Tipping Points in Schelling Segregation
NNoname manuscript No. (will be inserted by the editor)
Tipping Points in Schelling Segregation
George Barmpalias · Richard Elwes · Andy Lewis-PyeAbstract
Thomas Schelling’s spacial proximity model illustrated how racial segregation can emerge,unwanted, from the actions of citizens acting in accordance with their individual local preferences. One ofthe earliest agent-based models, it is closely related both to the spin-1 models of statistical physics, andto cascading phenomena on networks. Here a 1-dimensional unperturbed variant of the model is studied,which is open in the sense that agents may enter and exit the model. Following the authors’ previouswork [1] and that of Brandt, Immorlica, Kamath, and Kleinberg in [4], rigorous asymptotic results areestablished.This model’s openness allows either race to take over almost everywhere. Tipping points are identifiedbetween the regions of takeover and staticity. In a significant generalization of the models considered in [1]and [4], the model’s parameters comprise the initial proportions of the two races, along with independentvalues of the tolerance for each race.
Keywords
Schelling Segregation · Algorithmic Game Theory · Complex Systems · Non-linear Dynamics · Ising model · Spin Glass · Network Science
The game theorist Thomas Schelling proposed two models of racial segregation, which have both provedhighly influential as computational/mathematical approaches to understanding social phenomena, andwhich contributed to his receiving the Nobel Memorial Prize in 2005. In each case, the model comprisesa finite number of agents of two races, which we shall take to be green and red. The spacial proximity or checkerboard model entails agents taking up positions on a graph or grid (see [19], [20], [22]). Seg-regated regions may then appear as agents swap to neighbourhoods whose racial make-up is more totheir liking. Subsequently, numerous authors ([23,6,18,8,7]) have observed the structural similarity be-tween this model and variants of the Ising model considered in statistical mechanics for the analysis ofphase-transitions.Schelling’s second bounded neighbourhood or tipping model (see [19], [20], [21], [22]) is a non-spacialmodel, in which a number of agents share a single neighbourhood, and where the initial proportions andpreferences of the two races can give rise to total takeover by one race or the other. The simplest caseis when no agent wishes to be in the minority, and move out when they are, to be replaced by an agentof the other race. In this case whichever race has more agents initially will take over totally, and thusthe tipping point is at the 50% mark. Since Schelling’s insights, tipping points have become a focus ofinterest in both the academic literature and popular culture. Models closely related to Schelling’s havesubsequently been investigated from a number of angles, notably in the work of Granovetter [11] and aspopularised by Gladwell [10]. George Barmpalias: · Richard Elwes · Andy Lewis-Pye a r X i v : . [ c s . G T ] J un George Barmpalias et al.
More immediately, the current paper has its roots in the work of Brandt, Immorlica, Kamath, andKleinberg [4], which represented a major departure from the numerous previous studies of Schellingsegregation, for the first time providing a rigorous mathematical analysis of an unperturbed
Schelling ring(which is to say a 1-dimensional spacial proximity model). Earlier work, such as that by Young [26] andZhang ([27], [28], [29]), had concentrated on perturbed models, where agents have a small probability( ε ) of acting against their own interests. The introduction of this tiny amount of noise ensures that theresulting Markov process is reversible, and thus considerably easier to analyse. Yet it was established in[4] that the unperturbed model may give rise to dramatically different patterns of segregation from thelimiting case (letting ε →
0) of the corresponding perturbed model.The breakthrough in [4] was built upon in [1] by the present authors, which also provided a thoroughanalysis of an unperturbed Schelling ring, but over a much larger range of parameters. The present model,which we shall describe shortly, continues in the same vein in again providing rigorous mathematicalanalyses of unperturbed 1-dimensional models, but represents a significant generalization again in termsof their parameters. The major innovation is that we enrich the model with the introduction of two extraparameters which break the symmetry between the two races, in two ways. Firstly, we allow the tworaces to exist in different numbers from one another initially. Secondly the two races may now exhibitunequal levels of racial tolerance. Both of these are natural extensions of previous models, and indeedwere proposed by Schelling (see for instance [20] p 152).Another important difference from the work of [4] and [1] is that the current model will be open ,meaning that at each time step, a single unhappy agent is selected uniformly at random and replaced byone of the opposite race, if doing so will cause the new agent to be happy. Most previous versions of themodel are closed , and the dynamic has involved two unhappy agents swapping at each time step. We thusassume the presence of an unlimited number of agents of both races outside the model who are readyto move in, given the opportunity. We remark that the openness of the current model brings it closer tothe standard spin-1 models of statistical physics (although the closed variant also has a counterpart insystems employing Kawasaki dynamics). Progress has also recently been made by the authors in analysingclosed models, similarly enriched with more parameters than [4] and [1]. See [2].We shall follow tradition in discussing our model in terms of racial segregation. However, as has oftenbeen remarked, it is equally applicable to any other geographical division of people along binary lines.Examples from [19] include women from men, students from faculty, or officers from enlisted men. What ismore, the open model analysed here can equally well be interpreted in terms of static agents who modifysome personal attribute in response to their neighbours, as happens in the class of voter models (see forinstance [15]). In a voter model an agent will alter its state to mimic a randomly selected neighbour. In oursetting, it is the proportions of its neighbours occupying the two states which determines its action. Thuswe might propose an interpretation as a model of peer pressure, with two asymmetric states. Relevantexamples may include whether or not to smoke [5], whether or not to clap at the end of a performance[16], or preferences for rival musical subcultures such as hip-hop or heavy metal [25].Viewing the model from this perspective places it squarely within the category of cascading behaviourwithin networks. A central topic of study in this area is a general threshold model . The setting here isa graph, in which every node v is equipped with two things: a function g v which assigns a value g v ( N )to every subset N of the neighbours of v , and a threshold τ v ∈ [0 , active while others are not. At each time step t , every inactive node v computes g v ( A tv ) where A tv is the set ofneighbours of v which are active at time t . Then v activates at time t + 1 if and only if g v ( A tv ) ≥ τ v . Theprimary question of interest here is to find conditions (on the set of initially active nodes, the topologyof the graph, the functions g v , and the thresholds τ v ) which guarantee that the whole graph (or most ofit) will eventually become activated, or alternatively that the cascade will quickly fizzle out. See [14] fora good survey of this area.The parallels with the study of Schelling segregation are striking. One major difference, however, isthat while general threshold models evolve according to a synchronous dynamic (every agent that maychange will do so at each time-step), the literature on Schelling segregation traditionally has one agent(or pair of agents) changing at each time-step. In the current work we shall consider variants in ourSchelling model’s dynamics including introducing synchronicity, and see that in many cases (but not all)our conclusions are unaffected by the choice between them.Although our model is an instance of Schelling’s spacial proximity model rather than any kind ofhybrid or unified model, we nevertheless identify interesting phenomena in the spirit of his tipping orbounded neighbourhood models. That is to say, we shall identify thresholds in parameter-space on oneone side of which one race takes over, and on the other side of which the other does. This behaviour is ipping Points in Schelling Segregation 3 of course only possible in an open model, and furthermore is is only visible from our current asymptotic perspective: we shall prove precise results concerning the ring’s final configuration which are valid as theneighbourhood radius ( w ) grows large, and the ring size ( n ) grows large relative to w . More precisely,depending on the initial parameters, one of three conclusions will usually follow in the long run: either thering will remain essentially static or one race or the other will take over. The asymptotic interpretationis critical here because takeover in this setting need not entail the complete absence of the other race,but rather takeover almost everywhere in a measure-theoretic sense, meaning that in the ring’s finalconfiguration the majority race will almost certainly outnumber the other by any required margin forlarge enough values of w and n . Thus takeover or staticity may not be apparent in simulations involvingsmall w and n . We will identify boundaries between these three regions within the parameter-space ofthe model.While the results of [1] were somewhat counterintuitive (and perhaps politically discouraging) in thatincreased tolerance was seen in certain situations to lead to increased segregation, our results (describedbelow) on the open model suggest the maxim “tolerance wins out”. Loosely speaking more tolerantgroups thrive at the expense of their less tolerant neighbours, although we emphasise that the details arehighly sensitive to the initial proportions of the two groups. We also identify two very different regionsof staticity, in which very few people move. These occur at the extremes: in one case a city comprisingonly very tolerant individuals in which almost everyone is happy with their neighbourhood. We think ofthis as the region of contentment . In contrast, the region of frustration comprises people so intolerantthat, although almost all are unhappy with their neighbourhood, they are also unable to find anyone elseprepared to move in, and thus are forced to stay put.In more detail, and as already mentioned, the parameters in question represent a considerable gener-alisation of those from [1] in two directions. Firstly, we will no longer assume that the initial distributionis symmetric between the two races. Instead, each site will be occupied initially by a green agent withprobability ρ , and by a red agent with probability 1 − ρ . Thus it might be that our model describesa racially homogeneous red region, into which a few green individuals have recently moved (meaning asmall value of ρ ). It is clearly of interest to be able to predict whether the newcomers will eventually takeover the region, or will themselves be squeezed out.Secondly, we drop the assumption that the preferences of the two races are simple mirror images,and allow the two groups to exhibit different tolerances. This is in line with social research, which hassuggested in the past, for example, that black US citizens are happier in integrated neighbourhoods thantheir white compatriots (see for instance [24]). Thus we introduce two independent parameters τ g and τ r representing the tolerance of green and red agents, respectively.The model runs as follows. First we fix the parameters n, w ∈ N and ρ, τ g , τ r ∈ (0 , n −
1. We arrange these in a circle, meaning that addresses are computedmod n in everything that follows. Initially we populate the ring with agents of two races (red and green),with the colour of each node decided independently according to the toss of a biased coin, each node beinggreen initially with a probability of ρ and red with a probability of 1 − ρ . At each time-step, a node x issolely concerned with its own neighbourhood of size 2 w + 1, meaning the interval N ( x ) = [ x − w, x + w ](understood mod n as usual). If x is a green (red) node, it will be happy so long as the proportion of green(red) nodes in its neighbourhood is at least τ g (2 w + 1) (respectively τ r (2 w + 1)), and unhappy otherwise.We say that an unhappy node is hopeful if a change of colour would cause it to be happy.Now we introduce three possible dynamics by which the model may evolve: • Our primary object of study will be the selective model . Here, at each time-step a hopeful node isselected uniformly at random and its colour changed. • The incremental model is similar: at each time-step an unhappy node is selected uniformly at randomand its colour changed (regardless of whether this will make it happy). • In the synchronous model , at each time-step every currently unhappy node alters its colour (again,regardless of the effect on their happiness).In all cases, the process continues until no further changes are possible, at which stage we say thering (or process) has finished . (We shall establish in Lemma 1.13 that this is guaranteed to occur forthe selective model, and we shall later establish this in certain other cases.) Our principal concern is tofind the probability that a randomly selected node is green in the finished ring. We shall show that thisprobability is usually close to either ρ , 0, or 1. George Barmpalias et al.
We shall describe these tipping phenomena in terms of numerical relationships between ρ , τ g , and τ r .In particular, for any ρ there exist thresholds κ ρg & κ ρr and µ ρg = 1 − κ ρr & µ ρr = 1 − κ ρg as illustrated inFigure 1. ρκ ρg κ ρr ρµ ρg µ ρr Fig. 1
The thresholds κ ρg & κ ρr and µ ρg = 1 − κ ρr & µ ρr = 1 − κ ρg The following give more details: • For ρ ≤ we have κ ρg < κ ρr = = µ ρg < µ ρr . • For < ρ < we have κ ρg < κ ρr < < µ ρg < µ ρr . • For ρ = we have κ ρg = κ ρr ≈ . κ found in [1], and µ ρg = µ ρr =1 − κ ≈ . • For < ρ < we have κ ρr < κ ρg < < µ ρr < µ ρg . • For ρ ≥ we have κ ρr < κ ρg = = µ ρr < µ ρg .As the pair ( τ g , τ r ) ranges across the unit square, we shall find that the final configuration of the ringdepends principally on where τ g stands in relation to the thresholds κ ρg , , & µ ρg and where τ r standsregarding κ ρr , , & µ ρr , dividing the unit square into up to 16 open regions. We shall be able to analyseseveral of these simultaneously, but in some cases we shall find a more delicate dependency between τ g and τ r . Throughout we shall leave open the intriguing question of what happens when the parametersexactly coincide with the thresholds. (We remark in passing, that the parameters of the models in [4]and [1] both constitute threshold cases of the current situation.) We shall also leave open the outcomesof the process in two small open regions of the parameter space (see Questions 1.4 and 1.8 below). Weencourage others to investigate these matters.As already mentioned, our results are asymptotic in nature, and we will use the shorthand “all n (cid:29) w (cid:29)
0” which carries the meaning “all sufficiently large w , and all n sufficiently large compared to w ”. By a scenario we mean the class of all rings with fixed values of ρ , τ g , and τ r , but w and n varying.We will identify a scenario with its signature triple ( ρ, τ g , τ r ), and say that a value of ρ admits scenariossatisfying some property X , if there exist τ g and τ r such that X holds for ( ρ, τ g , τ r ). Our conclusions arethen of three types: • A scenario is static almost everywhere if for every ε > n (cid:29) w (cid:29) x chosen uniformlyat random has a probability below ε of having changed colour at any stage before the ring finishes. • A scenario suffers green (red) takeover almost everywhere if for every ε > n (cid:29) w (cid:29) x chosen uniformly at random has a probability exceeding 1 − ε of being green (red) in the finishedring.In some situations we will be able to strengthen the conclusion, and say that green (red) takes overtotally if for every ε > n (cid:29) w (cid:29) − ε .We now state our main results, and some open questions. (These depend on the existence of κ ρg & κ ρr and µ ρg & µ ρr , which will be established rigorously in sections 2 and 6 respectively.) Theorems 1.1 -1.9 which follow are encapsulated (for the cases ρ = 0 .
42 and ρ = 0 .
3) by Figures 2 and 3. Although thedetails of the diagrams are specific to ρ = 0 .
42 and ρ = 0 .
30, the major features apply more generally. ipping Points in Schelling Segregation 5
Points coloured grey correspond to scenarios static almost everywhere, while green (red) points indicategreen (red) takeover. Purple open regions represent those scenarios, other than those on the thresholds,whose outcome remains unclear. (Notice that there are no such regions in Figure 2, which is not unusual.See Questions 1.4 and 1.8.) Similar diagrams for other values of ρ are given in Figure 4. κ ρr µ ρr − ρ τ r κ ρg ρµ ρg τ g Fig. 2
The landscape for ρ = 0 .
42 under the selective dynamic κ ρr µ ρr − ρ τ r κ ρg ρµ ρg τ g Fig. 3
The landscape for ρ = 0 . In each case the roles of red and green may be interchanged by swapping the relevant words, exchanging ρ with 1 − ρ , and κ ρg with κ ρr , and µ ρg with µ ρr . Our first clutch of results (Theorem 1.1 - 1.5) apply to allthree dynamics, in the situation that at least one of τ g , τ r < . Theorem 1.1
Under all three dynamics, if τ g < κ ρg and τ r < κ ρr then the scenario will be static almosteverywhere. George Barmpalias et al.
Scenarios where κ ρr < τ r < and τ g < exhibit a more intricate dependency on ( ρ, τ g , τ r ). To resolvematters here we require the following numerical condition (Definition 3.1 below). A scenario ( ρ, τ g , τ r )where ρ ∈ (0 ,
1) and τ g , τ r ∈ (0 ,
1) and τ g + τ r (cid:54) = 1 is red dominating if ρ · (cid:18) τ τg − τg − τr g (cid:19) (1 − τ g ) − τg − τg − τr < (1 − ρ ) · (cid:18) τ τr − τg − τr r (cid:19) (1 − τ r ) − τr − τg − τr . It is green dominating if the reverse strict inequality holds.
