The Naming Game on the complete graph
Eric Foxall

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 agents suggest that when no words initially exist, the time to consensus is of order , assuming each agent speaks at a constant rate. We show rigorously that the time to consensus is at least , and that it is at most constant times when only two words remain. In order to do so we develop sample path estimates for quasi-left continuous semimartingales with bounded jumps.
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Taxonomy
TopicsOpinion Dynamics and Social Influence · Game Theory and Applications · Complex Network Analysis Techniques
The Naming Game on the complete graph
Eric Foxall
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 agents suggest that when no words initially exist, the time to consensus is of order , assuming each agent speaks at a constant rate. We show rigorously that the time to consensus is at least , and that it is at most constant times when only two words remain. In order to do so we develop sample path estimates for quasi-left continuous semimartingales with bounded jumps.
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 . 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 with for each , where is the collection of finite subsets of . Thus, for each vertex , we have a process whose state space consists of all finite subsets of the vertex set and which is defined as
[TABLE]
The process evolves as follows: For each , at the times of an independent Poisson process with rate one, chooses a listener uniformly at random from the set ; say this occurs at time .
- •
If is empty then speaks word to , so that and .
- •
If is non-empty then chooses a uniform random word from and speaks it to .
- –
If then .
- –
If then is unchanged and .
If 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 . If is the complete graph on vertices, i.e.,
[TABLE]
numerical studies and heuristic computations [2] indicate three distinct phases.
Early phase: rises from [math] to about in about time. 2. 2.
Middle phase: remains fairly constant up till about time. 3. 3.
Late phase: falls sharply to 1 within about time.
In this article we consider the early and middle phases, and what we call the final phase, where we assume that is initially equal to , and track the dynamics until it goes to . 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 , from arbitrary initial data. We assume the vertices are totally ordered according to some fixed order. Given , let be an independent and identically distributed sequence, with each exponentially distributed with mean one and each independent of and uniform on , and for , let . Then, the set of points
[TABLE]
defines what we call an augmented Poisson point process with intensity , since are the jump times of a Poisson process with intensity and each point comes equipped with an independent uniform random variable to help with the decision-making process.
Let denote the set of directed edges , and associate to each directed edge an independent augmented Poisson point process with intensity . Suppose that and , with labelled in increasing order.
- •
If then speaks word to at time .
- •
If , then speaks word to at time if and only if
[TABLE]
We then follow the rules as described above to determine . If is a finite graph, then since the intensity of the union is finite, its points are well-ordered in time with probability 1, and so can be determined from the initial state and the points by updating sequentially in time. If is an infinite graph, one needs to ensure that for each spacetime point , a finite number of events suffices to determine . Although this is not hard to do, we will ignore it since from here on we focus on the case where is the complete graph on vertices and thus finite for any .
Recall that denotes the set of words in existence at time . The following result gives estimates of in the middle phase of the process.
Theorem 1**.**
For any , let and . Then as
[TABLE]
The result is proved in two main steps.
First, we show that are ever created, and within time. 2. 2.
Then, we show that words are deleted in time.
The proof relies on approximating the size of the cluster corresponding to a given word 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 at time . We also need to control the correlation between distinct clusters . 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 denote the configuration at time when the initial configuration is , and let . If then clearly
[TABLE]
that is, if each vertex has initially a non-empty vocabulary consisting of words in , the same is true at later times. In particular, if for a pair of words , then each vertex has one of the three types and . We note that, starting from for all , 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 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 denote the number of sites at time with respective types and , the process is a continuous-time Markov chain. Since the states and 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 . Use for the law of the process with initial configuration .
Theorem 2**.**
Let , and define the stopping time
[TABLE]
Then, for any ,
[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 . The reason it is much faster here is because, once a majority of type or 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 words are created in about time, using Chebyshev’s inequality and a coupon-collecting argument, respectively. In Section 4.2 we show that words are deleted in 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 satisfying the “usual conditions” as described in [6]. Processes are assumed to be optional. Given , is the left continuous process and is the process of jumps. denotes the compensator and the continuous martingale part, when they exist.
A semimartingale is a process (on unless specified otherwise) that can be written as , where is an -measurable random variable, is a local martingale and has locally finite variation. Using I.3.17, a semimartingale is special if it can be written as
[TABLE]
where is the compensator of and is a uniquely defined local martingale satisfying . If is a semimartingale with bounded jumps, that is, for some then by I.4.24, is a special semimartingale and . If is also quasi-left continuous, that is, a.s. on , for any predictable time , then using I.2.35 in the proof of I.4.24, we obtain the slightly stronger estimate .
