# Complete Moment Convergence for Arrays of Rowwiseφ-Mixing Random Variables

Authors: Lulu Zheng; Yanfang Liu; Xuejun Wang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We investigate the complete moment convergence for maximal partial sum of arrays of rowwiseφ-mixing random variables under some more general conditions. The results obtained in the paper generalize and improve some known ones.


---

## Body

## 1. Introduction

Let{
X
n
,
n
≥
1
} be a sequence of random variables defined on a fixed probability space (
Ω
,
ℱ
,
P
). Let n and m be positive integers. Write ℱ
n
m
=
σ
(
X
i
,
n
≤
i
≤
m
). Given σ-algebras ℬ
,
ℛ in ℱ, let
(1)
φ
(
ℬ
,
ℛ
)
=
sup
⁡
A
∈
ℬ
,
B
∈
ℛ
,
P
(
A
)
>
0
|
P
(
B
∣
A
)
-
P
(
B
)
|
.
Define the φ-mixing coefficients by
(2)
φ
(
n
)
=
sup
⁡
k
≥
1
φ
(
ℱ
1
k
,
ℱ
k
+
n
∞
)
,
n
≥
0
.A random variable sequence{
X
n
,
n
≥
1
} is said to be φ-mixing if φ
(
n
)
↓
0 as n
→
∞. φ
(
n
) is called mixing coefficient. A triangular array of random variables {
X
n
k
,
k
≥
1
,
n
≥
1
} is said to be an array of rowwise φ-mixing random variables if, for every n
≥
1, {
X
n
k
,
k
≥
1
} is a φ-mixing sequence of random variables. The notion of φ-mixing random variables was introduced by Dobrushin [1] and many applications have been found. See, for example, Utev [2] for central limit theorem, Gan and Chen [3] for limit theorem, Peligrad [4] for weak invariance principle, Shao [5] for almost sure invariance principles, Chen and Wang [6], Shen et al. [7, 8], Wu [9], and Wang et al. [10] for complete convergence, Hu and Wang [11] for large deviations, and so forth. When these are compared with corresponding results of independent random variable sequences, there still remains much to be desired.Definition 1.
A sequence of random variables{
U
n
,
n
≥
1
} is said to converge completely to a constant a if, for any ɛ
>
0,
(3)
∑
n
=
1
∞
‍
P
(
|
U
n
-
a
|
>
ɛ
)
<
∞
.
In this case, one writes U
n
→
a completely. This notion was given first by Hsu and Robbins [12].Definition 2.
Let{
Z
n
,
n
≥
1
} be a sequence of random variables and a
n
>
0, b
n
>
0, and q
>
0. If
(4)
∑
n
=
1
∞
‍
a
n
E
{
b
n
-
1
|
Z
n
|
-
ɛ
}
+
q
<
∞
∀
ɛ
>
0
,
then the above result was called the complete moment convergence by Chow [13].Let{
X
n
k
,
k
≥
1
,
n
≥
1
} be an array of rowwise φ-mixing random variables with mixing coefficients {
φ
(
n
)
,
n
≥
1
} in each row, let {
a
n
,
n
≥
1
} be a sequence of positive real numbers such that a
n
↑
∞, and let {
Ψ
k
(
t
)
,
k
≥
1
} be a sequence of positive even functions such that
(5)
Ψ
k
(
|
t
|
)
|
t
|
q
↑
,
Ψ
k
(
|
t
|
)
|
t
|
p
↓
as
|
t
|
↑
for some 1
≤
q
<
p and each k
≥
1. In order to prove our results, we mention the following conditions:
(6)
E
X
n
k
=
0
,
k
≥
1
,
n
≥
1
,
(7)
∑
n
=
1
∞
∑
k
=
1
n
‍
E
Ψ
k
(
X
n
k
)
Ψ
k
(
a
n
)
<
∞
,
(8)
∑
n
=
1
∞
‍
(
∑
k
=
1
n
‍
E
(
X
n
k
a
n
)
2
)
v
/
2
<
∞
,
where v
≥
p is a positive integer.The following are examples of functionΨ
k
(
t
) satisfying assumption (5): Ψ
k
(
t
)
=
|
t
|
β for some q
<
β
<
p or Ψ
k
(
t
)
=
|
t
|
q
log
⁡
(
1
+
|
t
|
p
-
q
) for t
∈
(
-
∞
,
+
∞
). Note that these functions are nonmonotone on t
∈
(
-
∞
,
+
∞
), while it is simple to show that, under condition (5), the function Ψ
k
(
t
) is an increasing function for t
>
0. In fact, Ψ
k
(
t
)
=
(
Ψ
k
(
t
)
/
|
t
|
q
)
·
|
t
|
q, t
>
0, and |
t
|
q
↑ as |
t
|
↑; then we have Ψ
k
(
t
)
↑.Recently Gan et al. [14] obtained the following complete convergence for φ-mixing random variables.Theorem A.
Let{
X
n
,
n
≥
1
} be a sequence of φ-mixing mean zero random variables with ∑
n
=
1
∞
‍
φ
1
/
2
(
n
)
<
∞, let {
a
n
,
n
≥
1
} be a sequence of positive real numbers with a
n
↑
∞, and let {
Ψ
n
(
t
)
,
n
≥
1
} be a sequence of nonnegative even functions such that Ψ
n
(
t
)
>
0 as t
>
0 and (
Ψ
n
(
|
t
|
)
/
|
t
|
)
↑ and (
Ψ
n
(
|
t
|
)
/
|
t
|
p
)
↓ as |
t
|
↑
∞, where p
≥
2. If the following conditions are satisfied:
(9)
∑
n
=
1
∞
∑
k
=
1
n
‍
E
Ψ
k
(
X
k
)
Ψ
k
(
a
n
)
<
∞
,
(10)
∑
n
=
1
∞
‍
[
∑
k
=
1
n
‍
E
|
X
k
|
r
a
n
r
]
s
<
∞
,
where 0
<
r
≤
2, s
>
0, then
(11)
1
a
n
max
⁡
1
≤
j
≤
n
|
∑
k
=
1
j
‍
X
k
|
⟶
0
c
o
m
p
l
e
t
e
l
y
.For more details about this type of complete convergence, one can refer to Gan and Chen [3],  Wu et al. [15], Wu [16], Huang et al. [17], Shen [18], Shen et al. [19, 20], and so on. The purpose of this paper is extending Theorem A to the complete moment convergence, which is a more general version of the complete convergence, and making some improvements such that the conditions are more general. In this work, the symbol C always stands for a generic positive constant, which may vary from one place to another.