Theorem 1.2
Under all three dynamics, if τ g < and κ ρr < τ r < and ( ρ, τ g , τ r ) is green dominating,then green will take over almost everywhere. Theorem 1.3
Under all three dynamics, if τ g < κ ρg and κ ρr < τ r < , where the scenario is red domi-nating and τ g > ρ , then the scenario is static almost everywhere. Open Question 1.4
What is the outcome, under any dynamic, if τ g < κ ρg and κ ρr < τ r < , the scenariois red dominating and τ g ≤ ρ ? We shall discuss this further at the end of Section 4, but remark that this problematic region onlyexists for a limited range of values of ρ , namely < ρ < . . < ρ < .)Notice that the following result has no dependency on ρ : Theorem 1.5
Under all three dynamics, if τ g < < τ r green will take over totally. For the second batch of results (Theorems 1.6 - 1.9), we turn our attention to the case where both τ g , τ r > . Here, the different dynamics diverge, and our focus will be on the selective case: Theorem 1.6
Under the selective dynamic, if < τ g < µ ρg and < τ r , and if ( ρ, τ g , τ r ) is greendominating, then green will take over almost everywhere. Theorem 1.7
Under the selective dynamic, if < τ g < µ ρg and µ ρr < τ r , where the scenario is reddominating and additionally τ r < − ρ , then the scenario is static almost everywhere. Open Question 1.8
What is the outcome, under the selective dynamic, of scenarios where < τ g < µ ρg and µ ρr < τ r , the scenario is red dominating and τ r ≥ − ρ ? The range of ρ for which this mysterious region exists is the same as that for Question 1.4 above.Our analysis of the selective dynamic culminates in a region of frustration as discussed earlier: Theorem 1.9
Under the selective dynamic, if µ ρg < τ g and µ ρr < τ r , then the scenario is static almosteverywhere. For the incremental and synchronous dynamics, we shall leave the case τ g , τ r > largely open.However, we make the following conjecture: Conjecture 1.10
Under the incremental and synchronous dynamics, suppose that < τ g < τ r . Thengreen will take over totally.If < τ g = τ r then for any fixed ρ the probability of both red and green total takeover tends to as w, n → ∞ . Some intuition and partial results towards this conjecture are established in section 8, along with arelated discussion of the process’ run-time. ipping Points in Schelling Segregation 7 τ g < , amounts to comparing the relative probabilities of unhappy nodes of each colour along with thoseof stable intervals of each colour, in the initial configuration.The significance of initially unhappy red nodes is that they are likely to spark the growth of greenfirewalls , which is to say runs of ≥ w +1 successive green nodes. When τ g < , such a firewall is guaranteedto grow until it hits a stably red interval , meaning an interval of length w + 1 containing enough red nodes(specifically ≥ τ r (2 w + 1) many) to ensure that all remain perpetually happy. It is not difficult to seethat stably red intervals stop the growth of green firewalls, and that they are the only things to do so.Given this picture, it is natural that the relative frequencies of stable intervals and unhappy nodes inthe initial configuration should be important, and indeed we shall establish that such considerations aredecisive. The various thresholds we identify within the region τ g < can be understood as follows: • For τ r > , stably red intervals cannot occur for all large enough w . Thus if τ g < < τ r as in Theorem1.5, stable green intervals will likely exist in large enough rings (and will serve to prevent total redtakeover), as will unhappy nodes of both colours, while stable red intervals will not, suggesting that,eventually, total green takeover cannot be resisted. • τ g = κ ρg is the point below which stably green intervals become more likely than unhappy green nodes.Thus if τ g < κ ρg , stably green intervals are more numerous than unhappy green nodes, making redtakeover unlikely. • Green domination will correspond to unhappy red nodes being more common than unhappy greennodes. Hence, if τ g < and κ ρr < τ r < hold alongside green domination, as in Theorem 1.2, therewill be many more unhappy red nodes than green. Since stably red intervals are also infrequent, itfollows that many more nodes will be consumed by green firewalls than by red.When τ g , τ r > , under the selective dynamic, similar considerations apply, with a couple of changes:in the place of unhappy nodes we consider hopeful nodes. (Recall these are unhappy nodes for whomchanging colour would produce happiness. This is only automatic, for large enough w , when τ g + τ r < intractability , which similarly obstructs the growth of firewalls (butcannot occur when τ g , τ r < ).An interval J of length w + 1 is green intractable if it contains so few green nodes (specifically < τ g (2 w + 1) − ( w + 1) many), that no red node inside J can ever become hopeful, no matter whatoccurs outside J . Thus no red node within J will ever change colour. Such an interval is therefore, in thissetting, the only thing which can halt the growth of a green firewall. We may now interpret the remainingthresholds: • If τ g > µ ρg , then green intractable intervals are more likely than hopeful red nodes, making greentakeover improbable. • Green domination has an alternative characterisation when τ g , τ r > , as saying that hopeful red nodesare more likely than hopeful green nodes. If this holds, along with the assumptions that < τ g < µ ρg and < τ r as in Theorem 1.6, then more green firewalls will start than red ones, and since there arefew green intractable intervals to impede them, we may expect many more nodes to end up green thanred.1.12 Arguing that the ring finishesAs a final step before we launch into an analysis of segregation patterns, we address the question ofwhether the process is guaranteed to finish. For the selective dynamic, the following result, whose proofis included in Appendix A, is sufficient for our purposes: Lemma 1.13
For any scenario ( ρ, τ g , τ r ) and for all large enough w , the selective dynamic guaranteesthat the process will finish. The observant reader will notice from the proof that the requirement that w be large is not necessaryin scenarios where τ g , τ r < . Indeed, we expect that it could be dropped in all cases, although this doesnot follow from the current argument. George Barmpalias et al.
Of course, this implies that a ring under an incremental dynamic and with τ g , τ r < will also finish.We shall establish certain other cases as consequences of the results in sections 5 and 8, however we donot have complete answers for the incremental and synchronous dynamics when τ g , τ r > . We expect(indeed it is implicit in Conjecture 1.10) that for any scenario, for any ε >
0, and for all large enough n (cid:29) w (cid:29)
0, the probability that the ring will finish eventually exceeds 1 − ε . κ ρr We begin with some notation. Given a node a , we shall write N ( a ) := [ a − w, a + w ] for a ’s neighbourhood.Given some collection A of nodes and a time t , we write G t ( A ) := |{ x ∈ A : x is green at time t }| . Wesimilarly define R t ( A ), U t ( A ), F t ( A ), UG t ( A ), UR t ( A ), HG t ( A ), HR t ( A ), FG t ( A ) and FR t ( A ) to be thenumber of red, unhappy, hopeful, unhappy green, unhappy red, happy green, happy red nodes, hopefulgreen, and hopeful red nodes in A at time t respectively, and will omit t when its meaning is understoodfrom context. Thus a green node a is happy if G ( N ( a )) ≥ τ g (2 w + 1) and an unhappy red node b ishopeful if G ( N ( b )) ≥ τ g (2 w + 1) −
1. We will also apply this in the case that A = { a } is a singleton.Abusing notation slightly, G ( a ) can be thought of as the green characteristic function of a , taking values0 or 1.Similarly, an interval [ a, a + w ] of length w +1 is stably green if G [ a, a + w ] ≥ τ g (2 w +1). We mentionedin 1.11 that stably green intervals have the ability to halt the growth of red firewalls (stretches of at least w + 1 consecutive red nodes). We make this precise: Lemma 2.1
Suppose that u and u are nodes such that at time t = 0 each of u and u lie in (possiblydifferent) stably green intervals, and there is no unhappy green node in [ u , u ] . Then every green nodein [ u , u ] will remain perpetually and happily green.Proof Suppose not. Then let v in [ u , u ] be the first green node to become unhappy. Now v may onlybecome unhappy once some other green node v (cid:48) ∈ N ( v ) has turned red. Since v (cid:48) (cid:54)∈ [ u , u ], either v (cid:48) < u or v (cid:48) > u . We assume without loss of generality that v (cid:48) < u . Then v (cid:48) ∈ N ( u ). Now by assumption, u ∈ [ a, a + w ] for some stably green interval [ a, a + w ]. By stability, we cannot have v (cid:48) ∈ [ a, a + w ]. Thus v (cid:48) ∈ [ u − w, a − a, a + w ] ⊆ N ( v ) meaning, by stability, that N ( v ) containsenough green nodes to keep v happy, which is a contradiction.Notice that the possibility of stably green intervals in arbitrarily large rings requires that τ g ≤ . Infact we shall assume that τ g < throughout this section, unless stated otherwise. Our goal is to provethe following: Proposition 2.2
For any ρ ∈ (0 , , we work in the initial condition, and let U g be the probability thata uniformly randomly selected green node is unhappy, and S g be that of a uniformly randomly selectednode lying within a stably green interval. Then there exists a threshold κ ρg satisfying ρ > κ ρg > ρ , definedas the unique root of the equation f ( s ) := (cid:0) − s (cid:1) (1 − s ) (1 − s ) − s ) = 12(1 − ρ ) . This is such such that for any τ g ∈ (0 , : • If τ g < κ ρg , there exists ζ ∈ (0 , so that U g < ζ w S g for all w . • If τ g > κ ρg , there exists ζ ∈ (0 , so that S g < ζ w U g for all w .Similarly, there exists a threshold κ ρr where (1 − ρ ) > κ ρr > (1 − ρ ) , defined as the unique root of f ( s ) = ρ such that corresponding statements about U r and S r hold. With threshold in place, we shall argue that when τ g < κ ρg , any randomly selected node is highly likelybe closer on each side to a stably green interval than to an unhappy green node in the initial configura-tion. This will establish that such a node can never turn red, and will be enough to establish Theorem 1.1.Before we proceed with the proof of 2.2, we recall two important probabilistic results. The first is aclassical result from [12]. (The full statement is more general, but this is the version which shall find mostuseful.) ipping Points in Schelling Segregation 9 Proposition 2.3 (Hoeffding’s inequality)
Let X , . . . , X N be independent random variables such that P ( X i = 1) = p and P ( X i = 0) = 1 − p . Then for any δ > we have P (cid:32) N (cid:88) i =1 X i ≥ ( p + δ ) N (cid:33) ≤ exp (cid:0) − N δ (cid:1) . Secondly, we shall require Theorem 1.1 from [3], which appears as Lemma 3.1 in [1], and which werestate:
Lemma 2.4
Suppose h : Z → Z and p ∈ (0 , are such that there exist k ∈ (0 , so that for all largeenough N , we have (cid:16) (cid:16) p − (cid:17) k (cid:17) h ( N ) > N ≥ h ( N ) > pN > . Then for all large enough N , if X N ∼ b ( N, p ) , we have P ( X N = h ( N )) ≤ P ( X N ≥ h ( N )) ≤ (cid:18) − k (cid:19) · P ( X N = h ( N )) . That is to say in asymptotic notation, P ( X N ≥ h ( N )) = Θ ( P ( X N = h ( N ))) . For current purposes, the appropriate asymptotic notion is weaker than Θ:
Remark 2.5 If f and g are functions of w , we shall write f ≈ g to mean that there are rational functions P and Q such that P ( w ) , Q ( w ) > and P ( w ) g ( w ) ≤ f ( w ) ≤ Q ( w ) g ( w ) for all large enough w .Proof (Proof of Proposition 2.2) We fix a scenario ( ρ, τ g , τ r ) and work always in the initial configuration. For some green node b , wewish to compare the probability U g that b is unhappy with the probability S g that [ b − i, b + w − i ] isstably green for some i where 0 ≤ i ≤ w . Our first step is to approximate S g by focusing on the case i = 0.Let S g be the probability that [ b, b + w ] is stably green. Then S g ≤ ( w + 1) S g , meaning that S g ≈ S g .We shall therefore work with S g in place of S g , and observe later that this introduces no problems.Hence the first probability we shall compute is that of the interval [ b, b + w ] nodes being stably green.Here, the relevant distribution is binomial: X ∼ b ( w, ρ ), describing the number of green elements otherthan b in the interval, and S g ≈ P ( X ≥ τ g (2 w + 1) − b being unhappy, it will be convenient to count thered nodes in N ( b ). This is given by the distribution Z ∼ b (2 w, − ρ ). Then U g = P ( Z > (1 − τ g )(2 w + 1)). Remark 2.6
The behaviour of U g S g is easy to determine in certain situations: • If τ g ≤ ρ then S g ≥ while U g → . • If τ g = ρ then S g → and U g → . • If τ g > ρ , then S g → and U g → .All limits are taken as w → ∞ . Furthermore, it is a straightforward consequence of Hoeffding’s inequality(Proposition 2.3) that the quantities tending to do so at an exponential rate in w , meaning that theyare bounded above by ν w for some ν ∈ (0 , . Hence for the remainder of this proof we shall concentrate on scenarios where ρ < τ g < ρ . We nowapply Lemma 2.4 in the current context, with N = w , p = ρ , and h ( N ) = (cid:100) (2 w + 1) τ g (cid:101) −
1. Furthermore,making the assumption that τ g > ρ we may find k where 1 > k > ρ − ρ · − τ g τ g . Thus we get S g ≈ ρ h (1 − ρ ) w − h ( w ) h. Similarly, assuming only that τ g < ρ , we may take N = 2 w , p = 1 − ρ , and choose k (cid:48) so that 1 > k (cid:48) > − ρρ · τ g − τ g , to get U g ≈ (1 − ρ ) h (cid:48) ρ w − h (cid:48) ( 2 ) wh (cid:48) (1)where h (cid:48) = (cid:98) (1 − τ g )(2 w + 1) (cid:99) + 1. Putting these two estimates together, so long as ρ < τ g < ρ , we find U g S g ≈ (1 − ρ ) h (cid:48) + h − w ρ w − h (cid:48) − h ( 2 ) wh (cid:48) ( w ) h . We now employ Stirling’s formula, that n ! ≈ n n + e − n . Then, the powers of e cancel and we see: U g S g ≈ (1 − ρ ) h (cid:48) + h − w ρ w − h (cid:48) − h (2 w ) w + ( h ) h + ( w − h ) w − h + ( h (cid:48) ) h (cid:48) + (2 w − h (cid:48) ) w − h (cid:48) + ( w ) w + . (2)Now we introduce the approximations 2 wτ g and 2 w (1 − τ g ) for h and h (cid:48) respectively. Notice that | h − wτ g |≤ | h (cid:48) − w (1 − τ g ) |≤
