On the number of zero increments of random walks with a barrier

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1 On the number of zero increments of random wals with a barrier Alex Isanov, Pavlo Negadajlov To cite this version: Alex Isanov, Pavlo Negadajlov. On the number of zero increments of random wals with a barrier. Roesler, Uwe. Fifth Colloquium on Mathematics and Computer Science, 2008, Kiel, Germany. Discrete Mathematics and Theoretical Computer Science, DMTCS Proceedings vol. AI, Fifth Colloquium on Mathematics and Computer Science, pp , 2008, DMTCS Proceedings. <hal > HAL Id: hal Submitted on 7 Sep 2015 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

2 Fifth Colloquium on Mathematics and Computer Science DMTCS proc. AI, 2008, On the number of zero increments of random wals with a barrier Alex Isanov 1 and Pavlo Negadajlov 1 1 Faculty of Cybernetics, National T. Shevcheno University of Kiev, Kiev-01033, Uraine This is the second submission of this document. Continuing the line of research initiated in Isanov and Möhle (2008 and Negadajlov (2008 we investigate the asymptotic (as n behaviour of V n the number of zero increments before the absorption in a random wal with the barrier n. In particular, when the step of the unrestricted random wal has a finite mean, we prove that the number of zero increments converges in distribution. We also establish a wea law of large numbers for V n under a regular variation assumption. Keywords: absorption time; recursion with random indices; random wal; undershoot. 1 Introduction and results Let {ξ : N} be independent copies of a random variable ξ with proper and non-degenerate distribution p := P{ξ = }, N, p 1 > 0. Fix n N. The sequence {R (n R (n : N 0 := {0, 1,...}} defined as follows: R (n 0 := 0 and := R (n 1 + ξ 1 {R (n 1 +ξ <n}, N, is called a random wal with the barrier n. The Marov chain {R (n : N 0 } is non-decreasing and has the unique absorbing state n 1. Denote by T n := inf{ N : R (n = n 1} = 1 (n {R l <n 1} + 1 the absorption time. Other interesting functionals acting on random wals with a barrier are the number of non-zero increments (jumps M n := #{ N : R (n 1 R(n l=1 } = 1 (n {R l +ξ l+1 <n} The research of both authors was supported by the German Scientific Foundation (project no. 436UKR 113/93/ c 2008 Discrete Mathematics and Theoretical Computer Science (DMTCS, Nancy, France

3 244 Alex Isanov and Pavlo Negadajlov and the number of zero increments before the absorption time V n := #{ [0, T n 1] : R (n 1 = R(n Tn 1 } = 1 {R (n l +ξ l+1 n}. Notice that the last summand of the latter sum always equals zero. We however do not replace the upper limit of the summation by T n 2 to avoid summation over the empty set. Two real-world examples where the random wal with a barrier naturally arises are given next. (a An enterprise can pay Euro per month as a salary to all called-in employees. Potential employees apply for salaries which are independent copies of a random variable ξ taing values 100, 200, 300,... Euro. Then T , M and V define the number of people which apply for a job, to be hired and to be turned down, respectively. (b A private company (an express, say specializes in transportation of goods and uses three lorries which altogether are able to transport 10 tones. Assume that requests for transportation are independent copies of a random variable ξ taing values in the set of multiples of 100 g. Then T 10001, M and V define the total number of requests, the number of requests to be satisfied and to be rejected, respectively. In [5] and [7] the wea convergence of properly normalized and centered M n and T n, respectively, was investigated. Assuming that the distribution of ξ belongs to the domain of attraction of a stable law, a complete characterization of normalizing constants and possible limiting laws was obtained. The class of limiting laws is comprised of stable laws and laws of the integrals e Xt dt, where {X 0 t : t 0} is a zero-drift subordinator with nown Lévy measure. Furthermore, in [7] it was proved that all the results obtained in [5] for M n remain valid with M n replaced by T n. From this last observation it follows that the number of zero increments V n = T n M n is essentially smaller than the number of non-zero increments M n. The aim of the paper is to investigate asymptotic behaviour of V n as n. In Theorem 1.1 it is proved that under the assumption m := Eξ < the sequence {V n : n N} converges in distribution. In Theorem 1.2 assuming that m = and that the tail of distribution of ξ is regularly varying at, a wea law of large numbers is established. All the proofs rely upon the following observation: marginal distributions of the sequence {V n : n N} satisfy the recursions (6 which we call recursions with random indices. It is well-nown that recursive structure is typical for many objects arising in the Analysis of Algorithms, in the theory of random graphs, in the coalescent theory, etc. In this paper, we exhibit a relatively new model where recursions appear in a natural way and exploiting them turns out to be a right way of thining. Before formulating the main results, notice that Hinderer and Wal [4] investigated processes more general than random wals with a barrier but the circle of problems they considered was different from ours. As far as we now, the term random wal with a barrier was introduced in [5]. Theorem 1.1 If Eξ <, then, as n, where Y is a random variable with distribution V n d V := V Y {Y =1}, P{Y = } = (Eξ 1 P{ξ }, N, which is independent of {V n : n N}. In particular, P{V = 0} = (Eξ 1.

