ON CONDITIONS FOR THE STRONG LAW OF LARGE NUMBERS IN GENERAL BANACH SPACES

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)intoᏮis said to be a random element.

IfEX<∞, then the expectationEXis 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], Woyczy´nski [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 condi- tions of classical probability theory or on finding conditions on the random elements which ensure the SLLN. It is known (Woyczy´nski [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.

Lemma1(cf. Yurinskii [20]). LetX1,X2,...,Xnbe independentᏮ-valued random el- ements withEXi<∞, i=1,2,...,n. LetᏲkbeσ-field generated by(X1,X2,...,Xk), k=1,2,...,nand letᏲ0= {Ω,∅}. Then for1≤k≤n,

ESn|Ᏺk

−ESn|Ᏺk−1≤Xk+EXk, whereSn= n i=1

Xi. (1)

Lemma2(cf. Choi and Sung [5]). Let{Xn, n≥1}be a sequence of independent,Ꮾ- valued random elements. Then

Sn

n →0 a.s.,n → ∞ (2)

if and only if S₂k

2^k →0 a.s.,k → ∞ and Sn

n

→P 0, n → ∞. (3)

Let a functionφ:R→R⁺be nonnegative, even, continuous and nondecreasing on (0,∞)with limx→∞φ(x)= ∞and such that

(a) φ(x)/xor (b) φ(x)/x,

andφ(x)/x^p,x→ ∞,for somep, 1< p≤2.

Theorem3. Let {Xn, n ≥ 1} be a sequence of independent Ꮾ-valued random elements. Suppose that in the case (a) for somep, 1< p≤2,

(A) _∞

j=2j^−pE ^φp_Xj

φ^p(j)+φ^p_Xjj−1

i=1i^pE ^φp_Xi

φ^p(i)+φ^p_Xi<∞, (B) n^−p_n

i=1i^pE ^φp_Xi

φ^p(i)+φ^p_Xi=o(1), (C) _∞

n=1P(Xn ≥an) <∞,

for some sequence{an, n≥1}of positive numbers such that (D) _∞

n=1φ^p(an)E ^φp_Xn

φ^2p(n)+φ^2p_Xn<∞ or in the case (b) for somep, 1< p≤2,

(A1) _∞

j=2j^−pE ^φ_Xj

φ(j)+φ_Xjj−1

i=1i^pE ^φ_Xi

φ(i)+φ_Xi<∞, (B1) n^−p_n

i=1i^pE ^φ_Xi

φ(i)+φ_Xi=o(1),

and (C) is satisfied for some sequence{an, n≥1}of positive numbers such that (D1) _∞

n=1φ(an)E ^φ_Xn

φ²(n)+φ²_Xn<∞.

Then

n⁻¹ n k=1

Xk−EX_k P

→0, n → ∞, (4) if and only if

n⁻¹ n k=1

Xk−EX_k

→0 a.s.,n → ∞, (5)

whereX_k=XkI[Xk ≤k].

Proof. Suppose that (4) holds.

Letr≥1. We note that ∞

n=1

P

X_n ≠Xn

= ∞ n=1

E

IXn≥n

·IXn≥an

+IXn≥n

·IXn< an

≤ ∞ n=1

PXn≥an +2

∞ n=1

E φ^2rXn

φ^2r(n)+φ^2rXnIXn< an (6)

≤ ∞ n=1

PXn≥an +2

∞ n=1

φ^r(an)E φ^rXn

φ^2r(n)+φ^2rXn<∞, where we putr=pin the case (a) andr=1 in the case (b).

Hence{X_n, n≥1}and{Xn, n≥1}are equivalent. Therefore, by (4), we have

n⁻¹

n i=1

X_i−EX_i

→^P 0, n → ∞. (7)

Write

X^∗_k=X_k−EX_k, k≥1, S_n= n k=1

X_k, S_n^∗= n k=1

X_k^∗. (8)

Define

Yn,i=ES_n^∗|Ᏺi

−ES_n^∗|Ᏺi−1

, (9)

whereᏲi=σ (X₁^∗,X₂^∗,...,X_i^∗)andᏲ0= {Ω,∅}.