2. It follows easily that h h + ≈ (2 wτ g ) wτ g + with ‘ ≈ ’ interpretedas in Remark 2.5. (We observe in passing that this estimate would not hold under the asymptotic notionΘ.) Similar remarks apply to the other terms in the estimate, allowing us to deduce the following: U g S g ≈ (1 − ρ ) w (2 w ) w + ( w (1 − τ g )) w (1 − τ g )+ (2 w (1 − τ g )) w (1 − τ g )+ ( w ) w + . Hence we obtain the following key estimate: U g S g ≈ (cid:32) ( − τ g ) (1 − τ g ) (1 − τ g ) − τ g ) · · (1 − ρ ) (cid:33) w . (3)The question now is whether the term inside the brackets in 3 is greater than or less than 1. In manycases there is a threshold, κ ρg , where it is equal to 1. That is, κ ρg is the root, if it exists, of the equation: f ( s ) := (cid:0) − s (cid:1) (1 − s ) (1 − s ) − s ) = 12(1 − ρ ) . (4)To establish the existence of this root we shall appeal to the intermediate value theorem, noticing firstthat for 0 < s < we have f (cid:48) ( s ) > ρ < τ g < ρ and τ g < . Now we claim the following:(i) For 0 < ρ < , we have f ( ρ ) > − ρ ) .(ii) For 0 < ρ < we have − ρ ) > f ( ρ ).To prove (i), define g ( ρ ) := (1 − ρ ) f ( ρ ) = (cid:16) − ρ − ρ (cid:17) − ρ . We shall show that g ( ρ ) > . Notice that g (0) = so it suffices to show that g (cid:48) ( ρ ) >
0. Taking logarithms and differentiating, we find that g ( ρ ) := g (cid:48) ( ρ ) g ( ρ ) = 2 ln(1 − ρ ) − − ρ ) − − ρ . Since g ( ρ ) > g ( ρ ) >
0. Well g (0) = 2 ln 2 − > g (cid:48) ( ρ ) > g ( ρ ) := (1 − ρ ) f (cid:0) ρ (cid:1) = (cid:0) (cid:1) − ρ (cid:16) − ρ − ρ (cid:17) − ρ . We’ll show that g ( ρ ) < by a similar argument. Notice that g (0) = , hence it will suffice to show that g (cid:48) ( ρ ) <
0. Again, wetake logarithms and differentiate, getting g ( ρ ) := g (cid:48) ( ρ ) g ( ρ ) = ln 2 + ln (cid:0) − ρ (cid:1) − ln (1 − ρ ) − − ρ . Again, g ( ρ ) > g ( ρ ) <
0. Well, g (0) = ln 2 − < g (cid:48) ( ρ ) < ≤ ρ < , it holds that f ( ) > − ρ ) , where we extend by continuity to take f (cid:0) (cid:1) = 2. Along with (i) and (ii), this allows us to apply the intermediate value theorem.Hence for any 0 < ρ < the threshold κ ρg exists. For τ g < κ ρg we will have S g (cid:29) U g for all largeenough w , while for τ g > κ ρg we will have U g (cid:29) S g .However, for ρ ≥ , we get f ( s ) < − ρ ) for all s < . Hence in this region S g (cid:29) U g , for all largeenough w , whatever the value of τ g < . On the other hand, for τ g > we have S g = 0 < U g . Thus itmakes sense to set κ ρg := in this case.We can similarly compute κ ρr , simply by replacing τ g with τ r and 1 − ρ by ρ in the above analysis,making κ ρr the root of f ( s ) = ρ . By symmetry, we find that for < ρ < κ ρr exists. Butwhen ρ ≤ , we have that S r (cid:29) U r for all large enough w whatever the value of τ r < .Finally, recall that at the start of the proof we made the approximation S g ≈ S g . Since estimate 3 isexponential in w , the asymptotic limits are unaffected by this move, meaning that κ ρg and κ ρr representexactly the thresholds we seek. Combining these observations with Remark 2.6 (and the impossibility ofstably green intervals when τ g > ) we have completed the proof of Proposition 2.2. ipping Points in Schelling Segregation 11 We may now build towards the proof of our first theorem, that a scenario where τ g < κ ρg and τ r < κ ρr will be static almost everywhere.We begin at a node u selected uniformly at random. Looking outwards from u in both directions,we may encounter unhappy nodes and/or stable intervals of both colours. We need to understand themost likely order in which we will meet these. It seems plausible, by Proposition 2.2, that we are morelikely to find green stable intervals before unhappy green nodes, and red stable intervals before unhappyred nodes. Establishing this will suffice, as Lemma 2.1 then guarantees that there can then be no way forthe influence of any unhappy node to reach u , which must therefore remain unchanged.We restate the following, which is Lemma 3.2 from [1], recalling that “the first node to the left” ofsome given node u satisfying some criterion means the first in the sequence u, u − , u − , · · · to satisfythe condition. Lemma 2.7
Let P ( u ) and Q ( u ) be events which only depend on the neighbourhood of u in the initialconfiguration, meaning that if the neighbourhood of v in the initial configuration is identical that of u (i.e.for all i ∈ [ − w, w ] , u + i is of the same type as v + i ), then P ( u ) holds if and only if P ( v ) holds andsimilarly for Q ( u ) and Q ( v ) . Suppose also that:(i) P ( P ( u )) (cid:54) = 0 and P ( Q ( u )) (cid:54) = 0 .(ii) For all k , for all sufficiently large w compared to k , P ( P ( u )) / P ( Q ( u )) > kw .For any u , let x u be the first node to the left of u such that either P ( x u ) or Q ( x u ) holds. For any ε > ,if (cid:28) w (cid:28) n then the following occurs with probability > − ε for u chosen uniformly at random: x u isdefined and for no node v in [ x u − w, x u ] does Q ( v ) hold.An analogous result holds when ‘left’ is replaced by ‘right’. We can now establish Theorem 1.1. We apply Lemma 2.7, interpreting P ( u ) as the event that the node u lies in a green stable interval and Q ( u ) as its being green and unhappy, with Proposition 2.2 providingthe necessary probabilistic bounds. This tells us that, for any ε (cid:48) > n (cid:29) w (cid:29)
0, ifwe pick a node u uniformly at random, then with probability > − ε (cid:48) the nearest stable green intervalsto u will be closer on both sides than the nearest unhappy green nodes. Thus by Lemma 2.1, u , if greenwill never turn red.Then we simply repeat the argument with the roles of red an green interchanged, noting that if twoevents each have probability tending to 1, then so must their conjunction. To analyse the case κ ρr < τ r < , we need to answer the following question: in the initial configuration,which colour is likely to yield more unhappy nodes? Definition 3.1
A scenario ( ρ, τ g , τ r ) where τ g + τ r (cid:54) = 1 is red dominating if ρ · (cid:18) τ τg − τg − τr g (cid:19) (1 − τ g ) − τg − τg − τr < (1 − ρ ) · (cid:18) τ τr − τg − τr r (cid:19) (1 − τ r ) − τr − τg − τr . It is green dominating if the reverse strict inequality holds.
Our choice of terminology will be justified below in Propositions 3.4 and 6.3. Firstly, however weestablish some facts about domination, deferring the proof until Appendix B:
Lemma 3.2
Let S := (0 , × (0 , . We divide S into the two triangles T := { ( x, y ) ∈ S : x + y < } and T := { ( x, y ) ∈ S : x + y > } and the line L = { ( x, y ) ∈ S : x + y = 1 } . Also define S := (cid:0) , (cid:1) × (cid:0) , (cid:1) and S := (cid:0) , (cid:1) × (cid:0) , (cid:1) . (Notice that S i ⊂ T i .) Then the following hold:1. Suppose that ( τ g , τ r ) , ( τ (cid:48) g , τ (cid:48) r ) ∈ T i and that ( ρ, τ g , τ r ) is red dominating. If τ (cid:48) g ≥ τ g , and τ r ≥ τ (cid:48) r then ( ρ, τ (cid:48) g , τ (cid:48) r ) is red dominating. Conversely, if ( ρ, τ (cid:48) g , τ (cid:48) r ) is green dominating, so too is ( ρ, τ g , τ r ) .2. For i ∈ { , } , every scenario where ρ ≤ (respectively ρ ≥ ) and ( τ g , τ r ) ∈ S i is red (green)dominating.3. Any value of ρ where < ρ < admits both red and green dominating scenarios in both S and S . In some cases, red domination is easy to determine:
Corollary 3.3
Suppose ( ρ, τ g , τ r ) is a scenario where τ g + τ r (cid:54) = 1 and τ g ≥ ρ and τ r ≤ − ρ . Then ( ρ, τ g , τ r ) is red dominating. (Similarly, green domination follows when both of the reverse inequalitieshold.) Again we defer the proof to Appendix B. The following result justifies our choice of terminology forscenarios where τ g , τ r < : Proposition 3.4
Suppose that τ g , τ r < . The scenario ( ρ, τ g , τ r ) is red dominating if and only if thereexists η ∈ (0 , so that for all w we have U r < η w U g .The same holds with the roles of red and green interchanged.Proof Suppose first that τ g ≥ ρ , then automatically 1 − τ r ≥ ρ , so we may apply Corollary 3.3 to establishred domination. Also, U g → τ g > ρ ) or U g → (if τ g = ρ ) as n (cid:29) w → ∞ . Meanwhile U r → w . Thus the result follows.By an identical argument, if τ r ≥ − ρ then automatically τ g < ρ and the result follows by Corollary3.3 with the roles of red and green exchanged.This leaves us with the case where τ g < ρ and τ r < − ρ . Under the assumption that τ g < ρ , in Equa-tion 1, we derived an asymptotic expression for U g . Applying Stirling’s formula and the approximation h (cid:48) ≈ w (1 − τ g ) as previously, it follows that U g ≈ (1 − ρ ) w (1 − τ g ) ρ wτ g (2 w ) w + (2 w (1 − τ g )) w (1 − τ g )+ (2 wτ g ) wτ g + . Of course, by interchanging ρ with 1 − ρ , as well as τ g with τ r , under the assumption that τ r < − ρ we may form an analogous expression for U r . We may now take the quotient of these two, to find U g U r ≈ (cid:18) (1 − τ r ) − τ r · τ τ r r (1 − τ g ) − τ g · τ τ g g · ρ τ g + τ r − · (1 − ρ ) − τ g − τ r (cid:19) w . (5)The term within the bracket is then > <
1) if and only if ( ρ, τ g , τ r ) is red (green)dominating, thus establishing the result.Figure 4 depicts the boundary between red and green domination for a few values of ρ . Other detailsare also shown (exactly as in Figure 2), in particular green (red) points on the plane represent scenarioswhich suffer green (red) takeover, while grey points represent static scenarios. Recall from the discussionfollowing 3 that for ρ ≤ our zone of current interest κ ρr < τ r < does not exist, with the fate ofeach scenario entirely determined by the value of τ r relative to and µ ρr and of τ g relative to κ ρg and .Nevertheless red/green domination still makes sense as a numerical condition.For ρ > , the zone κ ρr < τ r < does exist, and we observe a form of threshold at ρ = λ ≈ . τ λg ≈ . τ λr ≈ . λ , so let us brieflydiscuss it here. By definition, λ is such that ( , κ λg ) and ( κ λr , ) lie exactly on the boundary betweenred and green domination. The point of it, therefore, is that for τ r < and ρ < λ , green dominationautomatically implies that τ g < κ ρg .However, for λ < ρ ≤ green dominating scenarios are also admissible within the zone κ ρr < τ r < and κ ρg < τ g < . Notice that at ρ = , the threshold between red and green domination is simply theline τ g = τ r .It is clear from Figure 4, that for ρ > , red/green domination can play a decisive role in determiningthe fate of any scenario. We shall now prove this. κ ρr and We now wish to apply the results of the previous section to understand scenarios where κ ρr < τ r < .This automatically requires ρ > . We shall make both these assumptions throughout this section, andalso insist that τ g < , since cases where τ g > will be subsumed into Theorem 1.5. However some ofour lemmas will have weaker hypotheses which we shall state explicitly for later reuse.To begin with, we shall deal with scenarios where ( ρ, τ g , τ r ) is green dominating, establishing greentakeover almost everywhere, thus proving Theorem 1.2. An example of such a ring is illustrated in Figure5. (We briefly explain how to interpret such a figure: the initial configuration is shown as the innermost ipping Points in Schelling Segregation 13 ρ = 0 . ρ = 0 . ρ = 0 . ρ = λ ≈ . ρ = 0 . ρ = 0 . Fig. 4
Domination thresholds for various values of ρ . In each case τ r is plotted on the horizontal axis against τ g on thevertical. The marked vertical lines represent (in increasing order) κ ρr , , µ ρr , and 1 − ρ . The horizontal lines are ρ , κ ρg , , and µ ρg . The black line is the threshold between red and green domination. Regions of red (respectively green) takeoverare marked in red (green), and static regions are marked in grey.4 George Barmpalias et al. Fig. 5 ρ = 0 . τ g = 0 . τ r = 0 . w = 50, n = 100 , ring, the initially unhappy elements are depicted outside that, and the final configuration is shown in theoutermost ring. In between, the elements which change are shown in their new colour, with their distancefrom the centre proportional to their time of change.)Our proof will be a modification of Section 4 of [1], and indeed certain things will be simpler in thecurrent case. In outline, the proof will proceed by letting ε > u uniformly atrandom, and then seeking to establish that u will be green in the final configuration with probabilityexceeding 1 − ε for all n (cid:29) w (cid:29)
0. As discussed in 1.11, a key notion will be that of a green firewall,meaning a sequence of at least w + 1 consecutive green nodes. Recall that any firewall is guaranteed togrow in both directions until it hits a stable interval of the opposite colour. Our plan is thus to establishthat green firewalls are highly likely to form on both sides of u , with no stable red intervals or unhappygreen nodes (which may spawn stable red intervals) in positions to block their paths from merging andencompassing u .The first step, in Lemma 4.7, will be to identify a sequence of nodes l i stretching to the left of u and r i to the right. Essentially l will turn out to be the first node to the left of u whose neighbourhood issuch that it will be unhappy if red. Then l will be the first such node to the left of l − (2 w + 1), andso on, with the r i emerging similarly to the right.We shall then prove that each of the following statements holds with probability at least 1 − ε (cid:48) forarbitrary ε (cid:48) >
0, conditional on the previous statements holding. (We withhold the technicalities for now,including suppressing several intermediate notions.) • The l i and r i exist and satisfy various criteria including the absence of red stable intervals and unhappygreen nodes between them (Lemma 4.7). • The distribution of green nodes in the vicinities of each l i and r i is smooth , meaning that there are noawkward concentrations of red or green nodes nearby. (See Definition 4.8 and Corollary 4.11). • The vicinity of each l i and r i is likely to reach maturity without interference from beyond the l i or r i ,where red firewalls may be growing (Definition 4.12 and Lemma 4.13). Thus we can be confident thatwithin our region of interest all the changes that occur will consist of red nodes turning green, ratherthan vice versa. • Smoothness will then allow us to argue that each l i and r i stands a reasonable chance of originating agreen firewall (Corollary 4.18 and Lemma 4.19).Together, these will establish that green firewalls are highly likely to grow on both the left and rightof u , and furthermore there will be no red stable intervals in positions to block these firewalls fromeventually meeting and consuming u .We now begin the proof by recalling some notation from [1] which will be useful when we wish todivide some interval I into k pieces. The following definition addresses this situation when the length of I is not divisible by k : ipping Points in Schelling Segregation 15 Definition 4.1
