LARGE DEVIATIONS FOR SOLUTIONS TO STOCHASTIC RECURRENCE EQUATIONS UNDER KESTEN S CONDITION D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J.

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1 LARGE DEVIATIONS FOR SOLUTIONS TO STOCHASTIC RECURRENCE EQUATIONS UNDER KESTEN S CONDITION D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ Abstract. In this paper we prove large deviations results for partial sums constructed from the solution to a stochastic recurrence equation. We assume Kesten s condition [17] under which the solution of the stochastic recurrence equation has a marginal distribution with power law tails, while the noise sequence of the equations can have light tails. The results of the paper are analogs of those obtained by A.V. and S.V. Nagaev [21, 22] in the case of partial sums of iid random variables. In the latter case, the large deviation probabilities of the partial sums are essentially determined by the largest step size of the partial sum. For the solution to a stochastic recurrence equation, the magnitude of the large deviation probabilities is again given by the tail of the maximum summand, but the exact asymptotic tail behavior is also influenced by clusters of extreme values, due to dependencies in the sequence. We apply the large deviation results to study the asymptotic behavior of the ruin probabilities in the model. (1.1) 1. Introduction Through the last 40 years, the stochastic recurrence equation Y n = A n Y n 1 + B n, n Z, and its stationary solution have attracted much attention. Here (A i, B i ), i Z, is an iid sequence, A i > 0 a.s. and B i assumes real values. (In what follows, we write A, B, Y,..., for generic elements of the strictly stationary sequences (A i ), (B i ), (Y i ),..., and we also write c for any positive constant whose value is not of interest.) It is well known that if E log A < 0 and E log + B <, there exists a unique strictly stationary ergodic solution (Y i ) to the stochastic recurrence equation (1.1) with representation n Y n = A i+1 A n B i, n Z, i= where, as usual, we interpret the summand for i = n as B n. One of the most interesting results for the stationary solution (Y i ) to the stochastic recurrence equation (1.1) was discovered by Kesten [17]. He proved under general conditions that the marginal distributions of (Y i ) have power law tails. For later use, we formulate a version of this result due to Goldie [10]. Theorem 1.1. (Kesten [17], Goldie [10]) Assume that the following conditions hold: 2000 Mathematics Subject Classification. Primary 60F10; secondary 91B30, 60G70. Key words and phrases. Stochastic recurrence equation, large deviations, ruin probability. D. Buraczewski and E. Damek were partially supported by the NCN grant N N Thomas Mikosch s research is partly supported by the Danish Natural Science Research Council (FNU) Grants Point process modelling and statistical inference and Heavy tail phenomena: Modeling and estimation. J. Zienkiewicz was supported by the MNiSW grant N N D. Buraczewski was also supported by the European Commission via IEF Project (contract number PIEF-GA SCHREC). 1

2 2 D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ (1.2) There exists α > 0 such that EA α = 1. ρ = E(A α log A) and E B α are both finite. The law of log A is non-arithmetic. For every x, PAx + B = x < 1. Then Y is regularly varying with index α > 0. In particular, there exist constants c +, c 0 such that c + + c > 0 and (1.3) PY > x c + x α, and PY x c x α as x. Moreover, if B 1 a.s. then the constant c + takes on the form c := E[(1 + Y ) α Y α ]/(αρ), Goldie [10] also showed that similar results remain valid for the stationary solution to stochastic recurrence equations of the type Y n = f(y n 1, A n, B n ) for suitable functions f satisfying some contraction condition. The power law tails (1.3) stimulated research on the extremes of the sequence (Y i ). Indeed, if (Y i ) were iid with tail (1.3) and c + > 0, then the maximum sequence M n = max(y 1,..., Y n ) would satisfy the limit relation (1.4) lim n P(c+ n) 1/α M n x = e x α = Φ α (x), x > 0, where Φ α denotes the Fréchet distribution, i.e. one of the classical extreme value distributions; see Gnedenko [11]; cf. Embrechts et al. [6], Chapter 3. However, the stationary solution (Y i ) to (1.1) is not iid and therefore one needs to modify (1.4) as follows: the limit has to be replaced by Φ θ α for some constant θ (0, 1), the so-called extremal index of the sequence (Y i ); see de Haan et al. [12]; cf. [6], Section 8.4. The main objective of this paper is to derive another result which is a consequence of the power law tails of the marginal distribution of the sequence (Y i ): we will prove large deviation results for the partial sum sequence S n = Y Y n, n 1, S 0 = 0. This means we will derive exact asymptotic results for the left and right tails of the partial sums S n. Since we want to compare these results with those for an iid sequence we recall the corresponding classical results due to A.V. and S.V. Nagaev [21, 22] and Cline and Hsing [3]. Theorem 1.2. Assume that (Y i ) is an iid sequence with a regularly varying distribution, i.e. there exists an α > 0, constants p, q 0 with p + q = 1 and a slowly varying function L such that (1.5) PY > x p L(x) x α and PY x q L(x) x α as x. Then the following relations hold for α > 1 and suitable sequences b n : (1.6) lim sup PS n ES n > x n p n P Y > x = 0 and (1.7) x b n lim sup n x b n PS n ES n x n P Y > x q = 0. If α > 2 one can choose b n = an log n, where a > α 2, and for α (1, 2], b n = n δ+1/α for any δ > 0.

3 LARGE DEVIATIONS 3 For α (0, 1], (1.6) and (1.7) remain valid if the centering ES n is replaced by 0 and b n = n δ+1/α for any δ > 0. For α (0, 2] one can choose a smaller bound b n if one knows the slowly varying function L appearing in (1.5). A functional version of Theorem 1.2 with multivariate regularly varying summands was proved in Hult et al. [13] and the results were used to prove asymptotic results about multivariate ruin probabilities. Large deviation results for iid heavy-tailed summands are also known when the distribution of the summands is subexponential, including the case of regularly varying tails; see the recent paper by Denisov et al. [5] and the references therein. In this case, the regions where the large deviations hold very much depend on the decay rate of the tails of the summands. For semi-exponential tails (such as for the log-normal and the heavy-tailed Weibull distributions) the large deviation regions (b n, ) are much smaller than those for summands with regularly varying tails. In particular, x = n is not necessarily contained in (b n, ). The aim of this paper is to study large deviation probabilities for a particular dependent sequence (Y n ) as described in Kesten s Theorem 1.1. For dependent sequences (Y n ) much less is known about the large deviation probabilities for the partial sum process (S n ). Gantert [8] proved large deviation results of logarithmic type for mixing subexponential random variables. Davis and Hsing [4] and Jakubowski [14, 15] proved large deviation results of the following type: there exist sequences s n such that PS n > a n s n n PY > a n s n c α for suitable positive constants c α under the assumptions that Y is regularly varying with index α (0, 2), n P ( Y > a n ) 1 and (Y n ) satisfies some mixing conditions. Both Davis and Hsing [4] and Jakubowski [14, 15] could not specify the rate at which the sequence (s n ) grows to infinity, and an extension to α > 2 was not possible. These facts limit the applicability of these results, for example for deriving the asymptotics of ruin probabilities for the random walk (S n ). Large deviations results for particular stationary sequences (Y n ) with regularly varying finite-dimensional distributions were proved in Mikosch and Samorodnitsky [19] in the case of linear processes with iid regularly varying noise and in Konstantinides and Mikosch [18] for solutions (Y n ) to the stochastic recurrence equation (1.1), where B is regularly varying with index α > 1 and EA α < 1. This means that Kesten s condition (1.2) is not satisfied in this case and the regular variation of (Y n ) is due to the regular variation of B. For these processes, large deviation results and ruin bounds are easier to derive by applying the heavy-tail large deviation heuristics : a large value of S n happens in the most likely way, namely it is due to one very large value in the underlying regularly varying noise sequence, and the particular dependence structure of the sequence (Y n ) determines the clustering behavior of the large values of S n. This intuition fails when one deals with the partial sums S n under the conditions of Kesten s Theorem 1.1: here a large value of S n is not due to a single large value of the B n s or A n s but to large values of the products A 1 A n. The paper is organized as follows. In Section 2 we prove an analog to Theorem 1.2 for the partial sum sequence (S n ) constructed from the solution to the stochastic recurrence equation (1.1) under the conditions of Kesten s Theorem 1.1. The proof of this result is rather technical: it is given in Section 3 where we split the proof into a series of auxiliary results. There we treat the different cases α 1, α (1, 2] and α > 2 by different tools and methods. In particular, we will use exponential tail inequalities which are suited for the three distinct situations. In contrast to the iid situation described in Theorem 1.2, we will show that the x-region where the large deviations hold cannot be chosen as an infinite interval (b n, ) for a suitable lower bound b n, but one also needs upper bounds c n b n. In Section 4 we apply the large deviation results to get precise asymptotic bounds for the ruin probability related to the random walk (S n ). This ruin bound is an analog of

