Internat. J. Math. & Math. Sci.
Vol. 24, No. 1 (2000) 29–42
S0161171200002945
© Hindawi Publishing Corp.
ON CONDITIONS FOR THE STRONG LAW OF LARGE
NUMBERS IN GENERAL BANACH SPACES
ANNA KUCZMASZEWSKA and DOMINIK SZYNAL
(Received 4 August 1998 and in revised form 26 March 1999)
Abstract. We give Chung-Teicher type conditions for the SLLN in general Banach spaces
under the assumption that the weak law of large numbers holds. An example is provided
showing that these conditions can hold when some earlier known conditions fail.
Keywords and phrases. Random element, Banach spaces, weak law of large numbers,
strong law of large numbers.
2000 Mathematics Subject Classification. Primary 60F15, 60B12.
Let (Ꮾ, ) be a real, separable Banach space. A strongly measurable mapping X
from a probability space (Ω, Ᏺ, ᏼ) into Ꮾ is said to be a random element.
If EX < ∞, then the expectation EX is defined by the Bochner integral.
Strong laws of large numbers (SLLN) for random elements, i.e., (X1 +· · ·+Xn )/n → 0
a.s., n → ∞, were investigated by Mourier [14], Fortet and Mourier [8], Beck [3], Beck and
Giesy [4], Hoffman-Jørgensen and Pisier [10], Heinkel [9], Taylor [17], Woyczyński [19],
and Adler, Rosalsky, and Taylor [1]. Their efforts have concentrated on a complete
characterization of all those Banach spaces in which the SLLN holds under conditions of classical probability theory or on finding conditions on the random elements
which ensure the SLLN. It is known (Woyczyński [19]) that in certain Banach spaces
the Chung’s condition (Chung [7]) implies the SLLN for a sequence of independent
random elements.
Some handy conditions for the SLLN in Banach spaces were given by Kuelbs and
Zinn [13] and by Alt [2].
Extensions of the Chung-Teicher type conditions (cf. Chung [7], Teicher [18], Chow
and Teicher [6]) for the SLLN for sequences of independent random elements in Hilbert
space were found by Szynal and Kuczmaszewska [16], Choi and Sung [5], and Sung [15].
The aim of our paper is to give conditions for the SLLN in Banach spaces which
can be applied in more general cases than those of Choi and Sung [5], Sung [15],
Adler, Rosalsky, and Taylor [1] and Kuczmaszewska and Szynal [12]. We present also
an example showing that these conditions can be applied when some earlier known
conditions fail. We make use of the following lemmas.
Lemma 1 (cf. Yurinskii [20]). Let X1 , X2 , . . . , Xn be independent Ꮾ-valued random elements with EXi < ∞, i = 1, 2, . . . , n. Let Ᏺk be σ -field generated by (X1 , X2 , . . . , Xk ),
k = 1, 2, . . . , n and let Ᏺ0 = {Ω, ∅}. Then for 1 ≤ k ≤ n,
E Sn | Ᏺk − E Sn | Ᏺk−1 ≤ Xk + E Xk ,
where Sn =
n
i=1
Xi .
(1)
30
A. KUCZMASZEWSKA AND D. SZYNAL
Lemma 2 (cf. Choi and Sung [5]). Let {Xn , n ≥ 1} be a sequence of independent, Ꮾvalued random elements. Then
Sn
→ 0 a.s., n → ∞
(2)
n
if and only if
S 2k
→ 0
2k
a.s., k → ∞
Sn P
→
0,
n
and
n → ∞.
(3)
Let a function φ : R → R+ be nonnegative, even, continuous and nondecreasing on