Let I = [ a, b ] and suppose k ≥ . We define the subintervals I (1 : k ) := (cid:2) a, a + (cid:4) b − ak (cid:5)(cid:3) :=[ I (1 : k ) , I (1 : k ) ] and I ( j : k ) := (cid:20) a + (cid:22) ( j − b − a ) k (cid:23) + 1 , a + (cid:22) j ( b − a ) k (cid:23)(cid:21) := [ I ( j : k ) , I ( j : k ) ] for ≤ j ≤ k . It will sometimes be useful to count the subintervals from right to left:
Definition 4.2
Let I = [ a, b ] and suppose k ≥ . For ≤ j ≤ k we define I ( j : k ) − = I ( k − j + 1 : k ) , I ( j : k ) − = I ( k − j + 1 : k ) and I ( j : k ) − = I ( k − j + 1 : k ) . We now begin to analyse the scenario by picking a node u uniformly at random. The aim of the proofwill be to show that for any ε , in the finished ring u is green with probability > − ε for all n (cid:29) w (cid:29) − ρ ≤ τ r and 1 − ρ > τ r . We postpone the former situation,and begin with the case where γ := 1 − ρ − τ r >
0. Notice that under this assumption, green dominationimplies that ρ > τ g via Corollary 3.3. Remark 4.3
There are < η, ζ < so that U g < η w U r and S r < ζ w U r . This follows from the assumptions of green domination and τ r > κ ρr , by Propositions 2.2 and 3.4. Remark 4.4
Red nodes are unlikely to be unhappy in the initial configuration: U r ≤ exp (cid:0) − γ (2 w + 1) (cid:1) . To justify this, let u be a randomly selected red node. Then we think of R ( N ( u )) as a sum of 2 w + 1independent random variables taking the value 1 or 0. Clearly its expected value is (1 − ρ )(2 w + 1). Then U r = P (cid:0) R ( N ( u )) < τ r (2 w + 1) (cid:1) = P (cid:0) (1 − ρ )(2 w + 1) − R ( N ( u )) > γ (2 w + 1) (cid:1) . The remark then followsby Hoeffding’s inequality (Proposition 2.3). Definition 4.5
Let u be a node, and let θ ∈ [0 , . We say u has a local green density of θ or that GD θ ( u ) holds, if G ( N ( u )) |N ( u )) | = θ . We shall be particularly interested in the case GD θ ∗ ( u ) where θ ∗ is as follows: Definition 4.6
Let θ ∗ be minimal such that GD θ ∗ ( v ) implies that v is unhappy if red. That is: θ ∗ := min (cid:26) m w + 1 : m w + 1 > − τ r & m ∈ N (cid:27) . Clearly then, θ ∗ → − τ r as w → ∞ , and it follows from our standing assumption 1 − τ r − ρ > θ ∗ > ρ for all w . Definition of the l i and r i . We proceed recursively, with l := u . Now define l i +1 to be the first nodeto the left of l i − (2 w + 1) which is either unhappy, or satisfies GD θ ∗ , or belongs to a red stable interval,so long as this node lies within [ u − n ]. The r i are defined identically to the right. A little later we shallchoose a specific value of k , not depending on w . For now we keep it flexible. Lemma 4.7
For any k > and any ε (cid:48) > , there exists d > such that for all large enough w and n large enough relative to w , the following hold with probability > − ε (cid:48) l k , . . . , l , r , . . . , r k are all defined.2. l k , . . . , l , r , . . . , r k all satisfy GD θ ∗ .3. There are no unhappy green nodes in [ l k , r k ] .4. No node in [ l k , r k ] belongs to a stable red interval.5. For i ≥ , we have | l i − − l i | , | r i +1 − r i | , | r − l |≥ e dw .Proof The first four points follow from Remark 4.3 and Lemma 2.7. The fifth follows from Remark 4.4and the fact that for any interval I , we have P ( UR ( I ) > ≤ (cid:80) x ∈ I UR ( x ).With all this done, our goal is to show that there is a high chance that at least one of the l i and atleast one of the r i will originate a green firewall. Since it is very likely that there are no unhappy greennodes or red stable intervals lying between these nodes, we can then be confident that the two firewallswill merge, thereby encompassing u . To this end we adapt the following notion from [1]: Definition 4.8
Suppose that GD θ ( u ) holds for some θ . Let L = [ u − (3 w +1) , u ] and R = [ u, u +(3 w +1)] .Suppose that k > is a multiple of and ε (cid:48) > . For j ≤ k let R j = R ( j : k ) and L j = L ( j : k ) − . Weadditionally say that Smooth k,ε (cid:48) ( u ) holds if: • For ≤ j ≤ k , we have | G ( L j ) ||L j | , | G ( R j ) ||R j | ∈ [ θ − ε (cid:48) , θ + ε (cid:48) ] • For k < j ≤ k , we have | G ( L j ) ||L j | , | G ( R j ) ||R j | ∈ [ ρ − ε (cid:48) , ρ + ε (cid:48) ]Thus Smooth k,ε (cid:48) ( u ) asserts that the proportion of green nodes in N ( v ) smoothly moves from θ to ρ as v moves from u to u ± (2 w + 1). Corollary 4.9
We make no assumption on ( ρ, τ g , τ r ) or θ . For all multiples of three k > and ε (cid:48) > ,and for all sufficiently large w , P ( Smooth k,ε (cid:48) ( u ) | GD θ ( u )) > − ε (cid:48) . Proof
Select u uniformly at random from nodes such that GD θ ( u ) holds. We prove the first smoothnesscriterion first. The nodes in N ( u ) form a hypergeometric distribution. Since we consider fixed k and ε (cid:48) and take w large, it suffices to prove the result for given j with 1 ≤ j ≤ k . Here the result follows from anapplication of Chebyshev’s inequality and standard results for the mean and variance of a hypergeometricdistribution: P (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) G ( L j ) | L j | − θ (cid:12)(cid:12)(cid:12)(cid:12) > ε (cid:48) (cid:19) < | L j | − ε (cid:48)− Var( G ( L j )) = O (1) | L j | − . Noting that (cid:12)(cid:12) | L j |− (3 w + 1) /k (cid:12)(cid:12) ≤
1, the result follows.Now let u − = u − (2 w + 1) and u = u + (2 w + 1). The fact that GD θ ( u ) holds has no impact on thedistributions for N ( u − ) and N ( u ), where both E ( G ( N ( u ))) = E ( G ( N ( u − ))) = ρ . Thus the secondsmoothness criterion follows directly from the weak law of large numbers.Of course, the l i and r i are not selected randomly, so we may not simply apply Corollary 4.9 toestablish their smoothness. Nevertheless we shall be able to deduce it from the following result whosesomewhat technical proof is contained in Appendix C: Proposition 4.10
Fix a value of ρ and a value θ (cid:54) = ρ . For any node u let x u be the first node to the leftof u such that GD θ ( x u ) holds.Let Q ( u ) be a property of nodes which depends only on the vicinity of u in the initial configuration(which is to say it depends on [ u − C, u + C ] , for some C independent of n ).Suppose there exists p > such that for all sufficiently large w we have P ( Q ( u ) | GD θ ( u )) ≥ p . Thenthere exists p (cid:48) > such that for all n (cid:29) w (cid:29) we have P ( Q ( x u )) ≥ p (cid:48) for u selected uniformly atrandom.If additionally the hypothesis holds with p = 1 − ε (cid:48) for all ε (cid:48) > , then we may likewise take p (cid:48) = 1 − ε for any ε > . We shall appeal to Proposition 4.10 several times, starting with this:
Corollary 4.11
Let ε (cid:48) > and k > be a multiple of and let k > be fixed. Then for all sufficientlylarge n (cid:29) w (cid:29) , with probability > − ε (cid:48) we have that Smooth k,ε (cid:48) ( l i ) and Smooth k,ε (cid:48) ( r i ) hold for all i ≤ k .Proof Corollary 4.9 and Proposition 4.10 combine to tell us that for uniformly randomly selected u ,we know that Smooth k,ε (cid:48) ( x u ) holds with probability > − ε (cid:48) . Applying this to u = u directly tellsus that Smooth k,ε (cid:48) ( l ) holds with probability > − ε (cid:48) . Of course a symmetric argument applies to r .Proceeding inductively, suppose that we have established the result for l i and r i . Then the sequence ofnodes [ l i − D, l i − ( w + 1)], where D is any quantity which is small compared to n , is independent of[ l i − w, r i + w ]. Hence we may apply the same argument again taking u = l i − (2 w + 1) to deduce that Smooth k,ε (cid:48) ( l i +1 ) holds with probability > − ε (cid:48) . A symmetric argument works for r i +1 .Now, the following definition is valid under all three dynamics. Recall that a node is hopeful if itis unhappy but a change of colour would cause it to become happy, and that we denote the numberof hopeful, hopeful green, and hopeful red nodes in a set A at time t by F t ( A ), FG t ( A ) and FR t ( A )respectively. (In our current scenario where τ h , τ r < hopefulness is automatic for unhappy nodes.However we shall reuse this notion later in another context.) ipping Points in Schelling Segregation 17 Definition 4.12
We say that a node u green completes at stage s if • F s ( N ( u )) = 0 , but F t ( N ( u )) > for all t < s • FG t ( N ( u )) = 0 for all t ≤ s . For the nodes we consider, it will typically be the case that FR ( N ( u )) >
0, otherwise we may havethe trivial situation of green completion at stage 0.If u green completes it follows that G t ( N ( u )) is a monotonic increasing function for t ≤ s . We shallapply the following Lemma to l i and r i where i < k , but phrase it more generally for reuse later. Againthe most useful case will be when u itself is a hopeful red node. Lemma 4.13
Suppose that u is a node and that v and v (cid:48) are its nearest hopeful green nodes to the left andright respectively in the initial configuration. Assume that there exists d > so that | u − v | , | v (cid:48) − u | > e wd forall w (cid:29) . Then the following holds independently of all other facts about the ring’s initial configuration:in the selective model with any τ g , τ r or in the incremental model with τ g , τ r < , for any ε (cid:48) > we havethat u green completes with probability > − ε (cid:48) for all large enough w . In the synchronous model for τ g , τ r < we have instead that u green completes with probability .Proof We work with the selective/incremental model first. Let I = [ v, u ] and I = [ u, v (cid:48) ]. Let k be thegreatest such that, when 1 ≤ j ≤ k , I ( j : k ) and I ( j : k ) are of length ≥ w + 1. For 1 ≤ j ≤ (cid:98) k/ (cid:99) define: J j := I ( j : k ) ∪ I ( j : k ) − . For 1 ≤ j ≤ (cid:98) k/ (cid:99) , let P j be the event that R ( J j ) increases by 1, and note that P j +1 cannot occur until P j has occurred. Now the basic idea is that if green completion fails to occur, then the sequence of events P , ..., P (cid:98) k/ (cid:99) must occur before any stage when F ( N ( u )) = 0.We label certain stages as being a ‘step towards green completion’, and certain others as being a ‘steptowards failure of green completion’. Steps towards green completion . If F ( N ( u )) > N ( u ) changesfrom red to green as a step towards green completion . Once F ( N ( u )) = 0, we consider every step to be astep towards green completion. Steps towards failure of green completion . If 1 ≤ j < (cid:98) k/ (cid:99) is the greatest such that P j has occurredprior to stage s or no P j has occurred and j = 1, and if P j +1 occurs at stage s , then we label s a steptowards failure of green completion .We now adopt a modified stage count which counts only steps towards either green completion or itsfailure. (Once F ( N ( u )) = 0, every stage is counted.) Now at any stage s at which some P j for j ≤ (cid:98) k/ (cid:99) is yet to occur, and at which F ( N ( u )) >
0, the probability of s being a step towards failure of greencompletion is at most 2( w + 2) times the probability of it being a step towards green completion (sincethere are at most 2( w +2) times as many nodes which, if chosen to change, will cause a step towards failureof green completion, as those which will cause a step towards green completion). Choosing 0 < d (cid:48) < d weget that for all sufficiently large w , (cid:98) k/ (cid:99) > e d (cid:48) w . We may therefore consider the first e d (cid:48) w many stageswhich are steps either towards green completion or failure of completion and, for large w , consider theprobability that at most 2 w + 1 of these are steps towards green completion. By the law of large numbers,this probability tends to 0 as w → ∞ . What is more, by assumption on u , 2 w + 1 many such steps morethan suffice for its green completion.For the synchronous model, we simply have to note that since e d (cid:48) w (cid:29) w + 1 for large enough w , theinfluence of v or v (cid:48) cannot be felt in N ( u ) within the first 2 w + 1 time-steps. Thus green completion isinevitable.We shall say that a node u originates a green firewall if u green completes, at which time N ( u )contains a run of w + 1 consecutive green nodes. The final step in the proof is to show that each l i and r i originates a green firewall with reasonable probability. We have to do a little more work to establishthis, and again we express things more generally. First we establish something weaker, that a firewall getsstarted in the following sense: Definition 4.14
With no assumptions on our scenario, let α ∈ (0 , . We say that a node u α -sparks if u green completes, and at the moment of completion the interval K α := [ u − (cid:98) α · w (cid:99) , u + (cid:98) α · w (cid:99) ] iscompletely green. Our strategy will be to argue, under suitable conditions on α , that each l i and r i has a reasonablechance of α -sparking, and then to establish that such a spark will guarantee the emergence of a greenfirewall. First, however, we need to consider a technical matter which will become important: Lemma 4.15
For any θ, ρ ∈ (0 , , define Z ( θ, ρ ) := 1 + θ − θ + 3 θ ρ − θ ρ . Then(i) ∂Z∂θ < (ii) If θ < (1 + ρ ) then Z ( θ, ρ ) > .Proof To start with, ∂Z∂θ = 3 θ − θ + 6 θρ − θ ρ = 3 θ ( θ − ρ (2 − θ )) < θ ( θ − − θ ) = − θ < θ ∈ (0 , θ , we only need to check that Z (cid:18)
12 (1 + ρ ) , ρ (cid:19) = 38 − ρ + 38 ρ + 18 ρ − ρ > ρ ∈ (0 , Remark 4.16
Lemma 4.15 establishes in particular that Z ( θ ∗ , ρ ) > for all large enough w . To see why the remark holds, recall that the hypotheses of Theorem 1.2 include the assumption that τ r > κ ρr and we saw in Proposition 2.2, that κ ρr > (1 − ρ ). Since θ ∗ → − τ r as w → ∞ , it follows thatfor large enough w , we shall have θ ∗ < (1 + ρ ).The following will go most of the way to establishing that the l i and r i have a good chance of sparking: Lemma 4.17
Let Z be as defined in 4.15, and suppose θ is such that Z ( θ, ρ ) > . Suppose that u isuniformly randomly selected from nodes satisfying GD θ ( u ) .For any α ∈ (cid:0) , θ (cid:1) , define θ α := θ − α − α . Now fix α small enough that also Z ( θ α , ρ ) > .Then there exists δ > (depending on the scenario, θ , and α but not on w ) such that if u greencompletes, then it α -sparks with probability > δ for all w (cid:29) .This holds in both the selective/incremental and synchronous models.Proof We adopt a novel way of counting stages, defining spark stage s to be the first time (if it exists)that [ u − s, u + s ] is entirely green. We shall consider spark stages up to s = (cid:98) α · w (cid:99) . Of course, if thisfinal stage is reached, then a green spark has occurred. Now suppose, inductively, that spark stage s hasbeen reached. We shall estimate the probability of reaching spark stage s + 1. To do this, we compute alower bound for G ( N ( u + s + 1)) at stage s recursively: define M := G ( N ( u )) = θ (2 w + 1). Define M s +1 := M s + R ( u − s ) + R ( u + s ) − G ( u + s − w ) + G ( u + s + w + 1) . (By G ( a ) for an individual node a we shall mean G ( a ) throughout this proof and similarly for R ). Then G ( N ( u + s + 1)) ≥ M s at spark stage s .Now, we have made no assumption on the distribution of nodes outside N ( u ). Therefore it is legitimateto treat nodes in [ u + w + 1 , u + 3 w + 1] as independent identical random variables with a probability ρ ofbeing green. Let us briefly make the additional temporary assumption that in the initial configuration,nodes in [ u − w, u + w ] are independent random variables with a probability of θ of being green. Claim P ( M s +1 > M s ) > P ( M s +1 < M s ) Proof of claim
At each stage we have P ( M s +1 = M s − P (cid:16) G ( u − s ) = G ( u + s ) = G ( u + s − w ) = R ( u + s + w + 1) = 1 (cid:17) = θ (1 − ρ )Similarly P ( M s +1 = M s ) = θ ρ + 3 θ (1 − θ )(1 − ρ ). Thus P ( M s +1 > M s ) = 1 − θ − θ (1 − θ )(1 − ρ )and P ( M s +1 > M s ) − P ( M s +1 < M s ) = Z ( θ, ρ ). Hence the claim follows from our assumption that Z ( θ, ρ ) > QED Claim ipping Points in Schelling Segregation 19
We now consider M s as a biased random walk, omitting all steps where M s +1 = M s . The claimestablishes that the walk is more likely to increase than decrease at each spark stage. Call the probabilitythat it increases p := (1 + Z ( θ, ρ )) > . It follows then that, the probability that it ever drops below M is − pp by a standard result on biased random walks.Everything we have stated here applies equally to the mirror-image process M (cid:48) s +1 := M (cid:48) s + R ( u + s ) + R ( u − s ) − G ( u − s + w ) + G ( u − s − w − . Moreover G ( N ( u − s − ≥ M (cid:48) s at spark stage s . If M s ≥ M and M (cid:48) s ≥ M (cid:48) for all s ≤ (cid:98) α · w (cid:99) then u − s − u + s + 1 are guaranteed to be unhappy if red at stage s , meaning that K α is certain to begreen in the event of green completion.All that remains is to drop the false assumption of independence. Taking the above two processes M and M (cid:48) together, at each spark stage s , we see four new nodes in N ( u ) namely u − s , u + s , u + s − w , and u − s + w . Thus the number of unseen nodes is (2 w + 1) − s of which at least θ (2 w + 1) − s are green.Thus, the real probability of picking a green node from the remaining unseen nodes is θ (2 w +1) − s (2 w +1) − s > θ α for all s ≤ (cid:98) α · w (cid:99) .Hence, for all large enough w , working now with the true probabilities, we find that for all largeenough w and all s ≤ (cid:98) α · w (cid:99) , we have P ( M s +1 > M s ) − P ( M s +1 < M s ) > Z ( θ α , ρ ) > α .Thus setting p (cid:48) := (1 + Z ( θ α , ρ )) we find p (cid:48) > and dropping the assumption of independence, theactual probability that M s +1 > M s will exceed p (cid:48) for all large enough w and all spark stages s ≤ (cid:98) α · w (cid:99) ,no matter what has occurred at previous stages. Thus the probability that M s never drops below M ,and thus that an α -spark occurs, will be at least δ := 1 − − p (cid:48) p (cid:48) = Z ( θ α ,ρ )1+ Z ( θ α ,ρ ) . Corollary 4.18
Preserving the notation from 4.17, suppose that θ (cid:54) = ρ and α is such that Z ( θ α , ρ ) > .Then there exists δ (cid:48) > (depending on the scenario, θ , and α but not on w ) such that for each l i , if l i green completes, then it α -sparks with probability > δ (cid:48) for all w (cid:29) .This holds in both the selective/incremental and synchronous models.Proof This is simply a matter of applying Lemma 4.17 and Proposition 4.10, bearing in mind that wecan apply the conclusion to l i exactly as in the proof of Corollary 4.11.Corollary 4.18 at last allows to choose k and ε (cid:48) satisfying ε (cid:48) + (1 − δ (cid:48) ) k < ε , applying Lemma 4.13for our chosen value of ε (cid:48) .Notice that k is independent of w as promised and is selected to ensure that the probability that no l i or no r i α -sparks is < ε , for all large enough w .The final step is now to appeal to smoothness to show that a green spark will lead to a green firewall.We shall apply the following to those l i which α -spark: Lemma 4.19
Suppose that θ > − τ r and θ > τ g and θ < (1 + ρ ) , and that α is small enough that also Z ( θ α , ρ ) > as in Lemma 4.17.Now fix integers r large enough that r (1 + ρ − θ ) (cid:29) and k large enough that rk < α .Suppose now that u is a node satisfying GD θ ( u ) and Smooth k,ε (cid:48) ( u ) and suppose that u α -sparks. Then N ( u ) contains a green firewall at the moment of green completion.Proof We proceed inductively, assuming that (cid:83) sj =1 R j has become fully green, where R j is as defined inDefinition 4.8) and 1 ≤ s < k . We shall show that R s +1 will also become green. (The base case s = r holds by assumption since u α -sparks. The final case s = k will amount to [ u, u + w ] forming a greenfirewall.)Let v ∈ R s +1 . First we shall bound G ( N ( v )) below. Let v (cid:48) be the rightmost node in R s , and look at N ( v (cid:48) ). Firstly, of course, (cid:83) sj =1 R j ⊆ N ( v (cid:48) ), which makes a contribution of ≥ s · w +1 k many green nodes.To the left of that, we have (cid:83) k − sj =1 L j , which by smoothness contribute at least (cid:0) k − s (cid:1) · w +1 k · ( θ − ε (cid:48) )many green nodes.To the right, however, we have (cid:83) k j = s +1 R j contributing at least (cid:0) k − s (cid:1) · w +1 k · ( θ − ε (cid:48) ), followed by (cid:83) k + sj = k +1 R j which contributes at least s · w +1 k · ( ρ − ε (cid:48) ).Adding all these together, we get a lower bound for G ( v (cid:48) ) which may then adapt to a lower boundfor G ( v ), by observing that | v − v (cid:48) |≤ w +1 k . Hence we find our bound: G ( v ) ≥ w + 1 k · (cid:18) s + 2 (cid:18) k − s (cid:19) ( θ − ε (cid:48) ) + s · ( ρ − ε (cid:48) ) − (cid:19) . That is to say G ( v ) ≥ w ( θ − ε (cid:48) ) + 3 w + 1 k · ( − s · (1 − θ + ρ + ε (cid:48) ))Now s ≥ r , and by assumption on r we have r · (1 − θ + ρ + ε (cid:48) ) (cid:29)
1, from which it follows that G ( v ) ≥ (2 w + 1) θ , meaning that if v is red then it must be unhappy.This concludes the proof of Theorem 1.2 in the case (1 − ρ ) > τ r . Proof of Theorem 1.2 when (1 − ρ ) ≤ τ r Here we find that the probability that a randomly chosen red node is unhappy U r → − ρ ) < τ r )or U r → (if (1 − ρ ) = τ r ) as n, w → ∞ . In either case the foregoing proof goes through with only veryminor modifications.Suppose first that (1 − ρ ) < τ r . Again we define l := u and l i +1 to be the first node to the left of l i − (2 w + 1) which is either unhappy, or satisfies GD θ ∗ , or belongs to a red stable interval, so long asthis node lies within [ u − n ]. The r i are defined identically to the right. Let v (respectively v (cid:48) ) be thenearest unhappy green node to the left (right) of u . This time we find that the l i and r i are much closertogether. This is of no concern so long as they are far enough from v and v (cid:48) , and with the proof as beforewe obtain the following variant of Lemma 4.7: Lemma 4.20
For any k > and ε (cid:48) > there exists d > such that for all large enough w the followinghold with probability > − ε (cid:48) l k , . . . , l , r , . . . , r k are all defined.2. l k , . . . , l , r , . . . , r k all satisfy GD θ ∗ .3. No node in [ l k , r k ] lies in a stable red intervals.4. For i ≥ , we have | l i − v | , | r i − v (cid:48) |≥ e dw . Now, Corollary 4.11 applies again (technically of course the current l i are different from those involvedin the original statement, however the proof goes through without alteration). This gives us that for each ε (cid:48) > k ∈ N , Smooth k,ε (cid:48) ( l i ) and Smooth k,ε (cid:48) ( r i ) holds for all i ≤ k with probability > − ε (cid:48) for all w (cid:29) l i and r i all green complete with probability > − ε (cid:48) for all large enough w . Finally Corollary 4.18 also applies to the l i and r i and once again allowsus to choose k , such that the probability that no l i or no r i initiate a green firewall is < ε .In the case τ r = 1 − ρ there is a minor complication in that we cannot apply results such as Corollary4.11, owing to the hypothesis in various lemmas that θ (cid:54) = ρ . However we may get around this quitesimply, by letting τ (cid:48) r < τ r be such that τ (cid:48) r > κ ρr and ( ρ, τ g , τ (cid:48) r ) is green dominating. Then τ (cid:48) r < − ρ andwe proceed through this section’s main argument working with τ (cid:48) r in place of τ r throughout, beginningwith Definition 4.6.Clearly any red node which is τ (cid:48) r -unhappy is automatically τ r -unhappy, and thus we may deduce theexistence of green firewalls on either side of u as before. It may be that other green regions also grow,undetected by our analysis, around red nodes which are initially τ r -unhappy but not τ (cid:48) r -unhappy, butthis is unproblematic. This completes the proof of Theorem 1.2. Proof of Theorem 1.3
We now turn our attention to scenarios where τ r < κ ρr , while > τ g > κ ρg and ( ρ, τ g , τ r ) is greendominating, which we shall prove to be static almost everywhere under the additional assumption that τ r > (1 − ρ ). Interchanging the roles of red and green will establish Theorem 1.3. (We have made thisalteration to be able to apply our previous lemmas verbatim.) Figure 6 is instructive of what to expect(and also provides an example where large values of n and w are required for the essential staticity ofthe situation to be revealed).We begin by observing that the following follows from Propositions 3.4 and 2.2: ipping Points in Schelling Segregation 21 Fig. 6 ρ = 0 . τ g = 0 . τ r = 0 . w = 70, n = 5 , , Remark 4.21
There exist < η, ζ, ξ < so that for all w (cid:29) we have S g < ξ w U g , while U g < η w U r ,and in turn U r < ζ w S r . Again begin by picking a node u at random. Now we outline the general intuition. Let y (respectively y (cid:48) ) be the nearest node to the left (right) of u which are either unhappy or belong to a stable interval.By Remark 4.21 above and Lemma 2.7, for large enough w , both y and y (cid:48) are highly likely to belong tored stable intervals. This guarantees that u , if red, can never turn green.So suppose, for the remainder of this section, that u is green. In the initial configuration Remark4.21 tells us that unhappy green nodes are hugely more frequent than stable green intervals. Thus on firstsight, there appears to be a danger that u will be engulfed in a red firewall. However, unhappy red nodesare commoner still, and we shall show that these are highly likely to give rise to stable green intervals(or short firewalls) in positions protecting u .The argument runs much as previously. First we define l := u , and define l i +1 to be the first node tothe left of l i − (2 w + 1) which is either unhappy or satisfies GD θ ∗ (defined as in Definitions 4.5 and 4.6),so long as this node lies within [ u − n ]. The r i are defined identically to the right. As before Remark4.21 together with Lemma 2.7 give us the following. Lemma 4.22
For any k > and ε (cid:48) > there exists d > such that for all large enough w the followinghold with probability > − ε (cid:48) l k , . . . , l , r , . . . , r k are all defined.2. l k , . . . , l , r , . . . , r k all satisfy GD θ ∗ .3. There are no unhappy green nodes in [ l k , r k ] .4. For i ≥ , we have | l i +1 − l i | , | r i +1 − r i | , | r − l |≥ e dw . Notice that in this case, Remark 4.21 alone provides enough information to derive the fourth point.Now, Corollary 4.11 applies again. (Again our current l i are technically different from those mentionedthere but the proof remains valid.) Therefore, for each ε (cid:48) > k ∈ N , Smooth k,ε (cid:48) ( r i ) holds for all i ≤ k with probability > − ε (cid:48) for all w (cid:29) l i and r i all green complete with probability > − ε (cid:48) for all large enough w .Since we have stipulated that τ g > (1 − ρ ), we find that with θ ∗ (defined as in Definition 4.6) satisfies θ ∗ < (1 + ρ ) for all large enough w . Thus by Remark 4.16 Z ( θ ∗ , ρ ) > w . Hence wemay pick α as before and apply Corollary 4.18 to our current l i and r i to establish that each has a chanceof at least δ (cid:48) > α -sparking. Furthermore, if some l i or r i does spark then we may apply Lemma 4.19to establish that a green firewall will automatically ensue (again we are relying on our hypothesis that τ g > (1 − ρ )).This once again allows us to choose k , such that the probability that no l i or no r i initiate a greenfirewall is < ε . Now let l and r be the l i and r i nearest u which do initiate green firewalls. It is certainthat these green firewalls will spread towards u until they hit stable red intervals. Suppose this hashappened by stage s . At this stage, the vicinity of u has transformed to resemble the case of Theorem u we encounter red stable intervals before unhappy red nodes andgreen stable intervals unhappy green nodes. Thus with probability > − ε , the node u will not changecolour. Discussion of Question 1.4
It is clear that the bulk of the foregoing argument does not extend to cases where τ r ≤ (1 − ρ ) (thiscorresponds to Question 1.4 with the roles of red and green interchanged). For values of ρ approximately0 . , such scenarios are possible. For example, the scenario ( ρ, τ g , τ r ) = (0 . , . , .
13) satisfiesall the hypotheses of this section: τ r < κ r ≈ .
21 and κ . g ≈ . < τ g < . τ r = 0 . < (1 − ρ ) = 0 . ρ, τ g , τ r ) = (0 . , . , .
49) which is locatedin the lower purple region of Figure 3.In this scenario, at least we have Z ( θ ∗ , ρ ) > w . But there are even scenarioswhere this weaker condition fails. For example the scenario comprising ( ρ, τ g , τ r ) = (0 . , . , . τ r < κ . r ≈ .
186 and κ . g ≈ . < τ g < . τ r = 0 . < (1 − ρ ) = 0 .
13. Moreover, as w → ∞ we have θ ∗ → − τ r = 0 .
93, and wefind Z ( θ ∗ , ρ ) → − .
