The distribution of second degrees in the Bucley Osthus random graph model Andrey Kupavsii, Liudmila Ostroumova, Dmitriy Shabanov, Prasad Tetali Abstract We consider a random graph process analogous to one suggested by Barabási and Albert Their model can be roughly described as follows At each step we add one vertex and one outgoing edge, and the probability that the new vertex is connected to a vertex v is proportional to the degree of v These types of processes serve as models of real-world networs In this paper we consider a well-nown generalization of the Barabási and Albert model the Bucley-Osthus model Bucley and Osthus proved that in this model the degree sequence has a power law distribution As a natural and arguably more interesting next step, we study the second degrees of vertices Roughly speaing, the second degree of a vertex is the number of vertices at distance two from this vertex The distribution of second degrees is of interest because it is a good approximation of PageRan, where the importance of a vertex is measured by taing into account the popularity of its neighbors We prove that the second degrees also obey a power law More precisely, we estimate the expectation of the number of vertices with the second degree greater than or equal to and prove the concentration of this random variable around its expectation using the now-famous Talagrand concentration inequality over product spaces As far as we now this is the only application of Talagrand inequality to random web graphs, where the preferential attachment edges are not defined over a product distribution, maing the application nontrivial, and requiring certain novelty Keywords: random graphs, preferential attachment, power law distribution, second degrees 1 Introduction In this paper we consider some properties of random graphs The standard random graph model Gn, m was introduced by Erdős and Rényi in [12] In this model we randomly choose Yandex, Department of theoretical and practical research; Department of Discrete Mathematics, Moscow Institute of Physics and Technology; Moscow State University, Faculty of Mechanics and Mathematics, Department of Number Theory Yandex, Department of theoretical and practical research; Moscow State University, Mechanics and Mathematics Faculty, Department of Mathematical Statistics and Random Processes Yandex, Department of theoretical and practical research; Department of Discrete Mathematics, Moscow Institute of Physics and Technology; Moscow State University, Faculty of Mechanics and Mathematics, Department of Probability Theory School of Mathematics and School of Computer Science, Georgia Institute of Technology, Atlanta, GA, USA 1
one graph from all graphs with n vertices and m edges The similar model Gn, p was suggested by Gilbert in [14]: we have an n-vertex set and we join each pair of vertices independently with probability 0 < p < 1 Many papers deal with the classical models Fundamental results about these models can be found in [5], [13], [16] Recently there has been an increasing interest in modeling complex real-world networs It is well understood that real structures differ from standard random graphs Many models of real-world networs and main results can be found in [6] For example, a basic characteristic of random graphs is their degree sequence In many real-world structures the degree sequence obeys a power law distribution However, standard random graph models do not have this property In 1999, Barabási and Albert [3] suggested the so-called preferential attachment model that has a desired degree distribution Later Bollobás and Riordan [7] gave a more precise definition of this model In this model the probability that a new vertex is connected to some previous vertex v is proportional to the degree of v Bollobás and Riordan also proved that the degree sequence has a power law distribution with exponent equal to 3 Naturally one would not expect that this constant will suit all or even most of the real networs In order to mae the model more flexible, two groups of authors see [10] and [11] proposed to add one more parameter an initial attractiveness of a node which is a positive constant that does not depend on the degree In [9], Bucley and Osthus gave an explicit construction of that model Many papers deal with different variations of preferential attachment We mention here the paper by Rudas, Tóth and Valo see [18] The authors consider quite a generic model of a random tree and prove some interesting results concerning a neighborhood structure of a random vertex Also one can find a neighborhood analysis in preferential attachment models in the preprint [4] on the wea graph limit This paper deals with the Bucley Osthus model, which we now describe Let n be a number of vertices in our graph, m N and a R be fixed parameters We begin with the case m 1 We inductively construct a random graph Ha,1 n The graph Ha,1 1 consists of one vertex and one loop we can also start with Ha,1, 0 which is the empty graph Assume that we have already constructed the graph Ha,1 t 1 At the next step we add one vertex t and one edge between vertices t and i, where i is chosen randomly with Pi s d H t 1 s a,1 if 1 s t 1, a1t 1 a if s t a1t 1 Here d H t a,1 s is the degree of the vertex s in Ha,1 t We will also use the notation ds : d H n a,1 s To construct Ha,m n with m > 1 we start from Ha,1 mn Then we identify the vertices 1,, m to form the first vertex; we identify the vertices m 1,, 2m to form the second vertex, etc As for the edges, if the edge e connects vertices im and jm l, 1, l m, in the graph Ha,1 mn then we draw an edge e between vertices i 1 and j 1 in Ha,m n Note that we have a one-to-one correspondence between the edges of Ha,1 mn and Ha,m, n so there may be multiple edges and multiple loops between vertices in Ha,m n Denote by H n a,m the probability space of constructed graphs In [9] Bucley and Osthus proved that the degree sequence of Ha,m n has a power law with exponent 2 a if a is a natural number Recently Grechniov substantially improved this 2
