On the Strong Law of Large Numbers for Sequences of Pairwise Independent Random Variables

0.1 MB·

by  Valery Korchevsky

On the Strong Law of Large Numbers
for Sequences of Pairwise Independent
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1
0 Random Variables
2

n Valery Korchevsky∗
a
J

4

] Abstract
R
P We establish new sufficient conditions for the applicability of the strong law of
. largenumbers(SLLN)forsequencesofpairwiseindependentnon-identicallydistributed
h
randomvariables. TheseresultsgeneralizeEtemadi’sextensionofKolmogorov’sSLLN
t
a foridenticallydistributedrandomvariables. Someoftheobtainedresultsholdwithan
m arbitrary norming sequence in place of theclassical normalization.
[

1 Keywords: strong law of large numbers, pairwise independent random variables.
v
4
3 1. Introduction
2
2 Let {X }∞ be a sequence of random variables defined on the same probability space and
n n=1
0 put S = n X .
. n k=1 k
1 The clPassical Kolmogorov’s theorem states that if {X }∞ is a sequence of indepen-
n n=1
0 dent identically distributed random variables and E|X | < ∞ then S /n → EX almost
1 n 1
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surely. Etemadi[4]generalizedtheKolmogorovtheoremreplacingthemutualindependence
1
assumption by the pairwise independence assumption.
:
v
i TheoremA([4]). Let{X }∞ beasequenceofpairwiseindependentidenticallydistributed
X n n=1
random variables. If E|X |<∞ then S /n→EX almost surely.
1 n 1
r
a
Onecanfindfurther extensionsofthe Kolmogorovtheoremtowideclassesofdependent
random variables in the papers [6] and [7].
In the present work we generalize Theorem A to non-identically distributed random
variables. This problem was considered in the several papers. Chandra and Goswami [3]
established the following result.
Theorem B ([3]). Let {X }∞ be a sequence of pairwise independent random variables
n n=1
and put G(x)=sup P(|X |>x) for x≥0. If
n≥1 n
∞
G(x)dx<∞,
Z
0
∗Saint-PetersburgState UniversityofAerospaceInstrumentation, Saint-Petersburg.
E-mail: [email protected]
1
Then
n
1
c (X −EX )→0 a.s. (n→∞)
k k k
n
X
k=1
for each bounded sequence {c }.
n
The following result was obtained by Bose and Chandra [1].
Theorem C ([1]). Let {X }∞ be a sequence of pairwise independent random variables.
n n=1
Suppose that
∞ 1 n
G(x)dx<∞ with G(x)=sup P(|X |>x) for all x≥0, (1)
Z n k
0 n≥1 kX=1
∞
P(|X |>n)<∞.
n
X
n=1
Then
S −ES
n n →0 a.s. (2)
n
Kruglov [5] proved the next generalization of Theorem A.
Theorem D ([5]). Let {X }∞ be a sequence of pairwise independent random variables.
n n=1
Assume that
supE|X |<∞. (3)
n
n≥1
If there exists a random variable X such that E|X|<∞ and
n
1
sup P(|X |>x)≤CP(|X|>x) for all x≥0,
k
n
n≥1 X
k=1
where C is a positive constant, then relation (2) holds.
TheaimofpresentworkistogeneralizeTheoremsCandD.Wepresentageneralization
of Theorem C using an arbitrary norming sequence in place of the classical normalization.
Furthermore we show that condition (3) in Theorems D can be dropped.
In order to prove the theorems in the present work, we use methods developed by Bose
and Chandra [1] (see also Chandra [2]).
2. Main results
Theorem1. Let{X }∞ beasequenceofpairwise independentrandom variables. Assume
n n=1
that {a }∞ is non-decreasing unbounded sequence of positive numbers. Suppose that
n n=1
∞ 1 n
G(x)dx<∞ with G(x)=sup P(|X |>x) for all x≥0, (4)
Z a k
0 n≥1 n Xk=1
2
∞
P(|X |>a )<∞. (5)
n n
X
n=1
Then
S −ES
n n →0 a.s.
a
n
Theorem 1 generalizes Theorem C, which corresponds to the case a =n for all n≥1.
n
Theorem 2. Let {X }∞ be a sequenceof pairwise independent random variables. If there
n n=1
exists function H(x) such that H(x) is non-increasing in the interval x≥0,
∞ 1 n
H(x)dx<∞, and sup P(|X |>x)≤H(x) for all x≥0, (6)
Z n k
0 n≥1 X
k=1
then
S −ES
n n →0 a.s.
n
As a consequence of Theorem 2 we immediately obtain the following result.
