aa r X i v : . [ m a t h . P R ] A ug BAUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD
RICH ´ARD BALKA AND TIBOR T ´OM ´ACS
Abstract.
Let { X n } n ≥ be either a sequence of arbitrary random variables,or a martingale difference sequence, or a centered sequence with a suitablelevel of negative dependence. We prove Baum–Katz type theorems by onlyassuming that the variables X n satisfy a uniform moment bound condition. Wealso prove that this condition is best possible even for sequences of centered,independent random variables. This leads to Marcinkiewicz–Zygmund typestrong laws of large numbers with estimate for the rate of convergence. Introduction
Motivation and related results.
Let { X n } n ≥ be a sequence of randomvariables, we always assume that they are defined on the same probability space.For all n ∈ N + let S n = P ni =1 X i and M n = max ≤ i ≤ n | S i | . For some positiveparameters p, r consider the statement(M) ∞ X n =1 n p/r − P ( M n > εn /r ) < ∞ for all ε > , and the weaker claim(S) ∞ X n =1 n p/r − P ( | S n | > εn /r ) < ∞ for all ε > . The main goal of the paper is to prove (M) or (S) under different conditions. Wemay assume that 0 < r < p ≥ r . Indeed, if p < r then (M) trivially holds.If p ≥ r ≥ ε > { X n } n ≥ is an i.i.d. sequence with mean zero and finite variance.We cite the known results in this subsection, for the new ones see Subsection 1.2.First consider the classical results for i.i.d. random variables. Following Hsu andRobbins [13] we say that a sequence { X n } n ≥ converges completely to 0 if ∞ X n =1 P ( | X n | > ε ) < ∞ for all ε > . By the Borel–Cantelli Lemma this implies that X n → { X n } n ≥ is a centered i.i.d. sequence of randomvariables then S n /n → S n /n converge completely to 0? Hsu and Robbins [13] showedthat E ( X ) < ∞ is sufficient, and Erd˝os [10, 11] proved that it is necessary. Mathematics Subject Classification.
Primary: 60F15, 60G42, 60G50; Secondary: 60F10.
Key words and phrases. complete convergence, Marcinkiewicz–Zygmund strong law of largenumbers, rate of convergence, independent random variables, martingale difference sequences.The first author was supported by the National Research, Development and Innovation Office–NKFIH, Grant 104178.
Theorem 1.1 (Hsu–Robbins–Erd˝os strong law of large numbers) . Let { X n } n ≥ bea sequence of centered i.i.d. random variables. Then the following are equivalent:(i) E ( X ) < ∞ ,(ii) P ∞ n =1 P ( | S n | > εn ) < ∞ for all ε > . A more general classical theorem is the following.
Theorem 1.2 (Baum–Katz, Chow) . Let { X n } n ≥ be a sequence of i.i.d. randomvariables. Let < r < and let p ≥ r . The following statements are equivalent: E ( | X | p ) < ∞ and if p ≥ then E ( X ) = 0 , (1) ∞ X n =1 n p/r − P ( | S n | > εn /r ) < ∞ for all ε > , (2) ∞ X n =1 n p/r − P ( M n > εn /r ) < ∞ for all ε > . (3)The equivalence of (1) and (2) in the case r = p = 1 is due to Spitzer [24], thecase r = 1, p = 2 is the Hsu–Robbins–Erd˝os strong law, while the general case isdue to Baum and Katz [2]. For the equivalence of (1) and (3) see Chow [5].The next theorem is the Marcinkiewicz–Zygmund strong law of large numbers,see [20]. Note that the case p = 1 dates back to Kolmogorov [15] and includes theclassical strong law of large numbers. Theorem 1.3 (Marcinkiewicz–Zygmund strong law of large numbers) . For an i.i.d.sequence of random variables { X n } n ≥ and < p < the following are equivalent:(i) E ( | X | p ) < ∞ and if p ≥ then E ( X ) = 0 ,(ii) lim n →∞ n − /p S n = 0 almost surely. The following statement explains the connection between Theorem 1.2 and theMarcinkiewicz–Zygmund strong law of large numbers and its rate of convergence. Itis formulated for arbitrary sequences of random variables, see [7, Remarks 1 and 2]and see also [17, Lemma 4] for the proof of part (ii).
Statement 1.4.
Let { X n } n ≥ be an arbitrary sequence of random variables, let < r < and let p ≥ r . Assume that (M) holds.(i) If p = r then for all ε > we have ∞ X n =1 P ( M n > ε n/p ) < ∞ , which implies that lim n →∞ n − /p S n = 0 almost surely.(ii) If p > r then for all ε > we have ∞ X n =1 n p/r − P (cid:18) sup k ≥ n k − /r | S k | > ε (cid:19) < ∞ . Since the above probabilities are non-increasing, we obtain that P (cid:18) sup k ≥ n k − /r | S k | > ε (cid:19) = o ( n − p/r ) as n → ∞ . AUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD 3
In contrast to Theorems 1.2 and 1.3, we will not assume independence or iden-tical distributions in the following. Now we summarize the known results in thisdirection, which requires some technical definitions.The following theorem partly generalizes the Marcinkiewicz–Zygmund stronglaw of large numbers, see Stout [27, Theorem 3.3.9] and [27, Corollary 3.3.5]. It isbased on Chung [6] in the case of independent variables and is implicitly containedin Lo`eve [19]. Its proof uses a conditional three series theorem [27, Theorem 2.8.8].
Theorem 1.5 (Stout) . Let f : [0 , ∞ ) → R + be a non-decreasing function with ∞ X n =1 f (2 n ) < ∞ . Let < p < . Let { X n } n ≥ be(i) an arbitrary sequence of random variables if < p < and suppose that x x p − f ( x ) is non-increasing,(ii) a martingale difference sequence if ≤ p < and assume that x x p − f ( x ) is non-increasing.If sup n ≥ E ( | X n | p f ( | X n | )) < ∞ then lim n →∞ n − /p S n = 0 almost surely. Definition 1.6.
We say that the sequence { X n } n ≥ is weakly dominated by arandom variable X if there is a constant C ∈ R + such that for all n ∈ N + and x > P ( | X n | > x ) ≤ C P ( | X | > x ) , and weakly mean dominated if(WMD) 1 n n X k =1 P ( | X k | > x ) ≤ C P ( | X | > x )for some C ∈ R + and for all n ∈ N + and x > Definition 1.7.
A finite family of random variables { X i : 1 ≤ i ≤ n } is called negatively associated (NA) if for every pair of disjoint subsets A , A ⊂ { , . . . , n } we have Cov( f ( X i : i ∈ A ) , f ( X j : j ∈ A )) ≤ f and f for which the covarianceexists. An infinite family of random variables is NA if every finite subfamily is NA.The following definition is due to Lehmann [18]. Definition 1.8.
Two random variables X and Y are called negatively quadrantdependent (NQD) if for all x, y ∈ R we have P ( X ≤ x, Y ≤ y ) ≤ P ( X ≤ x ) P ( Y ≤ y ) . Every independent sequence is NA, and each pairwise independent sequence ispairwise NQD. It is proved in [14] that every NA sequence is pairwise NQD. Forthe next theorem see Kuczmaszewska [16, Theorem 2.1] and its proof.
RICH´ARD BALKA AND TIBOR T ´OM´ACS
Theorem 1.9 (Kuczmaszewska) . Let < r < and p > r . Assume that thesequence { X n } n ≥ is weakly mean dominated by a random variable X satisfying E ( | X | p ) < ∞ . If p > assume that { X n } n ≥ is centered and NA. Then (M) holds. Remark 1.10. If p = 1 we obtain the above theorem by applying the Markovinequality P ( Z > t ) ≤ E ( Z q ) /t q at the beginning of [16, (2.4)] for some q > Theorem 1.11 (Wu, Tan–Wang–Zhang, Gan–Chen) . Let < r ≤ p and ≤ p < .Let { X n } n ≥ be a centered, pairwise NQD sequence which is weakly dominated bya random variable X with E ( | X | p ) < ∞ . Then (S) holds. If r = p then (M) holds. Now we state the last two results of this subsection, which consider martingaledifference sequences (MDS). Miao, Yang, and Stoica [21, Theorems 2.1 (1), 2.3]proved the following theorem about the case 1 ≤ p < Theorem 1.12 (Miao–Yang–Stoica) . Assume that < r ≤ p < . Let { X n } n ≥ be a MDS which is weakly mean dominated by X . Property (M) holds if(i) r = p = 1 and E ( | X | log + | X | ) < ∞ ,(ii) < p < and E ( | X | p ) < ∞ . Remark 1.13.