4 4 D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ the celebrated result by Embrechts and Veraverbeke [7] in the case of a random walk with iid step sizes. 2. Main result The following is the main result of this paper. It is an analog of the well known large deviation result of Theorem 1.2. Theorem 2.1. Assume that the conditions of Theorem 1.1 are satisfied and additionally there exists ε > 0 such that EA α+ε and E B α+ε are finite. Then the following relations hold: (1) For α (0, 2], M > 2, (2.1) sup n sup n 1/α (log n) M x PS n d n > x n P Y > x <, If additionally e sn n 1/α (log n) M and lim n s n /n = 0 then (2.2) lim sup PS n d n > x n n P Y > x c+ c c + + c = 0, n 1/α (log n) M xe sn where d n = 0 or d n = ES n according as α (0, 1] or α (1, 2]. (2) For α > 2 and any c n, (2.3) sup n sup c n n 0.5 log nx If additionally c n n 0.5 log n e sn (2.4) lim sup n c n n 0.5 log nxe sn PS n ES n > x n P Y > x <. and lim n s n /n = 0 then PS n ES n > x c+ c n P Y > x c + + c = 0. Clearly, if we exchange the variables B n by B n in the above results we obtain the corresponding asymptotics for the left tail of S n. For example, for α > 1 the following relation holds uniformly for the x-regions indicated above: PS n ney x lim = c c n n P Y > x c + + c. Remark 2.2. The deviations of Theorem 2.1 from the iid case (see Theorem 1.2) are two-fold. First, the extremal clustering in the sequence (Y n ) manifests in the presence of the additional constants c and c ±. Second, the precise large deviation bounds (2.2) and (2.4) are proved for x-regions bounded from above by a sequence e sn for some s n with s n /n 0. Mikosch and Wintenberger [20] extended Theorem 2.1 to more general classes of stationary sequences (Y t ). In particular, they proved similar results for stationary Markov chains with regularly varying finitedimensional distributions, satisfying a drift condition. The solution (Y t ) to (1.1) is a special case of this setting if the distributions of A, B satisfy some additional conditions. Mikosch and Wintenberger [20] use a regeneration argument to explain that the large deviation results do not hold uniformly in the unbounded x-regions (b n, ) for suitable sequences (b n ), b n. 3. Proof of the main result 3.1. Basic decompositions. In what follows, it will be convenient to use the following notation Ai A Π ij = j i j and Π 1 otherwise j = Π 1j,

5 LARGE DEVIATIONS 5 and Ỹ i = Π 2i B 1 + Π 3i B Π ii B i 1 + B i, i 1. Since Y i = Π i Y 0 + Ỹi the following decomposition is straightforward: n n (3.1) S n = Y 0 Π i + Ỹ i =: Y 0 η n + S n, where (3.2) Sn = Ỹ1 + + Ỹn and η n = Π Π n, n 1. In view of (3.1) and Lemma 3.1 below it suffices to bound the ratios P S n d n > x n P Y > x uniformly for the considered x-regions, where d n = E S n for α > 1 and d n = 0 for α 1. The proof of the following bound is given at the end of this subsection. Lemma 3.1. Let (s n ) be a sequence such that s n /n 0. Then for any sequence (b n ) with b n the following relations hold: lim n sup P Y 0 η n > x b nxe sn n P Y > x = 0 and lim sup n sup b nx P Y 0 η n > x n P Y > x <, Before we further decompose S n we introduce some notation to be used throughout the proof: for any x in the considered large deviation regions, (3.3) m = [(log x) 0.5+σ ] for some positive number σ < 1/4, where [ ] denotes the integer part. n 0 = [ρ 1 log x], where ρ = E(A α log A). n 1 = n 0 m and n 2 = n 0 + m For α > 1, let D be the smallest integer such that D log EA > α 1. Notice that the latter inequality makes sense since EA < 1 due to (1.2) and the convexity of the function ψ(h) = EA h, h > 0. For α 1, fix some β < α and let D be the smallest integer such that D log EA β > α β where, by the same remark as above, EA β < 1. Let n 3 be the smallest integer satisfying D log x n 3, x > 1. Notice that since the function Ψ(h) = log ψ(h) is convex, putting β = 1 if α > 1, by the choice of D we have 1 D < Ψ(α) Ψ(β) α β < Ψ (α) = ρ, therefore n 2 < n 3 if x is sufficiently large. For fixed n, we change the indices i j = n i + 1 and, abusing notation and suppressing the dependence on n, we reuse the notation Ỹ j = B j + Π jj B j Π j,n 1 B n. Writing n 4 = min(j + n 3, n), we further decompose Ỹj: (3.4) Ỹ j = Ũj + W j = B j + Π jj B j Π j,n4 1B n4 + W j. Clearly, W j vanishes if j n n 3 and therefore the following lemma is nontrivial only for n > n 3. The proof is given at the end of this subsection.

6 6 D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ Lemma 3.2. For any small δ > 0, there exists a constant c > 0 such that n (3.5) P ( W j c j ) > x c n x α δ, x > 1, j=1 where c j = 0 or c j = E W j according as α 1 or α > 1. By virtue of (3.5) and (3.4) it suffices to study the probabilities P a j = 0 for α 1 and a j = EŨj for α > 1. We further decompose Ũi into (3.6) Ũ i = X i + S i + Z i, where for i n n 3, (3.7) n X i = B i + Π ii B i Π i,i+n1 2B i+n1 1, S i = Π i,i+n1 1B i+n1 + + Π i,i+n2 1B i+n2, Z i = Π i,i+n2 B i+n Π i,i+n3 1B i+n3. j=1 (Ũj a j ) > x, where For i > n n 3, define X i, S i, Z i as follows: For n 2 < n i < n 3 choose X i, S i as above and Z i = Π i,i+n2 B i+n Π i,n 1 B n. For n 1 n i n 2, choose Z i = 0, Xi as before and Finally, for n i < n 1, define S i = 0, Z i = 0 and S i = Π i,i+n1 1B i+n1 + + Π i,n 1 B n. X i = B i + Π ii B i Π i,n 1 B n. Let p 1, p, p 3 be the largest integers such that p 1 n 1 n n 1 + 1, pn 1 n n 2 and p 3 n 1 n n 3, respectively. We study the asymptotic tail behavior of the corresponding block sums given by (3.8) X j = jn 1 i=(j 1)n 1+1 X i, S j = jn 1 i=(j 1)n 1+1 where j is less or equal p 1, p, p 3 respectively. The remaining steps of the proof are organized as follows. S i, Z j = jn 1 i=(j 1)n 1+1 Section 3.2. We show that the X j s and Z j s do not contribute to the considered large deviation probabilities. This is the content of Lemmas 3.4 and 3.5. Section 3.3. We provide bounds for the tail probabilities of S j ; see Proposition 3.6 and Lemma 3.8. These bounds are the main ingredients in the proof of the large deviation result. Section 3.4. In Proposition 3.9 we combine the bounds provided in the previous subsections. Section 3.5: We apply Proposition 3.9 to prove the main result. Proof of Lemma 3.1. The infinite series η = i=0 Π i has the distribution of the stationary solution to the stochastic recurrence equation (1.1) with B 1 a.s. and therefore, by Theorem 1.1, P (η > x) c x α, x. It follows from a slight modification of Jessen and Mikosch [16], Lemma 4.1(4), and the independence of Y 0 and η that (3.9) P Y 0 η > x c x α log x, x. Z i,