## 2. Preliminary Lemmas

In this section, we give the following lemma which will be used to prove our main results.Lemma 3 (cf. Wang et al. [10]).
Let{
X
n
,
n
≥
1
} be a sequence of φ-mixing random variables satisfying ∑
n
=
1
∞
‍
φ
1
/
2
(
n
)
<
∞, p
≥
2. Assume that E
X
n
=
0, and E
|
X
n
|
p
<
∞, for each n
≥
1. Then there exists a constant C depending only on p and φ
(
·
) such that
(12)
E
(
max
⁡
1
≤
j
≤
n
|
∑
i
=
a
+
1
a
+
j
‍
X
i
|
p
)
≤
C
[
∑
i
=
a
+
1
a
+
n
‍
E
|
X
i
|
p
+
(
∑
i
=
a
+
1
a
+
n
‍
E
X
i
2
)
p
/
2
]
,
for every a
≥
0 and n
≥
1. In particular, one has
(13)
E
(
max
⁡
1
≤
j
≤
n
|
∑
i
=
1
j
‍
X
i
|
p
)
≤
C
[
∑
i
=
1
n
‍
E
|
X
i
|
p
+
(
∑
i
=
1
n
‍
E
X
i
2
)
p
/
2
]
,
for every n
≥
1.


## 3. Main Results and Their Proofs

Let{
X
n
k
,
k
≥
1
,
n
≥
1
} be an array of rowwise φ-mixing random variables and let φ
n
(
·
) be the mixing coefficient of {
X
n
k
,
k
≥
1
} for any n
≥
1. Our main results are as follows.Theorem 4.
Let{
X
n
k
,
k
≥
1
,
n
≥
1
} be an array of rowwise φ-mixing random variables satisfying sup
⁡
n
≥
1
∑
k
=
1
∞
‍
φ
n
1
/
2
(
k
)
<
∞ and let {
a
n
,
n
≥
1
} be a sequence of positive real numbers such that a
n
↑
∞. Also, let {
Ψ
k
(
t
)
,
k
≥
1
} be a positive even function satisfying (5) for 1
≤
q
<
p
≤
2. Then under conditions (6) and (7), one has
(14)
∑
n
=
1
∞
‍
a
n
-
q
E
{
max
⁡
1
≤
j
≤
n
|
∑
k
=
1
j
‍
X
n
k
|
-
ɛ
a
n
}
+
q
<
∞
,
∀
ɛ
>
0
.Proof.
Firstly, let us prove the following statements from conditions (5) and (7).
(i) Forr
≥
1, 0
<
u
≤
q,
(15)
∑
n
=
1
∞
‍
(
∑
k
=
1
n
‍
E
|
X
n
k
|
u
I
(
|
X
n
k
|
>
a
n
)
a
n
u
)
r
≤
∑
n
=
1
∞
‍
(
∑
k
=
1
n
‍
E
|
X
n
k
|
q
I
(
|
X
n
k
|
>
a
n
)
a
n
q
)
r
≤
∑
n
=
1
∞
‍
(
∑
k
=
1
n
‍
E
Ψ
k
(
X
n
k
)
Ψ
k
(
a
n
)
)
r
≤
(
∑
n
=
1
∞
∑
k
=
1
n
‍
E
Ψ
k
(
X
n
k
)
Ψ
k
(
a
n
)
)
r
<
∞
.
(ii) Forv
≥
p,
(16)
∑
n
=
1
∞
∑
k
=
1
n
‍
E
|
X
n
k
|
v
I
(
|
X
n
k
|
≤
a
n
)
a
n
v
≤
∑
n
=
1
∞
∑
k
=
1
n
‍
