aa r X i v : . [ m a t h . P R ] N ov A RANDOM MODEL OF PUBLICATION ACTIVITY ´AGNES BACKHAUSZ AND TAM ´AS F. M ´ORIDepartment of Probability Theory and Statistics,E¨otv¨os Lor´and UniversityP´azm´any P. s. 1/C, H-1117 Budapest, Hungary
E-mail address: [email protected], [email protected]
Abstract.
We examine a random structure consisting of objectswith positive weights and evolving in discrete time steps. It gener-alizes certain random graph models. We prove almost sure conver-gence for the weight distribution and show scale-free asymptoticbehaviour. Martingale theory and renewal-like equations are usedin the proofs. Introduction
In this paper we examine a dynamic model inspired by scientificpublication activity and networks of coauthors. However, the modelcontains many simplifying assumptions that are not valid in reality.We still use the terminology of publications for sake of simplicity.The model consists of a sequence of researchers. Each of them has apositive weight which is increasing in discrete time steps. The weightsreflect the number and importance of the researcher’s publications.One can think of cumulative impact factor for instance.We start with a single researcher having a random positive weight.At the n th step a new publication is born. The number of its authors israndomly chosen. Then we select the authors, that is, one of the groupsof that size; the probability that a given group is chosen is proportionalto the sum of the weights of its members. After that the weights ofthe authors of the new publication are increased by random bonuses.Finally, a new researcher is added to the system with a random initialweight.This is a preferential attachment model; one can see that authorswith higher weights have larger chance to be chosen and increase theirweights when the new publication is born. Date : 7 November 2014.2000
Mathematics Subject Classification.
Key words and phrases.
Scale free, random graphs, martingales, renewalequation.The European Union and the European Social Fund have provided financialsupport to the project under the grant agreement no. T ´AMOP 4.2.1./B-09/KMR-2010-0003. ´AGNES BACKHAUSZ AND TAM ´AS F. M ´ORI
We are interested in the weight distribution of the model. That is,for fixed t >
0, we consider the ratio of authors of weight larger than t , and study the asymptotic behaviour of this quantity as the numberof steps goes to infinity.Our main results (Section 3) include the almost sure convergence ofthe ratio of authors of weight larger than t under suitable conditions;first, when all weights are integer valued, then assuming that these ran-dom variables have continuous distribution. In both cases we describethe limiting sequence or function and determine its asymptotics. Theyare polynomially decaying under suitable conditions, thus our modelshows scale-free behaviour.The proofs of the almost sure convergence are based on the methodsof martingale theory, while the polynomial decay of the asymptoticweight distribution follows from the results of [1] about renewal-likeequations. See Section 4 for the details.This model generalizes some random graph models. To see this,assume that every publication has only one author, and at each step,when a publication is born, connect its author to the new one with anedge. We get a random tree evolving in time.In the particular case where the initial weights and author’s bonusesare always equal to 1, we get the Albert–Barab´asi random tree [2]. Theneighbour of the new vertex is chosen with probabilities proportional tothe degrees of the old vertices. Similarly, if the initial weights and thebonuses are fixed, but they are not necessarily equal to each other, weget random trees with linear weights [8], sometimes called generalizedplane oriented recursive trees. In these cases the asymptotic degreedistribution is well-known.2. Notations and assumptions
Notations.
Let the label of the only researcher being present inthe beginning be 0; the label of the researcher coming in the n th stepis n . X i is the initial weight of researcher i for i = 0 , , . . . . We sup-pose that X , X , . . . are independent, identically distributed positiverandom variables. ν n is the number of coauthors at step n . This is an integer valuedrandom variable for each n . Obviously ν n ≤ n must hold for all n ≥ ν n ≥ n ≥
1. Since the authors’ weights are not necessarily increased,this may be supposed without loss of generality.Given that ν n = k , a group of size k is chosen randomly from re-searchers 0 , . . . , n −
1. The probability that a given group is chosen isproportional to the total weight of the group. The selected researcherswill be the authors of the n th paper. RANDOM MODEL OF PUBLICATION ACTIVITY 3
Let Y n, , Y n, , . . . , Y n,ν n be nonnegative random variables. These arethe authors’ bonuses at step n . That is, the weight of the i th coauthorof the n th paper is increased by Y n,i . The order of the coauthors is thenatural order of the labels.Let Z n be the total weight of the n th paper; that is, Z n = Y n, + Y n, + . . . + Y n,ν n for n ≥ W ( n, i ) denotes the weight of author i after step n for i = 0 , . . . , n .This is equal to X i plus the sum of all bonuses Y j,ℓ for which author i is the ℓ th author of the j th paper ( ℓ = 1 , . . . , ν j , j = 1 , , . . . , n ).Let S n be the total weight after n steps; namely, S n = W ( n,
0) + . . . + W ( n, n ) = X + . . . + X n + Z + . . . + Z n .X, ν, Y n , Y and Z are random variables. X is equal to X indistribution, and Y n is equal to Y n, in distribution for n ≥
1. Theother random variables will be determined later by the assumptions.Finally, F n is the σ -algebra generated by the first n steps; F + n = σ {F n , ν n +1 } .Throughout this paper I ( A ) denotes the indicator of event A . Wesay that two sequences ( a n ), ( b n ) are asymptotically equal ( a n ∼ b n ),if they are positive except finitely many terms, and a n /b n → n → ∞ . A sequence ( a n ) is exponentially small if | a n | ≤ q n holds forall sufficiently large n ∈ N for some 0 < q < Assumptions.
