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On strong law of large numbers and growth rate for a class of random variables

Abstract

In this paper, we study the strong law of large numbers for a class of random variables satisfying the maximal moment inequality with exponent 2. Our results embrace the Kolmogorov strong law of large numbers and the Marcinkiewicz strong law of large numbers for this class of random variables. In addition, strong growth rate for weighted sums of this class of random variables is presented.

MSC:60F15.

1 Introduction

Let { X n ,n1} be a sequence of random variables defined on a fixed probability space (Ω,F,P). S n = i = 1 n X i , n1, S 0 =0. Let { a n ,n1} and { b n ,n1} be sequences of constant with 0< b n . Then { a n X n ,n1} is said to obey the general strong law of large numbers (SLLN) with norming constant { b n ,n1} if the normed weighted sums

1 b n i = 1 n a i ( X i EX i )0almost surely (a.s.)
(1.1)

holds. Note that the SLLN of the form (1.1) embraces the Kolmogorov SLLN ( b n =n, a n =1) and the Marcinkiewicz SLLN ( b n = n 1 / r , a n =1, r>0). When b n = i = 1 n a i , fundamental results for the SLLN were obtained.

Under an independent assumption, many SLLNs for the weighted sums are obtained. One can refer to Adler and Rosalsky [1], Chow and Teicher [2], Fernholz and Teicher [3], Jamison et al. [4] and Teicher [5].

Under a pairwise independent assumption, Rosalsky [6] obtained some SLLNs for weighted sums of pairwise independent and identically distributed random variables. Sung [7] obtained sufficient conditions for (1.1) if { X n ,n1} is a sequence of pairwise independent random variables satisfying 0 x p 1 sup n 1 P(| X n |>x)dx<. Sung [8] presented the following result: 1 a n i = 1 n ( X i EX i I(| X i | a i ))0 a.s., where { a n } is a sequence of positive constants with a n n and { X n } is a sequence of pairwise independent and identically distributed random variables.

For more details about strong limit theorems for dependent case, one can refer to Wu [9], Wu and Jiang [10], Hu et al. [11], Shen et al. [12], Zhou et al. [13] and Zhou [14], and so forth.

Recently Sung [15] gave the following definition.

Definition 1.1 (Sung [15])

A random variable sequence { X n ,n1} is said to satisfy the maximal moment inequality with exponent 2 if for all nm1, there exists a constant C independent of n and m such that

E ( max m k n | i = m k X i | 2 ) C i = m n EX i 2 .
(1.2)

We can see that a wide class of mean zero random variables satisfies (1.2). Inspired by Sung [7, 15], we establish SLLN of the form (1.1) for a class of random variables satisfying the maximal moment inequality with exponent 2.

The rest of the paper is organized as follows. In Section 2, some preliminary definition and lemmas are presented. In Section 3, main results and their proofs are provided.

Throughout the paper, let I(A) be the indicator function of the set A. C denotes a positive constant not depending on n, which may be different in various places. Let { a n ,n1} and { b n ,n1} be sequences of positive numbers, a n b n represents that there exists a constant C>0 such that a n C b n for all n.

2 Preliminaries

The following lemmas and definition will be needed in this paper.

Lemma 2.1 (Sung [7])

Let { X n ,n1} be a sequence of random variables and put G(x)= sup n 1 P(| X n |>x) for x0. Assume that 0 x p 1 G(x)dx< for some 1p<2. Then

  1. (i)

    n = 1 P(| X n |> n 1 / p )<.

  2. (ii)

    n = 1 EX n 2 I(| X n | n 1 / p )/ n 2 / p <.

  3. (iii)

    EX n I(| X n |> c n )0 for any sequence { c n ,n1} satisfying c n .

Lemma 2.2 (Sung [15])

Let { X n ,n1} be a sequence of random variables satisfying the maximal moment inequality with exponent 2. If n = 1 EX n 2 <, then n = 1 X n converges almost surely.