Note that (i) is optimal: Elton [9] proved that if X is a centeredrandom variable with E ( | X | log + | X | ) = ∞ then there is a martingale differencesequence { X n } n ≥ such that X n and X have the same distribution for all n and S n /n → ∞ as n → ∞ almost surely. Thus (M) cannot hold by Statement 1.4 (i).The above theorem generalizes a result of Dedecker and Merlev´ede [7, Theorem 5]for real valued random variables. In the case p ≥ Theorem 1.14 (Miao–Yang–Stoica) . Assume that < r < ≤ p and q ( r, p ) =2( p − r ) / (2 − r ) . Let { X n } n ≥ be a MDS such that sup n ≥ E ( | X n | q ) < ∞ .(i) If q > q ( r, p ) then (M) holds.(ii) If q = q ( p, r ) then there is a MDS { X n } n ≥ which is weakly dominated by arandom variable X satisfying E ( | X | q ) < ∞ such that (S) does not hold. The results of the paper.
The goal of the paper is to investigate statements(M) and (S) for arbitrary random variables, martingale difference sequences, andcentered sequences with a certain level of negative dependence. We will deduce(M) by assuming only a uniform moment condition sup n ≥ E ( | X n | q f ( | X n | )) < ∞ ,so in contrast to Theorems 1.9, 1.11, and 1.12 properties (WD) and (WMD) willnot be assumed. We will find the smallest possible suitable constant q = q ( p, r )which we call the critical exponent . In particular, we generalize Theorems 1.14 and1.5. We will be also able to determine the precise smaller order term f . Similarlyto Theorem 1.5 the function f : [0 , ∞ ) → R + might be any non-decreasing functionsatisfying P ∞ n =1 /f (2 n ) < ∞ , and the finiteness of the sum is really necessary evenfor (S). By Corollary 3.2 we may assume that f ( n ) = n o (1) as n → ∞ , see alsoRemark 1.18 for the least possible order of magnitude of f . Stoica claimed similartheorems for martingale difference sequences in [25] and [26], but those results areincorrect. He stated in [25] that if 0 < r < < p and { X n } n ≥ is a MDS with AUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD 5 sup n ≥ E ( | X n | p ) < ∞ then (S) holds. This was disproved in [24], see Theorem 1.14.Theorems 1 and 2 in [26] state that if 1 ≤ r ≤ p < { X n } n ≥ is a MDS withsup n ≥ E ( | X n | p log + | X n | ) < ∞ then (S) holds. Theorem 6.5 below witnesses thatthis is not true even for independent, centered sequences of random variables.The following theorem is one of the most important results in the paper. Theorem 1.15.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) < ∞ . Let < r < and let p ≥ r . Let { X n } n ≥ be a(i) sequence of arbitrary random variables if < r ≤ p ≤ and r < ,(ii) martingale difference sequence if < p ≤ or r = p = 1 ,(iii) centered, negatively associated sequence of random variables if p ≥ .If sup n ≥ E ( | X n | p f ( | X n | )) < ∞ then for all ε > we have ∞ X n =1 n p/r − P ( M n > εn /r ) < ∞ . We will prove the above theorem in several steps. Theorem 4.1 implies (i),Theorems 5.1 and 5.2 yield (ii), and (iii) is stated as Theorem 6.1.The next theorem is analogous to Theorem 1.11 with a similar proof. Thus,instead of proving it, we suggest the reader to follow [28, Theorems 1.1, 1.2].
Theorem 1.16.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) < ∞ . Let < r ≤ p and ≤ p < . If { X n } n ≥ is a centered, pairwise NQD sequencewith sup n ≥ E ( | X n | p f ( | X n | )) < ∞ then (S) holds. If r = p then (M) holds, too. The following theorem shows that the moment conditions above are sharp evenfor independent, centered random variables, even if r = p and we want to obtainonly lim n →∞ n − /p S n = 0 almost surely, recall Statement 1.4 (i). Theorem 6.5.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) = ∞ . Let < r < and let p ≥ r . Then there exists a sequence of independent, centeredrandom variables { X n } n ≥ such that sup n ≥ E ( | X n | p f ( | X n | )) < ∞ and ∞ X n =1 n p/r − P ( | S n | > n /r ) = ∞ . Moreover, if r = p then lim sup n →∞ n − /p S n ≥ almost surely. RICH´ARD BALKA AND TIBOR T ´OM´ACS
First consider Theorem 1.15 (i). In the case of arbitrary random variables weneed to suppose that r <
1. Indeed, for 1 ≤ r ≤ p let X n ≡ n . Thensup n ≥ E ( | X n | q ) = 1 for all q > ∞ X n =1 n p/r − P ( | S n | > (1 / n /r ) = ∞ X n =1 n p/r − ≥ ∞ X n =1 n − = ∞ . Theorem 1.15 (i) easily follows from the following, more general theorem.
Theorem 4.1.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) < ∞ . Let < r < and let p ≥ r , and define q = q ( r, p ) = max { p, ( p − r ) / (1 − r ) } . Assumethat { X n } n ≥ is a sequence of random variables with sup n ≥ E ( | X n | q f ( | X n | )) < ∞ .Then for all ε > we have ∞ X n =1 n p/r − P ( M n > εn /r ) < ∞ . We prove that the above theorem is sharp. By Theorem 6.5 it is enough toconsider the case p ≥ Theorem 4.2.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) = ∞ . Let < r < ≤ p and let q = q ( r, p ) = ( p − r ) / (1 − r ) . Then there is a sequenceof random variables { X n } n ≥ such that sup n ≥ E ( | X n | q f ( | X n | )) < ∞ and ∞ X n =1 n p/r − P ( | S n | > n /r ) = ∞ . We prove Theorem 1.15 (ii) for p < p ≥ p = 2. Theorem 5.2.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) < ∞ . Let < r < ≤ p and let q = q ( r, p ) = 2( p − r ) / (2 − r ) . Let { X n } n ≥ be amartingale difference sequence such that sup n ≥ E ( | X n | q f ( | X n | )) < ∞ . Then forall ε > we have ∞ X n =1 n p/r − P ( M n > εn /r ) < ∞ . The following theorem witnesses that the above result is best possible.
AUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD 7
Theorem 5.4.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) = ∞ . Let < r < ≤ p and let q = q ( r, p ) = 2( p − r ) / (2 − r ) . Then there is a martingaledifference sequence { X n } n ≥ such that sup n ≥ E ( | X n | q f ( | X n | )) < ∞ and ∞ X n =1 n p/r − P ( | S n | > n /r ) = ∞ . Miao, Yang, and Stoica proved that the threshold in Theorems 5.2 and 5.4 is at q ( r, p ) = 2( p − r ) / (2 − r ), recall Theorem 1.14. We will improve their methods inorder to find the precise smaller order term.Theorem 1.15 (iii) is stated as Theorem 6.1, which will simply follow from aninequality of Shao [23]. If 0 < r < { X n } n ≥ is centered from Theorems 1.16 and 6.1. Corollary 6.3.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) < ∞ . Let < r < ≤ p , and let { X n } n ≥ be a sequence of(1) pairwise NQD random variables if ≤ p < ,(2) negatively associated random variables if p ≥ .Assume that sup n ≥ E ( | X n | p f ( | X n | )) < ∞ . Then for all ε > we have ∞ X n =1 n p/r − P ( M n > εn /r ) < ∞ . The above theorems witness that if 0 < r < < p then the critical exponents forindependent centered sequences and martingale difference sequences are different,since p < p − r ) / (2 − r ). See Table 1 for the values of the critical exponents. Table 1.
The critical exponents for different intervals of p andtypes of sequences. ICS and MDS denote independent, centeredsequences, and martingale difference sequences, respectively.ICS MDS Arbitrary Sequences p ≤ p p p if r < < p ≤ p p ( p − r ) / (1 − r ) if r < p > p p − r ) / (2 − r ) ( p − r ) / (1 − r ) if r < p ≥ Problem 1.17.
Let < r < ≤ p and let k ≥ be an integer. Let { X n } n ≥ be asequence of k -wise independent, centered random variables. Do there exist resultssimilar to Theorem 5.2 (replace M n by | S n | if necessary) and Theorem 5.4 withsome q = q ( r, p, k ) ? If yes, is it true that q ( r, p, k ) = p for all r, p, k ? RICH´ARD BALKA AND TIBOR T ´OM´ACS
In fact, for some values of r, p, k we can show that q ( r, p, k ) = p is the critical ex-ponent for (S). The following remark is about the least possible order of magnitudeof f in the above theorems. Remark 1.18.
Let log + ( x ) = max { , log x } for x > + (0) = 1. For k ∈ N + let log + k ( x ) denote the k th iteration of log + ( x ). For m ∈ N + and ε > f m , f m,ε : [0 , ∞ ) → R + as f m ( x ) = m Y k =1 log + k ( x ) ,f m,ε ( x ) = f m ( x ) (cid:0) log + m ( x ) (cid:1) ε . It is easy to see that for all m ∈ N + and ε > ∞ X n =1 f m (2 n ) = ∞ and ∞ X n =1 f m,ε (2 n ) < ∞ . In Section 2 we recall some definitions and easy facts. In Section 3 we prove anumber of technical lemmas. Section 4 is devoted to arbitrary random variables. InSection 5 we prove our theorems about martingale difference sequences. Finally, inSection 6 we verify Theorems 6.1 and 6.5, and Corollary 6.3. Note that the proofsafter Section 3 can be read independently of each other.2.
