Volume 2008, Article ID 325845, 6 pagesdoi:10.1155/2008/325845 Research Article Recurring Mean Inequality of Random Variables Mingjin Wang Department of Applied Mathematics, Jiangsu Poly
Trang 1Volume 2008, Article ID 325845, 6 pages
doi:10.1155/2008/325845
Research Article
Recurring Mean Inequality of Random Variables
Mingjin Wang
Department of Applied Mathematics, Jiangsu Polytechnic University, Changzhou, Jiangsu 213164, China
Correspondence should be addressed to Mingjin Wang, wang197913@126.com
Received 16 August 2007; Revised 25 February 2008; Accepted 9 May 2008
Recommended by Jewgeni Dshalalow
A multidimensional recurring mean inequality is shown Furthermore, we prove some new inequalities, which can be considered to be the extensions of those established inequalities, including, for example, the Polya-Szeg ¨o and Kantorovich inequalities
Copyright q 2008 Mingjin Wang This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1 Introduction
The theory of means and their inequalities is fundamental to many fields including mathematics, statistics, physics, and economics.This is certainly true in the area of probability and statistics There are large amounts of work available in the literature For example, some useful results have been given by Shaked and Tong 1, Shaked and Shanthikumar
2, Shaked et al 3, and Tong 4, 5 Motivated by different concerns, there are numerous ways to introduce mean values In probability and statistics, the most commonly used mean
is expectation In 6, the author proves the mean inequality of two random variables The purpose of the present paper is to establish a recurring mean inequality, which generalizes the mean inequality of two random variables ton random variables This result can, in turn, be
extended to establish other new inequalities, which include generalizations of the Polya-Szeg ¨o and Kantorovich inequalities7
We begin by introducing some preliminary concepts and known results which can also
be found in6
Definition 1.1 The supremum and infimum of the random variable ξ are defined as inf x {x : Pξ ≤ x 1} and sup x {x : Pξ ≥ x 1}, respectively, and denoted by sup ξ and inf ξ.
Definition 1.2 If ξ is bounded, the arithmetic mean of the random variable ξ, Aξ, is given by
Aξ supξ inf ξ
Trang 2In addition, if infξ ≥ 0, one defines the geometric mean of the random variable ξ, Gξ, to be
Definition 1.3 If ξ1, , ξ n are bounded random variables, the independent arithmetic mean of the product of random variables ξ1, , ξ n , Aξ1, , ξ n is given by
Aξ1, , ξ n
n
i1 supξ in i1 infξ
Definition 1.4 If ξ1, , ξ nare bounded random variables with infξ i ≥ 0, i 1, , n, one defines the independent geometric mean of the product of random variables ξ1, , ξ nto be
Gξ1, , ξ n
n
i1
Remark 1.5 If ξ1, , ξ nare independent, then
Aξ1, , ξ n
A n
i1
ξ i
,
Gξ1, , ξ n
G n
i1
ξ i
.
1.5
The mean inequality of two random variables6
Theorem 1.6 Let ξ and η be bounded random variables If inf ξ > 0 and inf η > 0, then
Eξ2·Eη2
E2ξη ≤
A2ξ, η
Equality holds if and only if
P η ξ B a η A b
1,
Gη2
Eξ2 Gξ2
Eη2
1.7
for A sup ξ, B sup η, a inf ξ, b inf η.
2 Main results
Our main results are given by the following theorem
Theorem 2.1 Suppose that ξ1, , ξ n , ξ n1 are bounded random variables, inf ξ i > 0, i 1, , n 1 Let {Un} be a sequence of real numbers If
n
i1 Eξ2
i
E2n
then
n1
i1 Eξ2
i
E2n1
i1 ξ i ≤ A
2
ξ1, , ξ n1
G2ξ1, , ξ n1 Un. 2.2
Trang 3Proof Let A i sup ξ i , a i inf ξ i , i 1, , n 1 We have
Pξ1· · · ξ n A n1 − a1· · · a n ξ n1
A1· · · A n ξ n1 − ξ1· · · ξ n a n1
≥ 0 1. 2.3 So
PA1· · · A n1 a1· · · a n1
ξ1· · · ξ n1 ≥ A1a1· · · A n a n ξ2
n1 A n1 a n1 ξ2
1· · · ξ2
n
1, 2.4 which implies that
A1· · · A n1 a1· · · a n1
Eξ1· · · ξ n1
≥ A1a1· · · A n a n Eξ2
n1
A n1 a n1 Eξ2
1· · · ξ2
n
. 2.5 Using the Jensen inequality7 and assumption 2.1, we get
A1· · · A n1 a1· · · a n1
Eξ1· · · ξ n1
≥ A1a1· · · A n a n Eξ2
n1
A n1 a n1 E2
ξ1· · · ξ n
≥ A1a1· · · A n a n Eξ2
n1
A n1 a n1 Eξ2
1· · · Eξ2
n
Un
≥ 2
A1a1· · · A n a n Eξ2
n1
A n1 a n1 Eξ2
1· · · Eξ2
n
Un
1/2
.