Corollary 1. Let{X }∞ beasequenceofpairwise independent randomvariables. Ifthere
n n=1
exists a random variable X such that E|X|<∞ and
n
1
sup P(|X |>x)≤CP(|X|>x) for all x≥0,
k
n
n≥1 X
k=1
where C is a positive constant, then
S −ES
n n →0 a.s.
n
Corollary 1 shows that we can omit condition (3) in Theorem D.
3. Proofs
To proveTheorems 1 we need the following propositionthat is a consequenceof Theorem1
in [3].
Lemma 1. Let {X }∞ be a sequence of non-negative random variables with finite vari-
n n=1
ances. Assume that {a }∞ is non-decreasing unbounded sequence of positive numbers.
n n=1
Suppose that
n
Var(S )≤C Var(X ) for all n≥1, (7)
n k
X
k=1
where C is a positive constant,
∞
Var(X )
n
<∞, (8)
a2
nX=1 n
3
n
1
sup EX <∞.
n
a
n≥1 n X
k=1
Then
S −ES
n n →0 a.s.
a
n
Proof of Theorem 1. Note that for pairwise independent random variables X , the positive
n
parts X+ :=max{0,X } are pairwise independent. Likewise, the X− :=max{0,−X } are
n n n n
pairwise independent. Thus it is enough to prove the theorem separately for the positive
and negative parts. So we can assume that X ≥0 for all n≥1.
n
Let Y = X I , T = n Y for every n ≥ 1. To prove the theorem , it is
n n {Xn≤an} n k=1 k
sufficient to show that P
ES −ET
n n
→0 (n→∞), (9)
a
n
T −ET
n n
→0 a.s., (10)
a
n
S −T
n n
→0 a.s. (11)
a
n
Note that for any non-negative random variable Z and a>0
∞
E(ZI )=aP(Z >a)+ P(Z >x)dx (12)
{Z>a} Z
a
and
a
E(ZI )≤ P(Z >x)dx. (13)
{Z≤a} Z
0
Fix an integer N ≥1. Then, using (4) and (12), for n>N we obtain
n
ES −ET = E(X I )
n n k {Xk>ak}
X
k=1
n n ∞
= a P(X >a )+ P(X >x)dx
k k k Z k
kX=1 Xk=1 ak
n N ∞ n ∞
= a P(X >a )+ P(X >x)dx+ P(X >x)dx
k k k Z k Z k
Xk=1 kX=1 ak k=XN+1 ak
n N ∞ n ∞
≤ a P(X >a )+ P(X >x)dx+ P(X >x)dx
k k k Z k Z k
Xk=1 Xk=1 0 Xk=1 aN
n ∞ ∞
≤ a P(X >a )+a G(x)dx+a G(x)dx.
k k k NZ nZ
kX=1 0 aN
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Condition (5) and Kronecker’s lemma (see, for example, [8]) imply that
n
1
a P(X >a )→0 (n→∞).
k k k
a
n X
k=1
Thus for each N ≥1 we have
1 ∞
limsup (ES −ET )≤ G(x)dx
a n n Z
n→∞ n aN
so we get assertion (9) by letting N →∞.
To establish (10) we shall prove that conditions of Lemma 1 are satisfied for sequence
{Y }∞ . It follows from pairwise independence of random variables Y that assertion (7)
n n=1 n
is satisfied for sequence {Y }∞ .
n n=1
Using (4) and (13), we obtain
∞ Var(Y ) ∞ E X2I ∞ 1 a2n
n ≤ n {Xn≤an} ≤ P(X >x1/2)dx
a2 (cid:0) a2 (cid:1) a2 Z n
nX=1 n nX=1 n nX=1 n 0
∞ 1 an ∞ 1 [an]+1
≤2 yP(X >y)dy ≤4 yP(X >y)dy
a2 Z n ([a ]+1)2 Z n
nX=1 n 0 nX=1 n 0
∞ ∞ 1 [an]+1 k
≤8 yP(X >y)dy
j3 Z n
nX=1j=X[an]+1 kX=1 k−1
∞ ∞ 1 j k
≤8 yP(X >y)dy
j3 Z n
nX=1j=X[an]+1 kX=1 k−1
∞ 1 j k
≤8 yP(X >y)dy
j3 Z n
j=X[a1]+1n:Xan≤j kX=1 k−1
≤8 ∞ j 1 k y n:an≤jP(Xn >y)dy
j2 Z P j
Xj=1Xk=1 k−1
∞ j 1 k ∞ ∞ 1 k
≤8 yG(y)dy =8 yG(y)dy
j2 Z j2 Z
Xj=1Xk=1 k−1 Xk=1Xj=k k−1
∞ 1 k ∞ k ∞
≤16 yG(y)dy ≤16 G(y)dy =16 G(y)dy <∞.
k Z Z Z
Xk=1 k−1 Xk=1 k−1 0
where [a ] is the integer part of a . Hence condition (8) is satisfied for sequence {Y }∞ .
n n n n=1
For each n≥1 we have
1 n 1 n 1 n ∞ ∞
EY ≤ EX = P(X >x)dx≤ G(x)dx<∞.
a k a k a Z k Z
n kX=1 n kX=1 n kX=1 0 0
5
Thus the sequence of random variables {Y }∞ satisfies conditions of Lemma 1, so rela-
n n=1
tion (10) holds.
To complete the proof it remains to verify assertion (11). Using (5), we obtain
∞ ∞
P(X 6=Y )= P(X >a )<∞.
n n n n
X X
n=1 n=1
From Borel–Cantelli lemma and relation (10) it follows that
S −ET
n n
→0 a.s. (14)
a
n
Thus (11) follows from (10) and (14).
Proof of Theorem 2. AsH(x)isnon-increasingintheintervalx≥0,condition ∞H(x)dx<
0
∞ implies that ∞ 2kH(2k)<∞. Thus using (6) we have R
k=0
P
∞ ∞ 2k+1
P(|X |>n)= P(|X |>n)
n n
nX=2 kX=0n=X2k+1
∞ 2k+1 ∞ 2k+1 ∞
≤ P(|X |>2k)≤ P(|X |>2k)≤ 2k+1H(2k)<∞.
n n
kX=0n=X2k+1 kX=0nX=1 kX=0
Now,takingintoaccountthat(6)implies(1),wecanconcludethatthedesiredresultfollows
from Theorem C.
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