemma 16**.**

As in Lemma 15, let $b_{t}=z_{t}-z^{*}$,

[TABLE]

Let

[TABLE]

For $C>0$, if $\epsilon\leq\min(C^{1/2}/6,1/60)$ and $n\geq\max(20,64(1+C^{1/2})^{2})$ then

[TABLE]

Also, for $C_{2}>0$,

[TABLE]

Proof.

Notice that $\mu(u)=uz+\rho_{1}\mathbf{1}(u=0)+\rho_{2}\mathbf{1}(u\leq n^{-1})$ where $0\leq\rho_{1},\rho_{2}\leq 1$. Using transitions 5 and 6 from Table 2,

[TABLE]

If $|z-z^{*}|<3\epsilon$ and $3\epsilon,n^{-1}\leq 1/20$, then since $z^{*}>1/5$, it follows that $z-n^{-1}>1/10$ and $z(z-n^{-1})>1/100$, so that $\mu(u^{2})\geq n^{-1}/25$.

For $C>0$, let $\tau_{3}=\tau_{1}\wedge\inf\{t:u^{2}\geq Cn^{-1}\}$. If $u<C^{1/2}n^{-1/2}$ then since $c_{\Delta}(u)\leq 2n^{-1}$,

[TABLE]

so we can take $\lambda\in(0,n^{3/2}/2C_{1})$ and $c=C_{1}^{2}n^{-5/2}$. From the bound on the jump size, we find that $u_{\tau_{3}}^{2}\leq Cn^{-1}+C_{1}n^{-3/2}$. If $\tau_{3}>T$ then noting $u_{0}^{2}\geq 0$ it follows that

[TABLE]

Taking $T=25(3+C)$ and $\lambda=C_{1}^{-2}n^{3/2}/50$, $c\lambda T=2n^{-1}$ and $n^{-1}T/25=(3+C)n^{-1}$, so

[TABLE]

If $n\geq(2C_{1})^{2}$ this is at least $n^{-1}/2$, so taking $a=n^{-1}/2$ and $\bullet=-$ gives the estimate

[TABLE]

Next, let $c_{1}=z^{*}-3\epsilon$ and define $h$ by $h_{t}=e^{-c_{1}t}u_{t}$ so that $\mu_{t}(h)\geq 0$ and $-M_{t}(h)\geq u_{0}-h_{t}$ for $t<\tau_{1}$. Here we take $c=c(t)$ that depends on time, such that $\sigma^{2}_{t}(h)\leq c(t)$, and use the more general inequality $\langle h\rangle_{t}\leq\int_{0}^{t}c(s)ds$. Since $|\Delta_{i}(u)|\leq 2n^{-1}$ for any $u,i$, $|\Delta_{i}(h_{t})|\leq 2e^{-c_{1}t}n^{-1}\leq 2n^{-1}$, so we can take $\lambda\in(0,n/4)$ and $c(t)=4n^{-1}e^{-2c_{1}t}$. If $u_{0}\geq C^{1/2}n^{-1/2}$ and $\tau_{1}>T$ then

[TABLE]

Letting $T=(2c_{1})^{-1}\log n$ and bounding the integral by $1/2c_{1}$ we obtain

[TABLE]

If $C\leq(3/2c_{1})^{2}n$, then taking $\lambda=C^{1/2}n^{1/2}c_{1}/6$ and $\epsilon\leq C^{1/2}/6$, and taking $a=C^{1/2}n^{-1/2}/3$ and $\bullet=-$ we find that

[TABLE]

To get a matching lower bound on $\tau_{1}$ we need an upper bound on $\mu(u)$. If $u>n^{-1}$ and $|z-z^{*}|\leq 3\epsilon$ then letting $c_{2}=z^{*}+3\epsilon$, $\mu(u)\leq c_{2}u$. Let $\tau_{4}=\tau_{1}\wedge\inf\{t:u_{t}\leq n^{-1}\}$. If $u_{0}=C_{2}n^{-1/2}$ for $C_{2}>0$ and $u_{\tau_{4}}\leq n^{-1}$ then as before, for $\lambda\leq n/4$ we find

[TABLE]

If $u_{0}>n^{-1}$ then $\tau_{4}\neq\tau_{1}$ if and only if $u_{\tau_{4}}\leq n^{-1}$. Taking $\lambda=n^{1/2}$, $a=C_{2}n^{-1/2}-n^{-1}-2n^{-1/2}/c_{1}$ and using $\bullet=-$ we find that

