On the Degree Sequence and its Critical Phenomenon of an Evolving Random Graph Process

On the Degree Sequence and its Critical Phenomenon of an Ev olving Random Graph Pro cess Xian-Y uan W u 1 ∗ , Zhao Dong 2 † , Ke Liu 2 ‡ , and Kai-Y uan Cai 3 § 1 School of Ma thematical S ciences, Capital Normal Univ ersity , Beiji ng, 100037, China. Email: wuxy@mail. cnu.edu.cn 2 Academy of Mathematics and System Sciences, Chinese Academy of Sciences, Beijing, 100190, China. Email: dz hao@amss.ac.cn; kliu@amss.ac.c n 3 Department of Automatic Con trol, Beijing Universit y of Aeronautics and Astronautics, Bei- jing, 100083, China. Email: ky cai@buaa.edu.cn Abstract : In this pap er we focus on the problem of the d egree sequence for the follo wing rand om graph process. A t an y time-step t , one of the fo llowi ng three substeps is executed: with probability α 1 , a new vertex x t and m edges incident with x t are added; or, with probability α − α 1 , m edges are added; or fi nally , with probabilit y 1 − α , m random edges are deleted. Note th at in any case edges are added in the manner of pr efer ential attachment . we pro ve t h at there exists a critical p oint α c satisfying: 1) if α 1 < α c , then the mo del has p o wer law degree sequence; 2) if α 1 > α c , then the mod el has exp onential degree sequence; and 3) if α 1 = α c , then the mo del has a degree sequence lying b etw een the ab ov e tw o cases. 1 In tro duction and statemen t of the results Recently there has b een much interest in studying la rge-sca le real-world netw ork s and attempting to mo del their pro p e rties. F or a genera l int ro duction to this topic, r eaders can refer to Alber t and AMS classi fication: 05C 07. 05C 80. Key words and phrases: degree sequence; p ow er l a w; critical phenomenon; real-world net works. ∗ Supported i n part by the Natural Science F oundation of China † Supported i n part by the Natural Science F oundation of China under grants 10671197 and 10721101 ‡ Pa rtiall y supp orted by the Natural Science F oundation of China under grants 60674082, 70221001 and 70731003. § Supported by the Natural Science F oun dation of Chi na and Mi croSoft Research Asi a under grant 60633010 1 Barab´ asi [1], Aiello, Chung and Lu [3], Bo llo b´ as and Riorda n [9], Hayes [17], Newma n [22] and W a tts [26]. Although the study of real-world netw orks as g raphs can b e traced back to long time ag o such as the classica l mo del pr op osed by Erd¨ os a nd R´ enyi [14] a nd Grilbe r t [16], recent influential activity per haps star ted with the work o f W atts a nd Strog a tz ab out the ‘small-world phenomenon’ published in 1998 [27]. Ano ther influential work may be due to the sca le-free mo del pro po sed by Bollo b´ as and Alber t in 1999 [5 ]. Since then v arious forms o f sca le-free pheno menon have b een widely revealed. In particular, p ow er la w degree distributions ha ve b een extensively in vestigated. Many new mo dels hav e b een introduced to circumv ent the sho r tcomings of the clas s ical mo dels introduced by Erd¨ os and R´ en yi [14] and Grilb ert [1 6]. One class of these new mo dels w as aimed to expla in the under lying causes for the emergence o f power law degr ee distributions. This ca n be obse r ved in ‘LCD mo del’ [10] and its gener alization due to Buckley and Osthus [8], ‘copying’ mo dels of Kumar et al . [19], the very general mo dels defined by Copp er and F rieze [1 2] a nd the o ther mo del with random deletions defined by Copp er , F riez e and V era [13] etc . F or the r eal-world netw ork of W orld Wide W eb/Internet, exp erimental studies by Albert, Ba rab´ asi and Jeong [2 ], Br