E
|
X
n
k
|
p
I
(
|
X
n
k
|
≤
a
n
)
a
n
p
≤
∑
n
=
1
∞
∑
k
=
1
n
‍
E
Ψ
k
(
X
n
k
)
Ψ
k
(
a
n
)
<
∞
.
For n
≥
1, denote M
n
(
X
)
=
max
⁡
1
≤
j
≤
n
|
∑
k
=
1
j
‍
X
n
k
|. It is easy to check that
(17)
∑
n
=
1
∞
‍
a
n
-
q
E
{
M
n
(
X
)
-
ɛ
a
n
}
+
q
=
∑
n
=
1
∞
‍
a
n
-
q
∫
0
∞
‍
P
{
M
n
(
X
)
-
ɛ
a
n
>
t
1
/
q
}
d
t
=
∑
n
=
1
∞
‍
a
n
-
q
(
∫
0
a
n
q
‍
P
{
M
n
(
X
)
>
ɛ
a
n
+
t
1
/
q
}
d
t
+
∫
a
n
q
∞
‍
P
{
M
n
(
X
)
>
ɛ
a
n
+
t
1
/
q
}
d
t
∫
0
a
n
q
‍
P
{
M
n
(
X
)
>
ɛ
a
n
+
t
1
/
q
}
d
t
)
≤
∑
n
=
1
∞
‍
P
{
M
n
(
X
)
>
ɛ
a
n
}
+
∑
n
=
1
∞
‍
a
n
-
q
∫
a
n
q
∞
‍
P
{
M
n
(
X
)
>
t
1
/
q
}
d
t
≐
I
1
+
I
2
.
To prove (14), it suffices to prove that I
1
<
∞ and I
2
<
∞. Now let us prove them step by step. Firstly, we prove that I
1
<
∞.
For alln
≥
1, define
(18)
X
k
(
n
)
=
X
n
k
I
(
|
X
n
k
|
≤
a
n
)
,
T
j
(
n
)
=
1
a
n
∑
k
=
1
j
‍
(
X
k
(
n
)
-
E
X
k
(
n
)
)
,
then for all ɛ
>
0, it is easy to have
(19)
P
(
max
⁡
1
≤
j
≤
n
|
1
a
n
∑
k
=
1
j
‍
X
n
k
|
>
ɛ
)
≤
P
(
max
⁡
1
≤
j
≤
n
|
X
n
k
|
>
a
n
)
+
P
(
max
⁡
1
≤
j
≤
n
|
T
j
(
n
)
|
>
ɛ
-
max
⁡
1
≤
j
≤
n
|
1
a
n
∑
k
=
1
j
‍
E
X
k
(
n
)
|
)
.
By (5), (6), (7), and (15) we have
(20)
max
⁡
1
≤
j
≤
n
|
1
a
n
∑
k
=
1
j
‍
E
X
k
(
n
)
|
=
max
⁡
1
≤
j
≤
n
|
1
a
n
∑
k
=
1
j
‍
E
X
n
k
I
(
|
X
n
k
|
≤
a
n
)
|
=
max
⁡
1
≤
j
≤
n
|
1
a
n
∑
k
=
1
j
‍
E
X
n
k
I
(
|
X
n
k
|
>
a
n
)
|
≤
∑
k
=
1
n
‍
E
|
X
n
k
|
I
(
|
X
n
k
|
>
a
n
)
a
n
⟶
0
as
n
⟶
∞
.
From (19) and (20), it follows that, for n large enough,
(21)
P
(
max
⁡
1
≤
j
≤
n
|
1
a
n
∑
k
=
1
j
‍
X
n
k
|
>
ɛ
)
≤
∑
k
=
1
n
‍
P
(
|
X
n
k
|
>
a
n
)
+
P
(
max
⁡
1
≤
j
≤
n
|
T
j
(
n
)
|
>
ɛ
2
)
.
Hence we only need to prove that
(22)
I
≐
∑
n
=
1
∞
∑
k
=
1
n
‍
P
(
|
X
n
k
|
>
a
n
)
<
∞
,
I
I
≐
∑
n
=
1
∞
‍
P
(
max
⁡
1
≤
j
≤
n
|
T
j
(
n
)
|
>
ɛ
2
)
<
∞
.
For I, it follows by (15) that
(23)
I
=
∑
n
=
1
∞
∑
k
=
1
n
‍
E
I
(
|
X
n
k
|
>
a
n
)
≤
∑
n
=
1
∞
∑
k
=
1
n
‍
E
|
X
n
k
|
q
I
(
|
X
n
k
|
>