06 approximately.Although our previous arguments fail in such cases, needless to say, it does not follow that no l i evolvesinto a green stable interval protecting u . It seems that a more detailed analysis is needed to resolve suchsituations. τ r > > τ g We begin with an observation whose proof we leave to the reader:
Lemma 5.1
The following are equivalent:(i) τ g + τ r ≤ (ii) For all n (cid:29) w (cid:29) , there can exist happy adjacent nodes of opposite colours.(iii) For all n (cid:29) w (cid:29) , all unhappy nodes are hopeful. It follows that when τ g + τ r >
1, the selective and incremental dynamics have the possibility to differ,as indeed will be the case. Having said this, in this section we may establish Theorem 1.5 which appliesunder every dynamic, and states that if τ g < < τ r then green will take over totally, as illustrated inFigure 7. This will also establish that, for all large enough n , the initial configuration will very likely besuch that the process is guaranteed to finish, under both the incremental and synchronous dynamics. Fig. 7 ρ = 0 . τ g = 0 . τ r = 0 . w = 40, n = 100 , Proof (Proof of Theorem 1.5)
We work with w large enough that w +12 w +1 < τ r . Suppose that at some stage [ a, b ] is a green firewall oflength at least w , and that a − b + 1 are red. Then every node in [ a, b ] is happy and stably so. Onthe other hand, a − b + 1 are unhappy (and indeed hopeful) and will remain so as long as they arered. Hence the firewall cannot shrink and will eventually grow to encompass a − b + 1.It follows that as soon as we have a green firewall of length ≥ w , total green takeover is inevitableunder every dynamic. There are various possible arguments for how such a firewall may emerge in smallerrings. However, for all large enough n , we may make the cheap observation that such a firewall will appearin the initial configuration with probability > − ε by the weak law of large numbers.Theorem 1.5 is necessarily probabilistic; it is not true that every ring will suffer green takeover. Forexample given any w and tolerances satisfying τ r > ≥ τ g where (cid:100) (2 w + 1) τ r (cid:101) + (cid:100) (2 w + 1) τ g (cid:101) ≤ w + 1, aring comprising alternating blocks of (cid:100) (2 w + 1) τ r (cid:101) many red nodes followed by (cid:100) (2 w + 1) τ g (cid:101) many greennodes will be totally static throughout under all three dynamics. In this section our focus will be entirely on the selective model where τ g , τ r > . Recall that an unhappyred node x is hopeful if G ( N ( x )) ≥ τ g (2 w +1) −
1. (Notice that when τ g , τ r > unhappiness automaticallyfollows from this second condition, for large enough w .) In this situation, a new type of stable intervalemerges: Definition 6.1
An interval J of length w + 1 is green intractable if G ( J ) < τ g (2 w + 1) − ( w + 1) . We define red intractability analogously. The point of this definition is that, regardless of the situationoutside J , no red node inside a green intractable interval can be hopeful, and thus can never turngreen. Hence green firewalls cannot spread through green intractable intervals. Now we shall compare theprobabilities of hopeful individuals versus intractable intervals.The probability F r that, in the initial configuration, a randomly selected red node is hopeful is thesame as the probability that a green node in that position would be happy. Setting X ∼ b (2 w, ρ ) we have F r = P ( X ≥ h ) where h := (cid:100) τ g (2 w + 1) (cid:101) − J be an interval of length w + 1, and let T g be the probability that such aninterval, selected uniformly at random is green intractable. We now introduce an approximation for T g as follows. Set Y ∼ b ( w + 1 , ρ ), and set T (cid:48) g = P ( Y ≤ (cid:96) ) where (cid:96) := (cid:100) ( τ g − )(2 w + 1) (cid:101) −
1. It is easy tosee that T g ≈ T (cid:48) g , and thus we may work with T (cid:48) g in place of T g in all that follows.We wish to understand the ratio F r T g . First notice that if τ g ≤ ρ then T g → w , while F r ≥ . Hence F r T g → ∞ .Similarly, if τ g ≥ ρ +12 then F r → w , while T g ≥ . Hence F r T g → ρ < τ g < ρ +12 , and in this region we may derive some estimates fromLemma 2.4 as before. Firstly, taking p = ρ , with h as above, along with N = 2 w , and some k such that1 > k > − τ g τ g · ρ − ρ (which is possible since ρ < τ g ), we find that: F r ≈ ρ h (1 − ρ ) w − h ( 2 ) wh. (6)Similarly, working with (cid:96) above in place of h and taking N = w + 1 as well as some k (cid:48) where1 > k (cid:48) > − τ g )2 τ g − · ρ − ρ (which is possible since τ g > ρ +12 ), we find T g ≈ ρ (cid:96) (1 − ρ ) w +1 − (cid:96) ( w ) + 1 (cid:96). Taking the ratio of these two, we find F r T g ≈ ρ h − (cid:96) (1 − ρ ) w − (cid:96) − h ( 2 ) w h ( w ) + 1 (cid:96) . Now we use Stirling’s approximation, which gives us that F r T g ≈ ρ h − (cid:96) (1 − ρ ) w − (cid:96) − h (2 w ) w + (cid:96) (cid:96) + ( w + 1 − (cid:96) ) w + − (cid:96) ( w + 1) w + h h + (2 w − h ) w − h + . Next we introduce the approximations h ≈ wτ g and (cid:96) ≈ w (cid:0) τ g − (cid:1) , and noticing that w − (cid:96) ≈ w − h ≈ w (1 − τ g ), we get F r T g ≈ Q ( w ) · ρ w (2 w ) w (2 w ) w ( τ g − ) (cid:0) τ g − (cid:1) w ( τ g − ) w w (2 w ) wτ g τ wτ g g for some polynomial Q ( w ). Hence F r T g ≈ (cid:32) ρ (cid:0) τ g − (cid:1) τ g − τ τ g g (cid:33) w . (7)Thus we deduce the existence of the thresholds µ ρg as the root, when it exists, of g ( x ) := ( x − ) x − x x = 12 ρ . Similarly, µ ρr is the root of g ( x ) = − ρ ) . Comparing this with Equation 4, and noticing that g ( x ) = f (1 − x ), we deduce that µ ρr = 1 − κ ρg and similarly µ ρg = 1 − κ ρr . Thus we have arrived at: Proposition 6.2
For any ρ ∈ (0 , , we work in the initial configuration and interpret F r as the probabil-ity that a uniformly randomly selected red node is hopeful, and T g as that of a uniformly randomly selectednode lying within a green intractable interval. Then there exists a threshold µ ρg where ρ < µ ρg < (1 + ρ ) ,such that for any τ g > : • If τ g < µ ρg , there exists ζ ∈ (0 , so that T g < ζ w F r for all w . • If τ g > µ ρg , there exists ζ ∈ (0 , so that F r < ζ w T g for all w .(Similarly, there exists a threshold µ ρr where − ρ < µ ρr < − ρ such that corresponding statementsabout F g and T r hold.) The thresholds µ ρr and µ ρg are illustrated in Figure 1. Notice that < τ g < µ ρg is only possible when ρ > and similarly < τ r < µ ρr requires that ρ < .Besides understanding the relative frequency of intractable intervals and hopeful nodes, we shall alsoneed to know whether red or green hopeful nodes are more numerous in a given scenario. Happily, we donot need to introduce yet another threshold: Proposition 6.3 If τ g , τ r > , the scenario ( ρ, τ g , τ r ) is red dominating if and only if there exists η ∈ (0 , so that for all w , we have F r < η w F g . The same holds with the roles of red and green interchanged.Proof If ρ ≤ − τ r , then automatically ρ < τ g . Also F g → ρ ≤ − τ r ) or F g → (if ρ = 1 − τ r ) as w → ∞ . At the same time, F r → w . Thus the result holds by Corollary 3.3.Similarly, if 1 − ρ ≤ − τ g , then 1 − ρ < τ r , and result holds by Corollary 3.3 with the roles of redand green interchanged.Thus we are left with the case 1 − τ r < ρ < τ g , where the argument amounts to applying Stirling’sapproximation to 6 and to the equivalent criterion for F g and taking the ratio of the two expressions,exactly as in Proposition 3.4. We leave the details to the reader.Recall also that in Lemma 3.2 we have already established some useful facts about red/green domi-nation in our current region of interest τ g , τ r > . τ r > In this section we limit ourselves to the selective model in scenarios where τ g , τ r > , and shall proveTheorems 1.6 - 1.9. First we assume that < τ g < µ ρg and τ r > , and that ( ρ, τ g , τ r ) is green dominating.Notice that this implies that τ g < (1 + ρ ) by Proposition 6.2.We shall establish green takeover, thus proving Theorem 1.6, using much the same machinery as insection 4. An example is illustrated in Figure 8.As usual, we begin by picking a node u uniformly at random and aim to establish that u will begreen in the finished ring with probability > − ε . We postpone the case τ g ≤ ρ and assume that τ g > ρ .Thus, by Corollary 3.3 it also follows that 1 − τ r < ρ . Via Propositions 6.2 and 6.3, our hypotheses implythe following: ipping Points in Schelling Segregation 25 Fig. 8 ρ = 0 . τ g = 0 . < µ . g ≈ . τ r = 0 . > µ . r ≈ . w = 70, n = 1 , , Remark 7.1
There exist < η, ζ < so that for all w (cid:29) we have T g < η w F r and F g < ζ w F r . We also know, by Hoeffding’s inequality (Proposition 2.3), that red nodes are unlikely to be hopeful: F r < exp (cid:0) − w + 1)( τ g − ρ ) (cid:1) . (8)Let θ ∗ := min (cid:26) m w + 1 : m w + 1 > τ g & m ∈ N (cid:27) . (9)Then θ ∗ → τ g as w → ∞ and thus by assumption ρ < θ ∗ < (1 + ρ ) for large enough w .Now we set l := u and define l i +1 to be the first node to the left of l i − (2 w + 1) which is eitherhopeful, or satisfies GD θ ∗ , or belongs to a green intractable interval, so long as this node lies within[ u − n ]. The r i are defined identically to the right. Again we shall choose a specific value of k in duecourse. As before we derive the following from Remark 7.1, Lemma 2.7, and Bound 8: Lemma 7.2
For any k > and ε (cid:48) > , there exists d > such that for all large enough w the followinghold with probability > − ε (cid:48) l k , . . . , l , r , . . . , r k are all defined.2. l k , . . . , l , r , . . . , r k all satisfy GD θ ∗ .3. There are no hopeful green nodes in [ l k , r k ] .4. No node in [ l k , r k ] belongs to a green intractable interval.5. For i ≥ , we have | l i +1 − l i | , | r i +1 − r i | , | r − l |≥ e dw . Next, we can apply Corollary 4.11 once again to conclude that for any ε (cid:48) > k ≥ Smooth k,ε (cid:48) ( l i ) and Smooth k,ε (cid:48) ( r i ) hold for all i with probability > − ε (cid:48) for all n (cid:29) w (cid:29)
0. Furthermoreby Lemma 4.13, we know that the l i and r i are each very likely to green complete. Next, since θ ∗ < (1+ ρ )for large enough w , we may apply Corollary 4.18 for some α so that Z ( θ α , ρ ) >
0. This establishes thatat least one of the l i and at least one of the r i will α -spark with some probability δ (cid:48) , independent of w .Finally, noting also that θ ∗ → τ g > − τ r as w → ∞ we may apply Lemma 4.19 to establish that those l i and r i which do spark will initiate green firewalls.This again allows us to pick k guaranteeing that u will be engulfed in a green firewall with proba-bility > − ε . Proof of Theorem 1.6 when τ g ≤ ρ Here we find that the probability that a randomly chosen red node is hopeful F r → τ g < ρ ) or F r → (if τ g = ρ ) as n, w → ∞ .Suppose first that τ g < ρ . Once again we define l := u and l i +1 to be the first node to the left of l i − (2 w + 1) which is either hopeful, or satisfies GD θ ∗ , or belongs to a red intractable interval, so long as this node lies within [ u − n ]. The r i are defined identically to the right. Let v (respectively v (cid:48) ) be thenearest hopeful green node to the left (right) of u . Again, the l i and r i are close together but far from v and v (cid:48) : Lemma 7.3
For any k > and ε (cid:48) > there exists d > such that for all large enough w the followinghold with probability > − ε (cid:48) l k , . . . , l , r , . . . , r k are all defined.2. l k , . . . , l , r , . . . , r k all satisfy GD θ ∗ .3. No node in [ l k , r k ] lies in an intractable red intervals.4. For i ≥ , we have | l i − v | , | r i − v (cid:48) |≥ e dw . Now, Corollary 4.11 applies yet again, giving us that for each ε (cid:48) > k ∈ N , Smooth k,ε (cid:48) ( l i ) and Smooth k,ε (cid:48) ( r i ) holds for all i ≤ k with probability > − ε (cid:48) for all w (cid:29) l i and r i all green complete with probability > − ε (cid:48) for all large enough w . Finally Corollary 4.18 and Lemma 4.19 also apply to the l i and r i andonce again allow us to choose k , such that the probability that no l i or no r i initiate a green firewall is < ε .Finally we address the case τ g = ρ . Attempting to apply our previous results directly brings us intoconflict with the hypothesis, in various places, that θ (cid:54) = ρ . Again, we may get around this straightforwardlyby picking τ (cid:48) g > τ g where τ (cid:48) g < µ ρg such that ( ρ, τ (cid:48) g , τ r ) is green dominating. Then τ (cid:48) g > ρ , and we proceedthrough this section’s main argument with τ (cid:48) g in place of τ g throughout, starting with Remark 7.1. Sinceany unhappy red node which is τ (cid:48) g -hopeful is automatically τ g -hopeful, we deduce the existence of greenfirewalls exactly as previously. This concludes the proof of Theorem 1.6. Proof of Theorem 1.7
As before we shall reverse the roles of red and green for convenience, and thus aim to show that if < τ r < µ ρr and µ ρg < τ g , if the scenario is green-dominating and additionally if τ g < (1 + ρ ), then thescenario is static almost everywhere. An example is illustrated in Figure 9. Fig. 9 ρ = 0 . τ g = 0 . > µ . g ≈ . τ r = 0 . < µ . r ≈ . w = 100, n = 5 , , First we define l := u , and define l i +1 to be the first node to the left of l i − (2 w + 1) which is eitherhopeful or satisfies GD θ ∗ (defined as in Equation 9), so long as this node lies within [ u − n ]. The r i aredefined identically to the right. Exactly as before, Remark 7.1 gives us the following: Lemma 7.4
For any k > and ε (cid:48) > , there exists d > such that for all large enough w the followinghold with probability > − ε (cid:48) l k , . . . , l , r , . . . , r k are all defined.2. l k , . . . , l , r , . . . , r k all satisfy GD θ ∗ . ipping Points in Schelling Segregation 27
3. There are no hopeful green nodes in [ l k , r k ] .4. For i ≥ , we have | l i +1 − l i | , | r i +1 − r i | , | r − l |≥ e dw . Notice that τ g > µ ρg > ρ , meaning that for all large enough w , we have θ ∗ (cid:54) = ρ . Hence Corollary 4.11applies again, giving us that for each ε (cid:48) > k ∈ N , Smooth k,ε (cid:48) ( l i ) and Smooth k,ε (cid:48) ( r i ) holds for all i ≤ k with probability > − ε (cid:48) for all w (cid:29) l i and r i all green complete with probability > − ε (cid:48) for all large enough w . Finally noting that θ ∗ → τ g < (1 + ρ ), we may finish off by applyingCorollary 4.18 for some suitable α and Lemma 4.19 to the l i and r i , once again allowing us to choose k ,such that the probability that no l i or no r i initiate a green firewall is < ε .Let l and r be the l i and r i nearest u which do initiate green firewalls. It is certain that these greenfirewalls will spread towards u until they hit green intractable intervals. Suppose this has happened bystage s . At this stage, looking away from u we encounter intractable intervals of both colours beforehopeful nodes of either colour. Thus with probability > − ε , the node u will not change colour. Discussion of Question 1.8
Similar remarks apply here as in the case of Question 1.4, and we expect that any technique which resolvesthat question will apply here too. We may also repackage the counterexamples we mentioned in that case:( ρ, τ g , τ r ) = (0 . , . , . τ g > µ . g ≈ .
79 while < τ r < µ . g ≈ .
52, and the scenario is greendominating. However also τ g > (1 + ρ ) = 0 . ρ, τ g , τ r ) = (0 . , . , .
87) which is locatedwithin the upper purple region of Figure 3.Again we can find cases where Z ( θ ∗ , ρ ) > ρ, τ g , τ r ) = (0 . , . , . τ g > µ . g ≈ .
81 and < τ r < µ . g ≈ .
503 while also Z ( θ ∗ , ρ ) → − . θ ∗ → τ g . Proof of Theorem 1.9
To complete our analysis of the selective model, we turn to the case where both µ ρg < τ g and µ ρr < τ r .Recall that Theorem 1.9 asserts that such a scenario is static almost everywhere, as illustrated in Figure10. The proof of this follows from Lemma 2.7 applied twice, by interpreting P ( u ) as the event that u lies in a green (respectively red) intractable interval and Q ( u ) as u being green (red) and hopeful. Thenecessary probabilistic bounds are given in Proposition 6.2. Fig. 10 ρ = 0 . τ g = 0 . > µ . g ≈ . τ r = 0 . > µ . r ≈ . w = 50, n = 100 , Finally, we turn our attention to scenarios where both τ r , τ g > under the incremental and synchronousdynamics. In the case of a synchronous model, we can mimic Theorem 1.5 with the following proposition,whose proof is deferred to Appendix D. This also serves to establish that, under the given conditions on τ g and τ r and for all large enough n , the initial configuration is highly likely to be such that the processis guaranteed to finish. Proposition 8.1
In the synchronous model, suppose that < τ g < and τ g < τ r . Then green takesover totally. Recall that Conjecture 1.10 generalises Proposition 8.1, asserting that if < τ g < τ r , then green willtake over totally under both the incremental and synchronous dynamics.Towards this conjecture, we briefly make the some observations about a perturbed version of ourmodel. The initial configuration is set up exactly as previously described. But now, for any 1 > ε > ε -perturbed model as follows: at each time-step with probability 1 − ε we proceed as in theprevious incremental model, but with probability ε we pick a node at random and alter its colour. Thus ε can roughly be thought of as the probability of an error at each stage. We remark that this process isa regular perturbed Markov process in the sense of Section 3.4 of [26].The advantage of working with such a perturbed process is that for each ε the Markov process isirreducible: any state of the ring is accessible from any other in a finite number of steps. Thus, followingYoung and notably in the works of Zhang ([27], [28], [29]), it has become common practice to analyseSchelling segregation via perturbed models of this sort, and to examine the limit as ε →
0. It is particularlyof interest to identify the stochastically stable states, which are the states most likely to emerge in thelong run, as ε →
0. They are defined as follows: for each ε >
0, Markov chain theory guarantees thatthere will be a unique stationary distribution µ ε on the state-space. A state s is stochastically stable if µ ( s ) := lim ε → µ ε ( s ) > G and R representing totally green and red rings respectively. By Young’s Theorem (Theorem 3.1 of[26]) whether or not these are stochastically stable will depend on their stochastic potential . We referthe interested reader to [26] for the formal definition, however in the current context its meaning isstraightforward: the stochastic potential of the state G is the minimum number of errors required toreach it from the opposite recurrence class R .That is to say, the stochastic potential of G is simply the minimum number of green nodes whichhave to be artificially inserted into an otherwise entirely red ring in order to generate one unhappy redelement. (Notice that if these green nodes are inserted in consecutive positions, the remainder of thetransformation from R to G may then take place error-free.) This number is (cid:98) (1 − τ r )(2 w + 1) (cid:99) + 1.Similarly the stochastic potential of state R is (cid:98) (1 − τ g )(2 w + 1) (cid:99) + 1. Thus we have the following result,which supports, but does not formally imply, the incremental case of Conjecture 1.10: Theorem 8.2
In the perturbed model, suppose that < τ g < τ r . Then total green takeover representsthe only stochastically stable state.If < τ g = τ r , total takeover by either colour is stochastically stable. Now, there is a strong sense in which Conjecture 1.10 and Theorem 8.2 fail to give the entire story.So we finish with some remarks on the run-time of the process, and hypothesise the existence of anotherimportant tipping point in each of the incremental and synchronous models. The following are easy tosee from our analysis so far: • In the selective model, the the expected run-time is at most linear in n . • In the incremental model, if either τ g , τ r < , the expected run-time is at most linear in n . • In the synchronous model, if either τ g , τ r < , the run-time will be at most linear in n with probability > − ε for all large enough n and w .We further conjecture the following: Conjecture 8.3 • In the incremental model, if either τ g , τ r < , the expected run-time is at most linear in n . ipping Points in Schelling Segregation 29 • In the incremental model, if both τ g , τ r > , the process will finish, but the expected run-time is super-polynomial in n . • In the synchronous model, if both τ g , τ r > , the expected run-time is superpolynomial in n (whichincludes the possibility of never finishing). We briefly discuss the intuition behind this conjecture in the incremental case. Consider a greenfirewall [ a, c ] where c − a (cid:29) w + 1 and c + 1 is red. Let b be the rightmost happy element within thisfirewall. We assume that b < c . Now consider the interval I = [ b + 1 , b + w ]. Since b is happy and b + 1is not, I must contain exactly (cid:100) τ g (2 w + 1) (cid:101) − ( w + 1) many green nodes which will all be unhappy. Thequestion of interest is whether the happy firewall, which currently ends at b , is more likely to advance orretreat.If τ g > , then UG ( I ) > w , and irrespective of the remaining nodes in I , the happy firewall ismore likely to retreat. However, if τ g < , then UG ( I ) < w for all large enough w . At the same time, UR ( I ) ≥ min { w − UG ( I ) , (cid:100) τ r (2 w + 1) (cid:101) − ( w + 1) } , and presuming τ g < τ r , the happy firewall is morelikely to advance.The situation where τ g , τ r > is redolent of the classic Ehrenfest Urn, a simple model of a thermo-dynamic process proposed by T. & P. Ehrenfest in [9]. An urn is filled with a fixed number ( w ) of balls,divided in some proportion between red green. At each time step, a ball is selected uniformly at randomfrom the urn and replaced with a ball of the opposite colour. It is fairly clear that the model’s limitingdistribution as t → ∞ is b ( w, ), regardless of the starting configuration. The celebrated analysis of Kacin [13] also established that if the urn begins in an all-green state (and subject to the technical provisothat w is even) the expected time until this state recurs is exponential in w . Acknowledgements
Lewis-Pye was was supported by a Royal Society University Research Fellowship.Barmpalias was supported by the Research Fund for International Young Scientists from the National Natural ScienceFoundation of China, grant numbers 613501-10236 and 613501-10535, and an International Young Scientist Fellowship fromthe Chinese Academy of Sciences; support was also received from the project Network Algorithms and Digital Informationfrom the Institute of Software, Chinese Academy of Sciences and a Marsden grant of New Zealand.
