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Analytic solution for preferential attachment probabilities scheme Andrea Monsellato 7 1 January 20, 2017 0 2 n Abstract a J Preferentialattachmentprobabilitiesschemeappearinthecontextofscalefreerandomgraphs 9 [1],[2]. In this work we present preferential attachment probabilities scheme as a sequence 1 of dependent Bernoulli random variables and we give an explicit expression of the transition probabilities. ] R P 1 Introduction . h t Preferential attachment probabilities schemes arise in the context of scale free random graphs, fol- a lowing the description used in [2]: m [ The preferential linking arises in this simple model not because of some special rule including a 1 function of vertex degree as in [1],[3] but quite naturally. Indeed, in the model that we consider v here, the probability that a vertex has a randomly chosen edge attached to it is equal to the ratio of 1 the degree k of the vertex and the total number of edges, 2t−1. 7 3 we want to mimic this dynamic using a sequence of dependent Bernoulli random variables. 5 0 1. Let {Ym(n)}n>0 a sequence of r.v. such that Ym(n) ∼ Ber Pni=2−mn1−Y1m(i) , with Ym(1) = 1. The 0 sequence {Ym(n)}n>0, represent the trails to attach a new (cid:16)arcs at node(cid:17)m when a new node n 7 is generated and jointed to the network, obviously n > m. The degree of the node m will be 1 X (n)= n Y (i). : m i=m m v i P X r a 1 We recover transition probabilities using elementary probability arguments, then P(Xm(n)=k)=P(Xm(n)=k|Xm(n−1)=k−1)P(Xm(n−1)=k−1)+ +P(Xm(n)=k|Xm(n−1)=k)P(Xm(n−1)=k)= =P(Xm(n−1)+Ym(n)=k|Xm(n−1)=k−1)P(Xm(n−1)=k−1)+ +P(Xm(n−1)+Ym(n)=k|Xm(n−1)=k)P(Xm(n−1)=k)= =P(Ym(n)=1|Xm(n−1)=k−1)P(Xm(n−1)=k−1)+ +P(Ym(n)=0|Xm(n−1)=k)P(Xm(n−1)=k)= k−1 2n−1−k = 2n−1P(Xm(n−1)=k−1)+ 2n−1 P(Xm(n−1)=k) Now set pm =P(X (n)=k), we have that n,k m k−1 2n−1−k pm =pm +pm , 0≤m<n, 0≤k ≤n−m (1) n,k n−1,k−12n−1 n−1,k 2n−1 2 Recovering explicit expression of transition probabilities Choosing m=0 in (1), i.e. we consider the probabilities transition equations for the ﬁrst node, we have p =1, p =0, ∀k >n (cid:26) pn1,,1k = 2kn−−11pnn,k−1,k−1+ 2n2−n−1−1kpn−1,k, ∀k ≤n (2) Observe that (1) is time invariant for all nodes scaling properly the quantity n−m. For (2) holds that: 2(n−1) pn,1 = 2n−1 pn−1,1 ∀n≥2 i.e. 2n−1(n−1)! pn,1 = Qn (2i−1) ∀n≥1 (3) i=1 also n−1 pn,n= 2n−1pn−1,n−1 ∀n≥2 i.e. (n−1)! pn,n = Qn (2i−1) ∀n≥1 i=1 2 Let 2n−k n (2i−1) a = i=1 p (4) n,k n,k (n−1)! Q from (2),(3) and (4) we have that a =22(n−1), a =1, a =0 ∀k >n n,1 n,n n,k (5) (cid:26) an,k = n−11[(k−1)an−1,k−1+2(2n−1−k)an−1,k] ∀k =1,...,n−1 n>1 Now we give an analytic solution of (5). Theorem 1. Let {a } as in (5), then n,k a =1 n,n a =22(n−k) 1+(k−1) n−k−1 2−2(j+1)(k+2j)! ∀k =1,...,n−1 n>1 (6) ( n,k j=0 (j+1)!