Any (right-continuous) Markov chain with values in 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 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 is defined only up to some predictable time that may be finite; in this case, information about can be recovered from the stopped processes defined by , where is an announcing sequence for , i.e., an increasing sequence of stopping times with limit .
If is a local martingale satisfying and for some , by I.4.1, is locally square-integrable, so by I.4.3, has a compensator, denoted and called the predictable quadratic variation. Relative to the decomposition (I.4.18) into continuous and discontinuous martingale parts,
[TABLE]
and is the compensator of , the quadratic variation of .
Lemma 1**.**
Let be a quasi-left continuous local martingale with and for some . Then,
[TABLE]
is a local supermartingale with initial value .
Proof.
Let be a continuous predictable process with locally finite variation, satisfying , and let . Applying Itô’s formula I.4.57 using the function ,
[TABLE]
Noting that when , 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 can be taken continuous (i.e. it has a continuous version; see I.4.3). Since it is the compensator of , is also predictable and has locally finite variation. Letting , the same is true for . With this choice of , , which implies is a local supermartingale. ∎
Lemma 2**.**
Let be a special semimartingale with locally square-integrable martingale part . Then, is quasi-left continuous iff and are continuous.
Proof.
If is quasi-left continuous (qlc), then by I.2.35 its predictable projection . Since the operation is linear (which follows from uniqueness and property (ii) in I.2.28), and since , it follows that
[TABLE]
Since is special it has a compensator , and by I.3.21, . By the above, this is [math], i.e., is continuous. Using (1), which implies that is qlc, by definition of qlc. Since is locally square-integrable, by I.4.3, is continuous.
On the other hand, if is continuous then by I.4.3, is qlc. If in addition is continuous then which implies is qlc, by definition of qlc. ∎
Lemma 3**.**
Let be a quasi-left continuous semimartingale such that for some . Then for and ,
[TABLE]
Proof.
Notice that and that for and , . As noted just above (1), since has bounded jumps it is special and by Lemma 2, is qlc. Take in Lemma 1, which has , and use Doob’s inequality to find
[TABLE]
∎
For practicality’s sake we’ll use a slightly cruder version of (3). Since , if then , so from (3) it follows that for and ,
[TABLE]
Using Lemma 2 as inspiration, say that a special semimartingale with locally square-integrable martingale part is quasi-absolutely continuous (qac) if both and are absolutely continuous. In this case define the drift and the diffusivity for Lebesgue-a.e. by
[TABLE]
For deterministic processes, qac is equivalent to absolute continuity, since , and absolute continuity implies locally finite variation. For Markov chains on with jump measure , if qac holds then and are given by functions
[TABLE]
i.e., and . Conversely, if and the total intensity of the jump measure is bounded on compact subintervals of , then is qac up to the first explosion time , and .
Lemma 4**.**
[Product rule] Suppose are qac semimartingales on a common filtered probability space. Then exists and is absolutely continuous, and exists, is absolutely continuous, and is given by
[TABLE]
where .
Proof.
By definition of quadratic variation,
[TABLE]
Since and have locally finite variation, . Since are locally square-int, has compensator , so has compensator
[TABLE]
The result will follow if we can show is absolutely continuous. For any , applying the Cauchy-Schwarz inequality to the symmetric, bilinear and semidefinite map gives
[TABLE]
Absolutely continuity of means that for any there is so that if then . Using the Cauchy-Schwarz inequality to obtain the second line,
[TABLE]
which shows that is absolutely continuous. ∎
4 Early and middle phases
In this section we consider the behaviour of for . Define
[TABLE]
respectively the number of words created up to time , and the number of words created and then deleted by time . Theorem 1 is implied by the following two propositions, whose proof is the objective of this section.
Proposition 1**.**
For each ,
Proposition 2**.**
For each ,
In words, in order to estimate we obtain good control on , then show that is not too big. We begin with .
4.1 Creation of vocabulary
Our first task is to prove Proposition 1, and to do so we show that rises from [math] to within time, then remains constant. For a vertex let denote the size of the vocabulary of individual , and let
[TABLE]
be the first time that every individual knows at least one word. Clearly is non-decreasing as a set, so exists and . Once everyone knows a word, no new words are created, so for . Proposition 1 is implied by the following two lemmas, in which we estimate and .
