Asymptotic normality of the size of the giant component via a random walk
aa r X i v : . [ m a t h . P R ] A p r Asymptotic normality of the size of the giantcomponent via a random walk
B´ela Bollob´as ∗† Oliver Riordan ‡ October 21, 2010; revised April 14, 2011
Abstract
In this paper we give a simple new proof of a result of Pittel andWormald concerning the asymptotic value and (suitably rescaled) limitingdistribution of the number of vertices in the giant component of G ( n, p )above the scaling window of the phase transition. Nachmias and Peresused martingale arguments to study Karp’s exploration process, obtaininga simple proof of a weak form of this result. We use slightly differentmartingale arguments to obtain a much sharper result with little extrawork. The component of a random graph containing a given vertex may be ‘explored’by a step-by-step process that is by now well known, described in detail below.A key feature of this process is that vertices are ‘examined’ one at a time,and tested for edges to ‘new’ vertices. This means that the behaviour of theexploration is closely connected to that of a certain random walk. In the contextof random graphs, this process was introduced by Karp [4] in 1990; slightlyearlier, Martin-L¨of [5] used essentially the same process in a different context,namely the study of epidemics, where it arises even more naturally. Somewhatlater, Aldous [1] introduced a variant of the process adapted to explore all components of a random graph; recently, analyzing this latter exploration withmartingale techniques related to those in [5], Nachmias and Peres [6] gave asimple proof that in the weakly supercritical range, i.e., when p = (1 + ε ) /n where ε = ε ( n ) satisfies ε → ε n → ∞ , the largest component of G ( n, p ) ∗ Department of Pure Mathematics and Mathematical Statistics, Wilberforce Road, Cam-bridge CB3 0WB, UK and Department of Mathematical Sciences, University of Memphis,Memphis TN 38152, USA. E-mail: [email protected] . † Research supported in part by NSF grants CNS-0721983, CCF-0728928 and DMS-0906634, and ARO grant W911NF-06-1-0076 ‡ Mathematical Institute, University of Oxford, 24–29 St Giles’, Oxford OX1 3LB, UK andDepartment of Mathematical Sciences, University of Memphis, Memphis TN 38152, USA.E-mail: [email protected] . εn + o p ( εn ) vertices. (They also studied the weakly subcritical case,which we shall not discuss further here.)Here we shall analyze the same process more carefully, obtaining a sim-ple new proof of the following asymptotic normality result due to Pittel andWormald [8]. Let ρ = ρ λ denote the survival probability of the Galton–Watsonbranching process in which the number of offspring of each individual has aPoisson distribution with mean λ . For λ > ρ λ as the uniquepositive solution to 1 − ρ = e − λρ . (1)When λ > λ ∗ for λ (1 − ρ λ ); this is often known as the dual branchingprocess parameter to λ , and satisfies λ ∗ < λ ∗ e − λ ∗ = λe − λ . (The corre-sponding Poisson branching process provides an approximation of the randomgraph in the vicinity of a generic vertex outside the giant component.) Theorem 1.
Let p = λ/n where λ = λ ( n ) satisfies λ = O (1) and ( λ − n → ∞ as n → ∞ , and let L denote the number of vertices in the largest componentof G ( n, p ) . Then L − ρnσ d → N (0 , , where d → denotes convergence in distribution, N (0 , is the standard normaldistribution, ρ = ρ λ > is defined by (1) , and σ = ρ (1 − ρ )(1 − λ ∗ ) n. The special case of this result in which λ is constant goes back to Stepanov [9](see also Pittel [7]); the form above is due to Pittel and Wormald [8], who provedmuch more, including asymptotic joint normality of the sizes of the largestcomponent and of its 2-core.Specializing to the barely supercritical case, the formulae above simplifyconsiderably. Indeed, it is easy to check that if λ = 1 + ε and ε →
0, then ρ λ = 2 ε + O ( ε ), and λ ∗ = 1 − ε + O ( ε ). Thus Theorem 1 has the followingcorollary. Corollary 2.
