Limit theorems for discounted convergent perpetuities
aa r X i v : . [ m a t h . P R ] F e b Limit theorems for discounted convergent perpetuities
Alexander Iksanov ∗ Anatolii Nikitin † Igor Samoilenko ‡ Abstract
Let ( ξ , η ), ( ξ , η ) , . . . be independent identically distributed R -valued random vectors.We prove a strong law of large numbers, a functional central limit theorem and a law ofthe iterated logarithm for convergent perpetuities P k ≥ b ξ + ... + ξ k η k +1 as b → − . Underthe standard actuarial interpretation, these results correspond to the situation when theactuarial market is close to the customer-friendly scenario of no risk. Key words: cluster set; functional central limit theorem; law of the iterated logarithm; perpe-tuity; strong law of large numbers2000 Mathematics Subject Classification: Primary: 60F15,60F17Secondary: 60G50
Let ( ξ , η ), ( ξ , η ) , . . . be independent copies of an R -valued random vector ( ξ, η ) with ar-bitrarily dependent components. Denote by ( S k ) k ∈ N (as usual, N := N ∪ { } ) the standardrandom walk with jumps ξ k defined by S := 0 and S k := ξ + . . . + ξ k for k ∈ N . Whenever arandom series P k ≥ e − S k η k +1 converges a.s., its sum is called perpetuity because of the followingactuarial application. Assuming, for the time being, that ξ and η are a.s. positive, we can inter-pret η k and e − ξ k as the planned payment and the discount factor (risk) for year k , respectively.Then P k ≥ e − S k η k +1 can be thought of as ‘the present value of a permanent commitment tomake a payment ... annually into the future forever’ (the phrase borrowed from p. 1196 in[13]). When studying the aforementioned random series from purely mathematical viewpoint,the one-sided assumptions are normally omitted whereas the term ‘perpetuity’ is still used. Seethe books [7] and [16] for surveys of the area of perpetuities from two different perspectives.In the present paper we investigate the asymptotic behavior as b → − of the convergentseries P k ≥ b S k η k +1 that we call discounted convergent perpetuity . We intend to prove thebasic limit theorems for the discounted convergent perpetuities: a strong law of large numbers,a functional central limit theorem and a law of the iterated logarithm. Getting back to theactuarial interpretation, these results describe the fluctuations of the present value when theactuarial market is close to the customer-friendly scenario of no risk.A sufficient condition for the almost sure (a.s.) absolute convergence of the random series P k ≥ b S k η k +1 with fixed b ∈ (0 ,
1) is E ξ ∈ (0 , ∞ ) and E log + | η | < ∞ , see, for instance, Theorem2.1 in [13]. This sufficient condition holds, that is, the discounted perpetuity is well-defined forall b ∈ (0 , ∗ Faculty of Computer Science and Cybernetics, Taras Shevchenko National University of Kyiv, Ukraine; e-mailaddress: [email protected] † Faculty of Natural Sciences, Jan Kochanowski University of Kielce; e-mail address: [email protected] ‡ Faculty of Computer Science and Cybernetics, Taras Shevchenko National University of Kyiv, Ukraine; e-mailaddress: [email protected]
1e start with a strong law of large numbers.
Theorem 1.1.
Assume that µ := E ξ ∈ (0 , ∞ ) and E | η | < ∞ . Then lim b → − (1 − b ) X k ≥ b S k η k +1 = µ − m a.s. , (1) where m := E η . Throughout the paper we write P → to denote convergence in probability, and ⇒ and d −→ to denote weak convergence in a function space and weak convergence of one-dimensional dis-tributions, respectively. Also, we denote by D (0 , ∞ ) the Skorokhod space of right-continuousfunctions defined on (0 , ∞ ) with finite limits from the left at positive points. We proceed bygiving a functional central limit theorem. Theorem 1.2.
Assume that µ = E ξ ∈ (0 , ∞ ) , E η = 0 and s := Var η ∈ (0 , ∞ ) . Then, as b → − , (cid:16) (1 − b ) / X k ≥ b uS k η k +1 (cid:17) u> ⇒ (2 s µ − ) / Z [0 , ∞ ) e − uy d B ( y ) ! u> (2) in the J -topology on D (0 , ∞ ) , where ( B ( t )) t ≥ is a standard Brownian motion.Remark . The limit process in Theorem 1.2 is an a.s. continuous Gaussian process on (0 , ∞ )with covariance E Z [0 , ∞ ) e − uy d B ( y ) Z [0 , ∞ ) e − vy d B ( y ) = 1 u + v , u, v > . (3)Such a process has appeared in the recent articles [8], [17] and [18]. The latter paper providesadditional references.Putting in (2) u = 1 and using (3) with u = v = 1 we obtain a one-dimensional central limittheorem. Corollary 1.4.
Under the assumptions of Theorem 1.2, as b → − , (1 − b ) / X k ≥ b S k η k +1 d −→ ( s µ − ) / Normal(0 , , where Normal(0 , denotes a random variable with the standard normal distribution. Finally, we are interested in the rate of a.s. convergence in Theorem 1.1 when m = 0 whichis expressed by a law of the iterated logarithm. A hint concerning the form of this law is givenby the central limit theorem, Corollary 1.4. For a family ( x t ) we denote by C (( x t )) the set ofits limit points. Theorem 1.5.
Assume that µ = E ξ ∈ (0 , ∞ ) , E η = 0 and s = Var η ∈ (0 , ∞ ) . Then lim sup (lim inf) b → − (cid:16) − b log log − b (cid:17) / X k ≥ b S k η k +1 = +( − )(2 s µ − ) / a.s. (4) In particular, C (cid:18)(cid:18)(cid:16) − b s µ − log log − b (cid:17) / X k ≥ b S k η k +1 : b ∈ ((1 − e − ) / , (cid:19)(cid:19) = [ − ,
1] a.s.2
Related literature
Random power series . The random power (or geometric) series P k ≥ b k η k +1 for b ∈ (0 ,
1) is arather particular case of a discounted convergent perpetuity which corresponds to the degeneraterandom walk S k = k for k ∈ N . In this section we first discuss known counterparts of our mainresults for the random power series. Law of large numbers . Under the assumption E | η | < ∞ , the following strong law of largenumbers can be found in Theorem 1 of [19]lim b → − (1 − b ) X k ≥ b k η k +1 = m a.s. , (5)where m = E η . Central limit theorem . Under the assumption E | η | < ∞ Theorem 1 in [12] proves a Berry-Ess´eeninequality which entails(1 − b ) / (cid:16) X k ≥ b k η k +1 − m − b (cid:17) d −→ s Normal(0 , , b → − , where s = Var η ∈ (0 , ∞ ). Theorem 4.1 in [25] is a functional limit theorem in the Skorokhodspace for the process ( P ⌊ (1 − b ) − t ⌋ k =0 b k η k +1 ) t ≥ , properly normalized and centered, as b → − .Here and hereafter, ⌊ x ⌋ denotes the integer part of real x . The corresponding limit process is atime-changed Brownian motion. Law of iterated logarithm . It was proved in Theorem 3 of [11] thatlim sup b → − (cid:16) − b log log − b (cid:17) / X k ≥ b k η k +1 = 2 / s for centered bounded η k with variance s . In Theorem 2 of [19] this limit relation was statedwithout proof, for not necessarily bounded η k . Our Theorem 1.5 is an analogue of Theorem1.1 in [5] dealing with the random power series. In Theorem 1.1 of [22] the sequence ( η k ) k ∈ N isstationary, conditionally centered and ergodic with E η < ∞ . In this more general setting theauthors prove a counterpart of (4) for the corresponding random power series. Another proofin both settings based on a strong approximation result is given in Theorem 2.1 of [26]. See also[10] and [23] for related results.Although the random power series is a toy example of perpetuities, transferring results fromthe former to the latter may be a challenge. To justify this claim, we only mention that whilenecessary and sufficient conditions for the a.s. convergence of random power series can be easilyobtained (just use the Cauchy root test in combination with the Borel-Cantelli lemma), thecorresponding result for perpetuities is highly non-trivial, see Theorem 2.1 in [13] and its proof.The reason is clear: the random power series is a weighted sum of independent random variables,whereas it is not the case for perpetuities.Investigation of (general) weighted sums of independent identically distributed random vari-ables has been and still is a rather popular trend of research. We refrain from giving a surveyand only mention recent contributions [1, 2] in which a random Dirichlet series is analyzed. Discounted perpetuities . As far as we know, Theorems 1.1, 1.2 and 1.5 are new. Under theadditional assumption E ξ < ∞ (we only require E ξ ∈ (0 , ∞ )) our Corollary 1.4 follows fromTheorem 6.1 in [24] which we state as Proposition 2.1 for reader’s convenience.3 roposition 2.1. Assume that µ = E ξ ∈ (0 , ∞ ) , σ = Var ξ ∈ [0 , ∞ ) , s = Var η ∈ [0 , ∞ ) , σ + s > . Then α − / (cid:16) X k ≥ e − αS k − η k − α m µ − (cid:17) d −→ v Normal(0 , , α → ∞ , where Normal(0 , denotes a random variable with the standard normal distribution, m = E η , v := 2 − σ µ − m + γ m µ − + 2 − σ µ − and γ := E ξη − µ m ∈ R . We stress that our idea of proof of Theorem 1.2 is different from Vervaat’s. Also, we notethat in Theorem 2 of [9] the method of moments is employed for proving a (one-dimensional)central limit theorem for P k ≥ b S k as b → − under the assumptions ξ ≥ E ξ p < ∞ for all p > We shall use a fragment of Theorem 5 on p. 49 in [14] that we give in a form adapted to oursetting.