o der et al . [7] and F aloutso s, F a loutsos and F a loutsos [15] de mo nstrated that the prop ortion o f vertices o f a given degree follows a n approximate inv erse p ow er law, i.e., the pro po rtion of vertices of degree k is approximately C k − α for some constants C and α . How ever other forms of the degree distributions can also b e observed in real-world netw orks (see [4] and [25]). F or exa mple, Guassian distributions can be observed in the acqua int ance netw ork of Mor mons [6]; expo nential distribution can b e observed in the p ow ergr id of so uthern California [27]. On the other hand, the degree distribution of the netw ork of world airp orts [4] interpolates b e t ween Gaussian and exp onential distributions, whereas the degr ee dis tr ibution o f the citation netw or k in high ener gy physics [20] int erp ola tes b etw een exp onential and p ow er law distributions. F or mo re for ms o f degre e distributions, readers can refer to [2 4]. Different mo dels often lea d to different fo rms of degree distr ibutio ns. An interesting pr oblem arises natur ally: do es it exist some dynamically evolving r andom gra ph pro cess which brings forth v arious degree distributions by contin uous changing of its par ameters only? This phenomeno n ha s bee n numerically in vestigated in reference [28]: F o r a genera l mo del of co llab oratio n netw or ks in [28] , Zhou et al. indicate that, while a relev ant para meter α increa ses from 0 to 1 . 5, four kinds o f degre e distributions a pp e ar as exp onential, arsy-varsy, semi-p ower law and p ower law in turn. Note that the ab ov e classification is r a ther roug h as no unambiguous b orderline b etw een tw o neig hboring pa tterns is determined. How ever, to the b es t of our knowledge, it seems that the pr oblem and its answer ha ve no t bee n formulated in a mathematica lly rigo rous manner . In this pap er we fo cus on a mo de l with edge deletions and provide pr e cise ana lysis, while a parameter v aries, the mo del exhibits v arious deg ree 2 distributions. Now, we begin to intro duce our mo del and then state our main res ults. Cons ide r the following pro cess which g e nerates a sequence of g raphs G t = ( V t , E t ), t ≥ 1. W rite v t = | V t | a nd e t = | E t | . Time-Step 1. L e t G 1 consist o f a n isola ted vertex x 1 . Time-Step t ≥ 2. 1, With probability α 1 > 0 w e add a vertex x t to G t − 1 . W e then add m rando m edges inciden t with x t . In the c a se o f e t − 1 > 0 , the m r a ndom neighbo urs w 1 , w 2 , . . . , w m are c hose n indep endently . F or 1 ≤ i ≤ m and w ∈ V t − 1 , P ( w i = w ) = d w ( t − 1) 2 e t − 1 , (1.1) where d w ( t − 1 ) denotes the degree of vertex w at the beginning of substep t . Thu s neighbours are chosen by pr efer ential attachment. In cas e of e t − 1 = 0 , then we a dd a new vertex x t and join it to a randomly chosen vertex in V t − 1 . 2, With pro bability α − α 1 ≥ 0 we add m ra ndom edges to existing vertices. If e t − 1 > 0, then bo th endpo ints are chosen independently with the same proba bilities as in (1.1). Otherwise, we do nothing. 