Stochastic Aalysis ad Applicatios, 25: 89 03, 2007 Copyright Taylor & Fracis Group, LLC ISSN 0736-2994 prit/532-9356 olie DOI: 0.080/073629906005997 Geeralized Law of the Iterated Logarithm ad Its Covergece Rate Pigya Che Departmet of Mathematics, Jia Uiversity, Guagzhou, P.R. Chia Yogcheg Qi Departmet of Mathematics ad Statistics, Uiversity of Miesota Duluth, Duluth, Miesota, USA Abstract: This article cosiders the partial sums from a sequece of idepedet ad idetically distributed radom variables. It is well-kow that the Hartma- Witer law of the iterated logarithm holds if ad oly if the secod momet exists. This article studies the geeralized law of the iterated logarithm for the partial sums whe they are ormalized by a sequece of costats that are regularly varyig with idex /2. As a result, two equivalet coditios for the law are obtaied. Keywords: Complete covergece; Idepedet radom variables; Laws of the iterated logarithm; Sums. Mathematics Subject Classificatio 2000: 60F5; 60G50. Received December 5, 2005; Accepted August 5, 2006 The authors thak the referee for a careful readig of the mauscript ad helpful commets. Che s research was supported by the Natioal Nature Sciece Foudatio of Chia, ad Qi s research was partially supported by NSF grat DMS-060476. Address correspodece to Yogcheg Qi, Departmet of Mathematics ad Statistics, Uiversity of Miesota Duluth, SCC 40, 7 Uiversity Drive, Duluth, Miesota 5582, USA; E-mail: yqi@d.um.edu
90 Che ad Qi. INTRODUCTION Throughout the article, let X X be a sequece of idepedet ad idetically distributed radom variables, ad defie the partial sum by S = i= X i for. I the past cetury the partial sum S has bee oe of the most popular ad fudametal topics i probability. The well-kow law of the iterated logarithm due to Hartma ad Witer [] states that if the lim sup E X = 0 ad E X 2 = S 2 log log = ad lim if S 2 log log = a.s. 2 The coverse was proved by Strasse [2]. Differet proofs for the Hartma-Witer law of the iterated logarithm ad its coverse were ivestigated i Strasse [3], Heyde [4], Egorov [5], Teicher [6], Rosalsky [7], Martikaie [8], Csörgő ad Révész [9], de Acosta [0], Feller [], Heyde [2], ad Steiger ad Zaremba [3]. The covergece rate of the law of the iterated logarithm was also studied by may authors. It has bee proved that is equivalet to the followig statemet: =3 P S > { < if > 2 log log = if 0 < < 3 See, for example, Davis [4], Gut [5], ad Li et al. [6]. Ad the the equivalece of, 2, ad 3 was iterestigly established. Equatio 3 is just oe of may extesios of Hsu-Robbis complete covergece theorem. For some bibliographies see Li et al. [7]. The Hsu-Robbis theorem states that P S > < for all >0 if E X = 0 ad E X 2 <. See, for example, Hsu ad Robbis [8]. The coverse was proved by Erdős [9, 20]. A recet work by Pruss [2] studied ecessary ad sufficiet coditios for P S > a < for all >0 4 where ad a are some positive costats appropriately chose. For example, both ad a ca be take as regularly varyig fuctios of at ifiity with certai costraits.
