Asymptotic formulas for probabilities of large deviations of ladder heights

Asymptotic formulas for large-deviation probabilities of a ladder height in a random walk generated by a sequence of sums of i.i.d. random variables are deduced. Two cases are considered: a) the distribution F(x) of summands is normal with a zero mean. b) F(x) belongs to the domain of the normal...

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Asymptotic formulas for probabilities of large deviations of ladder heights / S.V. Nagaev // Theory of Stochastic Processes. — 2008. — Т. 14 (30), № 1. — С. 100–116. — Бібліогр.: 17 назв.— англ.
0321-3900
https://nasplib.isofts.kiev.ua/handle/123456789/4541
519.21
Asymptotic formulas for large-deviation probabilities of a ladder height in a random walk generated by a sequence of sums of i.i.d. random variables are deduced. Two cases are considered: a) the distribution F(x) of summands is normal with a zero mean. b) F(x) belongs to the domain of the normal attraction of a stable law with the exponent 0 < α < 1. The method of Laplace transforms is applied in proofs.
This article was partially supported by the Russian Foundation for Basic Research (grant 06-01-00069)
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Інститут математики НАН України
Asymptotic formulas for probabilities of large deviations of ladder heights
Article
published earlier
institution Digital Library of Periodicals of National Academy of Sciences of Ukraine
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title Asymptotic formulas for probabilities of large deviations of ladder heights
spellingShingle Asymptotic formulas for probabilities of large deviations of ladder heights
Nagaev, S.V.
title_short Asymptotic formulas for probabilities of large deviations of ladder heights
title_full Asymptotic formulas for probabilities of large deviations of ladder heights
title_fullStr Asymptotic formulas for probabilities of large deviations of ladder heights
title_full_unstemmed Asymptotic formulas for probabilities of large deviations of ladder heights
title_sort asymptotic formulas for probabilities of large deviations of ladder heights
author Nagaev, S.V.
author_facet Nagaev, S.V.
publishDate 2008
language English
publisher Інститут математики НАН України
format Article
description Asymptotic formulas for large-deviation probabilities of a ladder height in a random walk generated by a sequence of sums of i.i.d. random variables are deduced. Two cases are considered: a) the distribution F(x) of summands is normal with a zero mean. b) F(x) belongs to the domain of the normal attraction of a stable law with the exponent 0 < α < 1. The method of Laplace transforms is applied in proofs.
issn 0321-3900
url https://nasplib.isofts.kiev.ua/handle/123456789/4541
citation_txt Asymptotic formulas for probabilities of large deviations of ladder heights / S.V. Nagaev // Theory of Stochastic Processes. — 2008. — Т. 14 (30), № 1. — С. 100–116. — Бібліогр.: 17 назв.— англ.
work_keys_str_mv AT nagaevsv asymptoticformulasforprobabilitiesoflargedeviationsofladderheights
first_indexed 2025-11-24T03:25:34Z
last_indexed 2025-11-24T03:25:34Z