Let ε = ε ( n ) satisfy ε → and ε n → ∞ , and let L denote thenumber of vertices in the largest component of G ( n, (1 + ε ) /n ) . Then L − ρn √ ε − n d → N (0 , , (2) where ρ > is defined by (1) with λ = 1 + ε . (cid:3) Under the conditions of Corollary 2 we have ρ ∼ ε , while the standarddeviation √ ε − n is o ( εn ), so Corollary 2 implies in particular the result ofNachmias and Peres [6] mentioned earlier.2 The proof
We consider the component exploration process as in [6], itself based on those ofKarp [4], Martin-L¨of [5] and Aldous [1], although we shall use slightly differentterminology and initial conditions. At each step, every vertex will have one ofthree states, active , explored , or unseen . The exploration will take place in n steps, at times t = 1 , . . . , n , starting from the initial state at time 0, when everyvertex is unseen.Fix an order on the vertices. At step 1 ≤ t ≤ n (i.e., going from time t − t ) let v t be the first active vertex, if there are any; otherwise v t is thefirst unseen vertex. In the latter case we say that we ‘start a new component’at step t . Having defined v t , reveal all edges from v t to (other) unseen vertices;let η t be the number of such edges, and label the corresponding neighbours of v t as active; label v t itself as explored. After t steps of the process, exactly t vertices have been explored. We write A t and U t for the numbers of active andunseen vertices after 0 ≤ t ≤ n steps, so U t = n − t − A t , A = 0 and U = n .After n steps, it is very easy to see that the process has revealed a spanningforest in G , having first revealed a spanning tree of one component, then aspanning tree of another component (if there is more than one), and so on.Write C t for the number of components started by time t , and set X t = A t − C t . We claim that X t = A t − C t = t X i =1 ( η i − . (3)Indeed, if in step t we do not start a new component, then we explore an activevertex and then change η t vertices from unseen to active, so A t − A t − = η t − C t = C t − . If we do start a new component, which happens if and onlyif A t − = 0, then we explore an unseen vertex, so A t − A t − = A t = η t and C t − C t − = 1. This establishes (3).Let 0 = t < t < t < · · · < t k = n enumerate { t : A t = 0 } , i.e., theset of times at which there are no active vertices. We start exploring the i thcomponent at time t i − + 1 and finish at time t i , so L = max { t i − t i − : 1 ≤ i ≤ k } . (4)Since C t = i for t i − < t ≤ t i , recalling that X t = A t − C t we have t i = inf { t : X t = − i } . (5)Writing c ( G ) for the number of components of G = G ( n, p ), note that X n = − c ( G ), and that X t may decrease by at most one at each step, so the infimumis defined for all 1 ≤ i ≤ c ( G ).Let F t denote the sigma-field generated by η , . . . , η t ; in other words, F t isthe (finite, of course) sigma-field generated by all information revealed by step t . Set U ′ t = U t if A t > U ′ t = U t − U ′ t is the number3f edges tested at step t + 1. Hence, given F t , the random variable η t +1 has abinomial distribution with parameters U ′ t and p : P ( η t +1 = k | F t ) = (cid:18) U ′ t k (cid:19) p k (1 − p ) U ′ t − k . If we know the sequence ( η t ), then we know the entire outcome of the process,and in particular L . More precisely, we can use (3) to find ( X t ), then (5) tofind the t i (and thus ( C t ), ( A t ) and ( U t )), and finally (4) gives us L .So far we have been following (with minor modifications) the definitions andinitial analysis in [6]. But now our analysis takes a different route.Let us write D t for the expectation of η t − F t − , noting that D t israndom, and satisfies D t +1 = E ( η t +1 − | F t ) = pU ′ t − . Recalling that U t = n − t − A t = n − t − X t − C t , and noting that U ′ t = U t − ( C t +1 − C t ), this gives D t +1 = p ( n − t − X t − C t +1 ) − . (6)Our next aim is to approximate the process ( X t ) that we wish to study bya simpler process ( e X t ), consisting of a deterministic term plus a term closelyrelated to a martingale. Let ∆ t = η t − − D t , so E (∆ t | F t − ) = 0 by thedefinition of D t . From (3), (6) and η t +1 − D t +1 + ∆ t +1 we obtain therecurrence X t +1 = (1 − p ) X t + ∆ t +1 + p ( n − t ) − − pC t +1 . (7)Let x t = n − t − n (1 − p ) t , so x = 0 and x t +1 = (1 − p ) x t + p ( n − t ) − . (8)Subtracting (8) form (7) we see that X t +1 − x t +1 = (1 − p )( X t − x t ) + ∆ t +1 − pC t +1 , whence X t − x t = t X i =1 (1 − p ) t − i (∆ i − pC i ) . (9)With this in mind, we define our approximating process by e X t = x t + t X i =1 (1 − p ) t − i ∆ i . (10)4 emma 3. For any p > and any ≤ t ≤ n we have | X t − e X t | ≤ ptC t . Proof.