a
n
)
a
n
q
<
∞
.
For I
I, take r
≥
2. Since p
≤
2, r
≥
p, we have by Markov inequality, Lemma 3, C
r-inequality, and (16) that
(24)
I
I
≤
∑
n
=
1
∞
‍
(
ɛ
2
)
-
r
E
max
⁡
1
≤
j
≤
n
|
T
j
(
n
)
|
r
≤
C
∑
n
=
1
∞
‍
(
ɛ
2
)
-
r
1
a
n
r
[
∑
k
=
1
n
‍
E
|
X
k
(
n
)
|
r
+
(
∑
k
=
1
n
‍
E
|
X
k
(
n
)
|
2
)
r
/
2
]
≤
C
∑
n
=
1
∞
∑
k
=
1
n
‍
E
|
X
k
(
n
)
|
r
a
n
r
+
C
∑
n
=
1
∞
‍
(
∑
k
=
1
n
‍
E
|
X
k
(
n
)
|
2
a
n
2
)
r
/
2
≤
C
∑
n
=
1
∞
∑
k
=
1
n
‍
E
|
X
n
k
|
p
I
(
|
X
n
k
|
≤
a
n
)
a
n
p
+
C
∑
n
=
1
∞
‍
(
∑
k
=
1
n
‍
E
|
X
n
k
|
p
I
(
|
X
n
k
|
≤
a
n
)
a
n
p
)
r
/
2
≤
C
∑
n
=
1
∞
∑
k
=
1
n
‍
E
|
X
n
k
|
p
I
(
|
X
n
k
|
≤
a
n
)
a
n
p
+
C
(
∑
n
=
1
∞
∑
k
=
1
n
‍
E
|
X
n
k
|
p
I
(
|
X
n
k
|
≤
a
n
)
a
n
p
)
r
/
2
<
∞
.
Next we prove that I
2
<
∞. Denote Y
n
k
=
X
n
k
I
(
|
X
n
k
|
≤
t
1
/
q
), Z
n
k
=
X
n
k
-
Y
n
k, and M
n
(
Y
)
=
max
⁡
1
≤
j
≤
n
|
∑
k
=
1
j
‍
Y
n
k
|. Obviously,
(25)
P
{
M
n
(
X
)
>
t
1
/
q
}
≤
∑
k
=
1
n
‍
P
{
|
X
n
k
|
>
t
1
/
q
}
+
P
{
M
n
(
Y
)
>
t
1
/
q
}
.
Hence,
(26)
I
2
≤
∑
n
=
1
∞
∑
k
=
1
n
‍
a
n
-
q
∫
a
n
q
∞
‍
P
{
|
X
n
k
|
>
t
1
/
q
}
d
t
+
∑
n
=
1
∞
‍
a
n
-
q
∫
a
n
q
∞
‍
P
{
M
n
(
Y
)
>
t
1
/
q
}
d
t
≐
I
3
+
I
4
.
For I
3, by (15), we have
(27)
I
3
=
∑
n
=
1
∞
∑
k
=
1
n
‍
a
n
-
q
∫
a
n
q
∞
‍
P
{
|
X
n
k
|
I
(
|
X
n
k
|
>
a
n
)
>
t
1
/
q
}
d
t
≤
∑
n
=
1
∞
∑
k
=
1
n
‍
a
n
-
q
∫
0
∞
‍
P
{
|
X
n
k
|
I
(
|
X
n
k
|
>
a
n
)
>
t
1
/
q
}
d
t
=
∑
n
=
1
∞
∑
k
=
1
n
‍
E
|
X
n
k
|
q
I
(
|
X
n
k
|
>
a
n
)
a
n
q
<
∞
.
Now let us prove that I
4
<
∞. Firstly, it follows by (6) and (15) that
(28)
max
⁡
t
≥
a
n
q
max
⁡
1
≤
j
≤
n
t
-
1
/
q
|
∑
k
=
1
j
‍
E
Y
n
k
|
=
max
⁡
t
≥
a
n
q
max
⁡
1
≤
j
≤
n
t
-
1
/
q
|
∑
k
=
1
j
‍
E
Z
n
k
|
≤
max
⁡
t
≥
a
n
q
t
-
1
/
q
∑
k
=
1
n
‍
E
|
X
n
k
|
I
(
|
X
n
k
|
>
t
1
/
q
)
≤
∑
k
=
1
n
‍
a
n
-
1
E
|
X
n
k
|
I
(
|
X
n
k
|
>
a
n
)
≤
∑
k
=
1
n
‍
E
|
X
n
k
|
q
I
(
|
X
n
k
|
>
a
n
)
a
n
q
⟶
0
as
n
⟶
∞
.
Therefore, for n sufficiently large,