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fulltext Theory of Stochastic Processes Vol. 14 (30), no. 1, 2008, pp. 100–116 UDC 519.21 SERGEY V. NAGAEV ASYMPTOTIC FORMULAS FOR PROBABILITIES OF LARGE DEVIATIONS OF LADDER HEIGHTS Asymptotic formulas for large-deviation probabilities of a ladder height in a random walk generated by a sequence of sums of i.i.d. random variables are deduced. Two cases are considered: a) the distribution F (x) of summands is normal with a zero mean. b) F (x) belongs to the domain of the normal attraction of a stable law with the exponent 0 < α < 1. The method of Laplace transforms is applied in proofs. 1. Introduction Let X,X1, X2, . . . , Xn, . . . be i.i.d. variables with the distribution function F (x), not degenerate at zero. Put Sn = n∑ i=1 Xi, Fn(D) = P(Sn ∈ D). Introduce the notation N+ = min { n : Sn > 0 } , N− = min { n : Sn ≤ 0 } . Let Z+ := SN+ , Z− := SN− be respectively ascending and descending ladder heights, and F+(x) = P(Z+ < x), F−(x) = P(Z− > x). Denote, by H+(x), the renewal function corresponding to the distribution F+ of the ascending ladder height, H+(x) = ∞∑ n=0 F+ n (x), where F+ n , n ≥ 1, is the n-th convolution of F+, F0 is the degenerate distribution concentrated at zero. Similarly, the renewal function H−(x) is defined by F−. Notice that H+(x) = F0(x) + ∞∑ n=1 P ( min 0<k≤n−1 Sk > 0, 0 < Sn < x ) , H−(x) = F0(x) + ∞∑ n=1 P ( max 0≤k≤n−1 Sk ≤ 0, x < Sn ≤ 0 ) (see [1], Ch.12, § 2). 2000 AMS Mathematics Subject Classification. Primary 60F10. Key words and phrases. Characteristic function, harmonic renewal measure, Karamata’s criterion, ladder height, Laplace transform, slowly varying function, Spitzer series, Tauberian theorem. This article was partially supported by the Russian Foundation for Basic Research (grant 06-01-00069) 100 PROBABILITIES OF LARGE DEVIATIONS OF LADDER HEIGHTS 101 Hence, P(Z+ > x) = − ∫ 0+ −∞ P(X > x− y)dH−(y) = ∫ ∞ x H−(x− y)dF (y), (1) if x > 0, and P(Z− > x) = ∫ ∞ 0+ P(X < x− y)dH+(y) = ∫ x −∞ H+(x− y)dF (y), (2) if x < 0. The measure ν(D) = ∞∑ n=1 Fn(D) n (3) is called the harmonic renewal measure. Since Fn(x+ l)− Fn(x) < c(F ) l + 1√ n (4) (see, e.g., [2]), the measure ν is σ-finite. Harmonic renewal measures were studied in [3–7]. For every x > 0, put G+(x) = ν((0, x)). Evidently, G+(x) = ∞∑ n=1 n−1[Fn(x) − Fn(0+)] <∞. Similarly, define G−(x) = ν((x, 0]) for x ≤ 0. Harmonic renewal measures are of interest for us, first of all, because∫ ∞ 0 e−sxdH+(x) = exp { − ∫ ∞ 0+ e−sxdG+(x) } (5) and ∫ 0 −∞ esxdH−(x) = − exp { − ∫ 0+ −∞ esxdG−(x) } , (6) which provides the possibility of studying the asymptotic behaviour of H±(±x) as x→ ∞. Proposition. Let a := EX = 0, 0 < σ2 := EX2 <∞. (7) Then lim s↓0 (∫ ∞ 0− e−sxdG+(x) + ln s ) = Q− 1 2 ln σ2 2 , (8) where Q = ∞∑ n=1 n−1 [ P(Sn ≥ 0)− 1 2 ] is the Spitzer series, and lim s↓0 ( − ∫ 0+ −∞ esxdG−(x) + ln s ) = −Q− 1 2 ln σ2 2 . (9) Hence, by using the refinement of Karamata’s Tauberian theorem given in [8], we immediately obtain 102 SERGEY V. NAGAEV Corollary 1. If conditions (7) hold, then lim x→±∞ ( G±(x) − ln |x| ) = C0 ±Q− 1 2 ln σ2 2 , (10) where C0 is the Euler constant. This result is obtained in [7] by using direct probabilistic arguments. Laplace