From (9) and (10) we have X t − e X t = − t X i =1 (1 − p ) t − i pC i . The result follows immediately since there are t terms in the sum, each boundedby pC t .Let S t = t X i =1 (1 − p ) − i ∆ i , so ( S t ) is a martingale, and e X t = x t + (1 − p ) t S t . (11)As we shall see below, it is easy to obtain very precise results about the dis-tribution of ( e X t ); before turning to the details, let us indicate in rather vagueterms why this should be the case.The variance of each ∆ i is O (1), so S t and hence (1 − p ) t S t have variance O ( t )and size O p ( √ t ). It is true that the distribution of ∆ t depends on earlier valuesof X i in a way that is hard to evaluate exactly, but the dependence is weak: theconditional variance of ∆ t is simply p (1 − p ) U ′ t − , so if we can bound the earlier X i within an additive error of o ( n ), then we obtain a bound on the variance of∆ t accurate to within a factor 1 + o (1). This gives only a o p ( √ t ) additive errorin the martingale term, which is negligible compared to the random variation.(It will turn out that we hit the giant component before seeing many othercomponents, so additional ptC t error from Lemma 3 will be negligible.) Thisstrongly suggests that given that Theorem 1 is true, there should be a simpleproof based on the analysis of ( e X t ). As we shall see, this is indeed the case.From now on we assume that p = λ/n , where λ = λ ( n ) > λ < M for some constant M . Often, we write λ = 1 + ε ;we assume also that ε n → ∞ .For the moment, we study ( e X t ). Let us first start with a standard observa-tion; the second part is a special case of Doob’s maximal inequality [3, Ch. III,Theorem 2.1]. Lemma 4.