Note that{Yn,i,1≤i≤n}is a martingale difference for fixedn. Then we have Sn^∗−ESn^∗=

n i=1

Yn,i. (10)

Now we prove thatES_n^∗/n→0, n→ ∞. Using (9) and Lemma 1, we get PS_n^∗−ES_n^∗> nε

≤ε⁻²n⁻²ES_n^∗−ES_n^∗2

=ε⁻²n⁻²E n i=1

Yn,i

₂

=ε⁻²n⁻² n i=1

E Y_n,i²

≤ε⁻²n⁻² n i=1

EX_i^∗+EX_i^∗2

≤8ε⁻²n⁻²ⁿ

i=1

EX_i²≤8ε⁻²n^−pn

i=1

i^pEX_i^p i^p .

(11)

Hence in the case (a) by (B), we get PS_n^∗−ES_n^∗> nε

≤16ε⁻²n^−pn

i=1

i^pE φ^pXi

φ^p(i)+φ^pXi=o(1), (12) while in the case (b) by (B1)

PS_n^∗−ES_n^∗> nε

≤16ε⁻²n^−p n i=1

i^pE φXi

φ(i)+φXi=o(1). (13) Therefore, in the case (a) and (b) we have

n⁻¹S_n^∗−ES_n^∗→^P 0, n → ∞. (14)

Thus we conclude, fromSn^∗/n →^P 0, n→ ∞, and (14), that ES_n^∗

n →0, n→ ∞. (15)

Now we are going to prove thatSn^∗/n→0 a.s.,n→ ∞. By Lemma 2 it is enough to prove thatS₂^∗k/2^k→0 a.s.,k→ ∞or equivalently, asES₂^∗k/2^k→0, k→ ∞, that

2^−kS₂^∗k−ES₂^∗k →0 a.s.,k → ∞. (16) Taking into account the identity

S_n^∗−ES_n^∗2= n i=1

Y_n,i² +2 n i=2

Yn,i i−1

j=1

Yn,j, (17)

we see that

2⁻²ⁿS₂^∗n−ES₂^∗n²=2⁻²ⁿ

2ⁿ

i=1

Y₂²n,i+2

2ⁿ

i=2

Y2ⁿ,i i−1

j=1

Y2ⁿ,j

. (18)

Now, put

Z2ⁿ,i=Y₂²n,iIXi< ai

−E

Y₂²n,iIXi< ai

|Ᏺi−1

, 1≤i≤2ⁿ. (19) Then by Chebyshev’s inequality, equation (9) and Lemma 1, we have

∞ n=1

P





2ⁿ

i=1

Z2ⁿ,i

> ε2²ⁿ



≤ε⁻² ∞ n=1

2⁻⁴ⁿE

2ⁿ

i=1

Z2ⁿ,i

2

=ε⁻² ∞ n=1

2⁻⁴ⁿ

2ⁿ

i=1

E

Y₂⁴n,iIXi< ai

≤ε⁻² ∞ n=1

2⁻⁴ⁿ

2ⁿ

i=1

EX_i^∗+EX_i^∗4IXi< ai

≤ε⁻²2⁸ ∞ n=1

2⁻⁴ⁿ

2ⁿ

i=1

EX_i⁴IXi< ai

≤ε⁻² 2¹²

15 ∞

i=1

i⁻⁴EX_i⁴IXi< ai .

(20)

Hence we see that in the case (a) under the assumptions (C) and (D) we have ∞

n=1

P





2ⁿ

i=1

Z2ⁿi

> ε2²ⁿ



≤ε⁻² 2¹²

15 ∞

i=1

i^−2pEX_i^2pIXi< ai

≤ε 2¹²

15 ∞

i=1

Eφ^2pX_i

φ^2p(i) IXi< ai

≤ε⁻²2¹³ 15

∞ i=1

φ^p(ai)E φ^pXi

φ^2p(i)+φ^2pXi<∞.

