Small-time Behaviour of Lévy Processes

Transcription

1 E l e c r o n i c J o u r n a l o f P r o b a b i l i y Vol. 9 (24), Paper no. 8, pages Journal URL hp:// ejpecp/ Small-ime Behaviour of Lévy Processes R. A. Doney Deparmen of Mahemaics, Universiy of Mancheser Oford Road, Mancheser M3 9Pl, U.K. rad@ma.man.ac.uk Absrac. In his paper a necesary and sufficien condiion is esablished for he probabiliy ha a Lévy process is posiive a ime o end o as ends o. This condiion is epressed in erms of he characerisics of he process, and is also shown o be equivalen o wo probabilisic saemens abou he behaviour of he process for small. Keywords and phrases: Lévy processes; local behaviour; Spizer s condiion. AMS subjec classificaion (2): Primary 6G5, 6G7 Submied o EJP on July 29, 23. Final version acceped on January 7,

2 Inroducion The quaniy ρ() = P (X > ) where X = (X, ) is a Lévy process is of fundamenal imporance in flucuaion heory. For eample, combining resuls in [] and [2] shows ha, boh as and as, ρ() ρ [, ] () ρ(s)ds ρ {Xs>}ds d A ρ, where A ρ denoes a random variable wih an arc-sine law of parameer ρ if < ρ <, and a random variable degenerae a ρ if ρ =,. I would herefore be useful o find a necessary and sufficien condiion for () o hold, ideally epressed in erms of he characerisics of X, ha is is Lévy measure Π, is Brownian coefficien σ 2, and γ, he coefficien of he linear erm in he Lévy-Iô decomposiion (2) below. This problem is obviously difficul, and has so far only been solved for large in he special case ρ =,. This resul is in Theorem 3.3 of [5], and in an eended form in Theorem.3 in [4]. In boh cases he resuls are deduced from he corresponding resuls for random walks due o Kesen and Maller in [7] and [8]. Here we consider he corresponding quesion for small, where apparenly he large resuls will have no relevance, bu in fac i urns ou ha here is a sriking formal similariy, boh in he saemen and proof. Our Lévy process will be wrien as X = γ + σb + Y () + Y (2), (2) where B is a sandard BM, Y () is a pure jump maringale formed from he jumps whose absolue values are less han or equal o, Y (2) is a compound Poisson process formed from he jumps whose absolue values eceed, and B, Y (), and Y (2) are independen. We denoe he Lévy measure of X by Π, and inroduce he ail funcions N() = Π{(, )}, M() = Π{(, )}, >, (3) and he ail sum and difference T () = N() + M(), D() = N() M(), >. (4) 2

3 The rôles of runcaed firs and second momens are played by A() = γ + D() + D(y)dy, U() = σ yt (y)dy, >. (5) We will denoe he jump process of X by = ( s, s ), and pu () = sup s (6) s for he magniude of he larges negaive jump which occurs by ime. For he case, a sufficien condiion for () wih ρ = was given in Theorem 2.3 of [5]; he following, ogeher wih Lemma 5 shows ha he condiion given here is also necessary: Theorem Suppose ha he Lévy process X has σ =, Π(R) =, and M(+) > ; hen he following are equivalen. ρ = P (X > ) as ; (7) X () P as ; (8) for some deerminisic d which decreases o and X P is regularly varying of inde a, as ; (9) d() and A() U()M() + as. () Remark 2 None of he above assumpions are really resricive. Firs, if σ, i was shown in [5] ha P (X > ) /2. I was also shown here ha when M(+) =, i.e. X is specrally posiive, and N(+) =, hen (7) occurs iff X is a subordinaor iff A() for all small, and i is easy o see ha hese are equivalen o (9) in his case. (Of course (8) is no relevan here, as ().) Finally he case when Π(R) < is of no real ineres; hen X is a compound Poisson process plus linear drif, and he behaviour of ρ as is deermined by wheher he drif is posiive or no. 2