Now we list the assumptions on the model.
Assumption . X , X , . . . are independent, identically distributed. Theinitial weights X n , and the triplets ( F n − , ( Y n, , . . . , Y n,ν n ) , ν n ) are in-dependent ( n = 1 , , . . . ). Assumption . X has finite moment generating function. Assumption . ν n and ( Y n, , . . . , Y n,ν n ) are independent of F n − for n ≥ Assumption . ν n → ν in distribution as n → ∞ ; in addition, E ν n → E ν < ∞ and E ν n → E ν < ∞ hold.Recall that ν n ≤ n . Assumption 4 trivially holds if ν is a fixedrandom variable with finite second moment, and the distribution of ν n is identical to the distribution of min ( n, ν ), or to the conditionaldistribution of ν with respect to { ν ≤ n } . Assumption . The conditional distribution of ( Y n, , . . . , Y n,ν n ), given ν n = k , does not depend on n . Moreover, the components are condi-tionally interchangeable, given ν n = k . Assumption . Z n has finite expectation.Now we know that ( ν n , Y n , Z n ) → ( ν, Y, Z ) in distribution as n →∞ , where Y and Z are random variables. We need that they also havefinite moment generating functions, and they are not degenerate. ´AGNES BACKHAUSZ AND TAM ´AS F. M ´ORI Assumption . Y and Z have finite moment generating functions. Assumption . X n , Y n , X and Y are positive with positive probabilitiesfor every n = 1 , , . . . . In addition, if Y is integer-valued, then thegreatest common divisor of the set { i : P ( Y = i ) > } is equal to 1.The condition on the positivity of X n and Y n is not crucial. Thepositivity of X and Y implies that the same holds for X n and Y n if n islarge enough; we may assume this for all n without loss of generality.On the other hand, if ( Y n ) is identically equal to 0, that is, thereare no bonuses at all, then the model only consists of the sequenceof independent and identically distributed initial weights X n , and theproblem of empirical weight distribution becomes trivial. The last partof this assumption excludes periodicity.There are two important particular cases satisfying all of our con-ditions. In the first one the weight of the paper is equally distributedamong the authors. That is, Z , Z , . . . are independent identically dis-tributed random variables, and Y n, = . . . = Y n,ν n = Z n /ν n . The otheroption is that every author gets the total bonus, regardless the numberof coauthors. More precisely, Y , Y , . . . are independent and identicallydistributed, and Y n, = . . . = Y n,ν n = Y n , thus Z n = ν n Y n .3. Main results
Discrete weight distribution.
Suppose first that
X, Y , Y , . . . arenonnegative integer valued random variables. Let ξ n ( j ) denote thenumber of researchers of weight j after n steps, that is, ξ n ( j ) = (cid:12)(cid:12) { ≤ i ≤ n : W ( n, i ) = j } (cid:12)(cid:12) , j, n = 1 , , . . . . The first theorem is about the almost sure behaviour of this quantity.
Theorem 1. ξ n ( j ) n → x j almost surely as n → ∞ with positive con-stants x j , j = 1 , , . . . . The sequence ( x j ) satisfies the recursion (1) x j = j − P i =1 x j − i (cid:20) ( j − i ) P ( Y = i ) E X + E Z + E (cid:0) ( ν − I ( Y = i ) (cid:1)(cid:21) + P ( X = j ) αj + β + 1 , where α = P ( Y > E X + E Z , β = E (cid:0) ( ν − I ( Y > (cid:1) . The second theorem describes the asymptotic behaviour of the se-quence ( x j ). Theorem 2.