3, With pro bability 1 − α ≥ 0 we delete min { m, e t − 1 } randomly chosen edge s from E t − 1 . Remark 1.1 The defer enc e b etwe en our mo del and t he mo del intr o duc e d in [13] is t hat, in our setting, vertex deletions, lo op and mu lti-e dge er asur es ar e forbidden, which makes { e t : t ≥ 1 } Markovian and makes it p ossible for us to give exact estimation to e t . In o rder to make the pro blem mea ningful, the following inequalities are natural a nd necessa ry: 1 / 2 < α ≤ 1; 0 < α 1 ≤ α. (1.2) F or given α and α 1 satisfying (1.2), define α c := 4 α − 2 , η := α c m/ 2 , (1.3) and choo se ǫ = ǫ ( α, α 1 ) ∈ (0 , η ) such that ρ ǫ := max  m ( α c − α 1 ) 2( η − ǫ ) , 1 2  < 1 . (1.4) Note that in case of α 1 ≥ α c , ρ ǫ = 1 2 . Let β = α c α c − α 1 ; γ = 1 − α 1 − α c 2(1 − α ) ; θ = 2 α c − α 1 2 α c ; µ = α c 2(1 − α ) . (1.5) 3 Obviously , β is well defined when α 1 6 = α c and 0 < γ < 1 when α 1 > α c . T o get o ur main results, bes ides (1.2 ), the following condition is nece s sary α 1 < 2 α c . (1.6) Now, Let D k ( t ) b e the num b er of vertices with de g ree k ≥ 0 in G t and let D k ( t ) b e the exp ectation of D k ( t ). The main results o f this pape r follow as Theorem 1.1 Assume that (1.2) and (1.6) hold. Then α c define d in (1.3) is a critic al p oint for the de gr e e se quenc e of the mo del satisfying: 1) if α 1 < α c , then ther e exists a c onstant C 1 = C 1 ( m, α, α 1 ) such t hat, for any ν ∈ (0 , 1 − ρ ǫ ) ,     D k ( t ) t − C 1 k − 1 − β     = O ( t ρ ǫ + ν − 1 ) + O ( k − 2 − β ); (1.7) 2) if α 1 > α c , then ther e exists a c onstant C 2 = C 2 ( m, α, α 1 ) such t hat     D k ( t ) t − C 2 γ k k − 1+ β     = O ( t − θ ) + O ( γ k k − 2+ β ); (1.8) 3) if α 1 = α c , then ther e exists a c onstant C c = C c ( m, α, α 1 ) such that, for any ν ∈ (0 , 1 2 ) ,     D k ( t ) t − C c u c ( k )     = O ( t − 1 2 + ν ) (1.9) uniformly in k . Wher e u c ( k ) = Z 1 0 t k − 1 e − µ 1 − t dt and β , γ , θ and µ ar e given in (1.5). Remark 1.2 The inte gr al u c ( k ) = Z 1 0 t k − 1 e − µ 1 − t dt c an b e r ewritten as u c ( k ) = " k − 2 X i =0 k − 2 − i X l =0  k − 1 i  ( k − i − l − 1)! ( k − i )! ( − 1) k − i − l − 1 # e − µ + " k − 1 X i =0  k − 1 i  µ k − i − 1 ( k − i )! # Z + ∞ 1 t − 2 e − µt dt. With help of c omputer c alculation, u c ( k ) satisfies lim k →∞ ln u c ( k ) / ( − k ) = lim k →∞ ( − ln k ) / ln u c ( k ) = 0 . Based on Theorem 1.1, we can obtain following tw o corolla ries, which provide a complete dis tinction with r esp ect to the parameter s b etw een the degree sequences for the pr esent mo del. 4 Corollary 1.2 If the p ar ameters satisfy that 1) α > 2 / 3 ; or 2) α ≤ 2 / 3 and α 1 < α c , then the pr esent r andom gr aph pr o c ess has the p ower law de gr e e se quenc e (1.7). Corollary 1.3 Assume α ≤ 2 / 3 . 1) If α c < α 1 < 2 α c , then the pr esent r andom gr aph pr o c ess has t he exp onential de gr e e se quenc e (1.8). 2) If α 1 = α c , then t he pr esent r andom gr aph pr o c ess has the critic al de gr e e se quenc e (1.9). Remark 1.3 When α > 2 / 3 , for any α 1 , t he ine quality α 1 ≤ α < α c = 4 α − 2 holds always, ther efor e, the p art 1) of Cor ol lary 1.2 fol lows fr om the p art 1) of The or e m 1.1. The p art 2) of Cor ol lary 1.2 and Cor ol lary 1.3 ar e str aightforwar d fr om The or em 1.1. Remark 1.4 A sp e cial c ase of t he p art 1) in Cor ol lary 1.2 is α = α 1 = 1 . In t his c ase, the m o del has a p ower law de gr e e se quenc e as C k − 3 , which c oincides with the r esult of [11]. F urthermor e, for any α ∈ (1 / 2 , 1] and α 1 = 2 α − 1 , the mo del has the de gr e e se quenc e C k − 3 . Remark 1.5 The r esults ar e uncle ar for t he fol lowing c ase: α ≤ 2 / 3 , 2 α c ≤ α 1 ≤ α . Cle arly , this c ase c an only app e ar when α ≤ 4 / 7 . It is natur al to c onje ctur e that the mo del p ossesses an exp onential de gr e e se quenc e in this c ase. The methodolo g y of the pro of for the main r esults follows the standard proce dur e which can b e found in [12] and [13]. The res t of the pap er is org anized as follows. In Section 2, we bo und the degr ee of vertex in G t . In Section 3, we esta blish the recurr e nc e for D k ( t ) and then derive the approximation of D k ( t ) b y a recurrence with resp ect to k . Finally , in section 4, w e solve the r ecurrence in k using Laplace’s metho d [18] and finish the pro of of Theorem 1 .1. 