4 Remark 3 Comparing he above resuls wih he large-ime resuls we see ha each of (7)-() has a formally similar counerpar a. (Acually he counerpar of (8) was omied in [4], bu i is easy o esablish.) One difference is ha (7) as implies ha X P, and of course his can happen as. A firs sigh he appearance of A in boh () and is counerpar a is surprising. However his can be undersood by realising ha A acs boh as a generalised mean a and a generalised drif a. To be precise, X P c, as or as is equivalen o T () and A() c as or as ; in he firs case if X has finie mean µ hen c = µ, and in he second if X has bounded variaion and drif δ hen c = δ. (See Theorems 2. and 3. of [5].) Remark 4 The srucure of he following proof also shows a srong similariy o he proof of he random walk resuls in [7] and [8]. There are of course differences in deail, and some simplificaions due o he advanages of working in coninuous ime and he abiliy o decompose X ino independen componens in various ways. There are also some era difficulies; for eample we need o esablish resuls relaed o he Cenal Limi Theorem which are sandard for random walks bu apparenly no previously wrien down for Lévy processs a zero. Also he case where Π has aoms presens echnical difficulies which are absen in he random walk siuaion; compare he argumen on page 499 of [7] o he upcoming Lemma 9. 2 Preliminary Resuls We sar by showing ha () can be replaced by he simpler A() M() Lemma 5 (i) If () holds hen () holds. (ii) If () holds hen and consequenly () holds. lim sup as. () U() A() 2, (2) 22

5 Proof For i) jus noe ha so ha lim inf A() M() U() = lim inf For (ii) we firs show ha where we wrie U () = 2yT (y)dy T () A() U() U()M() 2 M() lim U () M() 2yM(y)dy, U + () = 2ydy = 2 T (), (3) lim inf A() U()M(). =, (4) 2yN(y)dy, (5) so ha U() = U () + U + (). Given ε > ake > such ha εa() M() for (, ], and hence U () = 2yM(y)dy 2ε Noe also ha for < < we can wrie Then and similarly A(y)dy. A() = γ + A () A + (), where γ = γ + D(), A () = U () = = 2 M(y)dy, and A + () = U + () = 2 2yM(y)dy = 2 A (y)dy 2A (), A + (y)dy 2A + (). 23 N(y)dy. yda (y)

6 Thus for < < U () 2ε { γ + A (y) A + (y)}dy = ε{2 γ + 2A () + U () 2A + () U + ()} ε{2a() + U ()}. Since ε is arbirary, (4) follows. Also for < < ( ) U + () U () = 2 A + (y)dy A + () A (y)dy + A () ( ) = 2 A() A(y)dy 2A(). So U() 2A() + 2U () = 2A() + o{a()} as, and (2) follows. Since A() U()M() = A() M() A() U(), () is immediae. The main par of he proof consiss of showing ha () holds whenever ρ. We firs dispose of one siuaion where he argumen is sraighforward. Lemma 6 Le X be any Lévy process saisfying he assumpions of Theorem, having ρ as, and addiionally having M(+) <. Then () holds. Proof In his case we can wrie where X = X () + X (),, X () = s< s { s<} is a compound Poisson process which is independen of he specrally posiive process X (). Clearly P {X () = } = e M() as, 24

7 so we have P {X () > } as. Bu, as previously menioned, i was shown in [5] ha his happens iff X () is a subordinaor, i.e. i has bounded variaion, so ha Π() (d) = Π(d) <, N() as, and we can wrie X () = s< s { s>} + δ (), where he drif δ () is non-negaive. Comparing his o he represenaion (2) of X we see ha δ () = γ Π(d) + Π (d), where Π denoes he Lévy measure of X. If δ () > he alernaive epression A() = γ yπ(dy) + yπ (dy) + D(), (6) which resuls from (5) by inegraion by pars shows ha A() δ () as. Thus A()/{M()} δ () /{M(+)}. If δ () = he same conclusion follows because A() D(y)dy, since D(+) = N(+) M(+) =. The ne resul allows us o make some addiional assumpions abou X in he remaining case. Lemma 7 Le X # be any Lévy process wih no Brownian componen which has M # (+) = and ρ # = P (X # > ) as. Then here is a Lévy process X wih no Brownian componen such ha ρ = P (X > ) as whose Lévy measure can be chosen so ha (i) N() = M() =, M(+) = ; (ii) each of N and M is coninuous and sricly decreasing on (, c] for some c >. Moreover, (iii) () holds for X # if and only if i holds for X. Proof Noe ha if X () has he same characerisics as X # ecep ha Π # is replaced by Π () (d) = {Π # (d) + λ(d)} { <<}, 25