Preliminaries
Let { X n } n ≥ be a sequence of random variables defined on the probability space(Ω , F , P ). It is a martingale difference sequence if there is a filtration {F n } n ≥ suchthat F = {∅ , Ω } , X n is measurable with respect to F n , and E ( X n | F n − ) = 0 forall n ∈ N + . We may assume without loss of generality that F n = σ ( X , . . . , X n ) isthe σ -algebra generated by X , . . . , X n for all n ∈ N + . A random variable is called centered if E ( X ) = 0.Let E ⊂ R and let f : E → R . We say that f is non-decreasing (or increasing )if for all x, y ∈ E , x < y we have f ( x ) ≤ f ( y ) (or f ( x ) < f ( y )). We can similarlydefine the notions non-increasing and decreasing , and if E = N + then our definitionsextend to sequences as well. Let I ( A ) denote the indicator function of an event A . We use the notation a . b if a ≤ cb with some c ∈ R + , where c dependsonly on earlier fixed constants. The notation a n = o ( b n ) as n → ∞ means thatlim n →∞ a n /b n = 0. We need the following facts. Fact 2.1.
Let f : [0 , ∞ ) → R + be a non-decreasing function. Then the followingstatements are equivalent:(i) P ∞ n =1 /f (2 cn ) < ∞ for some c > ,(ii) P ∞ n =1 /f ( ε cn ) < ∞ for all ε, c > ,(iii) P ∞ n =1 / ( nf ( n c )) < ∞ for some c > ,(iv) P ∞ n =1 / ( nf ( εn c )) < ∞ for all ε, c > . The equivalence (i) ⇔ (iii) above follows from the equiconvergence of the series P ∞ n =1 a n and P ∞ n =1 n a n for any non-increasing, positive sequence { a n } n ≥ . Easycomparison implies the equivalences (i) ⇔ (ii) and (iii) ⇔ (iv). AUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD 9
Fact 2.2.
Let { X n } n ≥ be a sequence of random variables and let g, h : [0 , ∞ ) → R + be non-decreasing functions such that lim sup x →∞ h ( x ) g ( x ) < ∞ . Then sup n ≥ E ( g ( | X n | )) < ∞ implies that sup n ≥ E ( h ( | X n | )) < ∞ . The concept of martingale and the following inequality are due to J. L. Doob,see e.g. [8, Theorem 5.4.2].
Theorem 2.3 (Doob’s inequality) . Let { X i : 1 ≤ i ≤ n } be a finite martingaledifference sequence and let p ≥ . Then for all t > we have P ( M n ≥ t ) ≤ E ( | S n | p ) t p . Technical lemmas
Lemma 3.1.
Let { a n } n ≥ be a positive, non-increasing sequence such that ∞ X n =1 a n < ∞ . Then there is a non-increasing sequence { b n } n ≥ such that(1) b n ≥ a n for all n ≥ ,(2) lim n →∞ b n /b n − = 1 ,(3) P ∞ n =1 b n < ∞ .Proof. Fix a positive sequence { c k } k ≥ such that c k ր
1. We can choose anincreasing sequence of positive integers { n k } k ≥ such that for all k ∈ N + we have(3.1) ∞ X n = n k a n < − k (1 − c k +1 )and(3.2) c n k +1 − n k k ≤ − k (1 − c k +1 ) . We will construct b n recursively. Let b n = a n if n ≤ n . For every k ∈ N + and n k < n ≤ n k +1 define b n = max { a n , c k b n − } . Then clearly (1) holds and b n ≤ b n − for all n ≤ n . Assume n k < n ≤ n k +1 forsome k . Then a n ≤ a n − ≤ b n − and our definition imply that b n = max { a n , c k b n − } ≤ max { b n − , c k b n − } = b n − , so b n is non-increasing.Now we show (2). Assume that n > n k . Let n m < n ≤ n m +1 for some m ≥ k .As the sequence c k is monotone increasing, we have c k b n − ≤ c m b n − ≤ b n ≤ b n − , so c k ≤ b n /b n − ≤
1. Then c k ր n ≥ n define d n,n = a n and for i ≥ n +1 recursivelydefine d n,i = c k d n,i − if n k < i ≤ n k +1 . Let ℓ ≥ n be fixed. Let n = n ( ℓ ) be the largest integer such that n ≤ n ≤ ℓ and b n = a n . As b n = a n , we obtain that n exists. Then b ℓ = d n,ℓ by our definitions.As the map ℓ d n ( ℓ ) ,ℓ is clearly one-to-one, we have(3.3) X n ≥ n b n ≤ X n ≥ n ∞ X i = n d n,i . Fix k, n ∈ N + such that n k < n ≤ n k +1 . By definition n k +2 X i = n d n,i = a n n k +1 − n X i =0 c ik + c n k +1 − nk n k +2 − n k +1 X i =1 c ik +1 ! ≤ a n (cid:18) − c k + 11 − c k +1 (cid:19) ≤ a n − c k +1 . (3.4)For each j ≥ d n,i and (3.2) imply that n k + j +1 X i = n k + j +1 d n,i = a n c n k +1 − nk j − Y i =1 c n k + i +1 − n k + i k + i ! n k + j +1 − n k + j X ℓ =1 c ℓk + j ≤ a n c n k + j − n k + j − k + j − (1 − c k + j ) − ≤ a n − k − j +1 . (3.5)By (3.4) and (3.5) for all n ≥ n we obtain(3.6) ∞ X i = n d n,i ≤ a n − c k +1 + a n ∞ X j =2 − k − j +1 ≤ a n − c k +1 . Therefore (3.3), (3.6), and (3.1) imply that X n ≥ n b n ≤ ∞ X k =1 n k +1 X n = n k +1 ∞ X i = n d n,i ≤ ∞ X k =1 n k +1 X n = n k +1 a n − c k +1 ≤ ∞ X k =1 − k < ∞ . Thus (3) holds, and the proof is complete. (cid:3)
Corollary 3.2.
Let g : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 g (2 n ) < ∞ . Then there is a non-decreasing function f : [0 , ∞ ) → R + such that(i) P ∞ n =1 /f (2 n ) < ∞ ,(ii) lim n →∞ f (2 n +1 ) /f (2 n ) = 1 ,(iii) lim sup x →∞ f ( x ) /g ( x ) ≤ .Proof. First we define f (2 n ) for all n ∈ N + . We apply Lemma 3.1 for the sequence a n = 1 /g (2 n ), let { b n } n ≥ be a sequence satisfying properties (1)–(3) in Lemma 3.1.Define f (2 n ) = 1 /b n for all n ∈ N + . Then (i) holds by (3), and by (2) for all n wehave(3.7) lim n →∞ f (2 n +1 ) f (2 n ) = lim n →∞ b n b n +1 = 1 , AUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD 11 so (ii) is satisfied. Since b n is non-increasing, the sequence { f (2 n ) } n ≥ is non-decreasing. Let f : [0 , ∞ ) → R + be any non-decreasing function extending thesequence { f (2 n ) } n ≥ . Clearly for all n ∈ N + we have(3.8) f (2 n ) = 1 b n ≤ a n = g (2 n ) . Thus monotonicity, (3.8), and (3.7) imply that for all n ∈ N + and 2 n ≤ x ≤ n +1 we have f ( x ) g ( x ) ≤ f (2 n +1 ) g (2 n ) ≤ f (2 n +1 ) f (2 n ) → n → ∞ . The proof is complete. (cid:3) Lemma 3.3.