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A Deferred proofs from section 1
We deferred the proof of the following result from the introduction, and present it below:
Lemma 1.13
For any scenario ( ρ, τ g , τ r ) and for all large enough w , the selective dynamic guarantees that the processwill finish.Proof Our strategy is to define a harmony index for the whole ring, and establish that this quantity has a finite upperbound, but also increases (by at least some minimum positive amount) with each legitimate move. This will give the result.If τ g + τ r ≤
1, we start by picking χ such that 1 − τ g τ g ≥ χ ≥ τ r − τ r . If instead τ g + τ r >
1, then we require that w is large enough to allow us to choose χ where0 < − τ g + (cid:16) w +1 (cid:17) τ g − (cid:16) w +1 (cid:17) < χ < τ r − (cid:16) w +1 (cid:17) − τ r + (cid:16) w +1 (cid:17) . Now for a node x at time t , we’ll write G t ( x ) = 1 (respectively G t ( x ) = 0) if x is green (red) at time t , and define A t ( x ) := (cid:26) χ if G t ( x ) = 11 if G t ( x ) = 0 . Similarly define L t ( x ) := |{ y ∈ N ( x ) : G t ( y ) = G t ( x ) }| w + 1 . Now we define the following harmony index: S ( t ) := (cid:80) x A t ( x ) L t ( x ). Clearly this is bounded above by n · max { , χ } . Wewish to compare S ( t + 1) and S ( t ). Suppose that x is the node whose colour changes. Then L t +1 ( x ) = 1 − L t ( x ) + w +1 .Similarly for y ∈ N ( x ) with x (cid:54) = y and G t ( y ) = G t ( x ) we have L t +1 ( y ) = L t ( y ) − w +1 , and there are (2 w + 1) L t ( x ) − y . At the same time, for z ∈ N ( x ) with G t ( z ) (cid:54) = G t ( x ) we have L t +1 ( z ) = L t ( z ) + w +1 , and there are(1 − L t ( x ))(2 w + 1) many such z . Hence, S ( t + 1) = S ( t ) − A t ( x ) L t ( x ) + A t +1 ( x ) (cid:18) − L t ( x ) + 12 w + 1 (cid:19) − ((2 w + 1) L t ( x ) − A t ( x ) 12 w + 1 + (1 − L t ( x )) (2 w + 1) A t +1 ( x ) 12 w + 1 . Thus S ( t + 1) − S ( t ) = 2 A t +1 ( x ) − χ ) L t ( x ) + 1 + χ w + 1 . Hence it suffices to show that (1 + χ ) L t ( x ) < A t +1 ( x ) for which we check the four possible cases.Suppose first that τ g + τ r ≤
1. If G t ( x ) = 1 then, since x is unhappy L t ( x ) < τ g and (1 + χ ) τ g ≤ A t +1 ( x ) asrequired. On the other hand, if G t ( x ) = 0 then L t ( x ) < τ r and (1 + χ ) τ r ≤ χ = A t +1 ( x ), again as required.Suppose now that τ g + τ r >
1. This time if G t ( x ) = 1, then since x is hopeful 1 − L t ( x ) + w +1 ≥ τ r meaning(1 + χ ) L t ( x ) ≤ (1 + χ )(1 − τ r + w +1 ) < A t +1 ( x ) again by choice of χ . Finally, if G t ( x ) = 0 then hopefulness tells usthat 1 − L t ( x ) + w +1 ≥ τ g which gives us (1 + χ ) L t ( x ) ≤ (1 + χ )(1 − τ g + w +1 ) < χ = A t +1 ( x ).ipping Points in Schelling Segregation 31 B Deferred proofs from section 3
Here we present proofs of two of the more technical matters from section 3, starting with the following:
Lemma 3.2
Let S := (0 , × (0 , . We divide S into the two triangles T := { ( x, y ) ∈ S : x + y < } and T := { ( x, y ) ∈ S : x + y > } and the line L = { ( x, y ) ∈ S : x + y = 1 } . Also define S := (cid:0) , (cid:1) × (cid:0) , (cid:1) and S := (cid:0) , (cid:1) × (cid:0) , (cid:1) . (Noticethat S i ⊂ T i .) Then the following hold:1. Suppose that ( τ g , τ r ) , ( τ (cid:48) g , τ (cid:48) r ) ∈ T i and that ( ρ, τ g , τ r ) is red dominating. If τ (cid:48) g ≥ τ g , and τ r ≥ τ (cid:48) r then ( ρ, τ (cid:48) g , τ (cid:48) r ) is reddominating. Conversely, if ( ρ, τ (cid:48) g , τ (cid:48) r ) is green dominating, so too is ( ρ, τ g , τ r ) .2. For i ∈ { , } , every scenario where ρ ≤ (respectively ρ ≥ ) and ( τ g , τ r ) ∈ S i is red (green) dominating.3. Any value of ρ where < ρ < admits both red and green dominating scenarios in both S and S .Proof Define h : T ∪ T → R by h ( x, y ) := x (cid:16) x − x − y (cid:17) (1 − x ) (cid:16) − x − x − y (cid:17) y (cid:16) y − x − y (cid:17) (1 − y ) (cid:16) − y − x − y (cid:17) . Claim. ∂h∂x < ∂h∂y > T ∪ T . Proof of claim.
By differentiating ln h , we find that h ∂h∂x = k ( x,y )(1 − x − y ) where k ( x, y ) = (1 − y ) ln x + y ln(1 − x ) − y ln y − (1 − y ) ln(1 − y ) . Since h > T i it suffices to show that k ≤ S . Well ∂k∂x = 1 − x − yx (1 − x )whence ∂k∂x > T and ∂k∂x < T . Similarly ∂k∂y = ln(1 − x ) − ln y + ln(1 − y ) − ln x meaning that ∂k∂y > T and ∂k∂y < T . So we have established that k is monotonically strictly increasing in both x and y on T and monotonically strictly decreasing in both x and y on T . Along the line L we have k ( x, y ) = 0, hence itmust be that ∂h∂x < T and T as required. Since ( h ( x, y )) − = h ( y, x ), the result for ∂h∂y also follows. QED Claim
Statement 1 of the Lemma follows from the fact that, for ( τ g , τ r ) ∈ T ∪ T , the scenario ( ρ, τ g , τ r ) being red dominatingis equivalent to the assertion h ( τ g , τ r ) < − ρρ , with green domination equivalent to the reverse inequality.Now consider the restrictions h (cid:22) S i . Since lim ( x,y ) → (0 , ) h ( x, y ) = lim ( x,y ) → ( , h ( x, y ) = 4 and lim ( x,y ) → ( , h ( x, y ) =lim ( x,y ) → (1 , ) h ( x, y ) = , from which it follows that the restriction h : S i → ( ,
4) is surjective for i ∈ { , } .Thus for ρ ≥ we have h ( x, y ) > > − ρρ for any ( x, y ) ∈ S i . Similarly for ρ ≤ we have h ( x, y ) < < − ρρ , givingstatement 2.For statement 3, notice that if < ρ < then 4 > − ρρ > and the result again follows by the continuity andsurjectivity of h restricted to S i . Corollary 3.3
Suppose ( ρ, τ g , τ r ) is a scenario where τ g + τ r (cid:54) = 1 and τ g ≥ ρ and τ r ≤ − ρ . Then ( ρ, τ g , τ r ) is reddominating. (Similarly, green domination follows when both of the reverse inequalities hold.)Proof Let h be as in the proof of Lemma 3.2. We shall compute lim y ↑ (1 − ρ ) h ( ρ, y ). Well,ln h = ρ ln ρ + (1 − ρ ) ln(1 − ρ ) − y ln y − (1 − y ) ln(1 − y )1 − ρ − y . By L’Hˆopital’s rule, therefore lim y ↑ (1 − ρ ) ln h ( ρ, y ) = lim y ↑ (1 − ρ ) ln (cid:16) y − y (cid:17) = ln (cid:16) − ρρ (cid:17) . Hence, by the continuity of ln, wehave lim y ↑ (1 − ρ ) h ( ρ, y ) = − ρρ .By Lemma 3.2 (1) applied to T , the result therefore follows in the region τ g + τ r <
1. The case where τ g + τ r > T and compute lim x ↓ ρ h ( x, − ρ ) = − ρρ . C Deferred proofs from section 4
Throughout this appendix we work in a fixed scenario ( τ g , τ r , ρ ) and for some fixed θ (cid:54) = ρ . For any node u , we define x u tobe the first node to the left of u satisfying GD θ ( x u ). The following proposition plays a significant role in the current work: Proposition 4.10
Fix a value of ρ and a value θ (cid:54) = ρ . For any node u let x u be the first node to the left of u such that GD θ ( x u ) holds.Let Q ( u ) be a property of nodes which depends only on the vicinity of u in the initial configuration (which is to say itdepends on [ u − C, u + C ] , for some C independent of n ).Suppose there exists p > such that for all sufficiently large w we have P ( Q ( u ) | GD θ ( u )) ≥ p . Then there exists p (cid:48) > such that for all n (cid:29) w (cid:29) we have P ( Q ( x u )) ≥ p (cid:48) for u selected uniformly at random.If additionally the hypothesis holds with p = 1 − ε (cid:48) for all ε (cid:48) > , then we may likewise take p (cid:48) = 1 − ε for any ε > . p = 1 − ε (cid:48) for any ε (cid:48) > p = 1. Notice that even here we cannot simply apply Lemma 2.7, since we do not haveaccess to hypothesis (ii) there. Instead we shall perform some careful counting operations, working in the vicinity of somenode v satisfying GD θ ( v ), and bounding above the number of other such nodes that one can expect to find nearby. Beforecommencing this though, we mention a version of the law of large numbers which was shall use several times: Lemma C.1 (Strong law of large numbers)
Fix a scenario and a value of w . Let Q (cid:48) ( u ) be a property of nodes whichdepends only on the vicinity of u in the initial configuration (i.e. on [ u − C, u + C ] for some C independent of n ). Withprobability one, as n → ∞ the proportion of nodes u in the ring that satisfy Q (cid:48) ( u ) tends to P ( Q (cid:48) ( u )) . The proof of this can be found in [2]. In proving Proposition 4.10, the following definition will play an important role:
Definition C.2
We say that GD θ ( u, z ) holds if GD θ ( u ) holds and there are at most z many nodes v ∈ [ u − (2 w + 1) , u +(2 w + 1)] satisfying GD θ ( v ) . We shall show shortly that if θ (cid:54) = ρ , then we may choose z large enough that for all w (cid:29) GD θ ( u, z ) is highly likelyto follow from GD θ ( u ). To establish this, it will be helpful to introduce a weaker notion: Definition C.3
Given a node u and an integer k ≥ let N k ( u ) := [ u − (cid:100) w/k (cid:101) , u + (cid:100) w/k (cid:101) ] . For any z > , we say GD θ ( u, k, z ) holds if GD θ ( u ) holds and additionally there are at most z many nodes within N k ( u ) such that GD θ ( z ) holds. We remark that for probabilities p and p we shall use the notation p (cid:29) p to mean p p (cid:29)
0. We shall now show that GD θ ( u, k, z ) is likely to follow from GD θ ( u ): Lemma C.4
We make no assumption on ( ρ, τ g , τ r ) , supposing only that θ (cid:54) = ρ . Then for any ε (cid:48) > , all large enough z ,and all (cid:28) k (cid:28) w , we have P (cid:16) GD θ ( u, k, z ) | GD θ ( u ) (cid:17) > − ε (cid:48) . Proof
We assume first that θ > ρ . Again we start by selecting u uniformly at random from nodes such that GD θ ( u ) holds.First of all, we want to show that for sufficiently large z , if we step (cid:98) z/ (cid:99) many nodes to the right (or left) of u , thenwe will very probably reach a green density well below θ . To this end, let v = u + (cid:98) z/ (cid:99) and x = G ( N ( u ) \N ( v )). Then E ( x ) = θ (cid:98) z/ (cid:99) . By applying Chebyshev’s Inequality we conclude that for any ε (cid:48)(cid:48) > z , P ( | ( x / (cid:98) z/ (cid:99) ) − θ | > ε (cid:48)(cid:48) ) (cid:28) ε (cid:48) .Now consider x := G ( N ( v ) \N ( u )). The law of large numbers tells us that for any ε (cid:48)(cid:48) > z , P ( | ( x / (cid:98) z/ (cid:99) ) − ρ | > ε (cid:48)(cid:48) ) (cid:28) ε (cid:48) . Since G ( N ( v )) = G ( N ( u )) − x + x , we find that for any m > z , P ( G ( N ( u )) − G ( N ( v ))) < m ) (cid:28) ε (cid:48) .So far then, we have considered moving (cid:98) z/ (cid:99) many nodes to the right of u to a node v , and have concluded that G ( N ( v )) will very probably be well below G ( N ( u )) = θ (2 w + 1) (a similar argument also applies, of course, to the left).Now we have to show that as we move right from v to some node v (cid:48) , so long as v (cid:48) ∈ N k ( u ), the green node count G ( N ( v (cid:48) ))will very probably remain below θ (2 w + 1). In order to do this, we approximate G ( N ( v (cid:48) )), as v (cid:48) varies, by a biased randomwalk B ( v (cid:48) ).So let us briefly adopt the approximation that nodes in N ( u ) are independent identically distributed random variables,each with probability θ of being green. Then, for v (cid:48) ∈ N k ( u ) to the right of u , P (cid:16) B ( v (cid:48) + 1) = B ( v (cid:48) ) + 1 (cid:17) = ρ (1 − θ ), while P (cid:16) B ( v (cid:48) + 1) = B ( v (cid:48) ) (cid:17) = (1 − θ )(1 − ρ ) + θρ = 1 − θ + ρ + 2 θρ and P (cid:16) B ( v (cid:48) + 1) = B ( v (cid:48) ) − (cid:17) = (1 − ρ ) θ .Since θ > ρ , P (cid:16) B ( v (cid:48) + 1) = B ( v (cid:48) ) − (cid:17) > P (cid:16) B ( v (cid:48) + 1) = B ( v (cid:48) ) + 1 (cid:17) . Now removing those steps at which B ( v (cid:48) ) doesnot change, we get a biased random walk with probability say p > of going down at each step and (1 − p ) of goingup. Choose p with < p < p . Now, dropping the false assumption of independence, by taking k sufficiently large weensure that as we take successive steps right from v inside the interval N k ( u ), at each step, no matter what has occurredat previous steps, the probability of G ( N ( v (cid:48) )) increasing is less than (1 − p ) and the probability of G ( N ( v (cid:48) )) decreasing isgreater than p > . Thus by a standard fact about biased random walks, if G ( N ( u + (cid:98) z/ (cid:99) )) ≤ θ (2 w + 1) − m , then theprobability that any nodes v (cid:48) ∈ [ u + (cid:98) z/ (cid:99) , u + (cid:100) w/k (cid:101) ) satisfies G ( N ( v (cid:48) )) ≥ θ (2 w + 1) is less than (cid:16) − p p (cid:17) m .Finally, let m be such that (cid:16) − p p (cid:17) m (cid:28) ε (cid:48) , and let z be sufficiently large that, for v = u + (cid:98) z/ (cid:99) , P ( θ (2 w +1) − G ( N ( v ))