result Theorem 1 Grechniov, [15] Let a R If d dn m and ψn as n, then Bd m ma, a 2n Rd, n Bma, a 1 d 2 a n d 1 ψn, with probability tending to 1 as n Here Rd, n is the number of vertices in H n a,m with degree equal to d and Bx, y is the beta function In this paper we consider the so-called second degrees of vertices Roughly speaing, the second degree of a vertex is the number of vertices at distance two from the vertex We prove that the number of vertices Y n with the second degree at least decreases as a, where a is the initial attractiveness This means that the distribution of second degrees obeys power law To prove this we calculate the expectation of Y n and show the concentration of this random variable around its expectation using Talagrand s inequality The application of this inequality is nontrivial, in particular, we have to redefine the probability space of the Bucley-Osthus graph so that we obtain a product probability space After that we use Talagrand s inequality in asymmetric form But still it is difficult to verify the conditions of Talagrand s theorem, therefore we use some combinatorial constructions first This paper is organized as follows In Section 2 we give the main definitions and formulate the results In Sections 3 and 4 we prove the theorems from Section 2 2 Definitions and results 21 Definitions In this paper we study the random graph H n a,1 We shall write ij H n a,1 if H n a,1 contains the edge ij; we shall write t H n a,1 if t is a vertex of H n a,1 Given a vertex t H n a,1, the second degree of the vertex t is d 2 t #{ij : i t, j t, it H n a,1, tj H n a,1} In other words, the second degree of t is the number of edges adjacent to the neighbors of t except for the edges adjacent to the vertex t We say that a vertex t is a -vertex if d 2 t Let M 1 nd be the expectation of the number of vertices with degree d in H n a,1: Mnd 1 E #{t Ha,1 n : d H n a,1 t d} Let Y n denote the number of -vertices in H n a,1 In this paper we study second degrees of vertices in H n a,1 The main results are stated in Theorems 2 and 5 We also consider the variable X n equal to the number of vertices with second degree in H n a,1 Note that Y n i X ni 3
22 Expectation Theorem 2 For any > 1 we have a 1Γ2a 1 ln a1 EY n n 1 O O Γa 1 a n The easy consequence of Theorem 2 is Corollary 1 We have EY n Θ n for O n 1 a Using the same technique as in proof of Theorem 2 we can prove the following Theorem 3 For any 1 we have EX n a 1Γ2a 1n Γa a1 1 O Again, as a consequence we get Corollary 2 We have EX n Θ n for O n 1 ln a1 O n We need the following definition Let N n l, be the number of vertices in H n a,1 with degree l, with second degree, and without loops: N n l, #{t H n a,1 : dt l, d 2 t, tt / H n a,1} To prove Theorem 2 we need the following auxiliary theorem Theorem 4 In H n a,1 we have EN n l, cl, n θn, l,, where θn, l, < Cl The constants cl, are defined as follows: cl, 0 c0, 0, c1, c1, 1 a 1 3a 1 c a 1 3a 1, > 0, al 1 cl, cl, 1 l1 a 2a cl 1, l 2 a, > 0, l > 1 l1 a 2a Here c B,a2 Ba,a1 To prove these theorems we shall use two lemmas In [15] Grechniov obtained the following result Lemma 1 Let 1 be natural; then where θn, < C/ M 1 n B 1 a, a 2n Ba, a 1 4 θn,,
Denote by P n l, the number of vertices in H n a,1 with a loop, with degree l, and with second degree Lemma 2 For any n we have where EP n l, pl,, p2, 0 P, pl, 0 l 2 a pl 1, 0 l1 a 2 a, l 3, pl, al 2a 1 pl, 1 l1 a 1 a pl 1, l 2 a, l1 a 1 a l 3, 1 Here P is some constant For the other values of l and we have pl, 0 23 Concentration Theorem 5 Let δ > 0 and O n 1 2aδ Then for some ɛ > 0 we have P Y n EY n > EY n 1 ɛ ō1 It is a concentration result which means that the distribution of second degrees does, as the distribution of degrees, obey asymptotically a power law This theorem is a non-trivial application of Talagrand s inequality see [19] Instead of Talagrand s inequality it is possible to apply Azuma s inequality see [1], but as we show later the result would have been weaer with Azuma s inequality We can prove an analogous result for the value X n Theorem 6 Let δ > 0 and O n 1 4aδ Then for some ɛ > 0 we have P X n EX n > EX n 1 ɛ ō1 If we substitute a 1 in the Bucley Osthus model then we obtain the Bollobás Riordan model [8] The second degrees in this model were considered in [17] The concentration of second degrees in [17] was proved using Azuma s inequality This inequality provided the concentration of X n around its expectation for all O n 1 6δ with any positive δ As stated in Theorem 6 Talagrand s inequality gives the stronger result: for Bollobás Riordan model we obtain the concentration for all O n 1 5δ We obtain this improvement in spite of the fact that the proof of the concentration of X n in Theorem 6 uses the concentration of Y n from Theorem 5, so it is not optimal in this sense It is possible to generalize Theorem 5 and also Theorem 6 to the case of arbitrary m > 1 The only problem in this case is that we could not prove an analog of Theorem 2 or Corollary 1 for m > 1 since it demands even more calculations But one would expect that the following conjecture is true 5