2 Bounding the Degree F or times s a nd t with 1 ≤ s ≤ t , let d x s ( t ) be the degr ee o f v ertex x s in G t . If x s is not added in Time-Step s , i.e., at Time-Step s , one of the other t wo substeps is ex e c uted, put d x s ( t ) = 0. In this section, we will concentrate on the upp er b ound of d x s ( t ). F or the pr esent mo del, the estimation for v t is derived in [13] as | v t − α 1 t | ≤ ct 1 / 2 log t, qs , 5 for any constant c > 0. W e s ay an ev ent happ ens quite sure ly (qs) if the probability of the compli- men tar y set of the event is O ( t − K ) for any K > 0. F or the es timation of e t , it can b e derived by the same argument a s in [13] tha t | e t − η t | ≤ ct 1 / 2 log t, qs , (2.1) for a n y constant c > 0. By a standar d ar gument on larg e devia tion (see e.g . [2 1] a nd [23]), one further ha s: for a ny ǫ > 0, there exis ts c 1 , c 2 > 0 such that P ( e t ≤ ( η − ǫ ) t ) ≤ c 1 exp {− c 2 t } , (2.2) for a ll t ≥ 1 . The following is o ur b ounding for d x s ( t ), note that our r esult is based on the exact estimation (2.2) for e t . In our opinion, to b o und the degr ee of vertex effectively , afor ehand go o d estimations for e t are necessary . Lemma 2.1 F or any α ∈ (1 / 2 , 1] and α 1 ∈ (0 , α ] , d x s ( t ) ≤ ( t/s ) ρ ǫ (log t ) 3 qs , (2.3) where ρ ǫ is given in (1.4). Pr o of : Fix s ≤ t , s upp os e that x s is added in Time-Step s . Let X τ = d x s ( τ ) for τ = s, s + 1 , . . . , t and let λ = ( s/t ) ρ ǫ N M ǫ (log t + 1) , (2.4) where N b e lar ge enough and will b e de ter mined later, and M ǫ = 12 m 2 η − ǫ . Let Y b e the { 1 , 2 , 3 } -v alued random v ariable with P ( Y = 1 ) = α 1 , P ( Y = 2) = α − α 1 and P ( Y = 3) = 1 − α . Then c onditional on X τ = x and e τ ≥ m , we have X τ +1 = x + I { Y =1 } B ( m, x 2 e τ ) + I { Y =2 } B (2 m , x 2 e τ ) − I { Y =3 } S ( m, x e τ ) , (2.5) where B ( m, p ) is the Binomia l random v ariable with par ameter ( m, p ) and S ( m, x e τ ) is the sup er geometric r a ndom v ariable with par a meter ( e τ , x, m ). Noticing that λ is small eno ugh for large N , using the basic ine q uality e − y ≤ 1 − y + 2 y 2 for sma ll y > 0 6 and the fact that S ( m, x e τ ) ≤ m , (2 .5) implies E  E ( e λX τ +1 | X τ = x, e τ ) | e τ ≥ m  ≤ e λx ( α 1  1 + x 2 e τ ( e λ − 1)  m + ( α − α 1 )  1 + x 2 e τ ( e λ − 1)  2 m +(1 − α )  1 − λ E  S ( m, x e τ )  + 2 λ 2 m E  S ( m, x e τ )  . (2.6) Using the inequalities e y ≤ 1 + y + 2 y 2 for sma ll y > 0 and (1 + y ) m ≤ 1 + my + m 2 2 y 2 for sma ll y > 0 to the right hand side of (2.6) in turn, we get E  E ( e λX τ +1 | X τ = x, e τ ) | e τ ≥ m  ≤ e λx  1 + mλx 2 e τ [( α c − α 1 ) + 12 mλ ]  ≤ e λx  1 + mλx 2 e τ  max  ( α c − α 1 ) , η − ǫ m  + 12 mλ  (2.7) ≤ e λx ( 1 + mλx max  ( α c − α 1 ) , η − ǫ m  2 e τ (1 + M ǫ λ ) ) ≤ ex p ( λx " 1 + max  ( α c − α 1 ) , η − ǫ m  m 2 e τ (1 + M ǫ λ ) #) . (2.8) Now, we ex pr ess E ( e λX τ +1 | X τ = x ) as E  e λX τ +1 | X τ = x  = E  E ( e λX τ +1 | X τ = x, e τ )  = E  E ( e λX τ +1 | X τ = x, e τ ) | e τ < m  P ( e τ < m ) + E  E ( e λX τ +1 | X τ = x, e τ ) | e τ ≥ m  P ( e τ ≥ m ) =: I + I I . (2.9) On one hand, conditional on X τ = x and e τ < m , X τ +1 ≤ x + m ≤ e τ + m ≤ 2 m holds a lwa ys, so I ≤ e 2 m P ( e τ < m ) . (2.10) 7 On the other hand, I I can b e expressed a s I I = E  E ( e λX τ +1 | X τ = x, e τ ) | e τ ≥ m, e τ ≥ ( η − ǫ ) τ  × P ( e τ ≥ m, e τ ≥ ( η − ǫ ) τ ) + E  E ( e λX τ +1 | X τ = x, e τ ) | e τ ≥ m, e τ < ( η − ǫ ) τ  × P ( e τ ≥ m, e τ < ( η − ǫ ) τ ) , by (2.8) and the fact that x ≤ e τ , I I ≤ exp ( λx " 1 + max { ( α c − α 1 ) , η − ǫ m } m 2( η − ǫ ) τ (1 + M ǫ λ ) #) + e λ ( η − ǫ ) τ exp ( max { ( α c − α 1 ) , η − ǫ m } mλ 2 (1 + M ǫ λ ) ) P ( e τ < ( η − ǫ ) τ ) ≤ exp n λx h 1 + ρ ǫ τ (1 + M ǫ λ ) io + C ′ e λ ( η − ǫ ) τ P ( e τ < ( η − ǫ ) τ ) (2.11) for some constant C ′ = C ′ ( α, α 1 , ǫ, m ) > 0. By (2.2) a nd (2.4), choosing N la rge enough, then there exists co nstants c 3 , c 4 > 0 such that I I ≤ exp n λx h 1 + ρ ǫ τ (1 + M ǫ λ ) io + c 3 exp {− c 4 τ } . (2.12) Combining (2.9)-(2.12), using (2.2) aga in for (2.10), then there exis ts c onstants c 5 , c 6 > 0 such that E ( e λX τ +1 | X τ = x ) ≤ exp n λx h 1 + ρ ǫ τ (1 + M ǫ λ ) io + c 5 exp {− c 6 τ } . (2.13) Thu s E ( e λX τ +1 ) ≤ E  exp  X τ λ  1 + ρ ǫ (1 + M ǫ λ ) τ  + c 5 exp {− c 6 τ } . (2.14) Now, put λ t = λ and λ τ − 1 = λ τ (1 + ρ ǫ (1+ M ǫ λ τ ) τ ). Obviously , if λ s is small enough, then (2.14) holds for λ τ +1 , τ = s, s + 1 , . . . , t − 1. This will imply that E ( e λX t ) = E ( e λ t X t ) ≤ E ( e λ s X s ) + c 5 t X τ = s exp {− c 6 τ } ≤ e mλ s + C ′′ (2.15) for so me cons ta nt C ′′ > 0 . Let Λ = 10 N M ǫ (log t + 1) , note that Λ ca n b e taken sma ll enough unifor mly in t by taking N large enough. Now pr ovided λ τ ≤ Λ, we can wr ite λ τ − 1 ≤ λ τ  1 + ρ ǫ (1 + M ǫ Λ) τ  8 and then λ s ≤ λ t Y τ = s  1 + ρ ǫ (1 + M ǫ Λ) τ  ≤ 1 0 λ ( t/s ) ρ ǫ which is ≤ Λ by the definition of λ . Put u = ( t/s ) ρ ǫ (log t ) 3 , by (2.15) we get P ( X t ≥ u ) ≤ ( e mλ s + C ′′ ) e − λu = O ( t − K ) for a n y constant K > 0 a nd the Lemma follows.  Remark 2.1 F or any n lar ge enou gh, ρ ǫ c an b e r etake n as max  m ( α c − α 1 ) 2( η − ǫ ) , 1 n  , in fact, this c an b e done by enlar ging α c − α 1 to max  ( α c − α 1 ) , 2( η − ǫ ) nm  inste ad in (2.7) . Thus, in the c ase of α 1 ≥ α c , ρ ǫ c an b e t aken as 1 /n . Certainly, if this is done as ab ove, c onstants M ǫ and N should b e r etaken c orr esp ondingly. 3 The recur rence for D k ( t ) In this Section, we follow the bas ic pro cedur e s in [13] to establish the re currence for D k ( t ). Put D − 1 ( t ) = 0 for all t ≥ 1. F o r k ≥ 0, we have D k ( t + 1) = D k ( t ) +(2 α − α 1 ) m E  − k D k ( t ) 2 e t + ( k − 1 ) D k − 1 ( t ) 2 e t + O  ∆ t e t      e t > 0  P ( e t > 0 ) +(1 − α ) m E  ( k + 1) D k +1 ( t ) e t − k D k ( t ) e t + O  ∆ t e t      e t ≥ m  P ( e t ≥ m ) + α 1 I k = m P ( e t > 0) + O ( P ( e t = 0 )) + O ( P ( e t < m )) . (3.1) Here ∆ t denotes the maximum degree in G t and the term O  ∆ t e t  accounts for the probability that we create larger than one deg ree changes fo r some vertices at Time-Step t + 1. By (2.2) and Lemma 2.1, we hav e ∆ t e t ≤ O ( t ρ ǫ − 1 (log t ) 3 ) , q s. (3.2) 9 The term E  k D k ( t ) e t     e t > 0  can b e expres s ed as E  k D k ( t ) e t     e t > 0  = E  k D k ( t ) e t     | e t − η t | ≤ t 1 / 2 log t  P  | e t − η t | ≤ t 1 / 2 log t | e t > 0  + E  k D k ( t ) e t     | e t − η t | > t 1 / 2 log t, e t > 0  P ( | e t − η t | > t 1 / 2 log t | e t > 0 ) = E  k D k ( t ) | | e t − η t | ≤ t 1 / 2 log t  P ( | e t − η t | ≤ t 1 / 2 log t | e t > 0 ) η t × (1 + O ( t − 1 / 2 log t )) + O ( P ( | e t − η t | > t 1 / 2 log t | e t > 0 )) , (3.3) where we use d the fact that k D k ( t ) ≤ 2 e t to hand the se cond term. F or k ≥ 1, we hav e D k ( t ) = E ( D k ( t ) | e t > 0) P ( e t > 0 ), so E ( k D k ( t ) | | e t − η t | ≤ t 1 / 2 log t ) P ( | e t − η t | ≤ t 1 / 2 log t | e t > 0 ) = k D k ( t ) − E ( k D k ( t ) | | e t − η t | > t 1 / 2 log t, e t > 0) × P ( | e t − η t | > t 1 / 2 log t | e t > 0 ) = k D k ( t ) + O ( t · P ( | e t − η t | > t 1 / 2 log t | e t > 0 )) . (3.4) Thu s, using (2 .1 ), w e hav e for k ≥ 0 E  k D k ( t ) e t     e t > 0  = k D k ( t ) η t + O ( t − 1 / 2 log t ) . (3 .5) Similarly , E  k D k ( t ) e t     e t ≥ m  = k D k ( t ) η t + O ( t − 1 / 2 log t ) . (3.6) Substituting (3.2), (3.5) and (3.6 ) in to (3.1), using (2 .2 ) again to the other terms, w e derive the following appr oximate recur rence for D k ( t ): D − 1 ( t ) = 0 for all t > 0 and for k ≥ 0 D k ( t + 1) = D k ( t ) + ( A 2 ( k + 1 ) + B 2 ) D k +1 ( t ) t + ( A 1 k + B 1 + 1) D k ( t ) t +( A 0 ( k − 1 ) + B 0 ) D k − 1 ( t ) t + α 1 I k = m + O ( t ρ ǫ − 1 (log t ) 3 ) , (3.7) where A 2 = 1 − α 2 α − 1 ; A 1 = − 2 − α 1 2(2 α − 1) ; A 0 = 2 α − α 1 2(2 α − 1) ; B 2 = B 0 = 0 and B 1 = − 1 . 10 Note that the hidden constant, write as L , in term O ( t ρ ǫ − 1 (log t ) 3 ) o f (3.7) is unifor m in k , which follows from the fact that e t = O ( t ) and k D k ( t ) ≤ 2 e t = O ( t ) uniformly in k . If we heuristica lly put ¯ d k = D k ( t ) t and assume it is a cons tant, we get ¯ d k = ( A 2 ( k + 1) + B 2 ) ¯ d k +1 + ( A 1 k + B 1 + 1) ¯ d k +( A 0 ( k − 1 ) + B 0 ) ¯ d k − 1 + α 1 I k = m + O ( t ρ ǫ − 1 (log t ) 3 ) . This leads to the consideration o f the recurre nc e in k : d − 1 = 0 and for k ≥ − 1, ( A 2 ( k + 2) + B 2 ) d k +2 + ( A 1 ( k + 1 ) + B 1 ) d k +1 + ( A 0 k + B 0 ) d k = − α 1 I k = m − 1 . (3.8) The following Lemma s hows that, on certain conditions, (3.8) is a go o d approximation to (3.7). Note that our Lemma is a generaliz ation of Le mma 5 .1 in [13]. Lemma 3.1 L et d k b e a solut ion for (3.8 ) such that | d k | ≤ C k for k > 0 and a c onstant C . We have 1) if α 1 ≤ α c , then, for any ν ∈ (0 , 1 − ρ ǫ ) , ther e exist s a c onstant M 1 > 0 su ch that | D k ( t ) − td k | ≤ M 1 t ρ ǫ + ν , (3.9) for al l t ≥ 1 and k ≥ − 1 ; 2) if α c < α 1 < 2 α c , then ther e exists a c onst ant M 2 > 0 such that | D k ( t ) − td k | ≤ M 2 t 1 − θ , (3.10) for al l t ≥ 1 and k ≥ − 1 , wher e θ is given in (1.5). Pr o of . Let Θ k ( t ) = D k ( t ) − td k and k 0 = k 0 ( t ) = ⌊ t ρ ǫ (log t ) 3 ⌋ . Lemma 2.1 implies 0 ≤ D k ( t ) ≤ t − 10 for k ≥ k 0 ( t ) . (3.11) Pr o of of p art 1) : Equatio n (3.1 1) and d k ≤ C /k imply that (3.9) ho lds for k ≥ k 0 uniformly , i.e., there exis ts a constant N 1 > 0 , indep endent to k and t , such that | D k ( t ) − td k | = | Θ k ( t ) | ≤ N 1 t ρ ǫ for a ll k ≥ k 0 ( t ) a nd t ≥ 1. Recall that the hidden constant in O ( t ρ ǫ − 1 (log t ) 3 ) of (3.7) is deno ted by L . F or any ν ∈ (0 , 1 − ρ ǫ ), let R ≥ L satisfying Lt ρ ǫ − 1 (log t ) 3 ≤ R t ρ ǫ + ν − 1 11 for a ll t ≥ 1 . L e t N 2 = R ρ ǫ + ν + 1, take σ > 0 such that 1 − R N 2 − (1 + σ ) (1 − ρ ǫ − ν ) ≥ 0 , (3.12) and take δ ∈ (0 , 1) such that δ 1+ σ < e − 1 < δ. (3.13) Let t 1 > 0 b e an integer such that k 0 ( t ) ≤ − 1 A 1 t = 2(2 α − 1) 2 − α 1 t (3.14) and δ 1+ σ ≤  1 − 1 t + 1  t +1 ,  1 − 1 − R/l t + 1  t +1 1 − R/l ≤ δ (3.15) for a ll t ≥ t 1 and l ≥ N 2 . Now, for the ab ov e t 1 , le t N 3 ≥ N 1 satisfying | Θ k ( t ) | ≤ N 3 t ρ ǫ + ν for a ll 1 ≤ t ≤ t 1 and k ≥ − 1 . (3.16) T ake M 1 = ma x { N 2 , N 3 } . (3.17) W e will prove that (3.9) holds for the ab ov e M 1 by induction. Our inductive hyp othesis is H 1 t : | Θ k ( t ) | ≤ M 1 t ρ ǫ + ν for a ll k ≥ − 1 . Note that (3.16) and (3.1 7) imply that H 1 t holds for 1 ≤ t ≤ t 1 . It follows fr o m (3.7) and (3.8) that Θ k ( t + 1) = Θ k ( t ) + A 2 ( k + 1) Θ k +1 ( t ) t + ( A 1 k + B 1 + 1) Θ k ( t ) t + A 0 ( k − 1) Θ k − 1 t + O ( t ρ ǫ − 1 (log t ) 3 ) . (3.18) F or t ≥ t 1 , by (3.14), w e hav e t + A 1 k + B 1 + 1 ≥ 0 a nd then (3.18) implies | Θ k ( t + 1) | ≤ A 2 ( k + 1) | Θ k +1 ( t ) | t + ( t + A 1 k + B 1 + 1) | Θ k ( t ) | t + A 0 ( k − 1) | Θ k − 1 ( t ) | t + Rt ρ ǫ + ν − 1 ≤ ( t + A 2 ( k + 1 ) + A 1 k + B 1 + 1 + A 0 ( k − 1 )) M 1 t ρ ǫ + ν − 1 + Rt ρ ǫ + ν − 1 = ( t + A 2 + B 1 + 1 − A 0 ) M 1 t ρ ǫ + ν − 1 + Rt ρ ǫ + ν − 1 . 