(29)
max
⁡
1
≤
j
≤
n
|
∑
k
=
1
j
‍
E
Y
n
k
|
≤
t
1
/
q
2
,
t
≥
a
n
q
.
Then for n sufficiently large,
(30)
P
{
M
n
(
Y
)
>
t
1
/
q
}
≤
P
{
max
⁡
1
≤
j
≤
n
|
∑
k
=
1
j
‍
(
Y
n
k
-
E
Y
n
k
)
|
>
t
1
/
q
2
}
,
t
≥
a
n
q
.
Let d
n
=
[
a
n
]
+
1. By (30), Lemma 3, and C
r-inequality, we can see that
(31)
I
4
≤
C
∑
n
=
1
∞
‍
a
n
-
q
∫
a
n
q
∞
‍
t
-
2
/
q
E
(
max
⁡
1
≤
j
≤
n
|
∑
k
=
1
j
‍
(
Y
n
k
-
E
Y
n
k
)
|
)
2
d
t
≤
C
∑
n
=
1
∞
‍
a
n
-
q
∫
a
n
q
∞
‍
t
-
2
/
q
∑
k
=
1
n
‍
E
(
Y
n
k
-
E
Y
n
k
)
2
d
t
≤
C
∑
n
=
1
∞
‍
∑
k
=
1
n
‍
a
n
-
q
∫
a
n
q
∞
‍
t
-
2
/
q
E
Y
n
k
2
d
t
=
C
∑
n
=
1
∞
∑
k
=
1
n
‍
a
n
-
q
∫
a
n
q
∞
‍
t
-
2
/
q
E
X
n
k
2
I
(
|
X
n
k
|
≤
d
n
)
d
t
+
C
∑
n
=
1
∞
∑
k
=
1
n
‍
a
n
-
q
∫
d
n
q
∞
‍
t
-
2
/
q
E
X
n
k
2
I
(
d
n
<
|
X
n
k
|
≤
t
1
/
q
)
d
t
≐
I
41
+
I
42
.
For I
41, since q
<
2, we have
(32)
I
41
=
C
∑
n
=
1
∞
∑
k
=
1
n
‍
a
n
-
q
E
X
n
k
2
I
(
|
X
n
k
|
≤
d
n
)
∫
a
n
q
∞
‍
t
-
2
/
q
d
t
≤
C
∑
n
=
1
∞
∑
k
=
1
n
‍
E
X
n
k
2
I
(
|
X
n
k
|
≤
d
n
)
a
n
2
=
C
∑
n
=
1
∞
∑
k
=
1
n
‍
E
X
n
k
2
I
(
|
X
n
k
|
≤
a
n
)
a
n
2
+
C
∑
n
=
1
∞
∑
k
=
1
n
‍
E
X
n
k
2
I
(
a
n
<
|
X
n
k
|
≤
d
n
)
a
n
2
≐
I
41
′
+
I
41
′′
.
Since p
≤
2, by (16), it implies I
41
′
<
∞. Now we prove that I
41
′′
<
∞. Since q
<
2 and (
a
n
+
1
)
/
a
n
→
1 as n
→
∞, by (15) we have
(33)
I
41
′′
≤
C
∑
n
=
1
∞
∑
k
=
1
n
‍
d
n
2
-
q
a
n
2
E
|
X
n
k
|
q
I
(
a
n
<
|
X
n
k
|
≤
d
n
)
≤
C
∑
n
=
1
∞
∑
k
=
1
n
‍
(
a
n
+
1
a
n
)
2
-
q
E
|
X
n
k
|
q
I
(
|
X
n
k
|
>
a
n
)
a
n
q
≤
C
∑
n
=
1
∞
∑
k
=
1
n
‍
E
|
X
n
k
|
q
I
(
|
X
n
k
|
>
a
n
)
a
n
q
<
∞
.
Let t
=
u
q in I
42. Note that, for q
<
2,
(34)
∫
d
n
∞
‍
u
q
-
3
E
X
n
k
2
I
(
d
n
<
|
X
n
k
|
≤
u
)
d
u
=
∫
d
n
∞
‍
u
q
-
3
E
X
n
k
2
I
(
|
X
n
k
|
>
d
n
)
·
I
(
|
X
n
k
|
≤
u
)
d
u
=
E
[
X
n
k
2
I
(
|
X
n
k
|
>
d
n
)
∫
|
X
n
k
|
∞
‍
u
q
-
3
I
(
|
X
n
k
|
≤
u
)
d
u
]
=
E
[
X
n
k
2
I
(
|
X
n
k
|
>
d
n
)
∫
|
X
n
k
|
∞
‍
u
q
-
3
d
u
]
≤
C
E
|
X