4 On the number of zero increments of random wals with a barrier 245 Throughout the paper by ψ(x := Γ (x/γ(x we denote the logarithmic derivative of the gamma function. Also, set m(x := x 0 P{ξ > y}dy, x > 0. Theorem 1.2 Suppose that for some α (0, 1] and some L slowly varying at + Then, as n, p n α L(n, n. (1 =n V n EV n P 1, (2 and EV n (ψ(1 ψ(1 α 1 log n, if α (0, 1, and EV n log m(n, if α = 1. The rest of the paper is organized as follows: in Section 2 we discuss certain recursions with random indices which naturally arise in the context of this wor and in Section 3 the proofs of our main results are given. Notation: the record r( s( means that r( /s( 1, as an argument goes to + ; the record X n d (, P X means that the limit relation holds as n. 2 Recursions with random indices For fixed m, i N define and T m (i := R (m 0 (i := 0 and R (m (i := l=1 R (m 1 (i + ξ i+1 { b R (m 1 (i+ξ i+<m}, N, 1 { b R (m l (i<m 1}, V m (i := bt m(i 1 1 { b R (m l (i+ξ i+l+1 m}. For fixed n N and any i N marginal distributions of the sequences {R (n N 0 } are the same. Therefore, Introduce the notation S 0 := 0, S n := ξ ξ n and : N 0 } and {R (n (i : T n (i = d T n, AND V n (i = d V n. (3 N n := inf{ 1 : S n} AND Y n := n S Nn 1, n N. The sequence {S n : n N 0 } is an unrestricted random wal on the basis of which the random wal with barrier is constructed. The random variable Y n is called undershoot at n. With probability one, we have T n = (1 + T Yn (N n 1 {Yn 1} + N n 1 = T Yn (N n + N n 2 1 {Yn=1}, (4

