Distances in random graphs with finite mean and infinite variance degrees

Transcription

1 E e c t r o n i c J o u r n a o f P r o b a b i i t y Vo , Paper no. 25, pages Journa URL Distances in rando graphs with finite ean and infinite variance degrees Reco van der Hofstad Gerard Hooghiestra and Ditri Znaensi Abstract In this paper we study typica distances in rando graphs with i.i.d. degrees of which the tai of the coon distribution function is reguary varying with exponent 1 τ. Depending on the vaue of the paraeter τ we can distinct three cases: i τ >, where the degrees have finite variance, ii τ 2,, where the degrees have infinite variance, but finite ean, and iii τ 1, 2, where the degrees have infinite ean. The distances between two randoy chosen nodes beonging to the sae connected coponent, for τ > and τ 1, 2, have been studied in previous pubications, and we survey these resuts here. When τ 2,, the graph distance centers around 2 og og N/ ogτ 2. We present a fu proof of this resut, and study the fuctuations around this asyptotic eans, by describing the asyptotic distribution. The resuts presented here iprove upon resuts of Reittu and Norros, who prove an upper bound ony. The rando graphs studied here can serve as odes for copex networs where degree power aws are observed; this is iustrated by coparing the typica distance in this ode to Internet data, where a degree power aw with exponent τ 2.2 is observed for the socaed Autonoous Systes AS graph. Key words: Branching processes, configuration ode, couping, graph distance. Departent of Matheatics and Coputer Science, Eindhoven University of Technoogy, P.O. Box 51, 5600 MB Eindhoven, The Netherands. E-ai: [email protected] Supported in part by Netherands Organization for Scientific Research NWO. Deft University of Technoogy, Eectrica Engineering, Matheatics and Coputer Science, P.O. Box 501, 2600 GA Deft, The Netherands. E-ai: [email protected] EURANDOM, P.O. Box 51, 5600 MB Eindhoven, The Netherands. E-ai: [email protected] 70

2 AMS 2000 Subect Cassification: Priary 05C80; Secondary: 05C12, 60J80. Subitted to EJP on February 27, 2007, fina version accepted Apri 10,

3 1 Introduction Copex networs are encountered in a wide variety of discipines. A rough cassification has been given by Newan 18 and consists of: i Technoogica networs, e.g. eectrica power grids and the Internet, ii Inforation networs, such as the Word Wide Web, iii Socia networs, ie coaboration networs and iv Bioogica networs ie neura networs and protein interaction networs. What any of the above exapes have in coon is that the typica distance between two nodes in these networs are sa, a phenoenon that is dubbed the sa-word phenoenon. A second ey phenoenon shared by any of those networs is their scae-free nature; eaning that these networs have so-caed power-aw degree sequences, i.e., the nuber of nodes with degree fas of as an inverse power of. We refer to 1; 18; 25 and the references therein for a further introduction to copex networs and any ore exapes where the above two properties hod. A rando graph ode where both the above ey features are present is the configuration ode appied to an i.i.d. sequence of degrees with a power-aw degree distribution. In this ode we start by saping the degree sequence fro a power aw and subsequenty connect nodes with the saped degree purey at rando. This ode autoaticay satisfies the power aw degree sequence and it is therefore of interest to rigorousy derive the typica distances that occur. Together with two previous papers 10; 14, the current paper describes the rando fuctuations of the graph distance between two arbitrary nodes in the configuration ode, where the i.i.d. degrees foow a power aw of the for PD > = τ+1 L, where L denotes a sowy varying function and the exponent τ satisfies τ 1. To obtain a copete picture we incude a discussion and a heuristic proof of the resuts in 10 for τ 1,2, and those in 14 for τ >. However, the ain goa of this paper is the copete description, incuding a fu proof of the case where τ 2,. Apart fro the critica cases τ = 2 and τ =, which depend on the behavior of the sowy varying function L see 10, Section 4.2 when τ = 2, we have thus given a copete anaysis for a possibe vaues of τ 1. This section is organized as foows. In Section 1.1, we start by introducing the ode, in Section 1.2 we state our ain resuts. Section 1. is devoted to reated wor, and in Section 1.4, we describe soe siuations for a better understanding of our ain resuts. Finay, Section 1.5 describes the organization of the paper. 1.1 Mode definition Fix an integer N. Consider an i.i.d. sequence D 1,D 2,...,D N. We wi construct an undirected graph with N nodes where node has degree D. We assue that = N =1 D is even. If is odd, then we increase D N by 1. This singe change wi ae hardy any difference in what foows, and we wi ignore this effect. We wi ater specify the distribution of D 1. To construct the graph, we have N separate nodes and incident to node, we have D stubs or haf-edges. A stubs need to be connected to buid the graph. The stubs are nubered in any given order fro 1 to. We start by connecting at rando the first stub with one of the 705

4 1 reaining stubs. Once paired, two stubs haf-edges for a singe edge of the graph. Hence, a stub can be seen as the eft or the right haf of an edge. We continue the procedure of randoy choosing and pairing the stubs unti a stubs are connected. Unfortunatey, nodes having sef-oops ay occur. However, sef-oops are scarce when N, as shown in 5. The above ode is a variant of the configuration ode, which, given a degree sequence, is the rando graph with that given degree sequence. The degree sequence of a graph is the vector of which the th coordinate equas the proportion of nodes with degree. In our ode, by the aw of arge nubers, the degree sequence is cose to the probabiity ass function of the noda degree D of which D 1,...,D N are independent copies. The probabiity ass function and the distribution function of the noda degree aw are denoted by x PD 1 = = f, = 1,2,..., and Fx = f, 1.1 where x is the argest integer saer than or equa to x. We consider distributions of the for =1 1 Fx = x τ+1 Lx, 1.2 where τ > 1 and L is sowy varying at infinity. This eans that the rando variabes D obey a power aw, and the factor L is eant to generaize the ode. We assue the foowing ore specific conditions, spitting between the cases τ 1, 2, τ 2, and τ >. Assuption 1.1. i For τ 1, 2, we assue 1.2. ii For τ 2,, we assue that there exists γ 0, 1 and C > 0 such that x τ+1 Cog xγ 1 1 Fx x τ+1+cog xγ 1, for arge x. 1. iii For τ >, we assue that there exists a constant c > 0 such that and that ν > 1, where ν is given by 1 Fx cx τ+1, for a x 1, 1.4 ν = ED 1D 1 1]. 1.5 ED 1 ] Distributions satisfying 1.4 incude distributions which have a ighter tai than a power aw, and 1.4 is ony sighty stronger than assuing finite variance. The condition in 1. is sighty stronger than Main resuts We define the graph distance H N between the nodes 1 and 2 as the iniu nuber of edges that for a path fro 1 to 2. By convention, the distance equas if 1 and 2 are not connected. Observe that the distance between two randoy chosen nodes is equa in distribution to H N, 706

5 because the nodes are exchangeabe. In order to state the ain resut concerning H N, we define the centering constant { 2 og og N ogτ 2, for τ 2,, τ,n = 1.6 og ν N, for τ >. The paraeter τ,n describes the asyptotic growth of H N as N. A ore precise resut incuding the rando fuctuations around τ,n is foruated in the foowing theore. Theore 1.2 The fuctuations of the graph distance. When Assuption 1.1 hods, then i for τ 1,2, where p = p F 0,1. i PH N = 2 = 1 i PH N = = p, 1.7 N N ii for τ 2, or τ > there exist rando variabes R τ,a a 1,0], such that as N, P H N = τ,n + H N < = PR τ,an = + o1, 1.8 where a N = { og og N og og N ogτ 2 ogτ 2, for τ 2,, og ν N og ν N, for τ >. We see that for τ 1,2, the iit distribution exists and concentrates on the two points 2 and. For τ 2, or τ > the iit behavior is ore invoved. In these cases the iit distribution does not exist, caused by the fact that the correct centering constants, 2og og N/ ogτ 2, for τ 2, and og ν N, for τ >, are in genera not integer, whereas H N is with probabiity 1 concentrated on the integers. The above theore cais that for τ 2, or τ > and arge N, we have H N = τ,n + O p 1, with τ,n specified in 1.6 and where O p 1 is a rando contribution, which is tight on R. The specific for of this rando contribution is specified in Theore 1.5 beow. In Theore 1.2, we condition on H N <. In the course of the proof, here and in 14, we aso investigate the probabiity of this event, and prove that PH N < = q 2 + o1, 1.9 where q is the surviva probabiity of an appropriate branching process. Coroary 1. Convergence in distribution aong subsequences. For τ 2, or τ >, and when Assuption 1.1 is fufied, we have that, for, H N τ,n H N < 1.10 converges in distribution to R τ,a, aong subsequences N where a N converges to a. A siuation for τ 2, iustrating the wea convergence in Coroary 1. is discussed in Section

6 Coroary 1.4 Concentration of the hopcount. For τ 2, or τ >, and when Assuption 1.1 is fufied, we have that the rando variabes H N τ,n, given that H N <, for a tight sequence, i.e., i K i sup P H N τ,n K N H N < = We next describe the aws of the rando variabes R τ,a a 1,0]. For this, we need soe further notation fro branching processes. For τ > 2, we introduce a deayed branching process {Z } 1, where in the first generation the offspring distribution is chosen according to 1.1 and in the second and further generations the offspring is chosen in accordance to g given by g = + 1f +1, = 0,1,..., where µ = ED 1 ] µ When τ 2,, the branching process {Z } has infinite expectation. Under Assuption 1.1, it is proved in 8 that i τ n 2n ogz n 1 = Y, a.s., 1.1 where x y denotes the axiu of x and y. When τ >, the process {Z n /µν n 1 } n 1 is a non-negative artingae and consequenty i n Z n = W, a.s µνn 1 The constant q appearing in 1.9 is the surviva probabiity of the branching process {Z } 1. We can identify the iit aws of R τ,a a 1,0] in ters of the iit rando variabes in 1.1 and 1.14 as foows: Theore 1.5 The iit aws. When Assuption 1.1 hods, then i for τ 2, and for a 1,0], PR τ,a > = P in τ 2 s Y 1 +τ 2 s c Y 2] τ 2 /2 +a Y 1 Y 2 > 0, 1.15 s Z where c = 1 if is even and zero otherwise, and Y 1,Y 2 are two independent copies of the iit rando variabe in 1.1. ii for τ > and for a 1,0], PR τ,a > = E exp{ κν a+ W 1 W 2 } W 1 W 2 > 0 ], 1.16 where W 1 and W 2 are two independent copies of the iit rando variabe W in 1.14 and where κ = µν 1 1. The above resuts prove that the scaing in these rando graphs is quite sensitive to the degree exponent τ. The scaing of the distance between pairs of nodes is proved for a τ 1, except for the critica cases τ = 2 and τ =. The resut for τ 1,2, and the case τ = 1, where P H N 2, are both proved in 10, the resut for τ > is proved in 14. In Section 2 we wi present heuristic proofs for a three cases, and in Section 4 a fu proof for the case where 708

7 τ 2,. Theores quantify the sa-word phenoenon for the configuration ode, and expicity divide the scaing of the graph distances into three distinct regies In Rears 4.2 and A.1.5 beow, we wi expain that our resuts aso appy to the usua configuration ode, where the nuber of nodes with a given degree is deterinistic, when we study the graph distance between two unifory chosen nodes, and the degree distribution satisfied certain conditions. For the precise conditions, see Rear A.1.5 beow. 1. Reated wor There are any papers on scae-free graphs and we refer to reviews such as the ones by Abert and Barabási 1, Newan 18 and the recent boo by Durrett 9 for an introduction; we refer to 2; ; 17 for an introduction to cassica rando graphs. Papers invoving distances for the case where the degree distribution F see 1.2, has exponent τ 2, are not so wide spread. In this discussion we wi focus on the case where τ 2,. For reated wor on distances for the cases τ 1,2 and τ > we refer to 10, Section 1.4 and 14, Section 1.4, respectivey. The ode investigated in this paper with τ 2, was first studied in 21, where it was shown that with probabiity converging to 1, H N is ess than τ,n 1 + o1. We iprove the resuts in 21 by deriving the asyptotic distribution of the rando fuctuations of the graph distance around τ,n. Note that these resuts are in contrast to 19, Section II.F, beow Equation 56, where it was suggested that if τ <, then an exponentia cut-off is necessary to ae the graph distance between an arbitrary pair of nodes we-defined. The probe of the ean graph distance between an arbitrary pair of nodes was aso studied non-rigorousy in 7, where aso the behavior when τ = and x Lx is the constant function, is incuded. In the atter case, og N ogog N the graph distance scaes ie. A reated ode to the one studied here can be found in 20, where a Poissonian graph process is defined by adding and reoving edges. In 20, the authors prove siiar resuts as in 21 for this reated ode. For τ 2,, in 15, it was further shown that the diaeter of the configuration ode is bounded beow by a constant ties og N, when f 1 + f 2 > 0, and bounded above by a constant ties og og N, when f 1 + f 2 = 0. A second reated ode can be found in 6, where edges between nodes i and are present with probabiity equa to w i w / w for soe expected degree vector w = w 1,...,w N. It is assued that ax i wi 2 < i w i, so that w i w / w are probabiities. In 6, w i is often taen as w i = ci 1 1 τ 1, where c is a function of N proportiona to N τ 1. In this case, the degrees obey a power aw with exponent τ. Chung and Lu 6 show that in this case, the graph distance between two unifory chosen nodes is with probabiity converging to 1 proportiona to og N1 + o1 og og N when τ >, and to 2 ogτ o1 when τ 2,. The difference between this ode and ours is that the nodes are not exchangeabe in 6, but the observed phenoena are siiar. This resut can be heuristicay understood as foows. Firsty, the actua degree vector in 6 shoud be cose to the expected degree vector. Secondy, for the expected degree vector, we can copute that the nuber of nodes for which the degree is at east equas {i : w i } = {i : ci 1 τ 1 } τ+1. Thus, one expects that the nuber of nodes with degree at east decreases as τ+1, siiary as in our ode. The ost genera version of this ode can be found in 4. A these odes 709

8 Figure 1: Histogras of the AS-count and graph distance in the configuration ode with N = 10,940, where the degrees have generating function f τ s in 1.18, for which the power aw exponent τ taes the vaue τ = The AS-data is ighty shaded, the siuation is dary shaded. assue soe for of conditiona independence of the edges, which resuts in asyptotic degree sequences that are given by ixed Poisson distributions see e.g. 5. In the configuration ode, instead, the degrees are independent. 1.4 Deonstration of Coroary 1. Our otivation to study the above version of the configuration ode is to describe the topoogy of the Internet at a fixed tie instant. In a seina paper 12, Faoutsos et a. have shown that the degree distribution in Internet foows a power aw with exponent τ Thus, the power aw rando graph with this vaue of τ can possiby ead to a good Internet ode. In 24, and inspired by the observed power aw degree sequence in 12, the power aw rando graph is proposed as a ode for the networ of autonoous systes. In this graph, the nodes are the autonoous systes in the Internet, i.e., the parts of the Internet controed by a singe party such as a university, copany or provider, and the edges represent the physica connections between the different autonoous systes. The wor of Faoutsos et a. in 12 was aong others on this graph which at that tie had size approxiatey 10,000. In 24, it is argued on a quaitative basis that the power aw rando graph serves as a better ode for the Internet topoogy than the currenty used topoogy generators. Our resuts can be seen as a step towards the quantitative understanding of whether the AS-count in Internet is described we by the graph distance in the configuration ode. The AS-count gives the nuber of physica ins connecting the various autonoous doains between two randoy chosen doains. To vaidate the ode studied here, we copare a siuation of the distribution of the distance between pairs of nodes in the configuration ode with the sae vaue of N and τ to extensive easureents of the AS-count in Internet. In Figure 1, we see that the graph distance in the ode with the predicted vaue of τ = 2.25 and the vaue of N fro the data set fits the AS-count data rearaby we. 710