n
k
|
q
I
(
|
X
n
k
|
>
d
n
)
.
Then by (15) and d
n
>
a
n, we have
(35)
I
42
=
C
∑
n
=
1
∞
‍
∑
k
=
1
n
‍
a
n
-
q
∫
d
n
∞
‍
u
q
-
3
E
X
n
k
2
I
(
d
n
<
|
X
n
k
|
≤
u
)
d
u
≤
C
∑
n
=
1
∞
∑
k
=
1
n
‍
a
n
-
q
E
|
X
n
k
|
q
I
(
|
X
n
k
|
>
a
n
)
<
∞
.
This completes the proof of Theorem 4.Theorem 5.
Let{
X
n
k
,
k
≥
1
,
n
≥
1
} be an array of rowwise φ-mixing random variables satisfying sup
⁡
n
≥
1
∑
k
=
1
∞
‍
φ
n
1
/
2
(
k
)
<
∞ and let {
a
n
,
n
≥
1
} be a sequence of positive real numbers such that a
n
↑
∞. Also, let {
Ψ
k
(
t
)
,
k
≥
1
} be a positive even function satisfying (5) for 1
≤
q
<
p and p
>
2. Then conditions (6)–(8) imply (14).Proof.
Following the notation, by a similar argument as in the proof of Theorem4, we can easily prove that I
1
<
∞, I
3
<
∞ and that (19) and (20) hold. To complete the proof, we only need to prove that I
4
<
∞.
Letδ
≥
p and d
n
=
[
a
n
]
+
1. By (30), Markov inequality, Lemma 3, and the C
r-inequality we can get
(36)
I
4
≤
C
∑
n
=
1
∞
‍
a
n
-
q
∫
a
n
q
∞
‍
t
-
δ
/
q
E
max
⁡
1
≤
j
≤
n
|
∑
k
=
1
j
‍
(
Y
n
k
-
E
Y
n
k
)
|
δ
d
t
≤
C
∑
n
=
1
∞
‍
a
n
-
q
∫
a
n
q
∞
‍
t
-
δ
/
q
[
∑
k
=
1
n
‍
E
|
Y
n
k
|
δ
+
(
∑
k
=
1
n
‍
E
Y
n
k
2
)
δ
/
2
]
d
t
=
C
∑
n
=
1
∞
‍
∑
k
=
1
n
‍
a
n
-
q
∫
a
n
q
∞
‍
t
-
δ
/
q
E
|
Y
n
k
|
δ
d
t
+
C
∑
n
=
1
∞
‍
a
n
-
q
∫
a
n
q
∞
‍
t
-
δ
/
q
(
∑
k
=
1
n
‍
E
Y
n
k
2
)
δ
/
2
d
t
≐
I
43
+
I
44
.
For I
43, we have
(37)
I
43
=
C
∑
n
=
1
∞
∑
k
=
1
n
‍
a
n
-
q
∫
a
n
q
∞
‍
t
-
δ
/
q
E
|
X
n
k
|
δ
I
(
|
X
n
k
|
≤
d
n
)
d
t
+
C
∑
n
=
1
∞
∑
k
=
1
n
‍
a
n
-
q
∫
d
n
q
∞
‍
t
-
δ
/
q
E
|
X
n
k
|
δ
I
(
d
n
<
|
X
n
k
|
≤
t
1
/
q
)
d
t
≐
I
43
′
+
I
43
′
′
.
By a similar argument as in the proof of I
41
<
∞ and I
42
<
∞ (replacing the exponent 2 by δ), we can get I
43
′
<
∞ and I
43
′
′
<
∞.
ForI
44, since δ
>
2, we can see that
(38)
I
44
=
C
∑
n
=
1
∞
‍
a
n
-
q
∫
a
n
q
∞
‍
t
-
δ
/
q
(
∑
k
=
1
n
‍
E
X
n
k
2
I
(
|
X
n
k
|
≤
a
n
)