5 246 Alex Isanov and Pavlo Negadajlov which can be checed as follows. If n R (n N = n S n 1 N n 1 = Y n = 1, then T n = N n 1, and if Y n 1, then R (n N n = R (n N = S n 1 N n 1 < n 1. Therefore, ( T n = N n {Yn 1} = N n 1 + (1 + T Yn (N n 1 {Yn 1}. l=1 1 {R (n Nn+l <n 1} Similarly, with probability one we have V n = ( V Yn (N n + 11 {Yn 1} = V Yn (N n {Yn=1}, (5 which can be verified as follows. If n R (n N = n S n 1 N n 1 = Y n = 1, then V n = 0, and if Y n 1, then the first non-zero increment of the random wal with barrier is R (n N n R (n N, and T n 1 n N n + 1. Therefore, T n 1 V n = (1 + 1 (n {R l +ξ l+1 n} 1 {Y n 1} = l=n n BY (4 = 1 + bt Yn (N n 1 ( 1 + T n N n 1 1 { R b(yn l (N n+ξ Nn+l+1 Y n} From (3 and (5 we obtain a recursion with random indices 1 {R (n Nn+l +ξ Nn+l+1 n} 1 {Yn 1} = 1 {Yn 1} = (1 + V Yn (N n 1 {Yn 1}. V 1 := 1, V n = d V Y n {Yn=1}, (6 where Y n is independent of the sequence {V : N} which is a copy of {V : N}. 3 Proofs Letting n in the equality of distributions (6 and taing into account the convergence Y n d Y (see, for example, p. 371 in [3] the result of Theorem 1.1 follows. To prove Theorem 1.2 we need an auxiliary result. Lemma 3.1 If (1 holds with α [0, 1, then, as n, E log Y n = log n (ψ(1 ψ(1 α + o(1 AND (7 E log 2 Y n = log 2 n 2(ψ(1 ψ(1 α log n + o(log n. (8 If (1 holds with α = 1, then, as n, E log m(y n = log m(n 1 + o(1 AND E log 2 m(y n = log 2 m(n 2 log m(n + o(log m(n.

6 On the number of zero increments of random wals with a barrier 247 Proof: If (1 holds with α [0, 1, then according, for example, to Theorem in [1] log n log Y n d ( log η α, where a random variable η α, α (0, 1 has the beta distribution with parameters 1 α and α, i.e. sin πα P{η α dx} = π x α (1 x α 1 dx, x (0, 1, and P{η 0 = 1} = 1. If we can show that for each δ (0, 1 α sup E(n/Y n δ = sup Ef (log (n/y n <, N, (9 n 1 n 1 where f (x := exp(δx 1/, then by a Vallée-Poussin theorem (see [6] for each N the sequence {(log n log Y n : n N} is uniformly integrable. If it is true then lim n (log n E log Y n = E( log η α = ψ(1 ψ(1 α < which proves (7, and lim n E(log n log Y n 2 = E log 2 η α = (ψ(1 α ψ(1 2 + ψ (1 α ψ (1 <, which together with (7 proves (8. Let us establish (9. For j N 0 set u j := j =0 P{S = j}. Then n 1 EYn δ = (n δ P{ξ n }u. =0 The condition (1 implies that n =0 u 1 Γ(1 αγ(1+α The following four relations hold by Corollary in [1]: U(s := V (s := n=1 n=1 n α L(n (see, for example, Theorem in [1]. s n 1 1 u n Γ(1 α L((1 s 1 AS s 1; (1 s α n=0 s n n δ P{ξ n} Γ(1 α δl((1 s 1 (1 s 1 α δ AS s 1; s n EY δ n = U(sV (s n =1 Γ(1 α δ Γ(1 α EY δ Γ(1 α δ Γ(1 αγ(2 α n1 δ. 1 AS s 1; (1 s 1 δ The last equivalence entails (9. Assume now that (1 holds with α = 1. By Theorem 6 in [2] and its proof, log m(n log m(y n d ( log R, where R is a random variable with the uniform distribution on [0, 1]. Therefore, the stated result will be proved if, for example, we can show that the sequences {(log m(n log m(y n : n N}, = 1, 2, are uniformly integrable. This latter fact will follows from the following relation: for each ɛ (0, 1 ( ɛ m(n sup E <. (10 n 1 m(y n

