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(2008) pp. 147–154

http://www.ektf.hu/ami

Convergence rate in the strong law of large numbers for mixingales and superadditive

structures

Tibor Tómács

Department of Applied Mathematics Eszterházy Károly College, Eger, Hungary

Submitted 15 September 2008; Accepted 30 October 2008

Abstract

In this paper we study convergence rates in the strong laws of large num- bers for mixingales and superadditive structures, by using a general method.

Keywords: convergence rate, strong law of large numbers,Lr mixingale, se- quence with superadditive moment function

MSC:60F15

1. Introduction

Sung, Hu and Volodin [8] introduced a new method for obtaining convergence rate in the strong law of large numbers (SLLN), by using the approach of Fazekas and Klesov [2]. This result generalizes and sharpens the method of Hu and Hu [5].

Tómács [9] gave a general method by using a Hájek–Rényi type inequality (see Há- jek and Rényi [3]) for the probabilities, which sharpens the result of Sung, Hu and Volodin [8]. In this paper we apply this method for mixingales and superadditive structures.

The concept of L2 mixingales was introduced by McLeish [6], and generalized toLrmixingales by Andrews [1]. The definition of superadditive moment function is due to Móricz [7].

Fazekas and Klesov [2, Theorem 6.1 and 6.2] proved SLLN’s for mixingales.

In Section 3 we shall give the convergence rates in these SLLN’s. Hu and Hu [5, Theorem 2.1] obtained convergence rate in SLLN under the superadditivity property. In Section 4 we shall generalize this result.

We use the following notation. LetNbe the set of the positive integers andR the set of real numbers. Ifa1, a2, . . .∈Rthen in caseA=∅letmaxk∈Aak= 0and

147

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P

kAak = 0. In this paper let {Xk, k ∈N} be a sequence of random variables defined on a fixed probability space (Ω,F,P), Sn = Pn

k=1Xk for all n∈ N and S0 = 0. Finally in this paper let {bk, k ∈ N} be a nondecreasing unbounded sequence of positive real numbers.

2. A general method to obtain the rate of conver- gence in the SLLN

Definition 2.1. LetΘr (r >0) denote the set of functions ϑ: [0,∞)→Rwhich are nondecreasing, continuous at 0,ϑ(0) = 0,ϑ(x)>0for allx >0 and

X

n=1

n−2ϑ−r(n−1)<∞.

Remark 2.2. It is easy to see that if0 < δ < 1 and ϑ(x) = xδ/r (x> 0), then ϑ∈Θr.

Theorem 2.3 (Tómács [9], Theorem 3.4). Let{αk, k∈N}be a sequence of non- negative real numbers,r >0 and

βn = max

k6n bkϑ

X

i=k

αib−ri

!

, where ϑ∈Θr. (2.1) If

X

k=1

αkb−rk <∞ (2.2)

and there existsc >0 such that for anyn∈Nand any ε >0 P

maxk6n|Sk|>ε 6cεr

n

X

k=1

αk, (2.3)

then

n→∞lim βn

bn

= 0 and Sn

bn

=O βn

bn

almost surely (a.s.).

Lemma 2.4. Let {αk, k∈N} be a sequence of nonnegative real numbers,r >0, 0< δ <1,ϑ(x) =xδ/r for all x>0,bk =k1/r for all k∈Nand let βn be defined by (2.1). If there exist c >0 and 0< γ <1 such that P

i=kαi/i6cP

i=ki−1−γ for allk∈N, then

βn

n1/r =O 1

nγδ/r

.

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Proof. SinceP

i=ki1γ6R

k−1x1γdx=γ1(k−1)γ for allk>2, hence we get

X

i=k

αi

i 6 c

γ(k−1)γ for all k>2 (2.4) and

X

i=1

αi

i 6c

X

i=1

i−1−γ=c+c

X

i=2

i−1−γ 6c+ c

γ(2−1)γ = c

γ(γ+ 1).

It follows that β1=

X

i=1

αi

i

!δ/r

6 c

γ(γ+ 1) δ/r

6 c2γ

γ (γ+ 1) δ/r

. (2.5)

On the other hand ifn>2then (2.4) implies

26k6nmax k1/r

X

i=k

αi

i

!δ/r

6 max

26k6nk1/r c

γ(k−1)γ δ/r

6 max

26k6n

c2γ γ

δ/r

k(1−γδ)/r= c2γ

γ δ/r

n(1−γδ)/r.

This inequality, (2.5) andlimn→∞n(1−γδ)/r=∞imply forn∈Nlarge enough βn6const.maxn

(γ+ 1)δ/r, n(1−γδ)/ro

=const.n(1−γδ)/r.

So βnn−1/r6const.n−γδ/r for n∈Nlarge enough, which implies the statement.