Conjecture 1 For any m > 1 and O n min{ 1 2a, 2a} 1 we have EYn m Θ n, where a Yn m is the number of -vertices in Ha,m n We can generalize Theorem 5 in the following way Theorem 7 Suppose Conjecture 1 is true Let m N, δ > 0 and O Then for some ɛ > 0 we have P Yn m EYn m > EYn m 1 ɛ ō1 n min{ 1 2aδ, 1 2aδ} In Subsections 31 34 we prove Theorem 5 using Corollary 1 In Subsection 35 we prove Theorem 6 using Corollary 2 In Subsection 36 we present the setch of the proof of Theorem 7 In Section 4 we prove results from Subsection 22 Theorem 2, Theorem 4, and Lemma 2 in Subsections 42, 41 and 43 respectively Finally, we prove Theorem 3 in Subsection 44 3 Concentration 31 Interpretation of the Bucley Osthus model in terms of independent variables We consider the following sequence: 1, ξ 1, 2, ξ 2,, n, ξ n, where ξ 1,, ξ n are mutually independent random variables For every i, we have ξ i : Ω i {1,, 2i 1} here Ω i, F i, P i is some probability space and P i ξ i 2j 1 a a 1i 1 j 1,, i, P i ξ i 2j 1 a 1i 1 j 1,, i 1 We can interpret the sequence in the following way Each i is a vertex of a graph Each ξ i is an endpoint of the edge that goes from the vertex i If ξ i 2j 1, then the edge goes to the vertex j If ξ i 2j, then we say that the edge from the vertex i goes to the same vertex as the edge from the vertex j The value of the variable ξ j can also be even say ξ j 2j 1, for some integer j 1, then the edge from the vertex i is again redirected according to the variable ξ j1 Finally this process stops at some odd value 2v 1 and we say that ξ i as well as ξ j and ξ j1 leads to the vertex v We also say that ξ i leads to ξ j It is not hard to chec that the graph model we obtained is exactly the Bucley Osthus model Indeed, at each time step i the in-degree of each vertex j {1,, i 1} is equal to the number of variables that lead directly or indirectly to the vertex j Let us give yet another interpretation of the model described above Consider a vertex v from the obtained graph We can thin of all the variables that lead to v as connected as a rooted tree, with v as the root Let X {ξ i1,, ξ id } be the set of variables that lead to v 6
We inductively construct the corresponding tree on d vertices i 1,, i d First consider those variables ξ i 1 1,, ξ i 1 l1 from X that lead to v directly The corresponding vertices i 1 1,, i 1 l 1 are adjacent to v in the tree Suppose we choose all the vertices at distance s from v and i s 1,, i s l s are the vertices at distance s from v Consider the set {ξ i s1,, ξ 1 i s1 } of variables l s1 that lead to some of ξ i s 1,, ξ i s ls We join each of the vertices i s1 1,, i s1 l s1 to the corresponding vertex from {i s 1,, i s l s } We thus obtain the set of vertices at distance s 1 from v 32 Decreasing the number of -vertices We fix a value x x 1,, x n of the random vector ξ ξ 1,, ξ n from the probability space Ω n i1 Ω i The quantity Y n is a function from Ω to N We discuss the following question How can the value Y n Y n, x decrease, if we change one coordinate x i of the vector x? In other words, we want to find ci, x max x Y n, x Y n, x, where x is an arbitrary vector that differs from x in exactly the ith coordinate Lemma 3 For any x x 1,, x n and i {1,, n} we have ci, x 2 1 Proof It is convenient to thin about the tree interpretation of the random variables If we change a value x i of one random variable ξ i to some value x i, then all the variables that lead to x i are redirected to x i In terms of the tree interpretation, we pic the branch of the tree in which all the edges lead to x i : if x i is odd, then we lin the branch to the vertex with the number x i 1/2; if x i is even, then we lin the branch to the variable ξ x i /2 We want to interpret the change of one coordinate in terms of the graph Ha,1 n Suppose x i leads to a vertex v Then all the variables that lead to x i lead to v If we change x i to x i and x i leads to v, then we change the value of all such variables from v to v Or, in terms of Ha,1, n we tae a bundle of edges in the vertex v and move the bundle to the vertex v More precisely, if we had a bundle of edges i 1, v,, i d, v, then after the change we have the edges i 1, v,, i d, v All the rest stays the same Now we go on to the proof We should show that after the change of the ith coordinate, the number of -vertices we spoil does not exceed 2 1 Suppose we moved a bundle of edges i 1, v,, i d, v It is easy to see that we could spoil only the -vertices that have a common edge with v or v itself Note that we could not spoil the -vertices in the neighborhood of v We split the set N v of the vertices incident to v into two parts: I d {i j} and N v \I If N v \I 1, then after changing the edges from the bundle, all the -vertices from N v \I are still -vertices Indeed, all the edges in vertex v except for one are 2-incident edges for any neighbor of v, so there are at least such edges for every vertex from N v \I Similarly, if I 1, then no -vertices among i 1,, i d are spoiled except for at most one, since they are all adjacent to the vertex v The only case when some of i 1,, i d is spoiled is i j v and so we will not count the edges i 1, v,, i d, v in the second degree of i j Finally, the number of -vertices we spoil does not exceed min{ I, }min{ N v \I, }1 2 1 We now want to estimate the influence of each variable more accurately Suppose Y n, x q For each -vertex v j, j 1,, q, we consider a subset of coordinates K j K vj x {i j 1,, i j d j }, such that v j is a -vertex for any y that agrees with x on the coordinates from 7
K j It is worth noting that K j depends on x, but is not uniquely defined by it For any choice of the sets K 1,, K q, we denote their collection by K Kx Clearly, Y n, y q, for any y that agrees with x on all the coordinates from all K j K For each coordinate i, we define its multiplicity Ci, x, K {j : i K j } It is easy to see that for any x and any K one has ci, x Ci, x, K So we have ci, x min{2 1, Ci, x, K} : c i x, K We call a collection K stable, if for every -vertex v i we construct all the sets K i according to the following rule: if K i contains some of the coordinates that lead to a vertex w, then K i contains all of the coordinates that lead to w Now, it should not be surprising that for such set systems we can prove an analog of Lemma 3 Namely, consider a vector x, the corresponding -vertices v j, j 1,, q, and some stable collection K Let i {1,, n} and K \ {i} : {K j \ {i}, j 1,, q} Given i there exist at least q c i x, K such -vertices that are -vertices for any x with x s x s for all s K j \ {i}, j 1,, q To prove this fact one has to follow the proof of Lemma 3 and mae sure that the proof wors also for this case The only additional consideration needed is that the number of -vertices we loose does not exceed the multiplicity Cj, x, K Lemma 4 Let K be a stable collection as described above We have Y n, x Y n, x j J c jx, K for any vector x such that x i x i for all i {1,, n}\j Proof Suppose we change one coordinate j of x and obtain some vector ˆx Then we consider d j : Cj, x, K -vertices w 1,, w dj such that j K wi x We remove j from each of these sets and we chec for each i 1,, d j whether the obtained collection guarantees w i to be a -vertex or not If w i is a -vertex then we define K wi ˆx K wi x\{j} If w i is not a -vertex then we exclude the set K wi x from K At the end of this step we obtain a new collection ˆK To prove the lemma we need one consideration Namely, instead of changing the edges i 1, v,, i d, v to i 1, v,, i d, v we can create a new imaginary vertex w and change the edges to i 1, w,, i d, w We denote the obtained graph by G w We do not count w as a -vertex even if it has 2-incident edges It is easy to chec that for this graph the collection ˆK is stable The number of -vertices except for w in the graph G w is definitely not bigger than the same number for the graph corresponding to ˆx Moreover, the multiplicity of each coordinate in ˆK is less than or equal to the corresponding multiplicity in K We also have Y n, x Y n, G w c j x, K Similarly, if x differs from x in l coordinates, then the graph corresponding to x has at least as many -vertices as the graph G obtained by forming l imaginary vertices Moreover, at each step if we change the coordinate j and form the corresponding graph G we spoil at most min{2 1, Cj, x, K} -vertices and obtain a stable set system Consequently, we have Y n, x Y n, x Y n, x Y n, G j J c jx, K 33 Construction of a suitable set K Lemma 5 Suppose Y n, x q for some vector x in the corresponding graph G x Then we can construct a stable set system K {K 1,, K q } such that n i1 c ix, K 4 5q 8
Proof First consider the set V of vertices with degree at least 2 Put NV {u : u is a neighbor of v V } Note that a vertex from V can also belong to NV Assume that NV z All vertices from NV are -vertices Let BV be the set of vertices from NV which do not have an outcoming edge that goes to V We have BV z/ 1 since each vertex has at most one outcoming edge We denote by L v, L v {1,, n}, the set of coordinates that lead to v We also put LV v V L v and LBV v BV L v For any u NV we put K u LV LBV It is easy to see that for any x such that x i x i for every i LV LBV, the vertex u is -vertex in the graph corresponding to x For i LV LBV we estimate c i x, K by 2 1 Note that LV z z We add additional z variables because the vertices from V can have loops We have deg w 1 for w BV \V and BV \V z/ 1, therefore LBV \LV z So we have c i x, K 2 1 LBV \LV LV 2 1 2z z 4 5z i LV LBV Next we consider the set W of the remaining -vertices We have W q z By the definition, for any w W all the neighbors N w of w have degree less than or equal to 1 For each w W we consider V w {v 1,, v w }, V w N w, such that the number of edges adjacent to at least one of v i N w and not adjacent to w is between and 2 We can find such V w since w is a -vertex We can choose v i N w one by one, until the total number of 2-adjacent edges does not exceed But it cannot exceed 2 since deg v i 1 for v i N w Denote by LV w all the variables that lead to V w Now for each w W we put K w LV w L w Note that LV w 2 and for w W \V we have L w 1 Now we can mae the final estimate: n c i x, K c i x, K c i x, K 4 5z i1 i LV LBV LV w w W w W \V i {1,,n}\LV LBV L w 4 5z 2q z 1q z 4 5q The fact that K is a stable set system follows from the construction 34 Application of Talagrand s inequality First we briefly review Talagrand s inequality see eg, [1] Let Ω n i1 Ω i be a product probability space with product measure Suppose α α 1,, α n, n i1 α2 i 1 We define the following distance between a set A Ω and a point x Ω: dista, x max min α i, α y A i I xy where I xy {i : x i y i } For t > 0 we denote by A t the set {x : dista, x t} 9