9 Figure 2: Epirica surviva functions of the graph distance for τ = 2.8 and for the four vaues of N. Having otivated why we are interested to study distances in the configuration ode, we now expain by a siuation the reevance of Theore 1.2 and Coroary 1. for τ 2,. We have chosen to siuate the distribution 1.12 using the generating function: τ 2 g τ s = 1 1 s τ 2, for which g = 1 1 c τ 1, Defining it is iediate that f τ s = τ sτ 1 s, τ 2,, 1.18 τ 2 τ 2 g τ s = f τ s f τ 1, so that g = + 1f +1. µ For fixed τ, we can pic different vaues of the size of the siuated graph, so that for each two siuated vaues N and M we have a N = a M, i.e., N = M τ 2, for soe integer. For τ = 2.8, this induces, starting fro M = 1000, by taing for the successive vaues 1,2,, M = 1,000, N 1 = 5,624, N 2 = 48,697, N = 72,95. According to Coroary 1., the surviva functions of the hopcount H N, given by PH N > H N <, and for N = M τ 2, run approxiatey parae on distance 2 in the iit for N, since τ,n = τ,m + 2 for = 1,2,. In Section.1 beow we wi show that the distribution with generating function 1.18 satisfies Assuption 1.1ii. 711

10 1.5 Organization of the paper The paper is organized as foows. In Section 2 we heuristicay expain our resuts for the three different cases. The reevant iterature on branching processes with infinite ean. is reviewed in Section, where we aso describe the growth of shortest path graphs, and state couping resuts needed to prove our ain resuts, Theores in Section 4. In Section 5, we prove three technica eas used in Section 4. We finay prove the couping resuts in the Appendix. In the seque we wi write that event E occurs whp for the stateent that PE = 1 o1, as the tota nuber of nodes N. 2 Heuristic expanations of Theores 1.2 and 1.5 In this section, we present a heuristic expanation of Theores 1.2 and 1.5. When τ 1,2, the tota degree is the i.i.d. su of N rando variabes D 1,D 2,...,D N, with infinite ean. Fro extree vaue theory, it is we nown that then the bu of the contribution to coes fro a finite nuber of nodes which have giant degrees the so-caed giant nodes. Since these giant nodes have degree roughy N 1/τ 1, which is uch arger than N, they are a connected to each other, thus foring a copete graph of giant nodes. Each stub of node 1 or node 2 is with probabiity cose to 1 attached to a stub of soe giant node, and therefore, the distance between any two nodes is, whp, at ost. In fact, this distance equas 2 precisey when the two nodes are attached to the sae giant node, and is otherwise. For τ = 1 the quotient M N /, where M N denotes the axiu of D 1,D 2,...,D N, converges to 1 in probabiity, and consequenty the asyptotic distance is 2 in this case, as basicay a nodes are attached to the unique giant node. As entioned before, fu proofs of these resuts can be found in 10. For τ 2, or τ > there are two basic ingredients underying the graph distance resuts. The first one is that for two disoint sets of stubs of sizes n and out of a tota of L, the probabiity that none of the stubs in the first set is attached to a stub in the second set, is approxiatey equa to n 1 i= L n 2i In fact, the product in 2.1 is precisey equa to the probabiity that none of the n stubs in the first set of stubs is attached to a stub in the second set, given that no two stubs in the first set are attached to one another. When n = ol, L, however, these two probabiities are asyptoticay equa. We approxiate 2.1 further as n 1 i=0 1 { n 1 exp L n 2i i=0 og n + 2i } e n L, 2.2 L L where the approxiation is vaid as ong as nn + = ol 2, when L. The shortest path graph SPG fro node 1 is the union of a shortest paths between node 1 and a other nodes {2,...,N}. We define the SPG fro node 2 in a siiar fashion. We appy the above heuristic asyptotics to the growth of the SPG s. Let Z 1,N denote the nuber of stubs 712

11 that are attached to nodes precisey 1 steps away fro node 1, and siiary for Z 2,N. We then appy 2.2 to n = Z 1,N, = Z 2,N and L =. Let Q, Z be the conditiona distribution given {Z s 1,N } s=1 and {Z2,N s } s=1. For = 0, we ony condition on {Z1,N s } s=1. For 1, we have the utipication rue see 14, Lea 4.1, PH N > = E +1 i=2 ] Q i/2, i/2 Z H N > i 1 H N > i 2, 2. where x is the saest integer greater than or equa to x and x the argest integer saer than or equa to x. Now fro 2.1 and 2.2 we find, } Q i/2, i/2 Z H N > i 1 H N > i 2 exp { Z1,N i/2 Z2,N i/ This asyptotic identity foows because the event {H N > i 1 H N > i 2} occurs precisey when none of the stubs Z 1,N i/2 attaches to one of those of Z2,N i/2. Consequenty we can approxiate PH N > E exp { 1 +1 A typica vaue of the hopcount H N is the vaue for which i=2 Z 1,N i/2 Z2,N i/2 }] i=2 Z 1,N i/2 Z2,N i/2 1. This is the first ingredient of the heuristic. The second ingredient is the connection to branching processes. Given any node i and a stub attached to this node, we attach the stub to a second stub to create an edge of the graph. This chosen stub is attached to a certain node, and we wish to investigate how any further stubs this node has these stubs are caed brother stubs of the chosen stub. The conditiona probabiity that this nuber of brother stubs equas n given D 1,...,D N, is approxiatey equa to the probabiity that a rando stub fro a = D D N stubs is attached to a node with in tota n + 1 stubs. Since there are precisey N =1 n + 11 {D =n+1} stubs that beong to a node with degree n + 1, we find for the atter probabiity g N n = n + 1 N 1 {D =n+1}, 2.6 =1 where 1 A denotes the indicator function of the event A. The above forua coes fro saping with repaceent, whereas in the SPG the saping is perfored without repaceent. Now, as we grow the SPG s fro nodes 1 and 2, of course the nuber of stubs that can sti be chosen decreases. However, when the size of both SPG s is uch saer than N, for instance at ost N, or sighty bigger, this dependence can be negected, and it is as if we choose each tie independenty and with repaceent. Thus, the growth of the SPG s is cosey reated to a branching process with offspring distribution {g n N } n=1. 71

12 When τ > 2, using the strong aw of arge nubers for N, N µ = ED 1], and 1 N N 1 {D =n+1} f n+1 = PD 1 = n + 1, =1 so that, aost surey, g N n n + 1f n+1 µ = g n, N. 2.7 Therefore, the growth of the shortest path graph shoud be we described by a branching process with offspring distribution {g n }, and we coe to the question what is a typica vaue of for which +1 i=2 Z 1 i/2 Z2 i/2 = µn, 2.8 where {Z 1 } and {Z 2 } denote two independent copies of a deayed branching process with offspring distribution {f n }, f n = PD = n, n = 1,2,..., in the first generation and offspring distribution {g n } in a further generations. To answer this question, we need to ae separate arguents depending on the vaue of τ. When τ >, then ν = n 1 ng n <. Assue aso that ν > 1, so that the branching process is supercritica. In this case, the branching process Z /µν 1 converges aost surey to a rando variabe W see Hence, for the two independent branching processes {Z i }, i = 1,2, that ocay describe the nuber of stubs attached to nodes on distance 1, we find that, for, Z i µν 1 W i. 2.9 This expains why the average vaue of Z i,n grows ie µν 1 = µ exp 1og ν, that is, exponentia in for ν > 1, so that a typica vaue of for which 2.8 hods satisfies µ ν 1 = N, or = og ν N/µ + 1. We can extend this arguent to describe the fuctuation around the asyptotic ean. Since 2.9 describes the fuctuations of Z i around the ean vaue µν 1, we are abe to describe the rando fuctuations of H N around og ν N. The detais of these proofs can be found in 14. When τ 2,, the branching processes {Z 1 } and {Z 2 } are we-defined, but they have infinite ean. Under certain conditions on the underying offspring distribution, which are ipied by Assuption 1.1ii, Davies 8 proves for this case that τ 2 ogz +1 converges aost surey, as, to soe rando variabe Y. Moreover, PY = 0 = 1 q, the extinction probabiity of {Z } =0. Therefore, aso τ 2 ogz 1 converges aost surey to Y. Since τ > 2, we sti have that µn. Furtherore by the doube exponentia behavior of Z i, the size of the eft-hand side of 2.8 is equa to the size of the ast ter, so that the typica vaue of for which 2.8 hods satisfies Z 1 +1/2 Z2 +1/2 µn, or ogz1 +1/2 1 + ogz2 +1/2 1 og N. This indicates that the typica vaue of is of order og og N 2 ogτ 2,

13 as foruated in Theore 1.2ii, since if for soe c 0,1 ogz 1 +1/2 1 cog N, ogz2 +1/2 1 1 cog N then + 1/2 = ogcog N/ ogτ 2, which induces the eading order of τ,n defined in 1.6. Again we stress that, since Davies resut 8 describes a distributiona iit, we are abe to describe the rando fuctuations of H N around τ,n. The detais of the proof are given in Section 4. The growth of the shortest path graph In this section we describe the growth of the shortest path graph SPG. This growth reies heaviy on branching processes BP s. We therefore start in Section.1 with a short review of the theory of BP s in the case where the expected vaue ean of the offspring distribution is infinite. In Section.2, we discuss the couping between these BP s and the SPG, and in Section., we give the bounds on the couping. Throughout the reaining sections of the seque we wi assue that τ 2,, and that F satisfies Assuption 1.1ii..1 Review of branching processes with infinite ean In this review of BP s with infinite ean we foow in particuar 8, and aso refer the readers to reated wor in 22; 2, and the references therein. For the fora definition of the BP we define a doube sequence {X n,i } n 0,i 1 of i.i.d. rando variabes each with distribution equa to the offspring distribution {g } given in 1.12 with distribution function Gx = x =0 g. The BP {Z n } is now defined by Z 0 = 1 and Z n Z n+1 = X n,i, n 0. i=1 In case of a deayed BP, we et X 0,1 have probabiity ass function {f }, independenty of {X n,i } n 1. In this section we restrict to the non-deayed case for sipicity. We foow Davies in 8, who gives the foowing sufficient conditions for convergence of τ 2 n og1 + Z n. Davies ain theore states that if there exists a non-negative, nonincreasing function γx, such that, i x ζ γx 1 Gx x ζ+γx, for arge x and 0 < ζ < 1, ii x γx is non-decreasing, iii 0 γe ex dx <, or, equivaenty, γy e y og y dy <, then ζ n og1 + Z n converges aost surey to a non-degenerate finite rando variabe Y with PY = 0 equa to the extinction probabiity of {Z n }, whereas Y Y > 0 adits a density on 0,. Therefore, aso ζ n ogz n 1 converges to Y aost surey. 715

14 The conditions of Davies quoted as i-iii sipify earier wor by Seneta 2. For exape, for {g } in 1.17, the above is vaid with ζ = τ 2 and γx = Cog x 1, where C is sufficienty arge. We prove in Lea A.1.1 beow that for F as in Assuption 1.1ii, and G the distribution function of {g } in 1.12, the conditions i-iii are satisfied with ζ = τ 2 and γx = Cog x γ 1, with γ < 1. Let Y 1 and Y 2 be two independent copies of the iit rando variabe Y. In the course of the proof of Theore 1.2, for τ 2,, we wi encounter the rando variabe U = in t Z κ t Y 1 + κ c t Y 2, for soe c {0,1}, and where κ = τ 2 1. The proof reies on the fact that, conditionay on Y 1 Y 2 > 0, U has a density. The proof of this fact is as foows. The function y 1,y 2 in t Z κ t y 1 + κ c t y 2 is discontinuous precisey in the points y 1,y 2 satisfying y 2 /y 1 = κ n 1 2 c, n Z, and, conditionay on Y 1 Y 2 > 0, the rando variabes Y 1 and Y 2 are independent continuous rando variabes. Therefore, conditionay on Y 1 Y 2 > 0, the rando variabe U = in t Z κ t Y 1 + κ c t Y 2 has a density..2 Couping of SPG to BP s In Section 2, it has been shown inforay that the growth of the SPG is cosey reated to a BP } with the rando offspring distribution {g N } given by 2.6; note that in the notation {Ẑ1,N Ẑ 1,N we do incude its dependence on N, whereas in 14, Section.1 this dependence on N was eft out for notationa convenience. The presentation in Section.2 is virtuay identica to the one in 14, Section. However, we have decided to incude ost of this ateria to eep the paper sef-contained. By the strong aw of arge nubers, g N + 1PD 1 = + 1/ED 1 ] = g, N. Therefore, the BP {Ẑ1,N }, with offspring distribution {g N }, is expected to be cose to the BP {Z 1 } with offspring distribution {g } given in So, in fact, the couping that we ae is two-fod. We first coupe the SPG to the N dependent branching process {Ẑ1,N }, and consecutivey we coupe {Ẑ1,N } to the BP {Z 1 }. In Section., we state bounds on these coupings, which aow us to prove Theores 1.2 and 1.5 of Section 1.2. The shortest path graph SPG fro node 1 consists of the shortest paths between node 1 and a other nodes {2,...,N}. As wi be shown beow, the SPG is not necessariy a tree because cyces ay occur. Reca that two stubs together for an edge. We define Z 1,N 1 = D 1 and, for 2, we denote by Z 1,N the nuber of stubs attached to nodes at distance 1 fro node 1, but are not part of an edge connected to a node at distance 2. We refer to such stubs as free stubs, since they have not yet been assigned to a second stub to fro an edge. Thus, Z 1,N is the nuber of outgoing stubs fro nodes at distance 1 fro node 1. By SPG 1 we denote the SPG up to eve 1, i.e., up to the oent we have Z 1,N free stubs attached to nodes on distance 1, and no stubs to nodes on distance. Since we copare Z 1,N to the th generation of the BP Ẑ1,N, we ca Z 1,N the stubs of eve. For the copete description of the SPG {Z 1,N }, we have introduced the concept of abes in 14, Section. These abes iustrate the resebances and the differences between the SPG {Z 1,N } and the BP {Ẑ1,N }. 716

15 SPG stubs with their abes Figure : Scheatic drawing of the growth of the SPG fro the node 1 with N = 9 and the updating of the abes. The stubs without a abe are understood to have abe 1. The first ine shows the N different nodes with their attached stubs. Initiay, a stubs have abe 1. The growth process starts by choosing the first stub of node 1 whose stubs are abeed by 2 as iustrated in the second ine, whie a the other stubs aintain the abe 1. Next, we unifory choose a stub with abe 1 or 2. In the exape in ine, this is the second stub fro node, whose stubs are abeed by 2 and the second stub by abe. The eft hand side coun visuaizes the growth of the SPG by the attachent of stub 2 of node to the first stub of node 1. Once an edge is estabished the paired stubs are abeed. In the next step, again a stub is chosen unifory out of those with abe 1 or 2. In the exape in ine 4, it is the first stub of the ast node that wi be attached to the second stub of node 1, the next in sequence to be paired. The ast ine exhibits the resut of creating a cyce when the first stub of node is chosen to be attached to the second stub of node 9 the ast node. This process is continued unti there are no ore stubs with abes 1 or 2. In this exape, we have Z 1,N 1 = and Z 1,N 2 = 6. Initiay, a stubs are abeed 1. At each stage of the growth of the SPG, we draw unifory at rando fro a stubs with abes 1 and 2. After each draw we wi update the reaization of the SPG according to three categories, which wi be abeed 1, 2 and. At any stage of the generation of the SPG, the abes have the foowing eaning: 1. Stubs with abe 1 are stubs beonging to a node that is not yet attached to the SPG. 2. Stubs with abe 2 are attached to the SPG because the corresponding node has been chosen, but not yet paired with another stub. These are the free stubs entioned above.. Stubs with abe in the SPG are paired with another stub to for an edge in the SPG. 717