8 where λ denoes Lebesgue measure, we have X () + Y () = X # + Y (2), where Y () is a compound Poisson process independen of X () and Y (2) is a compound Poisson process independen of X #. Since P (Y (i) = ) as for i =, 2 i is immediae ha P (X () > ) as. By consrucion N () () = M () () =, for < < boh N () () = N # () + N # () and M () () = M # ()+ M # () are sricly decreasing, and (6) shows ha A () () = A # () D # (). Thus M () () M # () as and we see ha () holds for X () if and only if i holds for X #. This esablishes (i), and allows us o assume in he remainder of he proof ha N # () = M # () =, and boh N # and M # are sricly decreasing on (, ]. For (ii) i remains only o show ha we can ake N and M o be coninuous. So suppose ha Π # has aoms of size a n and b n locaed a > 2 > > and y < y 2 < < respecively, for n =, 2, ; clearly if Π # has only finiely many aoms here is nohing o prove, and he case when he resricion of Π # o (, ] or [, ) has only finiely many aoms can be deal wih in a similar way o wha follows. Noe ha from 2 Π # (d) < we have a n 2 n + b n yn 2 <. (7) Now le Π (c) denoe he coninuous par of Π #, so ha Π # = Π (c) + a n δ( n ) + b n δ( y n ), where δ() denoes a uni mass a. Wih U[a, b] denoing a uniform probabiliy disribuion on [a, b] we inroduce he measure Π = Π (c) + a n U[ n, n + α n ] + b n U[ y n, y n + β n ]. We choose α n >, β n > o saisfy he following condiions; for n =, 2,, and n + α n < n+, y n + β n < y n+ ; (8) α n 3 n, β n y 3 n. (9) 26

9 Noe ha (7) and (9) imply ha c := 2 a n α n <, c := 2 b n β n <, and hence ( lim Π(d) ε ε and ε ) ( n ) Π # (d) = lim a k ( k + α n k n 2 ) a k k = c, (2) ( ε ε ) ( ) n lim Π(d) Π # (d) = lim b k (y k β n k ε n 2 ) + b k y k = c. (2) Now le X be a Lévy process wih Lévy measure Π, no Brownian componen, and having γ = γ # + c + c. Since T () = T # () = and we have go Π by moving some of he mass of Π # o he righ, we have, for each fied >, X = γ + lim ε ( <s< s { s >ε} ( P γ # + (c + c ) + lim ε ( lim Π(d) ε ε< < <s< ε< < ε< < Π(d) ) # s { # s >ε} ε< < ) Π # (d) = X #. Π # (d) Thus P (X > ), and o conclude we only need o show ha () holds for X # if and only if i holds for X. Again using D() = D # () = we have, for (, ), A() = γ N(y)dy + = γ # + c + c + M(y)dy N # (y)dy + {N # (y) N(y)}dy + 27 M # (y)dy {M(y) M # (y)}dy )