Definition 2.3 A random variable sequence { X n ,n1} is said to be stochastically dominated by a random variable X if there exists a constant C such that

P ( | X n | > x ) CP ( | X | > x )

for all x0 and n1.

Lemma 2.4 Let { X n ,n1} be a sequence of random variables which is stochastically dominated by a random variable X. For any α>0 and b>0, the following statement holds:

E| X n | α I ( | X n | b ) C { E | X | α I ( | X | b ) + b α P ( | X | > b ) } ,
(2.1)
E| X n | α I ( | X n | > b ) CE|X | α I ( | X | > b ) ,
(2.2)

where C is a positive constant.

Lemma 2.5 (Hu [16])

Let b 1 , b 2 , be a nondecreasing unbounded sequence of positive numbers. Let α 1 , α 2 , be nonnegative numbers, and Λ k = α 1 ++ α k for k1. Let r be a fixed positive number. Assume that for each n1,

E ( max 1 k n | S k | ) r C k = 1 n α k .
(2.3)

If

l = 1 Λ l ( 1 b l r 1 b l + 1 r ) <,
(2.4)
Λ n b n r is bounded,
(2.5)

then

lim n S n b n =0a.s.,
(2.6)

and with the growth rate

S n b n =O ( β n b n ) a.s.,
(2.7)

where

β n = max 1 k n b k ν k δ / r ,0<δ<1, v n = k = n α k b k r , lim n β n b n =0.
(2.8)

And

E ( max 1 l n | S l b l | ) r 4C l = 1 n α l b l r <,
(2.9)
E ( sup l 1 | S l b l | ) r 4C l = 1 α l b l r <.
(2.10)

If we further assume that α n >0 for infinitely many n, then

E ( sup l 1 | S l β l | ) r 4C l = 1 α l β l r <.
(2.11)

Proof It follows from Corollary 2.1.1 of Hu [16] that (2.6)-(2.8) hold. By (2.3) and Theorem 1.1 of Fazekas and Klesov [17], we have

E ( max 1 l n | S l b l | ) r 4C l = 1 n α l b l r 4C l = 1 α l b l r <.
(2.12)

Therefore

E ( sup l 1 | S l b l | ) r = lim n E ( max 1 l n | S l b l | ) r 4C l = 1 α l b l r <,
(2.13)

following from the monotone convergence theorem of Rao [18]. Equation (2.11) follows from the proof of Lemma 1.2 of Hu and Hu [19]. □

3 Main results

Theorem 3.1 Let { X n ,n1} be a sequence of random variables and put G(x)= sup n 1 P(| X n |>x) for x>0. Denote Y n = n 1 / p I( X n < n 1 / p )+ X n I(| X n | n 1 / p )+ n 1 / p I( X n > n 1 p ), n1, where p is a positive constant. Let { a n ,n1} and { b n ,n1} be sequences of positive numbers with b n . Suppose that { a n b n ( Y n E Y n ),n1} satisfies the maximal moment inequality with exponent 2. Assume that the following two conditions hold:

i = 1 n a i =O( b n ),
(3.1)
a n b n =O ( n 1 / p ) for some 1p<2.
(3.2)

If 0 x p 1 G(x)dx<, then

lim n 1 b n i = 1 n a i ( X i EX i )=0a.s.
(3.3)

Proof By Lemma 2.1(i),

n = 1 P( X n Y n )= n = 1 P ( | X n | > n 1 / p ) <.
(3.4)

Therefore P( X n Y n , i.o.)=0 follows from the Borel-Cantelli lemma and (3.4). Thus (3.3) is equivalent to the following:

lim n 1 b n i = 1 n a i ( Y i EX i )=0a.s.
(3.5)

So, in order to prove (3.3), we need only to prove

lim n 1 b n i = 1 n a i ( Y i E Y i )=0a.s.
(3.6)

and

lim n 1 b n i = 1 n a i ( EX i E Y i )=0.
(3.7)