[TABLE]

On the other hand, $\mu(u_{t})\leq c_{2}u_{t}$ for $t<\tau_{4}$, so $s_{t}=e^{-c_{2}(t\wedge\tau_{4})}u_{t\wedge\tau_{4}}$ is a supermartingale. Using non-negativity of $s_{t}$, the fact that $t\mapsto e^{-c_{2}t}$ is non-increasing, and optional stopping,

[TABLE]

Using Markov’s inequality,

[TABLE]

Letting $T=(2c_{2})^{-1}(\log n-C_{2})$, this is at most $(2\epsilon)^{-1}C_{2}e^{-C_{2}/2}$. The second statement then follows from a union bound. ∎

Proposition 9**.**

Let $\tau=\inf\{t:u_{t}\geq 2\epsilon\}$ as in Lemma 15,16. Then for any $\alpha>0$, there is $\epsilon_{0}>0$ so that for $\epsilon\in(0,\epsilon_{0}]$,

[TABLE]

Proof.

Given $\alpha>0$, let $0<\epsilon_{0}\leq 1/60$ be small enough that

[TABLE]

Recall that $b_{t}=z_{t}-z^{*}$ and let $\tau_{0}=\tau\wedge\inf\{t:|b_{t}|\leq\epsilon\}$ as in Lemma 15. Fix $\epsilon\in(0,\epsilon_{0}]$. Using the first result of Lemma 15,

[TABLE]

If $\tau=\tau_{0}$ and $\tau_{0}<2/\epsilon^{2}$ then in particular, $\tau\leq((2z^{*})^{-1}+\alpha)\log n$ for large enough $n$. If $\tau>\tau_{0}$ then $|b_{\tau_{0}}|<\epsilon$. Letting $N=n$ in the second result of Lemma 15 and using the strong Markov property,

[TABLE]

Then, letting $C=(\alpha/100)\log n$ in the first result of Lemma 16 and using again the strong Markov property,

[TABLE]

If $n$ is large enough then $\epsilon^{2}n/6>75+(\alpha/4)\log n+((2z^{*})^{-1}+\alpha/2)\log n$. Combining these results, we find that from any initial distribution, if $n$ is large enough then

[TABLE]

If $n$ is large enough then $2/\epsilon^{2}+75\leq(\alpha/4)\log n$ and the first statement follows. For the second statement, recall that $\tau_{1}=\tau\wedge\inf\{t:|b_{t}|\geq 3\epsilon\}$, and let $C_{2}=(\alpha/4c_{2})\log n$ to find that

[TABLE]

By definition, either $u_{\tau_{1}}\geq 2\epsilon$ or $|b_{\tau_{1}}|\geq 3\epsilon$. Combining with the second result of Lemma 15, if $\epsilon^{2}n/6>((2c_{1})^{-1}-\alpha/2)\log n$ then

[TABLE]

and the second statement follows. ∎

Next we show that if $u_{0}\geq 2\epsilon$ then there is a good chance $u_{t}\geq\epsilon$ for as long as we need.

Lemma 17**.**

If $n>1/\epsilon$ then for $T>0$,

[TABLE]

Proof.

We know that if $u>n^{-1}$ then $\mu(u)=uz\geq 0$. Since $c_{\Delta}(u)\leq 2n^{-1}$, we can take $\lambda\in(0,n/4)$ and $c=4n^{-1}$. Let $\tau_{0}=\inf\{t:u_{t}<\epsilon\}$. If $u_{0}\geq 2\epsilon$ and $\tau_{0}<T$ then $M_{\tau_{0}}(u)-c\lambda\tau_{0}\geq\epsilon-4\lambda n^{-1}T$, so using $\bullet=+$,

[TABLE]

Optimizing in $\lambda$ then gives the result. ∎

Lemma 18**.**

Let $\tau=\inf\{\,t\colon u_{t}<\epsilon\quad\hbox{or}\quad u_{t}>1-\epsilon\,\}$ and let

[TABLE]

If $\epsilon\leq 1/4$ and $n\geq 1/\epsilon$ then

[TABLE]

and for integer $N>0$,

[TABLE]

Proof.