16 The growth process as depicted in Figure starts by abeing a stubs by 1. Then, because we construct the SPG starting fro node 1 we reabe the D 1 stubs of node 1 with the abe 2. We note that Z 1,N 1 is equa to the nuber of stubs connected to node 1, and thus Z 1,N 1 = D 1. We next identify Z 1,N for > 1. Z 1,N is obtained by sequentiay growing the SPG fro the free stubs in generation Z 1,N 1. When a free stubs in generation 1 have chosen their connecting stub, Z 1,N is equa to the nuber of stubs abeed 2 i.e., free stubs attached to the SPG. Note that not necessariy each stub of Z 1,N 1 contributes to stubs of Z1,N, because a cyce ay swaow two free stubs. This is the case when a stub with abe 2 is chosen. After the choice of each stub, we update the abes as foows: 1. If the chosen stub has abe 1, we connect the present stub to the chosen stub to for an edge and attach the brother stubs of the chosen stub as chidren. We update the abes as foows. The present and chosen stub et together to for an edge and both are assigned abe. A brother stubs receive abe When we choose a stub with abe 2, which is aready connected to the SPG, a sef-oop is created if the chosen stub and present stub are brother stubs. If they are not brother stubs, then a cyce is fored. Neither a sef-oop nor a cyce changes the distances to the root in the SPG. The updating of the abes soey consists of changing the abe of the present and the chosen stubs fro 2 to. The above process stops in the th generation when there are no ore free stubs in generation 1 for the SPG, and then Z 1,N is the nuber of free stubs at this tie. We continue the above process of drawing stubs unti there are no ore stubs having abe 1 or 2, so that a stubs have abe. Then, the SPG fro node 1 is finaized, and we have generated the shortest path graph as seen fro node 1. We have thus obtained the structure of the shortest path graph, and now how any nodes there are at a given distance fro node 1. The above construction wi be perfored identicay fro node 2, and we denote the nuber of free stubs in the SPG of node 2 in generation by Z 2,N. This construction is cose to being independent, when the generation size is not too arge. In particuar, it is possibe to coupe the two SPG growth processes with two independent BP s. This is described in detai in 14, Section. We ae essentia use of the couping between the SPG s and the BP s, in particuar, of 14, Proposition A..1 in the appendix. This copetes the construction of the SPG s fro both node 1 and 2.. Bounds on the couping We now investigate the reationship between the SPG {Z i,n } and the BP {Z i } with aw g. These resuts are stated in Proposition.1,.2 and.4. In their stateent, we write, for i = 1,2, Y i,n = τ 2 ogz i,n 1 and Y i = τ 2 ogz i 1,.1 where {Z 1 } 1 and {Z 2 } 1 are two independent deayed BP s with offspring distribution {g } and where Z i 1 has aw {f }. Then the foowing proposition shows that the first eves of the SPG are cose to those of the BP: 718

17 Proposition.1 Couping at fixed tie. If F satisfies Assuption 1.1ii, then for every fixed, and for i = 1,2, there exist independent deayed BP s Z 1, Z 2, such that i,n i PY N = Y i = 1..2 In words, Proposition.1 states that at any fixed tie, the SPG s fro 1 and 2 can be couped to two independent BP s with offspring g, in such a way that the probabiity that the SPG differs fro the BP vanishes when N. In the stateent of the next proposition, we write, for i = 1,2, T i,n = T i,n ε = { > : Z i,n κ 1 ε 2 N = { > : κ Y i,n 1 ε2 τ 1 τ 1 } og N},. where we reca that κ = τ 2 1. We wi see that Z i,n grows super-exponentiay with as ong as T i,n. More precisey, Z i,n is cose to Z i,n κ, and thus, T i,n can be thought of as the generations for which the generation size is bounded by N 1 ε2 τ 1. The second ain resut of the couping is the foowing proposition: Proposition.2 Super-exponentia growth with base Y i,n Assuption 1.1ii, then, for i = 1,2, a P ε Y i,n ε 1, ax T i,n ε Y i,n for arge ties. If F satisfies Y i,n > ε = o N,,ε1,.4 b P ε Y i,n P ε Y i,n ε 1, T i,n ε : Z i,n 1 > Zi,N ε 1, T i,n ε : Z i,n > N 1 ε 4 τ 1 = o N,,ε1,.5 = o N,,ε1,.6 where o N,,ε1 denotes a quantity γ N,,ε that converges to zero when first N, then and finay ε 0. Rear.. Throughout the paper iits wi be taen in the above order, i.e., first we send N, then and finay ε 0. Proposition.2 a, i.e..4, is the ain couping resut used in this paper, and says that as ong as T i,n ε, we have that Y i,n is cose to Y i,n, which, in turn, by Proposition.1, is cose to Y i. This estabishes the couping between the SPG and the BP. Part b is a technica resut used in the proof. Equation.5 is a convenient resut, as it shows that, with high probabiity, Z i,n is onotonicay increasing. Equation.6 shows that with high 4 for a T i,n ε. probabiity Z i,n N 1 ε τ 1 stubs in generation sizes that are in T i,n ε, which aows us to bound the nuber of free We copete this section with a fina couping resut, which shows that for the first which is not in T i,n ε, the SPG has any free stubs: 719

18 Proposition.4 Lower bound on Z i,n +1 Then, P T i,n ε, + 1 T i,n ε,ε Y i,n i,n for +1 T ε. Let F satisfy Assuption 1.1ii. ε 1,Z i,n +1 N 1 ε τ 1 = o N,,ε1..7 Propositions.1,.2 and.4 wi be proved in the appendix. In Section 4 and 5, we wi prove the ain resuts in Theores 1.2 and 1.5 subect to Propositions.1,.2 and.4. 4 Proof of Theores 1.2 and 1.5 for τ 2, For convenience we cobine Theore 1.2 and Theore 1.5, in the case that τ 2,, in a singe theore that we wi prove in this section. Theore 4.1. Fix τ 2,. When Assuption 1.1ii hods, then there exist rando variabes R τ,a a 1,0], such that as N, where a N = by og og N P H N = 2 ogτ 2 + HN < = PR τ,an = + o1, 4.1 og og N ogτ 2 og og N ogτ 2 1,0]. The distribution of R τ,a, for a 1,0], is given PR τ,a > = P in τ 2 s Y 1 + τ 2 s c Y 2] τ 2 /2 +a Y 1 Y 2 > 0, s Z where c = 1 if is even, and zero otherwise, and Y 1,Y 2 are two independent copies of the iit rando variabe in Outine of the proof We start with an outine of the proof. The proof is divided into severa ey steps proved in 5 subsections, Sections In the first ey step of the proof, in Section 4.2, we spit the probabiity PH N > into separate parts depending on the vaues of Y i,n = τ 2 ogz i,n 1. We prove that PH N >,Y 1,N Y 2,N = 0 = 1 q 2 + o1, N, 4.2 where 1 q is the probabiity that the deayed BP {Z 1 } 1 dies at or before the th generation. When becoes arge, then q q, where q equas the surviva probabiity of {Z 1 } 1. This eaves us to deterine the contribution to PH N > for the cases where Y 1,N further show that for arge enough, and on the event that Y i,n for i = 1,2, where ε > 0 is sa. We denote the event where Y i,n E,N ε, and the event where ax N T ε Y i,n Y i,n Y 2,N > 0. We > 0, whp, Y i,n ε,ε 1 ], ε,ε 1 ], for i = 1,2, by ε for i = 1,2 by F,N ε. The events E,N ε and F,N ε are shown to occur whp, for F,N ε this foows fro Proposition.2a. 720

19 The second ey step in the proof, in Section 4., is to obtain an asyptotic forua for P{H N > } E,N ε. Indeed we prove that for 2 1, and any 1 with 1 1/2, ] P{H N > } E,N ε = E 1 E,N ε F,N εp, 1 + o N,,ε1, 4. where P, 1 is a product of conditiona probabiities of events of the for {H N > H N > 1}. Basicay this foows fro the utipication rue. The identity 4. is estabished in 4.2. In the third ey step, in Section 4.4, we show that, for = N, the ain contribution of the product P, 1 appearing on the right side of 4. is { exp λ N Z 1,N } in 1 +1 Z2,N N 1, B N where λ N = λ N N is in between 1 2 and 4 N, and where B N = B N ε, N defined in 4.51 is such that 1 B N ε, N precisey when T 1,N ε and N 1 T 2,N ε. Thus, by Proposition.2, it ipies that whp 1 +1 N 1 ε 4 Z 1,N τ 1 and Z 2,N N 1 N 1 ε 4 In turn, these bounds aow us to use Proposition.2a. Cobining 4. and 4.4, we estabish in Coroary 4.10, that for a and with og og N N = 2 +, 4.5 ogτ 2 we have { P{H N > N } E,N ε = E 1 E,N ε F,N ε exp λ N { = E 1 E,N ε F,N ε exp λ N where κ = τ 2 1 > 1. τ 1. Z 1,N }] in 1 +1 Z2,N N 1 + o N,,ε1 1 B N exp{κ 1+1 Y 1,N in κ N 1 Y 2,N N 1 }}] + o N,,ε1, 1 B N In the fina ey step, in Sections 4.5 and 4.6, the iniu occurring in 4.6, with the approxiations Y 1,N 1 +1 Y 1,N and Y 2,N N 1 Y 2,N, is anayzed. The ain idea in this anaysis is as foows. With the above approxiations, the right side of 4.6 can be rewritten as { ]}] E 1 E,N ε F,N ε exp λ N exp in κ 1+1 Y 1,N + κ N 1 Y 2,N og + o N,,ε1. 1 B N 4.7 The iniu appearing in the exponent of 4.7 is then rewritten see 4.7 and 4.75 as κ N/2 { in t Z κt Y 1,N + κ c t Y 2,N κ N/2 } og. Since κ N/2, the atter expression ony contributes to 4.7 when in t Z κt Y 1,N + κ c t Y 2,N κ N/2 og

20 Here it wi becoe apparent that the bounds 1 2 λ N 4 are sufficient. The expectation of the indicator of this event eads to the probabiity P in t Z κt Y 1 + κ c t Y 2 κ a N /2,Y 1 Y 2 > 0, with a N and c as defined in Theore 4.1. We copete the proof by showing that conditioning on the event that 1 and 2 are connected is asyptoticay equivaent to conditioning on Y 1 Y 2 > 0. Rear 4.2. In the course of the proof, we wi see that it is not necessary that the degrees of the nodes are i.i.d. In fact, in the proof beow, we need that Propositions.1.4 are vaid, as we as that is concentrated around its ean µn. In Rear A.1.5 in the appendix, we wi investigate what is needed in the proof of Propositions.1.4. In particuar, the proof appies aso to soe instances of the configuration ode where the nuber of nodes with degree is deterinistic for each, when we investigate the distance between two unifory chosen nodes. We now go through the detais of the proof. 4.2 A priory bounds on Y i,n We wish to copute the probabiity PH N >. To do so, we spit PH N > as PH N > = PH N >,Y 1,N Y 2,N = 0 + PH N >,Y 1,N Y 2,N > We wi now prove two eas, and use these to copute the first ter in the right-hand side of 4.8. Lea 4.. For any fixed, i N 1,N PY Y 2,N = 0 = 1 q, 2 where q = PY 1 > 0. Proof. The proof is iediate fro Proposition.1 and the independence of Y 1 and Y 2. The foowing ea shows that the probabiity that H N converges to zero for any fixed : Lea 4.4. For any fixed, i PH N = 0. N Proof. As observed above Theore 1.2, by exchangeabiity of the nodes {1,2,...,N}, PH N = P H N, 4.9 where H N is the hopcount between node 1 and a unifory chosen node unequa to 1. We spit, for any 0 < δ < 1, P H N = P H N, Z 1,N N δ + P H N, Z 1,N > N δ

21 The nuber of nodes at distance at ost fro node 1 is bounded fro above by Z1,N. The event { H N } can ony occur when the end node, which is unifory chosen in {2,...,N}, is in the SPG of node 1, so that P HN, Z 1,N N δ Nδ = o1, N N 1 Therefore, the first ter in 4.10 is o1, as required. We wi proceed with the second ter in By Proposition.1, whp, we have that Y 1,N = Y 1 for a. Therefore, we obtain, because Y 1,N = Y 1 ipies Z 1,N = Z 1, P HN, > N δ P > N δ = P > N δ + o1. Z 1,N Z 1,N However, when is fixed, the rando variabe Z1 is finite with probabiity 1, and therefore, i P Z 1,N > N δ = N This copetes the proof of Lea 4.4. Z 1 We now use Leas 4. and 4.4 to copute the first ter in 4.8. We spit PH N >,Y 1,N Y 2,N = 0 = PY 1,N Y 2,N = 0 PH N,Y 1,N Y 2,N = By Lea 4., the first ter is equa to 1 q 2 + o1. For the second ter, we note that when = 0 and H N <, then H N 1, so that Y 1,N Using Lea 4.4, we concude that PH N,Y 1,N Y 2,N = 0 PH N Coroary 4.5. For every fixed, and each N, possiby depending on N, i PH N >,Y 1,N Y 2,N = 0 = 1 q 2. N By Coroary 4.5 and 4.8, we are eft to copute PH N >,Y 1,N a ea that shows that if Y 1,N > 0, then whp Y 1,N ε,ε 1 ]: Lea 4.6. For i = 1,2, i sup ε 0 i sup i sup P0 < Y i,n N < ε = i sup ε 0 i sup Y 2,N > 0. We first prove i sup PY i,n > ε 1 = 0. N Proof. Fix, when N it foows fro Proposition.1 that Y i,n we obtain that = Y i, whp. Thus, i sup ε 0 i sup i sup P0 < Y i,n N 72 < ε = i sup ε 0 i sup P0 < Y i < ε,

22 and siiary for the second probabiity. The reainder of the proof of the ea foows because d Y i as, and because conditionay on Y i > 0 the rando variabe Y i adits Y i a density. Write E,N = E,N ε = {Y i,n ε,ε 1 ],i = 1,2}, 4.15 F,N = F,N ε = { ax Y i,n Y i,n ε,i = 1,2 } T N As a consequence of Lea 4.6, we obtain that so that In the seque, we copute ε PE c,n {Y 1,N Y 2,N > 0} = o N,,ε1, 4.17 PH N >,Y 1,N Y 2,N > 0 = P{H N > } E,N + o N,,ε and often we wi ae use of the fact that by Proposition.2, P{H N > } E,N, 4.19 PE,N F c,n = o N,,ε Asyptotics of P{H N > } E,N We next give a representation of P{H N > } E,N. In order to do so, we write Q i, Z, where i, 0, for the conditiona probabiity given {Z s 1,N } i s=1 and {Z2,N s } s=1 where, for = 0, we condition ony on {Z s 1,N } i s=1, and Ei, Z for its conditiona expectation. Furtherore, we say that a rando variabe 1 is Z -easurabe if 1 is easurabe with respect to the σ-agebra generated by {Z s 1,N and {Z2,N s. The ain rewrite is now in the foowing ea: } s=1 } s=1 Lea 4.7. For 2 1, ] P{H N > } E,N = E 1 E,N Q, Z H N > 2 1P, 1, 4.21 where, for any Z -easurabe 1, with 1 1/2, P, 1 = 2 1 i=2 Q i/2 +1, i/2 Z H N > i H N > i i=1 Q 1 +1, 1 +i Z H N > i H N > i 1. Proof. We start by conditioning on {Z s 1,N } s=1 and {Z2,N s } s=1, and note that 1 E,N is Z - easurabe, so that we obtain, for 2 1, ] P{H N > } E,N = E 1 E,N Q, Z H N > 4.2 ] = E 1 E,N Q, Z H N > 2 1Q, Z H N > H N >

23 Moreover, for i, such that i +, Q i, and, siiary, Z H N > H N > i = E i, Z Q i,+1 Z H N > H N > i + 1 ] = E i, Z Q i,+1 Z H N > i + H N > i + 1Q i,+1 Z H N > H N > i + ], Q i, Z H N > H N > i = E i, Z Q i+1, H N > i + H N > i + 1Q i+1, Z H N > H N > i + ]. In particuar, we obtain, for > 2 1, Q, Z Z H N > H N > 2 1 = E, Z Q +1, Z H N > 2 H N > ] Q +1, Z H N > H N > 2, so that, using that E,N is Z -easurabe and that EE, Z X]] = EX] for any rando variabe X, P{H N > } E,N 4.27 ] = E 1 E,N Q, Z H N > 2 1Q +1, Z H N > 2 H N > 2 1Q +1, Z H N > H N > 2. We now copute the conditiona probabiity by repeatedy appying 4.24 and 4.25, increasing i or as foows. For i + 2 1, we wi increase i and in turn by 1, and for 2 1 < i +, we wi ony increase the second coponent. This eads to Q, Z H N > H N > 2 1 = E, Z =1 i=2 = E, Z P, 1 ], Q i/2 +1, i/2 Z H N > i H N > i Q 1 +1, 1 + Z H N > H N > were we used that we can ove the expectations E i, Z outside, as in 4.27, so that these do not appear in the fina forua. Therefore, fro 4.2, 4.28, and since 1 E,N and Q, Z H N > 2 1 are Z -easurabe, ] P{H N > } E,N = E 1 E,N Q, Z H N > 2 1E, Z P, 1 ] ] = E E, Z 1 E,N Q, Z H N > 2 1P, 1 ] ] = E 1 E,N Q, Z H N > 2 1P, This proves ] 725