(21)

Similarly, in the case (b) under the assumptions (C) and (D1) we lead to ∞

n=1

P





2ⁿ

i=1

Z2ⁿ,i

> ε2²ⁿ



≤ε⁻² 2¹²

15 ∞

i=1

i^−2pEX_i^2pIXi< ai

≤ε⁻² 2¹³

15 ∞

i=1

φ ai

E φXi

φ²(i)+φ²Xi<∞.

(22)

Therefore, by the Borel-Cantelli lemma, we state that

(2ⁿ)⁻² ₂n

i=1

Y₂²n,iIXi< ai

−

2ⁿ

i=1

E

Y₂²n,iIXi< ai

|Ᏺi−1

→0 a.s.,n → ∞.

(23) Now, note that in the case (a) the assumption (B) implies

2ⁿ−22ⁿ i=1

E

Y₂²n,iIXi< ai

|Ᏺi−1

≤

2ⁿ₋₂²ⁿ

i=1

EX_i^∗+EX^∗_i²≤8

2ⁿ₋₂²ⁿ

i=1

EX_i²

≤16

2ⁿ−p2ⁿ i=1

i^pE φ^pXi

φ^p(i)+φ^pXi=o(1).

(24)

Similarly, in the case (b), by B1, we get 2ⁿ₋₂²ⁿ

i=1

E

Y₂²n,iIXi< ai

|Ᏺi−1

≤16

2ⁿ_−p²ⁿ

i=1

i^pE φXi

φ(i)+φXi=o(1). (25) Therefore, in the case (a) and (b), we obtain

2ⁿ−22ⁿ i=1

Y₂²n,iIXi< ai

→0 a.s.,n → ∞. (26)

Using the assumption (C), we get 2ⁿ₋₂²ⁿ

i=1

Y₂²n,iIXi≥ai

→0 a.s.,n → ∞ (27)

which proves in the end that 2ⁿ−22ⁿ

i=1

Y₂²n,i →0 a.s.,n→ ∞. (28)

Now we see that

Yn,i_i−1

j=1Yn,j, 2≤i≤n

is a martingale difference for fixedn.

Therefore, in the case (a), after using Chebyshev’s inequality, (9), and Lemma 1,

we get ∞ n=1

P



 2⁻²ⁿ

2ⁿ

i=2

Y2ⁿ,i i−1

j=1

Y2ⁿ,j

> ε





≤ε^−2∞

n=1

2⁻⁴ⁿ² n i=2

E Y2ⁿ,i i−1

j=1

Y2ⁿ,j

₂

≤ε⁻² ∞ n=1

2⁻⁴ⁿ

2ⁿ

i=2

E X_i^∗+EX_i^∗2 i−1

j=1

Y2ⁿ,j

2

≤2⁸ε⁻² ∞ n=1

2⁻⁴ⁿ

2ⁿ

i=2

EX_i²ⁱ⁻¹

j=1

EX_j² (29)

≤ 2¹²

15

ε⁻² ∞ i=1

i⁻⁴EX_i²ⁱ⁻¹

j=1

EX_j²

≤ 2¹²

15

ε⁻² ∞ i=1

i^−2pEX_i^pi−1

j=1

EX_j^p

≤ 2¹⁴

15

ε⁻² ∞ i=1

i^−pE φ^pXi φ^p(i)+φ^pXiⁱ⁻¹

j=1

j^pE φ^pXj

φ^p(j)+φ^pXj<∞.

Similarly, in the case (b), we have ∞

n=1

P



 2⁻²ⁿ

2ⁿ

i=2

Y2ⁿ,i i−1

j=1

Y2ⁿ,j

≥ε





≤ 2¹²

15

ε⁻² ∞ i=2

i^−2pEX_i^pi−1

j=1

EX_j^p

≤ 2¹⁴

15

ε⁻² ∞ i=2

i^−pE φXi φ(i)+φXiⁱ⁻¹

j=1

j^pE φXj

φ(j)+φXj<∞.

