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On complete convergence and complete moment convergence for weighted sums of
ρ
∗
$\rho^{*}$
-mixing random variables
On complete convergence and complete moment convergence for weighted sums of
ρ
∗...
Chen, Pingyan; Sung, Soo
2018-06-01 00:00:00
Department of Applied Let r ≥ 1, 1 ≤ p <2, and α, β >0 with 1/α +1/β =1/p.Let {a ,1 ≤ k ≤ n, n ≥ 1} be nk Mathematics, PaiChaiUniversity, –1 α Daejeon, South Korea an array of constants satisfying sup n |a | < ∞,and let {X , n ≥ 1} be a nk n n≥1 k=1 Full list of author information is sequence of identically distributed ρ -mixing random variables. For each of the three available at the end of the article cases α < rp, α = rp,and α > rp, we provide moment conditions under which ∞ m r–2 1/p n P max a X > εn < ∞, ∀ε >0. nk k 1≤m≤n n=1 k=1 We also provide moment conditions under which ∞ m q r–2–q/p 1/p n E max a X – εn < ∞, ∀ε >0, nk k 1≤m≤n n=1 k=1 where q > 0. Our results improve and generalize those of Sung (Discrete Dyn. Nat. Soc. 2010:630608, 2010) and Wu et al. (Stat. Probab. Lett. 127:55–66, 2017). MSC: 60F15 Keywords: ρ -mixing random variables; Complete convergence; Complete moment convergence; Weighted sum 1 Introduction Due to the estimation of least squares regression coeﬃcients in linear regression and non- parametric curve estimation, it is very interesting and meaningful to study the limit be- haviors for the weighted sums of random variables. We recall the concept of ρ -mixing random variables. Deﬁnition 1.1 Let {X , n ≥ 1} be a sequence of random variables deﬁned on a probability space (, F, P). For any S ⊂ N = {1,2,...},deﬁne F = σ (X , i ∈ S). Given two σ -algebras S i A and B in F,put EXY – EXEY ρ(A, B)= sup : X ∈ L (A), Y ∈ L (B) . 2 2 2 2 E(X – EX) E(Y – EY ) © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 2 of 16 Deﬁne the ρ -mixing coeﬃcients by ρ = sup ρ(F , F ): S, T ⊂ N with dist(S, T) ≥ n , S T ∗ ∗ ∗ where dist(S, T)= inf{|s – t| : s ∈ S, t ∈ T }.Obviously, 0 ≤ ρ ≤ ρ ≤ ρ = 1. Then the n+1 n 0 ∗ ∗ sequence {X , n ≥ 1} is called ρ -mixing if there exists k ∈ N such that ρ <1. A number of limit results for ρ -mixing sequences of random variables have been es- tablishedbymanyauthors.Werefer to Bradley[3] for the central limit theorem, Bryc and Smolenski [4], Peligrad and Gut [5], and Utev and Peligrad [6] for the moment inequalities, and Sung [1] for the complete convergence of weighted sums. Special cases for weighted sums have been studied by Bai and Cheng [7], Chen et al. [8], Choi and Sung [9], Chow [10], Cuzick [11], Sung [12], Thrum [13], and others. In this paper, we focus on the array weights {a ,1 ≤ k ≤ n, n ≥ 1} of real numbers satisfying nk –1 α sup n |a | < ∞ (1.1) nk n≥1 k=1 for some α > 0. In