Theorem 8 Talagrand s inequality For any t > 0 and A Ω we have PA1 PA t e t2 4 We use the inequality to derive the following theorem Theorem 9 For t > 0,, s N, and fs that satisfies the condition f 2 s > 2 14 5s we have P Y n s tfs P Y n s e t2 4 Proof The inequality is trivial for tfs > s, so we can assume wlog that tfs s Since ξ i are independent, we can apply Talagrand s inequality to the points x from the probability space Ω Actually, all we need to prove is that for any x such that Y n, x s we have x / A t, where A {y : Y n, y s tfs} Suppose Y n, x q s Given x we fix a set system K as in Lemma 5 Then by Lemma 4 for any y A we have q s tfs Y n, x Y n, y j I xy c j x, K c We define a suitable vector α αx Namely, α i i x,k n It is easy to see that c2 j x,k α 2 j 1 We have n n c 2 jx, K max c j x, K c j x, K 2 14 5q j In the last inequality we used Lemma 5 and the definition of c j x, K Now we show that i I xy α i > t for any y A i I xy α i i I xy c i x, K n c2 j x, K q s tfs 2 14 5q tfs 2 14 5s > t 1 The second inequality holds since for q s, tfs s we have q stfs tfs The last q s inequality follows from the statement of the theorem From 1 we obtain that dista, x > t, in other words, x / A t We apply Theorem 9 with t 2 ln n, s my n tey n 1 ε, fs EY n 1 ε Here my n is the median of Y n, and, consequently, P Y n s tfs 1/2 Since for any random variable Z we have mz 2EZ, it is easy to see that the conditions of Theorem 9 hold if EY n 1 2ε 122 1 2 ln n If ε is small enough then this inequality is a consequence of Corollary 1 and the conditions of Theorem 5 We obtain that P Y n my n 2 ln ney n 1 ε 2e t2 4 ō1/n, and since Y n n for all, we have EY n my n 2 ln ney n 1 ε ō1 10
Similarly we can derive that P Y n my n 2 ln ney n 1 ε 2e t2 4 ō1/n and EY n my n 2 ln ney n 1 ε ō1 Consequently, for some δ > 0 and all sufficiently large n we have EY n my n EY n 1 δ Therefore, for some ε > 0 P Y n EY n > EY n 1 ε ō1 This concludes the proof of Theorem 5 35 Proof of Theorem 6 We use the obvious fact that X n Y n Y n 1 Fix some ɛ > 0 First we want to apply Theorem 9 to Y n and Y n 1 We argue as after the proof of Theorem 9 We put fs n1 ɛ, t 2 ln n and s 1 my n tfs, s 2 my n, s 3 my n 1tfs, s 4 my n 1 We apply Theorem 9 to Y n with s 1 and s 2, and to Y n 1 with s 3 and s 4 and obtain ō1 P Y n EY n > n1 ɛ ō1, ō1 P Y n 1 EY n 1 > n1 ɛ ō1, provided n 2 2ɛ Θ n 2 a ln n 22a It is easy to see that this holds if the conditions of Theorem 6 are satisfied for some δ > 0 We have X n EX n Y n EY n Y n 1 EY n 1, so ō1 P X n EX n > n1 ɛ ō1 6 Since n1 ɛ ō1 EX n 1 ɛ for some ɛ > 0, this inequality completes the proof of Theorem 36 Generalization to the case of arbitrary m The proof of Theorem 9 can be modified to the case of the graph H n a,m We present only the setch of the argument Suppose m > 1 is fixed The number of variables changes from n to mn The interpretation in terms of independent variables wors for this case Lemmas 3, 4, 5 hold for m > 1, but with some minor changes 11
When we tae a bundle of edges from a vertex v and move it to some vertex v, we can spoil not only the neighborhood of v, but also the vertex v Namely, suppose we change edges v, w 1,, v, w l to v, w 1,, v, w l If v was a -vertex and edges v, w 1,, v, w l were counted in the second degree of v, then the second degree of v may decrease this is impossible in the graph H n a,1 since H n a,1 is a tree It is not difficult to see that we cannot spoil the other vertices Hence, we can formulate some analogs of Lemmas 3, 4 Lemma 6 For any x x 1,, x n and i {1,, n} we have ci, x 2 2 We also have ci, x 1 min{2 1, Ci, x, K} : c i x, K Lemma 7 Let K be a stable set collection We have Y n, x Y n, x j J c jx, K 1 for any vector x such that x i x i for all i {1,, n}\j Then Lemma 5 holds for c j x, K and m > 1 without any changes The only thing left is to modify the proof of Theorem 9 We put α i mn Finally, i I xy α i c j x, K 1 2 max c j x, K 2 j i I xy c i x, K 1 mn c jx, K 1 2 mn c i x,k1 mn c jx,k1 2 c j x, K mn 2 34 5q mn q s tfs 2 34 5q mn tfs 2 34 5s mn > t, if f 2 s > 2 34 5s mn So we can formulate an analog of Theorem 9 Theorem 10 For t > 0, m,, s N, and fs that satisfies the condition f 2 s > 2 34 5s mn we have P Yn m s tfs P Yn m s e t 2 Finally, arguing as after the proof of Theorem 9, one can see that Theorem 7 follows from Theorem 10 and Conjecture 1 4 4 Estimation of EY n We need the following notation Let X be a function on n the number of vertices, l the first degree we are interested in, the second degree we are interested in; then denote by θx some function on n, l, such that θx < X 12