We have x j ∼ C j − γ as j → ∞ , where C is a positiveconstant, and γ = E X + E Z E Y +1 . RANDOM MODEL OF PUBLICATION ACTIVITY 5
Continuous weight distribution.
Now we assume that the distribu-tion of X and the conditional distributions of Y n | ν n = k are continuousfor k = 1 , , . . . , n, n = 1 , , . . . . This implies that the distribution of Y n is continuous. Moreover, since the conditional distribution does notdepend on n according to Assumption 5, the distribution of Y is alsocontinuous.Let F ( t ) = P ( Y > t ), H ( t ) = E (cid:0) ( ν − I ( Y > t ) (cid:1) , and L ( t, s ) = sF ( s ) + t (1 − F ( s )) E X + E Z − H ( s ) , ≤ s ≤ t. It is clear that L ( t, s ) is continuous, and, being the difference of twoincreasing functions, it is of bounded variation for fixed t .This time ξ n ( t ) denotes the number of researchers with weight morethan t after n steps. ξ n ( t ) = (cid:12)(cid:12) { ≤ i ≤ n : W ( n, i ) > t } (cid:12)(cid:12) , t > , n = 1 , , . . . . Theorem 3. ξ n ( t ) n → G ( t ) almost surely, as n → ∞ , where G ( t ) isthe solution of the following integral equation. (2) G ( t ) = Z t G ( t − s ) d s L ( t, s ) + H ( t ) + P ( X > t ) t E X + E Z + E ν for t > , and G (0) = 1 . Adding some extra conditions we can obtain results on the asymp-totic behaviour of G . Theorem 4.
Suppose that the distribution of Y is absolutely contin-uous. Then we have G ( t ) ∼ C t − γ as t → ∞ , where C is a positiveconstant, and γ = E X + E Z E Y .
Remark 1.
The difference of the exponents in the discrete and con-tinuous cases is due to the difference in the definitions. Namely, in thefirst case ξ n denotes the weight distribution, while in the second case itstands for the complementary cumulative weight distribution function. Proofs
First we prove some propositions we will often use in the sequel.
Lemma 1.
Let ( F n ) be a filtration, ( ξ n ) a nonnegative adapted pro-cess. Let ( w n ) be a regularly varying sequence of positive numbers withexponent µ > − . Suppose that (3) E (cid:0) ( ξ n − ξ n − ) (cid:12)(cid:12) F n − (cid:1) = O (cid:0) n − δ +2 µ (cid:1) ´AGNES BACKHAUSZ AND TAM ´AS F. M ´ORI holds with some δ > . Let ( u n ) , ( v n ) be nonnegative predictable pro-cesses such that u n < n for all n ≥ . ( a ) Suppose that E ( ξ n | F n − ) ≤ (cid:16) − u n n (cid:17) ξ n − + v n , and lim n →∞ u n = u , lim sup n →∞ v n /w n ≤ v with some random vari-ables u > , v ≥ . Then lim sup n →∞ ξ n nw n ≤ vu + µ + 1 a.s. ( b ) Suppose that E ( ξ n | F n − ) ≥ (cid:16) − u n n (cid:17) ξ n − + v n , and lim n →∞ u n = u , lim inf n →∞ v n /w n ≥ v with some random variables u > , v ≥ . Then lim inf n →∞ ξ n nw n ≥ vu + µ + 1 a.s. This is a stochastic counterpart of a lemma of Chung and Lu [5]. Wewill often apply this proposition with the sequence w n ≡ µ = 0. Proof.