fact, under condition (1.1), many authors have studied the limit behav- iors for the weighted sums of random variables. Let {X, X , n ≥ 1} be a sequence of independent and identically distributed random vari- ables. When α =2, Chow [10] showed that the Kolmogorov strong law of large numbers –1 n a X → 0 a.s. (1.2) nk k k=1 holds if EX =0 and EX < ∞.Cuzick[11] generalized Chow’s result by showing that (1.2) also holds if EX =0 and E|X| < ∞ for β >0 with 1/α +1/β = 1. Bai and Cheng [7]proved that the Marcinkiewicz–Zygmund strong law of large numbers –1/p n a X → 0 a.s. (1.3) nk k k=1 holds if EX =0 and E|X| < ∞,where 1 ≤ p <2 and 1/α +1/β =1/p.Chenand Gan[14] showed that if 0 < p <1 and E|X| < ∞,then(1.3) still holds without the independent assumption. Under condition (1.1), a convergence rate in the strong law of large numbers is also discussed. Chen [15] showed that ∞ m r–2 1/p n P max a X > εn < ∞, ∀ε > 0, (1.4) nk k 1≤m≤n n=1 k=1 if {X, X , n ≥ 1} is a sequence of identically distributed negatively associated (NA) random (r–1)β variables with EX =0 and E|X| < ∞,where r >1, 1 ≤ p <2, 1/α +1/β =1/p,and α < rp. The main tool used in Chen [15] is the exponential inequality for NA random variables (see Theorem 3 in Shao [16]). Sung [1]proved(1.4) for a sequence of identically Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 3 of 16 ∗ rp distributed ρ -mixing random variables with EX =0 and E|X| < ∞,where α > rp,by using the Rosenthal moment inequality. Since the Rosenthal moment inequality for NA has been established by Shao [16], it is easy to see that Sung’s result also holds for NA random variables. However, for ρ -mixing random variables, we do not know whether the corresponding exponential inequality holds or not, and so the method of Chen [15] does not work for ρ -mixing random variables. On the other hand, the method of Sung [1] is complex and not applicable to the case α ≤ rp. In this paper, we show that (1.4) holds for a sequence of identically distributed ρ -mixing random variables with suitable moment conditions. The moment conditions for the cases α < rp and α > rp are optimal. The moment conditions for α = rp are nearly optimal. Al- though the main tool is the Rosenthal moment inequality for ρ -mixing random variables, our method is simpler than that of Sung [1]eveninthe case α > rp. We also extend (1.4) to complete moment convergence, that is, we provide moment conditions under which ∞ m r–2–q/p 1/p n E max a X – εn < ∞, ∀ε > 0, (1.5) nk k 1≤m≤n n=1 k=1 where q >0. Note that if (1.5) holds for some q >0, then (1.4)alsoholds.The proofiswellknown. Throughout this paper, C always stands for a positive constant that may diﬀer from one place to another. For events A and B,wedenote I(A, B)= I(A ∩ B), where I(A)is the