Let g : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 g (2 n ) < ∞ . Then there is a non-decreasing function f : [0 , ∞ ) → R + such that(i) P ∞ n =1 /f (2 n ) < ∞ ,(ii) lim n →∞ f (2 n +1 ) /f (2 n ) = 1 ,(iii) lim sup x →∞ f ( x ) /g ( x ) ≤ ,(iv) f has continuous second derivative on (0 , ∞ ) ,(v) f ′ ( x ) /f ( x ) = o (1 /x ) as x → ∞ ,(vi) f ′′ ( x ) /f ( x ) = o (1 /x ) as x → ∞ .Proof. By Corollary 3.2 we may assume that(3.9) lim n →∞ g (2 n +1 ) g (2 n ) = 1 . Let f (2 n ) = g (2 n ) for all n ∈ N + , then clearly (i) and (ii) hold. Each non-decreasingfunction f : [0 , ∞ ) → R + extending the sequence { f (2 n ) } n ≥ satisfies (iii). Indeed,monotonicity, f (2 n +1 ) = g (2 n +1 ), and (3.9) imply that for every n ∈ N + and2 n ≤ x ≤ n +1 we have f ( x ) g ( x ) ≤ f (2 n +1 ) g (2 n ) = g (2 n +1 ) g (2 n ) → n → ∞ .Let f ( x ) = f (2) for 0 ≤ x ≤
2, and let n ∈ N + be fixed. For 0 ≤ x ≤ n define r n ( x ) = q n (1 − cos(2 − n πx )) , and let us define f (2 n + x ) = f (2 n ) + Z x r n ( t ) d t. Then clearly f (2 n +1 ) − f (2 n ) = Z n r n ( t ) d t = 2 n q n , so(3.10) q n = 2 − n ( f (2 n +1 ) − f (2 n )) . Since r n ( x ) ≥ n ∈ N + and 0 ≤ x ≤ n , the function f is non-decreasing.As r n (0) = r n (2 n ) = 0, we obtain that f is continuously differentiable such that f ′ ( x ) = 0 if 0 < x ≤ f ′ (2 n + x ) = r n ( x ) for all n ∈ N + . It is easy to see that f ′′ (2 n ) = 0 for all n ∈ N + , so the formula of f ′ ( x ) implies that f ′ is continuouslydifferentiable, thus (iv) holds. Let n ∈ N + and 0 ≤ x ≤ n . Then r n ( x ) ≤ q n ,(3.10), and (ii) yield that f ′ (2 n + x ) f (2 n + x ) = r n ( x ) f (2 n + x ) ≤ − n f (2 n +1 ) − f (2 n ) f (2 n + x ) ≤ n + x f (2 n +1 ) − f (2 n ) f (2 n ) = o (cid:18) n + x (cid:19) as n → ∞ , hence (v) is satisfied. Let n ∈ N + and 0 ≤ x ≤ n . Clearly | r ′ n ( x ) | ≤ − n π ( f (2 n +1 ) − f (2 n )) , so (ii) implies that | f ′′ (2 n + x ) | f (2 n + x ) = | r ′ n ( x ) | f (2 n + x ) ≤ − n π f (2 n +1 ) − f (2 n ) f (2 n + x ) ≤ π (2 n + x ) f (2 n +1 ) − f (2 n ) f (2 n ) = o (cid:18) n + x ) (cid:19) as n → ∞ , so (vi) holds. The proof is complete. (cid:3) Corollary 3.4.
Let g : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 g (2 n ) < ∞ . Then there is a non-decreasing function f : [0 , ∞ ) → R + such that(1) P ∞ n =1 /f (2 n ) < ∞ ,(2) lim n →∞ f (2 n +1 ) /f (2 n ) = 1 ,(3) lim sup x →∞ f ( x ) /g ( x ) ≤ ,(4) for all c > there is an R c > such that the function h c ( x ) = x − c f ( x ) isdecreasing for x ≥ R c ,(5) for all < p < there is a concave increasing function g p : [0 , ∞ ) → R + and N p > such that g p ( x ) = x p f ( x ) for all x ≥ N p ,(6) for all q > there is a convex increasing function g q : [0 , ∞ ) → R + and N q > such that g q is affine on [0 , N q ] and g q ( x ) = x q f ( x ) for all x ≥ N q .Proof. Let us choose a non-decreasing function f : [0 , ∞ ) → R + for which properties(i)–(vi) of Lemma 3.3 hold. Then clearly f satisfies (1), (2), and (3). First we proveproperty (4). By (v) of Lemma 3.3 we have h ′ c ( x ) = − cx − c − f ( x ) + x − c f ′ ( x ) = x − c − f ( x )( − c + o (1)) < x ≥ R c with some constant R c >
0, which proves (4).Now we show (5). Let f p ( x ) = x p f ( x ), using (v) and (vi) of Lemma 3.3 weobtain that f ′′ p ( x ) = p ( p − x p − f ( x ) + 2 px p − f ′ ( x ) + x p f ′′ ( x )= x p − f ( x )( p ( p −
1) + o (1)) < x ≥ K p with some K p >
0. By (v) of Lemma 3.3 we obtain that(3.12) f ′ p ( x ) f p ( x ) = px p − f ( x ) + x p f ′ ( x ) x p f ( x ) = x p − f ( x )( p + o (1)) x p f ( x ) = p + o (1) x < x AUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD 13 and f ′ p ( x ) > x ≥ L p with some L p >
0. Let N p = max { K p , L p } . Define g p ( x ) = f p ( x ) if x ≥ N p and let g p be affine on [0 , N p ] with slope f ′ p ( N p ) >
0. By(3.11) we have f ′′ p ( x ) < x ≥ N p , so g p is increasing and concave. We onlyneed to show that g p (0) >
0. Indeed, by (3.12) we obtain that g p (0) = f p ( N p ) − f ′ p ( N p ) N p > . Thus (5) holds.Finally, we prove (6). Let f q ( x ) = x q f ( x ), similarly to (3.11) we obtain that f ′′ q ( x ) = x q − f ( x )( q ( q −
1) + o (1)) > f ′ q ( x ) > x ≥ N q with some N q >
0. Choose 0 < ε q < f ′ q ( N q ) such that f q ( N q ) − ε q N q >
0. Let g q ( x ) = f q ( x ) if x ≥ N q and let g q be affine on [0 , N q ]with slope ε q . Clearly g q is increasing and convex, so we only need to show that g q (0) >
0. Indeed, we have g q (0) = f q ( N q ) − ε q N q > . Hence (6) holds, and the proof is complete. (cid:3)
Lemma 3.5.
Let g : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 g (2 n ) < ∞ . Then there is a non-decreasing function f : [0 , ∞ ) → R + such that(i) P ∞ n =1 /f (2 n ) < ∞ ,(ii) lim n →∞ f (2 n +1 ) /f (2 n ) = 1 ,(iii) lim sup x →∞ f ( x ) /g ( x ) < ∞ ,(iv) h ( x ) = xf ( x ) is piecewise linear, increasing, and convex on [0 , ∞ ) .Proof. By Corollary 3.2 we may assume that(3.13) lim n →∞ g (2 n +1 ) g (2 n ) = 1 . Define a n = g (2 n ) for all n ∈ N + . Define the sequence { b n } n ≥ for all n ∈ N + suchthat b = a , b = a , and for all n ∈ N + we recursively define(3.14) b n +2 = max { a n +2 , (3 / b n +1 − (1 / b n } . The definition clearly implies that for all n ∈ N + we have(3.15) 2 b n +2 − b n +1 + b n ≥ . First we show that b n is non-decreasing. Indeed, b ≥ b and assume by inductionthat b n +1 ≥ b n for some n ≥
1, then b n +2 ≥ (3 / b n +1 − (1 / b n ≥ b n +1 . Now weprove that for all n ∈ N + we have(3.16) b n a n ≤ . Fix an arbitrary integer m ≥ k be the largest integer such that 2 ≤ k ≤ m and b k = a k . As b = a , we obtain that k exists. Let us define { c n } n ≥ such that c = b k − , c = b k , and for all n ∈ N let c n +2 = (3 / c n +1 − (1 / c n . Then clearly b m = c m − k +1 . Solving the linear recursion for c n and using that0 < c ≤ c we obtain for all n ∈ N that c n = 2 c − c + c − c n − < c . Thus b m = c m − k +1 < c = 2 b k = 2 a k . Therefore the monotonicity of the sequence { a n } n ≥ implies that(3.17) b m a m ≤ a k a m ≤ , so (3.16) holds.Let us define f : [0 , ∞ ) → R + as follows. Let f (2 n ) = b n for all n ∈ N + , and let f ( x ) = f (2) if 0 ≤ x ≤
2. If n ∈ N + and 0 ≤ x ≤ n then let f ( x + 2 n ) = b n + 2( b n +1 − b n ) xx + 2 n . Since { b n } n ≥ is non-decreasing, it is easy to see that f is non-decreasing andcontinuous. Then b n ≥ a n and P ∞ n =1 /a n < ∞ yield that (i) holds.Let us define e n = b n +1 /b n for all n ∈ N + and let E = lim sup n →∞ e n . Clearly e n ≥ n , so it is enough to show for (ii) that E ≤
1. By b n +1 ≥ a n +1 and(3.13) we obtain that(3.18) lim sup n →∞ a n +2 b n +1 ≤ lim sup n →∞ a n +1 b n +1 lim n →∞ a n +2 a n +1 ≤ . For all n ∈ N + we have(3.19) (3 / b n +1 − (1 / b n b n +1 = 32 − e n ≥ . Then (3.14), (3.18), and (3.19) yield that(3.20) E = lim sup n →∞ e n +1 ≤ lim sup n →∞ (cid:18) − e n (cid:19) = 32 −
12 lim sup n e n = 32 − E .