Law of Iterated Logarithm 9 It is worth metioig the differece betwee the complete covergece results 3 ad 4. Expressio 4 is coverget for all >0 while 3 has a diverget part for some. Due to this differece, 4 geerally implies lim S /a = 0 a.s., while complete covergece results such as 3 yield the law of the iterated logarithm or its aalogue. For illustratio, take = / ad a = log /2 i 4. Pruss [2] proved the followig result: P S > log /2 < for all >0 5 =2 if ad oly if all of the followig coditios hold: a EX = 0, EX 2 / log + X <, ad b /T =2 2 < for all >0, where T = EX 2 I X log /2. It follows from our mai result i ext sectio that 5 is equivalet to lim S / log /2 = 0 with probability oe while the followig coditio P { S > log /2 < if > 0 = if 0 < < 0 =2 is equivalet to the statemet that lim sup S / log /2 = 0 ad lim if S / log /2 = 0 a.s., where 0 is a fixed umber. This gives a egative aswer to a questio raised by Pruss [2] whether coditio a is sufficiet for lim S / log /2 = 0 sice coditio a does t imply coditio b by the example give by Pruss [2]. As a matter of fact, coditio a caot eve esure the upper limit is fiite. See Remark 4 i ext sectio. The purpose of the preset paper is to ivestigate ecessary ad sufficiet coditios for a aalogue of Hartma-Witer law of the iterated logarithm: lim sup S S = a 0 ad lim if = a 0 a.s. 6 where 0 is a costat. We will maily reveal two equivalet coditios for 6 i terms of momets ad series of complete covergece. I this article, we oly cosider regularly varyig ormalizatio costats a with idex. I fact, if > 2 ad 6 holds for some fiite 0, the 0 has to be 0. Therefore, we will assume = /2. It is ot our itetio to give a descriptio for existece of costats a for 6 to hold with some fiite costat 0. Such geeral aalogues of the law of the iterated logarithm
92 Che ad Qi whe oe does ot impose ay regularity coditios o a ca be foud, for example, i Pruitt [22] or i a survey paper by Bigham [23]. For a survey of recet developmet ad some ew results o coditios that esure 6 see Eimahl ad Li [24]. A recet work by Li et al. [25] is also closely related to the preset article. See Remark 3 i ext sectio. We should metio that whe oe does ot impose the regularity o a, the value of 0 i 6 is hard to determie i geeral. 2. MAIN RESULTS I this article, we assume that A x = x /2 l x, where l x is positive ad mootoe fuctio that is slowly varyig at ifiity ad lim x l x =. Let A deote the iverse of A. Set a = A ad T = E X 2 I X a for. For coveiece assume A x > 0 for all x>0. Theorem. Assume that EX = 0 EA X < 7 ad for some 0 0 { } { exp 2 a 2 < if > 0 2T = if 0 < < 0 8 The { < if > P S > a 0 = if 0 < < 0 9 { P max S < if > k > a 0 0 k = if 0 < < 0 ad 6 holds. Corollary. Theorem 2. equivalet: If 7 holds, the 6 holds with { 0 = if { } } exp 2 a 2 < 2T Assume 0 0. The the followig three statemets are E 7 ad 8 hold;
Law of Iterated Logarithm 93 { E2 < if > P S > a 0 = if 0 < < 0 S E3 6 holds, that is, lim sup S = a 0 ad lim if = a 0 a.s. Remark. Whe the geeralized law of the iterated logarithm 6 holds, the the limit poit set of S /a is 0 0. Remark 2. Set A x = 2x log log x, x>0, ad 0 =. The E2 ad E3 become 3 ad 2, respectively. It is easy to see that is equivalet to 7 ad 8. Therefore, our Theorem 2 yields the Hartma- Witer law of the iterated logarithm. Remark 3. Li ad Tomkis [26] also discussed the aalogous law of the iterated logarithm. See their Theorem 2. They gave some sufficiet coditios for 6 to hold. For some more restrictive slowly varyig fuctios l x, Eimahl ad Li [24] proved the ecessary ad sufficiet coditios for 6 ad provided some simpler way to determie the costat 0. Li et al. [25] obtaied the equivalece betwee E2 ad E3 as well uder some mild coditios. The followig example shows that for a large class of fuctios A there exist distributios such that 7 holds ad 8 holds with some 0 0. Hece, for these distributios o-trivial aalogous law of the iterated logarithm holds. Pruss [2] costructed a similar example for the special case = to show that coditio a itroduced i Sectio does ot imply coditio b. Example. Let K m = it exp { } 2 m e 2m for m, where it x deotes the iteger part of x. Let A t = t /2 log t /2 for t>2, where 0 < 2. Exted A liearly to the iterval 0 2 such that A 0 = 0. Let X be a symmetric radom variable such that P X = A K m = P X = A K m = 2 m Km ad P X = 0 = m= P X =A K m. It is plai to see that E A X = m= Note that a = A. For each, defie The as T = M m= M 2 m K m A K m 2 = 2 m K m K m = M = max m K m m= 2 m log K m 2 M log K M