indicator function of an event A. 2 Preliminary lemmas To prove the main results, we need the following lemmas. The ﬁrst one belongs to Utev andPeligrad[6]. Lemma 2.1 Let q ≥ 2, and let {X , n ≥ 1} be a sequence of ρ -mixing random variables with EX =0 and E|X | < ∞ for every n ≥ 1. Then for all n ≥ 1, n n q q/2 m n n q 2 E max X ≤ C E|X | + E|X | , k q k k 1≤m≤n k=1 k=1 k=1 where C >0 depends only on q and the ρ -mixing coeﬃcients. Remark 2.1 By the Hölder inequality, (1.1)implies that –1 s sup n |a | < ∞ nk n≥1 k=1 for any 0 < s ≤ α,and –q/α q sup n |a | < ∞ nk n≥1 k=1 for any q > α. These properties will be used in the proofs of the following lemmas and main results. Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 4 of 16 Lemma 2.2 Let r ≥ 1, 0 < p <2, α >0, β >0 with 1/α +1/β =1/p, and let X be a random variable. Let {a ,1 ≤ k ≤ n, n ≥ 1} be an array of constants satisfying (1.1). Then nk (r–1)β CE|X| if α < rp, ∞ n r–2 1/p (r–1)β n P |a X| > n ≤ (2.1) CE|X| log(1 + |X|) if α = rp, nk n=1 k=1 rp CE|X| if α > rp. Proof Case 1: α ≤ rp. We observe by the Markov inequality that, for any s >0, 1/p P |a X| > n nk 1/p 1/β 1/p 1/β = P |a X| > n , |X| > n + P |a X| > n , |X|≤ n nk nk –α/p α α 1/β –s/p s s 1/β ≤ n |a | E|X| I |X| > n + n |a | E|X| I |X|≤ n . (2.2) nk nk It is easy to show that ∞ n r–2 –α/p α α 1/β n · n |a | E|X| I |X| > n nk n=1 k=1 r–1–α/p α 1/β ≤ C n E|X| I |X| > n n=1 ⎨ (r–1)β CE|X| if α < rp, ≤ (2.3) ⎩ (r–1)β CE|X| log(1 + |X|)if α = rp. Taking s > max{α,(r –1)β},wehavethat ∞ n r–2 –s/p s s 1/β n · n |a | E|X| I |X|≤ n nk n=1 k=1 r–2–s/p+s/α s 1/β ≤ C n E|X| I |X|≤ n n=1 r–2–s/β s 1/β = C n E|X| I |X|≤ n n=1 (r–1)β ≤ CE|X| , (2.4) since s >(r –1)β.Then(2.1)holds by (2.2)–(2.4). Case 2: α > rp. The proof is similar to that of Case 1. However, we use a diﬀerent trun- cation for X. We observe by the Markov inequality that, for any t >0, 1/p P |a X| > n nk 1/p 1/p 1/p 1/p = P |a X| > n , |X| > n + P |a X| > n , |X|≤ n nk nk –t/p t t 1/p –α/p α α 1/p ≤ n |a | E|X| I |X| > n + n |a | E|X| I |X|≤ n . (2.5) nk nk Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 5 of 16 Taking 0 < t < rp,wehavethat ∞ n r–2 –t/p t t 1/p n · n |a | E|X| I |X| > n nk n=1 k=1 r–1–t/p t 1/p ≤ C n E|X| I |X| > n n=1 rp ≤ CE|X| . (2.6) It is easy to show that ∞ n r–2 –α/p α α 1/p n · n |a | E|X| I |X|≤ n nk n=1 k=1 r–1–α/p α 1/p ≤ C n E|X| I |X|≤ n n=1 rp ≤ CE|X| , (2.7) since α > rp.Then(2.1)holds by (2.5)–(2.7). Lemma 2.3 Let r ≥ 1, 0 < p <2, α >0, β >0 with 1/α +1/β =1/p, and let X be a random variable. Let {a ,1 ≤ k ≤ n, n ≥ 1} be an array of constants satisfying (1.1). Then, for any nk s > max{α,(r –1)β}, ∞ n r–2–s/p s 1/p n E|a X| I |a X|≤ n nk nk n=1 k=1 (r–1)β CE|X| if α < rp, (r–1)β ≤ (2.8) CE|X| log(1 + |X|) if α = rp, rp CE|X| if