(0, ∞) with limx→∞ φ(x) = ∞ and such that
(a) φ(x)/x or
(b) φ(x)/x ,
and φ(x)/x p , x → ∞, for some p, 1 < p ≤ 2.
Theorem 3. Let {Xn , n ≥ 1} be a sequence of independent Ꮾ-valued random
elements. Suppose thatin the
p, 1< p ≤ 2,
case (a) for some
j−1 p
∞ −p
φp Xj φp Xi i=1 i E
< ∞,
E p
(A)
j=2 j
φ (j)+φp Xj φp (i)+φp Xi n
φp Xi = o(1),
(B) n−p i=1 ip E p
φ (i)+φp Xi ∞
(C)
n=1 P (Xn ≥ an ) < ∞,
for some sequence {an , n ≥
1} of
positive numbers such that
p X ∞
φ
n
p
< ∞
(D)
n=1 φ (an )E 2p
2p φ
(n)+φ
Xn
or in the case (b) for some
p, 1 < p ≤ 2, i
∞ −p
φ Xj j−1 ip E φ X
< ∞,
E
(A1 )
j=2 j
i=1
φ(j)+φ Xj φ(i)+φ Xi n
φ X i = o(1),
(B1 ) n−p i=1 ip E
φ(i)+φ
Xi
and (C) is satisfied for some
{an , n ≥ 1} of positive numbers such that
sequence
∞
φ Xn (D1 )
n=1 φ(an )E 2
< ∞.
2 φ (n)+φ
Xn
Then
n−1
n
k=1
P
0,
Xk − EXk →
n → ∞,
(4)
a.s., n → ∞,
(5)
if and only if
n−1
n
k=1
Xk − EXk → 0
where Xk = Xk I[Xk ≤ k].
Proof. Suppose that (4) holds.
Let r ≥ 1. We note that
∞
∞
P Xn ≠ Xn =
E I Xn ≥ n · I Xn ≥ an + I Xn ≥ n · I Xn < an
n=1
n=1
≤
∞
n=1
∞
P Xn ≥ an + 2
E
n=1
φ2r Xn I Xn < an
φ2r (n) + φ2r Xn (6)
ON CONDITIONS FOR THE STRONG LAW OF LARGE NUMBERS . . .
≤
∞
∞
P Xn ≥ an + 2
φr (an )E
n=1
n=1
31
φr Xn < ∞,
φ2r (n) + φ2r Xn where we put r = p in the case (a) and r = 1 in the case (b).
Hence {Xn , n ≥ 1} and {Xn , n ≥ 1} are equivalent. Therefore, by (4), we have
n
P
0,
Xi − EXi →
n−1 n → ∞.
(7)
i=1
Write
Xk∗ = Xk − EXk ,
Sn =
k ≥ 1,
n
k=1
Xk ,
Sn∗ =
n
k=1
Xk∗ .
(8)
Define
Yn,i = E Sn∗ | Ᏺi − E Sn∗ | Ᏺi−1 ,
(9)
where Ᏺi = σ (X1∗ , X2∗ , . . . , Xi∗ ) and Ᏺ0 = {Ω, ∅}.
Note that {Yn,i , 1 ≤ i ≤ n} is a martingale difference for fixed n. Then we have
n
∗
S − E S ∗ =
Yn,i .
n
(10)
n
i=1
Now we prove that ESn∗ /n → 0, n → ∞. Using (9) and Lemma 1, we get
2
P Sn∗ − E Sn∗ > nε ≤ ε−2 n−2 E Sn∗ − E Sn∗ n
2
n
2 −2 −2
=ε n E
Yn,i = ε−2 n−2
E Yn,i
i=1
i=1
n
2
≤ ε−2 n−2
E X ∗ + E X ∗ i
i=1
≤ 8ε−2 n−2
(11)
i
p
n
n
X 2
i
E Xi ≤ 8ε−2 n−p
ip E
.
p
i
i=1
i=1
Hence in the case (a) by (B), we get
n
ip E
P Sn∗ − E Sn∗ > nε ≤ 16ε−2 n−p
i=1
φp Xi = o(1),
φp (i) + φp Xi (12)
φ Xi = o(1).
φ(i) + φ Xi (13)
while in the case (b) by (B1 )
n
ip E
P Sn∗ − E Sn∗ > nε ≤ 16ε−2 n−p
i=1
Therefore, in the case (a) and (b) we have
P
n−1 Sn∗ − E Sn∗ →
0,
n → ∞.
(14)
32
A. KUCZMASZEWSKA AND D. SZYNAL
P
Thus we conclude, from Sn∗ /n →
0, n → ∞, and (14), that
∗
E S n → 0, n → ∞.
n
(15)
Now we are going to prove that Sn∗ /n → 0 a.s., n → ∞. By Lemma 2 it is enough to
prove that S2∗k /2k → 0 a.s., k → ∞ or equivalently, as ES2∗k /2k → 0, k → ∞, that
2−k S2∗k − E S2∗k → 0
a.s., k → ∞.
(16)
Taking into account the identity
i−1
n
n
∗ 2
S − E S ∗ 2 =
Yn,i
+2
Yn,i
Yn,j ,
n
n
i=1
i=2
(17)
j=1
we see that
2
−2n
2n
i−1
2n
∗ 2
S n − E S ∗n 2 = 2−2n
Y2n ,i + 2
Y2n ,i
Y2n ,j .
2
2
i=1
i=2
(18)
j=1
Now, put
Z2n ,i = Y22n ,i I Xi < ai − E Y22n ,i I Xi < ai | Ᏺi−1 ,
1 ≤ i ≤ 2n .
(19)
Then by Chebyshev’s inequality, equation (9) and Lemma 1, we have
 n