Let ( Z t ) ∞ be a discrete-time martingale with filtration ( F t ) andmean Z = 0 . Write I t for the increment Z t − Z t − . Then Var( Z t ) = t X i =1 Var( I i ) = t X i =1 E (cid:0) Var( I i | F i − ) (cid:1) , (12) and for any M ≥ , P (max i ≤ t | Z i | ≥ M ) ≤ Var( Z t ) /M . roof. For the first statement, observe that E I i = 0 for all i and E Z t = 0, whilefor i < j we have E ( I i I j ) = E ( E ( I i I j | F j − )) = E (0) = 0. Hence Var( Z t ) = E Z t = E (cid:0)P t I i (cid:1) = P i E I i = P i Var( I i ). Also, E (Var( I i | F i − )) = E ( E ( I i |F i − )) = E I i , proving (12).For the second statement, apply Doob’s maximal inequality. Alternatively,simply modify the martingale if | Z i | ≥ M holds for any i : let T be the (random)first such i , or T = t if there is no such i , and set Z ′ j = Z j for j ≤ T and Z ′ j = Z T for j > T . Since T is a stopping time, the conditional distributionof I ′ i = Z ′ i − Z ′ i − given F i − is either the same as that of I i , or zero, so theconditional variances of the I ′ i are at most those of the I i . Hence, by (12),Var( Z ′ t ) ≤ Var( Z t ). Since max i ≤ t | Z i | ≥ M if and only if | Z ′ t | ≥ M , applyingChebyshev’s inequality gives the result.Let us write CBi( m, p ) for the centered binomial distribution obtained by sub-tracting mp from a random variable with binomial distribution Bi( m, p ). Notethat the variance of this distribution is mp (1 − p ). The conditional distributionof ∆ t given F t − is exactly that of a centered binomial CBi( U ′ t − , p ). (Previ-ously, we first subtracted one, and then centered, but of course this is the sameas centering directly.) It follows that the differences I i = S i − S i − = (1 − p ) − i ∆ i satisfy Var( I i | F i − ) = (1 − p ) − i U ′ i − p (1 − p ) , (13)so Var( I i | F i − ) ≤ (1 − p ) − n np ≤ (1 − M/n ) − n M = O (1) . For any (deterministic) function t = t ( n ), Lemma 4 thus givessup i ≤ t | S i | = O p ( √ t ) . (14)Let f ( t ) = f n ( t ) = n − t − ne − pt be the continuous-time form of the idealizedtrajectory of ( e X t ) (and hence of ( X t )). It is easy to check that | f ( t ) − x t | = O (1),uniformly in p ≤ M/n and 0 ≤ t ≤ n ; our next lemma shows that ( e X t ) remainsclose to f n ( t ). Lemma 5.
For any ≤ t = t ( n ) ≤ n we have sup i ≤ t | e X t − f n ( t ) | = O p ( √ t ) . Proof.
Immediate from (14), (11) and | f n ( t ) − x t | = O (1).Together, Lemmas 3 and 5 show that ( X t ) remains close to the idealizedtrajectory f ( t ), as long as C t is not too large. As in [6], the basic idea is now toconsider the solution t = ρn to f ( t ) = 0, and choose a suitable t . We shallshow that in the interval [ t , t − t ] the function f ( t ) is far enough away fromzero that X t remains positive, so no new component is started in this interval.Then we consider more precisely the time when X t crosses below its previousminimum level and use (5) to obtain Theorem 1.6e start by examining f . Note that f ′ ( t ) = − npe − pt = p ( n − t − f ( t )) − , (15)and that f ′′ ( t ) = − np e − pt is negative and uniformly bounded by M /n . Since f ′ (0) = np − ε , it follows that if t ≤ εn/ (2 M ), then f ′ ( t ) ≥ ε/ f ( t ) ≥ εt/ . (16)From now on let us pick a function ω = ω ( n ) tending to infinity slowly, inparticular with ω = o ( ε n ). Set σ = √ εn and t = ωσ /ε, ignoring, as usual, the irrelevant rounding to integers. Note for later that t = o ( εn ). Lemma 6.
Let Z = − inf { X t : t ≤ t } denote the number of componentscompletely explored by time t , and let T = inf { t : X t = − Z } be the timeat which we finish exploring the last such component. Then Z ≤ σ /ω and T ≤ σ / ( εω ) hold whp. Considering the initial trajectory of the process ( X t ), it is not hard to checkthat in fact Z = O p ( ε − ) and T = O p ( ε − ), but the weaker bounds abovesuffice. Proof.