41 Proof of Theorem 4 It follows from the definition of Ha,1 n that N n l, 0 N n 0, 0 Indeed, since we have no vertices of degree 0, we see that N n 0, 0 Since vertices with loops are not counted in N n l,, we have no vertices of second degree 0 and N n l, 0 0 Therefore, we have EN n l, 0 EN n 0, 0 We want to prove that there exists such constant C that EN n l, cl, n θn, l,, where θn, l, < Cl Let us demonstrate that EN n 1, c1, n θ C We shall use induction on For 0 there is nothing to prove Assume that for j < we have EN n 1, j c1, j n θ Cj Denote by N i l the number of vertices with degree l in H i a,1 We use induction on i and the equality EN i1 1, N i 1,, N i 1, 1, N i N i 1, 1 2a 1 an i1, 1 a 1i a a 1i a 1 a N i 2 a 1i a Let us explain this equality Suppose we constructed Ha,1 i We add one vertex and one edge There are N i 1, vertices with degree 1 and with second degree in Ha,1 i The probability 2a that we spoil one of these vertices is We also have N a1ia i1, 1 vertices with degree 1 and with second degree 1 The probability that one of these vertices has degree 1 and second degree in Ha,1 i1 is Finally, with probability equal to the vertex i 1 a1ia a1ia has necessary degrees in Ha,1 i1 From 2 we obtain EN i1 1, EN i 1, 1 2a 1 aen i1, 1 1 a EN i a 1i a a 1i a a 1i a 3 Note that if we have at least one vertex with first degree 1 and second degree in Ha,1, i then we have at least edges in this graph Therefore EN i 1, 0 when i < Consider the case i First, note that EN 1, c1, θ C with some C For a finite number of small we can find a constant C such that EN 1, c1, θ C 13
Using 3, Lemma 1, and the assumptions of the theorem we get EN 1, EN 1 1, 1 1 a a 1 M 1 1 1 a a 1 c1, 1 1 a 1 θ C 1 c 1 a 1 θ C1 a 1 a 1 3a 1 1 c1, c1, 1 1 a a 1 a 1 θ C 1 c 1 a a 1 θ C 1 3a 3a 1 a2 c1, c1, c1, 1 1 a a 1 a 1 θ C 1 c 1 a a 1 θ C 1 c1, 3a 3a 1 a2 a 1 c1, 1 a 1 3a 1 3a 3a 1 a2 a 1 C 1 a c c1, 1 a 1 3a 1 a 1 1 c 1 a a 1 C 1 Ca 1 c1, 3a 1 c1, Ca 1 1 3a 1 c This holds for big values of Indeed, 3a 3a 1 a 2 a 1 3a 1 C 1 a 1 C 3a 1, if and C are big enough Consider the case i Using 3, Lemma 1, and the inductive assumption we get EN i1 1, EN i 1, 1 2a EN i 1, 1 1 a a 1i a a 1i a M i 1 1 a a 1i a c1, i θ C 1 2a a 1i a c i θ 1 C1 c1, 1 i θ C 1 1 a a 1i a 1 a a 1i a i 3a 1 a c1, c1, θ C 1 2a a 1i a a 1i a c1, 1θ C 1 c1, i 1 c1, 1i 1 a a 1i a 1 a a 1i a ci 1 a a 1i a cθ 1 C1 a1 1 a a 1i a c1, i 1 c1, θ C 1 2a a 1i a 1 aac1, 1 a 1i a 3a 1 1 aac a 1i a 3a 1 c1, 1θ C 1 1 a a 1i a cθ 1 C1 a1 1 a a 1i a We want to prove that there exists a constant C such that c1, C 2a a 1i a 1 aac1, 1 a 1i a 3a 1 1 aac a 1i a 3a 1 14
c1, 1C 1 1 a a 1i a cc 1 1 a a 1i a It is sufficient to prove that the following inequalities hold: c1, 1C 2a 1 a a 1i a 3a 1 and cc 2a 1 a a 1i a 3a 1 Or Note that 1 aac1, 1 c1, 1C 1 a 1i a 3a 1 1 aac a 1i a 3a 1 cc 1 1 a a 1i a C 2a a C 1 3a 1, C 2a a C 1 3a 1 2a 1 3a 1 2a a aa 1 a 1 O a 2 3a1 2 For big values of there exists a constant C such that C 2a 1 3a 1 a 5a 2a 1 a O a 1 2 But we can not choose a constant C if 2a 1 3a 1 There is a finite number of with 5a2a1 a O a 1 0 For such we want to prove that 2 EN n 1, c1, n O f with some function f Using the method above we obtain the same inequalities: f 2a a f 1 3a 1, f 2a a C 1 3a 1 There exists a function f such that the inequalities hold This completes the proof for EN n 1, Consider the case l > 1 Assume that for all i < l, j < we have EN n i, j ci, j n θ Ci j We use the following equality, which is similar to 3: EN i1 l, EN i l, 1 l1 a a 1 a 1i a l 2 a EN il 1, a 1i a al 1EN il, 1 4 a 1i a Note that if we have at least one vertex with first degree l and second degree in H i a,1 without a loop, then we have at least l 1 edges in this graph Therefore EN i l, 0 when i < l 1 Consider the case i l 1 It is sufficient to prove that EN l 1 l, Ccl, l 15 1 a a 1i a
with some C For any finite number of small l and we can easily find a constant C such that Using 4, we get EN l 1 l, Ccl, l EN l 1 l, l 2 a EN l 2l 1, a 1l 2 a al 1EN l 2l, 1 a 1l 2 a l 2 a l 1 al 1 l 1 Ccl 1, Ccl, 1 a 1l 2 a a 1l 2 a l 2 al 1l Ccl 1, Ccl, 1al l al 2a l al 2a The last inequality holds if is big enough We also need to consider a finite number of small First we show that for any finite number of small we have cl, Ω Indeed, from the recurrent relation we obtain a 2 l 4 a1 1 a l l 2 a cl, 1 cl 1, 1 a 1l 1 a/a 1 Therefore Γl 1 a cl, 1 Ω a 1 l Γl 2 a/a 1 Ω a 2 l 3 a1 1 a l Here we used Statement 1 from Subsection 42 For 2 we have al 1 cl, cl, 1 l1 a 2a cl 1, l 2 a l1 a 2a It is sufficient to prove that there exists a positive function f such that a2 4 fl2al a1 f 1al 1l 5 a2 a1 fl 2aa1l 1 4 a2 a1, a2 4 fl1 a 2a l a1 l 1 4 a2 a1 f3a a 2 a2 4 2l 1 a1 a2 5 f 1al 1l a1 The last inequality holds for some positive function f So we want to prove that EN l 1 l, O a 2 l 3 a1 1 a l 16