2n
2
∞
∞
2
2n 

P Z2n ,i > ε2
2−4n E
Z2n ,i
≤ ε−2
i=1
n=1
n=1
i=1
= ε−2
∞
n
2−4n
n=1
≤ ε−2
∞
i=1
n
2−4n
n=1
≤ ε−2 28
≤ ε−2
2
E Y24n ,i I Xi < ai
2
4 E X ∗ + E X ∗ I Xi < ai
∞
i
(20)
n
2−4n
n=1
212
15
i
i=1
∞
i=1
2
4 E X I Xi < ai
i=1
i
4 i−4 E Xi I Xi < ai .
Hence we see that in the case (a) under the assumptions (C) and (D) we have
 n

12 ∞
∞
2
2p
2n
−2 2

n
P 
Z
i−2p E Xi I Xi < ai
>
ε2
≤
ε
2 i
15 i=1
i=1
n=1
12 ∞
φ2p Xi 2
≤ε
I Xi < ai
E
2p
15 i=1
φ (i)
13 ∞
φp Xi −2 2
p
< ∞.
φ (ai )E 2p
≤ε
φ (i) + φ2p Xi 15 i=1
(21)
ON CONDITIONS FOR THE STRONG LAW OF LARGE NUMBERS . . .
33
Similarly, in the case (b) under the assumptions (C) and (D1 ) we lead to
 n

12 ∞
2
2
2p
2n 
P 
Z2n ,i i−2p E Xi I Xi < ai
≤ ε−2
> ε2
15 i=1
i=1
n=1
13 ∞
φ Xi 2
< ∞.
≤ ε−2
φ ai E 2
15 i=1
φ (i) + φ2 Xi ∞
(22)
Therefore, by the Borel-Cantelli lemma, we state that
n −2
(2 )
2n
i=1
n
Y22n ,i I
2
Xi < ai −
E Y22n ,i I Xi < ai | Ᏺi−1
→ 0
a.s., n → ∞.
i=1
(23)
Now, note that in the case (a) the assumption (B) implies
n
2n
2
−2 E Y22n ,i I Xi < ai | Ᏺi−1
i=1
n
n
2
2
2
−2 2
−2 E Xi∗ + E Xi∗ ≤ 8 2n
E Xi ≤ 2n
i=1
(24)
i=1
n
2
−p ≤ 16 2n
ip E
i=1
φp Xi = o(1).
φp (i) + φp Xi Similarly, in the case (b), by B1 , we get
n
2n
n
2
2
−2 −p E Y22n ,i I Xi < ai | Ᏺi−1 ≤ 16 2n
ip E
i=1
i=1
φ Xi = o(1). (25)
φ(i) + φ Xi Therefore, in the case (a) and (b), we obtain
2
n −2
n
2
i=1
Y22n ,i I Xi < ai → 0
a.s., n → ∞.
(26)
a.s., n → ∞
(27)
Using the assumption (C), we get
n
2n
2
−2 i=1
Y22n ,i I Xi ≥ ai → 0
which proves in the end that
n
2
n −2 2
Y22n ,i → 0
a.s., n → ∞.
(28)
i=1
i−1
Now we see that Yn,i j=1 Yn,j , 2 ≤ i ≤ n is a martingale difference for fixed n.
Therefore, in the case (a), after using Chebyshev’s inequality, (9), and Lemma 1,
34
A. KUCZMASZEWSKA AND D. SZYNAL
we get