Lemma 5**.**
For ,
[TABLE]
Lemma 6**.**
Let be the number of words ever created. Then,
[TABLE]
Proof of Lemma 5.
Let denote mute vertices, those not yet knowing a word, and observe that is equivalent to . For each distinct ordered pair of vertices , at rate , the directed edge has an event, and both and are removed from , if either or both still belongs. If we let denote the number of mute vertices at time , it follows that is a Markov chain with and transitions
[TABLE]
We find that
[TABLE]
Letting , and taking expectations in the above, , which has the unique solution . Fix and let . Using Markov’s inequality,
[TABLE]
To get a lower bound we turn to , which has transitions
[TABLE]
so
[TABLE]
Letting , and taking expectations above,
[TABLE]
so letting , using and solving the above DE, we find
[TABLE]
As above let , then and for fixed ,
[TABLE]
so . Using Chebyshev’s inequality,
[TABLE]
The result follows by taking a union bound of both estimates. ∎
We note in passing that and increases by at rate . Heuristically, , so , for . This can be made precise using stochastic calculus, although we do not pursue it here.
Proof of Lemma 6.
Letting for each vertex be the Bernoulli random variable equal to one if and only if 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 and let be the event that the first edge becoming active starting from or is edge . Since there are edges starting from each vertex,
[TABLE]
In addition, the two vertices cannot both speak before listening when occurs whereas the two events are independent on the event 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 . This completes the proof. ∎
4.2 Maintenance of vocabulary
Next, we prove Proposition 2, that says that with probability tending to as ,
[TABLE]
Clearly , like , is non-decreasing, since once a word vanishes from the population, it does not come back. We first bound by a simpler quantity. Say that agreement upon word occurs at if
[TABLE]
If word is created at some time , then , and remains in individual ’s vocabulary at least until the first time that agreement occurs at . This implies
[TABLE]
In words, in order to delete a word from the population, it must at least be deleted from its source. Since each agreement contributes at most 2 to , it follows that
[TABLE]
In order to control we first define some useful observable quantities. For we recall the cluster of , that is, the set of individuals that know word at time :
[TABLE]
Recall that denotes the size of the vocabulary of individual , and let
[TABLE]
denote the rate at which word is spoken. Let denote the times at which speaks to , and let
[TABLE]
noting that and is a collection of independent Poisson processes with intensity 1. If we let
[TABLE]
then , and in particular,
[TABLE]
Let and , and let . Each site that knows word speaks it at rate to each of the other sites in . Letting
[TABLE]
so that , it follows that increases by at rate
[TABLE]
Since is the total speaking rate and , summing the above display over we find
[TABLE]
We have reduced the problem of controlling to that of controlling . The following becomes the goal of this subsection. Since its proof has a few parts, we call it a theorem.
Theorem 3**.**
For small ,
[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 , . Using Theorem 3, with probability as
[TABLE]
Since as it follows that with probability , and since , the same is true for . ∎
To begin the proof of Theorem 3 we introduce a modified construction to help us make a coupling. First, for each ordered triple let be the rate at which word is spoken by site to , let be the rate at which site hears word , and as above let be the rate at which word is spoken. We calculate
[TABLE]
Clearly for each and . Fix an ordering of and define an independent family of augmented Poisson point processes with intensity 1, that will correspond to listening events. For , and let
[TABLE]
noting that partitions . Then, if and , word is spoken by to , which defines the process. Using this construction and given we obtain upper bounds on for all , valid up to the time
[TABLE]
That is, we obtain for each a pair of processes with nice properties, such that and for pointwise on realizations of the process. Given , are non-decreasing and defined as follows. For let
[TABLE]
Define
[TABLE]
Let be such that . Initially, and . is defined as follows.
[TABLE]
Then, is defined as follows.
[TABLE]
We demonstrate the claimed comparison.
Lemma 7**.**
For each , and for .
Proof.