7 LARGE DEVIATIONS 7 Since s n /n 0 as n we have P Y 0 η n > x sup b nxe sn n P Y > x sup P Y 0 η > x b nxe sn n P Y > x 0. There exist c 0, x 0 > 0 such that P Y 0 > y c 0 y α for y > x 0. Therefore P Y 0 η n > x Px/η n x 0 + c 0 x α Eη α n1 x/ηn>x 0 cx α Eη α n. By Bartkiewicz et al. [1], Eη α n cn. Hence This concludes the proof. P Y 0 η n > x I n = sup b nx n P Y > x sup cx α Eηn α b nx n P Y > x <. Proof of Lemma 3.2. Assume first that α > 1. Since E W j is finite, D log EA > α 1 and D log x n 3, we have for some positive δ (3.10) E W j (EA)n3 1 EA E B c ed log x log EA c x (α 1) δ, and hence by Markov s inequality n n P ( W j E W j ) > x 2 x 1 E W j c n x α δ. j=1 If β < α 1 an application of Markov s inequality yields for some positive δ, j=1 j=1 j=1 n P W n j > x x β E W j β x β ne B β (EA β ) n3 (1 EA β ) cx β ne D log x log EAβ c n x α δ. In the last step we used the fact that D log EA β > α β. This concludes the proof of the lemma Bounds for PX j > x and PZ j > x. We will now study the tail behavior of the single block sums X 1, Z 1 defined in (3.8). We start with a useful auxiliary result. Lemma 3.3. Assume ψ(α+ɛ) = EA α+ɛ < for some ɛ > 0. Then there is a constant C = C(ɛ) > 0 such that ψ(α + γ) C e ργ for γ ɛ/2, where ρ = E(A α log A). Proof. By a Taylor expansion and since ψ(α) = 1, ψ (α) = ρ, we have for some θ (0, 1), (3.11) ψ(α + γ) = 1 + ργ + 0.5ψ (α + θγ)γ 2. If θγ < ɛ/2 then, by assumption, ψ (α + θγ) = EA α+θγ (log A) 2 is bounded by a constant c > 0. Therefore, ψ(α + γ) 1 + ργ + cγ 2 = e log(1+ργ+c γ2) C e ργ. The following lemma ensures that the X i s do not contribute to the considered large deviation probabilities.

8 8 D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ Lemma 3.4. There exist positive constants C 1, C 2, C 3 such that where PX 1 > x PX 1 > x C 1 x α e C2 (log x)c 3, x > 1, n 1 X 1 = ( B i + Π ii B i Π i,i+n1 2 B i+n1 1 ). Proof. We have X 1 = n 0 k=m+1 R k, where for m < k n 0, R k = Π 1,n0 k B n0 k Π i,i+n0 k 1 B i+n0 k + + Π n1,n 1+n 0 k 1 B n1+n 0 k. Notice that for x sufficiently large, n 0 k=m+1 R k > x n 0 k=m+1 R k > x/k 3. Indeed, on the set R k x/k 3, m < k n 0 we have for some c > 0 and sufficiently large x, by the definition of m = [(log x) 0.5+σ ], n 0 k=m+1 R k x m + 1 We conclude that, with I k = PR k > x/k 3, n 0 P k=m+1 k=1 1 k 2 c x (log x) 0.5+σ < x. R k > x n 0 k=m+1 Next we study the probabilities I k. Let δ = (log x) 0.5. By Markov s inequality, I k (x/k 3 ) (α+δ) ER α+δ k By Lemma 3.3 and the definition of n 0 = [ρ 1 log x], I k. (x/k 3 ) (α+δ) n α+δ 0 (EA α+δ ) n0 k E B α+δ. I k c (x/k 3 ) (α+δ) n α+δ 0 e (n0 k)ρδ c x α k 3(α+δ) n α+δ 0 e kρδ. Since k (log x) 0.5+σ m there are positive constants ζ 1, ζ 2 such that kδ k ζ1 (log x) ζ2 and therefore for sufficiently large x and appropriate positive constants C 1, C 2, C 3, n 0 k=m+1 This finishes the proof. I k c x α n α+δ 0 n 1 k=m+1 e ρ kζ 1 (log x) ζ 2 k 3(α+δ) C 1 x α e C2 (log x)c 3. The following lemma ensures that the Z i s do not contribute to the considered large deviation probabilities. Lemma 3.5. There exist positive constants C 4, C 5, C 6 such that where PZ 1 > x PZ 1 > x C 4 x α e C5 (log x)c 6, x > 1, n 1 Z 1 = (Π i,i+n2 B i+n Π i,i+n3 1 B i+n3 ).

9 LARGE DEVIATIONS 9 Proof. We have Z 1 = n 3 n 2 k=1 R k, where R k = Π 1,n2+k B n2+k Π i,i+n2+k 1 B i+n2+k + + Π n1,n 1+n 2+k 1 B n1+n 2+k. As in the proof of Lemma 3.4 we notice that, with J k = P R k > x/(n 2 + k) 3, for x sufficiently large n 3 n 2 P k=1 R k > x n 3 n 2 Next we study the probabilities J k. Choose δ = (n 2 + k) 0.5 < ɛ/2 with ɛ as in Lemma 3.3. By Markov s inequality, J k ((n 2 + k) 3 /x) α δ α δ E R k ((n 2 + k) 3 /x) α δ n1 α δ (EA α δ ) n2+k E B α δ. By Lemma 3.3 and since n 2 + k = n 0 + m + k, k=1 J k. (EA α δ ) n2+k c e δρ(n2+k) c x δ e δρ(m+k). There is ζ 3 > 0 such that δ(m + k) (log x + k) ζ3. Hence, for appropriate constants C 4, C 5, C 6 > 0, n 3 n 2 k=1 This finishes the proof. n 3 n 2 J k c x α n α δ 1 k=1 (n 2 + k) 3(α δ) e ρ(log x+k)ζ 3 C 4 x α e C5 (log x)c Bounds for PS j > x. The next proposition is a first major step towards the proof of the main result. For the formulation of the result and its proof, recall the definitions of S i and S i from (3.7) and (3.8), respectively. Proposition 3.6. Assume that c + > 0 and let (b n ) be any sequence such that b n. Then the following relation holds: lim sup PS 1 > x n x b n n 1 PY > x c (3.12) = 0. If c + = 0 then (3.13) lim sup PS 1 > x n x b n n 1 P Y > x = 0. The proof depends on the following auxiliary result whose proof is given in Appendix B. Lemma 3.7. Assume that Y and η k (defined in (3.2)) are independent and ψ(α +ɛ) = EA α+ɛ < for some ɛ > 0. Then for n 1 = n 0 m = [ρ 1 log x] [(log x) 0.5+σ ] for some σ < 1/4 and any sequences b n and r n the following relation holds: lim sup Pη k Y > x n k PY > x c = 0, provided c + > 0. If c + = 0 then r nkn 1,b nx lim sup n r nkn 1,b nx Pη k Y > x k P Y > x = 0.