12 Let ε 0 = A 2 + B 1 + 1 − A 0 = ( α 1 − α c ) /α c , no ticing that α 1 ≤ α c , we have ε 0 ≤ 0 . Then, combining (3.12), (3 .15) and (3.17), we have ( t + ε 0 ) M 1 t ρ ǫ + ν − 1 + Rt ρ ǫ + ν − 1 M 1 ( t + 1) ρ ǫ + ν ≤ M 1 t ρ ǫ + ν + Rt ρ ǫ + ν − 1 M 1 ( t + 1) ρ ǫ + ν = (  1 − 1 − R/M 1 t + 1  t +1 1 − R/ M 1 ) 1 − R/ M 1 t +1 (  1 − 1 t + 1  t +1 ) − (1 − ρ ǫ − ν ) t +1 ≤ δ 1 − R/ M 1 t +1 · ( δ 1+ σ ) − (1 − ρ ǫ − ν ) t +1 = δ (1 − R M 1 − (1+ σ )(1 − ρ ǫ − ν )) / ( t +1) ≤ 1 . The induction hypothesis H 1 t +1 has b een verified and the pr o of of part 1 ) is completed. Pr o of of p art 2) : In this case, we hav e α c < α 1 < 2 α c and then, for some ν ∈ (0 , 1 / 2), ε 0 ≤ ρ ǫ + ν < 1 − θ ( note that in this case ρ ǫ = 1 / 2). Same as what we have done for pa rt 1), for certain σ > 0 and δ ∈ ( e − 1 , 1), w e hav e ( t + ε 0 ) M 2 t − θ + Rt − θ M 2 ( t + 1) 1 − θ ≤ δ (1 − ε 0 − R M 2 − (1+ σ ) θ ) / ( t +1) ≤ 1 for sufficient larg e t and M 2 . This is eno ugh for a inductive pr o of of (3.10).  Remark 3.1 [Remark 5.2 in [13]] L emma 3.1 implies that if ther e is a solution for (3.8) such that d k ≤ C /k , then lim t →∞ D k ( t ) /t exists and e quals to d k . In p articular, it is shown that: if ther e exists a solution for (3.8) such that d k ≤ C /k , t hen the solution is u n ique. 4 Solving (3.8 ) and the pro of of Theorem 1.1 In order to solve (3.8), let us consider the following ho mogeneous equation ( A 2 ( k + 2) + B 2 ) f k +2 + ( A 1 ( k + 1) + B 1 ) f k +1 + ( A 0 k + B 0 ) f k = 0 , k ≥ 1 (4.1) which is solved by Laplace’s metho d as explained in [1 8]. F or k ≥ 1 , we construct function f k has the following form f k = Z b a t k − 1 v ( t ) dt, (4.2) where co nstants a and b , a nd function v ( t ) ar e to be determined later . Int egr ating by parts k f k = [ t k v ( t )] b a − Z b a t k v ′ ( t ) dt. (4.3) 13 Let φ 1 ( t ) = A 2 t 2 + A 1 t + A 0 , φ 0 ( t ) = B 2 t 2 + B 1 t + B 0 . Substituting (4.2 ) and (4.3) in to (4.1), we o btain [ t k φ 1 ( t ) v ( t )] b a − Z b a t k φ 1 ( t ) v ′ ( t ) dt + Z b a t k − 1 φ 0 ( t ) v ( t ) dt = 0 . (4.4) Equation (4.1 ) will b e satisfied if we hav e v ′ ( t ) v ( t ) = φ 0 ( t ) tφ 1 ( t ) , (4.5) and [ t k v ( t ) φ 1 ( t )] b a = 0 . (4 .6) Let a = 0 and b equal to a ro o t of v ( t ) φ 1 ( t ) = 0, the pa rameters a and b can b e determined satisfying (4.6). Obviously , φ 0 ( t ) a nd φ 1 ( t ) ca n b e rewritten as φ 0 ( t ) = − t ; φ 1 ( t ) = At 2 − ( A + B ) t + B = A ( t − 1)( t − B / A ) , (4.7) where A = 1 − α 2 α − 1 , B = 2 α − α 1 α c = A + α c − α 1 α c . (4.8) Now, we so lve the equation (4.1) in the fo llowing c ases: 1 ), α 1 < α c ; 2 ), α 1 > α c and 3), α 1 = α c resp ectively . F or case α 1 < α c , we hav e B > A , then the differential equation (4.5) is homogeneo us and can b e int egr ated to derive v ( t ) = ( t − 1) β ( t − B / A ) − β , (4.9 ) where β = 1 / ( B − A ) is given by (1.5). Since in this case β > 1, so b y (4.7), the equation v ( t ) φ 1 ( t ) = A ( t − 1) 1+ β ( t − B / A ) 1 − β = 0 (4.10) has a unique ro ot 1. Thus, the para meter b = 1 satisfies (4.6). Substituting the par ameter b and the function v ( t ) in to (4.2) a nd r emoving a co nstant mult iplicative factor, w e obtain a solution u 1 ( k ) to (4 .1) for k ≥ 1: u 1 ( k ) = Z 1 0 t k − 1  1 − t 1 − ζ t  β dt, (4.11) where ζ = A/B . The order of the function u 1 ( k ) with re s pe c t to k is given by the following Lemma. 