Firstly, we prove (3.6). In view of Lemma 2.1(i), (ii) and (3.2), we have

n = 1 E ( a n b n ( Y n E Y n ) ) 2 n = 1 a n 2 b n 2 E Y n 2 = n = 1 a n 2 b n 2 [ EX n 2 I ( | X n | n 1 / p ) + n 2 / p E I ( | X n | > n 1 / p ) ] n = 1 [ n 2 / p EX n 2 I ( | X n | n 1 / p ) + P ( | X n | > n 1 / p ) ] < .

Thus it follows by Lemma 2.2 that

n = 1 a n b n ( Y n E Y n )converges a.s.
(3.8)

By Kronecker’s lemma, we can obtain (3.6) immediately.

Secondly, we prove (3.7). By (3.1) and Lemma 2.1(iii), we can get

lim n 1 b n i = 1 n a i E X i I ( | X i | > i 1 / p ) =0.
(3.9)

By (3.2) and Lemma 2.1(i), we have

n = 1 a n b n n 1 / p P ( X n > n 1 / p ) n = 1 P ( X n > n 1 / p ) < n = 1 P ( | X n | > n 1 / p ) <

and

n = 1 a n b n n 1 / p P ( X n < n 1 / p ) <.

Thus it follows by Kronecker’s lemma that

lim n 1 b n i = 1 n a i i 1 / p P ( X i > i 1 / p ) =0
(3.10)

and

lim n 1 b n i = 1 n a i i 1 / p P ( X i < i 1 / p ) =0.
(3.11)

Therefore,

lim n 1 b n i = 1 n a i ( EX i E Y i ) = lim n [ 1 b n i = 1 n a i E X i I ( | X i | > i 1 / p ) + 1 b n i = 1 n a i i 1 / p P ( X i < i 1 / p ) 1 b n i = 1 n a i i 1 / p P ( X i > i 1 / p ) ] = 0

follows from (3.9)-(3.11). Hence the result is proved. □

Theorem 3.2 Let { X n ,n1} be a sequence of mean zero random variables, which is stochastically dominated by a random variable X. Let { a n ,n1} and { b n ,n1} be sequences of positive numbers with b n . Put c n = b n a n for n1, 1<r<2. Denote Y n = c n I( X n < c n )+ X n I(| X n | c n )+ c n I( X n > c n ), n1, and suppose that { a n b n ( Y n E Y n ),n1} satisfies the maximal moment inequality with exponent 2. Assume that the following two conditions hold:

E|X | r <,
(3.12)
N(n)=Card{i: c i n} n r ,n1.
(3.13)

Then

lim n 1 b n i = 1 n a i X i =0a.s.
(3.14)

Proof Let N(0)=0. By (3.13), we can see that c n as n. By (3.12) and (3.13),

n = 1 P ( X n Y n ) = n = 1 P ( | X n | > c n ) n = 1 P ( | X | > c n ) = n = 1 c n j < c n + 1 P ( | X | > c n ) j = 1 j 1 < c n j P ( | X | > j 1 ) = j = 1 ( N ( j ) N ( j 1 ) ) k = j P ( k 1 < | X | k ) = k = 1 P ( k 1 < | X | k ) j = 1 k ( N ( j ) N ( j 1 ) ) = k = 1 N ( k ) P ( k 1 < | X | k ) k = 1 k r P ( k 1 < | X | k ) E | X | r < ,
(3.15)

which implies P( X i Y i , i.o.)=0 from the Borel-Cantelli lemma. So, in order to prove (3.14), we need only to prove

lim n 1 b n i = 1 n a i Y i =0a.s.
(3.16)