94 Che ad Qi Notice that M = m if K m <K m+. For ay fixed umbers 0 < < < 2, we have for all large m ad K m+ =K m K m+ =K m K m+ e 2a2 / 2T < =K m K m+ e 2a2 / 2T > By usig that as t t =K m e x dx e t t we have as log y ad 0 that y e 2 log 2 m log K m e 2 2 log 2 m log K m 2 dx = / e x e log x x dx e log y log y / log y Sice for every >0 K we coclude that x e log x dx < e log < =K K x e log x dx e log e log K log K =K as log K ad 0. Now we set = 2 i 2m log K m. From the above equatio it is easy to see that e 2 i log 2 m log K m = o e 2 i log 2 m log K m =K m+ =K m Therefore, as m K m+ =K m e 2 i log 2 m log K m e 2 i log 2 m log K m =K m 2 i 2 m e 2 i 2m log K m 2 i e 2 i 2m
Law of Iterated Logarithm 95 The for ay >, choose a such that < <. The from ad the above equatio we have =K e 2a2 / T = m= K m+ =K m which is domiated, up to a costat scale, by m= e 2 2m Likewise, we ca show by usig 2 that =K e 2a2 < / 2T = e 2a2 / T for every 0. Therefore, the distributios defied i this example satisfy coditio 7 ad coditio 8 with 0 =. Remark 4. By redefiig K m i the example above, for ay 0 0, we ca fid a distributio such that both 7 ad 8 hold. For example, if we defie K m = it exp { } 2 m e m2m, the 7 holds, ad 8 holds with 0 =, that is { } exp 2 a 2 = 2T for all >0 3. PROOFS I this sectio, let x be the stadard ormal distributio fuctio, that is, x = 2 x e t2 /2 dt For sake of simplicity, we use C to deote some fiite costat that ca be differet as it appears at differet places. We preset several lemmas before we proceed to give the proofs for the two theorems i Sectio 2. Lemma. Assume 0 0. The 8 is equivalet to { a < if > 0 T = if 0 < < 0 3
96 Che ad Qi Proof. We oly prove the lemma whe 0 0. The proofs are similar whe 0 = 0or. Note that as x x 2 x e x2 2 = 2 e x2 2 log x Sice log x x 2 0asx, we have for ay three fixed costats 0 < < < 2 <, ifx is large eough, say x>x 0 e 2 2 x2 /2 < x < e 2 x2 /2 Thus, there exist costats C ad C 2 which deped o, 2, ad such that C e 2 2 x2 /2 < x < C 2 e 2 x2 /2 for all x>0 Hece, we have C exp{ 2 2 a2 / 2T } a < <C 2 T exp{ 2 a2 / 2T } from which Lemma follows. Lemma 2 Petrov [27], Theorem 5.6. Let Y Y be a sequece of idepedet idetically distributed radom variables with EY = 0 EY 2 = 2 > 0 E Y 3 < ad put W = i= Y i. The for all ad for all x P W <x x C E Y 3 3 + x 3 Lemma 3. For every s t > 0, P max S P S k >s+ t >t k max k P S S k < s Proof. Apply similar argumet as Lemma 6.2 of Ledoux ad Talagrad [28]. We omit the details. Lemma 4. Assume 7 holds. The S a 0 i probability
Law of Iterated Logarithm 97 Proof. Similar to that of Lemma i Pruss [2]. We omit the details. Lemma 5. If g x is regularly varyig with idex, the g x has the followig represetatio { x g x = c x x exp } b t dt t where lim x c x = c 0 ad lim x b x = 0. Proof. See Bigham et al. [29]. Lemma 6. For all k =k a 3 Cka 3 k Proof. Sice x 5/4 /A 3 x is regularly varyig with idex /4, we have from Lemma 5 that there exists a costat c such that x 5/4 A 3 x cy 5/4 A 3 y for all x y a. Therefore, 5/4 a 3 ck 5/4 a 3 k for all k. Thus we obtai for every k =k a 3 = =k 5/4 a 3 5/4 ck 5/4 a 3 k =k Cka 3 5/4 k Proof of Theorem. We oly prove the theorem for 0 0. Set W = i= X ii X i a E XI X a ad 2 = Var XI X a for. Defie a 0 = 0. For simplicity we assume 2 > 0. First we will prove that as a EXI X a 0 4 Sice x/a x is regularly varyig with idex, we have x/a x Ca / for all x>a by Lemma 5. Hece, as a EXI X a a E X I X >a We shall prove that for ay 0 P W > a = a EA X X A X I X >a CEA X I X >a 0 a < 5 2 /2