Since $c_{\Delta}(z)\leq 2n^{-1}$, we can take $\lambda\in(0,n/4)$ and $c=4n^{-1}$. Recall that

[TABLE]

so if $u\leq 1-\epsilon$, $z\leq\epsilon/4$ and $\epsilon\leq 1/4$ then $\mu(z)\geq\epsilon/4$. Minding the jump size, $z_{\tau_{2}}\leq\epsilon/4+2n^{-1}$. If $\tau_{2}>T$ and $\epsilon/4\geq 4n^{-1}\lambda$ then

[TABLE]

Taking $T=20$ and $\lambda=\epsilon n/32$, if $n\geq 1/\epsilon$ we have the lower bound $\epsilon/4$. Taking $a=\epsilon/4$ and $\bullet=-$ gives the first statement. Now, let $\tau_{4}=\tau\wedge\inf\{t:|z_{t}-\epsilon/6|\geq\epsilon/12\}$. Suppose that $|z_{0}-\epsilon/6|\leq n^{-1}$, then $\mu(z_{t})\geq\epsilon/4$ for $t<\tau_{4}$. If $z_{\tau_{4}}\leq\epsilon/12$ then

[TABLE]

Taking $\lambda=\epsilon n/16$ and $a=\epsilon/12-n^{-1}$, $\mathbb{P}(z_{\tau_{4}}\leq\epsilon/12)\leq e^{-\epsilon^{2}n/192+\epsilon/16}$. On the other hand, if $z_{\tau_{4}}\geq\epsilon/4$ and $\tau_{4}\leq T$ then since $\mu(z)\leq 1/2+2n^{-1}$,

[TABLE]

Taking $\lambda=n/4$, $T=\epsilon/48$ and $a=\epsilon/24-n^{-1}$, if $n\geq 4$ then

[TABLE]

Combining these,

[TABLE]

The result follows by stopping the process each time $|z-\epsilon/6|\leq n^{-1}$, using the strong Markov property, taking a union bound, and using $e^{-\epsilon n/92}\leq e^{-\epsilon^{2}n/192}$. ∎

Lemma 19**.**

Let $\tau=\inf\{t:u_{t}<\epsilon\,\,\hbox{or}\,\,u_{t}>1-\epsilon\}$ as in Lemma 18 and let $\tau_{5}=\tau\wedge\inf\{t:z_{t}<\epsilon/12\}$. Then,

[TABLE]

Proof.

If $u\geq\epsilon$ and $z\geq\epsilon/12$ then $\mu(u)\geq\epsilon^{2}/12$. Since $c_{\Delta}(u)\leq 2n^{-1}$, we can take $\lambda\leq n/4$ and $c=4n^{-1}$. If $\tau_{5}>T$ then since $u_{T}-u_{0}\leq 1$, if $\epsilon^{2}/12\geq 4\lambda n^{-1}$ then

[TABLE]

Taking $\lambda=-\epsilon^{2}n/96$, $T=48/\epsilon^{2}$, $a=1$ and $\bullet=-$, the result follows. ∎

Proposition 10**.**

Let $\tau=\inf\{t:u_{t}\leq\epsilon\,\,\hbox{or}\,\,u_{t}\geq 1-\epsilon\}$, as in Lemma 18. Then for any $\alpha>0$ and $\epsilon\in(0,1/4)$,

[TABLE]

Proof.

Taking $T=n^{1/2}$ in Lemma 17,

[TABLE]

Let $\tau_{2}=\tau\wedge\inf\{t\colon z_{t}>\epsilon/4\}$ as in Lemma 18. Using the first result of Lemma 18,

[TABLE]

Letting $N=n$ in the second result of Lemma 18 and using the strong Markov property,

[TABLE]

Then, using the strong Markov property and the result of Lemma 19,

[TABLE]

If $n$ is large enough then $\min(n^{1/2},\,\epsilon n/48,\,\alpha\log n)\geq 20+48/\epsilon^{2}$. Combining the estimates gives the result. ∎

Lemma 20**.**

Let $v=\max(x,y),\,w=\min(x,y)$ and define

[TABLE]

For any $T>0$ and $\epsilon\leq 1/4$,

[TABLE]

Also, if $n\geq 4/\epsilon$ then for $c>0$,

[TABLE]

Proof.