10 = A # () + c + c + = A # () + {N # (y) N(y)}dy + {N(y) N # (y)}dy + where we have used (2) o see ha ( {N(y) N # (y)}dy = lim Π(d) ε and similarly for c. Since (9) gives 2 {N(y) N # (y)}dy 2 n: n< ε a n 3 n 2 n: n< {M(y) M # (y)}dy {M(y) M # (y)}dy, (22) ε n: n< ) Π # (d) = c, a n α n a n 2 n = o(), and he same argumen applies o he second inegral in (22), we see ha A # () = A() + o() as. (23) Ne noe ha if / ( y n, y n + β n ], hen M() = M # (). On he oher hand, if = y n + θβ n, wih < θ, hen M # () = M() + ( θ)b n M() as, so (iii) follows. The ne piece of informaion we need is reminiscen of he Berry-Esseen Theorem; Lemma 8 Le µ be any Lévy measure, and wrie µ for he resricion of µ o he inerval [ b, b ], where b as. Suppose ha for each > Z has an infiniely divisible disribuion deermined by E(e iθz ) = ep { ( e iθ + iθ)µ (d)} := ψ (θ). (24) Pu R σ 2 = 2 µ (d) = EZ 2, R 28

11 and wrie Φ for he sandard Normal disribuion funcion. Then for any ε > here is a posiive consan W ε such ha for all for all saisfying P {Z σ } Φ() ε (25) σ b W ε. (26) Proof Noe firs ha EZ = and EZ 3 = R 3 µ (d) := ζ, and wrie ν = EZ3 6(EZ 2 ) 3 2 = ζ 6. σ 3 We will apply he inequaliy (3.3), p 52 of [6], wih fied, F () = P {Z σ }, and G() = Φ() ν ( 2 )φ(), where φ is he sandard Normal densiy funcion. From i we deduce ha for any T > he LHS of (25) is bounded above by Here ν ( 2 ) φ() + π m = sup T T ψ ( θ σ ) e θ2 2 ( + ν (iθ) 3 ) θ dθ + 24m πt. (27) G () = sup φ(){ + ν ( ) } M, where M is an absolue consan, for all saisfying ν. Bu if (26) holds we have ν b σ 2 6 σ 3 = b 6 σ 6W ε, (28) so his will hold provided 6W ε. Now fi T = 72M, so ha he hird πε erm in (27) is no greaer han ε/3. The same argumen shows ha he firs erm in (27) is also no greaer han ε/3 provided (26) holds and 3W ε /ε. Finally, o deal wih he middle erm we wrie θ = θ σ, and noe ha ψ ( θ) = e θ2 2 ep { (e i θ ( + i θ b 2 θ 2 2 )µ(d)} = e θ2 (i θ) 2 ep { b 3 µ(d)} ep { r(i θ)µ(d)} 6 b = e θ2 2 ep ν (iθ) 3 ep { r(i θ)µ(d)}, (29) b 29

12 where for some posiive consan c ε r(z) = e z ( + z + 2 z2 + 6 z3 ) ε z 3 (3) whenever z c ε. Since for θ T and b we have θ T b σ T W ε we see ha when (26) holds and W ε (T/c ε ) we can apply (3) o deduce ha for θ T ) ( b r(i θ)µ(d)} ε θ 3 3 µ(d) b ε θ 3 b σ 2 ε θ 3 b σ ε θ 3. I follows from his and (28) ha, increasing he value of W ε if necessary, we can make π T T ψ ( θ σ ) e θ2 2 +ν(iθ)3 ( + ep { b r(i θ)µ(d)}) dθ ε θ 6, and clearly we can also arrange ha π T T e θ2 2 {e ν(iθ)3 ν (iθ) 3 } dθ ε θ 6, whenever (26) holds. Puing hese bounds ino (27) finishes he proof. Ne we record a varian of he Lévy-Khinchine decomposiion (2) which is imporan for us: Lemma 9 If X is any Lévy process wih no Brownian componen and b, b (, ) and > are fied we can wrie where Y (,+) = lim ε Y (, ) = lim ε X = γ(b, b ) + Y (,+) { s {ε< s<b} s { s + Y (, ) + Y (2,+) + Y (2, ), (3) b ε Π(b) } ε s { b < s< ε} Π(d) b 22, Y (2,+) = s s { s b}, }, Y (2, ) = s s { s b },