3. Mixingales

Let{Fk, k ∈N} be a nondecreasing sequence of sub σ-fields of F, EmXk = E(Xk | Fm) denote the conditional expectation of Xk given Fm for m > 0 and EmXk= 0form60.

Definition 3.1 (McLeish [6], Andrews [1]). The sequence{(Xk,Fk), k∈N} is an Lrmixingaleif there exist nonnegative constants{ck, k>0}and{ψk, k>0}such that ψk ↓0 and for all nonnegative integerskandmwe have

||Ek−mXk||r6ckψm and ||Xk−Ek+mXk||r6ckψm+1, where||ξ||r= (E|ξ|r)1/r for any random variableξ.

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Lemma 3.2. If {(Xk,Fk), k∈N} isLr mixingale, wherer>2 andP

m=1ψm<

∞, then there exists c >0 such that for anyn∈Nand any ε >0

P

maxk6n|Sk|>ε 6cε−r

n

X

k=1

c2k

!r/2 .

Proof. Hansen [4] proved in Lemma 2 under these conditions, that there exists c >0such that for anyn∈N

E

maxk6n |Sk|r 6c

n

X

k=1

c2k

!r/2

.

Hence Markov’s inequality implies the statement.

Theorem 3.3. Let {(Xk,Fk), k ∈ N} be an Lr mixingale, where r > 2 and P

m=1ψm<∞. Letβn defined by (2.1)with αk =

k

X

i=1

c2i

!r/2

k1

X

i=1

c2i

!r/2 .

If

X

k=1

c2k brk

k

X

i=1

c2i

!r/2−1

<∞, (3.1)

then

n→∞lim βn

bn

= 0 and Sn

bn

=O βn

bn

a.s.

Proof. IfA=∅ thenP

i∈Ac2i = 0, henceα1=cr1. Since

n

X

k=1

αk =

n

X

i=1

c2i

!r/2

,

hence Lemma 3.2 implies (2.3). By the mean value theorem xr/22 −xr/21 6(x2−x1)r

2xr/2−12 for all 06x16x2. (3.2) Using (3.2) withx1=Pk−1

i=1 c2i and x2=Pk

i=1c2i we get αk =xr/22 −xr/21 6c2kr

2

k

X

i=1

c2i

!r/2−1 .

This inequality and (3.1) imply (2.2). Since every conditions of Theorem 2.3 are

satisfied, the statement is proved.

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Lemma 3.4. If {(Xk,Fk), k ∈ N} is an Lr mixingale, where 1 < r 6 2 and P

m=1ψm<∞, then there existsc >0such that for anyn∈Nand any ε >0 P

maxk6n |Sk|>ε 6cεr

n

X

k=1

crk.

Proof. Hansen [4] proved in Lemma 2 of Erratum under these conditions, that there exists c >0 such that for anyn∈N

E

maxk6n|Sk|r 6c

n

X

k=1

crk.

Hence Markov’s inequality implies the statement.

Theorem 3.5. Let {(Xk,Fk), k ∈ N} be an Lr mixingale, where 1 < r 62 and P

m=1ψm<∞. Letβn defined by (2.1)with αk =crk. If

X

k=1

crk

brk <∞, (3.3)

then

n→∞lim βn

bn

= 0 and Sn

bn

=O βn

bn

a.s.

Proof. The statement is a corollary of Lemma 3.4 and Theorem 2.3.

Corollary 3.6. Let {(Xk,Fk), k ∈N} be an Lr mixingale, where 1 < r62 and P

m=1ψm<∞. If there exist c >0 and0< γ <1 such thatck 6ck−γ/r for all k∈N, then for all0< δ <1

Sn

n1/r =O 1

nγδ/r

a.s.

Proof. Letbk=k1/rk =crk andϑ(x) =xδ/r(x>0), where0< δ <1is a fixed constant. Then for allk∈N

X

i=k

αi

i =

X

i=k

ci

bi

r

6cr

X

i=k

i1γ.

Hence using Theorem 3.5 and Lemma 2.4 we get Sn

n1/r =O βn

n1/r

=O 1

nγδ/r

a.s.

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4. Sequences with superadditive moment function

Definition 4.1 (Móricz [7]). {Xk, k∈N}is said to have ther-th (r >0)moment function of superadditive structureif there existsg: N∪{0} ×N→[0,∞)such that g(b, k) +g(b+k, l)6g(b, k+l) for all b∈N∪ {0}, k∈N, l∈N (4.1) and for some α >1

E|Sb+n−Sb|6gα(b, n) for all b∈N∪ {0}, n∈N. (4.2) We shall use the notation gn =g(0, n)(n ∈N) and g0 = 0. It is easy to see that gn6gn+1 for alln∈N∪ {0}.