Let
[TABLE]
then and for . For the remainder, assume and let be such that . By construction, is added to if
[TABLE]
and otherwise, does not increase. If then , and if then . So, from the second line of (12),
[TABLE]
If then . By definition of , if and then
[TABLE]
If and then , since this implies existence of a point in
[TABLE]
which is counted in but not in . If it follows that for each which implies the containment is preserved across transitions (13) that cause to increase. Since is non-decreasing and transitions are well-ordered this implies for . It remains to check for . But in this case, (10) and the previous argument give
[TABLE]
∎
Next we fix and examine assuming , and dropping the for neatness. Notice that is non-decreasing and increases by at rate at least , which implies . Since increases by one at a time, let be the order in which vertices are added to , with , and condition on . We track and which suffices to determine . Let denote the time at which is added to , and let be such that . For , is the least value of such that there is a point
[TABLE]
and in addition, this point belongs to . Using this and basic properties of exponential random variables, together with the thinning property of the Poisson process, we find that conditioned on ,
[TABLE]
is a Markov chain with the following transitions:
[TABLE]
In particular, is an i.i.d. collection of Poisson processes with intensity . Since the above does not depend on the choice of values for the same holds unconditionally. Thus can be viewed as follows: initially , then subject to the random environment determined by the , increases by at rate . Let
[TABLE]
and let denote the process with that increases by at rate . Since and is non-decreasing in , it follows that
[TABLE]
We can think of as a branching process with immigration rate , in which individual produces offspring at the time-decreasing rate . Two tasks lie ahead. The first is to estimate . The second is to estimate . We then combine the results to obtain Theorem 3. This is outlined as follows.
Proposition 3**.**
Let . For small , and ,
[TABLE]
Proposition 4**.**
For small , and ,
[TABLE]
Proof of Theorem 3.
Use Propositions 3 and 4 with and . ∎
4.2.1 Estimation of
Since does not appear in the definition of we may as well define it using an infinite sequence of Poisson processes with intensity . Clearly . Since will be chosen , we will have as , so throughout we assume .
We begin with a useful heuristic. Let . Replacing with its expectation , increases by at rate , which we approximate with the differential equation
[TABLE]
Let . The above equation is linear and has solution
[TABLE]
If is close to then grows just a bit faster than linearly in time. In order to analyze we break it up into two steps:
Up to a fixed time , when the are fairly small. 2. 2.
From time to , when the are fairly large.
The reason to do this is because the estimates that say are only effective once has had time to increase. The following is the main result of this subsection.
Proposition 5**.**
Let . There exist so that for and ,
[TABLE]
Recall . Using this result we can prove Proposition 3.
Proof of Proposition 3.
For each , using Lemma 7 and (14),
[TABLE]
Applying the result of Proposition 5 and taking a union bound over , if and then
[TABLE]
If then recalling that , for large . Since , , and if is small then . Letting , the probability is and since , it follows that
[TABLE]
uniformly in , as . ∎
We tackle the proof of Proposition 5 in a couple of steps.
Step 1. We obtain a somewhat crude upper bound on that has the virtue of being effective starting at time [math]. For let , define then define by
[TABLE]
In words, at the moment an individual is added to the process, the corresponding counting process is stopped, so that always contributes to . Since for , is dominated by . The next result controls .
Lemma 8**.**
There is so that for and ,
[TABLE]
Proof.
Begin by observing that has the concise description
[TABLE]
where the increment is independently sampled every time there is a jump. Our first task is to control the size of . We compute the drift:
[TABLE]
Let . Using Lemma 24 with and , for we find
[TABLE]
This translates to a bound on as follows. Since ,
[TABLE]
Since has transition rate and jump size exactly 1, . Taking in (4) (which satisfies ) while noting ,
[TABLE]
Combining with (16) and taking a union bound,
[TABLE]
Intuitively, grows roughly like . Let . Since is convex, the inequality goes in the wrong direction for an upper bound on . Anticipating our needs, we let in (35) to find
[TABLE]
Using the fact that and that probabilities are at most , then using (34),
[TABLE]
Also, if and then
[TABLE]
Let and let . Since decreases with , it follows that for . Recalling defined earlier, it follows that
[TABLE]
Since and , and since , it follows that
[TABLE]
If and then since , and
[TABLE]
To conclude, take , use (17) and note for . ∎
Step 2. Next, we do two things.
Lemma 9. We control the environment for . 2. 2.
Lemma 10. We use this to get an upper bound on for .
Fix , then let
[TABLE]
denote the last passage time of below the curve , and for let
[TABLE]
Note that for .
Lemma 9**.**
There is so that for and ,
[TABLE]
Proof.
For each , using Lemma 23 with and ,
[TABLE]
Let , so the right-hand side above is . Then, a union bound and the fact that gives
[TABLE]
For let
[TABLE]
Note that and uniformly for . Since is non-decreasing, if and then
[TABLE]
Taking a union bound over the estimate at times , gives
[TABLE]
The right-hand side is at most
[TABLE]
and using (33), this is at most
[TABLE]
Then note that and for large enough, uniformly for . ∎
Lemma 10**.**
Given let . There is so that for and ,
[TABLE]
Proof.