2.6
Hence,
G2
ξ1, , ξ n1
Eξ2
1· · · Eξ2
n1
Un
1/2
≤ Aξ1, , ξ n1
Eξ1· · · ξ n1
from which2.2 follows
Combining this result with Theorem 1.6, the following recurring inequalities are immediate
Corollary 2.2 Let ξ1, , ξ n be bounded random variables If inf ξ i > 0, i 1, , n, then
Eξ2
2
E2
ξ1ξ2
≤ A
2
ξ1, ξ2
G2ξ1, ξ2
,
Eξ2
3
E2
ξ1ξ2ξ3
≤ A
2
ξ1, ξ2, ξ3
G2ξ1, ξ2, ξ3
A
2
ξ1, ξ2
G2ξ1, ξ2
,
.
n
k1 Eξ2
k
E2n
k1 ξ k ≤ n
k2
A2ξ1, ξ k
G2ξ1, ξ k
2.8
Trang 43 Some applications
In this section, we exhibit some of the applications of the inequalities just obtained We make use of the following known lemma which we state here without proof
Lemma 3.1 If 0 < m2≤ m1≤ M1≤ M2, then
1/2m1 M1
m1M1
≤ 1/2
m2 M2
m2M2
Theorem 3.2 the extensions of the inequality of Polya-Szeg¨o Let a ij > 0, a i minj a ij , A i maxj a ij , for i 1, , n and j 1, , m Then,
n
i1
m
j1
a2
ij ≤ m n−2
4n−1
n
k2
a1· · · a k A1· · · A k2
a1· · · a k A1· · · A k
m
j1
n
i1
a ij
2
Proof This result is a consequence of inequality2.8 Let ξ1have the distribution
Pξ1 a1j
m1, j 1, , m. 3.3
We definen − 1 functions as follows:
f i
a1j
a ij , i 2, , n, j 1, , m. 3.4 Letξ i f i ξ1, i 2, , n Then,
Eξ2
i m1m
j1
a2
ij , i 1, , n,
Eξ1· · · ξ n 1
m
m
j1
n
i1
a ij ,
Aξ1, , ξ k
1 2
a1· · · a k A1· · · A k
, Gξ1, , ξ k
a1· · · a k A1· · · A k
3.5
Inequality2.8 then becomes
n
i1 1/mm j1 a2
ij
1/mm j1n i1 a ij2 ≤ n
k2
1/2a1· · · a k A1· · · A k2
a1· · · a k A1· · · A k2 , 3.6 from which our result follows
Remark 3.3 For n 2, we can get the inequality of Polya-Szeg¨o 7:
m
k1
a2
k
m
k1
b2
k
≤ 1 4
AB
ab
ab AB
2
m
k1
a k b k
2
wherea k , b k > 0, k 1, , m, a min a k , A max a k , b min b k, andB max b k
Trang 5Theorem 3.4 the extensions of Kantorovich’s inequality Let A be an m × m positive Hermitian
matrix Denote by λ1 and λ m the maximum and minimum eigenvalues of A, respectively For real
β1, , β n and β β1 · · · β n , and any vector 0 / x ∈ R m ,the following inequality is satisfied:
n
i1 x∗A β i x
x∗A β/2 x2 ≤
x∗xn−2
4n−1
n
k2
l1· · · l k L1· · · L k2
l1· · · l k L1· · · L k , 3.8
where
l i
⎧
⎨
⎩
λ β i /2
m , β i ≥ 0,
λ β i /2
1 , β i < 0, L i
⎧
⎨
⎩
λ β i /2
λ β i /2
m , β i < 0, i 1, , n. 3.9 Proof Let λ1 ≥ · · · ≥ λ mbe eigenvalues ofA and let Λ diagλ1, , λ m There is a Hermitian matrixU that satisfies
Let
y Ux y1, y2, , y mT
, p i y i2
m
i1 y i2, i 1, , m. 3.11 Then,
n
i1 x∗A β i x
x∗A β/2 x2
n
i1 x∗U∗Λβ i Ux
x∗U∗Λβ/2 Ux2
n
i1 y∗Λβ i y
y∗Λβ/2 y2
y∗yn−2n i1m k1 λ β i
k p k
m
k1 λ β/2 k p k2
x∗xn−2n
i1m
k1 λ β i
k p k
m
k1 λ β/2 k p k2 .
3.12
What remains to show is that
n
i1m
k1 λ β i
k p k
m
k1 λ β/2 k p k2 ≤ 1
4n−1
n
k2
l1· · · l k L1· · · L k2
l1· · · l k L1· · · L k , ∀p i ≥ 0, m
i1
p i 1. 3.13
We define the random variableζ, and assign Pζ λ i p i , i 1, , m Suppose ξ i
ζ β i /2 , i 1, , n Notice that λ1andλ nare the upper and lower bounds of the random variable
ζ, so l iandL iare the lower and upper bounds ofξ i According toLemma 3.1, we know that
A2ξ1, , ξ k
G2ξ1, , ξ k ≤
1/2l1· · · l k L1· · · L k2
l1· · · l k L1· · · L k2 . 3.14
Trang 6Noticing that
Eξ1· · · ξ n
Eζ β/2m
k1
we can use inequality2.8 to express inequality 3.13 as
Eξ2
1· · · Eξ2
n
E2
ξ1· · · ξ n ≤ n
k2
1/2l1· · · l k L1· · · L k2
l1· · · l k L1· · · L k2 . 3.16
Remark 3.5 If n 2, β1 1, and β2 −1, this inequality takes the form
x∗Axx∗A−1x
x∗x2 ≤
λ1 λ m2
which is Kantorovich’s inequality7
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