Let k = σ /ω . Note that by choice of ω we have k/ √ t → ∞ . Let A denote the event that sup t ≤ t | e X t − f ( t ) | < k/
2. Then by Lemma 5, A holdswhp.At time T we have X T = − Z . Noting that pt = o (1), we have pt ≤ / n is large enough, which we assume from now on. Since T ≤ t by definition,it follows that pT ≤ /
2. But then Lemma 3 gives | X T − e X T | ≤ pT C T ≤ Z/ , and thus e X T ≤ − Z/
2. Since f ( t ) ≥ t ≤ t < ρn , this gives | e X T − f ( T ) | ≥ Z/
2. Hence, whenever A holds, we have Z ≤ k , and the first statement follows.Turning to second statement, recall from (16) that f ( t ) ≥ εt/ t ≤ t = o ( εn ). Consider the interval I = [ σ / ( εω ) , t ]. In this interval we have f ( t ) ≥ σ / (2 ω ) = k/
2, so if A holds then e X t > t ∈ I . As shown above,we have e X T ≤ − Z/ ≤
0, so whenever A holds then T / ∈ I . Since T ≤ t bydefinition, this completes the proof.Let T = inf { t : X t = − Z − } . Then by the properties of the explorationprocess, there is a component with T − T vertices; we aim to show that thiscomponent has size close to the anticipated size of the giant component.7ince np = O (1), by Lemmas 3 and 6 we have thatsup t ≤ T | X t − e X t | ≤ σ / √ ω (17)holds whp.Let t = ρn , noting that t ∼ εn if ε →
0, and that t is the uniquepositive solution to f ( t ) = 0. Let t − = t − t and t +1 = t + t . Note that t +1 = O ( εn ) = O ( σ ). From (17) and Lemma 5 we have thatsup t ≤ min { T ,t +1 } | X t − f ( t ) | ≤ √ ωσ (18)holds whp.Let a = − f ′ ( t ), so from (15) and the definition of t we have a = − f ′ ( t ) = 1 − p ( n − t ) = 1 − λ (1 − ρ ) = 1 − λ ∗ , where λ ∗ is the dual branching process parameter to λ . In particular, a = Θ( ε ).Since f ( t ) = 0 and f ′′ ( t ) is uniformly O (1 /n ), recalling that t = o ( εn ) itfollows easily that f ( t − ) and f ( t +1 ) are both of order εt = ωσ . To be concrete,if n is large enough, then we certainly have f ( t − ) ≥ √ ωσ and f ( t +1 ) ≤ − √ ωσ , say. Since f ( t ) ≥ εt / ≥ √ ωσ and f is unimodal, we have inf t ≤ t ≤ t − f ( t ) ≥ √ ωσ . Let B denote the event described in (18). Then, whenever B holds,we have X t ≥ t ≤ t ≤ min { T , t − } . Since X T ≤ − Z − <
0, this implies T > t − .Recall from Lemma 6 that (crudely) Z ≤ σ whp. Suppose Z ≤ σ , B holds,and T > t +1 . Then from B and the bound on f ( t +1 ) we have X t +1 ≤ − √ ωσ < − Z , contradicting T > t +1 . It follows that T ≤ t +1 holds whp.At this point we have shown that | T − t | ≤ t holds whp, which gives | T − T − t | ≤ t . Since ω may tend to infinity arbitrarily slowly, this alreadyshows that T − T = t + O p ( σ /ε ) = ρn + O p ( √ ε − n ). To go further, we nextanalyze the distribution of X t more precisely.From Lemma 6 and the bound T > t − whp just proved, whp we have C t − = Z ≤ σ /ω . Noting that t = t − t − = o ( n ), it follows that E C t = o ( n ).Lemma 3 and Lemma 5 thus give | X t − f ( t ) | = o p ( n ), uniformly in t ≤ t . Since X t − f ( t ) is deterministically bounded by n , it follows that E | X t − f ( t ) | and hence E | X t + C t +1 − f ( t ) | are o ( n ), uniformly in t ≤ t . Let u t = n − t − f ( t ) = ne − pt .Since U ′ t = n − t − ( X t + C t +1 ), we have shown that E t − X t =0 | U ′ t − u t | = o ( t n ) = o ( εn ) . (19)8ote that p (1 − p ) t − X t =0 (1 − p ) − t u t ∼ p t − X t =0 e pt ne − pt ∼ n p Z ρ e λx d x = nλλ − ( e λρ −
1) = nρ/ (1 − ρ ) , (20)using e − λρ = 1 − ρ in the last step. Lemma 7.