10 10 D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ Proof of Proposition 3.6. For i n 1, consider S i + S i = Π i,n1 B n Π i,i+n1 2B i+n1 1 + S i + Π i,i+n2 B i+n Π i,n2+n 1 1B n2+n 1 = Π i,n1 (B n1+1 + A n1+1b n Π n1+1,n 2+n 1 1B n2+n 1 ). Notice that P S S n 1 > x n 1 P S 1 > x/n 1. Therefore and by virtue of Lemmas 3.4 and 3.5 there exist positive constants C 7, C 8, C 9 such that P S S n 1 > x C 7 x α e C8(log x)c 9, x 1. Therefore and since S 1 = n 1 S i it suffices for (3.12) to show that n1 S i > x We observe that where lim sup n PS 1 + x b n n 1 PY > x c = 0. n 1 S 1 + S i d =: UT 1 and T 1 + T 2 = Y, U = Π 1,n1 + Π 2,n1 + + Π n1,n 1, T 1 = B n1+1 + Π n1+1,n 1+1B n Π n1+1,n 2+n 1 1B n2+n 1, T 2 = Π n1+1,n 2+n 1 B n2+n Π n1+1,n 2+n 1+1B n2+n Since U = d η n1 and Y = d T 1 + T 2, in view of Lemma 3.7 we obtain provided c + > 0 or lim sup PU(T 1 + T 2 ) > x c = 0, n x b n n 1 PY > x lim sup PU(T 1 + T 2 ) > x = 0, n x b n n 1 P Y > x if c + = 0. Thus to prove the proposition it suffices to justify the existence of some positive constants C 10, C 11, C 12 such that (3.14) P UT 2 > x C 10 x α e C11 (log x)c 12, x > 1. For this purpose we use the same argument as in the proof of Lemma 3.4. First we write P UT 2 > x PU Π n1+1,n 1+n 2+k B n1+n 2+k+1 > x/(log x + k) 3. k=0 Write δ = (log x + k) 0.5. Then by Lemma 3.3, Markov s inequality and since n 2 = n 0 + m, PU Π n1+1,n 1+n 2+k B n1+n 2+k+1 > x/(log x + k) 3 (log x + k) 3(α δ) x (α δ) EU α δ (EA α δ ) n2+k E B α δ c (log x + k) 3(α δ) x (α δ) e (n2+k)ρδ c e (m+k)ρδ (log x + k) 3(α δ) x α.

11 LARGE DEVIATIONS 11 There is ζ > 0 such that (m + k)δ (log x + k) ζ and therefore, P UT 2 > x c x α k=0 c x α e (log x)ζ ρ/2. e (log x+k)ζρ (log x + k) 3(α δ) This proves (3.14) and the lemma. Observe that if i j > 2 then S i and S j are independent. For i j 2 we have the following bound: Lemma 3.8. The following relation holds for some constant c > 0: sup P S i > x, S j > x c n x α, x > 1. i 1, i j 2 Proof. Assume without loss of generality that i = 1 and j = 2, 3. Then we have S 1 ( Π 1,n1 + + Π n1,n 1 ) ( B n1+1 + Π n1+1,n 1+1 B n Π n1+1,n 1+n 2 1 B n2+n 1 ) =: U 1 T 1, S 2 ( Π n1+1,2n Π 2n1,2n 1 ) ( B 2n1+1 + Π 2n1+1,2n 1+1 B 2n Π 2n1+1,2n 1+n 2 1 B 2n1+n 2 ) =: U 2 T 2, S 3 ( Π 2n1+1,3n Π 3n1,3n 1 ) ( B 3n1+1 + Π 3n1+1,3n 1+1 B 3n Π 3n1+1,3n 1+n 2 1 B 3n1+n 2 ) =: U 3 T 3. d We observe that U 1 = ηn1, U i, i = 1, 2, 3, are independent, U i is independent of T i the T i s have power law tails with index α > 0. We conclude from (3.12) that for each i, and P S 1 > x, S 2 > x PT 1 > x n 1/(2α) 1 + PT 1 x n 1/(2α) 1, U 1 T 1 > x, U 2 T 2 > x c n x α + Pn 1/(2α) 1 U 1 > 1, U 2 T 2 > x c n x α + PU 1 > n 1/(2α) 1 PU 2 T 2 > x c n x α. In the same way we can bound P S 1 > t, S 3 > t. We omit details Semi-final steps in the proof of the main theorem. In the following proposition, we combine the various tail bounds derived in the previous sections. For this reason, recall the definitions of X i, S i and Z i from (3.8) and that p 1, p, p 3 are the largest integers such that p 1 n 1 n n 1 + 1, pn 1 n n 2 and p 3 n 1 n n 3, respectively. Proposition 3.9. Assume the conditions of Theorem 2.1. x-regions: In particular, consider the following Λ n = (n 1/α (log n) M, ) for α (0, 2], M > 2, (c n n 0.5 log n, ) for α > 2, c n,

12 12 D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ and introduce a sequence s n such that e sn Λ n and s n = o(n). Then the following relations hold: c + p c P j=1 c + + c lim sup sup (S j c j ) > x (3.15), n x Λ n n P Y > x p P j=1 0 = lim sup (S j c j ) > x c+ c (3.16) n x Λ n,log xs n n P Y > x c + + c, P p 1 0 = lim sup j=1 (X j e j ) > x (3.17), n x Λ n n P Y > x (3.18) P 0 = lim sup n x Λ n p 3 j=1 (Z j z j ) > x n P Y > x where c j = e j = z j = 0 for α 1 and c j = ES j, e j = EX j, z j = EZ j for α > 1. Proof. We split the proof into the different cases corresponding to α 1, α (1, 2] and α > 2. The case 1 < α 2. Step 1: Proof of (3.15) and (3.16). Since M > 2, we can choose ξ so small that (3.19) 2 + 4ξ < M and ξ < 1/(4α), and we write y = x/(log n) 2ξ. Consider the following disjoint partition of Ω: p Ω 1 = S j y, Ω 2 = Ω 3 = Then for A = p j=1 (S j c j ) > x, (3.20) j=1 1i<kp S i > y, S k > y, p S k > y, S i y for all i k. k=1 PA = PA Ω 1 + PA Ω 2 + PA Ω 3 =: I 1 + I 2 + I 3. Next we treat the terms I i, i = 1, 2, 3, separately. Step 1a: Bounds for I 2. We prove (3.21) lim n sup x Λ n (x α /n) I 2 = 0. We have I 2 1i<kp P S i > y, S k > y. For k i + 3, S k and S i are independent and then, by (3.12), P S i > y, S k > y = (P S 1 > y) 2 c (n 1 (y)) 2 y 2α, where n 1 (y) is defined in the same way as n 1 = n 1 (x) with x replaced by y. n 1 (y) n 1 (x). For k = i + 1 or i + 2, we have by Lemma 3.8 P S i > y, S k > y c (n 1 (y)) 0.5 y α., Also notice that