By (3.12), (3.13), Lemma 2.4, and the proof of (3.15), we have

n = 1 E ( a n b n ( Y n E Y n ) ) 2 n = 1 c n 2 E Y n 2 = n = 1 c n 2 [ EX n 2 I ( | X n | c n ) + E ( c n 2 I ( | X n | > c n ) ) ] n = 1 c n 2 EX 2 I ( | X | c n ) + n = 1 P ( | X | > c n ) = n = 1 c n j < c n + 1 c n 2 EX 2 I ( | X | c n ) + n = 1 P ( | X | > c n ) j = 1 j 1 < c n j c n 2 EX 2 I ( | X | j ) + C j = 2 ( N ( j ) N ( j 1 ) ) ( j 1 ) 2 k = 1 j EX 2 I ( k 1 < | X | k ) + C = j = 2 ( N ( j ) N ( j 1 ) ) ( j 1 ) 2 [ EX 2 I ( 0 < | X | 1 ) + k = 2 j EX 2 I ( k 1 < | X | k ) ] + C = j = 2 ( N ( j ) N ( j 1 ) ) ( j 1 ) 2 EX 2 I ( 0 < | X | 1 ) + k = 2 EX 2 I ( k 1 < | X | k ) j = k ( N ( j ) N ( j 1 ) ) ( j 1 ) 2 + C j = 2 N ( j ) ( ( j 1 ) 2 j 2 ) EX 2 I ( 0 < | X | 1 ) + k = 2 EX 2 I ( k 1 < | X | k ) j = k N ( j ) ( ( j 1 ) 2 j 2 ) + C j = 2 j r 3 + k = 2 EX 2 I ( k 1 < | X | k ) j = k j r 3 + C k = 2 k r 2 E ( | X | r k 2 r I ( k 1 < | X | k ) ) + C E | X | r + C < .
(3.17)

Combining Lemma 2.2, (3.17) and Kronecker’s lemma, we can get

lim n 1 b n i = 1 n a i ( Y i E Y i )=0a.s.
(3.18)

To complete the proof of (3.16), it suffices to show that

lim n 1 b n i = 1 n a i E Y i =0.
(3.19)

By (3.12), (3.13) and EX n =0, it follows that

n = 1 | a n b n E Y n | n = 1 c n 1 [ E | X n | I ( | X n | > c n ) + E ( c n I ( | X n | > c n ) ) ] = n = 1 c n 1 E | X n | I ( | X n | > c n ) + n = 1 P ( | X n | > c n ) .
(3.20)

Observe that

n = 1 c n 1 E | X n | I ( | X n | > c n ) C n = 1 c n 1 E | X | I ( | X | > c n ) = C n = 1 c n j < c n + 1 c n 1 E | X | I ( | X | > c n ) C j = 1 j 1 < c n j c n 1 E | X | I ( | X | > j 1 ) C j = 2 ( N ( j ) N ( j 1 ) ) ( j 1 ) 1 n = j 1 E | X | I ( n < | X | n + 1 ) = C n = 1 E | X | I ( n < | X | n + 1 ) j = 2 n + 1 ( N ( j ) N ( j 1 ) ) ( j 1 ) 1 C n = 1 E | X | I ( n < | X | n + 1 ) j = 2 n N ( j ) ( ( j 1 ) 1 j 1 ) + C n = 1 E | X | I ( n < | X | n + 1 ) N ( n + 1 ) n C n = 1 E | X | I ( n < | X | n + 1 ) j = 2 n j r 2 + C n = 1 n r 1 E | X | I ( n < | X | n + 1 ) C n = 1 n r 1 E | X | I ( n < | X | n + 1 ) C n = 1 E | X | r I ( n < | X | n + 1 ) C E | X | r < .
(3.21)

So, we can get

n = 1 | a n b n E Y n |<

from (3.15), (3.20) and (3.21). Consequently,

n = 1 a n b n E Y n converges,
(3.22)

which implies (3.19) from Kronecker’s lemma. We complete the proof of theorem. □

Theorem 3.3 Let { X n ,n1} be a sequence of mean zero random variables satisfying the maximal moment inequality with exponent 2. Denote Q n = max 1 k n EX k 2 , n1 and Q 0 =0. For 1p<2, assume that

n = 1 Q n n 2 / p <.
(3.23)