13 are independen, and γ(b, b ) = γ b Π(d) + b Π (d). Proof This is proved in he same way as (2), ecep we compensae over he inerval ( b, b) raher han (, ). Finally we are in a posiion o esablish he main echnical esimae we need in he proof of Theorem ; Proposiion Suppose ha X is a Lévy process wih no Brownian componen whose Lévy measure saisfies N(+) = M(+) =, and suppose d() and d () saisfy N(d()) = M(d ()) = (32) for all small enough >. Then here is a finie consan K such ha, for any λ >, ρ >, L here eiss C = C(X, λ, ρ, L) > wih P {X γ(d(λ), d (ρλ)) + Kd(λ) Ld (ρλ)} C (33) for all small enough. Proof We sar by noing ha if we use decomposiion (3) for each fied wih b and b replaced by d(λ) and d (ρλ), (32) gives P (Y (2,+) = ) = e N(d(λ)) = e λ. (34) Also Y (,+) has mean zero and since d(), because N(+) =, we can apply Lemma 8 o Y (,+) wih b = d(λ) and σ 2 = d(λ) 2 Π(d). Choosing = and wriing W for he W ε of Lemma 8 wih ε = /4 we conclude from (25) ha P {Y (,+) } whenever 4 σ2 {W d(λ)} 2. On he oher hand, if σ 2 {W d(λ)} 2 i follows from Chebychev s inequaliy ha P { Y (,+) > Kd(λ)} Thus in all cases we can fi K large enough ha σ 2 {Kd(λ)} 2 W 2 K 2. P {Y (,+) Kd(λ)} 4. (35) 22

14 An eacly similar argumen shows ha we can fi a finie K wih P {Y (, ) K d (ρλ)} 4. (36) Finally we noe ha if Z is a random variable wih a Poisson(/ρλ) disribuion P {Y (2, ) (K + L)d (ρλ)} P {Z (K + L)} >. (37) Combining (34)-(37) gives he required conclusion. 3 Proofs Proof of Theorem. Since we have demonsraed in Lemma 5 he equivalence of () and (), and Theorem 2.3 of [5] shows ha () implies (7) under our assumpions, we will firs show ha (7) implies (), and laer heir equivalence o (9) and (8) So assume (7), and also ha M(+) =, since Lemma 6 deals wih he conrary case. Then X saisfies he assumpions we made abou X # in Lemma 7, so ha resul allows us o save era noaion by assuming ha he conclusions (i) and (ii) of ha lemma apply o X. For he momen assume also ha N(+) =, so ha we can define d and d as he unique soluions of (32) on (, ] for some fied >. Our firs aim is o show ha γ(, ) lim inf N(). (38) To see his we use Proposiion wih L =, which since P (X ) implies ha for all sufficienly small γ(d(λ), d (λρ)) + Kd(λ). (39) Wriing ν() = yπ(dy) and ν () = yπ (dy) we have γ(, 2 ) = γ ν( ) + ν ( 2 ), and clearly γ(, 2 ) is a decreasing funcion of 2 for fied. Thus if d(λ) d (λρ) hen from (39) γ(d(λ), d(λ)) + Kd(λ) γ(d(λ), d (λρ)) + Kd(λ). 222

15 However if d (λρ) < d(λ) hen ν (d (λρ)) ν (d(λ)) = d(λ) d (λρ) yπ (dy) d(λ)m(d (λρ)) = d(λ) λρ. Consequenly in boh cases γ(d(λ), d(λ)) (Kλ + ρ )d(λ) λ = (Kλ + ρ )d(λ)n(d(λ)), and since Lemma 7 allows us o assume he coninuiy of d, we have γ(, ) lim inf N() (Kλ + ρ ). Bu in his we may choose λ arbirarily small and ρ arbirarily large, so (38) follows. Assume ne ha () fails, and recall ha A() = γ(, ) + D() = γ(, ) + N() M(). Then for some sequence k and some D < Le γ(, ) + N() DM() when =, 2. (4) k = K 2DM( k ), or equivalenly k = d ( 2D k K ), hen from (38) we have γ( k, k ) 2 kn( k ) for all large enough k, and hence, using (4), 2D k M( k ) k N( k ). Thus N( k ) 2DM( k ) = K k, and hence d(k k ) k. (4) 223