trans- forms are used in [3,4], however, in the case where the distribution F is concentrated on a semiaxis or stable. In paper [5], the representation for ν([−x, x]) is obtained under the condition that EX = 0, E|X |3 < ∞, and some convolution of F (x) has an absolutely continuous component. In that representation, the Spitzer series Q is absent, which is quite explainable since, by (8) and (9),∫ ∞ −∞ e−h|x|ν(dx) = − ( ln s+ ln σ2 2 ) . Instead of Laplace transforms, the generalized Fourier transforms are used in [5]. Combining (5) and (8) and then (5) and (9), we obtain Corollary 2. If conditions (7) are fulfilled, then lim s↓0 s ∫ ∞ 0− e−sxdH+(x) = √ 2 σ eQ (11) and lim s↓0 s ∫ 0+ −∞ e−sxdH−(x) = − √ 2 σ e−Q. (12) The Karamata’s Tauberian theorem makes it possible to obtain the asymptotics of H+(x) for x→∞, namely, Corollary 3. If conditions (7) hold, then lim x→±∞ |x| −1H±(x) = √ 2 σ e±Q. (13) It is known that, under conditions (7) and (8), EZ+ < ∞, EZ− < −∞ (see [9]). Therefore, by the renewal theorem, lim |x|→∞ |x|−1H±(x) = 1 |m±| , (14) where m± = EZ±. Comparing (13) and (14), we conclude that m± = ± σ√ 2 e−Q. (15) We say that the distribution F has a long right tail if, for any l > 0, lim x→∞ F (x+ l)− F (x) F (x) = 0. (16) Respectively F has a long left tail if lim x→−∞ F (x+ l)− F (x) F (x) = 0. (17) PROBABILITIES OF LARGE DEVIATIONS OF LADDER HEIGHTS 103 Theorem 1. If conditions (7) and (17) hold, then, for x→ −∞, P(Z− < x) ∼ ω− ∫ x −∞ F (y)dy, (18) where ω− = √ 2 σ eQ. If conditions (7) and (16) hold, then, for x→∞, P(Z+ > x) ∼ ω+ ∫ ∞ x (1− F (y))dy, (19) where ω+ = √ 2 σ e−Q. Here and below, a(x) ∼ b(x) means that lim x→∞ a(x)/b(x) = 1. We use c, c(·), c(·, ·) to denote constants which may be different in different contexts. An asymptotics of large deviation probabilities is studied in [10–13]. The formula P(Z+ > x) ∼ 1 m− ∫ ∞ x (1 − F (y))dy (20) is obtained, in particular, in [13], and is valid if A := ∞∑ n=1 1 n P(Sn ≤ x) =∞, E|Z−| <∞. As we have already noticed above, E|Z−| <∞ under conditions (7) and (8). In addition, A = ∞ in this case. On the other hand, as it is shown above (see (15)), m− = σ√ 2 e−Q. The additional information which is contained in (19) as compared with (20) consists namely in this fact. Notice also that (19) is deduced by the quite different method by comparison with (20). Theorem 2. Let, for x→∞, F (−x) ∼ q xα , 1− F (x) ∼ p xα , (21) where 0 < α < 1, p ≥ 0, q ≥ 0. Then there exists the slowly varying function L(x), x > 0, such that, for x→ −∞, H−(x) ∼ |x|γL(|x|), (22) where γ = α 2 − c(α, β) π , c(α, β) = arctan ( β tan πα 2 ) , β = p− q p+ q . The analogous result takes place for H+(x). We see that the greatest value γ = α and the least γ = 0 are achieved, respectively, for β = −1 and for β = 1. These values of β correspond to the extreme types of stable laws with the exponent α which F is attracted to. If β = 0, then evidently γ = α 2 . Letting α = 2 in the last equality, we obtain the value 1 for γ. It is not improbable that the function L(x) in Theorem 2 is in fact constant. This is the case if the distribution F is concentrated on the negative semiaxis (see [3]). The analysis of the proof of Theorem 2 shows that L(x) = const in the case of symmetric F . Theorem 3. If conditions (16) and (21) are fulfilled, then there exists the slowly varying function l(x) such that, for x→∞, 1− F+(x) ∼ xγ−αl(x), (23) where γ = α 2 − c(α, β) π , c(α, β) = arctan ( β tan πα 2 ) . 