24 We note that we can oit the ter Q, Z H N > 2 1 in 4.21 by introducing a sa error ter. Indeed, we can write Q, Z H N > 2 1 = 1 Q, Z H N Bounding 1 E,N P, 1 1, the contribution to 4.21 due to the second ter in the righthand side of 4.0 is according to Lea 4.4 bounded by ] E Q, Z H N 2 1 = PH N 2 1 = o N We concude fro 4.20, 4.21, and 4.1, that ] P{H N > } E,N = E 1 E,N F,N P, 1 + o N,,ε We continue with 4.2 by bounding the conditiona probabiities in P, 1 defined in Lea 4.8. For a integers i, 0, } exp { 4Z1,N i+1 Z2,N Q i+1, Z H N > i + H N > i + 1 exp The upper bound is aways vaid, the ower bound is vaid whenever i+1 Z s 1,N + s=1 s=1 Z 2,N s } { Z1,N i+1 Z2,N Proof. We start with the upper bound. We fix two sets of n 1 and n 2 stubs, and wi be interested in the probabiity that none of the n 1 stubs are connected to the n 2 stubs. We order the n 1 stubs in an arbitrary way, and connect the stubs iterativey to other stubs. Note that we ust connect at east n 1 /2 stubs, since any stub that is being connected reoves at ost 2 stubs fro the tota of n 1 stubs. The nuber n 1 /2 is reached for n 1 even precisey when a the n 1 stubs are connected with each other. Therefore, we obtain that the probabiity that the n 1 stubs are not connected to the n 2 stubs is bounded fro above by n 1 /2 t=1 1 n 2 2t + 1 n 1 /2 t=1 1 n Using the inequaity 1 x e x, x 0, we obtain that the probabiity that the n 1 stubs are not connected to the n 2 stubs is bounded fro above by e n 1/2 n 2 e n 1 n Appying the above bound to n 1 = Z 1,N i+1 and n 2 = Z 2,N, and noting that the probabiity that H N > i + given that H N > i + 1 is bounded fro above by the probabiity that none of the Z 1,N i+1 stubs are connected to the Z 2,N stubs eads to the upper bound in 4.. We again fix two sets of n 1 and n 2 stubs, and are again interested in the probabiity that none of the n 1 stubs are connected to the n 2 stubs. However, now we use these bounds repeatedy, and 726

25 we assue that in each step there reain to be at east L stubs avaiabe. We order the n 1 stubs in an arbitrary way, and connect the stubs iterativey to other stubs. We obtain a ower bound by further requiring that the n 1 stubs do not connect to each other. Therefore, the probabiity that the n 1 stubs are not connected to the n 2 stubs is bounded beow by n 1 t=1 1 n L 2t + 1 When L 2n 1 2 and 1 t n 1, we obtain that 1 n 2 L 2t+1 1 2n 2. Moreover, when x 1 2, we have that 1 x e 2x. Therefore, we obtain that when L 2n 1 2 and n 2 4, then the probabiity that the n 1 stubs are not connected to the n 2 stubs when there are sti at east L stubs avaiabe is bounded beow by n 1 t=1 1 n 2 L 2t + 1 n 1 t=1 e 4n 2 = e 4n 1 n The event H N > i + conditionay on H N > i + 1 precisey occurs when none of the Z 1,N i+1 stubs are connected to the Z 2,N stubs. We wi assue that 4.4 hods. We have that L = 2 i s=1 Z1,N s 2 s=1 Z2,N s, and n 1 = Z 1,N i+1, n 2 = Z 2,N. Thus, L 2n 1 2 happens precisey when i+1 L 2n 1 = 2 Z s 1,N 2 s=1 s=1 Z 2,N s This foows fro the assued bound in 4.4. Aso, when n 2 = Z 2,N, n 2 4 is ipied by 4.4. Thus, we are aowed to use the bound in 4.8. This eads to { } Q i+1, Z H N > i + H N > i + 1 exp 4Z1,N i+1 Z2,N, 4.40 which copetes the proof of Lea The ain contribution to P{H N > } E,N We rewrite the expression in 4.2 in a ore convenient for, using Lea 4.8. We derive an upper and a ower bound. For the upper bound, we bound a ters appearing on the right-hand side of 4.22 by 1, except for the ter Q 1 +1, 1 Z H N > H N > 1, which arises when i = 2 1, in the second product. Using the upper bound in Lea 4.8, we thus obtain that P, 1 exp { Z1,N 1 +1 Z2,N 1 2 } The atter inequaity is true for any Z -easurabe 1, with 1 1/2. To derive the ower bound, we next assue that 1 +1 s=1 1 Z s 1,N + s=1 727 Z 2,N s 4, 4.42

26 so that 4.4 is satisfied for a i in We write, recaing., { B 1 N ε, = 1/2 : + 1 T 1,N 2,N ε, T }. ε 4.4 We restrict ourseves to 1 B 1 N ε,, if B 1 N ε,. When 1 B 1 N ε,, we are aowed to use the bounds in Proposition.2. Note that { 1 B 1 N ε,} is Z -easurabe. Moreover, it foows fro Proposition.2 that if 1 B 1 N ε,, that then, with probabiity converging to 1 as first N and then, Z 1,N s N 1 ε 4 τ 1, < s 1 + 1, and Z s 2,N When 1 B 1 N ε,, we have 1 +1 s=1 1 Z s 1,N + s=1 Z 2,N s = ON 1 τ 1 = on = oln, N 1 ε 4 τ 1, < s as ong as = on τ 2 τ 1. Since throughout the paper = Oog og N see e.g. Theore 1.2, > 0, the Assuption 4.42 wi aways be fufied. and τ 2 τ 1 Thus, on the event E,N { 1 B 1 N ε,}, using.5 in Proposition.2 and the ower bound in Lea 4.8, with probabiity 1 o N,,ε1, and for a i {2,...,2 1 1}, Q i/2 +1, i/2 Z and, for 1 i 2 1, Q 1 +1, 1 +i Z H N > i H N > i 1 exp { 4Z1,N i/2 +1 Z2,N i/2 H N > i H N > i 1 exp { 4Z1,N } exp { 4Z 1,N 1 +1 Z2,N 1 }, Z2,N 1 +i } exp { 4Z 1,N 1 +1 Z2,N 1 } Therefore, by Lea 4.7, and using the above bounds for each of the in tota 2+1 ters, we obtain that when 1 B 1 N ε,, and with probabiity 1 o N,,ε1, P, 1 exp { 4 Z1,N 1 +1 Z2,N 1 } 2+1 exp { 4 Z1,N 1 +1 Z2,N 1 } We next use the syetry for the nodes 1 and 2. Denote { B 2 N ε, = 1/2 : + 1 T 2,N 1,N ε, T }. ε 4.48 Tae = 1, so that 1/2 1, and thus { B 2 N ε, = 1/2 1 : + 1 T 1,N ε, } T 2,N ε Then, since the nodes 1 and 2 are exchangeabe, we obtain fro 4.47, when 1 B 2 N ε,, and with probabiity 1 o N,,ε1, P, 1 exp { 4 Z1,N 1 +1 Z2,N 1 }

27 We define B N ε, = B 1 N ε, B 2 N ε,, which is equa to { B N ε, = 1 : + 1 T 1,N 2,N ε, T }. ε 4.51 We can suarize the obtained resuts by writing that with probabiity 1 o N,,ε1, and when B N ε,, we have for a 1 B N ε,, where λ N = λ N satisfies P, 1 = exp { λ N Z 1,N 1 +1 Z2,N 1 }, λ N Reation 4.52 is true for any 1 B N ε,. However, our couping fais when Z 1,N 1 +1 grows too arge, since we can ony coupe Z i,n with Ẑi,N up to the point where Z i,n N 1 ε Therefore, we next tae the axia vaue over 1 B N ε, to arrive at the fact that, with probabiity 1 o N,,ε1, on the event that B N ε,, P, 1 = ax exp { Z λ N 1 B N ε, 1,N 1 +1 Z2,N 1 } { = exp λ N Fro here on we tae = N as in 4.5 with a fixed integer. or Z2,N 1 2 τ 1. Z 1,N } in 1 +1 Z2,N B N ε, In Section 5, we prove the foowing ea that shows that, apart fro an event of probabiity 1 o N,,ε1, we ay assue that B N ε, N : Lea 4.9. For a, with N as in 4.5, i sup ε 0 i sup i sup P{H N > N } E,N {B N ε, N = } = 0. N Fro now on, we wi abbreviate B N concude that, = B N ε, N. Using 4.2, 4.54 and Lea 4.9, we Coroary For a, with N as in 4.5, P { {H N > N } E,N = E 1 E,N F,N exp λ N Z 1,N }] in 1 +1 Z2,N N 1 + o N,,ε1, 1 B N where 1 2 λ N N 4 N. 4.5 Appication of the couping resuts In this section, we use the couping resuts in Section.. Before doing so, we investigate the iniu of the function t κ t y 1 + κ n t y 2, where the iniu is taen over the discrete set {0,1,...,n}, and where we reca that κ = τ

28 Lea Suppose that y 1 > y 2 > 0, and κ = τ 2 1 > 1. Fix an integer n, satisfying n > ogy 2/y 1 og κ, then t = argin t {1,2,...,n} κ t y 1 + κ n t n y 2 = round 2 + ogy 2/y 1, 2og κ where roundx is x rounded off to the nearest integer. In particuar, { κ t } y 1 ax, κn t y 2 κ. κ n t y 2 κ t y 1 Proof. Consider, for rea-vaued t 0,n], the function Then, ψt = κ t y 1 + κ n t y 2. ψ t = κ t y 1 κ n t y 2 og κ, ψ t = κ t y 1 + κ n t y 2 og 2 κ. In particuar, ψ t > 0, so that the function ψ is stricty convex. The unique iniu of ψ is attained at ˆt, satisfying ψ ˆt = 0, i.e., ˆt = n 2 + ogy 2/y 1 2og κ 0,n, because n > ogy 2 /y 1 /og κ. By convexity t = ˆt or t = ˆt. We wi show that t ˆt 1 2. Put t 1 = ˆt and t 2 = ˆt. We have Writing t i = ˆt + t i ˆt, we obtain for i = 1,2, κˆt y 1 = κ n ˆt y 2 = κ n 2 y1 y ψt i = κ n 2 y1 y 2 {κ t i ˆt + κˆt t i }. For 0 < x < 1, the function x κ x +κ x is increasing so ψt 1 ψt 2 if and ony if ˆt t 1 t 2 ˆt, or ˆt t 1 1 2, i.e., if ψt 1 ψt 2 and hence the iniu over the discrete set {0,1,...,n} is attained at t 1, then ˆt t On the other hand, if ψt 2 ψt 1, then by the ony if stateent we find t 2 ˆt 1 2. In both cases we have t ˆt 1 2. Finay, if t = t 1, then we obtain, using 4.55, 1 κn t y 2 κ t y 1 whie for t = t 2, we obtain 1 κt y 1 κ n t y 2 κ. = κˆt t 1 κ t 1 ˆt = κ2ˆt t 1 κ, We continue with our investigation of P {H N > N } E,N. We start fro Coroary 4.10, and substitute.1 to obtain, P {H N > N } E,N = E 1 E,N F,N exp { λ N exp in 1 B N κ 1 +1 Y 1,N κ N 1 Y 2,N N 1 og LN ]}] + o N,,ε1,

29 where we rewrite, using 4.51 and., { B N = 1 N 1 : κ 1+1 Y 1,N 1 ε2 τ 1 og N,κ N 1 Y 2,N 1 } ε2 τ 1 og N. Moreover, on F,N, we have that in 1 B N κ 1+1 Y 1,N κ N 1 Y 2,N N 1 is between and in κ 1 +1 Y 1,N ε + κ N 1 Y 2,N ε 1 B N in κ 1 +1 Y 1,N + ε + κ N 1 Y 2,N + ε. 1 B N To abbreviate the notation, we wi write, for i = 1,2, Define for ε > 0, { H,N = H,N ε = 4.57 Y i,n,+ = Y i,n + ε, Y i,n, = Y i,n ε in 0 1 N 1 κ 1 +1 Y 1,N, } + κ N 1 Y 2,N, 1 + ε 2 og N. On the copeent H c,n, the iniu over 0 1 N 1 of κ 1+1 Y 1,N, + κ N 1 Y 2,N, exceeds 1+ε 2 og N. Therefore, aso the iniu over the set B N of κ 1+1 Y 1,N, +κ N 1 Y 2,N, exceeds 1 + ε 2 og N, so that fro 4.56, Lea 4.8 and Proposition.2 and with error at ost o N,,ε1, P {H N > N } E,N H c,n { E 1 H c,n exp 1 2 exp E 1 H c,n exp in κ 1 +1 Y 1,N 1 B κ N 1 ]}] Y 2,N N 1 og LN N { 1 2 exp in κ 1 +1 Y 1,N, + 1 B κ N 1 ]}] Y 2,N, og LN N { E exp 1 2 exp 1 + ε 2 }] og N og e 1 2c Nε2 = o N,,ε1, 4.59 because cn, whp, as N. Cobining 4.59 with 4.18 yieds PH N >,Y 1,N Y 2,N > 0 = P{H N > } E,N H,N + o N,,ε Therefore, in the reainder of the proof, we assue that H,N hods. Lea With probabiity exceeding 1 o N,,ε1, in κ 1 +1 Y 1,N,+ + 1 B κ N 1 Y 2,N,+ = in κ 1 +1 Y 1,N,+ + N 0 1 < κ N 1 Y 2,N,+, 4.61 N and in κ 1 +1 Y 1,N, 1 B N + κ N 1 Y 2,N, = in κ 1 +1 Y 1,N, 0 1 < N + κ N 1 Y 2,N,

30 Proof. We start with 4.61, the proof of 4.62 is siiar, and, in fact, sighty siper, and is therefore oitted. To prove 4.61, we use Lea 4.11, with n = N +1, t = 1 +1, y 1 = Y 1,N,+ and y 2 = Y 2,N,+. Let t = argin t {1,2,...,n} κ t y 1 + κ n t y 2, and assue without restriction that κ t y 1 κ n t y 2. We have to show that t 1 B N. According to Lea 4.11, 1 κt Y 1,N,+ κ n t Y 2,N,+ = κt y 1 κ n t y 2 κ. 4.6 We define x = κ t Y 1,N,+ and y = κn t Y 2,N,+, so that x y. By definition, on H,N, κ t Y 1,N, Since, on E,N, we have that Y 1,N ε, Y 1,N,+ = Y 1,N, + κn t Y 2,N, 1 + ε2 og N ε Y 1,N, ε + ε 1,N ε εy, = 1 + ε2 1,N 1 ε2y,, 4.64 and iewise for Y 2,N,+. Therefore, we obtain that on E,N H,N, and with ε sufficienty sa, x + y 1 + ε2 κ t 1 ε 2 Y 1,N, Moreover, by 4.6, we have that + ] 1 + ε 2 2 κn t Y 2,N, 1 ε 2 og N 1 + εog N x y Hence, on E,N H,N, we have, with κ 1 = τ 2, x = x + y 1 + y x κ ε 1 + ε og N = og N, κ 1 τ 1 when ε > 0 is sufficienty sa. We cai that if note the difference with 4.67, x = κ t Y 1,N,+ 1 ε og N, 4.68 τ 1 then 1 = t 1 B N ε, N, so that 4.61 foows. Indeed, we use 4.68 to see that κ 1 +1 Y 1,N = κ t Y 1,N κ t Y 1,N,+ 1 ε og N, 4.69 τ 1 so that the first bound in 4.57 is satisfied. The second bound is satisfied, since κ N 1 Y 2,N = κ n t Y 2,N κ n t Y 2,N,+ = y x 1 ε og N, 4.70 τ 1 where we have used n = N + 1, and Thus indeed 1 B Nε, N. 72