i−1
2n
−2n 
P 2
Y2n ,i
Y2n ,j > ε
n=1
i=2
j=1
2
n
i−1
∞
2
≤ ε−2
2−4n
E Y2n ,i
Y2n ,j
∞
≤ε
−2
n=1
i=2
∞
2n
2
−4n
n=1
≤ 28 ε−2
∞
2 i−1
∗
X + E X ∗ E
i
i=2
i
2
2
2 i−1
E X E X i
i=2
∞
p
p i−1
212 −2 −2p ε
i
E Xi E Xj 15
i=1
j=1
(29)
j
j=1
∞
2
2 i−1
212 −2 −4 ε
i E Xi E Xj 15
i=1
j=1
≤
Y2n ,j
j=1
≤
2 n
2−4n
n=1
≤
j=1
∞
i−1
φp Xj φp Xi 214 −2 −p
p
< ∞.
ε
i E p
j
E
15
φ (i) + φp Xi j=1
φp (j) + φp Xj i=1
Similarly, in the case (b), we have


i−1
2n
∞
−2n 
P 2
Y2n ,i
Y2n ,j ≥ ε
n=1
i=2
j=1
≤
≤
∞
p
p i−1
212 −2 −2p ε
i
E Xi E Xj 15
i=2
j=1
(30)
∞
i−1
φ Xj φ Xi 214 −2 −p
p
< ∞.
ε
i E
j
E
15
φ(i) + φ Xi j=1
φ(j) + φ Xj i=2
Now using the Borel-Cantelli lemma we obtain
n
2−2n
2
Y2n ,i
i=2
i−1
Y2n ,j → 0
a.s., n → ∞.
(31)
j=1
Thus by (15) and (18) we see that
S2∗n
→ 0
2n
so we have
n
a.s., n → ∞,
n
Xi − EXi → 0
−1 a.s., n → ∞.
(32)
(33)
i=1
But {Xn , n ≥ 1} and {Xn , n ≥ 1} are equivalent, so that
∞
n−1 Xi − EXi → 0 a.s., n → ∞
n=1
which completes to proof of Theorem 3.
(34)
ON CONDITIONS FOR THE STRONG LAW OF LARGE NUMBERS . . .
35
Theorem 3 generalizes results of [2, 5, 10].
Before giving an example showing that the presented conditions under which WLLN
is equivalent to the SLLN can be applied when some earlier known ones fail we quote
the following results in this subject.
Theorem 4 [13]. Let {Xn , n ≥ 1} be a sequence of independent Ꮾ-valued random
variables such that
(a) Xj /j → 0 a.s., j → ∞,
(b) for some p ∈ [1, 2] and some r ∈ (0, ∞)