(30)

Now using the Borel-Cantelli lemma we obtain 2⁻²ⁿ

2ⁿ

i=2

Y2ⁿ,i i−1

j=1

Y2ⁿ,j →0 a.s.,n → ∞. (31) Thus by (15) and (18) we see that

S₂^∗n

2ⁿ →0 a.s.,n → ∞, (32)

so we have

n⁻¹

n i=1

X_i−EX_i

→0 a.s.,n → ∞. (33)

But{X_n, n≥1}and{Xn, n≥1}are equivalent, so that n⁻¹

∞ n=1

Xi−EX_i

→0 a.s.,n→ ∞ (34)

which completes to proof of Theorem 3.

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.

Theorem4[13]. Let{Xn, n≥1}be a sequence of independentᏮ-valued random variables such that

(a) Xj/j→0a.s.,j→ ∞,

(b) for somep∈[1,2]and somer∈(0,∞) ∞

n=1



 ² n+1 j=2ⁿ+1

EXj^p/2^(n+1)p





r

<∞. (35)

Then

Sn

n

→P 0 if and only if Sn

n →0a.s.,n→ ∞. (36)

Theorem5[13]. Let{Xn, n≥1}be a sequence of independentᏮ-valued random variables such that

(a) |Xj| ≤Mj/LLjfor some constantM <∞, whereLLj=log(log(j∨e^e)), and (b) _∞

n=1{−ε/Λ(n)}<∞for allε >0whereΛ(n)=₂ⁿ⁺¹

j=2ⁿ+1EXj²/2²⁽ⁿ⁺¹⁾. Then Sn

n

→P 0 if and only if Sn

n →0a.s.,n→ ∞. (37)

Theorem6[15]. Let{Xn, n≥1}be a sequence of independentB-valued random variables, and let{an}and{bn}be constants that0< bn. Suppose that

(i) _∞

i=2a²_iEφ_Xi

b⁴_iφ(ai)

_i−1

j=1

a²_jEφ_Xj

φ(a_j) <∞ (ii) _b¹2n

_n

i=1a²_iEφ_Xi

φ(ai) →0, (iii) _∞

i=1P(Xi> ai) <∞, (iv) _∞

i=1a⁴_iEφ_Xi

b⁴_iφ(a_i) <∞.

ThenSn/bn →P 0,if and only ifSn/bn→0a.s.,n→ ∞.

Example7. Let l²=

x=

xn, n≥1

∈R^∞, x²= ∞ n=1

|xn|²<∞

(38) and leteⁿ denotes the element having 1 for itsnth coordinate and 0 in the other coordinates.

Assume that{ξn, n≥1}is a sequence of independent random variables such that P

ξ1=0

=1, P

ξj= ± j (LLj)^2/(1+δ)

=(LLj)²+1 jlogj , P

ξj= ±j³

= 1

j^1+δ, P ξj=0

=1− 2

j^1+δ−2(LLj)²+2 jlogj ,

(39)

j≥1,0< δ <1, and defineXn=ξnen. We see that

∞ n=1



 ² n+1 j=2ⁿ+1

EXj^p 2^(n+1)p





r

=2 ∞ n=1



 ² n+1 j=2ⁿ+1

j^p/(LLj)^2p/(1+δ)·

(LLj)²+1

/jlogj+j^3p/j^(1+δ) 2^(n+1)p





r

≥2 ∞ n=1



 ² n+1 j=2ⁿ+1

j^3p−(1+δ) 2^(n+1)p





r

≥2 ∞ n=1

2ⁿ·2n(3p−(1+δ))

2^np2^p r

= 2 2^pr

∞ n=1

2^n(2p−δ)r= ∞.