Suppose that we have a graph on l 1 vertices and a vertex t has first degree l and second degree Then one edge from this vertex goes to the vertex 1, l 1 vertices send edges to t, l 2 and 2 vertices send edges to the neighbors of t There are ways to choose our 2 vertex t and its neighbors In each case the probability of these neighbors to send edges to the vertex t equals aa 1 a l 2 Γa l 1 a 2 O O O l 2 a1, 3a 24a 3 la 1 a a 1 l Γl 1 a/a 1 a 1 l so EN l 1 l, O a 2 l 4 a1 a 1 l This concludes the case i l 1 For i l 1 we have EN i1 l, EN i l, 1 al 1EN il, 1 a 1i a l1 a a 1 a 1i a cl, i θ Cl 1 l 2 a EN il 1, a 1i a l1 a a 1 a 1i a cl 1, i θ Cl 1 l 2 a a 1i a cl, 1 i θ Cl 1 al 1 a 1i a l1 a a 1 cl, i cl, i cl, θ Cl 1 a 1i a l 2 a cl 1, i a 1i a cl, 1i al 1 a 1i a l1 a a 1 a 1i a cl 1, θ Cl 1 l 2 a a 1i a cl, 1θ Cl 1 al 1 a 1i a il1 a i 2ia a cl, i 1 cl, a 1i a al 1 cl, 1i a 1i a cl, θ Cl l 2 a cl 1, i a 1i a 1 l1 a a 1 a 1i a cl 1, θ Cl 1 l 2 a a 1i a cl, 1θ Cl 1 al 1 a 1i a cl, i 1 a al 1cl, 1 a 1i al1 a 2a al 2 acl 1, a 1i al1 a 2a cl, θ Cl 1 l1 a a 1 a 1i a 17
cl 1, θ Cl 1 l 2 a a 1i a cl, 1θ Cl 1 al 1 a 1i a We want to prove the following inequality: Ccl, l l1 a a 1 a 1i a a al 1cl, 1 a 1i al1 a 2a al 2 acl 1, a 1i al1 a 2a Ccl 1, l 1 l 2 a a 1i a Ccl, 1 l 1 al 1 a 1i a It is sufficient to show that the following inequalities hold and In other words and l1 a a 1 al 1 Ccl, 1l a 1i al1 a 2a a al 1cl, 1 al 1 Ccl, 1l 1 a 1i al1 a 2a a 1i a l1 a a 1l 2 a Ccl 1, l a 1i al1 a 2a al 2 acl 1, l 2 a Ccl 1, l 1 a 1i al1 a 2a a 1i a Cl l1 a a 1 al 1 a al 1 Cl 1 al 1l1 a 2a Cl l1 a a 1l 2 a al 2 a Cl 1 l 2 al1 a 2a To prove both inequalities we mae the following transformations: l l1 a a 1 l 1 l1 a 2a l l1 a a 1 l 1 al a a1 a l a 1 2 O l a 2 l1 a 2a l 1 a 1 al a l1 a 2a a1 a l a 1 l1 a 2a O l a 2 l1 a 2a 2 l a 1 al 2 a1 a al 2al 2a l a 2 2 2a2 O l a 2 l2a 2 l a 1 1 a al 2 al 32 al 32 a 12 a2 l a 2 O l a 2 l1 a 2a 18
If l or is large enough, then there exists a constant C such that Cl a 1 al 2 al 32 al 32 a 12 a2 l a 2 O l a 2 l 2a a Finally, we need to consider the finite number of small l and We want to find some function fl, such that fl, cl, l l1 a a 1 a 1i a a al 1cl, 1 a 1i al1 a 2a al 2 acl 1, a 1i al1 a 2a l 2 a fl 1, cl 1, a 1i a fl, 1cl, 1 al 1 a 1i a Such function fl, exists This concludes the proof of Theorem 4 42 Proof of Theorem 2 In this proof we shall use the following statement Statement 1 For t > 0 and fixed a > 0 Γt a Γt t a 1 O1/t Proof From Stirling s formula we obtain Γt a t t a a t t a 1 1/t Γt t a e a t It is easy to chec that t ln 1 a a O1/t t So 1 a t t e a 1 O1/t We obtain Γt a Γt t t a t aa 1 O1/t t a 1 O1/t 19
421 Estimation of c1, Lemma 8 Proof As we now c Using the recurrent relation we obtain c1, B 1 a, a 2 Ba, a 1 Γ2a 11 O1/ Γa a1 a 1Γ2a 1Γ 1 a ΓaΓ 1 2a c1, c1, 1 a 1 3a 1 c a 1 3a 1 c1, a 1Γ2a 1 Γa a 1Γ2a 1Γa ΓaΓ 3a 2 cja j 1 a 1 j 3a 1 3a 1 Γj 1 aa j 1 a 1 Γj 1 2aj 3a 1 3a 1 a 1Γ2a 1Γa ΓaΓ 3a 2 Γj 3a 1 Γj 1 2a j a 1 O1/j Γ2a 1a1 1 O1/ Γa 2a2 Γ2a 11 O1/ Γa a1 422 Sum of cl, We want to estimate the sum l1 cl, First let us prove that the series l1 ln cl, converges for all N and The inequality p cl, C 1 q l holds for any p > 1 and q min{a, 1} p 1 Here we choose C so that p We need to prove that C p p p p 1 l al 2a 1 q l 1 q al 1 p l 2 a, l 1 q l 1 We mae some transformations: pl al 2a al 1 p1 ql 2 a, 20 c1, for any
pal a 2 al 1 pql 2 a, al a 2 min{a, 1}l 2 a The last inequality holds Therefore cl, C For l 2 and any c 0 we have Γl a c cl, l2a Γl a 1 Therefore, p 1q l and l1 ln cl, converges Γl a c Γl a c cl, 1 al 1cl 1, l 2a Γl a 1 Γl a 1 Γl a c cl, l2a Γl a 1 Γl a c cl, 1al 1 Γl a 1 l1 cl, Γl a c cl, ala c Γl a 1 Γl a c c 2 cl, 1al 1 c1, Γa Γl a 1 Γa Consider the function It is easy to see that We have 1 f c Γ a c Γ f c 1 f c a c 1 O1/ 1 a c Γl a c cl, jal j a c Γl a 1 f cj Γl a c cl, j 1al j 1 Γl a 1 f cj Γa c 2 c1, j f c j, Γa Γl a c cl, jal j a c Γl a 1 f cj Γl a c cl, jal j Γl a 1 f cj 1 a c j 1 Γa c 2 c1, j f c j, Γa Γl a c cl, al a c Γl a 1 f c Γl a c cl, j Γl a 1 ala c f c j j Γa c 2 c1, j f c j Γa Γl a c 1, Γl a 1 21
If c a then taing into account Lemma 8 and the above-mentioned asymptotics for f c j we have Hence, Γl a c cl, al a c Γl a 1 f c Γa c 2 c1, j f c j O1 Γa Γl a c cl, Γl a 1 f c O1/ We want to prove that for any 0 c < a 1 the following equality holds: Γl a c ln a c cl, Γl a 1 f c O 5 We have already proved this statement for a c < a 1 Suppose that for c 1 we have Then 1 cl, Γl a c Γl a 1 f c O ln a c cl, al a c 1 Γl a c 1 f c Γl a 1 1 cl, j Γl a c 1 ala c 1 ja c f c j Γl a 1 j 1 ln a c O O ln a c 1 j We proved 5 In particular, Γl a cl, Γl a 1 f 0 Put x cl, For l 2 c1, j Γa c 1 f c Γa 1j ln a cl, l a 1f 0 O 6 cl, l1 a 2a cl, 1al 1 cl 1, l 2 a So cl, l1 a 2a cl, 1al 1 cl, l 1 a, l1 cl, al a 1 cl, 1al 1 ac1,, 22
We have a 1x 1x 1 ac1, a lcl, 1 cl, a 1f 1 x 1f 1 x 1 af 1 c1, af 1 lcl, 1 cl,, 1 ja1f 1 jx j f 1 jja1x j af 1 jc1, j af 1 j lcl, j 1 cl, j, a 1f 1 x a f 1 a 1x a a f 1 jc1, j a 1 f 1 jc1, j aa 1 f 1 j lcl, j 1 cl, j f 1 j j lcl, j af 1 lcl, a1 Γ2a 1 j Γaj 1 O1/j 1 ln j a O O ln a a1 j Γ2a 1 1 ln j a a O Γa j Here we used 6 and Lemma 8 We obtain and x O ln a Γ2a 1 a Γa aγ2a 1 Γa a1 cl, c1, x l1 423 Estimation of EY n 1 O a 1Γ2a 1 Γa a1 ln a1 1 O 1 O ln a1 ln a1 Note that EN i1 l, j j EN i l, j j al 1EN i l, 1 a 1i a j j 1 a Mi 1 j a 1i a Therefore we obtain n 1 EN n l, j j i1 al 1EN i l, 1 a 1i a n 1 i1 j j 1 a Mi 1 j a 1i a Let us estimate the sum n 1 i1 j j 1 a Mi 1 j a 1i a 23