∞
n+1
2

n=1
r
p (n+1)p
 < ∞.
E Xj /2
(35)
j=2n +1
Then
Sn P
→
0
n
if and only if
Sn
→0 a.s., n → ∞.
n
(36)
Theorem 5 [13]. Let {Xn , n ≥ 1} be a sequence of independent Ꮾ-valued random
variables such that
(a) |Xj | ≤ Mj /LLj for some constant M < ∞, where LLj = log(log(j ∨ ee )), and
∞
2n+1
2
2(n+1)
(b)
. Then
n=1 {−ε/Λ(n)} < ∞ for all ε > 0 where Λ(n) =
j=2n +1 EXj /2
Sn P
→
0
n
if and only if
Sn
→0 a.s., n → ∞.
n
(37)
Theorem 6 [15]. Let {Xn , n ≥ 1} be a sequence of independent B-valued random
variables, and let
be constants
that 0 < bn . Suppose that
{an} and {bn }
∞ a2i Eφ Xi i−1 a2j Eφ Xj <∞
(i)
i=2
j=1
φ(aj )
bi4 φ(ai )
2
1 n a Eφ X
(ii) b2 i=1 i φ(a ) i → 0,
i
n∞
(iii)
i >
i=1 P (X
ai ) < ∞,
∞ a4i Eφ Xi (iv)
< ∞.
i=1
b4 φ(a )
P
i
i
0, if and only if Sn /bn →0 a.s., n → ∞.
Then Sn /bn →
Example 7. Let
∞
∞
2
2
l = x = xn , n ≥ 1 ∈ R , x =
|xn | < ∞
2
(38)
n=1
and let en denotes the element having 1 for its nth coordinate and 0 in the other
coordinates.
Assume that {ξn , n ≥ 1} is a sequence of independent random variables such that
P ξj = ±
P ξ1 = 0 = 1,
P ξj = ±j 3 =
1
j
,
1+δ
j
(LLj)2 + 1
,
=
(LLj)2/(1+δ)
j log j
2
2(LLj)2 + 2
,
P ξj = 0 = 1 − 1+δ −
j
j log j
(39)
36
A. KUCZMASZEWSKA AND D. SZYNAL
j ≥ 1, 0 < δ < 1, and define Xn = ξn en .
We see that
p r
Xj E


2(n+1)p
n=1 j=2n +1
r
 n+1 ∞
2
j p /(LLj)2p/(1+δ) · (LLj)2 + 1 /j log j + j 3p /j (1+δ)