104 SERGEY V. NAGAEV Notice that L(x) and l(x), generally speaking, do not satisfy the condition L(x) ∼ cl(x). However, if L(x) equals a constant, then it is true for l(x) as well. The result similar to Theorem 3 is also valid for F−(x). 2. Proof of Proposition Denote, by f(t), the characteristic function of the random variable X . The starting point is the next formula deduced in [14] (see also [15]) ∞∑ n=1 1 n ∫ ∞ 0+ e−hxdFn(x) + 1 2 ∞∑ n=1 (Fn(0+)− Fn(0))n−1 = −h π ∫ ∞ 0 ln |1− f(t)| h2 + t2 dt− 1 π ∫ ∞ 0 t arg (1− f(t)) h2 + t2 dt (24) with h π ∫ ∞ 0 ln |1− f(t)| h2 + t2 dt = −1 2 ∫ |x|�=0 e−h|x|dG(x), (25) 1 π ∫ ∞ 0 t arg (1− f(t)) h2 + t2 dt = 1 2 ∫ x<0 ehxdG(x) − 1 2 ∫ x>0 e−hxdG(x). (26) First, we show that the right-hand side of equality (26) goes to 1 2 ∞∑ n=1 (P(Sn < 0)−P(Sn > 0)) as h ↓ 0. We need several lemmas to proof it. Lemma 2.1. If a = 0 and σ2 <∞, then P(Sn > 0)−E { e−hSn ;Sn > 0 } ≤ hσ√n, (27) and P(Sn < 0)−E { ehSn ;Sn < 0 } < hσ √ n. (28) Proof. Applying the inequalities 1− e−x < x, x > 0, and E|Sn| ≤ σ√n, we have P(Sn > 0)−E { e−hSn ;Sn > 0 } = E { 1− e−hSn ;Sn > 0 } < hE { Sn;Sn > 0 } < hσ √ n. In just the same way, (28) is proved. � Put Δn(h) = ∫ x>0 e−hxdFn(x)− ∫ x<0 ehxdFn(x) + P(Sn < 0)−P(Sn > 0). (29) Lemma 2.2. Under conditions of Lemma 2.1, lim n→∞ sup h>0 ∣∣Δn(h) ∣∣ = 0. (30) Proof. Integrating by parts on the right-hand side of formula (29), we find that Δn(h) = h ∫ ∞ 0 (1 − Fn(x) − Fn(−x))e−hxdx. Consequently, |Δn(h)| < h sup x ∣∣∣1− Fn(x)− Fn(−x) ∣∣∣ ∫ ∞ 0 e−hxdx. By CLT, lim n→∞ ( 1− Fn(x)− Fn(−x) ) = 0. PROBABILITIES OF LARGE DEVIATIONS OF LADDER HEIGHTS 105 The assertion of the lemma follows from two previous relations. � Lemma 2.3. If the distribution F is not degenerate at zero, then E { e−hSn ;Sn > 0 } < c(F ) h √ n , (31) E { e−hSn ;Sn < 0 } < c(F ) h √ n . (32) Proof. We restrict ourselves to proving (31). Obviously,∫ ∞ 0+ e−hxdFn(x) < ∞∑ k=0 e−hkP(k < Sn ≤ k + 1). Since, by (4), P(k < Sn ≤ k + 1) < c(F )√ n ,∫ ∞ 0+ e−hxdFn(x) < c(F )√ n ∞∑ k=0 e−hk. On the other hand, ∞∑ k=0 e−hk = 1 1− e−h < 1 h . The desired result follows from these two inequalities. � Consider the series Σ(h) = ∞∑ n=1 n−1 ∣∣∣Δn(h) ∣∣∣ = ∑ n< ε2 h2 n−1 ∣∣∣Δn(h) ∣∣∣+ ∑ ε2 h2 <n≤ 1 h2ε2 n−1 ∣∣∣Δn(h) ∣∣∣+ n−1∑ n> 1 h2ε2 n−1 ∣∣∣Δn(h) ∣∣∣ =∑ 1 + ∑ 2 + ∑ 3 . (33) Applying Lemma 2.1, we find that∑ 1 < h ∑ n≤ ε2 h2 1√ n < 3ε. (34) Further, ∑ 2 < sup ε2 h2 <n≤ 1 h2ε2 Δn(h) ( h2 ε2 + ∫ 1 ε2h2 ε2 h2 dx x ) . Since ∫ 1 ε2h2 ε2 h2 dx x = −4 ln ε, we have, by (30), lim h↓0 ∑ 2 = 0. (35) Notice that, by Lemma 2.3, ∣∣∣Δn(h) ∣∣∣ < c(F ) h √ n + |2Fn(0)− 1|. 