ON THE WEAK LAW OF LARGE NUMBERS FOR SEQUENCES OF ...

Bull. Korean Math. Soc. 43 (2006), No. 3, pp. 551–558

ON THE WEAK LAW OF LARGE NUMBERS FOR SEQUENCES OF BANACH SPACE VALUED RANDOM ELEMENTS Nguyen Van Quang and Le Hong Son Abstract. We establish a weak law of large numbers for sequence of random elements with values in p-uniformly smooth Banach space. Our result is more general and stronger than some wellknown ones.

1. Introduction and notations Recently, the weak law of large numbers (w.l.l.n.) in Banach space has been studied by many authors (see [1], [3], [6]). The aim of this paper is to establish a weak law of large numbers for sequence of random elements in p-uniformly smooth Banach space. Our result is more general and stronger than some well-known ones in [2] (for details see below). Let us begin with some definitions and notations. A real separable Banach space X is said to be p-uniformly smooth (1 6 p 6 2) if kx + yk kx − yk + − 1; kxk = 1; kyk = τ } 6 Cτ p 2 2 for some constant C. ρ∗ (τ ) = sup{

Theorem 1.1. (see [7]) (Assouad, Hoffmann Jφrgensen) A real Banach space X is p-uniformly smooth if and only if there exists a positive K such that for all x, y ∈ X we have kx + ykp + kx − ykp 6 2kxkp + Kkykp . Theorem 1.2. (see [7]) (Assouad) A real separable Banach space X is p-uniformly smooth (1 6 p 6 2) if and only if all q > 1, there Received July 18, 2005. 2000 Mathematics Subject Classification: 60F05. Key words and phrases: weak law of large numbers, martingale, p-uniformly smooth Banach space. This work was supported in part by the National Science Council of Vietnam.

552

Nguyen Van Quang and Le Hong Son

exists a positive constant C such that for all X - valued martingale {Mn , Fn , n > 1} we have n X EkMn kq 6 CE( kdMi kp )q/p (with dMi = Mi − Mi−1 ). i=1

(Marcinkiewicz-Zygmund inequality) In [1], Adler, Rosalsky, and Volodin have taken notion about martingale type p Banach spaces: A real separable Banach space X is said to be martingale type p (1 6 p 6 2) if there exists a finite constant C such that for all martingale {Sn , n > 1} with values in X then p

sup EkSn k 6 C n>1

∞ X

EkSn − Sn−1 kp .

i=1

By Marcinkiewicz-Zygmund inequality we derive that a p-uniformly smooth Banach space is a martingale type p Banach space. In this paper we assume that X is a p-uniformly smooth Banach space (1 6 p 6 2), {Xn , n > 1} is a sequence of random elements with values in X , (Fn ) is a sequence of σ - algebras such that Xn - Fn measurable for all n = 1, 2, . . ..

2. Results The main aim of this paper is to prove the following result. P Theorem 2.1. Let (Sn = ni=1 Xi ) be a sequence of random elements with values in X and (bn ) a sequence of positive numbers with bn ↑ ∞ as n −→ ∞. Then writing Xni = Xi I(kXi k6bn ) , 1 6 i 6 n, we have that (2.1) (2.2)

P

→ 0 as n −→ ∞, if b−1 n Sn − n P P (kXi k > bn ) −→ 0 as n −→ ∞, i=1

(2.3)

b−1 n

(2.4)

b−p n

n P i=1 n P i=1

P

E(Xni /Fi−1 ) − → 0 as n −→ ∞, EkXni − E(Xni /Fi−1 )kp −→ 0 as n −→ ∞.

On the weak law of large numbers for sequences of Banach space

Proof. Let Snn = µ

n P i=1

553

Xni . Then we have

¶ Snn Sn P 6= bn bn Ãn ! [ 6P {Xni 6= Xi } i=1

6 =

n X i=1 n X

P (Xni 6= Xi ) P (kXi k > bn ) −→ 0, n −→ ∞ (by (2.2)),

i=1 P

and it suffices to show that b−1 → 0 as n −→ ∞. By (2.3), we have n Snn − b−1 n

n X

P

E(Xni /Fi−1 ) − → 0 as n −→ ∞.

i=1

So that it suffices to prove that b−1 n

n X

P

[Xni − E(Xni /Fi−1 )] − →0

as n −→ ∞.

i=1

Let Ynk =

n P i=1

[Xni − E(Xni /Fi−1 )] ,

1 6 k 6 n, n = 1, 2, . . . .

