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

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n 1 c (X − EX ) → 0 a.s. (n → ∞) n k k k kX=1 for each bounded sequence {cn}. The following result was obtained by Bose and Chandra [1].

∞ Theorem C ([1]). Let {Xn}n=1 be a sequence of pairwise independent random variables. Suppose that

∞ 1 n G(x) dx < ∞ with G(x) = sup P (|Xk| > x) for all x ≥ 0, (1) Z n≥1 n 0 kX=1 ∞ P (|Xn| >n) < ∞. nX=1 Then

S − ES n n → 0 a.s. (2) n Kruglov [5] proved the next generalization of Theorem A.

∞ Theorem D ([5]). Let {Xn}n=1 be a sequence of pairwise independent random variables. Assume that

sup E|Xn| < ∞. (3) n≥1 If there exists a random variable X such that E|X| < ∞ and

1 n sup P (|Xk| > x) ≤ CP (|X| > x) for all x ≥ 0, n≥1 n kX=1 where C is a positive constant, then relation (2) holds. The aim of present work is to generalize Theorems C and D. We present a generalization 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

∞ Theorem 1. Let {Xn}n=1 be a sequence of pairwise independent random variables. Assume ∞ that {an}n=1 is non-decreasing unbounded sequence of positive numbers. Suppose that

n ∞ 1 G(x) dx < ∞ with G(x) = sup P (|Xk| > x) for all x ≥ 0, (4) Z n≥1 an 0 kX=1

2 ∞ P (|Xn| >an) < ∞. (5) nX=1 Then

S − ES n n → 0 a.s. an

Theorem 1 generalizes Theorem C, which corresponds to the case an = n for all n ≥ 1.

∞ Theorem 2. Let {Xn}n=1 be a sequence of pairwise independent random variables. If there exists function H(x) such that H(x) is non-increasing in the interval x ≥ 0,

n ∞ 1 H(x) dx < ∞, and sup P (|Xk| > x) ≤ H(x) for all x ≥ 0, (6) Z n≥1 n 0 kX=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 {Xn}n=1 be a sequence of pairwise independent random variables. If there exists a random variable X such that E|X| < ∞ and

1 n sup P (|Xk| > x) ≤ CP (|X| > x) for all x ≥ 0, n≥1 n kX=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 prove Theorems 1 we need the following proposition that is a consequence of Theorem 1 in [3].

∞ Lemma 1. Let {Xn}n=1 be a sequence of non-negative random variables with ﬁnite vari- ∞ ances. Assume that {an}n=1 is non-decreasing unbounded sequence of positive numbers. Suppose that

n V ar(Sn) ≤ C V ar(Xk) for all n ≥ 1, (7) kX=1 where C is a positive constant,

∞ V ar(X ) n < ∞, (8) a2 nX=1 n

3 1 n sup EXn < ∞. n≥1 an kX=1 Then

S − ES n n → 0 a.s. an

Proof of Theorem 1. Note that for pairwise independent random variables Xn, the positive + − parts Xn := max{0,Xn} are pairwise independent. Likewise, the Xn := max{0, −Xn} are pairwise independent. Thus it is enough to prove the theorem separately for the positive and negative parts. So we can assume that Xn ≥ 0 for all n ≥ 1. I n Let Yn = Xn {Xn≤an}, Tn = k=1 Yk for every n ≥ 1. To prove the theorem , it is suﬃcient to show that P ES − ET n n → 0 (n → ∞), (9) an T − ET n n → 0 a.s., (10) an S − T n n → 0 a.s. (11) an Note that for any non-negative random variable Z and a> 0

∞ I E(Z {Z>a})= aP (Z>a)+ P (Z > x) dx (12) Za and

a I E(Z {Z≤a}) ≤ P (Z > x) dx. (13) Z0 Fix an integer N ≥ 1. Then, using (4) and (12), for n>N we obtain

n I ESn − ETn = E(Xk {Xk >ak }) kX=1 n n ∞ = a P (X >a )+ P (X > x) dx k k k Z k kX=1 kX=1 ak n N ∞ n ∞ = a P (X >a )+ P (X > x) dx + P (X > x) dx k k k Z k Z k kX=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 kX=1 kX=1 0 kX=1 aN n ∞ ∞ ≤ a P (X >a )+ a G(x) dx + a G(x) dx. k k k N Z n Z kX=1 0 aN

4 Condition (5) and Kronecker’s lemma (see, for example, [8]) imply that

n 1 akP (Xk >ak) → 0 (n → ∞). an kX=1 Thus for each N ≥ 1 we have

1 ∞ lim sup (ESn − ETn) ≤ G(x) dx n→∞ an ZaN so we get assertion (9) by letting N → ∞. To establish (10) we shall prove that conditions of Lemma 1 are satisﬁed for sequence ∞ {Yn}n=1. It follows from pairwise independence of random variables Yn that assertion (7) ∞ is satisﬁed for sequence {Yn}n=1. Using (4) and (13), we obtain

2 ∞ ∞ 2I ∞ an V ar(Y ) E X n n 1 n ≤ n {X ≤a } ≤ P (X > x1/2) dx a2 a2 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 [an]+1 ∞ ∞ 1 k ≤ 8 3 yP (Xn >y) dy j Zk−1 nX=1 j=[Xan]+1 kX=1 ∞ ∞ j 1 k ≤ 8 3 yP (Xn >y) dy j Zk−1 nX=1 j=[Xan]+1 kX=1 j ∞ 1 k ≤ 8 3 yP (Xn >y) dy j Zk−1 j=[Xa1]+1 n:Xan≤j kX=1

∞ j k 1 n P (Xn >y) ≤ 8 y n:a ≤j dy j2 Z P j Xj=1 kX=1 k−1 ∞ j ∞ ∞ 1 k 1 k ≤ 8 yG(y) dy =8 yG(y) dy j2 Z j2 Z Xj=1 kX=1 k−1 kX=1 Xj=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 [an] is the integer part of an. Hence condition (8) is satisﬁed for sequence {Yn}n=1. For each n ≥ 1 we have

1 n 1 n 1 n ∞ ∞ EYk ≤ EXk = P (Xk > x) dx ≤ G(x) dx < ∞. an an an Z Z kX=1 kX=1 kX=1 0 0

5 ∞ Thus the sequence of random variables {Yn}n=1 satisﬁes conditions of Lemma 1, so relation (10) holds. To complete the proof it remains to verify assertion (11). Using (5), we obtain

∞ ∞ P (Xn 6= Yn)= P (Xn >an) < ∞. nX=1 nX=1 From Borel–Cantelli lemma and relation (10) it follows that S − ET n n → 0 a.s. (14) an Thus (11) follows from (10) and (14). ∞ Proof of Theorem 2. As H(x) is non-increasing in the interval x ≥ 0, condition 0 H(x) dx < ∞ k k R ∞ implies that k=0 2 H(2 ) < ∞. Thus using (6) we have P

k ∞ ∞ 2 +1 P (|Xn| >n)= P (|Xn| >n) nX=2 kX=0 n=2Xk+1 k k ∞ 2 +1 ∞ 2 +1 ∞ k k k+1 k ≤ P (|Xn| > 2 ) ≤ P (|Xn| > 2 ) ≤ 2 H(2 ) < ∞. kX=0 n=2Xk+1 kX=0 nX=1 kX=0

Now, taking into account that (6) implies (1), we can conclude that the desired result follows from Theorem C.

References

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