Solving the above inequality implies that E ≤
1, so (ii) is satisfied.Monotonicity, (3.17), and (3.13) imply that for all n ∈ N + and 2 n ≤ z ≤ n +1 we have f ( z ) g ( z ) ≤ f (2 n +1 ) g (2 n ) = b n +1 a n ≤ a n +1 a n → n → ∞ , so (iii) holds.Finally, let us define h : [0 , ∞ ) → [0 , ∞ ) as h ( x ) = xf ( x ). Then h ( x ) = b x if0 ≤ x ≤
2, and for all n ∈ N + and 0 ≤ x ≤ n we have h ( x + 2 n ) = ( x + 2 n ) f ( x + 2 n ) = b n n + (2 b n +1 − b n ) x. Clearly h is continuous and increasing. We obtain that h is affine on [0 ,
2] withslope d := b , and for each n ∈ N + it is also affine on [2 n , n +1 ] with slope d n := 2 b n +1 − b n , so h is piecewise linear. In order to prove that h is convex, weneed to prove that the sequence { d n } n ≥ is non-decreasing. Clearly d ≥ d andby (3.15) for all n ∈ N + we have d n +1 − d n = 2 b n +2 − b n +1 + b n ≥ . Thus { d n } n ≥ is non-decreasing, so (iv) holds. The proof is complete. (cid:3) AUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD 15
Lemma 3.6.
Let g : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 g (2 n ) < ∞ . Let q ≥ . Then there is a non-decreasing function f : [0 , ∞ ) → R + such that(i) P ∞ n =1 /f (2 n ) < ∞ ,(ii) lim n →∞ f (2 n +1 ) /f (2 n ) = 1 ,(iii) lim sup x →∞ f ( x ) /g ( x ) < ∞ ,(iv) there is an increasing convex function f q : [0 , ∞ ) → R + and N q > such that f q is affine on [0 , N q ] and f q ( x ) = x q f ( √ x ) for x ≥ N q .Proof. Define g ∗ : [0 , ∞ ) → R + as g ∗ ( x ) = g ( √ x ). Clearly g ∗ is a non-decreasingfunction such that g ∗ ( x ) ≤ g ( x ) for x ≥
1. Fact 2.1 yields that P ∞ n =1 /g ∗ (2 n ) < ∞ . Corollary 3.4 and Lemma 3.5 imply that there is a non-decreasing function f ∗ : [0 , ∞ ) → R + such that(1) P ∞ n =1 /f ∗ (2 n ) < ∞ ,(2) lim n →∞ f ∗ (2 n +1 ) /f ∗ (2 n ) = 1,(3) lim sup x →∞ f ∗ ( x ) /g ∗ ( x ) < ∞ ,(4) there is an increasing convex function f ∗ q : [0 , ∞ ) → R + and N q > f ∗ q is affine on [0 , N q ] and f ∗ q ( x ) = x q f ∗ ( x ) for all x ≥ N q .Define f : [0 , ∞ ) → R + as f ( x ) = f ∗ ( x ). By (1) we have ∞ X n =1 f (2 n ) = ∞ X n =1 f ∗ (4 n ) < ∞ X n =1 f ∗ (2 n ) < ∞ , so (i) holds. By (2) we havelim n →∞ f (2 n +1 ) f (2 n ) = lim n →∞ f ∗ (4 n +1 ) f ∗ (4 n ) = lim n →∞ f ∗ (2 n +2 ) f ∗ (2 n +1 ) · lim n →∞ f ∗ (2 n +1 ) f ∗ (2 n ) = 1 , thus (ii) is satisfied. The definitions and (3) yield thatlim sup x →∞ f ( x ) g ( x ) = lim sup x →∞ f ∗ ( x ) g ∗ ( x ) = lim sup x →∞ f ∗ ( x ) g ∗ ( x ) < ∞ , so (iii) holds. Finally, define f q : [0 , ∞ ) → R + as f q ( x ) = f ∗ q ( x ), then (4) yieldsthat f q is an increasing convex function which is affine on [0 , N q ] and for all x ≥ N q we have f q ( x ) = f ∗ q ( x ) = x q f ∗ ( x ) = x q f ( √ x ) , hence (iv) holds. The proof is complete. (cid:3) Sequences of arbitrary random variables
The main goal of this section is to prove Theorems 4.1 and 4.2.
Theorem 4.1.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) < ∞ . Let < r < and let r ≤ p , and define q = q ( r, p ) = max { p, ( p − r ) / (1 − r ) } . Assumethat { X n } n ≥ is a sequence of random variables with sup n ≥ E ( | X n | q f ( | X n | )) < ∞ .Then for all ε > we have ∞ X n =1 n p/r − P ( M n > εn /r ) < ∞ . Proof.
Let ε > X i ≥ i ,otherwise we can replace X i by | X i | . Thus M n = S n for all n ∈ N + . First supposethat p <
1, then q = p . We may assume by Corollary 3.4 (5) and Fact 2.2 thatthere exists an N p > g p : [0 , ∞ ) → R + suchthat g p ( x ) = x p f ( x ) for all x ≥ N p . By Fact 2.2 we have(4.1) sup n ≥ E g p ( | X n | ) = C < ∞ . As g p is concave with g p (0) = 0, it is subadditive. Fix an integer n ≥ ( N p /ε ) r .Markov’s inequality, the fact that g p is increasing and subadditive, and (4.1) implythat for each ε > n ≥ n we have P ( M n > εn /r ) = P ( | S n | > εn /r )= P ( g p ( | S n | ) > g p ( εn /r )) ≤ E g p ( | S n | ) g p ( εn /r )= E g p ( | S n | )( εn /r ) p f ( εn /r ) ≤ P ni =1 E g p ( | X i | ) ε p n p/r f ( εn /r ) ≤ Cnε p n p/r f ( εn /r ) . n − p/r f ( εn /r ) . Therefore by Fact 2.1 we have X n ≥ n n p/r − P ( M n > εn /r ) . X n ≥ n nf ( εn /r ) < ∞ , which completes the proof for p < p ≥
1, then q = ( p − r ) / (1 − r ) ≥
1. We may assume byCorollary 3.4 (6), Lemma 3.5, and Fact 2.2 that there is an N q > g q : [0 , ∞ ) → R + such that g q ( x ) = x q f ( x ) for all x ≥ N q . ByFact 2.2 we have(4.2) sup n ≥ E g q ( | X n | ) = C < ∞ . Fix an integer n ≥ ( N q /ε ) r/ (1 − r ) . Markov’s inequality, the fact that g q is increasingand Jensen’s inequality holds for the convex g q , and (4.2) imply that for each ε > n ≥ n we have P ( M n > εn /r ) = P ( | S n | > εn /r ) ≤ P ( g q ( | S n | /n ) > g q ( εn /r − )) AUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD 17 ≤ E g q ( | S n | /n ) g q ( εn /r − )= E g q ( | S n | /n ) ε q n q (1 /r − f ( εn /r − ) ≤ (1 /n ) P ni =1 E g q ( | X i | ) ε q n p/r − f ( εn /r − ) ≤ Cε q n p/r − f ( εn /r − ) . n − p/r f ( εn /r − ) . Thus the above inequality and Fact 2.1 yields that X n ≥ n n p/r − P ( M n > εn /r ) . X n ≥ n nf ( εn /r − ) < ∞ . This concludes the proof. (cid:3)
Theorem 4.2.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) = ∞ . Let < r < ≤ p and let q = q ( r, p ) = ( p − r ) / (1 − r ) . Then there is a sequenceof random variables { X n } n ≥ such that sup n ≥ E ( | X n | q f ( | X n | )) < ∞ and ∞ X n =1 n p/r − P ( | S n | > n /r ) = ∞ . Proof.
For all k ∈ N + let p k = 4 k (1 − p/r ) f (4 k (1 /r − ) . Fix k ∈ N + such that for all k ≥ k we have p k <
1. Let X n ≡ n < k .Let k > k be fixed, then for all m, n ∈ { k − , . . . , k − } let X n = X m and let P ( X n = 4 k (1 /r − ) = p k and P ( X n = 0) = 1 − p k . Then sup n ≥ E ( | X n | q f ( | X n | )) = 4 q < ∞ , and for all 2 · k − ≤ n < k we have P ( | S n | > n /r ) ≥ P (4 k − X n ≥ k/r ) = P ( X n ≥ k (1 /r − ) = p k . The above inequality and Fact 2.1 imply that ∞ X n =1 n p/r − P ( | S n | > n /r ) ≥ X k>k X · k − ≤ n< k n p/r − P ( | S n | > n /r ) ≥ X k>k (2 · k − )(4 ( k − p/r ) − k ) p k = 2 − p/r − X k>k f (4 k (1 /r − ) = ∞ . The proof is complete. (cid:3) Martingale difference sequences
The cases < p < and r = p = 1 . The main goal of this subsection is toprove Theorem 5.1.