By noting that (Ynk , Fk ; 1 6 k 6 n) is a martingale on X , we have, for arbitrary ε > 0 Ã ! n X P kb−1 [Xni − E(Xni /Fi−1 )] k > ε n i=1

6 ε−p b−p n Ek

n X

[Xni − E(Xni /Fi−1 )] kp

(Markov’s inequality)

i=1 p = ε−p b−p n EkYnn k Ã n ! X 6 ε−p b−p kYni − Yn(i−1) kp (Marcinkiewicz-Zygmund inequality) n CE i=1

= Cε−p b−p n

n X

EkXni − E(Xni /Fi−1 )kp → 0 as n → ∞ (by (2.4)),

i=1

which completes the proof.

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In the case, when X = R then p = 2 and E|Xni − E(Xni /Fi−1 )|2 = 2 −E[E(X /F)]2 . The following corollary follows immediately from EXni ni theorem 2.1. P Corollary 2.2. Let (Sn = ni=1 Xi ) be a sequence of random variables and (bn ) a sequence of positive numbers with bn ↑ ∞ as n −→ ∞. Then writing Xni = Xi I(|Xi |6bn ) , 1 6 i 6 n, we have that P

(2.1)0

b−1 → 0 as n −→ ∞, if n Sn − n P P (|Xi | > bn ) −→ 0 as n −→ ∞,

(2.2)0

i=1

(2.3)0

b−1 n

(2.4)0

b−2 n

n P i=1

n P

P

E(Xni /Fi−1 ) − → 0 as n −→ ∞,

2 − E(X /F 2 {EXni ni i−1 ) } −→ 0 as n −→ ∞.

i=1

The below example shows that the above corollary is stronger than the theorem 2.13 in Pn[2] which considered the same problem under the assumption:(Sn = i=1 Xi , ; Fn ) is a martingale. Let (Yi ) be a sequence of independent and identically distributed random variables such that 1 P (Yi = −1) = P (Yi = 1) = . 2 Then EYi = 0 ( ∀i = 1, 2, . . .) and n 1X P Yi → 0 as n → ∞. n i=1

Put Xi = Yi + Then EXi = n X

1 i

1 . i

( ∀i = 1, 2 · · · ) and n

n

1X 1X1 P Xi = Yi + → 0 + 0 = 0 as n → ∞. n n i i=1 i=1 i=1 P Thus, (Sn = ni=1 Xi ) satisfies the condition (2.1)0 and it also satisfies the conditions (2.2)0 , (2.3)0 , (2.4)0 (with bn = n). (Because (Xi ) is the sequence of independent random variables and in this case, the conditions (2.2)0 , (2.3)0 , (2.4)0 are necessary asPwell as sufficient for the condition (2.1)0 (see [5], p. 290)). But (Sn = ni=1 Xi , ; Fn ) is not a martingale. (Fn denote the σ- algebra generated Pby (Xi ; 1 6 i 6 n)). It shows that the martingale condition of (Sn = ni=1 Xi ; Fn ) in the theorem 2.13 of [2] is too strong. 1 n


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Let (Xn ), X be random elements in X . The sequence {Xn , n > 1} is said to be stochastically dominated by X if there exists a constant C > 0 such that P {kXn k > t} 6 CP {kXk > t} for all nonnegative real numbers t and for all n > 1. In this case, we write (Xn ) ≺ X. Lemma 2.3. (see [4]) Assume fn : R −→ R+ satisfy : 0 6 fn 6 1 ; n = 1, 2, . . . and sup(xfn (x)) −→ 0 as x −→ ∞. Then 

n∈N Zy

1 sup  y n∈N



xfn (x)dx −→ 0 as y −→ ∞. 0

Corollary 2.4. Let (Sn = (Xn ) ≺ Y

n P i=1

Xi , Fn ) be a martingale on X . If

with Y ∈ L1 (Ω, F, P ), then P

n−1 Sn − → 0 as n −→ ∞.