13 LARGE DEVIATIONS 13 Summarizing the above estimates and observing that (3.19) holds, we obtain for x Λ n, This proves (3.21). Step 1b: Bounds for I 1. We will prove I 2 c [p 2 n 2 1y 2α + pn y α ] c n x α [ x α n (log n) 4ξα + (log n) 2ξα n1 0.5 ] c n x α[ (log n) (4ξ M)α + (log n) 2ξα 0.5]. (3.22) lim n sup x Λ n (x α /n) I 1 = 0. For this purpose, we write S y j = S j1 Sj y and notice that ES j = ES y j + ES j1 Sj >y. Elementary computations show that (3.23) S 1 α n max(α,1) 1 (2m + 1) max(α,1) E B α. Therefore by the Hölder and Minkowski inequalities, and by (3.12) ES j 1 Sj >y (E S j α ) 1/α (P S j > y) 1 1/α c (log x) 1.5+σ y α+1 (n 1 (y)) 1 1/α c (log x) 1.5+σ+2ξ(α 1) x α+1 n 1. Let now γ > 1/α and n 1/α (log n) M x n γ. Since p n 1 n and (3.19) holds, (3.24) If x > n γ then Hence (3.25) p ES j 1 Sj >y c (log x) 1.5+σ+2ξ(α 1) x α+1 n = o(x). x > (log x) M n 1/α and x α < (log x) Mα n 1. p ES j 1 Sj >y c x(log x) 1.5+σ+2ξ(α 1) (log x) Mα = o(x). Using the bounds (3.24) and (3.25), we see that for x sufficiently large, p I 1 P (S y j ESy j ) > 0.5 x (3.26) j=1 ( = P 1jp,j 1,4,7,... 3 P 1jp,j 1,4,7, jp,j 2,5,8,... + (S y j ESy j ) > x/6. 1jp,j 3,6,9,... )(S y j ESy j ) > 0.5 x In the last step, for the ease of presentation, we slightly abused notation since the number of summands in the 3 partial sums differs by a bounded number of terms which, however, do not contribute to the asymptotic tail behavior of I 1. Since the summands S y 1, Sy 4,... are iid and bounded, we may apply Prokhorov s inequality (A.1) to the random variables R k = S y k ESy 1 in (3.26) with y = x/(log n) 2ξ and B p = p var(s y 1 ). Then a n = x/(2y) = 0.5 (log n) 2ξ and since, in view of (3.23), var(s y 1 ) y2 α E S 1 α, ( p var(s y 1 I 1 c ) ) an ( c (log n) (1.5+σ) α+2ξ(α 1) 1) a n ( n ) an. xy x α Therefore, for x Λ n, (x α /n) I 1 c (log n) ((1.5+σ)α+2ξ(α 1))an Mα(an 1),

14 14 D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ which tends to zero if M > 2, ξ satisfies (3.19) and σ is sufficiently small. Step 1c: Bounds for I 3. Here we assume c + > 0. In this case, we can bound I 3 by using the following inequalities: for every ɛ > 0, there is n 0 such that for n n 0, uniformly for x Λ n and every fixed k 1, (3.27) (1 ɛ)c PA S k > y, S i y, i k n 1 PY > x (1 + ɛ)c. Write z = x/(log n) ξ and introduce the probabilities, for k 1, J k = P A j c j ) > z, S k > y, S i y, i k, j k(s (3.28) V k = P A j c j ) z, S k > y, S i y, i k. j k(s Write S = (S j c j ), where summation is taken over the set j : 1 j p, j k, k ± 1, k ± 2. For n sufficiently large J k is dominated by P S > z 8y, S k > y, S i y, i k P S k > y P S > 0.5 z, S i y, i k. Applying the Prokhorov inequality (A.1) in the same way as in step 1b, we see that P S > 0.5 z, S i y, i k c n z α c (log n) (M ξ)α, and by Markov s inequality, Therefore P S 1 > y c n 1(y) y α c n 1 y α. sup (x α /n 1 )J k c (log n) 3αξ Mα 0. x Λ n Thus it remains to bound the probabilities V k. We start with sandwich bounds for V k : (3.29) (3.30) PS k c k > x + z, S z 8y, S i y, i k V k PS k c k > x z, S z + 8y, S i y, i k. By (3.12), for every small ɛ > 0, n sufficiently large and uniformly for x Λ n, we have (3.31) (1 ɛ)c PS k c k > x + z n 1 PY > x PS k c k > x z n 1 PY > x (1 + ɛ)c, where we also used that lim n (x + z)/x = 1. Then the following upper bound is immediate from (3.30): From (3.29) we have V k n 1 PY > y PS k c k > x z (1 + ɛ)c. n 1 PY > x V k n 1 PY > y PS k c k > x + z L k. n 1 PY > x

15 LARGE DEVIATIONS 15 In view of the lower bound in (3.31), the first term on the right-hand side yields the desired lower bound if we can show that L k is negligible. Indeed, we have n 1 PY > xl k = PS k c k > x + z [ S > z 8y i k S i > y ] PS k c k > x + z, S > z 8y + i k PS k c k > x + z, S i > y [ ] PS k c k > x + z P S > z 8y + p P S 1 > y +c PS k c k > x + z, S i > y i k 2,i k Similar bounds as in the proofs above yield that the right-hand side is of the order o(n 1 /x α ), hence L k = o(1). We omit details. Thus we obtain (3.27). Step 1d: Final bounds. Now we are ready for the final steps in the proof of (3.16) and (3.15). Suppose first c + > 0 and log x s n. In view of the decomposition (3.20) and steps 1a and 1b we have as n and uniformly for x Λ n, P p j=1 (S j c j ) > x I 3 n PY > x n PY > x n 1 n p k=1 P p j=1 (S j ES j ) > x, S k > y, S j y, j k. n 1 PY > x In view of step 1c, in particular (3.27), the last expression is dominated from above by (p n 1 /n)(1 + ɛ)c (1 + ɛ)c and from below by n 1 p n (1 ɛ)c n n 2 n ( 1 (1 ɛ)c (1 ɛ)c 1 3s ) n. n nρ Letting first n and then ɛ 0, (3.15) follows and (3.16) is also satisfied provided the additional condition lim n s n /n = 0 holds. If c + = 0 then I 3 = o(n P Y > x). Let now x Λ n and recall the definition of V k from (3.28). Then for every small δ and sufficiently large x P p j=1 (S j c j ) > x n P Y > x and (3.15) follows from the second part of Lemma 3.7. I 3 n P Y > x n p 1 k=1 V k n n 1 P Y > x PS 1 > x(1 δ) c 1 sup x Λ n n 1 P Y > x Step 2: Proof of (3.17) and (3.18). We restrict ourselves to (3.17) since the proof of (3.18) is analogous. Write B = p 1 j=1 (X j e j ) > x. Then By Lemma 3.4, PB P B p 1 X k > y k=1 + P B X j y for all j p 1 = P 1 + P 2. P 1 p 1 P X 1 > y C 1 p 1 y α e C2(log y)c 3 = o(nx α ). For the estimation of P 2 consider the random variables X y j = X j1 Xj y and proceed as in step 1b.

16 16 D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ The case α > 2. The proof is analogous to α (1, 2]. We indicate differences in the main steps. Step 1b. First we bound the large deviation probabilities of the truncated sums (it is an analog of step 1b of Proposition 3.9). We assume without loss of generality that c n log n. Our aim now is to prove that for y = xc 0.5 n : (3.32) lim sup x α p n x Λ n n P (S j ES j ) > x, S j y for all j p = 0. j=1 We proceed as in the proof of (3.22) with the same notation S y j = S j1 Sj y. As in the proof mentioned, p ES j 1 Sj >y = o(x) and so we estimate the probability of interest by I := 3 P (S y (3.33) j ESy 1 ) > x/6 1jp,j 1,4,7,... The summands in the latter sum are independent and therefore one can apply the two-sided version of the Fuk-Nagaev inequality (A.3) to the random variables in (3.33): with a n = βx/y = c 0.5 n β and p var(s y 1 ) cpn2 1 cnn 1, (3.34) I (c p ) n(1.5+σ)α an 1 + exp xy α 1 (1 β)2 x 2 2e α. cnn 1 We suppose first that x n Then the first quantity in (3.34) multiplied by x α /n is dominated by ( ) c(log n) (1.5+σ)α c 0.5(α 1) an n (n/x α ) an 1 c 0.5an(1+α)+α (c(log n) (1.5+σ)α ) an n (n 0.5α 1 (log n) α ) an 1 0. The second quantity in (3.34) multiplied by x α /n is dominated by x α n exp (1 β)2 c 2 n(log n) 2 2e α n αγ 1 n c c2 n 0. cn 1 If x > n 0.75 then xn 0.5 > x δ for an appropriately small δ. Then the first quantity in (3.34) is dominated by ( c(log x) (1.5+σ)α) a nc 0.5a n(α 1) n The second quantity is dominated by which finishes the proof of (3.32). (n/x α ) an 1 c 0.5an(1+α)+α (c(log x) (1.5+σ)α ) an n (n 0.5α 1 x αδ ) an 1 0. x α n exp (1 β)2 c 2 nx 2δ (log n) 2 2e α cn 1 x α e cxδ 0, Step 1c. We prove that, for any ε (0, 1), sufficiently large n and fixed k 1, the following relation holds uniformly for x Λ n, (3.35) (1 ε)c P p j=1 (S j ES j ) > x, S k > y, S i y, i k (1 + ε)c. n 1 PY > x Let z Λ n be such that x/z. As for α (1, 2], one proves x α p P j ES j ) > x, n 1 j=1(s (S j ES j ) > z, S k > y, S j y, j k 0. j k