31 We concude that, in order to show that 4.61 hods with error at ost o N,,ε1, we have to show that the probabiity of the intersection of the events {H N > N } and { E,N = E,N ε = t : 1 ε τ 1 og N < κt Y 1,N,+ 1 + ε og N, 4.71 τ 1 κ t Y 1,N,+ + κn t Y 2,N,+ 1 + εog N }, is of order o N,,ε1. This is contained in Lea 4.1 beow. Lea 4.1. For N as in 4.5, i sup ε 0 i sup The proof of Lea 4.1 is deferred to Section 5. i sup PE,N ε E,N ε {H N > N } = 0. N Fro 4.56, Leas 4.12 and 4.1, we finay arrive at P {H N > N } E,N 4.72 { E 1 E,N exp λ N exp in κ 1 +1 Y 1,N, < κ N 1 ]}] Y 2,N, og LN + o N,,ε1, N and at a siiar ower bound where Y i,n i,n, is repaced by Y,+. Note that on the right-hand side of 4.72, we have repaced the intersection of 1 E,N F,N by 1 E,N, which is aowed, because of Evauating the iit The fina arguent starts fro 4.72 and the siiar ower bound, and consists of etting N and then. The arguent has to be perfored with Y i,n i,n,+ and Y, separatey, after which we et ε 0. Since the precise vaue of ε pays no roe in the derivation, we ony give the derivation for ε = 0. Observe that in κ 1+1 Y 1,N 0 1 < N + κ N 1 Y 2,N og = κ N/2 in κ 1+1 N /2 Y 1,N 0 1 < N = κ N/2 in N /2 +1 t< N /2 +1 κt Y 1,N + κ N/2 1 Y 2,N κ N/2 og + κ c t Y 2,N κ N/2 og, 4.7 where t = 1 +1 N /2, c N = c = /2 /2 +1 = 1 { is even}. We further rewrite, using the definition of a N in Theore 4.1, Cacuating, for Y i,n κ N/2 og = κ og og N og κ og og N og κ ε,ε 1 ], the iniu of κ t Y 1,N /2 og og N = κ a N /2 og og N κ c t Y 2,N, over a t Z, we concude that the arguent of the iniu is contained in the interva 1 2, ogε2 /2og κ]. Hence fro Lea 4.11, for N, n = c {0,1} and on the event E,N, in N /2 +1 t N /2 κt Y 1,N + κ c t Y 2,N = in t Z κt Y 1,N + κ c t Y 2,N

32 We define W,N = in t Z κt Y 1,N and, siiary we define W +,N and W,N, by repacing Y i,n The upper and ower bound in 4.72 now yied: E + κ c t Y 2,N κ a N /2 og og N, 4.76 by Y i,n,+ and Y i,n,, respectivey. 1 E,N exp λ N e κ N /2 W + ]],N + o N,,ε P{H N > N } E,N E 1 E,N exp λ N e κ N /2 W ]],N + o N,,ε1. We spit E 1 E,N exp λ N e κ N /2 W,N ]] = P G N E,N + I N + J N + K N + o N,,ε1, 4.78 where for ε > 0, F N = F N,ε = { W,N > ε }, GN = G N,ε = { W,N < ε }, 4.79 and where we define I N J N K N = E exp λ N e κ N /2 W,N ] ] 1 efn, 4.80 E,N ] 1 egn E,N = E = E exp λ N e κ N /2 W,N ] 1 exp λ N e κ N /2 W,N ] 1 ef c N G ec N E,N, 4.81 ] The spit 4.78 is correct since using the abbreviation exp W for exp λ N e κ N /2 W,N ], ] ] I N + J N + K N = E 1 E,N exp W{1 F e N + 1G e N + 1F e N c eg } 1 c e N G N = E 1 E,N exp W 1 egn ] ] Observe that = E 1 E,N exp W ] PE,N G N. 4.8 κ N/2 W,N 1 e F N > εκ N/2, κ N/2 W,N 1 e G N < εκ N/ We now show that I N,J N and K N are error ters, and then prove convergence of PE,N G N. We start by bounding I N. By the first bound in 4.84, for every ε > 0, and since λ N 1 2, i sup N I N i supexp { 1 N 2 exp{κ N/2 ε} } = Siiary, by the second bound in 4.84, for every ε > 0, and since λ N 4 N, we can bound J N as i sup J N i sup E 1 exp { 4 N exp{ κ N/2 ε} }] = N N Finay, we bound K N by K N P F c G c E N N,N, 4.87 and appy the foowing ea, whose proof is deferred to Section 5: 74

33 Lea For a, i sup ε 0 i sup The concusion fro is that: i sup P FN,ε c G N,ε c E,N ε = 0. N P{H N > N } E,N = P G N E,N + o N,,ε To copute the ain ter P G N E,N, we define and we wi show that Lea U = in t Z κt Y 1 + κ c t Y 2, 4.89 P G N E,N = P U κ a N /2 < 0,Y 1 Y 2 > 0 + o N,,ε Proof. Fro the definition of G N, G N E,N = { in t Z κt Y 1,N + κ c t Y 2,N κ a N /2 og U N og N } i,n < ε,y ε,ε 1 ] By Proposition.1 and the fact that = µn1 + o1, P G N E,N P in t Z κt Y 1 +κ c t Y 2 κ a N /2 < ε,y i ε,ε 1 ] = o N,,ε Since Y i converges to Y i aost surey, as, sup s Y s i Y i converges to 0 a.s. as. Therefore, P G N E,N P U κ a N /2 < ε,y i ε,ε 1 ] = o N,,ε1, 4.9 Moreover, since Y 1 has a density on 0, and an ato at 0 see 8, PY 1 ε,ε 1 ],Y 1 > 0 = o1, as ε 0. Reca fro Section.1 that for any fixed, and conditionay on Y 1 Y 2 > 0, the rando variabe U has a density. We denote this density by f 2 and the distribution function by F 2. Aso, κ a N /2 I = κ /2,κ /2 +1 ]. Then, P ε U κ a N /2 < 0 sup a I F 2 a F 2 a ε] The function F 2 is continuous on I, so that in fact F 2 is unifory continuous on I, and we concude that i sup supf 2 a F 2 a ε] = ε 0 a I This estabishes We suarize the resuts obtained sofar in the foowing coroary: 75

34 Coroary For a, with N as in 4.5, P {H N > N } E,N = q 2 P U κ a N /2 Y 1 Y 2 > 0 + o N,,ε1. Proof. By independence of Y 1 and Y 2, we obtain PY 1 Y 2 > 0 = q 2. Cobining this with 4.88 and 4.90, yieds P{H N > N } E,N = P U κ a N /2 < 0,Y 1 Y 2 > 0 + o N,,ε1 = q 2 P U κ a N /2 < 0 Y 1 Y 2 > 0 + o N,,ε Note that the change fro U < κ a N /2 to U κ a N /2 is aowed because F 2 adits a density. We now coe to the concusion of the proof of Theore 4.1. Coroary 4.16 yieds, with τ,n = 2, so that N = τ,n +, og og N ogτ 2 because P H N > τ,n + = P{H N > N } E,N + P{H N > N } E c,n = q 2 P U κ a N /2 Y 1 Y 2 > q 2 + o N,,ε1, P{H N > N } E c,n = P{H N > N } E c,n = P{H N > N } {Y 1,N 1,N {Y Y 2,N = 0} + P{H N > N } E c,n Y 2,N = 0} + P{H N > N } E c,n {Y 1,N = i ε 0 i 1 q2 + o N,,ε1 = 1 q 2 + o N,,ε1, where the second equaity foows fro {Y 1,N fro Coroary 4.5 and 4.17, respectivey. Y 2,N 1,N {Y Y 2,N Y 2,N > 0} > 0} = 0} E c,n, and the one but fina equaity Taing copeentary events, we obtain, P H N τ,n + = q 2 P U > κ a N /2 Y 1 Y 2 > 0 + o N,,ε1. Note that in the above equation the ters, except the error o N,,ε1, are independent of and ε, so that, in fact, we have, for N, P H N τ,n + = q 2 P U > κ a N /2 Y 1 Y 2 > 0 + o We cai that 4.97 ipies that, when N, P H N < = q 2 + o Indeed, to see 4.98, we prove upper and ower bounds. For the ower bound, we use that for any Z P H N < P H N τ,n +, 76

35 and et in 4.97, noting that κ a N /2 0 as. For the upper bound, we spit P H N < = P {H N < } {Y 1,N Y 2,N = 0} + P {H N < } {Y 1,N Y 2,N > 0}. For N, the first ter is bounded by PH N 1 = o1, by Lea 4.4. Siiary as N, the second ter is bounded fro above by, using Proposition.1, P {H N < } {Y 1,N Y 2,N > 0} P Y 1,N Y 2,N > 0 = q 2 + o1, 4.99 which converges to q 2 as. This proves We concude fro 4.97 and 4.98 that for N, P H N τ,n + H N < = P U > κ a N /2 Y 1 Y 2 > 0 + o Substituting κ = τ 2 1, and taing copeents in this yieds the cais in Theore Proofs of Leas 4.9, 4.1 and 4.14 In this section, we prove the three eas used in Section 4. The proofs are siiar in nature. Denote { T i,n } = { T i,n } { + 1 T i,n }. 5.1 We wi ae essentia use of the foowing consequences of Propositions.1 and.2: Lea 5.1. For any u > 0, and i = 1,2, i P { T i,n } E,N {Z i,n N u1 ε,n u1+ε ]} = o N,,ε1, 5.2 ii P { T i,n } E,N {Z i,n N 1 κτ 1 +ε } = o N,,ε1. 5. Proof. We start with the proof of i. In the course of this proof the stateent whp eans that the copeent of the invoved event has probabiity o N,,ε1. By Proposition.2, we have whp, for T i,n, and on the event E,N, that Y i,n Y i,n where the ast inequaity foows fro Y i,n Y i,n + ε Y i,n 1 + ε2, 5.4 ε. Therefore, aso Y i,n 1 2ε 2, 5.5 when ε is so sa that 1 + ε ε 2. In a siiar way, we concude that with T i,n, and on the event E,N, Furtherore, the event Z i,n Y i,n Y i,n 1 + 2ε N u1 ε,n u1+ε ] is equivaent to κ u1 εog N Y i,n κ u1 + εog N

36 Therefore, we obtain that, with u,n = uκ og N, and siiary, Y i,n 1 + 2ε εuκ og N 1 + 2εu,N, 5.8 Y i,n 1 2ε 2 1 εuκ og N 1 2εu,N. 5.9 We concude that whp the events T i,n ipy,ε Y i,n ε 1 and Z i,n N u1 ε,n u1+ε ] Y i,n u,n 1 2ε,1 + 2ε] u,n 1 2ε,u,N 1 + 2ε] Since ε Y i,n ε 1, we therefore ust aso have when ε is so sa that 1 2ε 1 2, u,n ε 2, 2 ] ε Therefore, P { T i,n } E,N {Z i,n N u1 ε,n u1+ε ]} sup P Y i,n x1 2ε,1 + 2ε] + o N,,ε1. x ε 2, 2 ε ] 5.12 Since, for N, Y i,n i sup ε 0 i sup i sup ε 0 i sup N i sup = Y i in probabiity, by Proposition.1, we arrive at P { T i,n } E,N {Z i,n N u1 ε,n u1+ε ]} 5.1 sup P Y i x1 2ε,1 + 2ε]. x ε 2, 2 ε ] We next use that Y i converges to Y i aost surey as to arrive at i sup i sup N sup P Y i x1 2ε,1 + 2ε] sup x ε 2, 2 ε ] x>0 P { T i,n } E,N {Z i,n N u1 ε,n u1+ε ]} 5.14 F 1 x1 + 2ε F 1 x1 2ε], where F 1 denotes the distribution function of Y i. Since F 1 is a proper distribution function on 0,, with an ato at 0 and aditting a density on 0,, we have i sup ε 0 supf 1 x1 + 2ε F 1 x1 2ε] = x>0 This is iediate fro unifor continuity of F 1 on each bounded subinterva of 0, and by the fact that F 1 = 1. The upper bound 5.14 together with 5.15 copetes the proof of the first stateent of the ea. We turn to the stateent ii. The event that T i,n Y i,n ipies 1 ε2 τ 1 κ +1 og N

37 By 5.6, we can therefore concude that whp, for T i,n, and on the event E,N, that, for ε so sa that 1 ε 2 /1 + 2ε 2 1 ε, which is equivaent to Therefore, Y i,n Z i,n 1 ε τ 1 κ +1 og N, 5.17 N 1 ε κτ 1 N 1 κτ 1 ε i sup ε 0 i sup ε 0 i sup i sup i sup N i sup N P { T i,n } E,N {Z i,n N 1 κτ 1 +ε } 5.19 P { T i,n } E,N {Z i,n N 1 κτ 1 ε,n 1 κτ 1 +ε ]} = 0, which foows fro the first stateent in Lea 5.1 with u = 1 κτ 1. Proof of Lea 4.9. By 4.20, it suffices to prove that P{H N > N } E,N F,N {B N ε, N = } = o N,,ε1, 5.20 which shows that in considering the event {H N > N } E,N F,N, we ay assue that B N ε, N. Observe that if B N ε, =, then B N ε, + 1 =. Indeed, if B N ε, + 1 and, then 1 B N ε,. If, on the other hand, B N ε, + 1 = {}, then aso B N ε,. We concude that the rando variabe, is we defined. Hence {B N ε, N = } = { N + 1} and we therefore have = sup{ : B N ε, } {B N ε, N = } = { < N } = { N 2} { = N 1} We dea with each of the two events separatey. We start with the first. Since the sets B N ε, are Z -easurabe, we obtain, as in 4.2, P{H N > N } E,N F,N { N 2} P{H N > + 2} E,N F,N 5.2 ] = E 1 E,N F,N P + 2, 1 + o N,,ε1. We then use 4.41 to bound P + 2, 1 exp { Z1,N 1 +1 Z2,N } Now, since B N ε,, we can pic 1 such that 1 1 B N ε,. Since B N ε, + 1 =, we have 1 1 / B N ε, + 1, ipying T 2,N and / T 2,N so that, by.7, Z 2,N +2 1 N 1 ε τ 1. 79

38 Siiary, since 1 B N ε, + 1 we have that 1 T 1,N by.7, Z 1,N 1 +1 N 1 ε τ 1. Therefore, since LN N, whp, Z 1,N 1 +1 Z2,N +2 1 and / T 1,N, so that, again N 21 ε τ 1 1, 5.25 and the exponent of N is stricty positive for τ 2, and ε > 0 sa enough. This bounds the contribution in 5.2 due to { N 2}. We proceed with the contribution due to { = N 1}. In this case, there exists a 1 with 1 1 B N ε, N 1, so that 1 T 1,N and N 1 T 2,N. On the other hand, B N ε, N =, which together with 1 1 B N ε, N 1 ipies that N 1 T 2,N, and N 1 +1 / T 2,N. Siiary, we obtain that 1 T 1,N that, whp, Z 1,N 1 +1 N 1 ε τ 1. and / T 1,N. Using Proposition.4, we concude We now distinguish two possibiities: a Z 2,N N 1 N τ 2 τ 1 +ε ; and b Z 2,N N 1 > N τ 2 τ 1 +ε. By 5. and the fact that N 1 T 2,N, case a has sa probabiity, so we need to investigate case b ony. In case b, we can write P {H N > N } E,N F,N { = N 1} {Z 2,N N 1 > N τ 2 τ 1 +ε } ] 5.26 = E 1 E,N F,N { = N 1}1 2,N {Z N >N τ 1 τ 2 +ε N, 1 } { 1 1 B N ε, N 1} 1 + o N,,ε1, where according to 4.41, we can bound P N, 1 exp We note that by Proposition.4 and siiary to 5.25, Z 1,N 1 +1 Z2,N N 1 } { Z1,N 1 +1 Z2,N N N 1 ε τ 2 τ 1 N τ 1 +ε 1 1 = N 1 τ 1 ε, 5.28 and again the exponent is stricty positive, so that, foowing the arguents in , we obtain that aso the contribution due to case b is sa. Proof of Lea 4.1. Reca that we have defined x = xt = κ t Y 1,N,+ and y = yt = κ n t Y 2,N,+, with n = N + 1, and that x y. The event E,N in 4.71 is equa to the existence of a t such that, 1 ε τ 1 og N x 1 + ε og N, and x + y 1 + εog N τ 1 Therefore, by 4.66, y x κ 2 1 ετ og N. 5.0 τ 1 740