106 SERGEY V. NAGAEV Therefore, ∑ 3 < c(F ) h ∑ n> 1 ε2h2 1 n3/2 + ∑ n> 1 (εh)2 n−1|2Fn(0)− 1|. Further, ∑ n> 1 (εh)2 1 n3/2 < h3ε3 + ∫ ∞ u>(εh)−2 du u3/2 < h3ε3 + 2εh. The series ∞∑ 1 n−1(2Fn(0)− 1) absolutely converges (see [16]). Thus, lim h↓0 ∑ 3 < 2ε. (36) It follows from (33) - (36) that lim h↓0 Σ(h) < 5ε. It means by (29) that lim h↓0 (∫ x>0 e−hxdG(x) − ∫ x<0 ehxdG(x) ) = ∞∑ n=1 n−1 ( P(Sn > 0)−P(Sn < 0) ) . (37) Proceed now to the left-hand side of equality (25). Choose δ in the partition∫ ∞ 0 ln |1− f(t)| t2 + h2 dt = (∫ δ 0 + ∫ ∞ δ ) ln |1− f(t)| t2 + h2 dt = I1(h) + I2(h) (38) in such a way as the function f(t) �= 1 in the interval (0, δ). Put f1(t) = 2 1− f(t) σ2t2 . Then I1(h) = ∫ δ 0 ln f1(t) t2 + h2 dt + ln σ2 2 ∫ δ 0 dt t2 + h2 + 2 ∫ δ 0 ln t t2 + h2 dt = I11(h) + ln (σ2 2 ) I12(h) + I13(h). (39) Since f1(0) = 1, lim h↓0 hI11(h) = 0. (40) Further, lim h↓0 hI12(h) = π 2 . (41) It is easily seen that lim h↓0 h ∫ ∞ δ ln t t2 + h2 dt = 0. Consequently, for h ↓ 0, hI13(h) = h ∫ ∞ 0 ln t t2 + h2 dt+ o(1). On the other hand,∫ ∞ 0 ln t t2 + h2 dt = lnh h ∫ ∞ 0 dt 1 + t2 + 1 h ∫ ∞ 0 ln t 1 + t2 dt = π 2h lnh. (42) PROBABILITIES OF LARGE DEVIATIONS OF LADDER HEIGHTS 107 We use the equality ∫ ∞ 0 ln t 1 + t2 dt = 0 (see [17], p. 546, Section 4.231, formula 8). Thus, for h ↓ 0, hI13(h) = π 2 lnh+ o(1). (43) It follows from (39) - (43) that hI1(h) = π ( lnh+ 1 2 ln σ2 2 ) + o(1). (44) Estimate now I2(h). First of all,∣∣∣ln |1− f(t)| ∣∣∣ < ∞∑ n=1 n−1|fn(t)|. Consequently, I2(h) < ∞∑ n=1 1 n ∫ ∞ δ |f(t)|n t2 dt. The next bound ∫ |t−v|≤0.65�σ2(L) �β3(L) |fn(t)| dt ≤ c σ̃(L) √ n (45) holds, where σ̃2(L) = ∫ |x|≤L x 2 dF̃ (x), β̃3(L) = ∫ |x|≤L |x|3 dF̃ (x), c < 7.61579, F̃ (x) is the symmetrization of X (see [14], Lemma 2.1.) Splitting the interval (δ,∞) into intervals of the length 1.3σ2(L) β3(L) and applying bound (45), it is not hard to show that ∫ ∞ δ |f(t)|n t2 dt < c(F )√ n . Consequently, uniformly in h > 0, I2(h) < c(F ). (46) Returning now to (38) and taking (44) and (45) into account, we obtain that, for h ↓ 0 h π ∫ ∞ 0 ln |1− f(t)| t2 + h2 dt = lnh+ 1 2 ln σ2 2 + o(1). (47) Combining (24), (37), and (47), we arrive at formula (8). Formula (9) is deduced in the same way. � 3. Proof of Theorem 1 Without loss of generality, we may assume that F (x) is continuous. By (13) for any x < x(ε) < 0, −(1− ε) < x σ√ 2 H−(x) < (1 + ε)x. (48) Using (1), we have P(Z+x) = ∫ ∞ x−x(ε) H−(x− y)dF (y) + ∫ x−x(ε) x H−(x− y)dF (y) = I1(x) + I2(x). (49) Obviously, by (48), (1− ε) σ√ 2 ∫ ∞ x−x(ε) (y − x)dF (y) < I1(x) < σ√ 2 ∫ ∞ x−x(ε) (y − x)dF (y)(1 + ε). (50) 108 SERGEY V. NAGAEV Further, I2(x) < H−(x(ε))(F (x(ε)) − F (x)). Hence, by condition (16) for x→∞, I2(x) = o(1 − F (x)). (51) By the same reason for x→∞,∫ x−x(ε) x dF (y) = o(1− F (x)). (52) Put F (x) = ∫ ∞ x (y − x)dF (y) = ∫ ∞ x (1− F (y))dy. It is easily seen that 1− F (x) = o(F (x)). (53) It follows from (51) and (53) that I2(x) = o(F (x)), (54) and from (52), (53) ∫ ∞ x−x(ε) (y − x)dF (y) ∼ F (x). (55) By (50) and (55), σ√ 2 (1 − ε) ≤ lim x→∞ inf I1(x)/F (x) ≤ lim x→∞ sup I2(x)/F (x) ≤ σ√ 2 (1 + ε). (56) Combining (49), (54), and (56), we get the desired result. � 4. Proof of Theorem 2 Previously, we prove several lemmas. Lemma 4.1. Let, for x→∞, 1− F (x) ∼ c xα , 0 < α < 1. Then for t ↓ 0∫ ∞ 0 sin(tx)dF (x) ∼ ctα ∫ ∞ 0 cosx xα dx. (57) Proof. Clearly,∫ ∞ 0 sin(tx)dF (x) = ∫ M/t 0 sin(tx)dF (x) + ∫ ∞ M/t sin(tx)dF (x). (58) There exists the constant K such that 1− F (x) < Kc xα . (59) Therefore, for any M > 0,∣∣∣ ∫ ∞ M/t sin(tx)dF (x) ∣∣∣ < 1− F (M/t) < Kc Mα tα. (60) Integrating by parts, we have∫ M/t 0 sin(tx)dF (x) = t ∫ M/t 0 (1− F (x)) cos(tx)dx + (1− F (M/t)) sin(M/t). Hence, by (59),∣∣∣ ∫ M/t 0 sin(tx)dF (x) − t ∫ M/t 0 (1− F (x)) cos(tx)dx ∣∣∣ < Kc Mα tα. (61) PROBABILITIES OF LARGE DEVIATIONS OF LADDER HEIGHTS 109 Further,∫ M/t 0 (1 − F (x)) cos(tx)dx = ∫ M/t ε/t (1 − F (x)) cos(tx)dx + ∫ ε/t 0 (1− F (x)) cos(tx)dx. Hence, by (59),∣∣∣ ∫ ε/t 0 (1 − F (x)) cos(tx)dx ∣∣∣ < Kc ∫ ε/t 0 dx xα < Kcε1−α 1− α tα−1. For M and ε fixed,∫ M/t ε/t (1− F (x)) cos(tx)dx = c t1−α (∫ M ε cosx xα dx+ o(1) ) . As a result, we obtain, for every fixed M and ε,∫ M/t 0 (1−F (x)) cos(tx)dx = ctα−1 (∫ M ε cosx xα dx+o(1) ) +θ Kcε1−α 1− α tα−1, |θ| ≤ 1. (62) Returning now to (61), we conclude that∫ M/t 0 sin(tx)dF (x) = ctα (∫ M ε cosx xα dx+ o(1) ) + θKCtα ( M−α + ε1−α 1− α ) . (63) The desired result follows from (58), (60), and (62). � Lemma 4.2. Under conditions of Lemma 4.1 for t→ 0,∫ ∞ 0 (1 − cos(tx))dF (x) ∼ c|t|α ∫ ∞ 0 sinx xα dx. (64) Proof. Without loss of generality, we may assume t > 0. Obviously,∫ ∞ 0 (1− cos(tx))dF (x) = ∫ M/t 0 (1− cos(tx))dF (x) + ∫ ∞ M/t (1− cos(tx))dF (x). (65) By (59) for M > 0, 1− F (M/t) < Kc Mα tα. (66) Hence, ∫ ∞ M/t (1 − cos(tx))dF (x) < Kc Mα tα. (67) Integrating by parts, we have∫ M/t 0 (1− cos(tx))dF (x) = t ∫ M/t 0 (1 − F (x)) sin(tx)dx + (1− F (M/t))(1 − cosM). Hence, by (67),∣∣∣ ∫ M/t 0 (1 − cos(tx))dF (x) − t ∫ M/t 0 (1− F (x)) sin(tx)dx ∣∣∣ < Kc Mα tα. (68) Further,∫ M/t 0 (1− F (x)) sin(tx)dx = ∫ ε/t 0 (1 − F (x)) sin(tx)dx + ∫ M/t ε/t (1− F (x)) sin(tx)dx. 110 SERGEY V. NAGAEV By (59), ∣∣∣∫ ε/t 0 (1− F (x)) sin(tx)dx ∣∣∣ < Kcε1−α 1− α tα−1. For M and ε fixed,∫ M/t ε/t (1− F (x)) sin(tx)dx = c t1−α (∫ M ε sinx xα dx+ o(1) ) . It follows from two last formulas that t ∫ M/t 0 (1− F (x)) sin(tx)dx = ctα (∫ M ε sinx xα dx+ o(1) ) + θ Kctα 1− αε 1−α, |θ| ≤ 1. (69) Combining (65)–(69), we obtain the assertion of Lemma 4.2. � Lemma 4.3. For any 0 < α < 1,∫ ∞ 0 eix xα dx = i1−αΓ(1− α). (70) Proof. By changing the contour of integration in accordance with the change of the variable x = iy, we find∫ ∞ 0 eix xα dx = i1−α ∫ ∞ 0 e−yy−αdy = i1−αΓ(1− α). � Lemma 4.4. Let F (x) satisfy conditions (21). Then, for t→ 0, 1− ∫ ∞ −∞ eitxdF (x) ∼ |t|α ( (p+ q) cos πα 2 + i(q − p) t|t| sin πα 2 ) Γ(1− α). (71) Proof. Obviously, 1− ∫ ∞ −∞ eitxdF (x) = ∫ ∞ −∞ (1− cos(tx))dF (x) − i ∫ ∞ −∞ sin(tx)dF (x). By (57) and (64) for t→ 0,∫ ∞ −∞ (1 − cos(tx))dF (x) ∼ (p+ q)|t|α ∫ ∞ 0 sinx xα dx and ∫ ∞ −∞ sin(tx)dF (x) ∼ (p− q)|t|α t |t| ∫ ∞ 0 cosx xα dx. Thus, 1− ∫ ∞ −∞ eitxdF (x) ∼ |t|α ( (p+ q) ∫ ∞ 0 sinx xα dx + i(q − p) t|t| ∫ ∞ 0 cosx xα dx ) . According to Lemma 4.3,∫ ∞ 0 cosx xα dx = Γ(1− α)Re i1−α = sin πα 2 ,∫ ∞ 0 sinx xα dx = Γ(1− α)Im i1−α = cos πα 2 . Substituting these values into the previous equality, we obtain the desired result. � Considering as before that F (x) satisfies conditions (21),we study the Laplace trans- form of the projection of the harmonic renewal measure (3) on the semiaxis (−∞, 0]. PROBABILITIES OF LARGE DEVIATIONS OF LADDER HEIGHTS 111 It is easily seen that ∫ 0− −∞ ehxdG−(x) = − ∫ ∞ 0+ e−hxdG−(−x). (72) Hence, ∫ 0+ −∞ e−hxdG−(x) = ∫ ∞ 0+ ehxdG−(x) + ν({0}). (73) By (24),∫ ∞ 0+ e−hxdG−(−x) + 1 2 ν({0}) = −h π ∫ ∞ 0 ln |1− f(−t)| h2 + t2 dt− 1 π ∫ ∞ 0 t arg (1 − f(−t)) h2 + t2 dt. (74) Using Lemma 4.4, it is easy to verify that, for t→ 0, |1− f(−t)| = |t|αΓ(1− α)(p2 + q2 − 2pq sinπα)1/2 + o(1). Consequently, for t→ 0, ln |1− f(−t)| = α ln |t|+ c(α, p, q) + o(1), where c(α, p, q) = 1 2 ln(p2 + q2 − 2pq sinπα) + ln Γ(1− α). Hence, by (42), h π ∫ ∞ 0 ln |1− f(−t)| h2 + t2 dt = αh π ∫ ∞ 0 ln t h2 + t2 dt+ ψ(h) = α 2 lnh+ ψ(h), (75) where limh→0 ψ(h) = c(α, p, q). According to (71), arg (1 − f(−t)) = t |t| arctan ( β tg πα 2 ) + ϕ(t), (76) where β = (p− q)/(p+ q), ϕ(t)→ 0 as t→ 0. Lemma 4.5. The function W (h) = exp {∫ 1 h tϕ(t) h2 + t2 dt } , (77) where ϕ(t)→ 0 for t→ 0, is slowly varying as h ↓ 0, i.e. for any c > 0, lim h↓0 W (ch) W (h) = 1. Proof. Changing the variable t = u−1, we have I(h) := ∫ 1 h tϕ(t) h2 + t2 dt = ∫ 1/h 1 ϕ(1/u) 1 + h2u2 du u = ∫ 1/h 1 ϕ(1/u) u du − h2 ∫ 1/h 1 uϕ(1/u) 1 + h2u2 du = I1(h) + I2(h). It is easily seen that∣∣∣ ∫ 1/h 1 uϕ(1/u) 1 + h2u2 du ∣∣∣ < ∫ 1/h 1 |ϕ(1/u)|udu = o(h−2). Consequently, lim h→0 I2(h) = 0. 112 SERGEY V. NAGAEV According to Karamata’s criterion, the function Z(x) = exp {∫ x 1 ϕ(1/u) u du } is slowly varying as x→∞. Hence, Z(1/h) is slowly varying as h ↓ 0. Since W (h) = Z(1/h) exp{I2(h)}, the function W (h) has the same property as well. � Lemma 4.6. Let a function ϕ(t) be continuous, and ϕ(0) = 0. Then lim h↓0 ∫ h −h tϕ(t) t2 + h2 dt = 0. Proof. The conclusion of the lemma follows from the inequalities∣∣∣ ∫ h −h tϕ(t) t2 + h2 dt ∣∣∣ < 2 sup |t|≤h |ϕ(t)| ∫ h 0 t t2 + h2 dt < sup |t|≤h |ϕ(t)|. � Lemma 4.7. For h ↓ 0,∫ 1 0 t arg(1− f(−t)) t2 + h2 dt = c(α, β) ln 1 h + lnW (h) + o(1) (78) Proof. Based on formula (76) and Lemmas 4.5 and 4.6, we can state that∫ 1 0 t arg (1 − f(−t)) t2 + h2 dt = c(α, β) ∫ 1 0 t t2 + h2 dt+ lnW (h) + o(1), where c(α, β) = arctan (β tan πα 2 ). Obviously, ∫ 1 0 t t2 + h2 dt = 1 2 ln(t2 + h2) ∣∣∣1 0 = ln 1 h + 1 2 ln(1 + h2). The conclusion of the lemma follows from last two formulas. � Consider the integral I(h) = 1 π ∫ ∞ 1 t arg (1− f(−t)) t2 + h2 dt. By (24) and (78), I(h) <∞ for h > 0. Lemma 4.8. For every distribution F , there exists the finite limit lim h→0 I(h) = I0. Proof. Arguing in the same way as in deducing formula (2.20) in [14], we are sure that 1 π ∫ ∞ 1 t(1− arg f(−t)) t2 + h2 dt = − ∞∑ n−1 n−1 ∫ ∞ 1 tImfn(−t) t2 + h2 dt. (79) Evidently, ∫ ∞ 1 tImfn(−t) t2 + h2 dt = − ∫ ∞ 1 tImfn(t) t2 + h2 dt. Further,∫ ∞ 1 tImfn(t) t2 + h2 dt = ∫ ∞ 1 t t2 + h2 ∫ ∞ −∞ sin(tx)dFn(x) = ∫ ∞ −∞ dFn(x) ∫ ∞ 1 t sin (tx) t2 + h2 dx. PROBABILITIES OF LARGE DEVIATIONS OF LADDER HEIGHTS 113 By Lemma 1.1 in [14] for α < h ≤ 1, ∣∣∣∫ ∞ 1 t sin (tx) t2 + h2 dt ∣∣∣ < { 2/|x|, |x| ≥ 1, 3, |x| < 1. It follows from last two relations that∣∣∣∫ ∞ 1 tIm fn(t) t2 + h2 dt ∣∣∣ < 3 ∫ |x|<n1/4 dFn(x) + 2 ∫ |x|>n1/4 dFn(x) |x| < c(F ) n1/4 . We have applied here a bound for the concentration function (4). Thus, series (79) converges uniformly in the interval (0, 1]. On the other hand, for any n, lim h↓0 ∫ ∞ 1 tIm fn(t) t2 + h2 dt = ∫ ∞ 1 Im fn(t) t dt. Consequently, lim h↓0 I(h) = 1 π ∞∑ n=1 ∫ ∞ 1 Im fn(t) t dt = I0. � It