With no assumption on ( ρ, τ g , τ r ) and θ (cid:54) = ρ , for any ε (cid:48) > , for all sufficiently large z , for all largeenough w , P ( GD θ ( u, z ) | GD θ ( u )) > − ε (cid:48) . Proof
Observe that if k is sufficiently large compared to k , if ε (cid:48) is sufficiently small, and if GD θ ( u, k , z ) and Smooth k ,ε (cid:48) ( u )both hold, then in the initial configuration there are at most z many nodes v ∈ [ u − (2 w + 1) , u + (2 w + 1)] where GD θ ( v )holds. Applying Lemmas C.4 and 4.9 therefore gives the result.The next result is another step towards Proposition 4.10 (with ¬ denoting logical negation). (We remark that z + 1 inpart (ii) could be replaced with many other expressions; however z + 1 will turn out to be a useful choice.)ipping Points in Schelling Segregation 33 Lemma C.6
Again we make no assumption on ( ρ, τ g , τ r ) , but fix θ (cid:54) = ρ . Let Q and p be as in Proposition 4.10. Define µ := P ( Q ( v ) | GD θ ( v, z )) P ( ¬ Q ( v ) | GD θ ( v, z )) . Then for any ε (cid:48) > , all sufficiently large z , and all sufficiently large w , we have(i) If p (cid:54) = 1 , then µ ≥ p − ε (cid:48) − p .(ii) If p = 1 , then µ ≥ z +1 ε (cid:48) .Proof Applying Corollary C.5, we require z large that, for all sufficiently large w , if we are given that GD θ ( v ) holds, then GD θ ( v, z ) fails with probability (cid:28) ε (cid:48) , i.e. putting ε = P ( ¬ GD θ ( v, z ) | GD θ ( v )), choose z so that ε (cid:28) ε (cid:48) for sufficiently large w . Also, by assumption on Q , we have that P ( ¬ Q ( v ) | GD θ ( v )) ≥ − p for all large enough w . Thus µ = P ( Q ( v ) ∧ GD θ ( v, z ) | GD θ ( v )) P ( ¬ Q ( v ) ∧ GD θ ( v, z ) | GD θ ( v ))) = 1 − P ( ¬ Q ( v ) ∨ ¬ GD θ ( v, z ) | GD θ ( v ))1 − P ( Q ( v ) ∨ ¬ GD θ ( v, z ) | GD θ ( v )))Thus, if p (cid:54) = 1, we get − (1 − p + ε )1 − p ≤ µ . Hence µ ≥ p − ε − p and the first statement follows by assumption on ε .If p = 1, then let ε (cid:48)(cid:48) > ε (cid:48) that − ε (cid:48)(cid:48) ε (cid:48)(cid:48) > z +1 ε (cid:48) . Then the the result obtained in part (i)gives us that µ ≥ − ε (cid:48)(cid:48) ε (cid:48)(cid:48) for all w (cid:29) ε (cid:48)(cid:48) . Lemma C.7
Again, our only assumption is that θ (cid:54) = ρ . Now for any node u , let x u be the first node to the left of u forwhich GD θ ( x u ) holds. For any ε (cid:48) > , for all large enough z and (cid:28) w (cid:28) n and for u chosen uniformly at random, x u is defined and GD θ ( x u , z ) holds with probability > − ε (cid:48) .Proof We work entirely in the initial configuration, and define an iteration which assigns colours to nodes as follows.Step 0. Pick a node t uniformly at random.Step s + 1. Let v s be the first node to the left of t s such that v s = x t s or such that s > v s = t . Carry out theinstructions for the first case below which applies:1. If there exists no such v s then terminate the iteration, and declare that it has ‘ended prematurely’.2. If v s = t and s > t s undefined and terminate the iteration.3. If | v s − t s | < w + 1, then colour t s pink.4. If ¬ GD θ ( v s , z ) holds, then colour t s black.5. If GD θ ( v s , z ) holds, then colour t s silver.In cases (3) - (5), define t s +1 = v s − (2 w + 1), unless t lies in the interval [ v s − (2 w + 1) , v s ), in which case terminatethe iteration. This completes the description of the iteration.The purpose of this construction is that every node u in the ring which satisfies GD θ ( u, z ) will lie in the interval I s := [ v s − w, v s ] for some s . Thus it will assist us in counting such nodes.First note that the probability that the iteration terminates prematurely can be made arbitrarily small by taking n large, and similarly that we may assume that t s of all colours exist. Also, by the assumption that θ (cid:54) = ρ , the proportion of t s coloured pink will be (cid:28) ε (cid:48) for all large enough w .Now let S be the greatest s such that t s is defined when the iteration terminates, and let ε := P ( ¬ GD θ ( u, z ) | GD θ ( u )).Then by Corollary C.5, we know that ε (cid:28) ε (cid:48) for all large enough w .Let χ := |{ u : GD θ ( u, z ) }||{ u : GD θ ( u ) ∧ ¬ GD θ ( u, z ) }| Then for large n , χ can be expected to be close to − ε ε by the strong law of large numbers (Lemma C.1). In order to findan upper bound for χ , first let us find an upper bound for |{ u : GD θ ( u, z ) }| .Let v s be as defined in the iteration. Then there can be at most z many nodes u in I s satisfying GD θ ( u, z ). Furthermore,since every node u satisfying GD θ ( u, z ) lies in some I s , we find an upper bound of |{ u : GD θ ( u, z ) }|≤ ( S + 1) z .Similarly we may find a lower bound for |{ u : GD θ ( u ) ∧ ¬ GD θ ( u, z ) }| . Let π be the proportion of the t s that arecoloured black. Now for each black t s we are guaranteed at least z many such nodes in I s . We therefore get a lower boundof ( S + 1) πz .Putting these two bounds together, we get χ ≤ z ( S + 1) π ( S + 1) z = 1 π . Since χ is close to − ε ε for large n and p (cid:28) ε (cid:48) , we infer that for all sufficiently large w , with probability tending to 1as n → ∞ , we have χ (cid:29) /ε (cid:48) , so that π (cid:28) ε (cid:48) . So, for sufficiently large w , with probability tending to 1 as n → ∞ , theproportion of the t s which are not coloured silver is (cid:28) ε (cid:48) .This concludes the proof, since as n → ∞ the proportion of the t s coloured silver will be less than or equal to theprobability of x u being defined and satisfying GD θ ( x u , z ) for u chosen uniformly at random (and indeed will converge tothis value).We may now complete the work of this appendix:4 George Barmpalias et al. Proof (Proof of Proposition 4.10)
Let 0 < ε (cid:48) (cid:28) ε (cid:28) p . Several of the previous Lemmas are stated for all sufficiently large z . Here we fix z large enoughto apply those results for our value of ε (cid:48) .We shall now extend the construction from the proof of Lemma C.7, and we preserve the notation introduced thereincluding the result of the colouring process (1)-(5). However, we further subdivide the silver nodes:(6) If GD θ ( v s , z ) ∧ ¬ Q ( v s ) holds, then colour t s silver and grey.(7) If GD θ ( v s , z ) ∧ Q ( v s ) holds, then colour t s silver and gold.Let π (cid:48) be the proportion of the t s which are coloured grey and Ξ the proportion coloured gold. We have alreadyestablished that Ξ > − π (cid:48) − ε (cid:48) with probability tending to 1.As in Lemma C.6, we define µ := P (cid:0) Smooth k,ε (cid:48) ( v ) ∧ GD θ ( v, z ) (cid:1) P (cid:0) ¬ Smooth k,ε (cid:48) ( v ) ∧ GD θ ( v, z ) (cid:1) and χ (cid:48) := |{ v : Q ( v ) ∧ GD θ ( v, z ) } ||{ v : ¬ Q ( v ) ∧ GD θ ( v, z ) } | . Now χ (cid:48) can be expected to be close to µ for large n , by the strong law of large numbers (Lemma C.1). Thus for alllarge enough w with probability approaching 1 we may apply the bounds for µ provided by Lemma C.6 also to χ (cid:48) .Now we form a new lower bound for χ (cid:48) in the same manner we did for χ . The numerator of the fraction is boundedabove by (1 − π (cid:48) )( S + 1) z . For the denominator, notice that if t s is coloured grey then we get at least one node u ∈ I s forwhich ¬ Q ( u ) ∧ GD θ ( u, z ) holds. Thus χ (cid:48) ≤ (1 − π (cid:48) )( S + 1) zπ (cid:48) ( S + 1) = (1 − π (cid:48) ) zπ (cid:48) . Suppose now that p (cid:54) = 1. Then by Lemma C.6, we have that p − ε (cid:48) − p ≤ (1 − π (cid:48) ) zπ (cid:48) with probability tending to 1. Rearrangingthis gives us 1 − π (cid:48) ≥ p − ε (cid:48) z (1 − p )+ p − ε (cid:48) .For this fixed value of z , since ε (cid:48) was chosen small enough, it follows that Ξ > pz with probability tending to 1. Nowwe may take p (cid:48) := pz . This concludes the proof in the case p (cid:54) = 1, since as w → ∞ , the value of Ξ will converge to theprobability of x u being defined and satisfying Q ( x u ) for u chosen uniformly at random. Thus this probability will alsoexceed p (cid:48) . Furthermore, the probability thatOn the other hand, if p = 1 then we know by Lemma C.6 that χ (cid:48) > z +1 ε (cid:48) with probability tending to 1. Thus (1 − π (cid:48) ) zπ (cid:48) > z +1 ε (cid:48) . With z fixed, by choice of ε (cid:48) it follows that Ξ > − ε for all large enough w with probability tending to1, which again is sufficient to establish the result. D Deferred proofs from section 8
In this appendix, we present the proof of the following result:
Proposition 8.1
In the synchronous model, suppose that < τ g < and τ g < τ r . Then green takes over totally.Proof Suppose that [ a, b ] is a happy green firewall at time t , meaning that each element of [ a, b ] is a happy green node, andwhere b − a ≥ w + 1. We suppose that [ a, b ] is maximal, meaning that a − b + 1 are not happy green nodes, andwe shall show that, for all large enough w , the happy firewall will advance (at both ends) either at time t + 1 or t + 2, andnever retreat. As in the proof of Theorem 1.4, we can then appeal to the law of large numbers to guarantee the existenceof such a firewall with probability > − ε (cid:48) , for large enough n , ensuring total takeover.Define k g := (cid:100) τ g (2 w + 1) (cid:101) − ( w + 1), and k r similarly. The conditions on τ g and τ r imply the following: k g < w and k r ≥ k g + 1, so long as w is large enough.Now we focus on the right end of [ a, b ]. (Similar arguments will apply to the left end.) Let J := [ b + 1 , b + w ]. Let b + c + 1 be the leftmost red node in J at time t , so that 0 ≤ c ≤ k g . Suppose first that c ≥
1. It follows immediately that G t ( J ) = UG t ( J ) = k g , since G t ( N ( x )) ∈ { G t ( N ( x − , G t ( N ( x − − } for x ∈ J .Define J := [ b +1 , b + c ]. By definition then, J is entirely green, which is to say G t ( J ) = c . Also let J := [ b + c +1 , b + w ].Then G t ( J ) = k g − c and R t ( J ) = w − k g . We now divide into two cases, depending on whether or not HR t ( J ) = 0.Suppose first that HR t ( J ) = 0. (This will hold, for instance, if w − k g ≤ k r .) Then UR t ( J ) = w − k g and at time t + 1we have G t +1 ( J ) = 0 and G t +1 ( J ) = w − k g . Notice that this implies G t +1 ( J ) ≥ k g , so the happy green firewall has notretreated. In this case, G t +1 ( N ( b + c + 1)) ≥ w − c + w − k g = 2 w − k g − c ≥ w − k g ≥ w + k g + 1. Hence b + c + 1 isa happy green node at time t + 1. Since R t +1 ( J ) = c < k r it follows that UR t +1 ( J ) = c , thus, moving on to time t + 2,we see that J is once again entirely green but is joined by b + c + 1. Thus G t +2 ( J ) ≥ k g + 1 and hence the happy greenfirewall has grown to encompass b + 1.On the other hand, if HR t ( J ) >
0, then we must have k r ≤ UR t ( J ) < w − k g (since counting from the left, we mustencounter k r unhappy red nodes before the first happy one). Thus at time t + 1 we have G t +1 ( J ) ≥ k r , ensuring thatthe happy firewall has not retreated. At the same time, UG t ( J ) = k g − c as before. At time t + 1 these will all becomeunhappy red nodes making UR t +1 ( J ) ≥ k g . But we cannot have equality here, since k g many unhappy red nodes are notenough to preserve the happiness of any red node happy at time t , at least one of which must therefore become unhappy.Thus UR t +1 ( J ) ≥ k g − c + 1. Moving on to time t + 2, we see J is again entirely green, and G t +2 ( J ) ≥ k g + 1, meaningthat the happy green firewall has grown.Now we deal with the case that c = 0, meaning that b + 1 is an unhappy red node at time t . Here we can say onlythat G t ( J ) ≥ k g . If G t ( J ) < w − k g , then UR t ( J ) ≥ k g + 1 (because UR t ( J ) = k r ≥ k g + 1), and thus at time t + 1, G t +1 ( J ) ≥ k g + 1, meaning that b + 1 will be a happy green node, and the happy firewall will have grown.ipping Points in Schelling Segregation 35Hence we can suppose that G t ( J ) ≥ w − k g . (Notice that this implies R t ( J ) ≤ k g < k r so that HR t ( J ) = 0.) Now let b + d be any green node in J where G t [ b + d, b + w ] ≥ w − k g . We claim that b + d is happy. Well, clearly d ≤ k g + 1. Hence G t [ b + d − w, b + d − ≥ w − k g . Equally, G t [ b + d, b + d + w ] ≥ w − k g . Hence G t ( N ( b + d )) ≥ w − k g ≥ k g asrequired.So, at time t if b + d is an unhappy green node, then G t [ b + d, b + w ] ≤ w − k g −
1. Clearly at most w − k g − UG t ( J ) ≤ w − k g −
1. Thus G t +1 ( J ) = HG t ( J ) + UR t ( J ) ≥ k g + 1, meaning that b + 1 willbe a happy green node at time tt