(2.5)

Proof. We’ll prove that (Xn ) satisfies all conditions of Theorem 2.1. Let Xni = Xi I(kXi k6n) , 1 6 i 6 n. At first we have n X (2.6) P (kXi k > n) 6 CnP (kY k > n) −→ 0 as n −→ ∞. i=1

Next, for arbitrary ε > 0 we have n X P (kn−1 E(Xni /Fi−1 )k > ε) i=1

6 ε−1 n−1 Ek

n X

E(Xni /Fi−1 )k

i=1

= ε−1 n−1 Ek

n X

£ ¤ E (Xi − Xi I(kXi k>n) /Fi−1 k

i=1

6 ε−1 n−1 E (2.7)

= ε−1 n−1

n X

E(kXi kI(kXi k>n) /Fi−1 )

i=1 n X

EkXi kI(kXi k>n) −→ 0 as

n −→ ∞.

i=1

¡

Z∞ ∵ EkXi k =

Z∞ P (kXi k > x)dx 6

0

0

¢ CP (kY k > x)dx = CEkY k < ∞ .

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At the end, we have n−p 6n

n X

EkXni − E(Xni /Fi−1 )kp

i=1 n X −p

E [ 2kXni kp + KE(kXni kp /Fi−1 )]

i=1

=n

−p

C

0

n X

EkXni kp

i=1

Zn = C 0 n−p

P (kXi kp > x)dx 0

Zn 6 C 0 n−p

pxp−1 P (kXi kp > x)dx 0

Zn 0 −p

6Cn

pxp−2 xP (kY k > x)dx

n 0

Zn (2.8)

6 C 0 pn−1

xP (kY k > x)dx −→ 0, as n −→ ∞ 0

(The last inequality follows from Lemma 2.2 with fn (x) = P (kY k > x), n = 1, 2, . . .). Combining (2.6), (2.7) and (2.8) we get (2.5) and which completes the proof. Corollary 2.5. Let (Sn = sup EkXn kp < ∞, then

n P i=1

Xi , Fn ) be a martingale on X . If

n

(2.9)

n

− 1q

P

Sn − → 0 as n −→ ∞ (with 0 < q < p).

Proof. We’ll prove that (Xn ) satisfies all conditions of Theorem 2.1 1 with bn = n q . Let Xni = Xi I 1 . Then we have q (kXi k6n )

n X i=1

1

P (kXi k > n q )


6

n X

n

− pq

557

EkXi kp

i=1

(2.10)

n X

=n

− pq

6n

i=1 − pq +1

EkXi kp

sup EkXn kp −→ 0 as n −→ ∞. n

For arbitrary ε > 0 we have Ã P

n

− 1q

n X

k

! E(Xni /Fi−1 )k > ε

i=1

6ε

−p

n

− pq

n

− pq

Ek

n X

E(Xni /Fi−1 kp

i=1

6ε

−p

E

n X

µ E kXi kp I

6 ε−p n

1− pq

1

(kXi k6n q )

i=1

(2.11)

¶

E(kXi kp I

1

(kXi k>n q

/Fi−1 P

/Fi−1 ) − → 0 as n −→ ∞.

At the end we have n X

− pq

n

EkXni − E(Xni /Fi−1 )kp

i=1

6n

− pq

C

n X

EkXni kp

i−1

− pq

= Cn (2.12)

− pq

6 Cn

n · X EkXi kp − EkXi kp I i=1 n X

¸ 1

(kXi k>n q )

EkXi kp

i=1 1− pq

6 Cn

µ ¶ p sup EkXn k −→ 0 as n −→ ∞. n

Combining (2.10), (2.11) and (2.12) we get (2.9) and which completes the proof.

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References [1] A. Adler, A Rosalsky, and A. I. Volodin, Weak laws with random indices for arrays of random elements in Rademacher type p Banach spaces. J. Theoret. Probab. (1997), no. 3, 605–623. [2] P. Hall and C. C. Heyde, Martingale limit theory and its application, Academic Press, New York, 1980. [3] D. H. Hong, M. Ord´ on ˜ ez Cabrera, S. H. Sung, and A. I. Volodin, On the weak law for randomly indexed partial sums for arrays of random elements in martingale type p Banach spaces, Statist. Probab. Lett. 46 (2000), no. 2, 177–185. [4] N. V. Hung and N. D. Tien, On the convergence of weighted sums of martingale differences, Acta Math. Vietnam. 13 (1988), no. 1, 43–53. [5] M. Lo`eve, Probability Theory I, 4th ed., Graduate Texts in Mathematics, Vol. 45, Springer-Verlag, Berlin and New York, 1977. [6] S. H. Sung, Weak law of large numbers for arrays of random variables, Statist. Probab. Lett. 42 (1998), no. 3, 293–298. [7] W. A. Woyczynski, Geometry and martingale in Banach spaces II. Independent increments, Marcel Dekker, Press New York, 1978. Department of Mathematics, University of Vinh, Vietnam E-mail : [email protected] [email protected]