Lemma 4.2. If{Xk, k∈N} hasr-th moment function of superadditive structure with r >0, α >1, then there exists a constant Ar,α depending only on r andα such that for anyn∈N and anyε >0

P

maxk6n |Sk|>ε

6Ar,αε−rgnα.

Proof. Móricz proved in [7] under these conditions, that there exists a constant Ar,α depending only onrandα, such that for anyn∈N

E

maxk6n|Sk|r

6Ar,αgnα.

Hence Markov’s inequality implies the statement.

Theorem 4.3. Assume that {Xk, k ∈ N} has r-th moment function of superad- ditive structure with r >0,α > 1. Let βn defined by (2.1) with αk =gαk −gk−1α . If

X

k=1

gαk −gk−1α

brk <∞, (4.3)

then

nlim→∞

βn

bn

= 0 and Sn

bn

=O βn

bn

a.s.

Proof. As gk increases, we getαk >0, thereby (4.3) implies (2.2). On the other hand Pn

k=1αk =gαn, so Lemma 4.2 implies (2.3). Now applying Theorem 2.3 we

get the statement.

Remark 4.4. Hu and Hu proved Theorem 4.3 in special case ϑ(x) =xδ/r (0 <

δ <1). (See Theorem 2.1 of Hu and Hu [5].)

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Corollary 4.5. Let0< γ <1,c >0,α >1andr >0. If for allb∈N∪{0}, n∈N E|Sb+n−Sb|6c

(b+n)(1−γ)/α−b(1−γ)/αα

,

then Sn

n1/r =O 1

nγδ/r

a.s. for all 0< δ <1.

Proof. Letg:N∪{0}×N∪{0} →[0,∞),g(i, j) =c1/α (i+j)(1−γ)/α−i(1−γ)/α . Then (4.1) and (4.2) are satisfied, hence {Xk, k ∈N} hasr-th moment function of superadditive structure.

Now letbk =k1/rfor allk∈N. Sincegiα=gα(0, i) =ci1γ for every nonnega- tive integeri, hence we get

X

i=k

giα−gαi−1 bri =

X

i=k

ci1−γ−c(i−1)1−γ i

=c

X

i=k

i1−γ 1

i − 1

i+ 1

−c(k−1)1−γ

k 6c

X

i=k

i−1−γ. (4.4) Since (4.4) implies (4.3), hence using Theorem 4.3 we have

Sn

n1/r =O βn

n1/r

a.s. (4.5)

Letϑ(x) =xδ/r, where 0< δ <1 is a fixed constant. Then (4.4) and Lemma 2.4 implyβn/n1/r=O 1/nγδ/r

. Hence we get the statement by (4.5).

Corollary 4.6. Let r >0,c >0and1< α <2. If for allb∈N∪ {0},n∈N E|Sb+n−Sb|6c√

b+n−√ bα

,

then Sn

n1/r =O

n(1α2)δ/r

a.s. for all 0< δ <1.

Proof. Apply Corollary 4.5 withγ= 1−α2.

References

[1] Andrews, D.W.K., Laws of large numbers for dependent nonidentically distributed random variables,Econometric Theory, 4 (1988) 458–467.

[2] Fazekas, I., Klesov, O., A general approach to the strong laws of large numbers, Theory of Probab. Appl., 45/3 (2000) 568–583.

[3] Hájek, J., Rényi, A., Generalization of an inequality of Kolmogorov, Acta Math.

Acad. Sci. Hungar., 6 no. 3–4 (1955) 281–283.

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[4] Hansen, B.E., Strong laws for dependent heterogeneous processes,Econometric The- ory, 7 (1991) 213–221; Erratum,Econometric Theory, 8 (1992) 421–422.

[5] Hu, S., Hu, M., A general approach rate to the strong law of large numbers,Stat. &

Prob. Letters, 76 (2006) 843–851.

[6] McLeish, D.L., A maximal inequality and dependent strong laws,Annals of Proba- bility, 3 (1975) 829–839.

[7] Móricz, F., Moment inequalities and the strong of large numbers, Z. Wahrschein- lichkeitstheorie verw. Gebiete, 35 (1976) 299-314.

[8] Sung, S.H., Hu, T.-C., Volodin, A., A note on the growth rate in the Fazekas- Klesov general law of large numbers and on the weak law of large numbers for tail series,Publicationes Mathematicae Debrecen, 73/1-2 (2008) 1–10.

[9] Tómács, T., A general method to obtain the rate of convergence in the strong law of large numbers,Annales Mathematicae et Informaticae, 34 (2007) 97–102.

Tibor Tómács

Department of Applied Mathematics Eszterházy Károly College

P.O. Box 43 H-3301 Eger Hungary

e-mail: [email protected]

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