Theorem 5.1.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) < ∞ . Let < p < and < r ≤ p or let r = p = 1 . Let { X n } n ≥ be a martingaledifference sequence such that sup n ≥ E ( | X n | p f ( | X n | )) < ∞ . Then for all ε > wehave ∞ X n =1 n p/r − P ( M n > εn /r ) < ∞ . Proof.
Let ε > n ≥ E ( | X n | p f ( | X n | )) = C < ∞ .For all k ∈ N + let F k ( t ) = P ( | X k | ≤ t )be the cumulative distribution function of | X k | . Define q = (2 − p ) /r > c = 2 − p >
0. By Corollary 3.4 (4) and Fact 2.2 we may assume that there existsan R c > x x − c f ( x ) is decreasing for x ≥ R c . For all n ∈ N + and for all k ∈ { , . . . , n } define Y k,n = X k I ( | X k | ≤ n /r ) − E ( X k I ( | X k | ≤ n /r ) | F k − ) ,Z k,n = X k I ( | X k | > n /r ) − E ( X k I ( | X k | > n /r ) | F k − ) . For all n ∈ N + define S ∗ k,n = k X i =1 Y i,n and M ∗ n = max ≤ k ≤ n | S ∗ k,n | ,S ∗∗ k,n = k X i =1 Z i,n and M ∗∗ n = max ≤ k ≤ n | S ∗∗ k,n | . Clearly for all n ∈ N + and 1 ≤ k ≤ n we have(5.1) E ( Y k,n | F k − ) = E ( Z k,n | F k − ) = 0 , so { S ∗ k,n } ≤ k ≤ n and { S ∗∗ k,n } ≤ k ≤ n are martingales. For all n ∈ N + and 1 ≤ k ≤ n we have S k = S ∗ k,n + S ∗∗ k,n , so for all n ∈ N + we have(5.2) M n ≤ M ∗ n + M ∗∗ n . By (5.1) we have E ( Y k,n Y ℓ,n ) = 0 for all 1 ≤ k < ℓ ≤ n , so applying Doob’sinequality for the martingale { S ∗ k,n } ≤ k ≤ n and the identity E ( X − E ( X | F )) = AUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD 19 E ( X ) − E ( E ( X | F )) ≤ E ( X ) implies that P ( M ∗ n > εn /r ) ≤ ε n − /r E (cid:0) ( S ∗ n,n ) (cid:1) = 1 ε n − /r n X k =1 E ( Y k,n ) ≤ ε n − /r n X k =1 E ( X k I ( | X k | ≤ n /r )) . Therefore ∞ X n =1 n p/r − P ( M ∗ n > εn /r ) . ∞ X n =1 n p/r − /r − n X k =1 E ( X k I ( | X k | ≤ n /r ))= ∞ X n =1 n − q − n X k =1 Z ∞ t I ( t ≤ n /r )) d F k ( t )= ∞ X k =1 Z ∞ t X n ≥ max { k,t r } n − q − d F k ( t ) . ∞ X k =1 Z ∞ t (max { k, t r } ) − q − d F k ( t )= ∞ X k =1 ( A k + B k + C k ) , where A k = Z R c t k − q − d F k ( t ) , B k = Z k /r R c t k − q − d F k ( t ) , C k = Z ∞ k /r t p − r d F k ( t ) . Clearly(5.3) ∞ X k =1 A k ≤ ∞ X k =1 R c k − q − < ∞ . Let k ≥ R rc . Using that x − c f ( x ) is non-increasing if x ≥ R c , and x r f ( x ) is non-decreasing, we obtain that B k + C k ≤ k − q − Z k /r R c t t p − f ( t ) k ( p − /r f ( k /r ) d F k ( t ) + Z ∞ k /r t p − r t r f ( t ) kf ( k /r ) d F k ( t ) ≤ kf ( k /r ) Z ∞ R c t p f ( t ) d F k ( t ) ≤ E ( | X k | p f ( | X k | )) kf ( k /r ) ≤ Ckf ( k /r ) . By Fact 2.2 we have A k + B k + C k ≤ E ( | X k | p − r ) < ∞ for all k , so the aboveinequality with (5.3) and Fact 2.1 imply that(5.4) ∞ X n =1 n p/r − P ( M ∗ n > εn /r ) . ∞ X k =1 ( A k + B k + C k ) < ∞ . Applying Doob’s inequality for the martingale { S ∗∗ k,n } ≤ k ≤ n , the triangle inequality,Jensen’s inequality (for the conditional expectation as well), and the law of total expectation in this order implies that P ( M ∗∗ n > εn /r ) . n − /r E ( | S ∗∗ n,n | ) ≤ n − /r n X k =1 E ( | Z k,n | ) ≤ n − /r n X k =1 E ( | X k | I ( | X k | > n /r )) ≤ n − /r n X k =1 E ( | X k || X k | p − f ( | X k | ) I ( | X k | > n /r )) n ( p − /r f ( n /r )= 2 n − p/r f ( n /r ) n X k =1 E ( | X k | p f ( | X k | )) ≤ C n − p/r f ( n /r ) . The above inequality and Fact 2.1 yield that(5.5) ∞ X n =1 n p/r − P ( M ∗∗ n > εn /r ) . ∞ X n =1 nf ( n /r ) < ∞ . Finally, (5.2), (5.4), and (5.5) imply that ∞ X n =1 n p/r − P ( M n > εn /r ) ≤ ∞ X n =1 n p/r − P ( M ∗ n > ( ε/ n /r )+ ∞ X n =1 n p/r − P ( M ∗∗ n > ( ε/ n /r ) < ∞ . The proof is complete. (cid:3)
The case p ≥ . The goal of the subsection is to prove Theorems 5.2 and 5.4.
Theorem 5.2.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) < ∞ . Let < r < ≤ p and let q = q ( r, p ) = 2( p − r ) / (2 − r ) . Let { X n } n ≥ be amartingale difference sequence such that sup n ≥ E ( | X n | q f ( | X n | )) < ∞ . Then forall ε > we have ∞ X n =1 n p/r − P ( M n > εn /r ) < ∞ . Before proving the above theorem we need the next inequality due to Burkholder,Davis, and Gundy, see [4, Theorem 1.1] or [3, Theorem 15.1].
Theorem 5.3 (Burkholder–Davis–Gundy Inequality) . Let g : [0 , ∞ ) → [0 , ∞ ) be aconvex function such that g (0) = 0 and there is a constant c ∈ R + such that (5.6) g (2 x ) ≤ cg ( x ) for all x > . AUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD 21
Then there exists a constant C ∈ R + depending only on c such that for everymartingale difference sequence { X i } i ≥ for all n ∈ N + we have E g ( M n ) ≤ C E g (cid:18)q X + · · · + X n (cid:19) . Proof of Theorem 5.2.
Let ε > q ≥
2, by Lemma 3.6 andFact 2.2 we may assume that(5.7) lim n →∞ f (2 n +1 ) f (2 n ) = 1and there is an increasing convex function f q/ : [0 , ∞ ) → [0 , ∞ ) and N, a ∈ R + such that f q/ (0) = 0 and f q/ is linear on [0 , N ], and for all x ≥ N we have f q/ ( x ) = x q/ f ( √ x ) − a. Define g q : [0 , ∞ ) → [0 , ∞ ) as g q ( x ) = f q/ ( x ). Since g q is a composition ofconvex increasing functions, it is increasing and convex. Clearly we have g q (0) = 0.Moreover, g q ( x ) = bx for x ∈ [0 , √ N ] with some constant b >
0, and for x ≥ N we obtain g q ( x ) = x q f ( x ) − a. Fact 2.2 yields that(5.8) sup n ≥ E g q ( | X n | ) = K < ∞ . Let h q : (0 , ∞ ) → R + be defined as h q ( x ) = g q (2 x ) g q ( x ) . As g q ( x ) > x >
0, the function h q is well defined, and the continuity of g q implies that h q is continuous, too. Since g q ( x ) = bx for x ∈ [0 , √ N ], we have h q ( x ) = 4 for x ∈ (0 , √ N/ x →∞ h q ( x ) < ∞ , sothe continuity of h q implies that h q is bounded. Therefore g q satisfies the growthcondition (5.6).Let us choose K ∈ R + such that ε q f ( εn /r − / ) > a for each n ≥ K . Define L = max (cid:26) K, (cid:16) √ N/ε (cid:17) r/ (2 − r ) (cid:27) . Applying that g q is increasing, Markov’s inequality, Theorem 5.3 for g q and thefinite martingale { S i / √ n } ≤ i ≤ n , Jensen’s inequality for f q/ , and (5.8) implies thatfor all n ≥ L we have P ( M n > εn /r ) = P ( g q ( M n / √ n ) > g q ( εn /r − / )) ≤ E g q ( M n / √ n ) g q ( εn /r − / ) ≤ C E (cid:16) g q (cid:16)p (1 /n ) P ni =1 X i (cid:17)(cid:17) g q ( εn /r − / )= C E (cid:0) f q/ (cid:0) (1 /n ) P ni =1 X i (cid:1)(cid:1) g q ( εn /r − / ) ≤ C (1 /n ) P ni =1 E f q/ ( X i ) g q ( εn /r − / )= C (1 /n ) P ni =1 E g q ( | X i | ) g q ( εn /r − / ) ≤ CKε q n q (1 /r − / f ( εn /r − / ) − a ≤ CKε − q n − p/r f ( εn /r − / ) . n − p/r f ( εn /r − / ) . Therefore the above inequality and Fact 2.1 imply that X n ≥ L n p/r − P ( M n > εn /r ) . X n ≥ L nf ( εn /r − / ) < ∞ . The proof is complete. (cid:3)
Theorem 5.4.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) = ∞ . Let < r < ≤ p and let q = q ( r, p ) = 2( p − r ) / (2 − r ) . Then there is a martingaledifference sequence { X n } n ≥ such that sup n ≥ E ( | X n | q f ( | X n | )) < ∞ and ∞ X n =1 n p/r − P ( | S n | > n /r ) = ∞ . Proof.