Lemma 3.1.
Let ( c k ( b )) k ∈ N and ( s k ) k ∈ N be sequences of real-valued functions defined on (0 , and real numbers, respectively. Assume that(i) P k ≥ | c k ( b ) | < ∞ for all b ∈ (0 , and that, for some b ∈ (0 , and some A > which doesnot depend on b , P k ≥ | c k ( b ) | ≤ A for all b ∈ ( b , ;(ii) lim b → − c k ( b ) = 0 for all k ∈ N ;(iii) lim b → − P k ≥ c k ( b ) = 1 .Then t ( b ) := P k ≥ c k ( b ) s k converges for all b ∈ (0 , . Furthermore, if lim n →∞ s n = s ∈ R ,then lim b → − t ( b ) = s .Proof of Theorem 1.1. We first prove thatlim b → − (1 − b ) X k ≥ b S k = µ − a.s. (6)For x ∈ R , put M ( x ) = { n ≥ S n ≤ x } . Since lim n →∞ S n = + ∞ a.s., we have M ( x ) < ∞ a.s. Furthermore, by Theorem B in [20], lim x →∞ x − M ( x ) = µ − a.s. Hence, given ε > x > | x − M ( x ) − µ − | ≤ ε whenever x ≥ x . Write X k ≥ b S k = X k ≥ b S k { S k ≤ x } + Z ( x , ∞ ) b x d M ( x ) . The number of summands in the sum on the right-hand side is a.s. finite, for it is equal to M ( x ),whence lim b → − P k ≥ b S k { S k ≤ x } = M ( x ) a.s. Integration by parts yields Z ( x , ∞ ) b x d M ( x ) + b x M ( x ) = | log b | Z ∞ x b x M ( x )d x ≤ ( µ − + ε ) b x (1 + | log b | x ) / | log b | . Thus, lim sup b → − (1 − b ) X k ≥ b S k ≤ µ − a.s.The proof of the converse inequality for the limit inferior is completely analogous.4assing to the proof of (1) we use summation by parts to obtain, for b ∈ (0 ,
1) and ℓ ∈ N , ℓ X k =1 b S k − η k = ℓ − X k =1 ( b S k − − b S k ) T k + b S ℓ − T ℓ , (7)where T := 0 and T k := η + . . . + η k for k ∈ N . We have lim ℓ →∞ b S ℓ − T ℓ = 0 a.s. because bythe strong law of large numbers the first factor decreases to zero exponentially fast, whereas thesecond factor exhibits at most linear growth. Hence, X k ≥ b S k − η k = X k ≥ k ( b S k − − b S k )( k − T k ) . We are going to apply Lemma 3.1 with c k ( b ) := µ (1 − b ) k ( b S k − − b S k ) for k ∈ N and b ∈ (0 , s k := k − T k for k ∈ N . While (ii) of Lemma 3.1 holds trivially (a.s.), (iii) is a consequence of P k ≥ c k ( b ) = µ (1 − b ) P k ≥ b S k and (6). Let us prove (i). By another appeal to the strong lawof large numbers, given ε ∈ (0 , µ ), there exists a random integer N such that b S k − ≤ b ( µ − ε )( k − whenever k ≥ N + 1. Fix any b ∈ (0 , k ≥ N + 1 and b ∈ ( b , | b S k − − b S k | ≤ max( b S k − , b S k ) | log b || ξ k | ≤ b − ( µ − ε )1 b ( µ − ε ) k | log b || ξ k | . (8)Using the inequality xe − x ≤ e − x/ for x ≥ k ≥ N + 1 and b ∈ ( b , k | b S k − − b S k | ≤ µ − ε ) − b − ( µ − ε )1 b ( µ − ε ) k/ | ξ k | =: cb ( µ − ε ) k/ | ξ k | . With this at hand, for b ∈ ( b , µ (1 − b )) − X k ≥ | c k ( b ) | = X k ≥ k | b S k − − b S k | ≤ N X k =1 k + X k ≥ N +1 k | b S k − − b S k | ≤ N ( N + 1)+ c X k ≥ b ( µ − ε ) k/ | ξ k | . In view of (5), lim b → − (1 − b ) P k ≥ b ( µ − ε ) k/ | ξ k | = 2 | E ξ | / ( µ − ε ) a.s. This justifies (i) in thepresent setting.By the strong law of large numbers lim k →∞ s k = lim k →∞ ( k − T k ) = m a.s. Invoking Lemma3.1 we arrive at (1). The proof of Theorem 1.1 is complete.Later on, we shall need the following result. Its proof is omitted, for it is analogous to theproof of Theorem 1.1. Lemma 3.2.
Assume that E | η | < ∞ . Let ( x n ) n ∈ N and ( y n ) n ∈ N be sequences of numbers in (0 , approaching as n → ∞ . Let λ > and M : (0 , → N be a function satisfying lim n →∞ M ( x n )(1 − y λn ) = a ∈ [0 , ∞ ] . If a = ∞ , then lim n →∞ − y λn y λM ( x n ) n X k ≥ M ( x n )+1 y λkn η k = m a.s. , where m = E η ; if a ∈ [0 , ∞ ) , then lim n →∞ (1 − y λn ) X k ≥ M ( x n )+1 y λkn η k = m e − a a.s. Clearly, these limit relations also hold if we put formally x n = y n = b and let b → − , thatis, if one passes to the limit continuously. Proof of Theorem 1.2
We shall prove weak convergence of the finite-dimensional distributions and then tightness. (2)
We shall use the Cram´er-Wold device. Namely, we intend to show that, for any ℓ ∈ N , any real α , . . . , α ℓ and any 0 < u < . . . < u ℓ < ∞ , as b → − ,(1 − b ) / ℓ X i =1 α i X k ≥ b u i S k η k +1 d −→ (2 s µ − ) / ℓ X i =1 α i Z [0 , ∞ ) e − u i y d B ( y ) . (9)For k ∈ N , denote by F k the σ -algebra generated by ( ξ j , η j ) ≤ j ≤ k . We shall write E k ( · ) for E ( ·|F k ). For each b ∈ (0 , (cid:16) (1 − b ) / ℓ X i =1 α i n − X k =0 b u i S k η k +1 , F n (cid:17) n ∈ N forms a martingale (the martingale is not necessarily integrable, for the situation that E b ξ = ∞ is not excluded). By the martingale central limit theorem (Theorem 2.5(a) in [15]), (9) followsif we can show that(1 − b ) X k ≥ E k (cid:16) ℓ X i =1 α i b u i S k η k +1 (cid:17) P → s µ − E (cid:16) ℓ X i =1 α i Z [0 , ∞ ) e − u i y d B ( y ) (cid:17) , b → − (10)and, for all ε > − b ) X k ≥ E k (cid:16) ℓ X i =1 α i b u i S k η k +1 (cid:17) { (1 − b ) / | P ℓi =1 α i b uiSk η k +1 | >ε } P → , b → − . (11)We start by proving (10):(1 − b ) X k ≥ E k (cid:16) ℓ X i =1 α i b u i S k η k +1 (cid:17) = s (1 − b ) (cid:16) ℓ X i =1 α i X k ≥ b u i S k + 2 X ≤ i
0, it suffices to show that, for all ε > u > − b ) X k ≥ E k ( b uS k η k +1 ) { (1 − b ) / b uSk | η k +1 | >ε } P → , b → − . T := sup { n ∈ N : S n ≤ } and note that T < ∞ a.s. as a consequence of lim n →∞ S n = + ∞ a.s. We infer(1 − b ) T X k =0 E k ( b uS k η k +1 ) { (1 − b ) / b uSk | η k +1 | >ε } ≤ s (1 − b ) T X k =0 b uS k → b → − . To proceed, observe that, for k ≥ T + 1, we have b uS k ≤
1, whence { (1 − b ) / b uS k | η k +1 | > ε } ⊆ {| η k +1 | > ε (1 − b ) − / } . This yields(1 − b ) X k ≥ T +1 E k ( b uS k η k +1 ) { (1 − b ) / b uSk | η k +1 | >ε } ≤ E η {| η | >ε (1 − b ) − / } (1 − b ) X k ≥ b uS k → b → − . The limit relation is justified by the fact that while the truncated second moment converges to0, lim b → − (1 − b ) P k ≥ b uS k = ( µu ) − a.s. by Theorem 1.1.For the proof of Proposition 5.7 we need the following one-dimensional central limit theorem. Lemma 4.1.