16 We now invoke Proposiion again, his ime choosing = k, λ = K, and ρ = 2D, o ge P {X γ(d(k ), d (2DK )) + Kd(K ) Ld (2DK )} C >, (42) whenever = k and k is large enough. However, in view of (4) he erm on he righ of he inequaliy is bounded above by k {γ + ν ( k ) ν( k )} + K k L k = k γ( k, k ) + (K L) k D k k M( k ) + (K L) k = ( 3 2 K L) k. If now we choose L = 2K we see ha (42) conradics (7); his conradicion implies ha () is in fac correc. We reached his conclusion making he addiional assumpion ha N(+) =, bu i is easy o see ha i also holds if N(+) <. In his case by an argumen we have used previously here is no loss of generaliy in aking N(+) =, so ha X is specrally negaive. We can hen repea he proof of Proposiion wih b(), he conclusion being ha for any L here eiss C = C(X, L) > wih P (X {[γ + ν (d ())] Ld ()}) C (43) for all sufficienly small. When (7) holds his clearly implies ha γ+ν () for all sufficienly small. If () were false, we would have γ + ν () DM() along some sequence k. Choosing k = /M( k ), or equivalenly k = d ( k ), (43) becomes Since P {X k { k [γ + ν ( k )] L k } C. γ + ν ( k ) L k k = γ + ν ( k ) L k M( k ) (D L) k M( k ), we again ge a conradicion by choosing L sufficienly large. This complees he proof of he equivalence of (7) and (). 224

17 As (9) obviously implies (7), our ne aim is o show he reverse implicaion, or in view of Lemma 5, ha () implies (9). Since M(+) > a firs consequence of () is ha here is a > wih A(y) > for < y, and a second is ha y A(y) as y. For δ define a funcion b δ () by b δ () = inf{ < y : A(y) y δ }. (44) Then b δ () as, and since A is coninuous, here is a > such ha A(b δ ()) = δb δ () for <. (45) Our firs aim is o show ha here is a slowly varying funcion f δ () which increases as and saisfies A(b δ ()) δ = b δ() Firs, using (5) we see ha for /2, := γ δ () f δ () for <. (46) γ δ (2) γ δ () say, where = A(b δ(2)) A(b δ ()) = + bδ (2) b δ () bδ (2) b δ () D(y)dy A(b δ ()) M(y)dy {b δ(2) b δ ()}M(b δ ()) A(b δ ()) A(b δ ()) ( ) bδ (2) b δ () = ε δ () (47) b δ () ε δ () = b δ()m(b δ ()) A(b δ ()) = M(b δ()). (48) δ From () we have ε δ () as for each fied δ, and i is also he case ha γ δ (2) γ δ () ε δ() for < 2. (49) To see his observe ha if b δ (2) b δ () b δ () hen his is immediae from (47), whereas if b δ (2) b δ () > b δ () hen γ δ (2) γ δ () = b δ(2)/2 b δ ()/ 225 = 2 bδ(2) b δ () >.

18 Ne, given < /2 choose k = k() such ha 2 (k+) < 2 k and k such ha ε δ (2 j ) /2 when j k. Applying (49) we ge for k() k γ δ () = b δ() b δ(2 k ) 2 (k+) = 2γ δ(2 k ) = 2 γ δ(2 k ) γ δ (2 (k ) ) γδ(2 (k ) ) γ δ (2 (k 2) ) γ δ (2 k) ) γ δ (2 (k ) ) γ δ(2 (k ) ) C δ k k ( ε δ (2 j )), where C δ = 2γ δ (2 (k ) ). If ln denoes he base 2 logarihm, we have k() ln /, so (46) holds wih f δ () = C δ ( ε δ (2 j )). k j ln / Clearly f δ () increases as, and if j = j() is he unique ineger wih we have ln / < j ln 2/ = + ln /, f δ () f δ (/2) = ( ε δ(2 j )), and we conclude ha f δ () is slowly varying as. (See [3], Prop.., p. 54.) If we now pu L δ () = (f δ ()) 2 when f δ (+) =, and L δ () = log / when f δ (+) <, we have ha L δ () is increasing and slowly varying as, and b δ () as. (5) L δ () Furhermore, since b δ () b () for δ, we auomaically have b δ () L () as for δ. (5) 226