17 LARGE DEVIATIONS 17 Apply the Fuk-Nagaev inequality (A.3) to bound P j ES j ) > j k(s z 2, S j y, j k. In the remainder of the proof one can follow the arguments of the proof in step 1c for α (1, 2]. The case α 1. The proof is analogous to the case 1 < α 2; instead of Prokhorov s inequality (A.1) we apply S.V. Nagaev s inequality (A.2). We omit further details Final steps in the proof of Theorem 2.1. We have for small ε > 0, (3.36) n P ( S i E S n i ) > x(1 + 2ε) P ( X i E X n i ) > xε P ( Z i E Z i ) > xε P S n d n > x n P ( S i E S n i ) > x(1 2ε) + P ( X i E X n i ) > xε + P ( Z i E Z i ) > xε. Divide the last two probabilities in the first and last lines by np Y > x. Then these ratios converge to zero for x Λ n, in view of (3.17), (3.18) and Lemmas 3.4 and 3.5. Now P n i=pn 1+1 ( S i E S i ) > x(1 2ε) =P n n 1 i=pn 1+1 P where S i = Π i,i+n1 1 B i+n1 + + Π i,i+n2 1 B i+n2. Notice that if i n n 2 then (for α > 1) n n 1 i=pn 1+1 ( S i E S i ) > x(1 2ε) ( S i > x(1 2ε) n n 1 i=pn 1+1 E S i, E S i = E S 1 and for n n 2 < i n n 1 Hence there is C such that E S i = (EA) n1 ( 1 + EA (EA) n i n1 )EB. n n 1 i=pn 1+1 E S i 2n 1 C

18 18 D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ and so by Proposition 3.6 P n n 1 i=pn 1+1 ( S i > x(1 2ε) n n 1 i=pn 1+1 E S n 1 i P ( S i > x(1 2ε) 2n 1C 2 2n 1 + P ( S i > x(1 2ε) 2n 1C, 2 i=n 1 ( ) C 1 n 1 x α = o np Y > x, provided lim n s n /n = 0. Taking into account (3.16) and the sandwich (3.36), we conclude that (2.2) holds. If the x-region is not bounded from above and n > n 1 (x) then the above calculations together with Lemma 3.1 show (2.1). If n n 1 (x), then and again (2.1) holds. PS n d n > x C 1 x α e C2(log x)c Ruin probabilities In this section we study the ruin probability related to the centered partial sum process T n = S n ES n, n 0, i.e. for given u > 0 and µ > 0 we consider the probability [ ψ(u) = Psup Tn µ n ] > u. n 1 We will work under the assumptions of Kesten s Theorem 1.1. Therefore the random variables Y i are regularly varying with index α > 0. Only for α > 1 the variable Y has finite expectation and therefore we will assume this condition throughout. Notice that the random walk (T n n µ) has dependent steps and negative drift. Since (Y n ) is ergodic we have n 1 (T n n µ) a.s. µ and in particular sup n 1 (T n n µ) < a.s. It is in general difficult to calculate ψ(u) for a given value u, and therefore most results on ruin study the asymptotic behavior of ψ(u) when u. If the sequence (Y i ) is iid it is well known (see Embrechts and Veraverbeke [7] for a special case of subexponential step distribution and Mikosch and Samorodnitsky [19] for a general regularly varying step distribution) that (4.1) ψ ind (u) u PY > u µ (α 1), u. (We write ψ ind to indicate that we are dealing with iid steps.) If the step distribution has exponential moments the ruin probability ψ ind (u) decays to zero at an exponential rate; see for example the celebrated Cramér-Lundberg bound in Embrechts et al. [6], Chapter 2. It is the main aim of this section to prove the following analog of the classical ruin bound (4.1): Theorem 4.1. Assume that the conditions of Theorem 1.1 are satisfied and additionally B 0 a.s. and there exists ε > 0 such that EA α+ε and EB α+ε are finite, α > 1 and c + > 0. The following asymptotic result for the ruin probability holds for fixed µ > 0, as u : (4.2) P sup(s n ES n nµ) > u n 1 c µ(α 1) u α+1 c c + u PY > u µ(α 1) Remark 4.2. We notice that the dependence in the sequence (Y t ) manifests in the constant c /c + in relation (4.2) which appears in contrast to the iid case; see (4.1)..

19 LARGE DEVIATIONS 19 To prove our result we proceed similarly as in the proof of Theorem 2.1. First notice that in view of (3.9), Psup(Y 0 η n E(Y 0 η n )) > u PY 0 η > u = o(u 1 α ). n 1 Thus, it is sufficient to prove u α 1 P sup( S n E S n nµ) > u c n 1 µ(α 1), for S n defined in (3.2). Next we change indices as indicated after (3.3). However, this time we cannot fix n and therefore we will proceed carefully; the details will be explained below. Then we further decompose S n into smaller pieces and study their asymptotic behavior. Proof of Theorem 4.1. The following lemma shows that the centered sums ( S n E S n ) n um for large M do not contribute to the asymptotic behavior of the ruin probability as u. Lemma 4.3. The following relation holds: lim M lim sup u α 1 P sup ( S n E S n nµ) > u = 0. u n>um Proof. Fix a large number M and define the sequence N l = um2 l, l 0. Assume for the ease of presentation that (N l ) constitutes a sequence of even integers; otherwise we can take N l = [um]2 l. Observe that P sup ( S n E S n nµ) > u n>um p l, where p l = P max n [Nl,N l+1 )( S n E S n nµ) > u. For every fixed l, in the events above we make the change of indices i j = N l+1 i and write, again abusing notation, With this notation, we have l=0 Ỹ j = B j + Π jj B j Π j,nl+1 2B Nl+1 1. p l = P max n (0,N l ] N l+1 1 i=n (Ỹi EỸi µ) > u. Using the decomposition (3.4) with the adjustment n 4 = min(j+n 3, N l+1 1), we write Ỹj = Ũj+ W j. Then, by Lemma 3.2, for small δ > 0, p l1 = P max N l+1 1 ( W i E W i µ/4) > u/4 n (0,N l ] P N l+1 1 i=n i=n l ( W i E W i ) + N l 1 W i > u/4 + N l µ/4 P N l+1 1 ( W i E W i ) > u/4 + N l (µ/4 E W 1 ) cn l+1 N α δ l c(um2 l ) 1 α δ. We conclude that for every M > 0, p l1 = o(u 1 α ) as u. l=0