Let M : (0 , → N satisfy lim b → − M ( b ) = ∞ . Under the assumptions of Theorem1.2, as b → − , (cid:16) M ( b ) X k =0 b µk (cid:17) − / M ( b ) X k =0 b S k η k +1 d −→ ( s µ − ) / Normal(0 , . After noting that P M ( b ) k =0 b S k ∼ P M ( b ) k =0 b µk a.s. as b → − by the strong law of large numbersfor random walks, a simplified version of the proof given above applies. We omit details. (2) Fix any c, d ∈ (0 , ∞ ), c < d . We have to prove tightness on [ c, d ].For each δ ∈ (0 , µ ) and k ∈ N , define the event R k ( δ ) := {| S k − µk | > δk } . We first checkthat lim b → − (1 − b ) / sup u ∈ [ c, d ] (cid:12)(cid:12)(cid:12) X k ≥ b uS k R k ( δ ) η k +1 (cid:12)(cid:12)(cid:12) = 0 a.s. (12)Indeed, the supremum does not exceed a.s. X k ≥ b cS k { S k ≥ } R k ( δ ) | η k +1 | + X k ≥ b dS k { S k < } R k ( δ ) | η k +1 | . Here, each summand converges a.s. as b → − to an a.s. finite random variable. Furthermore,the number of nonzero summands is a.s. finite in view of P k ≥ R k ( δ ) < ∞ a.s. which is aconsequence of the strong law of large numbers. Thus, (12) has been proved.Next, we intend to show that, for any u, v ∈ [ c, d ] and b < − b ) E (cid:16) X k ≥ ( b uS k − b vS k ) R ck ( δ ) η k +1 (cid:17) ≤ A ( u − v ) (13)7or a constant A which does not depend on u and v . Here, R ck ( δ ) denotes the complement of R k ( δ ), that is, R ck ( δ ) = {| S k − µk | ≤ δk } . To this end, we observe that R ck ( δ ) ⊆ { S k > } andthen invoking the mean value theorem for differentiable functions we obtain a.s. on R ck ( δ ) | b uS k − b vS k | ≤ b cS k | log b || u − v | S k ≤ ( µ + δ ) b c ( µ − δ ) k | log b || u − v | k ≤ µ + δ )( ce ( µ − δ )) − b ( c/ µ − δ ) k | u − v | . We have used the inequality sup x> | log b | xb x ≤ /e for the last step. It remains to note that E (cid:16) X k ≥ ( b uS k − b vS k ) R ck ( δ ) η k +1 (cid:17) = s E X k ≥ ( b uS k − b vS k ) R ck ( δ ) ≤ s ( µ + δ ) ( ce ( µ − δ )) − X k ≥ b c ( µ − δ ) k ( u − v ) and that lim b → − (1 − b ) X k ≥ b c ( µ − δ ) k = 2( c ( µ − δ )) − . Thus, (13) holds with A = 16 s ( µ + δ ) e − c − ( µ − δ ) − . By formula (12.51) on p. 95 in [3], thedistributions of (cid:16) (1 − b ) / X k ≥ b uS k R ck ( δ ) η k +1 (cid:17) u ∈ [ c, d ] are tight. The proof of Theorem 1.2 is complete. Our argument follows closely the paths of (slightly different) proofs of Theorem 1.1 in [5] andTheorem 1.1 in [22]. In the cited references S n = n , n ∈ N , that is, the random walk ( S n ) n ∈ N is deterministic. Of course, we know that in our setting, for large n , S n is approximately µn bythe strong law of large numbers. Thus, an additional effort is needed to justify the replacementof S n with µn .We start by proving an intermediate result. Proposition 5.1.
Under the assumptions of Theorem 1.5, lim sup b → − (cid:16) − b log log − b (cid:17) / X k ≥ b S k η k +1 ≤ (2 s µ − ) / a.s. (14) and lim inf b → − (cid:16) − b log log − b (cid:17) / X k ≥ b S k η k +1 ≥ − (2 s µ − ) / a.s. (15)We can and do assume that µ = s = 1. To see this, replace b S k − η k with b S k − /µ η k / s andnote that 1 − b µ ∼ µ (1 − b ) as b → − . Pick any δ ∈ (0 , b ∈ (0 ,
1) and such a δ , put N , δ ( b ) := j − b δ log 11 − b δ k b ∈ [(1 − e − ) / , f ( b ) := (cid:16) − b log log 11 − b (cid:17) − / . We prove Proposition 5.1 via a sequence of lemmas.
Lemma 5.2. lim b → − f ( b ) P k ≥ N , δ ( b ) b S k − η k = 0 a.s. Proof.
Pick any increasing sequence ( b n ) n ∈ N of positive numbers satisfying lim n →∞ b n = 1, b n +1 − b n ∼ c (1 − b n ) c , n → ∞ (16)for some c , c > X n ≥ n (1 − b n ) < ∞ (17)for some n ∈ N . One particular sequence satisfying these assumptions is given by b n = 1 − n − for n ∈ N (with c = 2 and c = 1 / n →∞ (1 − b n +1 ) / (1 − b n ) = 1 . Suppose we can prove that, for all ε > I := X n ≥ n P n sup b ∈ [ b n , b n +1 ] (cid:12)(cid:12)(cid:12) X k ≥ N , δ ( b ) b S k − η k (cid:12)(cid:12)(cid:12) > ε/f ( b n ) o < ∞ . Then, by the Borel–Cantelli lemma,sup b ∈ [ b n , b n +1 ] (cid:12)(cid:12)(cid:12) X k ≥ N , δ ( b ) b S k − η k (cid:12)(cid:12)(cid:12) ≤ ε/f ( b n )for n large enough a.s. Since f is nonnegative and decreasing on [(1 − e − ) / , n , (cid:12)(cid:12)(cid:12) X k ≥ N , δ ( b ) b S k − η k (cid:12)(cid:12)(cid:12) ≤ sup b ∈ [ b n , b n +1 ] (cid:12)(cid:12)(cid:12) X k ≥ N , δ ( b ) b S k − η k (cid:12)(cid:12)(cid:12) ≤ ε/f ( b n ) ≤ ε/f ( b )a.s. whenever b ∈ [ b n , b n +1 ]. Hence, lim sup b → − f ( b ) P k ≥ N , δ ( b ) b S k − η k ≤ ε a.s. which entailsthe claim.Since the function N , δ is nondecreasing on (0 ,
1) we obtainsup b ∈ [ b n , b n +1 ] (cid:12)(cid:12)(cid:12) X k ≥ N , δ ( b ) b S k − η k (cid:12)(cid:12)(cid:12) ≤ sup b ∈ [ b n , b n +1 ] X k ≥ N , δ ( b n ) b S k − | η k | . Further, by the strong law of large numbers, for large n , the latter is estimated from above bysup b ∈ [ b n , b n +1 ] X k ≥ N , δ ( b n ) b δ ( k − | η k | ≤ X k ≥ N , δ ( b n ) b δ ( k − n +1 E | η k | + X k ≥ N , δ ( b n ) b δ ( k − n +1 ( | η k | − E | η k | ) . Thus, noting that E | η k | ≤ I ≤ X n ≥ n { P k ≥ N , δ ( bn ) b δ ( k − n +1 >ε/ (2 f ( b n )) } + X n ≥ n P n X k ≥ N , δ ( b n ) b δ ( k − n +1 ( | η k | − E | η k | ) > ε/ (2 f ( b n )) o . (18)9sing (16) and − log x = (1 − x ) + O ((1 − x ) ) as x → − we obtain f ( b n ) X k ≥ N , δ ( b n ) b δ ( k − n +1 = f ( b n ) b δ ( N , δ ( b n ) − n +1 − b δn +1 ∼ f ( b n ) (1 − b δn ) / − b δn ∼ (1 − b n ) / (2 log log(1 / (1 − b n ))) / δ − / (1 − b n ) / → , n → ∞ . This proves that the first series on the right-hand side of (18) trivially converges, for it containsfinitely many nonzero summands. By Markov’s inequality and (16), the probability in the secondseries is upper bounded by4 ε − f ( b n ) X k ≥ N , δ ( b n ) b δ ( k − n +1 = 4 ε − f ( b n ) b δ ( N , δ ( b n ) − n +1 − b δn +1 ∼ ε − − b n log log(1 / (1 − b n )) , n → ∞ . In view of (17), this is the general term of a convergent series. Hence, the second series on theright-hand side of (18) converges. The proof of Lemma 5.2 is complete.For b ∈ (0 ,
1) close to 1, δ as above and θ >
0, put N , δ, θ ( b ) := j θ − b δ log log 11 − b δ k . Lemma 5.3. lim b → − f ( b ) P N , δ ( b ) k = N , δ, θ ( b )+1 b S k − η k = 0 a.s. Proof.