19 We are now in a posiion o prove (9). We use Lemma 9 wih b = b = b δ (). Replacing M{b δ ()} by, observing ha Y (2,+) b δ ()Z, where Z is a Poisson process wih rae N{b δ ()}, and combining Y (,+) and Y (, ) we deduce ha, a.s. for each fied, X A{b δ ()} + b δ () (Z N{b δ ()}) + Y () Here he Y s and Z are independen, and EY () =, V ary () = P {Y (2, ) b δ () 2 Π(d), = } = ep M{b δ ()}. + Y (2, ). I follows from () ha M{b δ ()} = o (A{b δ ()}/b δ ()) as, and A{b δ ()} = δb δ (), so for all sufficienly small we have P (Y (2, ) So for such Chebychev s inequaliy gives ) /δ. P (X δ 2 b δ()) δ + ( P A{b δ ()} + b δ () (Z N{b δ ()}) + Y () ( δ + P b δ () (Z N{b δ ()}) + Y () δ ) 2 b δ() δ δ 2 b δ(), Y (2 ) ) = + 4{V ar[y () + b δ ()Z ]} {δb δ ()} 2. (52) Ne we noe ha for all small enough V ar[y () + b δ ()Z ] = 2 Π(d) + {b δ ()} 2 N{b δ ()} b δ () b δ () 2 Π(d) + {b δ ()} 2 T {b δ ()} = U(b δ ()) 3b δ ()A(b δ ()) = 3δ{b δ ()} 2, 227

20 where we have used (2). Puing his ino (52) gives P (X δ 2 b δ()) δ + 2δ{b δ()} 2 {δb δ ()} 2 = 3 δ. Finally, using (5) we deduce from his ha, for arbirary K > and small enough P (X KL ()) P (X δ 2 b δ()) 3 δ. Leing, hen δ, we see ha (9) holds wih d() = L (). Finally (8) clearly implies (7). On he oher hand if (7) holds he above proof shows ha wih b() = b () as defined in (44) we have A(b()) = b() for <, and by (5) X b() X L () P as. (53) Since P ( () (8) follows from (53). b()) = ep M(b()) and M(b()) = b()m(b()) A(b()) A(b()) b() = b()m(b()) A(b()), References [] Beroin, J. An Inroducion o Lévy Processes. Cambridge Universiy Press, (996). [2] Beroin, J. and Doney. R.A. (997). Spizer s condiion for random walks and Lévy Processes. Ann. Ins. Henri Poincaré, 33, 67-78, (997). [3] Bingham, N. H., Goldie, C. M., and Teugels, J. L. Regular Variaion, Cambridge Universiy Press, (987). [4] Doney, R. A. A sochasic bound for Lévy processes. Ann. Probab., o appear, (24). [5] Doney, R. A. and Maller, R. A. Sabiliy and aracion o Normaliy for Lévy processes a zero and infiniy. J. Theoreical Probab., 5, , (22). 228

21 [6] Feller, W. E. An Inroducion o Probabiliy Theory and is Applicaions, vol. 2, 2nd ediion, Wiley, New York, (97). [7] Kesen, H. and Maller, R. A. Infinie limis and infinie limi poins for random walks and rimmed sums. Ann. Probab. 22, , (994). [8] Kesen, H. and Maller, R. A. Divergence of a random walk hrough deerminisic and random subsequences. J. Theoreical Probab.,, , (997) DEPARTMENT OF MATHEMATICS UNIVERSITY OF MANCHESTER MANCHESTER M3 9PL UNITED KINGDOM rad@ma.man.ac.uk 229