=2
2(n+1)p
n=1 j=2n +1
 n+1
r
r
∞
2
∞
3p−(1+δ)
n(3p−(1+δ))
j
n 2


≥2
≥2
2 ·
2(n+1)p
2np 2p
n=1 j=2n +1
n=1
∞

=
n+1
2
(40)
∞
2 n(2p−δ)r
2
= ∞.
pr
2 n=1
Moreover, we note that the condition (a) of Theorem 5 (see [13]) is not satisfied.
Therefore neither Theorem 4 nor Theorem 5 can be applied in this case.
Now we state that the series (i) with φ(x) = |x|1+δ , 0 < δ < 1, bn = n and an =
n/LLn in Theorem 6 (see [15]) can be written in the form
∞
a2j Eφ Xj j−1
Eφ Xi j 4 φ(aj ) i=1
φ(ai )
j=2
1+δ ∞
j 1+δ (LLj)2 · (LLj)2 + 1 j log j + j 3(1+δ)
j
j2
=
2
4 j 1+δ (LLj)1+δ
(LLj)
j
j=3
j−1
i2
×
(LLi)2
i=2
a2i
i1+δ (LLi)2 · (LLi)2 + 1 i log i + i3(1+δ) i1+δ
i1+δ (LLi)1+δ
≥
j−1
∞ 2
j (LLj)2 · j 2(1+δ) i2
i2(1+δ)
4 j 1+δ (LLj)1+δ
2 i1+δ (LLi)1+δ
j
(LLi)
j=3
i=1
≥
j−1
∞
∞
i3+δ
j δ−1
j δ−1
j 3+δ
j 2+2δ
≥
=
= ∞.
1+δ
1+δ
1−δ
1+δ
(LLj)
(LLi)
(LLj)
(LLj)
(LLj)2−2δ
j=3
i=1
j=3
j=3
(41)
∞
Thus we cannot also use Theorem 6.
But we can show that
3 are fulfiled as
assumptions of
the
Theorem
∞ −2
j−1 2
φ Xj φ X i j−1 2
1
1
−2
i=1 i E
≤ ∞
(A)
E
+ j (1+δ)
j=2 j
j=2 j
i=1 i
j log j
φ(j)+φ Xj φ(i)+φ Xi 1
1
× i log
< ∞,
+ i(1+δ)
i
φ Xi n
1
1
2
≤ n−2 n
(B) n−2 i=1 i2 E
+
= o(1),
(1+δ)
i=1 i
i log i
i
φ(i)+φ Xi ∞
∞
1
n
(C)
n=1 P Xn ≥ LLn =
n=1 n1+δ < ∞,
∞
φ Xn 2
1
≤ C ∞
(D)
n=1 φ(an )E
n=2 n log n(LLn)1+δ + n3(1+δ) (LLn)1+δ < ∞.
φ(n)+φ Xn ON CONDITIONS FOR THE STRONG LAW OF LARGE NUMBERS . . .
Now it is enough to see that (4) holds. Taking into account that
we need only to verify that
n
Xi − EXi ≥ nε → 0,
P ∞
n=1 P [Xn
n → ∞.
37
≠ Xn ] < ∞,
(42)
i=1
Using Chebyshev’s inequality and the fact that l2 is a space of the type 2, we have
n
2
n
E Xi P = o(1),
Xi − EXi ≥ nε ≤ C(ε)
n2
i=1
i=1
(43)
which completes the proof that
Sn
→ 0
n
a.s., n → ∞
(44)
Corollary 8. If (B) and (B1 ) are replaced by the condition
−1
n
φ Xi = o(1),
iE
φ(i)
i=1
n
(45)
and in the case (b) additionally EXk = 0, k ≥ 1, then
Sn P
→
0
n
a.s., n → ∞ if and only if
Sn
→ 0 a.s., n → ∞.
n
(46)
Proof. It is enough to show that the condition
n−1
n
i=1
iE
φ Xi = o(1)
φ(i)
(47)
implies
n
EXi I Xi < i → 0,
n−1 n → ∞.
(48)
i=1
Indeed in the case (a) we have
n
n
iEX I Xi < i i
EXi I Xi < i = n−1 n−1 i
i=1
i=1
n
φ Xi ≤ n−1
= o(1),
iE
φ(i)
i=1
(49)
while in the case (b),
n
n
n
iEX I Xi ≥ i i
−1 EXi I Xi < i = n i
i=1
i=1
n
φ Xi
≤ n−1
= o(1).
iE
φ(i)
i=1
−1 (50)
38
A. KUCZMASZEWSKA AND D. SZYNAL
Corollary 9. Let {Xn , n ≥ 1} be a sequence of independent Ꮾ-valued random
elements. If
∞
j
2 j−1
2
Xj Xi 2
E
i E
2
2 < ∞,
2
2
j + Xj i=1
i + Xi 2
n
Xi −2
2
i E
n
2 = o(1),
2
i + X i i=1
−2
j=2
∞
(51)
P Xn ≥ an < ∞
n=1
for some sequence {an , n ≥ 1} of positive numbers with
2
∞
Xn a2n E
4 < ∞,
n4 + Xn (52)
n=1
then
n−1
n
k=1
P
0,
Xk − EXk →
n → ∞
(53)
a.s., n → ∞.
(54)
if and only if
n−1
n
k=1
Xk − EXk → 0
To prove the above-given assertion it is enough to use in the case (b) of the
Theorem 3 the function φ(x) = x 2 .
Corollary 10. Let {Xn , n ≥ 1} be a sequence of independent random elements in
a Banach space of Ᏻα , 0 < α ≤ 1 [12]. Suppose that in the case (a) the conditions (A), (B),
(C) and (D) are satisfied with p = 1 + α, or in the case (b) the conditions (A1 ), (B1 ), (C),
and (D1 ) are satisfied with p = 1 + α. Then
n−1
n
k=1
Xk − EXk → 0
a.s., n → ∞.
Proof. It is enough to show that (4) holds. Indeed in the case (a), we get
n
2
n
n
2
−2 −2 E Xk Xk − EXk > nε ≤ ε n E
Xk − EXk ≤ 4ε−2 n−2
P
k=1
k=1
k=1
n
p φ
Xk
= o(1),
≤ 8ε−2 n−2
kp E p
p X φ
(k)
+
φ
k
k=1
and in the case (b), we have
n
p n
X k
−2 −p
p
P k E
Xk − EXk > nε ≤ 4ε n
kp
k=1
k=1
≤ 8ε
−2
−p
n
φ Xk = o(1).
k E
φ(k) + φ Xk k=1
n
p
(55)
(56)
(57)
ON CONDITIONS FOR THE STRONG LAW OF LARGE NUMBERS . . .
39
But{Xn , n ≥ 1} and {Xn , n ≥ 1} are equivalent, therefore we have (4) which completes the proof of Corollary 10.
Now we give a generalization of Theorem 3 replacing the condition (A) or (A1 ) by
less restrictive ones.
We need the following lemma (see [6, page 329]).
n
Lemma 11. If xj , 1 ≤ j ≤ n, are real numbers, Sn = j=1 xj and
Qk,n =
1≤i1 <i2 <···<ik ≤n
xi1 xi2 · · · · · xik ,
1 ≤ k ≤ n,
(58)
then for 2 ≤ k ≤ n,
Qk,n =
n
xj Qk−1,j−1
Snk = k! · Qk,n + ck ,
and
(59)
j=k
where ck is a generic designation for a finite linear combination (coefficients indepenm n
hj dent of n) of terms j=1
of order k, that is,
i=1 xi
m
hj = k,
1 ≤ hj ≤ k, 1 ≤ m ≤ k.
(60)
j=1
Using this lemma we can prove the following result.
Theorem 12. Let {Xn , n ≥ 1} be a sequence of independent Ꮾ-valued random elements. Suppose that in the case (a) for some p, 1 < p ≤ 2,
∞
jk−1
φp Xj φp Xj
−p(k−1)
p
k
k−1
· · ·
(A )
E p
j
·
E
jk =k jk
j
k−1
k−1=k−1
φ (jk )+φp
Xj
φp (jk−1 )+φp
Xj
k
k−1
p
j2 −1 p
φ
Xj
1
j1 =1 j1 E φp (j )+φp X < ∞ for k ≥ 2
1
j
1
(B) and (C)–(D) are satisfied, or the case (b) for some p, 1 < p ≤ 2,
jk −1
∞
φ Xj φ X j
−p(k−1)
p
k
k−1
· · ·
j
E
j
·
E
(A1 )
jk =k k
=k−1
k−1
j
k−1
φ(j )+φ Xj
φ(jk−1 )+φ Xj
k
k−1
k
j2 −1 p
φ Xj
1 j1 =1 j1 E φ(j )+φ X < ∞,
1
j
1
(B1 ), and (C)–(D1 ) are satisfied.
Then (4) holds if and only if (5) does.
Proof. Using Lemma 11 we get