Suppose first that v is strictly positive. Let F be the trivial σ -algebra, ξ = 0, and c n = n Y i =1 (cid:16) − u i i (cid:17) − , n ≥ . We have log c n = n X i =1 u i i (cid:0) o (1) (cid:1) = u n X i =1 o (1) i . Hence for all t > n →∞ (log c [ tn ] − log c n ) = u log t . Thatis, ( c n ) is regularly varying with exponent u . It is clear that(4) E (cid:0) c n ξ n (cid:12)(cid:12) F n − (cid:1) ≤ c n − ξ n − + c n v n . Therefore c n ξ n is a submartingale. Consider the Doob decomposition c n ξ n = M n + A n , where M n = n X i =1 (cid:0) c i ξ i − E (cid:0) c i ξ i (cid:12)(cid:12) F i − (cid:1)(cid:1) is a martingale, and A n = n X i =1 (cid:0) E (cid:0) c i ξ i (cid:12)(cid:12) F i − (cid:1) − c i − ξ i − (cid:1) . RANDOM MODEL OF PUBLICATION ACTIVITY 7
From inequality (4) it follows that A n ≤ n X i =1 c i v i . Consider the increasing process in the Doob decomposition of thesquare of the martingale ( M n ). Using condition (3) we get that B n = n X i =1 Var (cid:0) c i ξ i (cid:12)(cid:12) F i − (cid:1) = n X i =1 Var (cid:0) c i ( ξ i − ξ i − ) (cid:12)(cid:12) F i − (cid:1) ≤ n X i =1 c i E (cid:0) ( ξ i − ξ i − ) (cid:12)(cid:12) F i − (cid:1) = O n X i =1 i − δ +2 µ c i ! . Since n − δ +2 µ c n is still regularly varying with exponent 2 u + 1 − δ +2 µ , it follows that B n = O (cid:0) n − δ +2 µ c n (cid:1) (see e.g. [3, 4]). Hence, byPropositions VII-2-3 and VII-2-4 of [6], we have M n = O ( B / εn (cid:1) = O (cid:0) n (2 − δ +2 µ )(1 / ε ) c εn (cid:1) = o (cid:0) nc n w n (cid:1) a.s., for all 0 < ε < δ u + 1 + µ ) .On the other hand, using the fact u + µ > −
1, and the results of[3, 4] on regularly varying sequences we obtain that A n ≤ n X i =1 c i v i ≤ (cid:0) o (1) (cid:1) v n X i =1 c i w i ∼ v nc n w n u + µ + 1almost surely, as n → ∞ . This implies that c n ξ n ≤ (cid:0) o (1) (cid:1) vu + µ + 1 nc n w n , thus the proof of part ( a ) is complete for positive v .The general case of nonnegative v can be deduced from the positivecase by noticing that E ( ξ n | F n − ) ≤ (cid:16) − u n n (cid:17) ξ n − + max ( v n , ε )for arbitrary ε > A n ≥ n X i =1 c i v i ∼ vu + µ + 1 nc n w n , a.s. on the event { v > } . Hence, using c n ξ n ∼ A n , we get that c n ξ n ≥ vu + µ + 1 nc n w n (cid:0) o (1) (cid:1) . On the event { v = 0 } the inequality trivially holds. (cid:3) ´AGNES BACKHAUSZ AND TAM ´AS F. M ´ORI Lemma 2.
The conditional probability that an author of weight j ischosen, given F + n and ν n +1 = k , is equal to k − n + n + 1 − kn · jS n = k − n (cid:18) − jS n (cid:19) + jS n . Proof.
Consider those groups of size k ≥ i (0 ≤ i ≤ n ). There are (cid:0) nk − (cid:1) of them, because the total number ofresearchers is n + 1. Researcher i belongs to all of them, while the otherresearchers belong to (cid:0) n − k − (cid:1) of those groups. Therefore the total weightof these groups can be obtained in the following way. X H ⊂{ ,...,n }| H | = k, i ∈ H X j ∈ H W ( n, j ) = (cid:18) nk − (cid:19) W ( n, i ) + X j = i (cid:18) n − k − (cid:19) W ( n, j )= (cid:18) n − k − (cid:19) W ( n, i ) + (cid:18) n − k − (cid:19) S n . On the other hand, the total weight of all groups of size k is givenby (cid:18) nk − (cid:19) S n . Hence the conditional probability that researcher i participates inthe ( n + 1)st paper given that it has k authors is equal to k − n + n − k + 1 n · W ( n, i ) S n = k − n (cid:18) − W ( n, i ) S n (cid:19) + W ( n, i ) S n . This obviously holds for k = 1 as well. (cid:3) Proof of Theorem 1.