Similarly to (7), summation by parts yields N , δ ( b ) X k = N , δ, θ ( b )+1 b S k − η k = N , δ ( b ) − X k = N , δ, θ ( b )+1 ( b S k − − b S k ) T k + b S N , δ ( b ) − T N , δ ( b ) − b S N , δ, θ ( b ) T N , δ, θ ( b ) , where, as in the proof of Theorem 1.1, T k = η + . . . + η k for k ∈ N . By the strong law of largenumbers, for b close to 1, | b S N , δ ( b ) − T N , δ ( b ) − b S N , δ, θ ( b ) T N , δ, θ ( b ) | ≤ b δ ( N , δ ( b ) − | T N , δ ( b ) | + b δN , δ, θ ( b ) | T N , δ, θ ( b ) | . One can check that b δN , δ, θ ( b ) ∼ (cid:16) log 11 − b δ (cid:17) − (1+ θ ) / and b δN , δ ( b ) ∼ (1 − b δ ) / a.s. as b → − . (19)Further, recall that, as ℓ → ∞ , | T ℓ | ≤ sup k ≤ ℓ | T k | = O (cid:0) ( ℓ log log ℓ ) / (cid:1) a.s. (20)by the law of the iterated logarithm for standard random walks. Using this limit relation weinfer f ( b ) | b S N , δ ( b ) − T N , δ ( b ) − b S N , δ, θ ( b ) T N , δ, θ ( b ) | = O (((1 − b ) log(1 / (1 − b ))) / )+ O (cid:16)(cid:16) log log(1 / (1 − b ))(log(1 / (1 − b ))) θ (cid:17) / (cid:17) → b → − . (21)10ccording to (8), for b close to 1, (cid:12)(cid:12)(cid:12) N , δ ( b ) − X k = N , δ, θ ( b )+1 ( b S k − − b S k ) T k (cid:12)(cid:12)(cid:12) ≤ const | log b | ( sup k ≤ N , δ ( b ) | T k | ) X k ≥ N , δ, θ ( b )+1 b δk | ξ k | . With the help of (19) we obtain X k ≥ N , δ, θ ( b )+1 b δk | ξ k | ∼ E | ξ | b δN , δ, θ ( b ) − b δ ∼ E | ξ | (1 − b δ )(log(1 / (1 − b δ ))) (1+ θ ) / a.s. as b → − by an application of Lemma 3.2 with η = | ξ | and a = ∞ . This in combination with (20) yields f ( b ) (cid:12)(cid:12)(cid:12) N , δ ( b ) − X k = N , δ, θ ( b )+1 ( b S k − − b S k ) T k (cid:12)(cid:12)(cid:12) = O (cid:16) / (1 − b ))) θ/ (cid:17) → b → − . (22)The proof of Lemma 5.3 is complete.For b ∈ (0 ,
1) close to 1, put N ( b ) := j − b log 11 − b k . We claim that lim b → − f ( b ) N , δ ( b ) X k = N ( b )+1 b S k − η k = 0 a . s . For the most part, this follows by repeating the proof of Lemma 5.3 with N ( b ) replacing N , δ, θ ( b ), the only changes being that the second summand on the right-hand side of (21) andthe right-hand side of (22) are O (((1 − b ) δ log(1 / (1 − b ))) / ) as b → − . The last centeredformula in combination with Lemma 5.2 enable us to conclude thatlim b → − f ( b ) X k ≥ N ( b )+1 b S k − η k = 0 a . s . (23)This limit relation will be used in the proof of Proposition 5.7.Denote by F the trivial σ -algebra and recall that, for k ∈ N , F k denotes the σ -algebragenerated by ( ξ j , η j ) ≤ j ≤ k and that, for k ∈ N , we write E k ( · ) for E ( ·|F k ). Lemma 5.4.
For all ρ > , lim b → − f ( b ) N , δ, θ ( b ) X k =1 b S k − η k S k ( b ) = 0 a.s. (24) and lim b → − f ( b ) N , δ, θ ( b ) X k =1 b S k − E k − ( η k S k ( b ) ) = 0 a.s. , (25) where S k ( b ) := {| η k | > ρb − S k − ((1 − b δ ) log log(1 / (1 − b δ ))) − / } . roof. We only give a detailed proof of (24) and then explain which modifications are neededfor a proof of (25).
Proof of (24). For b ∈ (0 ,
1) and the same δ ∈ (0 ,
1) as before, put N δ ( b ) := j − b δ k . Plainly, for all ρ > b → − f ( b ) X k =1 b S k − η k {| η k | >ρb − Sk − ((1 − b δ ) log log(1 / (1 − b δ ))) − / } = 0 a.s.We first show that, for k ≥ b close to 1 and ε > b − S k − (cid:16) − b δ ) log log(1 / (1 − b δ )) (cid:17) / ≥ e − ε (cid:16) k log log k (cid:17) / a.s. (26)Let 3 ≤ k ≤ N δ ( b ). The function x x/ log log x is increasing for large x , whence1(1 − b δ ) log log(1 / (1 − b δ )) ≥ k log log k . Further, for 1 ≤ k ≤ N δ ( b ), b − S k − ≥ e ( − log b ) S k − ≥ e ( − log b ) inf ≤ i ≤ Nδ ( b ) − S i a.s.Since lim n →∞ S n = + ∞ a.s., we infer | inf i ≥ S i | < ∞ a.s. and thereuponlim b → − (log b ) inf ≤ i ≤ N δ ( b ) S i = 0 a.s.Thus, given ε > b ∗ such that b − S k − ≥ e − ε whenever 3 ≤ k ≤ N δ ( b ) and b ∈ ( b ∗ ,
1) (of course, b − S k − ≥ k ∈ N provided that ξ ≥ k .Let k ≥ N δ ( b ) + 1. By the strong law of large numbers, b S k − ≤ b δ ( k − a.s. for b close to 1.Put α k ( b ) := b δ ( k − (cid:16) k log log k (cid:17) / . We claim that the sequence ( α k ( b )) k ≥ N δ ( b )+1 is nonincreasing. Indeed, α k +1 ( b ) /α k ( b ) ≤ b δ (1 + 1 /k ) / ≤ b δ (1 + 1 / (2 k )) ≤ b δ (1 + (1 − b δ ) / ≤ . We have used max b ∈ [0 , (3 b δ − b δ ) = 2 for the last step. Hence, for b close to 1, b S k − (cid:16) k log log k (cid:17) / ≤ b δ ( k − (cid:16) k log log k (cid:17) / ≤ b δN δ ( b ) (cid:16) N δ ( b )log log N δ ( b ) (cid:17) / ≤ b δ (1 − b δ ) − (cid:16) − b δ ) log log(1 / (1 − b δ )) (cid:17) / ≤ e ε (cid:16) − b δ ) log log(1 / (1 − b δ )) (cid:17) / having utilized lim b → − b δ (1 − b δ ) − = e − / for the last inequality. The proof of (26) is complete.12or b ∈ (0 , K ( b ) be positive integers satisfying lim b → − K ( b ) = ∞ . In view of (26),for b close to 1, (cid:12)(cid:12)(cid:12) K ( b ) X k =3 b S k − η k S k ( b ) (cid:12)(cid:12)(cid:12) ≤ e ε K ( b ) X k =3 | η k | {| η k | >ρe − ε ( k/ log log k ) / } a.s. (27)and K ( b ) X k =3 | η k | S k ( b ) ≤ K ( b ) X k =3 | η k | {| η k | >ρe − ε ( k/ log log k ) / } a.s. (28)It is shown in the proof of Lemma 2.3 in [5] that X k ≥ ( k log log k ) − / E ( | η k | {| η k | >ρe − ε ( k/ log log k ) / } ) < ∞ (29)which particularly entails X k ≥ ( k log log k ) − / | η k | {| η k | >ρe − ε ( k/ log log k ) / } < ∞ a.s.By Kronecker’s lemma, we obtainlim b → − ( K ( b ) log log K ( b )) − / K ( b ) X k =3 | η k | {| η k | >ρe − ε ( k/ log log k ) / } = 0 a.s. (30)We treat the sums P N δ ( b ) k =3 and P N , δ, θ ( b ) k = N δ ( b )+1 separately. Relation (30) with K ( b ) = N δ ( b )implies that, for all ρ >