First we compute F t j j 1 a M 1 t j Let us prove by induction on that a 1Γ2a 1Γ a C 1 F n n 1 θ Γa 1Γ 2a n with some constant C For 1 and 2 we have F n 1 j 1j 1 a M 1 nj n1 a, F n 2 F n 1 a Mn1 1 1 a n1 a an 2a 1 O 1 1 a2 n 1 O 1/n 2a 1 For 3 we have M 1 i1j M 1 i j 1 j 1 a Mi 1 j 1 j 2 a a 1i a a 1i a We multiply this equality by j 1 a and sum over all j : F i1 j j 1 a M 1 i1j j j 1 am 1 i j j Mi 1 j 1 aj 1 a j M 1 j aj 1 a i j a 1i a a 1i a j 1 F i M 1 j 1 a i j a 1i a M i 1 1 j 1 F i 1 F i 1 F i a 1i a F i 1 2 a 1 a F i 1 a 1i a a 1i a 1 a 2 a a 1i a 1 a a 1i a Note that for i 1 < 1 we have F i1 0 Consider i 1 1 Using the inductive assumption we get a 1Γ2a 1Γ a F i1 i 1 2 a C 1 1 θ Γa 1Γ 2a a 1i a i a 1Γ2a 1Γ 1 ai 1 a C 2 1 θ Γa 1Γ 1 2a a 1i a i a 1Γ2a 1Γ a a i 1 Γa 1Γ 2a a 1i a i 1 2 a C 1 θ a 1i a i 24
1 2ai a 1i a θ And we need to show that for some constant C C 2 i a 1 2a a 1i a a 1i a C 2 2 a a 1i a C 1 This inequality holds for sufficiently large C We have n 1 i1 n 1 F i a 1i a Let us estimate the sum We start with the sum It is easy to see that i1 Γ2a 1Γ a 1 O Γa 1Γ 2a Γ2a 1Γ a Γa 1Γ 2a n 1 O n 1 i1 n al 1EN i l, 1 a 1i a al 1EN i l, 1 EN i l, Ocl, i i To verify this, one can follow the proof of Theorem 4 and mae sure that it wors for the inequality EN i l, < Ccl, a 1i a with some constant C note that the analog of Lemma 1 is also needed Therefore ln a i al 1EN i l, 1 O al 1cl, 1i O Using 5 we obtain EN i l, 1 cl, 1i 1 O l /i a 1Γ2a 1 ln a1 i 1 O O Γa a i Here we used the following estimate: cl, 1l O 2a cl, 1 cl, 1l OO1 O l l1 25 a1
So we have n 1 i1 al 1EN i l, 1 a 1i a Γ2a 1 n 1 O Γa a ln a1 O n Hence EN n l, j j aγ2a 1 Γa 1 n 1 O a a 1Γ2a 1 n 1 O Γa 1 a Consider vertices with loops For 0, we have ln a1 Γ2a 1Γ a Γa 1Γ 2a n ln a1 O n 1 O n EP n l, j j 0 n i1 a 1 ai 1 Oln n For 2, we have EP i1 l, j j Therefore, we obtain EP i l, j j al 2a 1EP i l, 1 a 1i a n 1 EP n l, j j i1 al 2a 1EP i l, 1 a 1i a n 1 i1 al 2a 1pl, 1 a 1i and From the recurrent relation for pl, it follows that 1 pl, O, l 2 pl, O l 3 To obtain the second estimate consider ql, pl, / For 1 we have 1al 2a 1 ql, l al 1 a ql, 1 ql, l al 1 a ql, 1 al 2a 2 Thus, ql, Oql, where qll a 1 al 2a 1 ql 1l 2 a ql 1, l 2 a, al 2a 1 ql 1, l 2 a 26
From this equality it follows that ql O 1 l 3 We can estimate the following sum: al 2a 1pl, 1 a 1 O Hence EP n l, j O ln n j Now we are ready to estimate EY n : EY n EN n l, j EP n l, j j a 1Γ2a 1 n 1 O Γa 1 a a 1Γ2a 1 n Γa 1 a This concludes the proof of Theorem 2 43 Proof of Lemma 2 ln a1 O O ln n n ln a1 1 O O n It is easy to see that EP n 0, EP n 1, 0 For all > 0 we have EP n 2, 0 For 0 we have n a n 1 aj 2 a n a Γ n 1 a1 Γ i a a1 EP n 2, 0 a 1i 1 1 aj 1 a 1i 1 Γ n Γ i 1 i1 ji1 1 n 1 O1/n n i1 i1 ai 1 O1/i O1 a 1i 1 The rest of the proof is by induction Consider l 3, 1 Suppose that for i < l and j < we have EP n i, j pi, j We use the following equality EP i1 l, EP i l, 1 a a1 la 1 a 1 al 2a 1 EP i l, 1 a 1i a a 1i a EP i l 1, l 2 a a 1i a 7 Note that if we have at least one vertex with a loop, with first degree l and second degree in the graph H i a,1, then we have at least l 1 edges in this graph Therefore EP i l, 0 if i < l 1 Consider the case i l 1 Using 7, we get for 1 a1 EP l 1 l, l 2 a EP l 2l 1, a 1l 2 a al 2a 1EP l 2l, 1 a 1l 2 a 27
l 2 apl 1, al 2a 1pl, 1 a 1l 2 a a 1l 2 a l al 1 apl, a 1l 2 a pl, The last inequality holds for 1/a Consider the case < 1/a As in proof of Theorem 4, at first we estimate pl, : pl, Ω a 2 l a1 1 a l For 0 we have Therefore, l 2 a pl, 0 pl 1, 0 1 al 1 1 a1 l a 2 a1 pl, 0 Ω 1 a l For 1 we have al 2a 1 pl, pl, 1 l1 a 1 a pl 1, l 2 a l1 a 1 a Again, it is sufficient to prove that there exists a positive function f such that for big l a2 fl 1 al a1 f 1al 2a 1l a2 a1 1 a2 fl 2aa1l 1 a1, a2 fl1 a 1 a l a1 l 1 a2 a1 f a 2 a2 1l 1 a1 a2 f 1al 2a 1l a1 1 The last inequality holds for some function f We want to prove that EP l 1 l, O a 2 l a1 1 a l There are l possible graphs on l 1 vertices with some vertex of first degree l, second degree, and without a loop And this vertex is exactly the vertex 1 The probability of this vertex to be a vertex with first degree l and second degree equals This concludes the case i l 1 l a 1 a l 2 O a 2 l 1a 1 1 O a 2 l a1 a 1 l 28
If i l 1, then EP i1 l, EP i l, 1 la 1 a 1 a 1i a al 2a 1 EP i l, 1 EP i l 1, l 2 a a 1i a a 1i a Using the recurrent relation for pl, and induction on i it is easy to prove that EP n l, pl, This concludes the proof of Lemma 2 44 Proof of Theorem 3 We estimate the expectation of X n as follows: EX n EN n l, EP n l, cl, no cl, l O pl, l1 References l1 a 1Γ2a 1n Γa a1 a 1Γ2a 1n Γa a1 l1 1 O 1 O l1 ln a1 O1 O1 ln a1 O n [1] N Alon, J H Spencer, Probabilistic method, Wiley Interscience, New Yor, 2002 [2] K Azuma, Weighted sums of certain dependent variables, Tôhou Math J, 19, 357-367, 1967 [3] A-L Barabási, R Albert, Emergence of scaling in random networs, Science, 286, 509-512, 1999 [4] N Berger, C Borgs, J T Chayes, and A Saberi, Wea local limits for preferential attachment graphs, 2009 [5] B Bollobás, Random Graphs, Second Edition, Cambridge Univ Press, 2001 [6] B Bollobás, O M Riordan, Mathematical results on scale-free random graphs, Handboo of graphs and networs, Wiley-VCH, Weinheim, 2003 [7] B Bollobás, O M Riordan, The degree sequence of a scale-free random graph process, Random Structures and Algorithms, 18:3, 279-290, 2001 [8] B Bollobás, O M Riordan, The diameter of a scale-free random graph, Combinatorica, 24:1, 5-34, 2004 l1 29
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