14 Lemma 4.1 [Lemma 6.1 in [13]] L et k ≥ 1 . Then u 1 ( k ) = (1 + O ( k − 1 )) D 1 k − (1+ β ) (4.12) for D 1 = D 1 ( α, α 1 ) a fixe d c onstant. In Case of α 1 > α c , we have B < A , and equation (4.5) ha s the same solution as (4 .9). In addition, under the co nditions (1.2) and (1.6), one further has β < − 1, and then the e quation (4.10) has a unique ro ot γ := B / A as given in (1.5). So we can take b = γ to s a tisfy (4.6). Thus u 2 ( k ) = Z γ 0 t k − 1  γ − t 1 − t  − β dt = γ k − β Z 1 0 t k − 1  1 − t 1 − γ t  − β dt is a solution to (4 .1) for k ≥ 1. By Lemma 4.1, we hav e u 2 ( k ) = (1 + O ( k − 1 )) D 2 γ k k − 1+ β (4.13) for so me fixed constant D 2 = D 2 ( α, α 1 ). Finally , we consider the ca se of α 1 = α c . In this case A = B and the equa tion (4.5) c a n b e int egr ated to derive v ( t ) = e − µ/ (1 − t ) with µ = 1 / A given in (1.5 ). With same argument as in cases 1) and 2), ta ke b = 1 and define u c ( k ) = Z 1 0 t k − 1 e − µ 1 − t dt, then u c is a solution to (4 .1) for k ≥ 1. Crudely , u c ( k ) ≤ Z 1 0 t k − 1 dt = 1 /k . (4.14 ) The precio us repr esentation of u c ( k ) can b e found in Rema rk 1.2. Note that in all the three cases, u 1 , u 2 and u c do not sa tisfy equation (4.1) when k = 0. In fa ct, as ca lculated in [13], fo r i = 1 , 2 o r c , we alwa ys have 2 A 2 u i (2) + ( A 1 + B 1 ) u i (1) = [ φ 1 ( t ) v ( t )] b a = − φ 1 (0) v (0) 6 = 0 . (4.15) Now, we are going to solve (3.8). By Remar k 3.1, we o nly need to constr uc t a so lution for (3.8) which satisfies the re quirements of Lemma 3.1. Actually , w e will construct such a solution based on the s olution of (4.1) given ab ove. Denote b y g the solution for (4.1), i.e., g = u 1 , u 2 or u c in the three cases resp ectively . 15 F or m > 1, de fine w k = 0 for k ≥ m , w m − 1 = − α 1 / [( m − 1) A 0 ] and for j = m − 2 , m − 3 , . . . , 1, let w j be such that A 2 ( j + 2) w j +2 + ( A 1 ( j + 1 ) + B 1 ) w j +1 + A 0 j w j = 0 . Then w k satisfies (3.8) for k ≥ 1. There fo re, any linea r combination of g and w is a solution of (3.8) for k ≥ 1. Now, let D = − 2 A 2 w 2 + ( A 1 + B 1 ) w 1 2 A 2 g (2) + ( A 1 + B 1 ) g (1) and d = − A 2 ( D g (1) + w 1 ) B 1 . Note that D and d depend on g = u 1 , u 2 and n c resp ectively . By (4.15), D is well-defined. Define d k =        0 , if k = − 1 d, if k = 0 D g ( k ) + w k , otherwise . It is stra ightforw ard to c heck that d k given ab ov e is the solution o f (3.8), by (4.12), (4.13) and (4.1 4), we know that d k satisfies the r e quirements o f Lemma 3.1. F or m = 1, we can take D = − α 1 2 A 2 g (2) + ( A 1 + B 1 ) g (1) , d = − D A 2 g (1) B 1 and direc tly define d k =        0 , if k = − 1 d, if k = 0 . D g ( k ) , other wise Similarly , in this ca s e d k is also a solution to (3.8) which satisfies the co nditio n o f Lemma 3.1. Pr o of of The or em 1.1 : By the c onstruction of the solution d k and Lemma 3.1, the theorem fo llows immediately by tak ing C i =          − (2 A 2 w 2 + ( A 1 + B 1 ) w 1 ) D i 2 A 2 u i (2) + ( A 1 + B 1 ) u i (1) , for m > 1 − α 1 D i 2 A 2 u i (2) + ( A 1 + B 1 ) u i (1) , for m = 1 for i = 1 , 2 , c, where D 1 and D 2 are given in (4.12) and (4 .13), D c = 1 .  Ac kno wledgemen ts This work was b egun when one of us (Xian-Y uan W u) was visiting Institute of Mathematics , Academia Sinica. He is thankful to the proba bilit y gr oup of IM-AS for ho spitality . 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