39 On the other hand, by the bounds in 5.29, y 1 + εog N x 1 + εog N 1 ε τ 1 og N = τ τ ε og N. 5.1 τ 2 τ 1 Therefore, by utipying the bounds on x and y, we obtain 1 ε 2 τ 2 τ 1 2 og2 N κ N+1 Y 1,N,+ Y 2,N,+ 1 + ε and thus τ τ 2 PE,N E,N {H N > N } P 1 ε 2 κ N+1 cog 2 N Y 1,N,+ Y 2,N,+ 1 + ε 1 + ε τ 2 τ 1 2 og2 N, 5.2 τ τ ε, 5. where we abbreviate c = τ 2. Since κ N +1 is bounded away fro 0 and, we concude that τ 1 2 c og 2 N the right-hand side of 5. is o N,,ε1, anaogousy to the fina part of the proof of Lea 5.1i. Proof of Lea We reca that U = in t Z κ t Y 1 + κ c t Y 2, and repeat the arguents eading to to see that, as first N and then, P F c G c E N N,N P ε U κ a N /2 ε,y 1 Y 2 > 0 + o N,1 5.4 = q 2 P ε U κ a N /2 ε Y 1 Y 2 > 0 + o N,1. Reca fro Section.1 that, conditionay on Y 1 Y 2 > 0, the rando variabe U has a density, and that we denoted the distribution function of U given Y 1 Y 2 > 0 by F 2. Furtherore, κ a N /2 I = κ /2,κ /2 +1 ], so that, unifory in N, P ε U κ a N /2 ε Y 1 Y 2 > 0 sup u I F 2 u + ε F 2 u ε] = 0, where the concusion foows by repeating the arguent eading to This copetes the proof of Lea A Proof of Propositions.1,.2 and.4 The appendix is organized as foows. In Section A.1, we prove three eas that are used in Section A.2 to prove Proposition.1. In Section A., we continue with preparations for the proofs of Proposition.2 and.4. In this section we foruate ey Proposition A..2, which wi be proved in Section A.4. In Section A.5, we end the appendix with the proofs of Proposition.2 and.4. As in the ain body of the paper, we wi assue throughout the appendix that τ 2,, so if we refer to Assuption 1.1, we ean Assuption 1.1ii. 741

40 A.1 Soe preparatory eas In order to prove Proposition.1, we ae essentia use of three eas, that aso pay a ey roe in Section A.4 beow. The first of these three eas investigates the tai behaviour of 1 Gx under Assuption 1.1. Reca that G is the distribution function of the probabiity ass function {g }, defined in Lea A.1.1. If F satisfies Assuption 1.1ii then there exists K τ > 0 such that for x arge enough x 2 τ Kτγx 1 Gx x 2 τ+kτγx, A.1.1 where γx = og x γ 1, γ 0,1. Proof. Using 1.12 we rewrite 1 Gx as + 1f +1 1 Gx = = 1 x + 21 Fx + 1] + µ µ =x+1 =x+2 1 F]. Then we use 1, Theore 1, p. 281, together with the fact that 1 Fx is reguary varying with exponent 1 τ 1, to deduce that there exists a constant c = c τ > 0 such that =x+2 Hence, if F satisfies Assuption 1.1ii, then 1 F] c τ x + 21 Fx + 2]. 1 Gx 1 µ x + 21 Fx + 1] x2 τ Kτγx, 1 Gx 1 µ c + 1x + 21 Fx + 1] x2 τ+kτγx, for soe K τ > 0 and arge enough x. Rear A.1.2. It foows fro Assuption 1.1ii and Lea A.1.1, that for each ε > 0 and sufficienty arge x, x 1 τ ε 1 Fx x 1 τ+ε, a A.1.2 x 2 τ ε 1 Gx x 2 τ+ε. b We wi often use A.1.2 with ε repaced by ε 6. Let us define for ε > 0, and the auxiiary event F ε by α = 1 ε5 τ 1, h = ε6, A.1. F ε = { 1 x N α : Gx G N x N h 1 Gx]}, A.1.4 where G N is the rando distribution function of {g N n }, defined in

41 Lea A.1.. For ε sa enough, and N sufficienty arge, PF c ε N h. A.1.5 Proof. First, we rewrite 1 G N x, for x N {0}, in the foowing way: Writing 1 G N x = we thus end up with n=x+1 = 1 g N n = 1 D N =1 =1 N =1 n=x+1 1 {D x+2} = 1 B N y = D 1 {D =n+1} = 1 =1 =1 N D 1 {D x+2} =1 N 1 {D x+2 }. A.1.6 N 1 {D y}, A.1.7 =1 1 G N x = 1 We have a siiar expression for 1 Gx that reads =1 B N x+2. A Gx = 1 µ PD 1 x + 2. =1 A.1.9 Therefore, with we can write Gx G N x] = β = 1 h 1 + 2h, and χ = τ 1 τ 1, Nµ 1 1 Gx] + 1 N β =1 + 1 N χ B N =N β =N χ +1 x+2 NP D 1 x + 2 ] B N x+2 NP D 1 x + 2 ] B N x+2 NP D 1 x + 2 ]. A.1.10 Hence, for arge enough N and x N α < N β < N χ, we can bound R N x Gx G N x Nµ 1 1 Gx] a + 1 N β =1 B N x+2 NP D 1 x + 2 b + 1 N χ =N β +1 B N NP D 1 c + 1 =N χ +1 BN d + 1 =N χ +1 NP D 1. e A

42 We use A.1.2b to concude that, in order to prove PFε c N h, it suffices to show that P { R N x > C g N h x 2 τ h} N h, A x N α for arge enough N, and for soe C g, depending on distribution function G. We wi define an auxiiary event A N,ε, such that R N x is ore easy to bound on A N,ε and such that PA c N,ε is sufficienty sa. Indeed, we define, with A = β + 2h, { } A N,εa = Nµ 1 N h, a and A N,εb = {ax 1 N D N χ }, b A N,εc = { B x N NPD 1 x } Aog NNPD 1 x, c 1 x N β A N,ε = A N,εa A N,εb A N,εc. A.1.1 By intersecting with A N,ε and its copeent, we have P { R N x > C g N h x 2 τ h } 1 x N α { } A.1.14 P A N,ε { R N x > C g N h x 2 τ h } + PA c N,ε. 1 x N α We wi prove that PA c N,ε N h, and that on the event A N,ε, and for each 1 x N α, the right-hand side of A.1.11 can be bounded by C g N h x 2 τ h. We start with the atter stateent. Consider the right-hand side of A Ceary, on A N,εa, the first ter of R N x is bounded by N h 1 Gx] C g N h x 2 τ+h C g N h x 2 τ h, where the one but ast inequaity foows fro A.1.2b, and the ast since x N α < N so that x 2h < N 2h. Since for > N χ and each, 1 N, we have that {D > } is the epty set on A N,εb, the one but ast ter of R N x vanishes on A N,εb. The ast ter of R N x can, for N arge, be bounded, using the inequaity N and A.1.2a, 1 =N χ +1 NPD 1 =N χ +1 1 τ+h Nχ2 τ+h τ 2 < C g N h+α2 τ+h C g N h x 2 τ h, for a x N α, and where we aso used that for ε sufficienty sa and τ > 2, We bound the third ter of R N x by χ2 τ + h < h + α2 τ + h. 1 N χ =N β +1 B N NPD 1 1 N N χ B N =N β +1 + NPD 1 ] N χ N 1 B N N β + PD 1 N β ]. A

43 We note that due to A.1.2a, for arge enough N, so that b N = PD 1 N β N β1 τ h, Aog NNPD 1 N β NPD 1 N β. A.1.16 A.1.17 Therefore, on A N,εc, we obtain that B N N β 2NPD 1 N β, A.1.18 for ε sa enough and arge enough N. Furtherore as ε 0, N χ+β1 τ+h < C g N h+α2 τ h C g N h x 2 τ h, for x N α, 2 τ h < 0, because after utipying by τ 1 and dividing by ε 5 χ + β1 τ + h < h + α2 τ h, or ε2 + 2τ h < τ 2 + h, as ε is sufficienty sa. Thus, the third ter of R N x satisfies the required bound. We bound the second ter of R N x on A N,εc, using again N, by 1 N N β =1 Aog NNP D 1 x + 2 Aog N = N Nβ =1 P D 1 x + 2. A.1.19 Let c be a constant such that PD 1 > x 1 2 cx 1 τ+h/2, then for a 1 x N α, 1 B N x+2 NP D 1 x + 2 c Aog N Nβ 1 τ+h/2 x + 2 N N β =1 c Aog N N =1 x τ+h/2 + N β τ+h/2] 2c Aog N N β τ+h/2 N N h 1/2 N β τ+h/2 < C g N h N α2 τ h C g N h x 2 τ h, A.1.20 because h 1/2 + β τ + h/2 < h + α2 τ h, or h5τ 4 h < 2ε 5 τ 2 + h, for ε sa enough and τ 2,. We have shown that for 1 x N α, N sufficienty arge, and on the event A N,ε, R N x C g N h x 2 τ h. A.1.21 It reains to prove that PA c N,ε N h. We use that PA c N,ε PA N,εa c + PA N,εb c + PA N,εc c, A.1.22 and we bound each of the three ters separatey by N h/. 745

44 Using the Marov inequaity foowed by the Marciniewicz-Zygund inequaity, see e.g. 11, Coroary 8.2 in Section, we obtain, with 1 < r < τ 1, and again using that N, PA N,εa c 1 = P N N D µ > N h /N =1 A.1.2 N = P D µ r > N 1 h /N r C r N 1 h r NE D 1 µ r ] 1 N h, =1 by choosing h sufficienty sa depending on r. The bound on PA N,εb c is a trivia estiate using A.1.2a. Indeed, for N arge, PA N,εb c = P ax D > N χ NPD 1 N χ N χ1 τ+h N N h, A.1.24 for sa enough ε, because τ > 2 + h. For PA N,εc c, we wi use a bound given by Janson 16, which states that for a binoia rando variabe X with paraeters N and p, and a t > 0, { P X Np t 2exp t 2 2Np + t/ }. A.1.25 We wi appy A.1.25 with t = b N x = Aog NNPD 1 x, and obtain that unifory in x N α, P B x N NPD 1 x > b N x { b N x 2 } 2exp 2NPD 1 x + b N x/ { } Aog N 2exp N A/, A.1.26 Aog N/NPD1 N α because og N NPD 1 N α as N. Thus, A.1.26 yieds, using A = β + 2h, PA N,εc c P B x N N β x=1 This copetes the proof of the ea. og N 0, N1+ατ 1 h NPD 1 x > b N x 2N β A/ = 2N 2h 1 N h. A.1.27 For the third ea we introduce soe further notation. For any x N, define Ŝ N x = x i=1 ˆX N N i, ˆV x = ax 1 i x ˆX N i, where { N ˆX i } x i=1 have the sae aw, say ĤN, but are not necessariy independent. 746

45 Lea A.1.4 Sus and axia with aw ĤN on the good event. i If Ĥ N satisfies 1 ĤN z] 1 + 2N h ]1 Gz], z y, A.1.28 then for a x N, there exists a constant b, such that: ŜN P x y b x1 + 2N h ] 1 G y ]. A.1.29 ii If Ĥ N satisfies and { ˆX N i } x i=1 1 ĤN y] 1 2N h ]1 Gy], A.1.0 are independent, then for a x N, N P ˆV x y ŜN Proof. We first bound P x y. We write ŜN P x y 1 1 2N h ]1 Gy] x. A.1.1 ŜN N P x y, ˆV x y + P ˆV N x > y. A.1.2 Due to A.1.28, the second ter is bounded by xp ˆXN 1 > y = x 1 ĤN y ] x1 + 2N h ] 1 G y ]. A.1. We use the Marov inequaity and A.1.28 to bound the first ter on the right-hand side of A.1.2 by ŜN N P x y, ˆV x y 1 ŜN y E x 1 {ˆV x N x y} y E ˆXN 1 1 { ˆX N 1 y} x y 1 y ĤN i] x y y 1 + 2N h ] 1 Gi]. A.1.4 i=1 For the atter su, we use 1, Theore 1b, p. 281, together with the fact that 1 Gy is reguary varying with exponent 2 τ 1, to deduce that there exists a constant c 1 such that y 1 Gi] c 1 y1 Gy]. A.1.5 i=1 i=1 Cobining A.1.2, A.1., A.1.4 and A.1.5, we concude that ŜN P x y b x1 + 2N h ] 1 G y ], A.1.6 where b = c This copetes the proof of Lea A.1.4i. N ˆX i } x i=1 For the proof of ii, we use independence of { N P ˆV ĤN x y = y = 1 x Hence, A.1.1 hods. 1 ĤN y, and condition A.1.0, to concude that ] x x 1 1 2N h ]1 Gy]. 747

46 Rear A.1.5. In the proofs in the appendix, we wi ony use that i the event F ε hods whp; ii that is concentrated around its ean; iii that, whp, the axia degree is bounded by N χ for any χ > 1/τ 1. Moreover, the proof of Proposition.1 reies on 14, Proposition A..1, and in its proof it was further used that iv p N N α 2, whp, for any α 2 > 0, where p N is the tota variation distance between g and g N, i.e., p N = 1 g n g n N. A Therefore, if instead of taing the degrees i.i.d. with distribution F, we woud tae the degrees in an exchangeabe way such that the above restrictions hod, then the proof carries on verbati. In particuar, this ipies that our resuts aso hod for the usua configuration ode, where the degrees are fixed, as ong as the above restrictions are satisfied. n A.2 Proof of Proposition.1 The proof aes use of 14, Proposition A..1, which proves the stateent in Proposition.1 under an additiona condition. In order to state this condition et {Ẑi,N } 1, i = 1,2, be two independent copies of the deayed BP, where Ẑi,N 1 has aw {f n } given in 1.1, and where the offspring of any individua in generation with > 1 has aw {g n N }, where g n N is defined in 2.6. Then, the concusion of Proposition.1 foows fro 14, Proposition A..1, for any such that, for any η > 0, and i = 1,2, P =1 Ẑ i,n N η = o1., N A.2.1 By exchangeabiity it suffices to prove A.2.1 for i = 1 ony, we can therefore sipify notation and write further ẐN instead of Ẑi,N. We turn to the proof of A.2.1. By Lea A.1. and A.1.2b, respectivey, for every η > 0, there exists a c η > 0, such that whp for a x N α, We ca a generation 1 good, when 1 G N x 1 + 2N h ]1 Gx] c η x 2 τ+η. A.2.2 Ẑ N and bad otherwise, where as aways ẐN 0 = 1. We further write ẐN 1 1 og N τ 2 η, A.2. H = {generations 1,..., are good}. A