0, lim b → − f ( b ) N δ ( b ) X k =1 b S k − η k S k ( b ) = 0 a.s. (31)To deal with the second sum, we write, for b close to 1, (cid:12)(cid:12)(cid:12) N , δ, θ ( b ) X k = N δ ( b )+1 b S k − η k S k ( b ) (cid:12)(cid:12)(cid:12) ≤ N , δ, θ ( b ) X k = N δ ( b )+1 b δ ( k − | η k | S k ( b ) = − b δN δ ( b ) N δ ( b ) X k =1 | η k | S k ( b ) + b δ ( N , δ, θ ( b ) − N , δ, θ ( b ) X k =1 | η k | S k ( b ) +(1 − b δ ) N , δ, θ ( b ) − X k = N δ ( b )+1 b δ ( k − k X j =1 | η j | S j ( b ) =: I ( b )+ I ( b )+ I ( b )having utilized the strong law of large numbers for the inequality. Analysis of I . The limit relation lim b → − b δN δ ( b ) = e − / together with (28) and (30) in whichwe take K ( b ) = N δ ( b ) proves lim b → − f ( b ) I ( b ) = 0 a.s. Analysis of I . Using (28) and (30) with K ( b ) = N , δ, θ ( b ) we inferlim b → − ( N , δ, θ ( b ) log log N , δ, θ ( b )) − / N , δ, θ ( b ) X k =1 | η k | S k ( b ) = 0 a.s.Combining this with the first part of (19) we obtain lim b → − f ( b ) I ( b ) = 0 a.s.13 nalysis of I . Write I ( b ) ≤ (1 − b δ )( sup N δ ( b )+1 ≤ k ≤ N , δ, θ ( b ) − T k ( b )) X k ≥ b δ ( k − ( k log log k ) / ∼ ( π / / (cid:16) log log(1 / (1 − b δ ))1 − b δ (cid:17) / ( sup N δ ( b )+1 ≤ k ≤ N , δ, θ ( b ) − T k ( b )) , b → − , where T k ( b ) := ( k log log k ) − / k X j =1 | η j | S j ( b ) . We have used Corollary 1.7.3 in [4] for the asymptotic equivalence. In view of (30) with K ( b ) = N δ ( b )+1 and (28), lim b → − sup N δ ( b )+1 ≤ k ≤ N , δ θ ( b ) − T k ( b ) = 0 a.s., whence lim b → − f ( b ) I ( b ) = 0a.s. The proof of (24) is complete. Proof of (25). Similarly to (31), we obtain with the help of (cid:12)(cid:12)(cid:12) K ( b ) X k =3 b S k − E k − ( η k S k ( b ) ) (cid:12)(cid:12)(cid:12) ≤ e ε K ( b ) X k =3 E | η k | {| η k | >ρe − ε ( k/ log log k ) / } a.s.(a counterpart of (27)) and (29) thatlim b → − f ( b ) N δ ( b ) X k =1 b S k − E k − ( η k S k ( b ) ) = 0 a.s.By the same reasoning, we also conclude that lim b → − f ( b ) I ∗ ℓ ( b ) = 0 a.s., ℓ = 1 , ,
3, where I ∗ ℓ ( b )is a counterpart of I ℓ ( b ) in which | η k | S k ( b ) is replaced with E k − ( | η k | S k ( b ) ).The proof of Lemma 5.4 is complete.As usual, S ck ( b ) will denote the complement of S k ( b ), that is, S ck ( b ) = {| η k | ≤ ρb − S k − ((1 − b δ ) log log(1 / (1 − b δ ))) − / } . Denote by B the class of increasing sequences ( b n ) n ∈ N of positive numbers satisfying the followingproperties:(a) lim n →∞ b n = 1 and lim n →∞ − b n +1 − b n = 1;(b) lim n →∞ b n +1 − b n − b n (cid:16) log log − b n (cid:17) / = 0;(c) for all ε > P n ≥ (cid:16) log (cid:16) − b n (cid:17)(cid:17) − − ε < ∞ .One can check that any increasing sequence ( b n ) n ∈ N of positive numbers satisfying b n =exp( − (1 − (log n ) − ) n ) for large n belongs to the class B . For instance, for the so defined b n wehave b n +1 − b n − b n ∼ (log n ) − and log log 11 − b n ∼ log n, n → ∞ which verifies the property (b).Recall that ‘i.o.’ is a shorthand for ‘infinitely often’ and that, for a sequence of sets A , A , . . . , { A n i.o. } := {∪ n ≥ ∩ k ≥ n A k } . emma 5.5. Let ( b n ) n ∈ N ∈ B . Then, for all ε > , P n N , δ, θ ( b n ) X k =1 b S k − n ˜ η k ( b n ) > (cid:0) (2 + ε ) N ( b n ) log log N ( b n ) (cid:1) / i.o. o = 0 , where ˜ η k ( b ) := η k S ck ( b ) − E k − ( η k S ck ( b ) ) for k ∈ N and N ( b ) := (1 − b ) − for b ∈ (0 , .Proof. The proof below follows the path of the proof of Lemma 3.6 in [22].We start by showing thatlim sup b → − (1 − b δ ) N , δ, θ ( b ) X k =1 b S k − E k − (˜ η k ( b )) ≤ δ a.s. (32)(recall that µ = 1 by convention). Indeed, E k − (˜ η k ( b )) ≤ E η k = 1, whence N , δ, θ ( b ) X k =1 b S k − E k − (˜ η k ( b )) ≤ X k ≥ b S k a.s.By Theorem 1.1 with η = 1 a.s., lim b → − (1 − b δ ) X k ≥ b S k = δ a.s.which entails (32).For n ∈ N , put t n := (cid:16) (2 + ε ) log log N ( b n ) N ( b n ) (cid:17) / and define the event B n := n N , δ, θ ( b n ) X k =1 b S k − n ˜ η k ( b n ) > (cid:0) (2 + ε ) N ( b n ) log log N ( b n ) (cid:1) / o . Equivalently, B n := n t n N , δ, θ ( b n ) X k =1 b S k − n ˜ η k ( b n ) − ( t n / e ρ (1+ ε ) N , δ, θ ( b n ) X k =1 b S k − n E k − (˜ η k ( b n )) > t n N ( b n ) (cid:16) − ( e ρ (1+ ε ) / (2 N ( b n ))) N , δ, θ ( b n ) X k =1 b S k − n E k − (˜ η k ( b n )) (cid:17)o . In view of (32), given β > B n ⊆ A n := n t n N , δ, θ ( b n ) X k =1 b S k − n ˜ η k ( b n ) − ( t n / e ρ (1+ ε ) N , δ, θ ( b n ) X k =1 b S k − n E k − (˜ η k ( b n )) > t n N ( b n )(1 − ( e ρ (1+ ε ) / δ + β )) o n . Thus, by the Borel-Cantelli lemma, Lemma 5.5 follows if we can check that X n ≥ P ( A n ) < ∞ . (33)As a preparation for this matter, we intend to show that E τ n =: E exp (cid:16) t n N , δ, θ ( b n ) X k =1 b S k − n ˜ η k ( b n ) − ( t n / e ρ (1+ ε ) N , δ, θ ( b n ) X k =1 b S k − n E k − (˜ η k ( b n )) (cid:17) ≤ . (34)Using e x ≤ x + ( x / e | x | for x ∈ R and E k − ˜ η k ( b n ) = 0 we infer E k − exp( t n b S k − n ˜ η k ( b n )) ≤ t n / b S k − n E k − (˜ η k ( b n ) exp( t n b S k − n | ˜ η k ( b n ) | )) a.s.Further, t n b S k − n | ˜ η k ( b n ) | ≤ ρ (2 + ε ) / ≤ ρ (1 + ε ) a.s.This in combination with e x ≥ x for x ≥ E k − (cid:0) exp( t n b S k − n ˜ η k ( b n )) (cid:1) exp( − ( t n / e ρ (1+ ε ) b S k − n E k − (˜ η k ( b n )) ≤ E N , δ, θ ( b n ) − τ n = exp (cid:16) t n N , δ, θ ( b n ) − X k =1 b S k − n ˜ η k ( b n ) − ( t n / e ρ (1+ ε ) N , δ, θ ( b n ) − X k =1 b S k − n E k − (˜ η k ( b n )) (cid:17) × E N , δ, θ ( b n ) − (cid:0) exp( t n b S k − n ˜ η N , δ, θ ( b n ) ( b n )) (cid:1) exp( − ( t n / e ρ (1+ ε ) b S k − n E N , δ, θ ( b n ) − (˜ η N , δ, θ ( b n ) ( b n )) ≤ exp (cid:16) t n N , δ, θ ( b n ) − X k =1 b S k − n ˜ η k ( b n ) − ( t n / e ρ (1+ ε ) N , δ, θ ( b n ) − X k =1 b S k − n E k − (˜ η k ( b n )) (cid:17) a.s.Further, E N , δ, θ ( b n ) − τ n = E N , δ, θ ( b n ) − ( E N , δ, θ ( b n ) − τ n ) ≤ exp (cid:16) t n N , δ, θ ( b n ) − X k =1 b S k − n ˜ η k ( b n ) − ( t n / e ρ (1+ ε ) N , δ, θ ( b n ) − X k =1 b S k − n E k − (˜ η k ( b n )) (cid:17) a.s.Repeating this argument N , δ, θ ( b n ) times we arrive at (34).We are ready to prove (33). By Markov’s inequality and (34), P ( A n ) ≤ exp (cid:0) − t n N ( b n )(1 − ( e ρ (1+ ε ) / δ + β )) (cid:1) E τ n ≤ exp (cid:0) − t n N ( b n )(1 − ( e ρ (1+ ε ) / δ + θ )) (cid:1) . Given small enough ε >
0, and δ ∈ (0 ,
1) and β > δ + β ∈ (0 ,
1) we can find ρ > ε )(1 − ( e ρ (1+ ε ) / )( δ + β )) >
1. This together with the property (c) of B ensures X n ≥ exp( − t n N ( b n )((1 − ( e ρ (1+ ε ) / δ + θ )))) < ∞ , and (33) follows. The proof of Lemma 5.5 is complete.16 emma 5.6. Let ( b n ) n ∈ N ∈ B . Then lim n →∞ sup b ∈ [ b n , b n +1 ] (cid:12)(cid:12)(cid:12) f ( b ) N , δ, θ ( b ) X k =1 b S k − η k − f ( b n ) N , δ, θ ( b n ) X k =1 b S k − n η k (cid:12)(cid:12)(cid:12) = 0 a.s. Proof.