20 20 D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ As in (3.7) we further decompose Ũi = X i + S i + Z i, making the definitions precise in what follows. Let p be the smallest integer such that pn 1 N l+1 1 for n 1 = n 1 (u). For i = 1,..., p 1 define X i as in (3.8), and X p = N l+1 1 i=(p 1)n 1+1 X i. Now consider p l2 = P max N l+1 1 ( X i E X i µ/4) > u/4 n (0,N l ] i=n P N l+1 1 N ( X i E X l 1 i ) + max ( X i E X i ) > u/4 + N l µ/4 n (0,N l ] i=n l i=n P p max (X i EX i ) > u/8 + N l µ/8 kp/2 i=k +P max kp/2 i=k = p l21 + p l22. p (X i EX i ) u/8 + N l µ/8, max max kn 1 kp/2 1j<n 1 i=kn 1 j The second quantity is estimated by using Lemma 3.4 as follows for fixed M > 0 p l22 c p P n1 ( X i E X i ) > u/8 + N l µ/8 X i > u/8 + N l µ/8 C 1 p N α l e C2(log N l) C 3 = o(u 1 α )2 (α 1)l, where C i, i = 1, 2, 3, are some positive constants. Therefore for every fixed M, p l22 = o(u 1 α ), as u. l=0 Next we treat p l21. We observe that X i and X j are independent for i j > 1. Splitting summation in p l21 into the subsets of even and odd integers, we obtain an estimate of the following type p l21 c 1 P max (X 2i EX 2i ) > c 2 (u + N l ), kp/2 k2ip where the summands are now independent. By the law of large numbers, for any ɛ (0, 1), k p/2, large l, P (X 2i EX 2i ) > ɛc 2 (u + N l ) 1/2. 2ik An application of the maximal inequality (A.5) in the Appendix and an adaptation of Proposition 3.9 yield p l21 2P (X 2i EX 2i ) > (1 ɛ)c 2 (u + N l ) cn 1 α 2ip l. Using the latter bound and summarizing the above estimates, we finally proved that lim M lim sup u α 1 u l=0 p l2 = 0. Similar arguments show that the sums involving the S i s and Z i s are negligible as well. This proves the lemma.

21 LARGE DEVIATIONS 21 In view of Lemma 4.3 it suffices to study the following probabilities for sufficiently large M > 0: P max nmu ( S n E S n nµ) > u. Write N 0 = Mu, change again indices i j = N 0 i + 1 and write, abusing notation, Ỹ j = B j + Π jj B j Π j,n0 1B N0. Then we decompose Ỹj as in the proof of Lemma 4.3. Reasoning in the same way as above, one proves that the probabilities related to the quantities W i, Xi and Z i are of lower order than u 1 α as u and, thus, it remains to study the probabilities (4.3) P max N 0 nn 0 i=n ( S i E S i µ) > u, where S i were defined in (3.7). Take n 1 = log N 0 /ρ, p = N 0 /n 1 and denote by S i the sums of n 1 consecutive S i s as defined in (3.8). Observe that for any n such that n 1 (k 1) + 1 n kn 1, k 1 p we have N 0 i=n ( S i E S i µ) 2n 1 (E S 1 + µ) + 2n 1 (E S 1 + µ) + (p+1)n 1 i=(k 1)n 1+1 p i=k 1 ( S i E S i µ) (S i ES i n 1 µ) and N 0 i=n ( S i E S i µ) 2n 1 (E S 1 + µ) + pn 1 i=kn 1 ( S i E S i µ) 2n 1 (E S p µ) + (S i ES i n 1 µ). i=k Therefore and since n 1 is of order log u, instead of the probabilities (4.3) it suffices to study ψ p (u) = P max np i=n p (S i ES i n 1 µ) > u. Choose q = [M/ε α+1 1 ] + 1 for some small ε 1 and large M. Then the random variables R k = kq 3 i=(k 1)q S i, k = 1,..., r = p/q,

22 22 D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ are independent and we have ψ p (u) P max nqr +P max niqr i kq 2,kq 1 jr k=j +P max +P jr k=j max qr<n<p i=n (S i ES i n 1 µ) > u(1 3ε 1 ) r (S kq 2 ES kq 2 n 1 µ) > ε 1 u r (S kq 1 ES kq 1 n 1 µ) > ε 1 u p (S i ES i n 1 µ) > ε 1 u =: 4 ψ (i) p (u). The quantities ψ p (i) (u), i = 2, 3, can be estimated in the same way; we focus on ψ p (2) (u). Applying Petrov s inequality (A.4) and Proposition 3.9, we obtain for some constant c not depending on ε 1, ψ p (2) (u) P max jr k=j r (S kq 2 ES kq 2 ) > ε 1 u r c P (S kq 2 ES kq 2 ) > ε 1 u/2 k=1 c r n 1 (ε 1 u) α c ε 1 u α+1. Hence we obtain lim ε1 0 lim sup u u α 1 ψ (2) p (u) = 0. By (3.15), for some constant c, ψ (4) p (u) c qn 1 (ε 1 u) α c Mu r(ε 1 u) α. Since r q > M/ε α+1 1 for large u we conclude for such u that r 1 M 1 ε α+1 1 and therefore ψ p (4) (u) cε 1 u 1 α and lim ε1 0 lim sup u u α 1 ψ p (4) (u) = 0. Since A and B are non-negative we have for large u with µ 0 = µ (q 2), ψ p (1) (u) P max jr P max jr j (R i ER i µ 0 n 1 ) > u(1 3ε 1 ) qn 1 (ES 1 + µ) j (R i ER i µ 0 n 1 ) > u(1 4ε 1 ). Combining the bounds above we proved for large u, small ε 1 and some constant c > 0 independent of ε 1 that j ψ p (u) P max (R i ER i µ 0 n 1 ) > u(1 4ε 1 ) + c ε 1 u α+1. jr Similar arguments as above show that ψ p (u) P max jr j (R i ER i µ 0 n 1 ) > u(1 + 4ε 1 ) c ε 1 u α+1.

23 LARGE DEVIATIONS 23 Thus we reduced the problem to study an expression consisting of independent random variables R i and the proof of the theorem is finished if we can show the following result. Write j Ω r = max (R i ER i n 1 µ 0 ) > u. Lemma 4.4. The following relation holds lim lim sup u α 1 PΩ r c c + = 0. M u µ(α 1) jr Proof. Fix some ε 0 > 0 and choose some large M. Define C 0 = (q 2)c c +. Reasoning as in the proof of (3.16), we obtain for any integers 0 j < k r and large u (4.4) 1 ε 0 u P k ( ) α i=j+1 Ri ER i > u 1 + ε 0 n 1 (k j)c 0 Choose ε, δ > 0 small to be determined later in dependence on ε 0. Eventually, both ε, δ > 0 will become arbitrarily small when ε 0 converges to zero. Define the sequence k 0 = 0, k l = [δn 1 1 (1 + ε) l 1 u], l 1. Without loss of generality we will assume k l0 = Mun 1 1 for some integer number l 0. For η > ε 0 (2l 0 ) 1 consider the independent events j D l = max (R i ER i ) > 2ηu, l = 0,..., l 0 1. Define the disjoint sets We will show that (4.5) k l <jk l+1 i=k l +1 W l = Ω r D l m l D c m, l = 0,..., l 0 1. PΩ r l 0 1 l=0 PW l o(u1 α ), u. First we observe that Ω r l 0 1 l=0 D l. Indeed, on l 0 1 l=0 Dc l we have j max jr (R i ER i n 1 µ 0 ) l 0 2ηu ε 0 u, and therefore Ω r cannot hold for small ε 0. Next we prove that (4.6) P m l(d m D l ) = o(u 1 α ), u. Then (4.5) will follow. First apply Petrov s inequality (A.4) to PD l with q 0 arbitrarily close to one and power p 0 (1, α). Notice that E R i p0 is of the order qn 1, hence m p0 is of the order δεqu cδεmε α 1 1 u. Next apply (4.4). Then one obtains for sufficiently large u, and small ε, δ, and some constant c depending on ε, δ, ε 0, ε 1, kl+1 PD l q0 1 P i=k l +1 (R i ER i ) > ηu q 1 0 n 1(k l+1 k l )(1 + ε 0 )C 0 (ηu) α c u 1 α. Hence P m l (D l D m ) = O(u 2(1 α) ) as desired for (4.6) if all the parameters ε, δ, ε 0, ε 1 are fixed.