Recall that in Theorem 1 we assumed that
X, Y , Y , . . . are integer valued random variables. Let us introduce H ( i ) = E (( ν − I ( Y = i )) , then β = P ∞ i =1 H ( i ).We prove the theorem by induction on j . The following argument isvalid for all j = 1 , , . . . . For j > j may change due tothe following events. • A given author of weight j is chosen and he gets positive bonus. • A given author of weight j − i is chosen and his bonus is equalto i . • The initial weight of the new author is j . RANDOM MODEL OF PUBLICATION ACTIVITY 9
Therefore Lemma 2 implies that(5) E (cid:0) ξ n ( j ) (cid:12)(cid:12) F + n − (cid:1) = ξ n − ( j ) (cid:20) − P (cid:0) Y n > (cid:12)(cid:12) F + n − (cid:1)(cid:16) ν n − n − n − ν n n − · jS n − (cid:17)(cid:21) + j − X i =1 ξ n − ( j − i ) P (cid:0) Y n = i (cid:12)(cid:12) F + n − (cid:1)(cid:16) ν n − n − n − ν n n − · j − iS n − (cid:17) + P ( X n = j ) . Recall that ν n ≥ H n ( i ) = E (cid:0) ( ν n − I ( Y n = i ) (cid:1) ; β n = n X i =1 H n ( i ) = E (cid:0) ( ν n − I ( Y n > (cid:1) . Let us take conditional expectation given F n − in both sides of (5).Then we get that(6) E (cid:0) ξ n ( j ) (cid:12)(cid:12) F n − (cid:1) = ξ n − ( j ) (cid:20) − β n n − − (cid:16) P ( Y n > − β n n − (cid:17) jS n − (cid:21) + j − X i =1 ξ n − ( j − i ) (cid:20) H n ( i ) n − (cid:16) P ( Y n = i ) − H n ( i ) n − (cid:17) j − iS n − (cid:21) + P ( X n = j ) ( j, n = 1 , , . . . ) . We are going to apply Lemma 1 to the sequence ( ξ n ( j )) with w n ≡ µ = 0. It is clear that | ξ n ( j ) − ξ n − ( j ) | ≤ ν n + 1, hence E (cid:0) ( ξ n ( j ) − ξ n − ( j )) (cid:12)(cid:12) F n − (cid:1) ≤ E ( ν n + 1) = O (1) . Thus, condition (3) on the differences of the sequence ξ n ( j ) is satisfied.Moreover, as n → ∞ , we have u n = n (cid:20) β n n − (cid:16) P ( Y n > − β n n − (cid:17) jS n − (cid:21) → β + αj. Note that α > Z , Z , . . . are not necessarily identi-cally distributed, they satisfy the following conditions. ∞ X n =1 Var ( Z n ) n < ∞ , lim n →∞ n n X i =1 E Z i = E Z. Therefore Kolmogorov’s theorem (Theorem 6.7. in [7]) can be ap-plied. We get that S n ∼ n ( E X + E Z ) almost surely as n → ∞ . Using this, and also the induction hypothesis when j >
1, we conclude that v n = j − X i =1 ξ n − ( j − i ) (cid:20) H n ( i ) n − (cid:16) P ( Y n = i ) − H n ( i ) n − (cid:17) j − iS n − (cid:21) + P ( X n = j ) → j − X i =1 x j − i (cid:20) H ( i ) + P ( Y = i ) j − i E X + E Z (cid:21) + P ( X = j ) , as n → ∞ .From equations (5) and (6) one can see that ( u n ) and ( v n ) are non-negative predictable processes. Moreover, u n < n if n is large enough,because then ν n < n and j < S n − . We have also seen that the limit of( u n ) is positive. Hence, by Lemma 1, the induction step and the proofof Theorem 1 is complete. (cid:3) Proof of Theorem 2.
Write recursion (1) in the following form. x j = j − X i =1 w j,i x j − i + r j , where for i, j ≥ w j,i = (cid:20) ( j − i ) P ( Y = i ) E X + E Z + E (cid:0) ( ν − I ( Y = i ) (cid:1)(cid:21) αj + β + 1 , and r j = P ( X = j ) αj + β + 1 . In order to apply Theorem 1 of [1] we try to find sequences ( a i ), ( b i ),( c j, i ) such that w j,i = a i + b i j + c j, i holds, then we have to check that a i , b i , c i,j , r i satisfy the following conditions.(i) a i ≥ i ≥