Since, as noted before, for , it follows that for and conditioned on , is dominated by the process with that increases by at rate . We proceed as in the proof of Lemma 8. We have
[TABLE]
For let , where , and let . Using Lemma 24 with , for we find
[TABLE]
Recall . Using (34) with ,
[TABLE]
Let , which is finite for and and decreases with . Let , so that for . Combining and noting ,
[TABLE]
Using that , we obtain the complementary bound . In this way
[TABLE]
Using the more generous lower bound and noting and , we find
[TABLE]
Let and rearrange terms in the formula for to complete the proof. ∎
Proof of Proposition 5.
We note the result of Lemma 8 applies to since it is dominated by . Fix in Lemma 9 and 10. Fix and let and . Let
[TABLE]
be the complement of the event from, respectively, Lemma 8, 9 and 10. On ,
[TABLE]
In particular, , so using Lemma 10, for and large enough,
[TABLE]
Using our choice of , on we find
[TABLE]
Since , on the inequality holds for all . Taking , since ,
[TABLE]
Using Lemmas 8 and 9, for and some , we find and
[TABLE]
If is large enough uniformly for the above and (19) show
, so a union bound gives
[TABLE]
Choose so that for from the statement of the Proposition, . It suffices to check that on . But on , noting that and on the second line,
[TABLE]
Since and by assumption, , it suffices that
[TABLE]
for , which is true for large enough, uniformly for . ∎
4.2.2 Estimation of
Write , where
[TABLE]
Proposition 6**.**
Let . If and then
[TABLE]
Proposition 7**.**
For each and , if 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 , each is dominated by , taking a union bound we find
[TABLE]
For a given function ,
[TABLE]
Taking in Proposition 5,
[TABLE]
We have the trivial bound , and it follows from the bound on given in the proof of Lemma 10 that
[TABLE]
Let . Taking , if is large enough uniformly for then for so using Lemma 9 and the above bounds, for and large we find
[TABLE]
if is large enough. If , then noting and for large , if then for large ,
[TABLE]
which approaches as . It follows that
[TABLE]
and a similar estimate shows that . 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 we control contributions to . Let denote the rate at which increases. Let be an independent collection of Poisson processes with intensity 1, let and let be an independent Poisson process with intensity . Let and let
[TABLE]
Let and . Then for any ,
[TABLE]
In the next lemma we control .
Lemma 11**.**
For , and ,
[TABLE]
Proof.
We first control the value of , then of for , then take a union bound to control the value over the interval . Let and fix . Using (4) with , , , , and and noting that ,
[TABLE]
since . Using the same result except with , , , and , and taking a union bound,
[TABLE]
Let be the jump times of , and label them in increasing order as . On the complement of both events above, and for , and so , and this gives
[TABLE]
To see that this also bounds for , replace with and use in the above to obtain the bound
[TABLE]
which has the same upper bound, assuming so that . The same works for and to give
[TABLE]
since . Taking a union bound over and noting gives the result. ∎
It remains to prove the following result.
Proposition 8**.**
For small and any , and ,
[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 . Note that if whp then whp.
Proof of Proposition 7.
We want to show that whp. Since , if then . Moreover
[TABLE]
Proposition 8 says that whp, so it is enough to show that if and then whp, or equivalently that
[TABLE]
In Lemma 11 take , and to find that
[TABLE]
Then, using (21) and Proposition 8 and taking a union bound over the possible values of ,
[TABLE]
The result then follows from (20) and the fact that for large . ∎
By taking a union bound over and noting the probability does not depend on , to obtain Proposition 8 it is sufficient to show that for any and small ,
[TABLE]
noting that the probability is the same for any . There are three ways that increases:
a site already in is added to , 2. 2.
agreement occurs at a site already in , or 3. 3.
a site is added simultaneously to and .
Let denote the rate of each event, so that . Since each site in is added to at rate ,
[TABLE]
Since there are sites in that can agree on word , and each word is spoken at rate at most to each site,
[TABLE]
recalling that is the size of the largest cluster. Each time a person speaks, the probability that agreement occurs is at most . Since increases at rate , it follows that
[TABLE]
The reader may think that should be [math], since a new addition to a cluster does not yet know the word. However, the upper bound cluster can grow when in the process itself, a word other than is being spoken. Using Proposition 5 we control and , which is two thirds of Proposition 8.
Lemma 12**.**
For each , small , and ,
[TABLE]
Proof.