Let { Y n , Z k } n,k ≥ be independent random variables such that for all n ∈ N + we have P ( Y n = 1) = P ( Y n = −
1) = 12 . For all k ∈ N + let p k = 4 k (1 − p/r ) f (4 k (1 /r − / ) . Fix k ≥ k ≥ k we have p k <
1. We define Z k ≡ k ≤ k andfor k > k let P ( Z k = 4 k (1 /r − / ) = p k and P ( Z k = 0) = 1 − p k . For all k ∈ N + and 4 k − ≤ n < k let us define X n = Y n Z k . Clearly we havesup n ≥ E ( | X n | q f ( | X n | ) = 1. Assume that X i : Ω → R are random variables on theprobability space (Ω , F , P ). Let F = {∅ , Ω } and let F n = σ ( X , . . . , X n ) for all n ∈ N + . We show that { X n } n ≥ is a martingale difference sequence with respectto the natural filtration {F n } n ≥ . Fix n, k ∈ N + with 4 k − ≤ n < k . Indeed, as Y n is independent of { Z , . . . , Z k , Y , . . . Y n − } , it is independent of σ ( Z k , F n − ), soa property of conditional expectation implies that for all n ∈ N + we have E ( X n | F n − ) = E ( Y n Z k | F n − ) = E ( Y n ) E ( Z k | F n − ) = 0 , AUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD 23 so { X n } n ≥ is really a martingale difference sequence. By the central limit theoremthere is an absolute constant c > k > k and 2 · k − ≤ n < k we have(5.9) P ( Y k − + · · · + Y n ≥ k ) = P ( Y k − + · · · + Y n ≤ − k ) ≥ c. We will prove that for all fixed k > k and 2 · k − ≤ n < k we have(5.10) P ( | S n | > n /r ) ≥ cp k . Let us use the notation S = S k − − and fix an arbitrary x ∈ R with P ( S = x ) > P ( | S n | > n /r | S = x ) ≥ cp k . As 4 k/r > n /r , either x + 4 k/r > n /r or x − k/r < − n /r . We may assume bysymmetry that x + 4 k/r > n /r . It is clear from the definition that S and S n − S are independent, and the independence of Z k and { Y k − , . . . , Y n } , and (5.9) yieldthat P ( | S n | > n /r | S = x ) ≥ P ( S n − S ≥ k/r | S = x )= P ( S n − S ≥ k/r ) ≥ P ( Y k − + · · · + Y n ≥ k , Z k = 4 k (1 /r − / )= P ( Y k − + · · · + Y n ≥ k ) P ( Z k = 4 k (1 /r − / ) ≥ cp k . This implies (5.11), so (5.10) holds. Inequality (5.10) and Fact 2.1 yield that ∞ X n =1 n p/r − P ( | S n | > n /r ) ≥ X k>k X · k − ≤ n< k n p/r − P ( | S n | > n /r ) ≥ X k>k (2 · k − )(4 ( k − p/r ) − k ) cp k = c − p/r − X k>k f (4 k (1 /r − / ) = ∞ . The proof is complete. (cid:3) Independent, negatively associated, and pairwise NQD randomvariables
The main goal of this section is to prove Theorems 6.1 and 6.5.
Theorem 6.1.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) < ∞ . Let < r < ≤ p and let { X n } n ≥ be a sequence of negatively associated, centeredrandom variables such that sup n ≥ E ( | X n | p f ( | X n | )) < ∞ . For all ε > we have ∞ X n =1 n p/r − P ( M n > εn /r ) < ∞ . First we need the following inequality of Shao [23, Theorem 3].
Theorem 6.2 (Shao) . Let { X i : 1 ≤ i ≤ n } be a centered, negatively associatedsequence of random variables with finite second moments. Let M n = max ≤ k ≤ n | S k | and B n = P ni =1 E ( X i ) . Then for all x > , a > , and < α < we have P ( M n ≥ x ) ≤ P (cid:18) max ≤ i ≤ n | X i | > a (cid:19) + 21 − α exp (cid:18) − x α ax + B n ) (cid:18) (cid:18) axB n (cid:19)(cid:19)(cid:19) . Proof of Theorem 6.1.
Fix ε >
0. By Fact 2.2 we have sup n ≥ E ( X n ) = C < ∞ .Thus B n = P ni =1 E ( X i ) ≤ Cn for all n . Let N = 8 p/ (2 − r ). Applying Theorem 6.2for n ∈ N + , x = εn /r , a = x/N , and α = 1 / P ( M n > εn /r ) ≤ a n + b n , where a n = 2 P (cid:18) max ≤ i ≤ n | X i | > εn /r /N (cid:19) ,b n = 4 exp (cid:18) − ε n /r ε n /r /N + Cn ) (cid:18) (cid:18) ε n /r N Cn (cid:19)(cid:19)(cid:19)
Let c = ε/N , by Markov’s inequality we have a n ≤ n X i =1 P ( | X i | > cn /r ) ≤ n X i =1 P ( | X i | p f ( | X i | ) ≥ c p n p/r f ( cn /r )) ≤ n X i =1 E ( | X i | p f ( | X i | )) c p n p/r f ( cn /r ) . n − p/r f ( cn /r ) , thus Fact 2.1 implies that(6.2) ∞ X n =1 n p/r − a n . ∞ X n =1 nf ( cn /r ) < ∞ . Since 2 /r >
1, easy calculation shows that b n = 4 exp (cid:18)(cid:18) − N (cid:18) r − (cid:19) + o (1) (cid:19) log n (cid:19) as n → ∞ , so for all large enough n we have b n ≤ exp( − ( N/ /r −
1) log n ) = n − N (2 − r ) / (8 r ) = n − p/r . Therefore(6.3) ∞ X n =1 n p/r − b n < ∞ . Clearly (6.1), (6.2), and (6.3) complete the proof. (cid:3)
AUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD 25
Corollary 6.3.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) < ∞ . Let < r < ≤ p , and let { X n } n ≥ be a sequence of(1) pairwise NQD random variables if ≤ p < ,(2) negatively associated random variables if p ≥ .Assume that sup n ≥ E ( | X n | p f ( | X n | )) < ∞ . Then for all ε > we have ∞ X n =1 n p/r − P ( M n > εn /r ) < ∞ . Proof.
Fix ε > X n ≥ n ,otherwise we replace X n by | X n | . Thus M n = S n . Fact 2.2 and p ≥ n ≥ E ( X n ) = K < ∞ . For all n ∈ N + define Y n = X n − E ( X n ) . Clearly if { X n } n ≥ is pairwise NQD/negatively associated then { Y n } n ≥ is alsopairwise NQD/negatively associated. As | Y n | ≤ max { K, X n } , the monotonicity ofthe function x x p f ( x ) implies that almost surely for all n ∈ N + we have | Y n | p f ( | Y n | ) ≤ K p f ( K ) + X pn f ( X n ) , so sup n ≥ E ( | Y n | p f ( | Y n | )) ≤ K p f ( K ) + sup n ≥ E ( X pn f ( X n )) < ∞ . Define T n = P ni =1 Y i . For all n ≥ (2 K/ε ) r/ (1 − r ) we have P ( M n > εn /r ) = P ( S n > εn /r ) ≤ P ( T n > εn /r − Kn ) ≤ P ( T n > ( ε/ n /r ) ≤ P ( | T n | > ( ε/ n /r ) . (6.4)Applying Theorem 1.15 (i) if p = 1, Theorem 1.16 if 1 < p <
2, and Theorem 6.1if p ≥ { Y n } n ≥ yields that ∞ X n =1 n p/r − P ( | T n | > ( ε/ n /r ) < ∞ , so (6.4) finishes the proof. (cid:3) The following lemma is due to Nash [22], which gives a necessary and sufficientcondition for P (lim sup n →∞ A n ) = 1 in terms of conditional probabilities. Lemma 6.4 (Nash) . Let { A n } n ≥ be events and define H ⊂ { , } N + such that H = { ( α , α , . . . ) : α n = 1 only for finitely many n and P ( I ( A ) = α , . . . , I ( A n ) = α n ) > for all n } . Then P (lim sup n →∞ A n ) = 1 if and only if for all ( α , α , . . . ) ∈ H we have ∞ X n =2 P ( A n | I ( A ) = α , . . . , I ( A n − ) = α n − ) = ∞ . The construction in the following theorem dates back to Chung [6, Theorem 2],but our proof is more involved.