47 We wi prove that when H hods, then are a good, then, for a, Ẑ N =1 ẐN N η. Indeed, when generations 1,..., og N P i=1 τ 2 η i. A.2.5 Therefore, =1 Ẑ N og N P i=1 τ 2 η i og N τ 2 η 2 τ 2 η 1 1 N η, for any η > 0, when N is sufficienty arge. We concude that A.2.6 P Ẑ N > N η PH, c A.2.7 =1 and Proposition.1 foows if we show that PH c = o1. In order to do so, we write 1 PH c = PH1 c + PH+1 c H. =1 A.2.8 For the first ter, we use A.1.2a to deduce that PH c 1 = P D 1 > og N 1 τ 2 η og N τ 1 η τ 2 η og N 1. A.2.9 For 1, we have ẐN =1 ẐN, and using A.2.6, =1 Ẑ N og N τ 2 η 2 τ 2 η 1 1 W N. A.2.10 Using Lea A.1.4i with ĤN = G N, x = and y = v N = og N τ 2 η, where A.1.28 foows fro A.2.2, we obtain that W N PH+1 c H Furtherore by A.1.2b, =1 ẐN P +1 v N ŜN ax P 1 W v N N ẐN = PẐN = b ax 1 + 2N h ] 1 Gv N ]. A W N 1 ax 1 Gv N ] ax v N 2 τ+η = og N 1. 1 W N 1 W N A.2.12 This copetes the proof of Proposition

48 A. Soe further preparations Before we can prove Propositions.2 and.4, we state a ea that was proved in 14. We introduce soe notation. Suppose we have L obects divided into N groups of sizes d 1,...,d N, so that L = N i=1 d i. Suppose we draw an obect at rando. This gives a distribution g d, i.e., g d n = 1 N d i 1 L {di =n+1}, n = 0,1,... A..1 Ceary, g N = g D, where D = D 1,...,D N. We further write i=1 G d x = x n=0 g d n. A..2 We next abe M of the L obects in an arbitrary way, and suppose that the distribution G d M x is obtained in a siiar way fro drawing conditionay on drawing an unabeed obect. More precisey, we reove the abeed obects fro a obects thus creating new d 1,...,d N, and we et G d M x = G d x. Even though this is not indicated, the aw G d M depends on what obects have been abeed. Lea A..1 beow shows that the aw G d M can be stochasticay bounded above and beow by two specific ways of abeing obects. Before we can state the ea, we need to describe those specific abeings. For a vector d, we denote by d 1 d 2... d N the ordered coordinates. Then the aws G d M and G d M, respectivey, are defined by successivey decreasing d N and d 1, respectivey, by one. Thus, N 1 G d x = 1 1 d i 1 L 1 {di x+1} + d N 1 L 1 1 {d N 1 x+1}, G d 1 x = 1 L 1 i=1 N i=2 d i 1 {di x+1} + d 1 1 L 1 1 {d 1 1 x+1}. A.. A..4 For G d M and G d M, respectivey, we perfor the above change M ties, and after each repetition we reorder the groups. Here we note that when d N = 1 in which case d i = 1, for a i, and for G d 1 we decrease d N by one, that we ony eep d 1,...,d N 1. A siiar rue appies when d 1 = 1 and for G d 1 we decrease d 1 by one. Thus, in these cases, the nuber of groups of obects, indicated by N, is decreased by 1. Appying the above procedure to d = D 1,...,D N we obtain that, for a x 1, G N M x G D M x M x G D x G N M 1 M 1 M N D i 1 {Di x+1} = i=1 N i=1 ] D i 1 {Di x+1} M = M GN x, A..5 1 ] G N x M, A..6 M 750

49 where equaity is achieved precisey when D N x + M, and #{i : D i = 1} M, respectivey. Finay, for two distribution functions F,G, we write that F G when Fx Gx for a x. Siiary, we write that X Y when for the distribution functions F X,F Y we have that F X F Y. We next prove stochastic bounds on the distribution G d M x that are unifor in the choice of the M abeed obects. The proof of Lea A..1 can be found in 14. Lea A..1. For a choices of M abeed obects G d M G d M G d M. A..7 Moreover, when X 1,...,X are draws fro G d M 1,...,G d M, where the ony dependence between the X i resides in the abeed obects, then X i X i i=1 i=1 X i, i=1 A..8 where {X i } i=1 and {X i} i=1, respectivey, are i.i.d. copies of X and X with aws GN M for M = ax 1 i M i, respectivey. and G N M We wi appy Lea A..1 to G D = G N. A..1 The inductive step Our ey resut, which wi yied the proofs of Proposition.2 and.4, is Proposition A..2 beow. This proposition wi be proved in Section A.4. For its foruation we need soe ore notation. As before we sipify notation and write further on Z N instead of Z i,n. Siiary, we write Z instead of Z i and T N ε instead of T i,n ε. Reca that we have defined previousy κ = 1 1 ε5 > 1 and α = τ 2 τ 1. In the seque we wor with Y N > ε, for arge enough, i.e., we wor with Z N > e εκ > 1, due to definition.1. Hence, we can treat these definitions as Y N = κ ogz N and Y = κ ogz. A..9 With γ defined in the Assuption 1.1ii, and 0 < ε < τ, we tae ε sufficienty arge to have τ 2 + ε 1 γ ε and 2 ε/2. A..10 = ε = ε For any ε <, we denote M N = =1 Z N, and M = Z. =1 A

50 As defined in Section of 14 we spea of free stubs at eve, as the free stubs connected to nodes at distance 1 fro the root; the tota nuber of free stubs, obtained iediatey after pairing of a stubs at eve 1 equas Z N see aso Section.2 above. For any 1 and 1 x Z N 1, et ZN denote the nuber of constructed free stubs at eve after pairing of the first x stubs of Z N 1. Note that for x = ZN 1, we obtain ZN = Z N. For genera x, the quantity Z N is oosey speaing the su of the nuber of chidren of the first x stubs at eve 1, and according to the couping at fixed ties Proposition.1 this nuber is for fixed, whp equa to the nuber of chidren of the first x individuas in generation 1 of the BP {Z } 1. We introduce the event ˆF, ε, ˆF, ε = { T N ε} a { < 1 : Y N Y N ε } b {ε Y N ε 1 } c {M N 2Z N }. d A..12 We denote by X N i, 1 the nuber of brother stubs of a stub attached to the ith stub of SPG 1. In the proof of Proposition A..2 we copare the quantity Z N to the su x i=1 XN i, 1 for part a and to ax 1 i x X N i, 1 for part b. We then coupe XN i, 1 to XN i, 1 for part a and to X N i, 1 for part b. Aong other things, the event ˆF, ε ensures that these coupings hod. Proposition A..2 Inductive step. Let F satisfy Assuption 1.1ii. For ε > 0 sufficienty sa and c γ sufficienty arge, there exist a constant b = bτ,ε > 0 such that, for x = Z N 1 N 1 ε/2 κτ 1, P ˆF, ε { Z N x κ+cγγx} b, a P ˆF, ε { Z N x κ cγγx } b. b The proof of Proposition A..2 is quite technica and is given in Section A.4. In this section we give a short overview of the proof. For 1, et SPG denote the shortest path graph containing a nodes on distance 1, and incuding a stubs at eve, i.e., the oent we have Z N free stubs at eve. As before, we denote by X N i, 1, i {1,...,x}, the nuber of brother stubs of a stub attached to the i th stub of SPG 1 see Figure A..1. Because Z N is the nuber of free stubs at eve after the pairing of the first x stubs, one woud expect that x Z N X N i, 1, A..1 i=1 where denotes that we have an uncontroed error ter. Indeed, the intuition behind A..1 is that oops or cyces shoud be rare for sa. Furtherore, when M N 1 is uch saer than N, then the aw of X N i, 1 shoud be quite cose to the aw GN, which, in turn, by Lea A.1. is cose to G. If X N i, 1 woud have distribution Gx, then we coud use the theory of sus of rando variabes with infinite expectation, as we as extree vaue theory, to obtain the inequaities of Proposition A

51 a b -2-th eve -1-th eve -2-th eve th eve N Z th eve 2 2 c d -2-th eve -2-th eve -1-th eve -th eve th eve -th eve Figure 4: The buiding of the th eve of SPG. The ast paired stubs are ared by thic ines, the brother stubs by dashed ines. In a the 1 st eve is copeted, in b the pairing with a new node is described, in c the pairing within the 1 st eve is described, and in d the pairing with aready existing node at th eve is described. In order to ae the above intuition rigorous, we use upper and ower bounds. We note that the right-hand side of A..1 is a vaid upper bound for Z N. We show beow that XN i, 1 have the sae aw, and we wish to appy Lea A.1.4i. For this, we need to contro the aw X N for which we use Lea A..1 to bound each X N i, 1. This couping aes sense ony on the good event where G N G N M i, 1, fro above by a rando variabe with aw M is sufficienty cose to G. For the ower bound, we have to do ore wor. The basic idea fro the theory of sus of rando variabes with infinite ean is that the su has the sae order as the axia suand. Therefore, we bound fro beow Z N Z N x. A..14 where Z N = ax 1 i x XN i, 1. A..15 However, we wi see that this ower bound is ony vaid when the chosen stub is not part of the shortest path graph up to that point. We show in Lea A..4 beow that the chosen stub has 75

52 abe 1 when Z N > 2M N 1. In this case, A..14 foows since the x 1 reaining stubs can eat up at ost x 1 x stubs. To proceed with the ower bound, we bound X N 1, 1,...,XN 1 stochasticay fro beow, using Lea A..1, by an i.i.d. sequence of rando variabes with aws G N M, where M is chosen appropriatey and serves as an upper bound on the nuber of stubs with abe. Again on the good event, G N M is sufficienty cose to G. Therefore, we are now faced with the probe of studying the axiu of a nuber of rando variabes with a aw cose to G. Here we can use Lea A.1.4ii, and we concude in the proof of Proposition A..2a that Z N is to eading order equa to x κ, when x = Z N 1 N 1 ε/2 κτ 1. For this choice of x, we aso see that Z N is of bigger order than M N 2, so that the basic assuption in the above heuristic is satisfied. This copetes the overview of the proof. We now state and prove the Leas A.. and A..4. The proof of Proposition A..2 then foows in Section A.4. We define the good event entioned above by F ε,m = N α x=1 { 1 2N h ]1 Gx] 1 G N M } x 1 G N x 1 + 2N h ]1 Gx]. A..16 The foowing ea says that for M N α, the probabiity of the good event is cose to one. Lea A... Let F satisfy Assuption 1.1ii. Then, for ε > 0 sufficienty sa, PF c ε,n α N h, for arge N. Proof. Due to Lea A.1. it suffices to show that for ε sa enough, and N sufficienty we have F c ε,n α F ε c. A..17 We wi prove the equivaent stateent that M F ε F ε,n α. A..18 It foows fro A..5 and A..6 that for every M and x and, in particuar, that for M N α, 1 G N M x 1 G N x 1 G N M x, A G N x] 1 GN x] M M M M ONα 1. A..20 Then we use A.1.2b to obtain that for a x N α, ε sa enough, and N sufficienty arge, ON α 1 N α 1+h = N 1 ε5 τ 1 1+ε6 < N 2ε6 N 1 ε5 τ 1 2 τ ε6 = N 2h N α2 τ h N 2h x 2 τ h N h 1 Gx]. A..21 Therefore, for M N α and with the above choices of ε, α and h, we have, unifory for x N α and on F ε, 1 G N M x] 1 G N x + 1 G N x] 1 G N M x] 1 + 2N h ]1 Gx], M 1 G N M x] 1 G N x 1 G N x] + 1 GN x] 1 2N h ]1 Gx], M 754 M

53 i.e., we have A..16, so that indeed F ε F ε,n α. For the couping of X N i, 1 foowing ea. Reca the definition of M N with the rando variabes with aws GN M given in A..11. x and G N x we need the Lea A..4. For any 1 there are at ost 2M N stubs with abe in SPG +1, whie the nuber of stubs with abe 2 is by definition equa to Z N Proof. The proof is by induction on. There are Z N 1 free stubs in SPG 1. Soe of these stubs wi be paired with stubs with abe 2 or, others wi be paired to stubs with abe 1 see Figure A..1. This gives us at ost 2Z N 1 stubs with abe in SPG 2. This initiaizes the induction. We next advance the induction. Suppose that for soe 1 there are at ost 2M N stubs with abe in SPG +1. There are Z N +1 free stubs with abe 2 in SPG +1. Soe of these stubs wi be paired with stubs with abe 2 or, others wi be ined with stubs with abe 1 again see Figure A..1. This gives us at ost 2Z N +1 new stubs with abe in SPG +2. Hence the tota nuber of these stubs is at ost 2M N + 2Z N +1 = 2MN +1. This advances the induction hypothesis, and proves the cai. +1. M A.4 The proof of Proposition A..2 We state and prove soe consequences of the event ˆF, ε, defined in A..12. We refer to the outine of the proof of Proposition A..2, to expain where we use these consequences. Lea A.4.1. The event ˆF, ε ipies, for sufficienty arge N, the foowing bounds: a M N 1 < N 1 ε 4 /4 κτ 1, b for any δ > 0, N δ, c d κ 1 ε ε og Z N 1 κ 1 ε 1 + ε, for 1, M N 1 2ZN 1, for 1. A.4.1 Proof. Assue that A..12a-d hods. We start by showing A.4.1b, which is evident if we show the foowing cai: og 1 ε 2 ετ 1 og N, A.4.2 og κ for N arge enough. In order to prove A.4.2, we note that if T N ε then, due to definition., κ 1 ε2 τ 1 og N og Z N where the atter inequaity foows fro Y N ogariths on both sides yieds A.4.2. We now turn to A.4.1a. Since 1 M N 1 = Z N =1 < 1 ε2 ετ 1 κ og N, A.4. > ε and A..9. Mutipying by κ and taing ax 1 1 ZN, 755

54 the inequaity A.4.1a foows fro A.4.2, when we show that for any 1, Z N N 1 ε 4 κτ 1. A.4.4 Observe that for < we have that, due to A..9,A..12c and A..12d, for any ε > 0 and fixed and by taing N sufficienty arge, Z N M N 2ZN ε 1 2eκ < N 1 ε 4 κτ 1. A.4.5 Consider 1. Due to A..9, inequaity A.4.4 is equivaent to κ +1 Y N 1 ε4 τ 1 og N. A.4.6 To obtain A.4.6 we wi need two inequaities. Firsty, A..12a and + 1 ipy that κ +1 Y N 1 ε2 τ 1 og N. A.4.7 Given A.4.7 and A..12b, we obtain, when Y N ε, and for 1, κ +1 Y N κ +1 Y N 1 ε2 1+ε 2 τ 1 + ε κ +1 Y N 1 + ε 2 A.4.8 og N = 1 ε4 τ 1 og N. Hence we have A.4.6 or equivaenty A.4.4 for 1. The bound in A.4.1c is an iediate consequence of A..9 and A..12b,c that ipy for 1 >, ε ε Y N 1 ε 1 + ε. We copete the proof by estabishing A.4.1d. We use induction to prove that for a, the bound M N 2Z N hods. The initiaization of the induction hypothesis for = foows fro A..12d. So assue that for soe < 1 the inequaity M N 2Z N hods, then M N +1 = ZN +1 + MN Z N ZN, A.4.9 so that it suffices to bound 2Z N Y N +1 Y N Y N +1 Y N + Y N by Z N +1. We note that ˆF, ε ipies that Y N 2ε ε 2 Y N +1, A.4.10 where in the ast inequaity we used that Y N +1 Y N ε ε ε > 2 ε, as ε 0. Therefore, 2Z N = 2e κ Y N 2e 1+ε2 κ Y N +1 = 2 Z N 1+ε 2 κ 1 Z N +1, A.4.11 when ε > 0 is so sa that ω = 1 + ε 2 κ 1 < 1 and where we tae arge enough to ensure that for, the ower bound Z N +1 = exp{κ+1 Y N +1 } > exp{κ+1 ε} > ω is satisfied