The distribution of S t is asymptotically normal with mean 0 andvariance nρ/ (1 − ρ ) .Proof. Recall that ( S t ) is a martingale with S = 0, and that the conditionaldistribution of the i th difference (1 − p ) − i ∆ i is (1 − p ) − i times a centered binomialCBi( U ′ i − , p ), and has conditional variance given by (13). The result followseasily by a standard martingale central limit theorem such Brown [2, Theorem2]. Note that here the differences are not uniformly bounded. However, we canwrite ∆ i as the sum of a random number U ′ i − of CBi(1 , p ) random variables,plus n − U ′ i − zero variables. We can take the new variables multiplied by(1 − p ) − i as the differences of a martingale ( S ′ j ) with the property that S t = S ′ nt .In this way we obtain a martingale with the same (random) final value in whichthe differences are bounded by (1 − p ) − n = O (1). The (random) sum of the(old or new) conditional variances is exactly s = P t − t =0 (1 − p ) − t U ′ t − p (1 − p ).By (19) and (20) the ratio of s to nρ/ (1 − ρ ) converges to 1 in probability, asrequired for the martingale central limit theorem.To relate the distribution of T to that of X t (or e X t ) we use the factthat ( X t ) has slope approximately − a near t ; a similar argument was given byMartin-L¨of [5]. Lemma 8.
We have sup | t − t |≤ t | e X t − e X t − a ( t − t ) | = o p ( σ ) . Proof.
From (11) we may write e X t − e X t as x t − x t +(1 − p ) t S t − (1 − p ) t S t = ( f ( t ) − f ( t ))+(1 − p ) t S t − (1 − p ) t S t + O (1) . Recalling that f ′ ( t ) = − a and f ′′ ( t ) = O (1 /n ) uniformly in t , the differencebetween the first term and a ( t − t ) is O ( | t − t | /n ) = O ( t /n ) = o ( σ ). Forthe rest, note that | (1 − p ) t − (1 − p ) t | ≤ | − (1 − p ) | t − t | | ≤ p | t − t | ≤ pt . Since S t = O p ( √ t ) and pt √ t = O ( n − ωσ ε − √ εn ) = o ( σ ), it thus sufficesto show that sup | t − t |≤ t | S t − S t | = o p ( σ ). But this follows easily by applyingLemma 4 to the martingale ( S t − S t − ) t +1 t = t − , which has final variance O ( t ) = o ( σ ). 9 roof of Theorem 1. Recall from Lemma 6 that Z , the number of componentsexplored by time t , satisfies Z = o p ( σ ). We have shown above that whp T = inf { t : X t = − Z − } lies between t − and t +1 . From (17), X t is within o p ( σ ) of e X t at least until T . It follows that at time T , we have e X t = o p ( σ ).Since a = Θ( ε ), Lemma 8 thus gives T = t + e X t /a + o p ( σ /ε ) . (21)From Lemma 7, (11) and the fact that f ( t ) = 0, we have that e X t is asymp-totically normal with mean 0 and variance(1 − p ) t nρ/ (1 − ρ ) ∼ e − λρ nρ/ (1 − ρ ) = nρ (1 − ρ ) . Hence e X t /a is asymptotically normal with mean 0 and variance nρ (1 − ρ ) /a = σ . Since this variance is of order ε − n = ε − σ , the o p ( σ /ε ) error term in (21)is irrelevant, and T is asymptotically normal with mean t = ρn and variance σ . Finally, from Lemma 6 we have T = o p ( σ /ε ). It follows that T − T isasymptotically normal with the parameters claimed in the theorem.This shows the existence of a component with the claimed size. As shown byNachmias and Peres [6], it is easy to check that the rest of the graph correspondsto a subcritical random graph, and whp will not contain a larger component. Acknowledgement.
We are grateful to an anonymous referee for several sug-gestions improving the presentation of the paper.
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