From (23) and (25) and the choice of , it suffices to show that
[TABLE]
The result of Proposition 3 holds with the probability being – to see this, take in the proof. This gives
[TABLE]
while is still , uniformly in as . The desired result for then follows from (25), since . To get the result for recall from the beginning of this section that , the number of agreements up to time , and from (11) that . Using the above bound on , with probability ,
[TABLE]
for large . From Lemma 22, , so it follows that
[TABLE]
∎
It remains to control . We’ll make use of the estimates from Lemma 11, namely that for a Poisson process 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 and any are not strongly correlated. Since we only randomize the location of “excess” events that expand , the reader may verify that up to a random permutation of certain vertices, the marginal distribution of each , and its domination of , are unchanged.
To carry out the modification, make the doubly-augmented, that is, each is again a Poisson point process with intensity 1, but on instead of . is defined in the same way as before, and is defined as follows.
[TABLE]
In other words,
- •
if was about to include , then will too, and
- •
if increases when does not, then with respect to
what other clusters are doing, it does so as randomly as possible.
We now control the size of , for any . “wp” is shorthand for “with probability”.
Lemma 13**.**
For any , if then
[TABLE]
Proof.
Let . There are three ways can increase.
acquires a site that belongs to some (possibly many) , , 2. 2.
some , acquires a site that belongs to , and 3. 3.
and some , simultaneously acquire the same site.
It suffices to show the contribution to from each item is wp . For item 1, the increase is at most , while for items 2,3 the increase is at most , since at each transition, increases for at most one , and by at most . Let denote the rate of each event. Then,
[TABLE]
Since for and each , and since each , using (26) with and a union bound,
[TABLE]
For , , so wp , the contribution from item 1 is at most which if is at most for any fixed wp . Similarly, but more simply since the increase per transition is 1, the contribution from item 3 is at most which is at most wp .
For item 2, note that and that for , is dominated by . Applying Proposition 5, bounding by for and using the trivial but convenient for , we find that for large enough,
[TABLE]
Taking the probability above is . Thus the contribution from item 2 is at most , where
[TABLE]
using and . If then for any , for large . Therefore
[TABLE]
It follows as before that wp , and the proof is complete. ∎
Combining this with the other term in (24) we control .
Lemma 14**.**
For any and small , each and ,
[TABLE]
Proof.
From the proof of Lemma 12 we know that . Using this, (24), and Lemma 13,
[TABLE]
If is large then 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 which, as we will see later, is a key quantity in our analysis in Table 2. Note that we have rescaled to . Also note that and are respectively the change in quantity and the transition rate at the transition.
Note that for . To get an idea of what to expect, notice that as , sample paths approach solutions to the ODE system with and
[TABLE]
that has the invariant set
[TABLE]
The subset is also invariant, since if then . Adding the and equations, the dynamics on is described by
[TABLE]
that has the stable fixed point . Thus, (27) has the equilibrium point
[TABLE]
whose stable manifold contains . For the dynamics off , let as defined above, taking values in . From (27), we derive
[TABLE]
We see that is non-decreasing, so is well-defined. If then , and if in addition then according to (28), has a positive limit, which contradicts , therefore we must have .
To see the connection to Theorem 2, note that if , then since the eigenvalues of the linearization near and are both non-zero, it should take about constant times amount of time for to exceed . Next, we delve into the land of martingales to make this intuition precise.
5.2 Controlling sample paths
Defining the process by , 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 but the same holds for and other functions of the state variables. For efficiency of notation, we’ll allow the function to change depending on the variable, so is different from and . Also, instead of the compensator we’ll use the predictor which includes the initial value, and we’ll denote by , and simply by . Then, , and and can be written
[TABLE]
Define the jump size . From (4), if , and then
[TABLE]
Defining the maximum transition rate , we have the basic inequality and we obtain the corollary
[TABLE]
For any quantity , we always have , since there are directed edges each ringing at rate , and for most quantities of interest, for a smallish integer , giving , allowing us to take equal to a small multiple of while still keeping . When the context is clear, we omit the variable and simply write .
The workflow of estimates is as follows. For any , we find so that the following holds with probability as . Item numbers correspond to the Lemmas where they are proved.
So long as , get within constant time and keep for time. 2. 2.
So long as , get within time, and find initial conditions so that for at least time. 3. 3.
Once , keep for amount of time. 4. 4.
So long as , get within constant time and keep for time. 5. 5.