24 24 D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ Thus we showed (4.5) and it remains to find suitable bounds for the probabilities PW l. On the set W l we have j j max (R i ER i µ 0 n 1 ) max (R i ER i ) 2ηlu ε 0 u, jk l jk l max j k l+1 <jr i=k l +1 (R i ER i ) 2ηl 0 u ε 0 u. Therefore for small ε 0 and large u on the event W l, j j max (R i ER i µ 0 n 1 ) = max (R i ER i µ 0 n 1 ) jr k l <jr k l (R i ER i µ 0 n 1 ) + max 2ε 0 u k l µ 0 n 1 + max j k l <jk l+1 i=k l +1 j k l <jk l+1 i=k l +1 (R i ER i ). (R i ER i ) + max j k l+1 <jr i=k l+1 +1 Petrov s inequality (A.4) and (4.4) imply for l 1 and large u that j PW l P max (R i ER i ) (1 2ε 0 )u + µ 0 n 1 k l k l <jk l+1 i=k l +1 kl+1 q0 1 P i=k l +1 (R i ER i ) (1 3ε 0 )u + µ 0 n 1 k l (R i ER i ) q 1 0 (k l+1 k l )n 1 (1 + ε 0 )C 0 ((1 3ε 0 ) + µ 0 δ(1 + ε) l 1 ) α u α = q 1 0 δε(1 + ε) l 1 (1 + ε 0 )C 0 ((1 3ε 0 ) + µ 0 δ(1 + ε) l 1 ) α u1 α. For l = 0 we use exactly the same arguments, but in this case (k 1 k 0 )n 1 = δu and k 0 = 0. Thus we arrive at the upper bound l 0 1 ( l0 1 PW l q0 1 (1 + ε δ 0)C 0 (1 3ε 0 ) α + δε(1 + ε) l 1 ) ((1 3ε 0 ) + µ 0 δ(1 + ε) l 1 ) α u 1 α (4.7) l=0 = q 1 0 (1 + ε 0) A(ε, δ, ε 0, l 0 )u 1 α. To estimate PW l from below first notice that on W l, for large u, max jr j (R i ER i µ 0 n 1 ) k l+1 (R i ER i µ 0 n 1 ) k l+1 i=k l +1 k l+1 i=k l +1 (R i ER i ) k l+1 µ 0 n 1 k l ER 1 (R i ER i ) k l+1 µ 0 n 1 ε 0 u.

25 LARGE DEVIATIONS 25 By (4.6) and (4.4), we have for l 1 and as u, Hence k l+1 PW l P l 0 1 l=0 Thus we proved that P i=k l +1 k l+1 1 (R i ER i ) > (1 + ε 0 )u + µ 0 n 1 k l+1 D l Dm c m l i=k l (R i ER i ) > (1 + ε 0 )u + µ 0 n 1 k l+1 P D l D m m l (k l+1 k l )n 1 (1 ε 0 )C 0 ((1 + ε 0 )u + µ 0 k l+1 n 1 ) α o(u1 α ) (1 2ε 0)C 0 δ(1 + ε) l 1 ε ((1 + ε 0 ) + µ 0 δ(1 + ε) l ) α u1 α. ( l0 1 δ PW l (1 2ε 0 )C 0 (1 + ε 0 + µ 0 δ) α + = (1 2ε 0 )C 0 B(ε, δ, ε 0, l 0 )u 1 α. l=1 δ(1 + ε) l 1 ) ε ((1 + ε 0 ) + µ 0 δ(1 + ε) l ) α u 1 α (4.8) l 0 1 (1 2ε 0 )B(ε, δ, ε 0, l 0 ) lim inf u C 1 0 uα 1 PW l l=0 l 0 1 lim sup C0 1 uα 1 PW l q0 1 (1 + ε 0)A(ε, δ, ε 0, l 0 ). u l=0 Finally, we will justify that the upper and lower bounds are close for small ε, δ, ε 0, large M and q 0 close to 1. For a real number s which is small in absolute value define the functions f s (x) = (1 + s + µ 0 x) α and F s (x) = (1 + s + µ 0 x) f s (x) on [δ, M]. Let x l = δ(1 + ε) l 1, l = 1,..., l 0. Since x l+1 x l = δε(1 + ε) l 1 are uniformly bounded by εm and f s is Riemann integrable on [0, ), choosing ε small, we have A(ε, δ, ε 0, l 0 ) = Thus we obtain the bound l 0 1 l=1 M δ f 3ε0 (x l )(x l+1 x l ) f 3ε0 (x) dx = F 3ε 0 (δ) F 3ε0 (M) + ε 0 µ 0 (α 1). (4.9) lim lim lim lim q 0 1 ε 0 0 M δ 0 q 1 0 (1 + ε 0)A(ε, δ, ε 0, l 0 ) = (µ 0 (α 1)) 1. Proceeding in a similar way, B(ε, δ, ε 0, l 0 ) M δ f ε0 (x)dx = F ε 0 (δ) F ε0 (M) ε 0 µ 0 (α 1) The right-hand side converges to (µ 0 (α 1)) 1 by letting δ 0, M and ε 0 0. The latter limit relation in combination with (4.8) and (4.9) proves the lemma..

26 26 D. BURACZEWSKI, E. DAMEK, T. MIKOSCH AND J. ZIENKIEWICZ Appendix A. Inequalities for sums of independent random variables In this section, we consider a sequence (X n ) of independent random variables and their partial sums R n = X X n. We always write B n = var(r n ) and m p = n j=1 E X j p for p > 0. First, we collect some of the classical tail estimates for R n. Lemma A.1. The following inequalities hold. Prokhorov s inequality; cf. Petrov [23], p. 77: Assume that the X n s are centered, X n y for all n 1 and some y > 0. Then PR n x exp x 2 y arsinh( xy ) (A.1), x > 0. 2 B n S. V. Nagaev s inequality; see [22]: Assume m p < for some p > 0. Then n ( e mp ) x/y (A.2) PR n > x PX j > y +, x, y > 0. xy p 1 j=1 Fuk-Nagaev inequality; cf. Petrov [23], p. 78: Assume that the X n s are centered, p > 2, β = p/(p + 2) and m p <. Then n ( mp ) βx/y PR n > x PX j > y + + exp βxy p 1 (1 β)2 x 2 (A.3) 2e p, x, y > 0. B n j=1 Petrov s inequality; cf. Petrov [23], p. 81: Assume that the X n s are centered and m p < for some p (1, 2]. Then for every q 0 < 1, with L = 1 for p = 2 and L = 2 for p (1, 2), (A.4) Pmax in R i > x q 1 0 PR n > x [(L/(1 q 0 )) 1 m p ] 1/p, x R. Lévy-Ottaviani-Skorokhod inequality; cf. Petrov [23], Theorem 2.3 on p. 51. If PR n R k c) q, k = 1,..., n 1, for some constants c 0 and q > 0, then (A.5) Pmax in R i > x q 1 PR n > x c, x R. Appendix B. Proof of Lemma 3.7 Assume first c + > 0. We have by independence of Y and η k, for any k 1, x > 0 and r > 0, Pη k Y > x ( ) PY > x/z k PY > x (0,x/r] = + [x/r, ) k PY > x dp(η k z) = I 1 + I 2. For every ε (0, 1) there is r > 0 such that for x r and z x/r, PY > x/z PY > x Hence for sufficiently large x, z α [1 ε, 1 + ε] and PY > xx α c + ε. I 1 k 1 Eη α k 1 ηk x/r[1 ε, 1 + ε] and I 2 c k 1 x α Pη k > x/r c k 1 Eη α k 1 ηk >x/r. We have I 1 (k 1 Eη α k k 1 Eη α k 1 ηk >x/r)[1 ε, 1 + ε] and by virtue of Bartkiewicz et al. [1], lim k k 1 Eη α k = c. Therefore it is enough to prove that (B.1) lim sup n r nkn 1,b nx k 1 Eη α k 1 ηk >x = 0. By the Hölder and Markov inequalities we have for ɛ > 0, (B.2) Eηk α 1 ηk >x (Eη α+ɛ k ) α/(α+ɛ)( Pη k > x ) ɛ/(α+ɛ) x ɛ Eη α+ɛ k.