7 248 Alex Isanov and Pavlo Negadajlov The proof that follows is similar to the previous one. Therefore, we only give a setch. As before, all the unexplained equivalences connected with asymptotic behaviour of the sequences and corresponding generating functions can be justified by an appeal to Corollary in [1]. Since n =0 u n/m(n (see p. 266 in [2], then K(s := s n 1 u n (m((1 s 1 (1 s n=0 AS s 1. Fix any ɛ (0, 1. The function r ɛ (x := x 0 m ɛ (yp{ξ y}dy is slowly varying at, and n =1 m ɛ (P{ξ } r ɛ (n. Consequently, Hence, as s 1, and Z(s := n=1 s n m ɛ (np{ξ n} r ɛ ((1 s 1 AS s 1. n=1 s n Em ɛ r ɛ ((1 s 1 (Y n = K(sZ(s (m((1 s 1 (1 s, n =1 Em ɛ (Y nr ɛ(n m(n. (11 By L Hospital s rule, r ɛ (n/m(n (1 ɛ 1 m ɛ (n. Therefore, (11 implies (10. Proof of Theorem 1.2. Suppose (1 holds with α (0, 1. Set a n := EV n and b n := EVn 2. From (6 we obtain a 1 = 1, ( n 1 1 a n = a P{Y n = } + 1 2P{Y n = 1}, n = 2, 3,... (12 1 P{Y n = n} and b 1 = 1, Let us chec that =1 ( n 1 1 b n = b P{Y n = } + 2a n 1. (13 1 P{Y n = n} =1 b n a 2 n 2 α log 2 n, (14 where α := (ψ(1 ψ(1 α 1. Suppose the condition lim n log n α does not hold. Then there exists an ɛ > 0 such that the inequalities a n > ( α + ɛ log n hold for infinitely many values of n. It is possible to decrease ɛ so that the inequality a n > ( α + ɛ log n + c holds infinitely often for any fixed positive c. Thus we can define n c := inf{n 1 : a n > ( α + ɛ log n + c}. Then From (12 it follows that a n ( α + ɛ log n + c for all n {1, 2,..., n c 1}. ( α + ɛ log n c + c < a n

8 On the number of zero increments of random wals with a barrier 249 ( nc 1 1 < (( α + ɛ log i + cp{y nc = i} + 1 2P{Y nc = 1} = 1 P{Y nc = n c } i=1 ( = c + ( α + ɛ E log Y nc + P{Y n c = n c } 1 P{Y nc = n c } (E log Y n c log n c P{Y nc = n c } (1 2P{Y n c = 1} = (n c as c ; therefore, using (7, the equality P{Y n = 1} = u n 1 and the fact that by the elementary renewal theorem lim n u n = 0, we obtain = c + ( α + ɛ log n c ( α + ɛ 1 α o(1. Letting c go to leads to ɛ 0, which is a contradiction. Thus, we have already checed that lim n a n log n α. A symmetric argument proves the converse inequality for the lower bound. Therefore, a n α log n. The asymptotic relation (14 for b n can be checed exactly in the same way by using the recursion (13. Thus, we already now that EV 2 n (EV n 2. Therefore, by Chebyshev s inequality, for any δ > 0 P{ V n /EV n 1 > δ} DV n (δev n 2 0, which proves (2. In the case when (1 holds with α = 1, the proof of the fact that EVn 2 (EV n 2 (log m(n 2 is absolutely analogous to the previous one. Thus it is omitted. Certainly, instead of the first part of Lemma 3.1 one has to use the second one. References [1] BINGHAM N. H., GOLDIE C. M. AND TEUGELS, J. L. (1989. Regular variation. Cambridge: Cambridge University Press. [2] ERICKSON, K. B. (1970. Strong renewal theorems with infinite mean. Trans. Amer. Math. Soc. 151, [3] FELLER, W. (1966. An introduction to probability theory and its applications. Vol. 2. New Yor: John Wiley & Sons, Inc. [4] HINDERER, K. AND WALK, H. (1972. Anwendungen von Erneuerungstheoremen und Taubersätzen für eine Verallgemeinerung der Erneuerungsprozesse. Math. Z. 126, [5] IKSANOV, A. AND MÖHLE, M. (2008. On the number of jumps of random wals with a barrier. Adv.Appl. Prob. 40,

9 250 Alex Isanov and Pavlo Negadajlov [6] MEYER, P. A. (1966. Probability and potentials. Mass.-Toronto-London: Blaisdell Publishing Company. [7] NEGADAJLOV, P. (2008. Asymptotic results concernning the absorption times of random wals with a barrier. Prob. Theory and Math. Statist., in press (in Urainian.