(k+j)! h i P Proof If k =1 obvious. If k =n−1 and ∀n≥2, then (5) will be 1 an,n−1 = n−1[(n−2)an−1,n−2+2nan−1,n−1] (7) now substituting (6) in (7) we have that 0 2−2(j+1)(n−1+2j)! 22 1+(n−2) =  (j+1)!(n−1+j)!  j=0 X   n−2 n−1−n+2−12−2(j+1)(n−2+2j)! 2n = 22(n−1−n+2) 1+(n−3) + n−1  (j+1)!(n−2+j)!  n−1 j=0 X   n−2 2n 22+(n−2)= 22 1+(n−3)2−2 + n−1 n−1 (cid:2) (cid:3) 22(n−1)−2n=2n−4 −4=−4 3 Let k =2,...,n−2 substituting we have that n−k−12−2(j+1)(k+2j)! n−k−12−2(j+1)(k−1+2j)! (n−1) 1+(k−1) =(k−1) 1+(k−2)  (j+1)!(k+j)!   (j+1)!(k−1+j)!  j=0 j=0 X X     n−k−12−2(j+1)(k+2j)! 1 n−k−22−2(j+1)(k+2j)! (n−1) 1+(k−1) = (2n−1−k) 1+(k−1)  (j+1)!(k+j)!  2  (j+1)!(k+j)!  j=0 j=0 X X     n−k−12−2(j+1)(k+2j)! 1 n−k−22−2(j+1)(k+2j)! (n−1) 1+(k−1) = (n−1)− (k−1) 1+(k−1)  (j+1)!(k+j)!  2  (j+1)!(k+j)!  j=0 (cid:20) (cid:21) j=0 X X    (8)  we recover that (8) is equivalent to n−k−22−2(j+1) (k−1+2j)! (k+2j)! (2n−k−3)! 2(k−2) −(k−1) =2−2(n−k−1) −1 (j+1)! (k−1+j)! (k+j)! (n−k−1)!(n−2)! j=0 (cid:20) (cid:21) X (9) Now let n=k+2+r, r ≥0 the equality (9) become r 2−2(j+1) (k−1+2j)! (k+2j)! (k+1+2r)! 2(k−2) −(k−1) =2−2(r+1) −1 (10) (j+1)! (k−1+j)! (k+j)! (r+1)!(k+r)! j=0 (cid:20) (cid:21) X By induction we prove equality (10): For r =0 we have 2−2 2−2(k+1)! [2(k−2)−(k−1)]= −1 1! 1!k! i.e. 2−2(k−3)=2−2(k−3). By induction hypothesis we prove equality (10) for r+1, supposing it holds for generic r, then: (k+1+2r)! 2−2(r+2) (k+1+2r)! (k+2+2r)! (k+3+2r)! 2−2(r+1) + 2(k−2) −(k−1) =2−2(r+2) (r+1)!(k+r)! (r+2)! (k+r)! (k+1+r)! (r+2)!(k+r+1)! (cid:20) (cid:21) (k+1+2r)! (k+1+2r)! (k+2+2r)! (k+3+2r)! 22(r+2) +2(k−2) −(k−1) = (k+r)! (k+r)! (k+1+r)! (k+r+1)! 4 i.e. 2(r+2)(k+r+1)+2(k−2)(k+r+1)−(k−1)(k+2+2r)=(k+3+2r)(k+2+2r) from obvious arrangement w.r.t. k the thesis follow. Now consider that n (2i−1)= (2n)!, for k =1,...,n−1 holds that i=1 2nn! Q (n−1)! (n−1)! p = a = 2nn!a n,k 2n−k n (2i−1) n,k 2n−k(2n)! n,k i=1 i.e. Q n!(n−1)! n−k−12−2(j+1)(k+2j)! p = 22n−k 1+(k−1) n,k (2n)!  (j+1)! (k+j)!  j=0 X   References [1] R.,Albert, A.L., Barabsi, Statistical mechanics of complex networks, Rev. Mod. Phys, 2002 [2] S.N., Dorogovtsev,J.F.F., Mendes, Evolution of networks, Adv. Phys., 2002, 51, 1079 [3] S.N., Dorogovtsev, J.F.F., Mendes, A.N., Samukhin, Structure of Growing Networks with Preferential Linking, Physical Review Letters, 2000,21 5