We have $v=u+w$ and $2w+z=1-u$. Recall that if $u>n^{-1}$ then $\mu(u)=uz$, so $\mu(2w+z)=(w-v)z$. Since $w\leq v$, $\mu(2w+z)\leq 0$. Since $c_{\Delta}(2w+z)\leq 2n^{-1}$, we can take $\lambda\leq n/4$ and $c=4n^{-1}$. If $2w_{0}+z_{0}\leq\epsilon$, $2w_{\tau}+z_{\tau}\geq 2\epsilon$ and $\tau\leq T$ then $M_{\tau}(2w+z)-c\lambda\tau\geq\epsilon-4\lambda n^{-1}T$. Taking $\lambda=n/8T$ and $\bullet=+$, the first statement follows. On the other hand, we have always $v\leq 1$, and if $t<\tau$ then $w_{t}\leq\epsilon$, $z_{t}\leq 2\epsilon$ and $v_{t}\geq 1-2\epsilon$. Looking to Table 2, ignoring the $5^{th}$ and $8^{th}$ transitions, ignoring some increases, and bounding the rates in the right direction it is easy to check that for $t\leq\tau$, $(w_{t},z_{t})$ dominates the process $(\tilde{w}_{t},\tilde{z}_{t})$ with initial value $(w_{0},z_{0})$ and the following transitions:

[TABLE]

Note the transition rates are linear. We easily compute

[TABLE]

so that if $w_{0}$ is deterministic, we solve to obtain

[TABLE]

Combining, $\operatorname{Var}(\tilde{w}_{t})\leq n^{-1}w_{0}e^{-(1+\epsilon)t}$ and so

[TABLE]

Similarly,

[TABLE]

so that if $z_{0}$ is deterministic,

[TABLE]

and since the above fraction is at most $2$,

[TABLE]

If $2w_{0}+z_{0}\geq a$ then $\max(w_{0},z_{0})\geq a/3$, so for $T>0$

[TABLE]

Letting $a=\epsilon-2n^{-1}$ and $T=C\log n$, if $n\geq 4/\epsilon$ then $a\geq\epsilon/2$ and the second statement follows. ∎

Lemma 21**.**

Let $v,w$ and $\tau$ be as in Lemma 20. If $\epsilon<1/6$, $n\geq 1/\epsilon$ and $C>0$ then

[TABLE]

Proof.

Let $\psi_{t}=(w_{t},z_{t})^{\top}$. We may assume $2w+z\leq 2\epsilon$ so that $v=1-(w+z)\geq 1-2\epsilon$. Define the $2\times 2$ matrices $Q=(-1,\,0\,;\,2,\,-1)$ and $B=(1,\,1\,;\,1,\,1)$. Computing,

[TABLE]

So, if $n\geq 1/\epsilon$ then letting $\delta=2\epsilon/(1-2\epsilon)$,

[TABLE]

Time-change by $v^{-1}$ so that $\mu(\psi)\leq Q_{\delta}\psi$ with $Q_{\delta}:=Q+\delta B$. Let $s_{t}=e^{-Q_{\delta}t}\psi_{t}$ so that $s_{t\wedge\tau}$ is a non-negative supermartingale. Using non-negativity and optional stopping, we see that

[TABLE]

and so

[TABLE]

Now, $Q_{\delta}$ has eigenvalues $-1+\delta\pm 2\delta^{2}$ and corresponding eigenvectors $(1,\pm 2\delta)^{\top}$. Thus $Q_{\delta}=SAS^{-1}$ with $S=(1,\,1\,;\,2\delta,\,-2\delta)$ and $S^{-1}=(1/2\delta,\,1/4\,;\,1/2\delta,\,-1/4)$. If $\delta\leq 1$, then $|S|\leq 1$, $|S^{-1}|\leq(2\delta)^{-1}$ and $|A|\leq-1+\delta+2\delta^{2}$ and so

[TABLE]

If $\epsilon\leq 1/6$ then $1-2\epsilon\geq 2/3$, $\delta\leq 3\epsilon\leq 1/2$ and $1-\delta(1+2\delta)\geq 1-6\epsilon$. If $w_{0},z_{0}\leq\epsilon$ then

[TABLE]

Recalling the time change and noting $v^{-1}\leq(1-2\epsilon)^{-1}$, then letting $t=(1-6\epsilon)^{-1}(1-2\epsilon)^{-1}(\log n+C)$ and using the fact that $1-2\epsilon\leq 1$ gives the result. ∎

Proposition 11**.**

Let $\tau=\inf\{t:u_{t}=1\,\,\hbox{or}\,\,u_{t}\leq 1-2\epsilon\}$. For $\alpha>0$, there is $\epsilon_{0}>0$ so that for $\epsilon\in(0,\epsilon_{0}]$,

[TABLE]

and

[TABLE]

Proof.