98 Che ad Qi I fact, for every fixed 0 we have from Lemma 2 P W a > a 2 /2 = P W a a 2 /2 C a 3 E XI X a EXI X a 3 C a 3 E X 3 I X a Ca 3 a 3 k P a k < X a k k= which coupled with Lemma 5 yields P W a > a 2 /2 C a 3 k P a k < X a k k= a 3 k= = C a 3 k P a k < X a k C =k a 3 kp k <A X k CE A X < k= Note that 2 T as. Uder assumptios 7 ad 8 we have from 5 ad Lemma that { < if > P W > a 0 = if 0 < < 0 which is equivalet to { < P X i I X i a > a = i= if > 0 if 0 < < 0 6 by virtue of 4. Uder coditio 7, 9, ad 6 are equivalet sice P X i I X i a > a P S > a P X >a i= ad P X i I X i a > a P S > a i= P X >a = E A X <
Law of Iterated Logarithm 99 The divergece part i 0 follows trivially from that i 9. For ay fixed > 0, set = 0 /2. By usig Lemmas 3 ad 4 we have for all large P max S k a 2P S + 0 a k Thus, the covergece part i 0 also follows from 9. Now we shall prove 6. We prove first It suffices to show that ad lim sup lim sup lim sup S a = 0 a.s. 7 S a 0 a.s. 8 S a 0 a.s. 9 To prove 8, by Borel Catelli lemma, it suffices to show that for every >0, there is a icreasig sequece m m of itegers such that m= S P max m < m a 0 + < I fact, by settig m = it + /2 m/3 we have for some iteger m 0, + 4 /3 m m + /3 m ad a m + /3 a m for all m m 0. Hece, we get from 0 that > P max S k 0 + /3 a k =2 = P max S k 0 + /3 a k m= m < m m P max S k 0 + /3 a m m 2 <k m = 4 m=m 0 + m < m m=m 0 + m=m 0 P So we have obtaied 8. m m m P max S k 0 + a m 2 m 2 <k m S max m < m a 0 +
00 Che ad Qi Now we prove 9. By Theorems 7.5 ad 7.6 i Petrov [27], to prove 2.8, it suffices to show that for every >0, there exists a odecreasig sequece m m with m such that P S m S m 0 a m = 20 m=2 Now fix 0. Let be a large umber to be chose later. For each let I deote the set of all itegers betwee ad, that is, I = i <i. Let j be the iteger i I such that P S j 0 /2 a j = max j I P S j 0 /2 a j. The from the divergece part of 9 we have P S j 0 /2 a j = from which we kow either or P S j2+ 0 /2 a j2+ = 2 P S j2 0 /2 a j2 = Without loss of geerality, assume 2 holds. Set m = j 2m+ for m. The m m. From the proof of 8 we ca easily coclude that for all large, m= P S m > 0 + a m <, which implies m= P S > m 0 + /2 a m < sice a m /2 a. Choose a m large 4 + 2 2, ad thus + /2 /2. The we coclude 20 from 2. This proves 9. By substitutig X for X i 7, we have lim if S a = 0 a.s. This completes the proof of Theorem. Proof of Theorem 2. From Theorem, we have proved that E E3. By substitutig S for S i 9 we also have E E2. To complete the proof of Theorem 2, we eed to show that E2 E ad E3 E. I fact, from Theorem, it suffices to prove 7 for each case sice 8 is ecessary for E2 or E3 to hold with the same 0 uder coditio 7.
Law of Iterated Logarithm 0 First assume E3 holds. We have lim sup X a lim sup S a + lim sup S a a a 2 0 By Borel Catelli lemma, this implies P X > 3 0 a < which is equivalet to EA X < from regularity of A. Furthermore, the fiiteess of EA X implies the fiiteess of all rth momets of S X with r 0 2. By the strog law of large umbers, E X a.s., while from E3, we have S 0 a.s. Therefore, we get EX = 0. This proves 7. Now assume E2 holds. Let m Y deote the media of radom variable Y. The it follows from Rogozi [30] that P max X k m X >x 8P S m S > x k 4 for all x>0, which together with the symmetrizatio iequality see, e.g., Lemma 6.6 i Petrov [27] yields P max X k m X >x 32P S > x k 8 for all x>0. Hece for all x>2 m X we have P max X k >x 32P k S > x 6 The E2 implies that P max k >c 0 a < k where c 0 = 64 0. Let p = P X >c 0 a. The p 0as, ad P max k X k >c 0 a = p. Easily it is verified that 2 p if p < 4 p 4 if p 4 from which oe ca coclude that P max k X k >c 0 a = p C mi p for some C>0 ad all. Hece, we have mi p <. This is equivalet to p < by Propositio of Pruss [2]. Thus, we have EA X <. The same argumet as i the proof from E3 to E applies to get EX = 0. The proof of the theorem is complete. a.s.
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