Throughout the proof we tacitly assume that the equalities and inequalities hold a.s. Westart by writing, for b ∈ [ b n , b n +1 ], f ( b ) N , δ, θ ( b ) X k =1 b S k − η k − f ( b n ) N , δ, θ ( b n ) X k =1 b S k − n η k = N , δ, θ ( b n ) X k =1 (cid:0) f ( b ) b S k − − f ( b n ) b S k − n (cid:1) η k + f ( b ) N , δ, θ ( b ) X k = N , δ, θ ( b n )+1 b S k − η k =: I n ( b ) + J n ( b ) . Summation by parts yields I n ( b ) = (cid:0) f ( b ) b S N , δ, θ ( b n ) − − f ( b n ) b S N , δ, θ ( b n ) − n (cid:1) T N , δ, θ ( b n ) + N , δ, θ ( b n ) − X k =1 (cid:0) f ( b )( b S k − − b S k ) − f ( b n )( b S k − n − b S k n ) (cid:1) T k =: I n, ( b ) + I n, ( b ) , where T k = η + . . . + η k for k ∈ N . For large enough n for which S N , δ, θ ( b n ) − ≥ δ ( N , δ, θ ( b n ) − ε > δ | log b n +1 | / (1 − b δn ) ≥ / − ε (35)(this is ensured by the property (a) of B ),sup b ∈ [ b n , b n +1 ] | I n, ( b ) | ≤ f ( b n ) b δ ( N , δ, θ ( b n ) − n +1 | T N , δ, θ ( b n ) |≤ b − δ f ( b n )(log(1 / (1 − b δn ))) − (1 / − ε )(1+ θ ) O (cid:0) ( N , δ, θ ( b n ) log log N , δ, θ ( b n )) / (cid:1) = O (cid:16) (log log(1 / (1 − b n ))) / (log(1 / (1 − b n ))) (1 / − ε )(1+ θ ) (cid:17) → n → ∞ . having utilized (20) for the inequality. We are now passing to the analysis of I n, ( b ). By thestrong law of large numbers, with the same δ ∈ (0 ,
1) there exists an a.s. finite τ such thatmax( δk, ≤ S k ≤ (2 − δ ) k for all k ≥ τ + 1. Since, for b ∈ [ b n , b n +1 ], (cid:12)(cid:12)(cid:12) τ X k =1 f ( b )( b S k − − b S k ) T k (cid:12)(cid:12)(cid:12) ≤ f ( b ) τ X k =1 ( b S k − + b S k ) | T k | ≤ f ( b n ) τ X k =1 ( b S k − n + b S k n ) | T k | we infer lim n →∞ sup b ∈ [ b n b n +1 ] (cid:12)(cid:12)(cid:12) τ X k =1 f ( b ) (cid:0) b S k − − b S k (cid:1) T k (cid:12)(cid:12)(cid:12) = 0 a.s.We need some preparation to treat the remaining part of the sum. Using the fact that when ξ k ≥ b f ( b )(1 − b ξ k ) is nonincreasing for b < { ξ k ≥ , τ ≤ k − } , for b ∈ [ b n , b n +1 ] and large n ∈ N , f ( b n )( b S k − n − b S k − n +1 )(1 − b ξ k n ) ≤ f ( b n ) (cid:0) b S k − n − b S k n (cid:1) − f ( b ) (cid:0) b S k − − b S k (cid:1) ≤ b S k − n (cid:0) f ( b n )(1 − b ξ k n ) − f ( b n +1 )(1 − b ξ k n +1 ) (cid:1) . { ξ k < , τ ≤ k − } we arrive at (cid:12)(cid:12) f ( b n ) (cid:0) b S k − n − b S k n (cid:1) − f ( b ) (cid:0) b S k − − b S k (cid:1)(cid:12)(cid:12) ≤ f ( b n ) (cid:0) b S k − n +1 − b S k − n (cid:1) | − b ξ k n | + b S k − n (cid:0) f ( b n ) | − b ξ k n | − f ( b n +1 ) | − b ξ k n +1 | (cid:1) for b and n as above. Thus, for b ∈ [ b n , b n +1 ] and large n ∈ N , (cid:12)(cid:12)(cid:12) N , δ, θ ( b n ) − X k = τ +1 (cid:0) f ( b )( b S k − − b S k ) − f ( b n )( b S k − n − b S k n ) (cid:1) T k (cid:12)(cid:12)(cid:12) ≤ N , δ, θ ( b n ) − X k = τ +1 (cid:12)(cid:12) f ( b )( b S k − − b S k ) − f ( b n )( b S k − n − b S k n ) (cid:12)(cid:12) | T k |≤ f ( b n ) N , δ, θ ( b n ) − X k = τ +1 (cid:0) b S k − n +1 − b S k − n (cid:1) | − b ξ k n || T k | + N , δ, θ ( b n ) − X k = τ +1 b S k − n (cid:0) f ( b n ) | − b ξ k n | − f ( b n +1 ) | − b ξ k n +1 | (cid:1) | T k | =: I n, ( b ) + I n, ( b ) . For all k ∈ N and all n ∈ N , | − b ξ k n | ≤ | log b n | ( ξ + k + ξ − k b ξ k n ) , (36)where, as usual, x + = max( x,
0) and x − = max( − x,
0) for x ∈ R . For k ≥ τ + 1 and n ∈ N , bythe mean value theorem for differentiable functions, b S k − n +1 − b S k − n ≤ S k − b S k − − n +1 ( b n +1 − b n ) ≤ b − S k − b S k − n +1 ( b n +1 − b n ) (37)and thereupon( b S k − n +1 − b S k − n ) | − b ξ k n | ≤ (2 − δ ) b − ( b n +1 − b n ) | log b n | k (cid:0) b S k − n +1 ξ + k + ( b n +1 / b n ) S k − b S k n ξ − k (cid:1) . Thus, (2 − δ ) − b N , δ, θ ( b n ) − X k = τ +1 (cid:0) b S k − n +1 − b S k − n (cid:1)(cid:12)(cid:12) − b ξ k n (cid:12)(cid:12) | T k |≤ ( b n +1 − b n ) | log b n | N , δ, θ ( b n ) X k =1 k | T k | (cid:0) b δ ( k − n +1 ξ + k + ( b n +1 / b n ) (2 − δ ) k b δkn ξ − k (cid:1) ≤ ( b n +1 − b n ) | log b n | N , δ, θ ( b n )( sup ≤ k ≤ N , δ, θ ( b n ) | T k | ) (cid:16) X k ≥ b δ ( k − n +1 ξ + k + ( b n +1 / b n ) (2 − δ ) N , δ, θ ( b n ) X k ≥ b δkn ξ − k (cid:17) . By Theorem 1.1, as n → ∞ , X k ≥ b δ ( k − n +1 ξ + k ∼ (1 − b δn +1 ) − E ξ + and X k ≥ b δkn ξ − k ∼ (1 − b δn ) − E ξ − a.s. (38)18sing (20) in combination with the property (a) of B for the first equality and the property (b)of B for the second we infer f ( b n )( b n +1 − b n ) | log b n | N , δ, θ ( b n )( sup ≤ k ≤ N , δ, θ ( b n ) | T k | ) X k ≥ b δ ( k − n +1 ξ + k = O (cid:16) b n +1 − b n − b n (cid:16) log log 11 − b n (cid:17) / (cid:17) = o (1) a.s. as n → ∞ . Invoking once again the property (b) of B we obtain lim n →∞ log( b n +1 / b n ) N , δ, θ ( b n ) = 0, whencelim n →∞ ( b n +1 / b n ) (2 − δ ) N , δ, θ ( b n ) = 1. With this at hand we can argue as before to conclude thata.s.lim n →∞ f ( b n )( b n +1 − b n ) | log b n | N , δ, θ ( b n )( b n +1 / b n ) (2 − δ ) N , δ, θ ( b n ) ( sup ≤ k ≤ N , δ, θ ( b n ) | T k | ) X k ≥ b δkn ξ − k = 0 . Thus, we have proved that lim n →∞ I n, ( b ) = 0 a.s.Further, I n, ( b ) = ( f ( b n ) − f ( b n +1 )) N , δ, θ ( b n ) − X k = τ +1 b S k − n | − b ξ k n || T k | + f ( b n +1 ) N , δ, θ ( b n ) − X k = τ +1 b S k − n (cid:0)(cid:12)(cid:12) − b ξ k n (cid:12)(cid:12) − (cid:12)(cid:12) − b ξ k n +1 (cid:12)(cid:12)(cid:1) | T k | . In view of (36), N , δ, θ ( b n ) − X k = τ +1 b S k − n | − b ξ k n || T k | ≤ | log b n | ( sup ≤ k ≤ N , δ, θ ( b n ) | T k | ) N , δ, θ ( b n ) − X k = τ +1 ( b S k − n ξ + k + b S k n ξ − k ) ≤ | log b n | ( sup ≤ k ≤ N , δ, θ ( b n ) | T k | ) X k ≥ ( b δ ( k − n ξ + k + b δkn ξ − k ) . Invoking (38) and (20) in combination with lim n →∞ N , δ, θ ( b n )( f ( b n )) = (1 + θ )(2 δ ) − weconclude that( f ( b n ) − f ( b n +1 )) N , δ, θ ( b n ) − X k = τ +1 b S k − n | − b ξ k n || T k | = O (cid:16) f ( b n ) − f ( b n +1 ) f ( b n ) (cid:16) log log 11 − b n (cid:17) / (cid:17) = o (1) a.s. as n → ∞ . The last equality is justified as follows. Using subadditivity of x x / on [0 , ∞ ) we obtain,for large n , (cid:16) f ( b n ) − f ( b n +1 ) f ( b n ) (cid:17) log log 11 − b n ≤ (cid:16) log log 11 − b n +1 − log log 11 − b n + b n +1 − b n − b n log log 11 − b n (cid:17) log log − b n log log − b n +1 . B entailslim n →∞ log(1 / (1 − b n +1 ))log(1 / (1 − b n )) = 1 and lim n →∞ log log(1 / (1 − b n +1 ))log log(1 / (1 − b n )) = 1 , (39)and the first of these ensureslim n →∞ (cid:16) log log 11 − b n +1 − log log 11 − b n (cid:17) = 0 . Finally, lim n →∞ b n +1 − b n − b n log log 11 − b n = 0is a consequence of the property (b) of B . Thus, the equality that we wanted to justify doesindeed hold.For the analysis of the second piece of I n, ( b ) we need an estimate similar to (37): for k, n ∈ N , (cid:12)(cid:12) − b ξ k n (cid:12)(cid:12) − (cid:12)(cid:12) − b ξ k n +1 (cid:12)(cid:12) = (cid:12)(cid:12) b ξ k n +1 − b ξ k n (cid:12)(cid:12) ≤ ( b n +1 − b n ) (cid:0) ξ + k b ξ k − n +1 + ξ − k b ξ k − n (cid:1) ≤ b − ( b n +1 − b n ) (cid:0) ξ + k + ξ − k b ξ k n (cid:1) . This implies that f ( b n +1 ) N , δ, θ ( b n ) − X k = τ +1 b S k − n (cid:0)(cid:12)(cid:12) − b ξ k n (cid:12)(cid:12) − (cid:12)(cid:12) − b ξ k n +1 (cid:12)(cid:12)(cid:1) | T k |≤ b − f ( b n )( b n +1 − b n )( sup ≤ k ≤ N , δ, θ ( b n ) | T k | ) X k ≥ ( b δ ( k − n ξ + k + b δkn ξ − k )= O (cid:16) b n +1 − b n − b n (cid:16) log log 11 − b n (cid:17) / (cid:17) = o (1) a.s. as n → ∞ . Here, while the first equality is ensured by (20) and (38), the second is a consequence of theproperty (b) of B . The proof of lim n →∞ I n ( b ) = 0 a.s. is complete.We proceed by analyzing J n ( b ): for b ∈ [ b n , b n +1 ], J n ( b ) = f ( b ) N , δ, θ ( b ) − X k = N , δ, θ ( b n ) ( b S k − − b S k ) T k + f ( b )( b S N , δ, θ ( b ) − T N , δ, θ ( b ) − b S N , δ, θ ( b n ) − T N , δ, θ ( b n ) )=: J n, ( b ) + J n, ( b ) . As before, appealing to the strong law of large numbers, we conclude thatsup b ∈ [ b n , b n +1 ] | J n, ( b ) | ≤ f ( b n ) b δ ( N , δ, θ ( b n ) − n +1 sup k ≤ N , δ, θ ( b n +1 ) | T k |≤ b − δ f ( b n )(log(1 / (1 − b δn ))) − (1 / − ε )(1+ θ ) O (cid:0) ( N , δ, θ ( b n +1 ) log log N , δ, θ ( b n +1 )) / (cid:1) = O (cid:16) (log log(1 / (1 − b n ))) / (log(1 / (1 − b n ))) (1 / − ε )(1+ θ ) (cid:17) → n → ∞ .
20e have used (20) and (35) for the inequality and the property (a) of B and its consequences(39) for the equality. Invoking (8) we obtain, for large n and appropriate constant C > b ∈ [ b n , b n +1 ] | J n, ( b ) | ≤ sup b ∈ [ b n , b n +1 ] f ( b ) N , δ, θ ( b ) − X k = N , δ, θ ( b n ) | b S k − − b S k || T k |≤ Cf ( b n ) | log b n | N , δ, θ ( b n +1 ) − X k = N , δ, θ ( b n ) b δkn +1 | ξ k | ( sup k ≤ N , δ, θ ( b n +1 ) | T k | ) ≤ Cf ( b n ) | log b n | X k ≥ N , δ, θ ( b n ) b δkn +1 | ξ k | O (cid:0) ( N , δ, θ ( b n +1 ) log log N , δ, θ ( b n +1 )) / (cid:1) . We use Lemma 3.2 with η = | ξ | , λ = δ , M ( b ) = N , δ, θ ( b ) − x n = b n and y n = b n +1 . Recallingthe property (a) of B we conclude that lim n →∞ N , δ, θ ( b n )(1 − b δn +1 ) = ∞ . Hence, an applicationof that lemma yields X k ≥ N , δ, θ ( b n ) b δkn +1 | ξ k | ∼ E | ξ | b N , δ, θ ( b n ) n +1 (1 − b δn +1 ) − a.s. as n → ∞ . Using once again the property (a) of B and (39) in combination with the estimate for b N , δ, θ ( b n ) n +1 which is implied by (35) we infersup b ∈ [ b n , b n +1 ] | J n, ( b ) | = O (cid:16) (log log(1 / (1 − b n ))) / (log(1 / (1 − b n ))) (1 / − ε )(1+ θ ) (cid:17) → n → ∞ . The proof of Lemma 5.6 is complete.We are ready to prove Proposition 5.1.
Proof of Proposition 5.1.
We only prove (14), for (15) is a consequence of (14) with − η k replac-ing η k .By Lemmas 5.2 and 5.3 and (24), (14) is equivalent tolim sup b → − f ( b ) N , δ, θ ( b ) X k =1 b S k − η k S ck ( b ) ≤ E k − ( η k S ck ( b ) ) = − E k − ( η k S k ( b ) ), and Lemma 5.6. Proposition 5.7.
Under the assumptions of Theorem 1.5, lim sup b → − (cid:16) − b log log − b (cid:17) / X k ≥ b S k η k +1 ≥ (2 s µ − ) / a.s. (40) and lim inf b → − (cid:16) − b log log − b (cid:17) / X k ≥ b S k η k +1 ≤ − (2 s µ − ) / a.s. (41)21ecall the notation: for b ∈ (0 ,
1) close to 1, N ( b ) = j − b log 11 − b k . Denote by B ∗ the class of increasing sequences ( b n ) n ∈ N of positive numbers satisfying the fol-lowing properties:(a) lim n →∞ b n = 1 and lim n →∞ (1 − b n ) log n = 0;(b) for large n , N n +1 ≥ N ( b n ), where N n := j log(1 − / log n )2 log b n k . (c) for all a ∈ (0 ,
1) and some n ∈ N , P n ≥ n (cid:16) log (cid:16) − b n (cid:17)(cid:17) − a = ∞ .It was shown in Section 3 of [6] (see also pp. 180,181 and 184 in [5]) that the sequence ( b n ) n ≥ given by b n := exp (cid:16) − n ! Q nj =2 (log j ) Q nk =3 log log k (cid:17) belongs to the class B ∗ .As in the proof of Proposition 5.1 we proceed via a sequence of lemmas. Lemma 5.8.
Under the assumptions of Theorem 1.5, lim n →∞ f ( b n ) N n X k =1 b S k − n η k = 0 a.s. Proof.