1, and the greatest common divisor of the set { i : a i > } is 1;(ii) r i is nonnegative, and not identically zero;(iii) there exists z > < ∞ X i =1 a i z i < ∞ , ∞ X i =1 | b i | z i < ∞ , ∞ X i =1 i − X j =1 | c i,j | z j < ∞ , ∞ X i =1 r i z i < ∞ . Therefore we set a i = lim j →∞ w j,i = P ( Y = i ) α ( E X + E Z ) = P ( Y = i | Y > , i = 1 , , . . . , RANDOM MODEL OF PUBLICATION ACTIVITY 11 then we define b i = lim j →∞ j ( w j,i − a i ) = 1 α h H ( i ) − ( αi + β + 1) a i i . Finally, we introduce c j,i = w j,i − a i − b i j = − b i · β + 1 j ( αj + β + 1) . Since ( a i ) is a probability distribution, for (iii) it suffices to showthat ( a i ), ( b i ), ( c j,i ), and ( r i ) are exponentially small.According to Assumption 7, Y has finite moment generating func-tion. This implies that ( a i ) is exponentially small. The same holds for( b i ), because ∞ X i =1 H ( i ) e εi = E (cid:0) ( ν − e εY ) (cid:1) ≤ h E ( ν − E (cid:0) e εY (cid:1)i / < ∞ if ε > ∞ X j =1 j − X i =1 | c j,i | e εi = ∞ X j =1 j − X i =1 | b i | β + 1 j ( αj + β + 1) e εi ≤ ∞ X j =1 β + 1 j ( αj + β + 1) ∞ X i =1 | b i | e εi < ∞ . The sequence ( r j ) is also exponentially small, because X has finitemoment generating function by Assumption 7. w j,i , a j , r j are nonnegative. Assumption 8 guarantees that the great-est common divisor of the set { j : a j > } is equal to 1, and r j > j .We have checked all conditions of Theorem 1 of [1]. Since X isnot identically 0, there exists a k with x k >
0. On the other hand, byAssumption 8, P ( Y = ℓ ) > ℓ . Now, one can see from the re-cursion that x k , x k + l , x k +2 l , . . . are all positive, hence the sequence ( x n )has infinitely many positive terms. Therefore, applying the theoremwe obtain that x j ∼ C j − γ as j → ∞ , where γ = − P ∞ i =1 b i P ∞ i =1 ia i . It is easy to see that ∞ X i =1 ia i = ∞ X i =1 i P ( Y = i | Y >
0) = E Y P ( Y >
0) ; − ∞ X i =1 b i = − βα + ∞ X i =1 ia i + β + 1 α = E X + E Z + E Y P ( Y > . Hence the statement of Theorem 2 follows. (cid:3)
Proof of Theorem 3.
We will use the results of the discrete part,namely, Theorem 1. Let h be sufficiently small positive number. Wewill consider limits as h → F n ( t ) = P ( Y n > t ) and H n ( t ) = E (cid:0) ( ν n − I ( Y n > t ) (cid:1) , as before.Furthermore, for a decreasing function ϕ let ∆ h ϕ ( t ) = ϕ ( t − h ) − ϕ ( t ).By Lemma 2, the conditional probability of the event that an authorof weight between t − ih and t − ( i − h is chosen, and his bonus is atleast ( i − h , given F + n − , is bounded from above by(7) (cid:20) t − ( i − hS n − + (cid:18) − t − ( i − hS n − (cid:19) ν n − n − (cid:21) P (cid:0) Y n > ( i − h (cid:12)(cid:12) F + n − (cid:1) . Hence the conditional probability with respect to F n − is at most u i := t − ( i − hS n − F n (cid:0) ( i − h (cid:1) + 1 n − (cid:18) − t − ( i − hS n − (cid:19) H n (cid:0) ( i − h (cid:1) . Note that u i depends on n , which is fixed at the moment. We get that E (cid:0) ξ n ( t ) (cid:12)(cid:12) F n − (cid:1) ≤ ξ n − ( t ) + ⌈ t/h ⌉ X i =1 h ξ n − ( t − ih ) − ξ n − (cid:0) t − ( i − h (cid:1)i u i + P ( X > t ) . After rearranging we obtain that(8) E (cid:0) ξ n ( t ) (cid:12)(cid:12) F n − (cid:1) ≤ ξ n − ( t )(1 − u )+ ⌈ t/h ⌉ X i =1 ξ n − ( t − ih )( u i − u i +1 ) + nu ⌈ t/h ⌉ +1 + P ( X > t ) . Here u = tS n − + 1 n − (cid:18) − tS n − (cid:19) E ( ν n − (cid:18) t E X + E Z + E ν − (cid:19) o (1) n , and u i − u i +1 = hS n − F n (cid:0) ( i − h (cid:1) + t − ihS n − ∆ h F n ( ih ) − n − hS n − H n (cid:0) ( i − h (cid:1) + 1 n − (cid:18) − t − ihS n − (cid:19) ∆ h H n ( ih ) . This implies that n ( u i − u i +1 ) → h E X + E Z F (cid:0) ( i − h (cid:1) + t − ih E X + E Z ∆ h F ( ih ) + ∆ h H ( ih ) , RANDOM MODEL OF PUBLICATION ACTIVITY 13 as n → ∞ . Finally, nu ⌈ t/h ⌉ +1 ≤ nn − (cid:16) hS n − (cid:17) H n ( t ) , hence lim sup n →∞ nu ⌈ t/h ⌉ +1 ≤ H ( t ) . Let G u ( t ) = lim sup n →∞ ξ n ( t ) n (subscript u stands for “upper”). G u ( t ) is a decreasing random func-tion, andlim sup n →∞ ⌈ t/h ⌉ X i =1 ξ n − ( t − ih )( u i − u i +1 ) ≤ ⌈ t/h ⌉ X i =1 G u ( t − ih ) (cid:20) F (cid:0) ( i − h (cid:1) E X + E Z h + t − ih E X + E Z ∆ h F ( ih ) + ∆ h H ( ih ) (cid:21) . Denote the sum on the right hand side by Σ u ( t, h ). We want to applyLemma 1 to the sequence ξ n ( t ). It satisfies (8), and, similarly to thediscrete case, E (cid:0) ( ξ n ( t ) − ξ n − ( t )) (cid:12)(cid:12) F n − (cid:1) ≤ E ( ν n + 1) = O (1)holds again. The other assumptions are also easy to check. Hence G u ( t ) ≤ h Σ u ( t, h ) + H ( t ) + P ( X > t ) i (cid:20) t E X + E Z + E ν (cid:21) − . One can readily verify that Σ u ( t, h ) converges to1 E X + E Z "Z t G u ( t − s ) F ( s ) ds − Z t G u ( t − s )( t − s ) dF ( s ) − Z t G u ( t − s ) dH ( s ) = Z t G u ( t − s ) d s L ( t, s )as h →
0, since the Riemann–Stieltjes integrals in the expression exist.This implies that(9) G u ( t ) ≤ (cid:20)Z t G u ( t − s ) d s L ( t, s )+ H ( t )+ P ( X > t ) (cid:21)(cid:20) t E X + E Z + E ν (cid:21) − . Therefore the solution of the corresponding integral equation (2)with initial condition G u (0) = 1 is an upper bound for G u ( t ). Thatis, G u ( t ) ≤ G ( t ), where G ( t ) is the deterministic function given in thetheorem.Now we give lower bounds by analogous argumentation. We estimate from below the conditional probability that an authorwith weight between t − ih and t − ( i − h is chosen and his bonusis at least ih , given F + n − . Similarly to (7), we have that it is greaterthan or equal to (cid:20) t − ihS n − + (cid:18) − t − ihS n − (cid:19) ν n − n − (cid:21) P (cid:0) Y n > ih (cid:12)(cid:12) F + n − (cid:1) . Hence the lower bound of the conditional probability with respect to F n − is the following. ℓ i := t − ihS n − F n ( ih ) + 1 n − (cid:18) − t − ihS n − (cid:19) H n ( ih ) . We obtain that E (cid:0) ξ n ( t ) (cid:12)(cid:12) F n − (cid:1) ≥ ξ n − ( t ) + ⌈ t/h ⌉ X i =1 h ξ n − ( t − ih ) − ξ n − (cid:0) t − ( i − h (cid:1)i ℓ i + P ( X > t ) . After rearranging we get a formula similar to (8).(10) E (cid:0) ξ n ( t ) (cid:12)(cid:12) F n − (cid:1) ≥ ξ n − ( t )(1 − ℓ )+ ⌈ t/h ⌉ X i =1 ξ n − ( t − ih )( ℓ i − ℓ i +1 ) + nℓ ⌈ t/h ⌉ +1 + P ( X > t ) . Here ℓ = t − hS n − F n ( h ) + 1 n − (cid:18) − t − hS n − (cid:19) H n ( h )= (cid:18) t − h E X + E Z F ( h ) + H ( h ) (cid:19) o (1) n , and ℓ i − ℓ i +1 = hS n − F n ( ih ) + t − ( i + 1) hS n − ∆ h F n (cid:0) ( i + 1) h (cid:1) − n − hS n − H n ( ih ) + 1 n − (cid:18) − t − ( i + 1) hS n − (cid:19) ∆ h H n (cid:0) ( i + 1) h (cid:1) . This implies that n ( ℓ i − ℓ i +1 ) converges to h E X + E Z F ( ih ) + t − ( i + 1) h E X + E Z ∆ h F (cid:0) ( i + 1) h (cid:1) + ∆ h H (cid:0) ( i + 1) h (cid:1) as n → ∞ . Finally, nℓ ⌈ t/h ⌉ +1 ≥ − nhS n − + H n ( t + 2 h ) , RANDOM MODEL OF PUBLICATION ACTIVITY 15 therefore lim inf n →∞ nℓ ⌈ t/h ⌉ +1 ≥ − h E X + E Z + H ( t + 2 h ) . Let G ℓ ( t ) = lim inf n →∞ ξ n ( t ) n ;then G ℓ ( t ) is also a decreasing random function. On the right handside of (10) we havelim inf n →∞ ⌈ t/h ⌉ X i =1 ξ n − ( t − ih )( ℓ i − ℓ i +1 ) ≥ Σ ℓ ( t, h ) , whereΣ ℓ ( t, h ) = ⌈ t/h ⌉ X i =1 G ℓ ( t − ih ) (cid:20) F ( ih ) E X + E Z h + t − ( i + 1) h E X + E Z ∆ h F (cid:0) ( i + 1) h (cid:1) + ∆ h H (cid:0) ( i + 1) h (cid:1)(cid:21) . Applying Lemma 1 we get that G ℓ ( t ) ≥ (cid:20) Σ ℓ ( t, h ) − h E X + E Z + H ( t + 2 h ) + P ( X > t ) (cid:21) × (cid:20) t − h E X + E Z F ( h ) + H ( h ) + 1 (cid:21) − . Let h go to zero again. The sum Σ ℓ ( t, h ) converges to the sameRiemann–Stieltjes integral as Σ u ( t, h ) does. Thus the right hand sideof the inequality above converges to the right hand side of (9). Hencewe obtain that G ℓ ( t ) ≥ G ( t ). This, together with the estimation for G u ( t ), implies the statement of the theorem. (cid:3) Proof of Theorem 4.