Since $2w+z=1-u$, the above definition of $\tau$ agrees with the one used in Lemmas 20-21. Given $\alpha>0$, $0<\epsilon_{0}<1/6$ be small enough that for $\epsilon\in(0,\epsilon_{0}]$,

[TABLE]

Letting $T=n^{1/2}$ and using the first result of Lemma 20,

[TABLE]

On the other hand, letting $C=(\alpha/2)\log n$ and using the first result of Lemma 21,

[TABLE]

Since $n^{1/2}>(1+\alpha)\log n$ for $n$ large enough, the first statement follows. For the second statement, letting $C=1-\alpha$ in Lemma 20,

[TABLE]

Combining with (32), the second statement follows. ∎

Proof of Theorem 2.

Let $\tau=\inf\{t:u_{t}=1\}$. Recall that $z^{*}=-2+\sqrt{5}$. To show the upper bound, for any $\alpha>0$ take $\epsilon>0$ small enough to satisfy all conditions, thenapply Propositions 9, 10 and 11 in sequence, stopping the process when $u_{t}\geq 2\epsilon$ and $u_{t}\geq 1-\epsilon$, to find that

[TABLE]

To show the lower bound, in Proposition 9 start from $(u,z)$ achieving the supremum, which is a maximum since the state space is finite. Apply the result of Proposition 9. Then, stop the process when $|u_{t}-(1-\epsilon-n^{-1})|\leq n^{-1}$, which occurs before $\tau$ since $u_{t}$ has jumps of size at most $2n^{-1}$. Apply Proposition 11. Combining the two, conclude that for any $\alpha>0$,

[TABLE]

∎

Acknowledgements

The author wishes to thank Nicolas Lanchier for suggesting the model, and for many helpful conversations while working on the paper.

Appendix

Miscellaneous estimates.

Since

[TABLE]

and $t\mapsto c-(1+a-\beta)t^{-\beta}$ increases with $t$, if $c-(1+a-\beta)x^{-\beta}>0$ we have the upper bound

[TABLE] 2. 2.

Factoring, using the fact that $|(1+\lambda)^{-1}\lambda^{1/2+\alpha}/2|\leq 1/2$ and $(1-x)^{-1}\leq 1+2x$ for $|x|\leq 1/2$, then using the fact that $\lambda^{1/2+\alpha}\leq(1+\lambda)^{1/2+\alpha}$,

[TABLE]

Lemma 22**.**

Let $X$ be Poisson with mean $\lambda$.

[TABLE]

Proof.

We have

[TABLE]

Also,

[TABLE]

Using Markov’s inequality,

[TABLE]

Optimizing in $\theta$ gives $\theta=\log c$ which is positive for $c>1$ and negative for $c<1$, and

[TABLE]

Expanding $\gamma(1+\delta)$ in an alternating Taylor series around $\delta=0$,

[TABLE]

using $\delta^{3}\leq\delta^{2}$ for $\delta\in[0,1)$ and $\frac{1}{2}+\frac{1}{6}=\frac{1}{3}$. (35) follows for $0<x<\lambda^{1/2}$ by letting $\delta=x\lambda^{-1/2}$. For $x=\lambda^{1/2}$ it follows by continuity of probability. ∎

Lemma 23**.**

Let $(N_{t})$ be a Poisson process with intensity $\lambda$. Fix $\alpha\in(0,1/2]$ and let

[TABLE]

denote the last passage time of $N_{t}$ above/below the curve $\lambda t\pm(\lambda t)^{1/2+\alpha}/2$, respectively. If $\lambda\geq 1$ and $t^{2\alpha}\geq 6$ then

[TABLE]

Proof.