Theorem 6.5.
Let f : [0 , ∞ ) → R + be a non-decreasing function such that ∞ X n =1 f (2 n ) = ∞ . Let < r < and let p ≥ r . Then there exists a sequence of independent, centeredrandom variables { X n } n ≥ such that sup n ≥ E ( | X n | p f ( | X n | )) < ∞ and ∞ X n =1 n p/r − P ( | S n | > n /r ) = ∞ . Moreover, if r = p then lim sup n →∞ n − /p S n ≥ almost surely.Proof. For all k ∈ N + let p k = 4 − kp/r f (4 k/r ) . Since 4 − kp/r ≤ − k , for c = exp( − /f (4 /r )) we can fix k ∈ N + such that for all k ≥ k we have p k < / − p k ) k ≥ c. We define a sequence of independent random variables { X n } n ≥ as follows. Let X n ≡ n < k . If 4 k − ≤ n < k for some integer k > k then let P ( X n = 4 k/r ) = P ( X n = − k/r ) = p k and P ( X n = 0) = 1 − p k . Then { X n } n ≥ is a sequence of independent, centered random variables such thatsup n ≥ E ( | X n | p f ( | X n | )) = 2. Fix k > k and 2 · k − ≤ n < k . We will prove that(6.6) P ( | S n | > n /r ) ≥ c k − p k . Let us use the notation S = S n − k − and fix an arbitrary x ∈ R with P ( S = x ) > P ( | S n | > n /r | S = x ) ≥ c k − p k . As 4 k/r > n /r , we have either x + 4 k/r > n /r or x − k/r < − n /r . By symmetrywe may assume that x + 4 k/r > n /r . Thus the independence of S and S n − S , and(6.5) yield that P ( | S n | > n /r | S = x ) ≥ P ( S n − S = 4 k/r | S = x ) = P ( S n − S = 4 k/r ) ≥ k − p k (1 − p k ) k − − ≥ c k − p k , where only the sequences with 4 k − − ∞ X n =1 n p/r − P ( | S n | > n /r ) ≥ X k>k X · k − ≤ n< k n p/r − P ( | S n | > n /r ) ≥ X k>k (2 · k − )(4 ( k − p/r ) − k ) c k − p k = c − p/r − X k>k f (4 k/r ) = ∞ . AUM–KATZ TYPE THEOREMS WITH EXACT THRESHOLD 27
This proves the first claim.Now assume that p = r . For all k ∈ N + let us define the event A k = { S k − ≥ k/p } . Fix arbitrary k > k and ( α , . . . , α k − ) ∈ { , } k − such that P ( I ( A ) = α , . . . , I ( A k − ) = α k − ) > . Repeating the argument of the proof of (6.7) for fixed values of X , . . . , X k − − and using the law of total probability for conditional probabilities we obtain that P ( A k | I ( A ) = α , . . . , I ( A k − ) = α k − ) ≥ · k − p k (1 − p k ) · k − − ≥ c k − p k & f (4 k/p ) . (6.8)Fact 2.1 implies that P k>k /f (4 k/p ) = ∞ , so (6.8) and Lemma 6.4 yield that P (lim sup k →∞ A k ) = 1. Thus lim sup n →∞ n − /p S n ≥ (cid:3) References [1] K. Alam, K. M. Lal Saxena, Positive dependence in multivariate distributions,
Comm. Statist.Theory Methods (1981), no. 12, 1183–1196.[2] L. E. Baum, M. Katz, Convergence rates in the law of large numbers, Trans. Amer. Math.Soc. (1965), 108–123.[3] D. L. Burkholder, Distribution function inequalities for martingales,
Ann. Probab. (1973),19–42.[4] D. L. Burkholder, B. J. Davis, R. F. Gundy, Integral inequalities for convex functions ofoperators on martingales. In Proceedings of the Sixth Berkeley Symposium on Mathemat-ical Statistics and Probability (Univ. California, Berkeley, Calif., 1970/1971), Volume II:Probability theory , 223–240, Univ. California Press, Berkeley, CA, 1972.[5] Y. S. Chow, Delayed sums and Borel summability of independent, identically distributedrandom variables,
Bull. Inst. Math. Acad. Sinica (1973), no. 2, 207–220.[6] K. L. Chung, Note on some strong laws of large numbers, Amer. J. Math. (1947), 189–192.[7] J. Dedecker, F. Merlev´ede, Convergence rates in the law of large numbers for Banach-valueddependent variables, Theory Probab. Appl. (2008), no. 3, 416–438.[8] R. Durrett, Probability: theory and examples , fourth edition, Cambridge University Press,Cambridge, 2010.[9] J. Elton, A law of large numbers for identically distributed martingale differences,
Ann.Probab. (1981), no. 3, 405–412.[10] P. Erd˝os, On a theorem of Hsu and Robbins, Ann. Math. Statistics (1949), 286–291.[11] P. Erd˝os, Remark on my paper “On a theorem of Hsu and Robbins”, Ann. Math. Statistics (1950), 138.[12] S. Gan, P. Chen, Some limit theorems for sequences of pairwise NQD random variables, ActaMath. Sci. Ser. B Engl. Ed. (2008), no. 2, 269–281.[13] P. L. Hsu, H. Robbins, Complete convergence and the law of large numbers, Proc. Natl. Acad.Sci. USA (1947), 25–31.[14] K. Joag-Dev, F. Proschan, Negative association of random variables, with applications, Ann.Statist. (1983), no. 1, 286–295.[15] A. Kolmogoroff, Uber die Summen durch den Zufall bestimmter unabh¨angiger Gr¨oen, (Ger-man) Math. Ann. (1928), no. 1, 309–319.[16] A. Kuczmaszewska, On complete convergence in Marcinkiewicz–Zygmund type SLLN fornegatively associated random variables, Acta Math. Hungar. (2010), no. 1–2, 116–130.[17] T. L. Lai, Convergence rates and r-quick versions of the strong law for stationary mixingsequences,
Ann. Probab. (1977), no. 5, 693–706.[18] E. L. Lehmann, Some concepts of dependence, Ann. Math. Statist. (1966), 1137–1153. [19] M. Lo`eve, On almost sure convergence, Proceedings of the Second Berkeley Symposium onMathematical Statistics and Probability , University of California Press, 1951, 279–303.[20] J. Marcinkiewicz, A. Zygmund, Sur les fonctions ind´ependantes,
Fund. Math. (1937),60–90.[21] Y. Miao, G. Yang, G. Stoica, On the rate of convergence in the strong law of large numbersfor martingales, Stochastics (2015), no. 2, 185–198.[22] S. W. Nash, An extension of the Borel–Cantelli lemma, Ann. Math. Statistics (1954),165–167.[23] Q. M. Shao, A comparison theorem on moment inequalities between negatively associatedand independent random variables, J. Theoret. Probab. (2000), no. 2, 343–356.[24] F. Spitzer, A combinatorial lemma and its application to probability theory, Trans. Amer.Math. Soc. (1956), 323–339.[25] G. Stoica, Baum–Katz–Nagaev type results for martingales, J. Math. Anal. Appl. (2007),no. 2, 1489–1492.[26] G. Stoica, A note on the rate of convergence in the strong law of large numbers for martingales,
J. Math. Anal. Appl. (2011), no. 2, 910–913.[27] W. F. Stout,
Almost sure convergence , Probability and Mathematical Statistics, Vol. 24.,Academic Press, New York-London, 1974.[28] X. Tan, H. Wang, Y. Zhang, Complete convergence of the non-identically distributed pairwiseNQD random sequences,
Comm. Statist. Theory Methods (2016), no. 9, 2626–2637.[29] Q. Y. Wu, Convergence properties of pairwise NQD random sequences, (Chinese) Acta Math.Sinica (Chin. Ser.) (2002), no. 3, 617–624. Department of Mathematics, University of British Columbia, and Pacific Institutefor the Mathematical Sciences, Vancouver, BC V6T 1Z2, Canada
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