(40)

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) withφ(x)= |x|^1+δ, 0< δ <1, bn=nandan= n/LLnin Theorem 6 (see [15]) can be written in the form

∞ j=2

a²_jEφXj j⁴φ(aj)

j−1

i=1

a²_iEφXi φ(ai)

= ∞ j=3

j² (LLj)²

j^1+δ (LLj)²

·

(LLj)²+1 jlogj

+

j^3(1+δ) j^1+δ j⁴

j^1+δ

(LLj)^1+δ

×

j−1

i=2

i² (LLi)²

i^1+δ (LLi)²

·

(LLi)²+1 ilogi

+ i^3(1+δ)

i^1+δ i^1+δ

(LLi)^1+δ

≥ ∞ j=3

j² (LLj)²

·j^2(1+δ) j⁴

j^1+δ

(LLj)^1+δ

j−1

i=1

i² (LLi)²

i^2(1+δ) i^1+δ

(LLi)^1+δ

≥ ∞ j=3

j^δ−1 (LLj)^1+δ

j−1

i=1

i^3+δ (LLi)^1+δ ≥

∞ j=3

j^δ−1 (LLj)^1−δ

j^3+δ (LLj)^1+δ =

∞ j=3

j^2+2δ (LLj)^2−2δ= ∞.

(41)

Thus we cannot also use Theorem 6.

But we can show that the assumptions of Theorem 3 are fulfiled as (A) _∞

j=2j⁻²E ^φ_Xj

φ(j)+φ_Xjj−1

i=1i²E ^φ_Xi

φ(i)+φ_Xi ≤ _∞

j=2j⁻² ₁

jlogj + _j(1+δ)¹

j−1 i=1i²

× ₁

ilogi+_i(1+δ)¹

<∞, (B) n⁻²_n

i=1i²E ^φ_Xi

φ(i)+φ

X_i≤n⁻²_n

i=1i² ₁

ilogi+_i(1+δ)¹

=o(1), (C) _∞

n=1P

Xn ≥_LLnⁿ

=_∞

n=1 1 n^1+δ <∞, (D) _∞

n=1φ(an)E ^φ_Xn

φ(n)+φ_Xn≤C_∞

n=2

₂

nlogn(LLn)^1+δ+_n3(1+δ)(LLn)^{1 1+δ}

<∞.

Now it is enough to see that (4) holds. Taking into account that_∞

n=1P[Xn≠Xn] <∞, we need only to verify that

P

n i=1

X_i−EX_i ≥nε

→0, n → ∞. (42)

Using Chebyshev’s inequality and the fact thatl²is a space of the type 2, we have

P

n i=1

X_i−EX_i ≥nε

≤C(ε)ⁿ

i=1

EX_i²

n² =o(1), (43)

which completes the proof that Sn

n →0 a.s.,n → ∞ (44)

Corollary8. If(B)and(B1)are replaced by the condition

n⁻¹ⁿ

i=1

iEφXi φ(i)

=o(1), (45)

and in the case (b) additionallyEXk=0, k≥1, then Sn

n

→P 0 a.s.,n → ∞if and only if Sn

n →0a.s.,n → ∞. (46) Proof. It is enough to show that the condition

n⁻¹ⁿ

i=1

iEφXi φ(i)

=o(1) (47)

implies

n⁻¹

n i=1

EXiIXi< i

→0, n → ∞. (48)

Indeed in the case (a) we have n⁻¹

n i=1

EXiIXi< i =n⁻¹

n i=1

iEXiIXi< i i

≤n⁻¹ n i=1

iE

φXi φ(i)

=o(1),

(49)

while in the case (b), n⁻¹

n i=1

EXiIXi< i =n⁻¹

n i=1

iEXiIXi≥i i

≤n⁻¹ n i=1

iE

φXi φ(i)

=o(1).