k
2n
−k
k
S ∗n − E S ∗n = 2−n
2n
Y2n ,i 
2
2
i=1
=
k−2
h=1

n
ch 2
−n
2
i=1
h
−k
Y2n ,i  Ak−h,2n + ck Ak,2n + k! 2n
Qk,2n ,
(61)
40
A. KUCZMASZEWSKA AND D. SZYNAL
where c1 , c2 , . . . , ck−2 are constants, 0 ≤ h ≤ k−2, Ak−h,2n is finite linear combination of
m
2 n h i
m
n −hi
i=1 [(2 )
i=1 hi = k − h
i=1 Y2n ,i ] satisfying hi ≥ 2 for 1 ≤ i ≤ i < m < k,
terms
and
Q0,2n = 1,
Qk,2n =
Y2n ,i1 · · · · · Y2n ,ik .
1≤i1 <···<ik ≤2n
Now note that, for hi ≥ 2,



hi /2
2n
m
2n
m n −hi n −2 hi 
2 
 2

2
Y2n ,j ≤
Y2n ,j
→ 0
i=1
i=1
j=1
j=1
(62)
a.s., n → ∞,
(63)
as
n
2n
2
−2 i=1
Y22n ,i → 0
a.s., n → ∞,
(64)
(cf. the proof of Theorem 3).
Thus for 0 ≤ h ≤ k − 2,
Ak−h,2n → 0
a.s., n → ∞.
(65)
Now we show that
n
2
jk−1
2
−kn
jk =k
Y2n ,jk
j2−1
jk−1 =k−1
Y2n ,jk−1 . . .
Y2n ,j1 → 0
a.s., n → ∞.
(66)
j1 =1
To prove (66) it is enough to see that in the case (a)
n