+
∑
k
=
1
n
‍
E
X
n
k
2
I
(
a
n

<

|
X
n
k
|

≤

t
1
/
q
)
)
δ
/
2
d
t


≤
C
∑
n
=
1
∞
‍
a
n
-
q
∫
a
n
q
∞
‍
t
-
δ
/
q
(
∑
k
=
1
n
‍
E
X
n
k
2
I
(
|
X
n
k
|
≤
a
n
)
)
δ
/
2
d
t

+
C
∑
n
=
1
∞
‍
a
n
-
q
∫
a
n
q
∞
‍
(
t
-
2
/
q
∑
k
=
1
n
‍
E
X
n
k
2
I
(
a
n

<

|
X
n
k
|

≤

t
1
/
q
)
)
δ
/
2
d
t


≐
I
44
′
+
I
44
′
′
.
Since δ
≥
p
>
q, from (8) we have
(39)
I
44
′
=
C
∑
n
=
1
∞
‍
a
n
-
q
(
∑
k
=
1
n
‍
E
X
n
k
2
I
(
|
X
n
k
|
≤
a
n
)
)
δ
/
2
∫
a
n
q
∞
‍
t
-
δ
/
q
d
t
≤
C
∑
n
=
1
∞
‍
(
∑
k
=
1
n
‍
E
X
n
k
2
I
(
|
X
n
k
|
≤
a
n
)
a
n
2
)
δ
/
2
≤
C
∑
n
=
1
∞
‍
(
∑
k
=
1
n
‍
E
X
n
k
2
a
n
2
)
δ
/
2
<
∞
.
Next we prove that I
44
′
′
<
∞. To start with, we consider the case 1
≤
q
≤
2. Since δ
>
2, by (15), we have
(40)
I
44
′
′
≤
C
∑
n
=
1
∞
‍
a
n
-
q
∫
a
n
q
∞
‍
(
t
-
1
∑
k
=
1
n
‍
E
|
X
n
k
|
q
I
(
a
n
<
|
X
n
k
|
≤
t
1
/
q
)
)
δ
/
2
d
t
≤
C
∑
n
=
1
∞
‍
a
n
-
q
∫
a
n
q
∞
‍
(
t
-
1
∑
k
=
1
n
‍
E
|
X
n
k
|
q
I
(
|
X
n
k
|
>
a
n
)
)
δ
/
2
d
t
=
C
∑
n
=
1
∞
‍
a
n
-
q
(
∑
k
=
1
n
‍
E
|
X
n
k
|
q
I
(
|
X
n
k
|
>
a
n
)
)
δ
/
2
∫
a
n
q
∞
‍
t
-
δ
/
2
d
t
≤
C
∑
n
=
1
∞
‍
(
∑
k
=
1
n
‍
E
|
X
n
k
|
q
I
(
|
X
n
k
|
>
a
n
)
a
n
q
)
δ
/
2
<
∞
.
Finally, we prove that I
44
′
′
<
∞ in the case 2
<
q
<
p. Since δ
>
q and δ
>
2, we have by (15) that
(41)
I
44
′
′
≤
C
∑
n
=
1
∞
‍
a
n
-
q
∫
a
n
q
∞
‍
(
t
-
2
/
q
∑
k
=
1
n
‍
E
X
n
k
2
I
(
|
X
n
k
|
>
a
n
)
)
δ
/
2
d
t
=
C
∑
n
=
1
∞
‍
a
n
-
q
(
∑
k
=
1
n
‍
E
X
n
k
2
I
(
|
X
n
k
|
>
a
n
)
)
δ
/
2
∫
a
n
q
∞
‍
t
-
δ
/
q
d
t
≤
C
∑
n
=
1
∞
‍
(
∑
k
=
1
n
‍
E
X
n
k
2
I
(
|
X
n
k
|
>
a
n
)
a
n
2
)
δ
/
2
<
∞
.
Thus we get the desired result immediately. The proof is completed.Corollary 6.
Let{
X
n
k
,
k
≥
1
,
n
≥
1
} be an array of rowwise φ-mixing mean zero random variables with sup
⁡
n
≥
1
∑
k
=
1
∞
‍
φ
n
1
/
2
(
k
)
<
∞, q
≥
1. If, for some α
>
0 and v
≥
2,
(42)
max
⁡
1
≤
k
≤
n
E
|
X
n
k
|
v
=
O
(
n
α
)
,
where (
v
/
q
)
-
α
>
max
⁡
{
v
/
2
,
2
}, v
≥
2, then, for any ɛ
>
0,
(43)
∑
n
=
1
∞
‍
n
-
1
E
{
max
⁡
1
≤
j
≤
n
|
∑
k
=
1
j
X
n
k
|
-
ɛ
n
1
/
q
}
+
q
<
∞
.Proof.
PutΨ
k
(
|
t
|
)
=
|
t
|
v, p
=
v