follows from Lemmas 4.7 and 4.8 that, for h ↓ 0,∫ ∞ 0 t arg (1− f(−t)) t2 + h2 dt = c(α, β) ln 1 h + lnW (h) + I0 + o(1). (80) Combining (72)–(75) and (80), we conclude that − ∫ 0+ −∞ ehxdG−(x) = ∫ ∞ 0+ e−hxdG−(−x) + ν({0}) = γ ln 1 h − π−1(lnW (h) + c0(F )) + 1 2 ν({0}), where c0(F ) = c(α, p, q) + I0, γ = α 2 − c(α, β) π . Applying now the Baxter identity (see, e.g., [1], Ch. 18, § 3), we find that∫ ∞ 0+ e−hxdH−(−x) = exp {∫ ∞ 0+ e−hxdG−(−x) } ∼ h−γ exp { − 1 π ( lnW (h) + c0(F ) ) + 1 2 ν({0}) } . (81) Using the Tauberian theorem for the Laplace transform (see [1], Ch. 13, § 5), we have H−(−x) ∼ xγL(x), (82) where L(x) = 1 Γ(1 + γ) exp { − 1 π ( lnW (x−1) + c0(F ) ) + 1 2 ν({0}) } , x > 0, which is equivalent to the assertion of the theorem. � 114 SERGEY V. NAGAEV 5. Proof of Theorem 3 Using formula (1), we have P(Z+ > x) ∼ p ∫ 0+ −∞ (x− y)−αdH−(y) = pα ∫ 0 −∞ (x− y)−α−1H−(y)dy. (83) We need several lemmas to find the asymptotics of the last integral. Lemma 5.1. For any x > 0,∫ 0 −√ x (x− y)−α−1H−(y)dy < c(ε)xγ/2−α+ε, (84) where ε is as small as one likes. Proof. By (82), there exists a constant c such that, for every x > 0,∫ 0 −√ x (x− y)−α−1H−(y)dy < c ∫ 0 −√ x (x+ |y|)−α−1|y|γL(|y|)dy = cxγ−α ∫ 0 − 1√ x (1 + |y|)−α−1|y|γL(x|y|)dy < c(ε)xγ/2−α+ε/2 (85) since L(x|y|) < c(ε)(x|y|)ε. � Lemma 5.2. As x→∞, Γ(1 + γ) ∫ −√ x −∞ (x − y)−α−1H−(y)dy ∼ xγ−α−1 ∫ −√ x −∞ (1 + |y|)−α−1|y|γL(x|y|)dy. (86) Proof. The assertion of the lemma follows from asymptotics (82). � Lemma 5.3. As x→∞, Γ(1 + γ) ∫ 0 −∞ (x− y)−α−1H−(y)dy ∼ xγ−α ∫ 0 −∞ (1 + |y|)−α−1|y|γL(|y|)dy. (87) Proof. Obviously,∫ −√ x −∞ (x− y)−α−1H−(y)dy = ∫ −√ x −∞ + ∫ 0 −√ x = I1 + I2. By Lemma 5.2, I1 > c(ε)xγ−α−ε. (88) It follows from (84) and (88) that I2 = o(I1). Hence, by (86), ∫ 0 −∞ (x − y)−α−1H−(y)dy ∼ ∫ −√ x −∞ (x− y)−α−1H−(y)dy ∼ xγ−α Γ(1 + γ) ∫ −√ x −∞ (1 + |y|)−α−1|y|γL(x|y|)dy. It remains to remark that, by (85) and (87),∫ −√ x −∞ (1 + |y|)−α−1|y|γL(x|y|)dy ∼ ∫ 0 −∞ (1 + |y|)−α−1|y|γL(x|y|)dy. � (89) PROBABILITIES OF LARGE DEVIATIONS OF LADDER HEIGHTS 115 Lemma 5.4. The function h(x) := ∫ 0 −∞ (1 + |y|)−α−1|y|γL(x|y|)dy (90) is slowly varying. Proof. By (89) for x→∞, h(cx) h(x) ∼ ∫ −√ x −∞ (1 + |y|)−α−1|y|γL(cx|y|)dy∫ −√ x −∞ (1 + |y|)−α−1|y|γL(x|y|)dy ∼ 1. � It follows from Lemmas 5.3 and 5.4 that∫ 0 −∞ (x− y)−α−1H−(y)dy ∼ xγ−α Γ(1 + γ) h(x), (91) where h(x) is a slowly varying function. Comparing (83) and (91), we find that P(Z+ > x) ∼ pα Γ(1 + γ) xγh(x). Hence, letting l(x) = pα Γ(1 + γ) , we obtain the conclusion of Theorem 3. � In conclusion, we remark that if the integral∫ 1 0 ϕ(t) t dt, where ϕ is defined by (76), is finite, then W (h) in (77) converges to some constant as h ↓ 0. Indeed, in this case for any η > h,∣∣∣∣∣ ∫ η h tϕ(t) t2 + h2 dt− ∫ η h ϕ(t) t dt ∣∣∣∣∣ ≤ sup 0≤t≤η |ϕ(t)| ∫ η h ( t t2 + h2 − 1 t ) dt. Obviously, ∫ η h ( t t2 + h2 − 1 t ) dt < h2 ∫ η h dt t3 < 1 3 . On the other hand, for every 0 < η < 1, lim h↓0 ∫ 1 η tϕ(t) t2 + h2 dt = ∫ 1 η ϕ(t) t dt. Thus, lim h↓0 ∫ 1 h tϕ(t) t2 + h2 dt = ∫ 1 0 ϕ(t) t dt. Further, if there exists the finite limit lim h↓0 W (h), then the same is true for L(x) in (82) as x→∞. 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