(50)

Corollary9. Let {Xn, n≥1} be a sequence of independent Ꮾ-valued random elements. If

∞ j=2

j⁻²E Xj² j²+Xj²

j−1

i=1

i²E Xi² i²+Xi²<∞, n⁻²

n i=1

i²E Xi²

i²+Xi²=o(1), ∞

n=1

PXn≥an

<∞

(51)

for some sequence{an, n≥1}of positive numbers with ∞

n=1

a²_nE Xn²

n⁴+Xn⁴<∞, (52)

then

n⁻¹ n k=1

Xk−EX_{k P}

→0, n → ∞ (53) if and only if

n⁻¹ n k=1

Xk−EX_k

→0 a.s.,n → ∞. (54)

To prove the above-given assertion it is enough to use in the case (b) of the Theorem 3 the functionφ(x)=x².

Corollary10. Let{Xn, n≥1}be a sequence of independent random elements in a Banach space ofᏳα,0< α≤1 [12]. Suppose that in the case (a) the conditions (A), (B), (C) and (D) are satisfied withp=1+α, or in the case (b) the conditions(A1),(B1),(C), and(D1)are satisfied withp=1+α. Then

n⁻¹ n k=1

Xk−EX_k

→0 a.s.,n → ∞. (55)

Proof. It is enough to show that (4) holds. Indeed in the case (a), we get P

n k=1

X_k−EX_k > nε

≤ε⁻²n⁻²E

n k=1

X_k−EX_k

2

≤4ε⁻²n⁻² n k=1

EX_k²

≤8ε⁻²n⁻² n k=1

k^pE φ^pXk

φ^p(k)+φ^pXk=o(1),

(56)

and in the case (b), we have P

n k=1

X_k−EX_k > nε

≤4ε⁻²n^−pn

k=1

k^pE X_k^p k^p

≤8ε⁻²n^−p n k=1

k^pE φXk

φ(k)+φXk=o(1).

(57)

But{Xn, n≥1}and{Xn, n≥1}are equivalent, therefore we have (4) which com- pletes 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]).

Lemma11. Ifxj,1≤j≤n, are real numbers,Sn=_n

j=1xjand

Qk,n=

1≤i1<i2<···<i_k≤n

xi1xi2· ··· ·xik, 1≤k≤n, (58)

then for2≤k≤n, Qk,n=

n j=k

xjQk−1,j−1 and S_n^k=k!·Qk,n+ck, (59)

whereckis a generic designation for a finite linear combination (coefficients indepen- dent ofn) of terms_m

j=1_n

i=1x_i^hj

of orderk, that is, m

j=1

hj=k, 1≤hj≤k,1≤m≤k. (60)

Using this lemma we can prove the following result.

Theorem12. Let{Xn, n≥1}be a sequence of independentᏮ-valued random ele- ments. Suppose that in the case (a) for somep, 1< p≤2,

(A) _∞

j_k=kj_k^−p(k−1)E ^φp_X

jk

φ^p(j_k)+φ^p_X

jkj_k−1

j_k−1=k−1j^p_k−1 · E ^φp_X

jk−1

φ^p(j_k−1)+φ^p_X

jk−1···

j₂−1

j1=1j₁^pE ^φp_Xj

1

φ^p(j1)+φ^p_Xj

1<∞fork≥2

(B) and (C)–(D) are satisfied, or the case (b) for somep,1< p≤2, (A₁) _∞

jk=kj_k^−p(k−1)E ^φ_X

jk

φ(jk)+φ_X

jkjk−1

j_k−1=k−1j_k−1^p · E ^φ_X

jk−1

φ(jk−1)+φ_X

jk−1···

_j2−1

j1=1j₁^pE ^φ_Xj

1

φ(j1)+φ_Xj

1<∞, (B1), and (C)–(D1) are satisfied.

Then (4) holds if and only if (5) does.

Proof. Using Lemma 11 we get 2ⁿ_−kS₂^∗n−ES₂^∗nk=



2⁻ⁿ

2ⁿ

i=1

Y2ⁿ,i





k

=

k−2

h=1

ch



2⁻ⁿ

2ⁿ

i=1

Y2ⁿ,i





h

Ak−h,2ⁿ+ckAk,2ⁿ+k!

2ⁿ_−k Qk,2ⁿ,

(61)