jk −1
j2 −1
∞
2
−kn 
P 2
Y2n ,jk
Y2n ,jk−1 · · ·
Y2n ,ji > ε
j =k
n=1
j
=k−1
=1
j
1
k
k−1
n
2
jk −1
j2 −1
∞
2
2−2kn E Y2n ,jk
Y2n ,jk−1 · · ·
Y2n ,j1 ≤ ε−2
n=1
j =k
j
=k−1
j =1
= ε−2
2−2kn
n=1
≤ ε−2 24k
2
jk =k
∞
2−2kn
n=1
EY22n ,jk
2n
jk =k
k
≤
∞
ε 2
22k − 1 j
k =k
jk −1
jk−1 =k−1
2
E Xjk ∞
ε−2 26k −p(k−1)
≤ 2k
j
E
2 − 1 j =k k
−2 7k
1
k−1
k
n
∞
p
X EY22n ,jk−1 · · ·
jk −1
jk−1 =k−1
jk
p
jk
jk −1
jk−1 =k−1
j2 −1
j1 =1
EY22n ,j1
j2 −1
2
2
E Xjk−1 · · ·
E Xj1 p
jk−1 E
X
jk−1
p
jk−1
φ Xjk −p(k−1)
jk
E p φ jk + φp Xj j1 =1
p
j2 −1
···
j1 =1
p
j1 E
p
X j1
p
j1
p
k
j2 −1
p
φp Xjk−1 φp Xj1 p
· · ·
< ∞,
jk−1 E p j
E
1
φ jk−1 + φp Xjk−1 φp (j1 ) + φp Xj1 =k−1
j =1
jk −1
×
jk−1
1
(67)
41
ON CONDITIONS FOR THE STRONG LAW OF LARGE NUMBERS . . .
while in the case (b)
n

jk−1
j2−1
2

P 2−kn Y2n ,jk
Y2n ,jk−1 · · ·
Y2n ,j1 >ε
j =k
n=1
jk−1 =k−1
j1 =1
k
p j
p
p
j2−1
∞
X
k−1
p Xj ε−2 26k −p(k−1) Xjk
jk−1
p
1
j
E
jk−1 E
···
j1 E
≤ 2k
p
p
p
2 − 1 j =k k
jk
jk−1
j1
jk−1 =k−1
j1 =1
k
∞
φ Xjk ε−2 27k −p(k−1)
≤ 2k
j
E (68)
2 − 1 j =k k
φ j k + φ X j k ∞

k
jk −1
×
jk−1 =k−1
p
jk−1 E
j2−1
p
φ Xjk−1 φ Xj1 · · ·
< ∞.
j1 E
φ jk−1 + φ Xjk−1 φ(j1 ) + φ Xj1 j =1
1
2 n
By (61), we see that 2−n i=1 Y2n ,i is a root of a kth degree polynomial in which the
leading coefficient is unity and the remaining coefficients (cf. (65) and (66)) converge
almost surely to zero. Therefore, the conclusion
n
2
2−n S2∗n − E S2∗n = 2−n
Y2n ,i → 0
a.s., n → ∞,
(69)
i=1
follows from the well-known relations between the roots and coefficients of a polynomial. Similarly, as in the proof of Theorem 3 we can complete the proof of Theorem 12.
Remark 13. The presented results extend also to random elements theorems of
[18] and [11].
Acknowledgement. We are very grateful to the referee for the suggestions allowing us to improve this paper.
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Anna Kuczmaszewska: Technical University of Lublin, Department of Applied Mathematics, Bernardyńska 13, 20-950 Lublin, Poland
Dominik Szynal: Department of Mathematics, Maria Curie-Skłodowska University,
Pl. M. Curie-Skłodowskiej 1, 20-031 Lublin, Poland