We start by noting that lim n →∞ f ( b n )( N n log log N n ) / = 0 (42)or, equivalently, lim n →∞ − b n log log(1 / (1 − b n )) N n log log N n = 0 . The latter is an immediate consequence of (1 − b n ) N n ∼ (log n ) − → n → ∞ andlim sup n →∞ log log N n log log(1 / (1 − b n )) ≤ . Formula (7) with ℓ = N n and b = b n reads N n X k =1 b S k − n η k = N n − X k =1 ( b S k − n − b S k n ) T k + b S Nn − n T N n , where T k = η + . . . + η k for k ∈ N . Using lim n →∞ N ( n ) log b n = 0 in combination with thestrong law of large numbers we inferlim n →∞ b S Nn − n = 1 and lim n →∞ b S Nn n = 1 a.s. (43)This together with the law of the iterated logarithm for standard random walks entailslim sup n →∞ b S Nn − n T N n ( N n log log N n ) / = 2 / a.s.22urther, with the same N as in (8) (we replace µ − ε with δ ),lim n →∞ N X k =1 ( b S k − n − b S k n ) T k = 0 a.s.According to (8), for large enough n and a constant c > (cid:12)(cid:12)(cid:12) N n − X k = N +1 ( b S k − n − b S k n ) T k (cid:12)(cid:12)(cid:12) ≤ N n − X k =1 | b S k − n − b S k n || T k | ≤ c | log b n | X k ≥ b δkn | ξ k | ( sup k ≤ N n | T k | )= O (( N n log log N n ) / ) a.s.The last equality is a consequence of Theorem 1.1 (which gives lim n →∞ | log b n | P k ≥ b δkn | ξ k | = δ − E | ξ | a.s.) and (20). An appeal to (42) completes the proof of Lemma 5.8. Lemma 5.9.
Under the assumptions of Theorem 1.5, for all ε ∈ (0 , , P n f ( b n ) N ( b n ) X k = N n +1 b S k − n η k > − ε i.o. o = 1 . (44) Proof.
Assume that we have already proved that, for all ε ∈ (0 , P {C n ( ε ) i.o. } = 1 , (45)where C n ( ε ) := n f ( b n ) b − S Nn n N ( b n ) X k = N n +1 b S k − n η k > − ε o , n ∈ N . Setting, for each ε ∈ (0 , D n ( ε ) := { b S Nn n > − ε } we conclude with the help of the secondequality in (43) that, for all ε ∈ (0 , P {D n ( ε ) eventually } = 1. This in combination with(45) yields, for all ε , ε ∈ (0 , P n C n ( ε ) \ D n ( ε ) i.o. o = 1 . Since C n ( ε ) \ D n ( ε ) ⊆ n f ( b n ) N ( b n ) X k = N n +1 b S k − n η k > (1 − ε )(1 − ε ) o , n ∈ N , we arrive at (44).By the property (b) of B ∗ , N n +1 ≥ N ( b n ) for large n which implies that, for large n , therandom variables b − S Nn n N ( b n ) X k = N n +1 b S k − n η k = η N n +1 + b ξ Nn +1 n η N n +2 + . . . + b ξ Nn +1 + ... + ξ N b n ) − n η N ( b n ) are independent. Hence, by the converse part of the Borel-Cantelli lemma, (45) is a consequenceof X n ≥ P n f ( b n ) b − S Nn n N ( b n ) X k = N n +1 b S k − n η k > − ε o = ∞ . (46)23e intend to prove (46). Fix any δ ∈ (0 , n ∈ N and t >
0, put q n ( t ) := 2 t − j log log 11 − b n k . For notational simplicity, we shall write q n for q n ( t ). Further, for each n ∈ N and each nonneg-ative integer k ≤ q n define numbers r k,n by r ,n := 0, r k,n := inf n j ≥ r k − ,n + 1 : b δr k − ,n n j − X k = r k − ,n b kn ≥ σ n q − n o , k ∈ N , k ≤ q n − , where σ n := P N ( b n ) − N n − k =0 b kn , and r q n ,n := N ( b n ) − N n + 1. One can check by a directcalculation that the numbers are well-defined and that, for k ∈ N , k ≤ q n , r k,n − r k − ,n − X k =0 b kn ∼ r k,n − r k − ,n ∼ σ n q n , n → ∞ . (47)For the latter we have used the fact that the relations lim n →∞ b N n n = 1 and lim n →∞ b N ( b n ) n = 0entail σ n ∼ (1 − b n ) − , n → ∞ . For each n ∈ N , k ∈ N , k ≤ q n and t >
0, put˜ Z k,n := q / n σ − n r k,n − X j = r k − ,n b S j n η j +1 and Z k,n := q / n σ − n b (1+ δ ) r k − ,n n r k,n − X j = r k − ,n b S j − S rk − ,n n η j +1 . Observe that the random variables Z ,n , . . . , Z q n ,n are independent, and Z k,n d = q / n σ − n b (1+ δ ) r k − ,n n r k,n − r k − ,n − X j =0 b S j n η j +1 , where d = denotes equality of distributions. Noting that lim n →∞ b (1+ δ ) r k − ,n n = 1 and then using(47) we infer with the help of Lemma 4.1 thatlim n →∞ P { Z k,n ≤ x } = P { Normal (0 , ≤ x } , x ∈ R . (48)In view of f ( b n ) ∼ ( σ n q / n t ) − := α n , n → ∞ ,
24t suffices to prove (46) with α n replacing f ( b n ). Then with r − ,n := 0 P n α n b − S Nn n N ( b n ) X k = N n +1 b S k − n η k > − ε o = P n α n N ( b n ) − N n − X k =0 b S k n η k +1 > − ε o = P n σ n q / n tα n q − n q n X k =1 ˜ Z k,n > t (1 − ε ) o ≥ P n ˜ Z k,n > t (1 − ε ) , ≤ k ≤ q n o ≥ P n ˜ Z k,n > t (1 − ε ) , S r k − ,n − S r k − ,n ≤ (1 + δ )( r k − ,n − r k − ,n ) , ≤ k ≤ q n o ≥ P n Z k,n > t (1 − ε ) , S r k − ,n − S r k − ,n ≤ (1 + δ )( r k − ,n − r k − ,n ) , ≤ k ≤ q n o = q n Y k =1 P n Z k,n > t (1 − ε ) , S r k − ,n − S r k − ,n ≤ (1 + δ )( r k − ,n − r k − ,n ) o . Using (48) and the weak law of large numbers for random walks we conclude that, uniformly in k ∈ N , k ≤ q n ,lim n →∞ P n Z k,n > t (1 − ε ) , S r k − ,n − S r k − ,n ≤ (1+ δ )( r k − ,n − r k − ,n ) o = P { Normal (0 , > t (1 − ε ) } . (49)Given constants c ∈ (0 ,
1) and ρ ∈ (0 ,
1) we can choose t so large that A ( t ) := 2 t − (log(1 /c ) + ρ + log( t (1 − ε ))) + (1 − ε ) < n ,log P n Z k,n > t (1 − ε ) , S r k − ,n − S r k − ,n ≤ (1 + δ )( r k − ,n − r k − ,n ) o ≥ log P { Normal (0 , > t (1 − ε ) }− ρ ≥ − (log(1 /c )+ ρ +2 − t (1 − ε ) +log( t (1 − ε ))) = 2 − t A ( t ) , where the first inequality is a consequence of (49), and the second inequality follows from Lemma12.9 on p. 349 in [21]. Hence, for c , ρ , t and n as above P n α n b − S Nn n N ( b n ) X k = N n +1 b S k − n η k > − ε o ≥ exp( − − t A ( t ) q n ) = exp (cid:16) − A ( t ) ⌊ log log(1 / (1 − b n )) ⌋ (cid:17) . This is the general term of a divergent series, hence (46) holds, because X n ≥ n (cid:16) log 11 − b n (cid:17) − A ( t ) = ∞ by the property (c) of B ∗ . The proof of Lemma 5.9 is complete.Now we can prove Proposition 5.7 and Theorem 1.5. Proof of Proposition 5.7.
Relation (40) is a consequence of formula (23) and Lemmas 5.8 and5.9. Replacing in (40) η k with − η k we obtain (41).25 roof of Theorem 1.5. Relation (4) follows from Propositions 5.1 and 5.7.Recalling our convention that µ = s = 1 it remains to prove that C (cid:18)(cid:18) f ( b ) X k ≥ b S k η k +1 : b ∈ ((1 − e − ) / , (cid:19)(cid:19) = [ − ,
1] a.s. (50)To this end, we first note that the random function b P k ≥ b S k − η k is a.s. continuous on[0 , b b S k − η k is a.s. continuous on [0 , , a ] for each a ∈ (0 ,
1) with probability one. This follows from theinequality b S k − ≤ b δ ( k − ≤ a δ ( k − which holds for large k and b ∈ [0 , a ] and the fact that E P k ≥ a δ ( k − | η k | < ∞ . Thus, the function b f ( b ) P k ≥ b S k − η k is a.s. continuous on ((1 − e − ) / ,
1) with lim sup b → − = 1 and lim inf b → − = −
1. This immediately entails (50) with thehelp of the intermediate value theorem for continuous functions.
Acknowledgement . A. Iksanov and I. Samoilenko were supported by the National ResearchFoundation of Ukraine (project 2020.02/0014 “Asymptotic regimes of perturbed random walks:on the edge of modern and classical probability”). A. Iksanov thanks Alexander Marynych fora useful discussion concerning the proof of Lemma 5.4.
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