Then

lim n S n n 1 / p =0a.s.,
(3.24)

and with the growth rate

S n n 1 / p =O ( β n n 1 / p ) a.s.,
(3.25)

where

β n = max 1 k n k 1 / p ν k δ / 2 , 0 < δ < 1 , v n = k = n α k k 2 / p , α k = C ( k Q k ( k 1 ) Q k 1 ) , k 1 , lim n β n n 1 / p = 0 .
(3.26)

And

E ( max 1 l n | S l l 1 / p | 2 ) 4 l = 1 n α l l 2 / p <,
(3.27)
E ( sup l 1 | S l l 1 / p | 2 ) 4 l = 1 α l l 2 / p <.
(3.28)

If we further assume that α n >0 for infinitely many n, then

E ( sup l 1 | S l β l | 2 ) 4 l = 1 α l β l 2 <.
(3.29)

In addition, for any 0<r<2,

E ( sup l 1 | S l l 1 / p | r ) 1+ 4 r 2 r l = 1 α l l 2 / p <.
(3.30)

Proof Since { X n ,n1} is a sequence of mean zero random variables satisfying the maximal moment inequality with exponent 2, we have

E ( max 1 j n | k = 1 j X k | 2 ) C k = 1 n EX k 2 Cn Q n = k = 1 n α k .
(3.31)

And we can obtain α k 0 for all k1 from its definition. Denote b n = n 1 / p and Λ n = k = 1 n α k , n1. By (3.23), we can get

l = 1 Λ l ( 1 b l 2 1 b l + 1 2 ) =C l = 1 l Q l ( 1 l 2 / p 1 ( l + 1 ) 2 / p ) 2 C p l = 1 Q l l 2 / p <.
(3.32)

Thus (2.4) holds. It follows from Remark 2.1 in [16] that (2.4) implies (2.5). By Lemma 2.5, we can get (3.24)-(3.29) immediately. It follows from (3.28) that

E ( sup l 1 | S l l 1 / p | r ) = 0 P ( sup l 1 | S l l 1 / p | r > t ) d t = 0 1 P ( sup l 1 | S l l 1 / p | r > t ) d t + 1 P ( sup l 1 | S l l 1 / p | r > t ) d t 1 + E ( sup l 1 | S l l 1 / p | 2 ) 1 t 2 / r d t 1 + 4 r 2 r l = 1 α l l 2 / p < .

The proof is completed. □

Remark 3.4 It is easy to see that a wide class of mean zero random variables satisfies the maximal moment inequality with exponent 2. Examples include independent random variables, negatively associated random variables (see Matula [20]), negatively superadditive dependent random variables (see Shen et al. [12]), φ-mixing random variables and AANA random variables (see Wang et al. [21, 22]), and ρ ˜ -mixing random variables (see Utev et al. [23]). So Theorems 3.1-3.3 hold for this wide class of random variables.

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Acknowledgements

The authors are most grateful to the editor and the anonymous referee for their careful reading and insightful comments. This work is supported by the National Natural Science Foundation of China (11171001, 11201001), Natural Science Foundation of Anhui Province (1208085QA03), Humanities and Social Sciences Project from Ministry of Education of China (12YJC91007), Key Program of Research and Development Foundation of Hefei University (13KY05ZD) and Doctoral Research Start-up Funds Projects of Anhui University.

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Shen, Y., Yang, J. & Hu, S. On strong law of large numbers and growth rate for a class of random variables. J Inequal Appl 2013, 563 (2013). https://doi.org/10.1186/1029-242X-2013-563

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Keywords

  • strong law of large numbers
  • weighted sums
  • with exponent 2