So long as and , get within constant times time. 6. 6.
Once , keep for time, and show that if then so long as , for at least time. 7. 7.
So long as , get within 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 and let
[TABLE]
If and then
[TABLE]
and for integer ,
[TABLE]
Proof.
Using Table 2, we find that
[TABLE]
Since and , , and using the product rule from Lemma 4 on ,
[TABLE]
If and then, since and ,
[TABLE]
so if in addition , then since ,
[TABLE]
and since , we find
[TABLE]
Moreover, if then for , , which implies that . So, we can take and . Choosing gives . If we find
[TABLE]
Taking gives a lower bound of . So, taking and gives the first statement. Next, let . Using (29), for . Thus, if , then
[TABLE]
Taking , . If then . Noting that , then taking and ,
[TABLE]
On the other hand, since and , . If , and then
[TABLE]
If then . Taking , . If then . Letting ,
[TABLE]
and taking and , it follows that
[TABLE]
The result follows by stopping the process each time , using (30) and (31), then using the Markov property and taking a union bound while noting that for . ∎
Lemma 16**.**
As in Lemma 15, let ,
[TABLE]
Let
[TABLE]
For , if and then
[TABLE]
Also, for ,
[TABLE]
Proof.
Notice that where . Using transitions 5 and 6 from Table 2,
[TABLE]
If and , then since , it follows that and , so that .
For , let . If then since ,
[TABLE]
so we can take and . From the bound on the jump size, we find that . If then noting it follows that
[TABLE]
Taking and , and , so
[TABLE]
If this is at least , so taking and gives the estimate
[TABLE]
Next, let and define by so that and for . Here we take that depends on time, such that , and use the more general inequality . Since for any , , so we can take and . If and then
[TABLE]
Letting and bounding the integral by we obtain
[TABLE]
If , then taking and , and taking and we find that
[TABLE]
To get a matching lower bound on we need an upper bound on . If and then letting , . Let . If for and then as before, for we find
[TABLE]
If then if and only if . Taking , and using we find that
[TABLE]
On the other hand, for , so is a supermartingale. Using non-negativity of , the fact that is non-increasing, and optional stopping,
[TABLE]
Using Markov’s inequality,
[TABLE]
Letting , this is at most . The second statement then follows from a union bound. ∎
Proposition 9**.**
Let as in Lemma 15,16. Then for any , there is so that for ,
[TABLE]
Proof.
Given , let be small enough that
[TABLE]
Recall that and let as in Lemma 15. Fix . Using the first result of Lemma 15,
[TABLE]
If and then in particular, for large enough . If then . Letting in the second result of Lemma 15 and using the strong Markov property,
[TABLE]
Then, letting in the first result of Lemma 16 and using again the strong Markov property,
[TABLE]
If is large enough then . Combining these results, we find that from any initial distribution, if is large enough then
[TABLE]
If is large enough then and the first statement follows. For the second statement, recall that , and let to find that
[TABLE]
By definition, either or . Combining with the second result of Lemma 15, if then
[TABLE]
and the second statement follows. ∎
Next we show that if then there is a good chance for as long as we need.
Lemma 17**.**
If then for ,
[TABLE]
Proof.
We know that if then . Since , we can take and . Let . If and then , so using ,
[TABLE]
Optimizing in then gives the result. ∎
Lemma 18**.**
Let and let
[TABLE]
If and then
[TABLE]
and for integer ,
[TABLE]
Proof.
Since , we can take and . Recall that
[TABLE]
so if , and then . Minding the jump size, . If and then
[TABLE]
Taking and , if we have the lower bound . Taking and gives the first statement. Now, let . Suppose that , then for . If then
[TABLE]
Taking and , . On the other hand, if and then since ,
[TABLE]
Taking , and , if then
[TABLE]
Combining these,
[TABLE]
The result follows by stopping the process each time , using the strong Markov property, taking a union bound, and using . ∎
Lemma 19**.**
Let as in Lemma 18 and let . Then,
[TABLE]
Proof.
If and then . Since , we can take and . If then since , if then
[TABLE]
Taking , , and , the result follows. ∎
Proposition 10**.**
Let , as in Lemma 18. Then for any and ,
[TABLE]
Proof.
Taking in Lemma 17,
[TABLE]
Let as in Lemma 18. Using the first result of Lemma 18,
[TABLE]
Letting 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 is large enough then . Combining the estimates gives the result. ∎
Lemma 20**.**
Let and define
[TABLE]
For any and ,
[TABLE]
Also, if then for ,
[TABLE]
Proof.