Let the density function of Y be denoted by f . From the absolute continuity of F the same follows for H . Let h bedefined by H ( t ) = Z ∞ t h ( s ) ds. Differentiating L with respect to s we obtain that ∂∂s L ( t, s ) = F ( s ) − sf ( s ) + tf ( s ) E X + E Z + h ( s ) (0 ≤ s ≤ t ) . Hence equation (2) may be written in the following form. G ( t ) = Z t G ( t − s ) w t,s ds + r ( t ) , where w t,s = F ( s ) + ( t − s ) f ( s ) E X + E Z + h ( s ) t E X + E Z + E ν = F ( s ) + ( t − s ) f ( s ) + h ( s ) ( E X + E Z ) t + ( E X + E Z ) E ν ; r ( t ) = H ( t ) + P ( X > t ) t E X + E Z + E ν . In order to apply Theorem 2 of [1] write w t,s in the following form. w t,s = f ( s ) + F ( s ) − ( s + ( E X + E Z ) E ν ) f ( s ) + h ( s ) ( E X + E Z ) t + ( E X + E Z ) E ν = f ( s ) + b ( s ) t + d , where b ( s ) = F ( s ) − (cid:0) s + ( E X + E Z ) E ν (cid:1) f ( s ) + h ( s ) ( E X + E Z ) ; d = ( E X + E Z ) E ν. Next we check that all assumptions required in [1] hold. Since f is aprobability density function, G is clearly decreasing and w is nonneg-ative, all we need is the following three facts.(i) d is a positive constant,(ii) r is a nonnegative, continuous function,(iii) there exists z > Z ∞ f ( t ) z t dt < ∞ , Z ∞ | b ( t ) | z t dt < ∞ , and r ( t ) z t is directly Riemann integrable on [0 , ∞ ).Here (i) follows from Assumption 8. From the continuity of F and H the same follows for r . Finally, the first part of condition (iii) eas-ily follows from Assumptions 2 and 7. In addition, using that r ismonotonically decreasing we get that ∞ X n =1 sup ≤ θ ≤ τ r ( t + nτ + θ ) z t + nτ + θ ≤ ∞ X n =1 (cid:2) r ( t + nτ ) z t + nτ (cid:3) z τ for z >
1. The right hand side is finite for almost all t , because R ∞ r ( s ) z s ds is finite. Therefore r ( t ) z t is directly Riemann integrable.Thus Theorem 4 follows from Theorem 2 of [1]. Using the continuityof G and the method of the discrete case it is easy to see that G is not RANDOM MODEL OF PUBLICATION ACTIVITY 17 identically 0 for large t , thus it is polynomially decaying. What is leftis to determine the exponent, that is, γ = − R ∞ b ( s ) ds R ∞ sf ( s ) ds . The denominator is equal to E Y . In the numerator we have Z ∞ b ( s ) ds = Z ∞ (cid:16) F ( s ) − (cid:0) s + ( E X + E Z ) E ν (cid:1) f ( s ) + h ( s ) ( E X + E Z ) (cid:17) ds = E Y − E Y − ( E X + E Z ) E ν + H (0) ( E X + E Z )= − ( E X + E Z ) E ν + E ( ν −
1) ( E X + E Z )= − ( E X + E Z ) . Therefore we got that γ = E X + E Z E Y , and the proof of Theorem 4 is complete. (cid:3)
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E-mail address : [email protected] Department of Probability Theory and Statistics, E¨otv¨os Lor´andUniversity, P´azm´any P. s. 1/C, H-1117 Budapest, Hungary
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