55 Proof of Proposition A..2a. Reca that α = 1 ε5 τ 1. We write P ˆF, ε { Z N x κ+c } γγx P N α ˆF, ε { Z N x κ+c } γγx + PF c α ε,n P N α ˆF, ε { Z N x κ+c } γγx +, A.4.12 where P M is the conditiona probabiity given that F ε,m hods, and where we have used Lea A.. with N h <. It reains to bound the first ter on the right-hand side of A For this bound we ai to use Lea A.1.4. Ceary because oops and cyces can occur, Z N x i=1 X N i, 1, A.4.1 where for 1 i x, X N i, 1 denotes the nuber of brother stubs of the ith -attached node. Since the free stubs of SPG 1 are exchangeabe, each free stub wi choose any stub with abe unequa to with the sae probabiity. Therefore, a X N i, 1 have the sae aw which we denote by HN. Then we observe that due to A..8, X N i, 1 can be couped with XN i, 1 having aw G N M, where M is equa to the nuber of stubs with abe at the oent we generate X N i, 1, which is at ost the nuber of stubs with abe in SPG pus 1. The ast nuber is due to Lea A..4 at ost 2M N By Lea A.4.1a, we have that 2M N 1 1 ε 4 / N κτ N 1 ε5 τ 1 = N α, A.4.14 and hence, due to A..8, we can tae as the argest possibe nuber M = N α. We now verify whether we can appy Lea A.1.4i. Observe that x N 1 ε/2 κτ 1 so that for N arge and each c γ, we have y = x κ+cγγx < N α, A.4.15 since by A.4.2, we can bound by Oog og N. Hence A.1.28 hods, because we condition on F ε,n α. We therefore can appy Lea A.1.4i, with ŜN x = x i=1 XN i, 1, ĤN = G N Nα, and, aso using the upper bound in A.1.1, we obtain, P N α ˆF, ε { Z N x κ+c γγx } b x1 + 2N h ] 1 G y ] if we show that 2b xy κ 1 +K τγy = 2b x x κ 1 +K τγyκ+c γγx b, c γ γx κ 1 + K τ γy + κk τ γy < 0. A.4.16 A.4.17 Inequaity A.4.17 hods, because γy = og y γ 1, γ 0,1, can be ade arbitrariy sa by taing y arge. The fact that y is arge foows fro A.4.1c and A.4.15, and since x exp{κ ε/2}, which can be ade arge by taing arge. Proof of Proposition A..2b. Siiary to A.4.12, we have P ˆF, ε { Z N x } κ cγγx P N α ˆF, ε { Z N x } κ cγγx +, 757 A.4.18

56 and it reains to bound the first ter on the right-hand side of A Reca that Z N = ax 1 i x XN i, 1, where, for 1 i x, X N i, 1 is the nuber of brother stubs of a stub attached to the ith free stub of SPG 1. Suppose we can bound the first ter on the right-hand side of A.4.18 by b, when Z N is repaced by Z N after adding an extra factor 2, e.g., suppose that P N α ˆF, ε { Z N 2 x κ cγγx } b. A.4.19 Then we bound P N α ˆF, ε { Z N x } κ cγγx P N α ˆF, ε { Z N + P N α 2 x κ cγγx } ˆF, ε { Z N x κ cγγx } { Z N > 2 x κ cγγx }. A.4.20 By assuption, the first ter is bounded by b, and we ust bound the second ter. We wi prove that the second ter in A.4.20 is equa to 0. For x sufficienty arge we obtain fro C og x, κ > 1, and γx 0, 2 x κ cγγx > 6x. A.4.21 Hence for x = Z N 1 > ε ε κ 1, it foows fro Lea A.4.1d, that Z N > 2 x induces Z N > 6Z N 1 2MN 1 + 2ZN 1. A.4.22 κ cγγx On the other hand, when x = N 1 ε/2 κτ 1 < Z N 1, then, by Lea A.4.1a, and where we use again C og x, κ > 1, and γx 0, 1 ε/2 Z N 2 x κ cγγx = 2 N κτ 1 κ c γγx > 2N 1 ε 4 /4 κτ 1 + 2N 1 ε/2 κτ 1 > 2M N 1 + 2x. A.4.2 We concude that in both cases we have that Z N 2M N 1 + 2x 2MN 2 + 2x. We cai that the event Z N > 2M N 2 + 2x ipies that Z N Z N x. A.4.24 Indeed, et i 0 {1,...,N} be the node such that D i0 = Z N + 1, 758

57 and suppose that i 0 SPG 1. Then D i0 is at ost the tota nuber of stubs with abes 2 and, i.e., at ost 2M N 2 + 2x. Hence ZN < D i0 2M N 2 + 2x, and this is a contradiction with the assuption that Z N > 2M N 2 + 2x. Since by definition i 0 SPG, we concude that i 0 SPG \ SPG 1, which is equivaent to saying that the chosen stub with Z N brother stubs had abe 1. Then, on Z N > 2M N 2 + 2x, we have A Indeed, the one stub fro eve 1 connected to i 0 gives us Z N free stubs at eve and the other x 1 stubs fro eve 1 can eat up at ost x stubs. We concude fro the above that P N α ˆF, ε { Z N x } κ cγγx { Z N > 2 x } κ cγγx A.4.25 P N α ˆF, ε { Z N x } κ cγγx { Z N > 2 x κ cγγx } x = 0, since A.4.21 ipies that 2 x κ cγγx x x κ cγγx. This copetes the proof that the second ter on the right-hand side of A.4.20 is 0. We are eft to prove that there exists a vaue of b such that A.4.19 hods, which we do in two steps. First we coupe {X N i, 1 }x i=1 with a sequence of i.i.d. rando variabes {XN i, 1 }x i=1 with aw G N N α, such that aost surey, and hence Then we appy Lea A.1.4ii with X N i, 1 XN i, 1, i = 1,2,...,x, A.4.26 Z N V x N def = ax 1 i x XN i, 1. ˆX N i = X N i, 1 and y = 2x/ κ cγγx. A.4.27 We use Lea A..1 to coupe {X N i, 1 }x i=1 with a sequence of i.i.d. rando variabes {XN i, 1 }x i=1, with aw G N N α. Indeed, Lea A..1 can be appied, when at a ties i = 1,2,...,x saping is perfored, excuding at ost N α stubs with abe. Since the nuber of stubs increases with i, we hence have to verify that M N α, when M is the axia possibe nuber of stubs with abe at the oent we generate X N 1. The nuber M is bounded fro above by 2M N 2 1 ε 4 /4 + 2x N κτ 1 + 2x N α, using A.4.1a for the first inequaity and x N 1 ε/2 κτ Nα for the second one. We finay restrict to x = Z N 1 N 1 ε/2 κτ 1, as required in Proposition A..2b. Note that y = 2x/ κ cγγx N α, so that F α,n α hods, which in turn ipies condition A.1.0. We N can therefore appy Lea A.1.4ii with ˆX i = X N i, 1, i = 1,2,...,x, ĤN = G N N α, and y = 2x/ κ cγγx, to obtain fro A.4.27, P N α ˆF, ε { Z N 2 x } κ cγγx P V x N y x 1 1 2N h ]1 Gy]. A

58 Fro the ower bound of A.1.1, 1 Gy] y κ 1 K τγy = 2 κ 1 K τγy x/ κ 1 K τγyκ c γγx 2x, A.4.29 because x/ > 1 and κ 1 c γ γx κk τ γy + c γ K τ γxγy c γ κ 1 γx κk τ γy 0, by choosing c γ arge and using γx γy. Cobining A.4.28 and A.4.29 and taing 1 2N h > 1 2, we concude that x x 1 1 2N h ]1 Gy] 1 e /4, A.4.0 4x because > and can be chosen arge. This yieds?? with b = 1. In the proof of Proposition.2, in Section A.5, we often use a coroary of Proposition A..2 that we foruate and prove beow. Coroary A.4.2. Let F satisfy Assuption 1.1ii. For any ε > 0 sufficienty sa, there exists an integer such that such that for any >, P ˆF, ε { Y N Y N 1 > τ 2 + ε1 γ } 2, A.4.1 for sufficienty arge N. Proof. We use that part a and part b of Proposition A..2 together ipy: P ˆF, ε { og Z N κog Z N 1 κog + c γ γ Z N 1 og } Z 1 N 2b. A.4.2 Indeed appying Proposition A..2, with = and x = Z N 1, and hence ZN x, = ZN, yieds: P ˆF, ε { Z N x κ+cγγx} b, A.4. P ˆF, ε { Z N x/ κ cγγx} b, A.4.4 and fro the identities {Z N {Z N we obtain A.4.2. x κ+cγγx } = {ogz N κogz N 1 og x κ+cγγx κog x}, x/ κ cγγx } = {ogz N κogz N 1 ogx/ κ+cγγx κog x}, Appying A.4.2 and A..9, we arrive at P ˆF, ε { Y N Y N 1 > τ 2 + ε1 γ } A.4.5 P ˆF, ε { κ κog + c γ γ Z N 1 og Z 1] N > τ 2 + ε 1 γ } + 2b. 760

59 Observe that, due to Lea A.4.1c, and since γx = og x γ 1, where 0 γ < 1, we have on ˆF, ε, κ κog + c γ γ Z N 1 og ] Z N 1 = κ κog + c γ og Z N 1 γ 1 og + og ] Z N κ κog + c γ og ] + c γ og Z N γ 1 κ c γ + κog + c γ κ 1 ε 1 + ε γ] κ 1 γ κ γ c γ + κog + c γ κ 1 ε 1 + ε γ] τ 2 + ε 1 γ, because, for arge, and since κ 1 = τ 2, 1 γ 1 γ τ 2 τ 2+ε κ γ c γ + κog ] 1 2, τ 2 τ 2+ε cγ κ 1 ε 1 + ε γ 1 2. We concude that the first ter on the right-hand side of A.4.5 is 0, for sufficienty arge, and the second ter is bounded by 2b 2, and hence the stateent of the coroary foows. 1 A.5 Proof of Proposition.2 and Proposition.4 Proof of Proposition.2a. We have to show that P ε Y i,n ε 1, ax Y i,n T i,n ε Y i,n > ε = o N,,ε1. Fix ε > 0, such that τ 2 + ε < 1. Then, tae = ε, such that A..10 hods, and increase, if necessary, unti A.4.1 hods. We use the incusion { ax Y N T N ε Y N > ε} { T N ε Y N Y N 1 > τ 2 + ε 1 γ}. A.5.1 If the event on the right-hand side of A.5.1 hods, then there ust be a T N Y N Y N 1 > τ 2 + ε1 γ, and therefore where we denote { ax T N ε Y N Y N > ε } T N ε G, 1 G c,, ε such that A.5.2 G, = G, ε = =+1 { Y N Y N 1 τ 2 + ε1 γ}. A

60 Since A..10 ipies that on G, 1 we have Y N Y N ε, < 1, we find that, { } { G, 1 G c, Y N Y N ε, : < 1 Y N Y N 1 > τ 2+ε1 γ}. A.5.4 Tae N sufficienty arge such that, by Proposition.1, P M N > 2ZN N,ε Y ε 1 P : Y N Y + P M > 2Z,ε Y ε 1 P M > 2Z,ε Y ε 1 + ε/4, A.5.5 where we reca the definition of M = Z in A..11. Next, we use that =1 i P M > 2Z,ε Y ε 1 = 0, A.5.6 since Y = τ 2 og Z converges a.s., so that when Y ε and is arge, M 1 is uch saer than Z, so that M = M 1 + Z > 2Z has sa probabiity, as is arge. Then we use A.5.1 A.5.6, together with A..12, to derive that P ε Y N ε 1, ax Y N Y N > ε P T N ε ε Y N ε 1, ax T N ε Y N Y N > ε, Y N = Y, A P : Y N Y P G, 1 G c N N, { T ε} {ε Y ε 1 } {Y N = Y, } + ε 2 > { P ˆF, ε Y N Y N 1 > τ 2 + ε1 γ} + ε < ε/2, > by Coroary A.4.2 and A..10. Proof of Proposition.2b. We first show.6, then.5. Due to Proposition.2a, and using that {Y N ε}, we find Y N Y N + ε Y N 1 + ε2, apart fro an event with probabiity o N,,ε1, for a T N. By A..9 and because T N, this is equivaent to which ipies.6. Z N Z N κ 1+ε 2 1 ε 2 N τ 1 1+ε2 = N 1 ε4 τ 1, We next show.5. Observe that T N ipies that either 1 T N, or 1 =. Hence, fro T N and Proposition.2a, we obtain, apart fro an event with probabiity o N,,ε1, Y N 1 Y N ε ε ε ε 2, A

61 for ε > 0 sufficienty sa, and Y N = Y N Y N + Y N By A..9 this is equivaent to Y N 1 + Y N 1 Y N 1 2ε Y N 1 1 4ε2. A.5.9 Z N Z N κ1 4ε 2 1 Z N 1, when ε > 0 is so sa that κ1 4ε 2 1, since τ 2,, and κ = τ 2 1. Proof of Proposition.4. We ust show that P T N ε,ε Y N ε 1,Z N +1 N 1 ε τ 1 = o N,,ε1, A.5.10 where we reca that { T N } = { T N } { + 1 T N }. In the proof, we wi ae repeated use of Propositions.1 and.2, whose proofs are now copete. According to the definition of ˆF, ε in A..12, P { T N ε} {ε Y N ε 1 } ˆF, ε c A.5.11 P ε Y N ε 1, ax Y N Y N > ε + P ε Y N ε 1,M N > 2Z N. T N ε In turn Propositions.2a, as we as A.5.5 A.5.6 ipy that both probabiities on the righthand side of A.5.11 are o N,,ε1. Therefore, it suffices to show that P { T N ε,ε Y N ε 1,Z N +1 N 1 ε τ 1 } ˆF, ε = P { + 1 T N ε,z N +1 N 1 ε τ 1 } ˆF, ε = o N,,ε1. A.5.12 Let x = N 1 ε/2 κτ 1, and define the event I N, = I N,a I N,b I N,c I N,d, where We spit I N,a = {M N 1 < N 1 ε 4 /4 κτ 1 }, A.5.1 I N,b = {x Z N }, A I N,c = {Z N N 1 ε τ 1 }, A.5.15 I N,d = {Z N +1 ZN x,+1 ZN }. A.5.16 P { + 1 T N ε,z N +1 N 1 ε τ 1 } ˆF, ε = P { + 1 T N ε,z N +1 N 1 ε τ 1 } ˆF, ε I N, + P { + 1 T N ε,zn +1 N 1 ε τ 1 } ˆF, ε IN, c. A

62 We cai that both probabiities are o N,,ε1, which woud copete the proof. We start to show that P { + 1 T N ε,zn +1 N 1 ε τ 1 } ˆF, ε I N, = o N,,ε1. A.5.18 Indeed, by A..12,.6, and Lea 5.1, with u = τ 1 1, for the second inequaity, P { + 1 / T N ε} {Z N N 1 ε τ 1 } ˆF, ε I N, A.5.19 P { T N ε} {ε Y N ε 1 } {Z N N 1 ε 1 ε 4 τ 1,N τ 1 ]} + o N,,ε1 = o N,,ε1. Therefore, we are eft to dea with the case where Z N we can use Proposition A..2b with x = N 1 ε/2 κτ 1 Z N that, whp, N 1 ε τ 1. For this, and assuing IN,, by I N,b, and = + 1 to obtain Z N +1 ZN x,+1 ZN x κ1 ε/2 N 1 ε 1 ε/2 2 τ 1 = N τ 1 N 1 ε τ 1 > N 1 ε τ 1, A.5.20 where we have used that when T N ε and Y N > ε, then we have cog og N, for soe c = cτ,ε, and hence, for N arge enough, This proves A κ cγγx x cγγx + 1 κ x cγγx x εκ/2. For the second probabiity on the right-hand side of A.5.17 it suffices to prove that P { + 1 T N ε} ˆF, ε I c N, = on,,ε1. A.5.21 In order to prove A.5.21, we prove that A.5.21 hods with IN, c repaced by each one of the four events IN, c a,...,ic N, d. For the intersection with the event Ic N, a, we appy Lea A.4.1a, which states that ˆF, ε IN, c a is the epty set. It foows fro. that if + 1 T N ε, then κ +1 Y N > 1 ε2 τ 1 og N. A.5.22 If ˆF, ε hods then by definition A..12, and Coroary A.4.2, whp, Y N Y N 1 Y N ε Y N 1 ε2. A.5.2 Hence, if ˆF, ε hods and + 1 T N ε, then, by A.5.22 A.5.2, whp, κogz N = κ +1 Y N 1 ε 2 κ +1 Y N 1 ε2 2 τ 1 og N, A.5.24 so that, whp, Z N x = N 1 ε/2 κτ 1, A

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