+
δ, δ
>
0, and a
n
=
n
1
/
q.
Sincev
≥
2, (
v
/
q
)
-
α
>
max
⁡
{
v
/
r
,
2
}, then
(44)
Ψ
k
(
|
t
|
)
|
t
|
q
=
|
t
|
v
-
q
↑
,
Ψ
k
(
|
t
|
)
|
t
|
p
=
|
t
|
v
|
t
|
p
=
1
|
t
|
δ
↓
as
|
t
|
↑
∞
.
It follows by (42) and (
v
/
q
)
-
α
>
2 that
(45)
∑
n
=
1
∞
∑
k
=
1
n
‍
E
Ψ
k
(
X
n
k
)
Ψ
k
(
a
n
)
=
∑
n
=
1
∞
∑
k
=
1
n
‍
E
|
X
n
k
|
v
n
v
/
q
≤
C
∑
n
=
1
∞
‍
1
n
(
v
/
q
)
-
α
-
1
<
∞
.
Since v
≥
2, by Jensen's inequality it follows that
(46)
∑
k
=
1
n
‍
E
|
X
n
k
|
2
n
2
/
q
≤
∑
k
=
1
n
‍
(
E
|
X
n
k
|
v
)
2
/
v
n
2
/
q
≤
C
1
n
(
2
/
q
)
-
(
2
α
/
v
)
-
1
.
Clearly (
2
/
q
)
-
(
2
α
/
v
)
-
1
>
0. Take s
>
p such that (
s
/
2
)
(
(
2
/
q
)
-
(
2
α
/
v
)
-
1
)
>
1. Therefore,
(47)
∑
n
=
1
∞
‍
[
∑
k
=
1
n
‍
E
|
X
n
k
|
2
n
2
/
q
]
s
/
2
<
∞
.
Combining Theorem 5 and (45)–(47), we can prove Corollary 6 immediately.Remark 7.
Noting that in this paper we consider the case1
≤
q
≤
p, which has a more wide scope than the case q
=
1, p
≥
2 in Gan et al. [14]. In addition, compared with φ-mixing random variables, the arrays of φ-mixing random variables not only have many related properties, but also have a wide range of application. So it is very significant to study it.Remark 8.
Under the condition of Theorem4, we have
(48)
∞
>
∑
n
=
1
∞
‍
a
n
-
q
E
{
max
⁡
1
≤
j
≤
n
|
∑
k
=
1
j
‍
X
n
k
|
-
ɛ
a
n
}
+
q
=
∑
n
=
1
∞
‍
a
n
-
q
∫
0
∞
‍
P
{
max
⁡
1
≤
j
≤
n
|
∑
k
=
1
j
‍
X
n
k
|
-
ɛ
a
n
>
t
1
/
q
}
d
t
≥
∑
n
=
1
∞
‍
a
n
-
q
∫
0
ɛ
q
a
n
q
‍
P
{
max
⁡
1
≤
j
≤
n
|
∑
k
=
1
j
‍
X
n
k
|
-
ɛ
a
n
>
ɛ
a
n
}
d
t
=
ɛ
q
∑
n
=
1
∞
‍
P
{
max
⁡
1
≤
j
≤
n
|
∑
k
=
1
j
‍
X
n
k
|
>
2
ɛ
a
n
}
.
Then we can obtain (11) directly. In this case, condition (10) is not needed. Especially, for p
=
2, the conditions of Theorem 4 are weaker than Theorem A. So Theorem 4 generalizes and improves it.Remark 9.
Note that Theorem A only considersq
=
1, while Theorem 5 considers q
≥
1. In addition, (14) implies (11), so Theorem 5 generalizes the corresponding result of Theorem A.



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*Source: 852137-2014-05-11.xml*