Let $f$ denote the function defined by $f(t)=\lambda t+(\lambda t)^{1/2+\alpha}$. Using Lemma 22, for each $t>0$,

[TABLE]

Since $|f^{\prime}(t)|\leq 2\lambda$ for any $t\geq 0$, $f$ is Lipschitz with constant $2\lambda$. Using this and the fact that $t\mapsto N_{t}$ is non-decreasing,

[TABLE]

so taking a union bound over $t\in\{T+1,T+2,\dots\}$,

[TABLE]

Using (33),

[TABLE]

If $\lambda\geq 1$ and $T^{2\alpha}\geq 6$, this is at most $6T^{1/2-\alpha}e^{-(\lambda T)^{1/2+\alpha}/3}$. An analogous estimate applies for the lower bound, giving $4$ instead of $6$ and $1/2$ instead of $1/3$ in the exponent, when $\lambda\geq 1$ and $T^{2\alpha}\geq 4$.

∎

Lemma 24**.**

Let $X$ be a non-decreasing quasi-absolutely continuous semimartingale on $\operatorname{\mathbb{R}}_{+}$ with jump size at most $c$ and defined for $t<\zeta$, where $\zeta=\sup_{r>0}\inf\{t:X_{t}\geq r\}$ is the first time of explosion. Suppose that

[TABLE]

for some locally integrable non-nonegative deterministic functions $b(t)$, $\ell(t)$. Let $m(t)=\exp(\int_{0}^{t}\ell(s)ds)$ and let $Y_{t}=X_{t}/(X_{0}m(t))-\int_{0}^{t}b(s)/m(s)ds$ denote the rescaled process. Let $\zeta^{\prime}=\zeta\wedge\inf\{t:m(t)=\infty\}$ and $\beta=\int_{0}^{\infty}b(t)/m(t)^{2}dt$, and assume $\beta<\infty$. Then, $\zeta\geq\zeta^{\prime}$ and for $y\geq 2$,

[TABLE]

Proof.

First we treat the case $X_{0}=1$, so that $Y_{t}=X_{t}/m(t)-\int_{0}^{t}b(s)/m(s)ds$. Given $y>0$ define $\tau(y)=\inf\{t:Y_{t}\geq y\}$, and note that $\tau(y)<\zeta^{\prime}$. Since $1/m(t)=e^{-\int_{0}^{t}\ell(s)ds}$, $(1/m(t))^{\prime}=-\ell(t)/m(t)$, so using linearity of the drift and Lemma 4,

[TABLE]

which implies $Y^{p}\leq 0$. Clearly $\sigma^{2}_{t}(Y)=(1/m(t))^{2}\sigma^{2}_{t}(X)$. Since $X$ is non-decreasing, it has finite variation, so in particular $X^{c}=0$, $X^{m}=X^{d}$ and $\langle X^{m}\rangle_{t}=(\sum_{s\leq t}(\Delta X_{s})^{2})^{p}$. In addition, $0\leq\Delta X_{s}\leq c$, so $(\Delta X_{s})^{2}\leq c\Delta X_{s}$. Using this and $\sum_{t\leq s\leq t+r}\Delta X_{s}\leq X_{t+r}-X_{t}$, for any $t,r$,

[TABLE]

which implies $\sigma^{2}_{t}(X)\leq c\mu_{t}(X)$. Using $\mu_{t}(X)\leq b(t)+\ell(t)X_{t}=b(t)+\ell(t)m(t)Y_{t}$,

[TABLE]

Since $Y_{t}<y$ for $t<\tau(y)$,

[TABLE]

the last equality defining $\alpha(t)$ and $\beta(t)$. Taking the antiderivative,

[TABLE]

Since $Y_{0}=1$, $Y_{\tau(y)}\geq y$ and $Y^{p}\leq 0$, it follows that for $\lambda>0$,

[TABLE]

Using (4) with $a=y-1-\lambda c(\beta+y)$, assuming $\lambda c\leq 1/2$ we find

[TABLE]

Optimizing $\lambda a$ gives $\lambda=(y-1)/(2c(y+\beta))$ and

[TABLE]

and if $y\geq 1$ the assumption $c\lambda\leq 1/2$ holds. For general $X_{0}$, first condition on $X_{0}$ and apply the above to $X_{t}/X_{0}$, which has jump size $c/X_{0}$. Then, integrate over $X_{0}$ to obtain the result.

To see that $\zeta\geq\zeta^{\prime}$, note that $\{\zeta\geq\zeta^{\prime}\}\supset\bigcup_{y}\{\sup_{t<\zeta^{\prime}}Y_{t}<y\}$ and that the above estimate implies the latter event has probability 1. ∎