We have and . Recall that if then , so . Since , . Since , we can take and . If , and then . Taking and , the first statement follows. On the other hand, we have always , and if then , and . Looking to Table 2, ignoring the and transitions, ignoring some increases, and bounding the rates in the right direction it is easy to check that for , dominates the process with initial value and the following transitions:
[TABLE]
Note the transition rates are linear. We easily compute
[TABLE]
so that if is deterministic, we solve to obtain
[TABLE]
Combining, and so
[TABLE]
Similarly,
[TABLE]
so that if is deterministic,
[TABLE]
and since the above fraction is at most ,
[TABLE]
If then , so for
[TABLE]
Letting and , if then and the second statement follows. ∎
Lemma 21**.**
Let and be as in Lemma 20. If , and then
[TABLE]
Proof.
Let . We may assume so that . Define the matrices and . Computing,
[TABLE]
So, if then letting ,
[TABLE]
Time-change by so that with . Let so that is a non-negative supermartingale. Using non-negativity and optional stopping, we see that
[TABLE]
and so
[TABLE]
Now, has eigenvalues and corresponding eigenvectors . Thus with and . If , then , and and so
[TABLE]
If then , and . If then
[TABLE]
Recalling the time change and noting , then letting and using the fact that gives the result. ∎
Proposition 11**.**
Let . For , there is so that for ,
[TABLE]
and
[TABLE]
Proof.
Since , the above definition of agrees with the one used in Lemmas 20-21. Given , be small enough that for ,
[TABLE]
Letting and using the first result of Lemma 20,
[TABLE]
On the other hand, letting and using the first result of Lemma 21,
[TABLE]
Since for large enough, the first statement follows. For the second statement, letting in Lemma 20,
[TABLE]
Combining with (32), the second statement follows. ∎
Proof of Theorem 2.
Let . Recall that . To show the upper bound, for any take small enough to satisfy all conditions, thenapply Propositions 9, 10 and 11 in sequence, stopping the process when and , to find that
[TABLE]
To show the lower bound, in Proposition 9 start from achieving the supremum, which is a maximum since the state space is finite. Apply the result of Proposition 9. Then, stop the process when , which occurs before since has jumps of size at most . Apply Proposition 11. Combining the two, conclude that for any ,
[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 increases with , if we have the upper bound
[TABLE] 2. 2.
Factoring, using the fact that and for , then using the fact that ,
[TABLE]
Lemma 22**.**
Let be Poisson with mean .
[TABLE]
Proof.
We have
[TABLE]
Also,
[TABLE]
Using Markov’s inequality,
[TABLE]
Optimizing in gives which is positive for and negative for , and
[TABLE]
Expanding in an alternating Taylor series around ,
[TABLE]
using for and . (35) follows for by letting . For it follows by continuity of probability. ∎
Lemma 23**.**
Let be a Poisson process with intensity . Fix and let
[TABLE]
denote the last passage time of above/below the curve , respectively. If and then
[TABLE]
Proof.
Let denote the function defined by . Using Lemma 22, for each ,
[TABLE]
Since for any , is Lipschitz with constant . Using this and the fact that is non-decreasing,
[TABLE]
so taking a union bound over ,
[TABLE]
Using (33),
[TABLE]
If and , this is at most . An analogous estimate applies for the lower bound, giving instead of and instead of in the exponent, when and .
∎
Lemma 24**.**
Let be a non-decreasing quasi-absolutely continuous semimartingale on with jump size at most and defined for , where is the first time of explosion. Suppose that
[TABLE]
for some locally integrable non-nonegative deterministic functions , . Let and let denote the rescaled process. Let and , and assume . Then, and for ,
[TABLE]
Proof.
First we treat the case , so that . Given define , and note that . Since , , so using linearity of the drift and Lemma 4,
[TABLE]
which implies . Clearly . Since is non-decreasing, it has finite variation, so in particular , and . In addition, , so . Using this and , for any ,
[TABLE]
which implies . Using ,
[TABLE]
Since for ,
[TABLE]
the last equality defining and . Taking the antiderivative,
[TABLE]
Since , and , it follows that for ,
[TABLE]
Using (4) with , assuming we find
[TABLE]
Optimizing gives and
[TABLE]
and if the assumption holds. For general , first condition on and apply the above to , which has jump size . Then, integrate over to obtain the result.
To see that , note that and that the above estimate implies the latter event has probability 1. ∎
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