On the conditions for the complete convergence in mean for double sums of independent random elements in banach spaces

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In this paper, we establish the condition for convergence of X∞ m=1 X∞ n=1 1 mn E kSmnk (mn) 1/p q for double arrays of independent random elements in Banach spaces following the type proposed by Li, Qi and Rosalsky.

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Nội dung Text: On the conditions for the complete convergence in mean for double sums of independent random elements in banach spaces

Trường Đại học Vinh Tạp chí khoa học, Tập 46, Số 2A (2017), tr. 31-42 ON THE CONDITIONS FOR THE COMPLETE CONVERGENCE IN MEAN FOR DOUBLE SUMS OF INDEPENDENT RANDOM ELEMENTS IN BANACH SPACES Vu Thi Ngoc Anh (1) , Nguyen Thi Thuy (2) of Mathematics, Hoa Lu University, Ninh Binh 2 Teacher of Mathematics, Thanh Chuong 3 High School, Nghe An. Received on 19/4/2017, accepted for publication on 20/10/2017 1 Department Abstract: In this paper, we establish the condition for convergence of ∞ X ∞ X 1 kSmn k q E for double arrays of independent random elements mn (mn)1/p m=1 n=1 in Banach spaces following the type proposed by Li, Qi and Rosalsky [6]. 1. Introduction and Preliminaries Throughout this paper, let (B, k.k) be a real separable Banach space. Li, Qi and Rosalsky [6] extended Theorem 2.4.1 in [1], Theorem 5 in [2] and Theorems 2.1 and 2.2 in [4] as follows: Let 0 0. Let {Xn , n ≥ 1} be a sequence of independent copies of a B-valued random element X. Set Sn = X1 + X2 + · · · + Xn , n ≥ 1. Then ∞ X 1 kSn k q E t) dt < ∞, if q < p   EkXkp ln(kXk + 1) < ∞,    EkXkq < ∞, if q = p if q > p 0 Based on that idea, we establish the condition for convergence of ∞ ∞ X X 1 kSmn k q E for double arrays of independent random elements in Banach 1/p mn (mn) m=1 n=1 spaces. 1) Email: anhyk86@gmail.com (V. T. N. Anh). 31 Vu Thi Ngoc Anh, Nguyen Thi Thuy/ On the conditions for the complete convergence in mean... Throughout this paper, the symbol C denotes a generic constant (0 < C < ∞) which might be different in different appearances. For a matrix A = (xij )m×n with the entries xij ∈ B, 1 ≤ i ≤ m, 1 ≤ j ≤ n, we denote π(A) = π ((xij )m×n ) = (x11 , x12 , . . . , x1n , . . . , xm1 , xm2 , . . . , xmn ) ∈ B mn . For a, b ∈ R, max{a, b} will be denoted by a ∨ b, min{a, b} will be denoted by a ∧ b. 2. Main Results Theorem 2.1. Let 0 0. Let {Xmn , m ≥ 1, n ≥ 1} be a double array of independent copies of a B-valued random element X. Set Smn = m X n X Xkl , m ≥ 1, n ≥ 1. k=1 l=1 Then ∞ X ∞ X 1 kSmn k q E p (3) The following three lemmas are used to prove Theorem 2.1. Lemma 2.1. Let {cmn , m ≥ 1, n ≥ 1} be a double array of nonnegative real numbers. Then i) For all k ≥ 1, l ≥ 1, we have k X l X cij ≤ i=1 j=1 ∞ X ∞ X cij . (4) ckl as n → ∞. (5) i=1 j=1 ii) We have n X n X ckl % k=1 l=1 ∞ X ∞ X k=1 l=1 iii) If {amn , m ≥ 1, n ≥ 1} is a double array of positive constants such that amn < m=1 n=1 ∞, then max n X n X 1≤k≤n 1≤l≤n i=k j=l 32 ∞ X ∞ X aij ckl % sup bkl ckl as n → ∞, k≥1, l≥1 (6) Trường Đại học Vinh where bkl = ∞ X ∞ X Tạp chí khoa học, Tập 46, Số 2A (2017), tr. 31-42 aij , k ≥ 1, l ≥ 1. i=k j=l Proof. i) For all m ∧ n ≥ k ∨ l, we have k X l X m X n X cij ≤ i=1 j=1 cij . i=1 j=1 We get (4) holds. ∞ P ∞ P ckl = ∞. Then for all M > 0, there exist n0 ∈ N∗ such that ii) Case 1: k=1 l=1 m X n X ckl > M for all m ∧ n ≥ n0 . k=1 l=1 Hence, n X n X ckl > M for all n ≥ n0 . k=1 l=1 We get (5) holds. ∞ P ∞ P Case 2: ckl = c < ∞. Then for all ε > 0, there exist n1 ∈ N∗ such that k=1 l=1 | m X n X ckl − c| < ε for all m ∧ n ≥ n1 . k=1 l=1 Hence, | n X n X ckl − c| < ε for all n ≥ n1 . k=1 l=1 We get (5) holds. iii) Case 1: sup bkl ckl = ∞. Let M > 0 be arbitrary. Then there exist kM , lM ∈ N∗ k≥1, l≥1 such that bkM lM ckM lM > 2M. By (5), we have lim n→∞ n X n X aij = ∞ X ∞ X i=1 i=1 aij < ∞. i=1 i=1 Hence, lim n→∞ X aij = 0. (7) i∨j>n We get there exist nM ∈ N∗ such that X aij ckM lM < M for all n ≥ nM . i∨j>n 33 Vu Thi Ngoc Anh, Nguyen Thi Thuy/ On the conditions for the complete convergence in mean... Thus, for n ≥ (kM ∨ lM ∨ nM ), we have n n X X aij ckM lM > M. i=kM j=lM Hence, max n X n X 1≤k≤n 1≤l≤n i=k j=l aij ckl ≥ M for all n ≥ (kM ∨ lM ∨ nM ). We get (6) holds. Case 2: sup bkl ckl = a < ∞. Let ε > 0 be arbitrary. Then there exist kε , lε ∈ N∗ such k≥1, l≥1 that ε bkε lε ckε lε ≥ a − . 2 By (7), we get there exists nε ∈ N∗ such that X ε aij ckε lε ≤ for all n ≥ nε . 2 i∨j>n Thus, for n ≥ (kε ∨ lε ∨ nε ), we have n X n X aij ckε lε ≥ a − ε. i=kε j=lε Hence, max n X n X 1≤k≤n 1≤l≤n i=k j=l aij ckl ≥ a − ε for all n ≥ (kε ∨ lε ∨ nε ). We get (6) holds. Lemma 2.2. Let q > 0 and let {amn , m ≥ 1, n ≥ 1} be a double array of positive constants ∞ X ∞ X such that amn < ∞. Let {Xmn , m ≥ 1, n ≥ 1} be a double array of independent m=1 n=1 symmetric random elements in B. Write bkl = ∞ X ∞ X aij , k ≥ 1, l ≥ 1, i=k j=l q k l X X T = akl Xij , i=1 j=1 k=1 l=1 ( if 0 < q ≤ 1 1, if 0 < q ≤ 1 and β = q−1 if q > 1 2 , if q > 1 ∞ X ∞ X ( 21−q , α= 1, 34 (8) Trường Đại học Vinh Tạp chí khoa học, Tập 46, Số 2A (2017), tr. 31-42 Then for all s, t, u ≥ 0, we have ! q sup bkl kXkl k > t P ≤ 2P k≥1, l≥1 t T > α (9) and P (T > s + t + u) ≤P s sup bkl kXkl kq > 2 β k≥1, l≥1 ! u t + 4P T > P T > . (10) αβ αβ 2 Furthermore, we have ! E sup bkl kXkl k q ≤ E(T ) (11) k≥1, l≥1 and ! E (T ) ≤ 6(α + β)3 E sup bkl kXkl kq + 6(α + β)3 c, (12) k≥1, l≥1 where c > 0 such that P (T > c) ≤ 24−1 (α + β)−3 . Proof. For n ≥ 1 and a matrix (xij )n×n with the entries xij ∈ B, 1 ≤ i ≤ n, 1 ≤ j ≤ n, we write q X n X n v X u X qnn (π((xij )n×n )) = auv xij . u=1 v=1 i=1 j=1 Then qnn x+y 2 ≤ α max(qnn (x), qnn (y)) for all x, y ∈ B nn and qnn (x + y) ≤ β(qnn (x) + qnn (y)) for all x, y ∈ B nn . Set nn Snn = qnn (π((Xij )n×n )), kl Enn = (Eij )n×n ( Xkl if i = k, j = l where Eij = , 1 ≤ k ≤ n, 1 ≤ l ≤ n, 0 otherwise kl kl Wnn = qnn π(Enn ) , 1 ≤ k ≤ n, 1 ≤ l ≤ n, kl Nnn = max Wnn . 1≤k≤n 1≤l≤n By Theorem 2.1 in [5], for all t ≥ 0, we have nn P (Nnn > t) ≤ 2P (Snn > t/α). (13) 35