α > rp. Proof Case 1: α ≤ rp.By(2.3)and (2.4)weget that ∞ n r–2–s/p s 1/p n E|a X| I |a X|≤ n nk nk n=1 k=1 ∞ n r–2–s/p s 1/p 1/β = n E|a X| I |a X|≤ n , |X| > n nk nk n=1 k=1 ∞ n r–2–s/p s 1/p 1/β + n E|a X| I |a X|≤ n , |X|≤ n nk nk n=1 k=1 ∞ n r–2–s/p (s–α)/p α 1/β ≤ n n E|a X| I |X| > n nk n=1 k=1 ∞ n r–2–s/p s 1/β + n E|a X| I |X|≤ n nk n=1 k=1 Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 6 of 16 (r–1)β CE|X| if α < rp, ⎩ (r–1)β CE|X| log(1 + |X|)if α = rp. Case 2: α > rp.Taking 0 < t < rp,wehaveby(2.6)and (2.7)that ∞ n r–2–s/p s 1/p n E|a X| I |a X|≤ n nk nk n=1 k=1 ∞ n r–2–s/p s 1/p 1/p = n E|a X| I |a X|≤ n , |X| > n nk nk n=1 k=1 ∞ n r–2–s/p s 1/p 1/p + n E|a X| I |a X|≤ n , |X|≤ n nk nk n=1 k=1 ∞ n r–2–s/p (s–t)/p t 1/p ≤ n n E|a X| I |X| > n nk n=1 k=1 ∞ n r–2–s/p (s–α)/p α 1/p + n n E|a X| I |X|≤ n nk n=1 k=1 rp ≤ CE|X| . Therefore (2.8)holds. The following lemma is a counterpart of Lemma 2.3. The truncation for |a X| is re- nk versed. Lemma 2.4 Let q >0, r ≥ 1, 0 < p <2, α >0, β >0 with 1/α +1/β =1/p, and let X be a random variable. Let {a ,1 ≤ k ≤ n, n ≥ 1} be an array of constants satisfying (1.1). Then nk the following statements hold. (1) If α < rp, then ∞ n r–2–q/p q 1/p n E|a X| I |a X| > n nk nk n=1 k=1 (r–1)β CE|X| if q <(r –1)β, (r–1)β ≤ (2.9) CE|X| log(1 + |X|) if q =(r –1)β, CE|X| if q >(r –1)β. (2) If α = rp, then ∞ n r–2–q/p q 1/p n E|a X| I |a X| > n nk nk n=1 k=1 ⎨ (r–1)β CE|X| log(1 + |X|) if q ≤ α = rp, ≤ (2.10) ⎩ q CE|X| if q > α = rp. Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 7 of 16 (3) If α > rp, then ∞ n r–2–q/p q 1/p n E|a X| I |a X| > n nk nk n=1 k=1 rp CE|X| if q < rp, rp ≤ (2.11) CE|X| log(1 + |X|) if q = rp, CE|X| if q > rp. –1 α Proof Without loss of generality, we may assume that n |a | ≤ 1 for all n ≥ 1. nk k=1 1/α From this we have that |a |≤ n for all 1 ≤ k ≤ n and n ≥ 1. nk (1) In this case, we have that α < rp <(r –1)β.If 0 < q < α,then ∞ n r–2–q/p q 1/p n E|a X| I |a X| > n nk nk n=1 k=1 ∞ n r–2–q/p –(α–q)/p α 1/p ≤ n n E|a X| I |a X| > n nk nk n=1 k=1 ∞ n r–2–q/p –(α–q)/p α 1/α 1/p ≤ n n E|a X| I n X > n nk n=1 k=1 r–1–α/p α 1/β ≤ C n E|X| I |X| > n n=1 ∞ i α 1/β 1/β r–1–α/p = C E|X| I i < |X|≤ (i +1) n i=1 n=1 (r–1)β ≤ CE|X| . (2.12) If q ≥ α,then ∞ n r–2–q/p q 1/p n E|a X| I |a X| > n nk nk n=1 k=1 ∞ n r–2–q/p q 1/α 1/p ≤ n E|a X| I n X > n nk n=1 k=1 ∞ n r–2–q/p q 1/β = n E|a X| I |X| > n nk n=1 k=1 r–2–q/p+q/α q 1/β ≤ C n E|X| I |X| > n n=1 ∞ i q 1/β 1/β r–2–q/β = C E|X| I i < |X|≤ (i +1) n i=1 n=1 (r–1)β CE|X| if α ≤ q <(r –1)β, (r–1)β ≤ (2.13) CE|X| log(1 + |X|)if q =(r –1)β, CE|X| if q >(r –1)β. Combining (2.12)and (2.13)gives (2.9). Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 8 of 16 (2) In this case, we have that α = rp =(r –1)β.If q ≤ α = rp =(r –1)β,then ∞ n r–2–q/p q 1/p n E|a X| I |a X| > n nk nk n=1 k=1 ∞ n r–2–q/p+(q–α)/p α 1/α 1/p ≤ n E|a X| I n X > n nk n=1 k=1 r–1–α/p α β ≤ C n E|X| I |X| > n n=1 –1 α β = C n E|X| I |X| > n n=1 ∞ i α β β –1 = C E|X| I i < |X|≤ (i +1) n i=1 n=1 (r–1)β ≤ CE|X| log 1+ |X| . (2.14) If q > α = rp =(r –1)β,then ∞ n r–2–q/p q 1/p n E|a X| I |a X| > n nk nk n=1 k=1 r–2–q/p q/α q q ≤ C n n E|X| ≤ CE|X| . (2.15) n=1 Combining (2.14)and (2.15)gives (2.10). (3) In this case, we have that (r –1)β < rp < α.If q ≤ rp,then ∞ n r–2–q/p q 1/p n E|a X| I |a X| > n nk nk n=1 k=1 ∞ n r–2–q/p q 1/p 1/p = n E|a X| I |a X| > n , |X| > n nk nk n=1 k=1 ∞ n r–2–q/p q 1/p 1/p + n E|a X| I |a X| > n , |X|≤ n nk nk n=1 k=1 ∞ n r–2–q/p q 1/p ≤ n E|a X| I |X| > n nk n=1 k=1 ∞ n r–2–q/p –(α–q)/p α 1/p + n n E|a X| I |X|≤ n nk n=1 k=1 r–1–q/p q 1/p ≤ C n E|X| I |X| > n n=1 r–1–α/p α 1/p + C n E|X| I |X|≤ n n=1 Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 9 of 16 ∞ i q 1/p 1/p r–1–q/p = C E|X| I i < |X|≤ (i +1) n i=1 n=1 ∞ ∞ α 1/p 1/p r–1–α/p + C E|X| I (i –1) < |X|≤ i n i=1 n=i rp CE|X| if q < rp, ≤ (2.16) ⎩ rp CE|X| log(1 + |X|)if q = rp. If rp < q < α,then ∞ n r–2–q/p q 1/p n E|a X| I |a X| > n nk nk n=1 k=1 r–1–q/p q q ≤ C n E|X| ≤ CE|X| . (2.17) n=1 If q ≥ α,then ∞ n r–2–q/p q 1/p n E|a X| I |a X| > n nk nk n=1 k=1 r–2–q/p q/α q q ≤ C n n E|X| ≤ CE|X| , (2.18) n=1 since q ≥ α >(r –1)β. Combining (2.16)–(2.18)gives (2.11). Lemma 2.5 Let 1 ≤ p <2, α >0, β >0 with 1/α +1/β =1/p, and let X be a random vari- able. Let {a ,1 ≤ k ≤ n, n ≥ 1} be an array of constants satisfying (1.1). If E|X| < ∞, then nk –1/p 1/p n E|a X|I |a X| > n → 0 (2.19) nk nk k=1 as n →∞, and hence, in addition, if EX =0, then –1/p 1/p n max a EXI |a X|≤ n → 0 (2.20) nk nk 1≤m≤n k=1 as n →∞. α –1 α 1/α Proof Denote A = sup n |a | .Then |a |≤ An for all 1 ≤ k ≤ n and n ≥ 1. nk nk n≥1 k=1 It follows that –1/p 1/p n E|a X|I |a X| > n nk nk k=1 –1 p 1/p ≤ n E|a X| I |a X| > n nk nk k=1 Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 10 of 16 –1 p p 1/β ≤ n |a | E|X| I |AX| > n nk k=1 p 1/β ≤ CE|X| I |AX| > n → 0 (2.21) as n →∞.Hence (2.19)holds. If, in addition, EX =0, then we get by (2.21)that –1/p 1/p n max a EXI |a X|≤ n nk nk 1≤m≤n k=1 –1/p 1/p = n max a EXI |a X| > n nk nk 1≤m≤n k=1 –1/p 1/p ≤ n E|a X|I |a X| > n → 0 nk nk k=1 as n →∞.Hence (2.20)holds. The following lemma shows that if 0 < p <1, then (2.20) holds without the condition EX =0. Lemma 2.6 Let 0< p <1, α >0, β >0 with 1/α+1/β =1/p, and let X be a random variable. Let {a ,1 ≤ k ≤ n, n ≥ 1} be an array of constants satisfying (1.1). If E|X| < ∞, then nk –1/p 1/p n E|a X|I |a X|≤ n → 0 nk nk k=1 as n →∞, and hence (2.20) holds. Proof Note that –1/p 1/p n E|a X|I |a X|≤ n nk nk k=1 –1/p 1/p 1/β = n E|a X|I |a X|≤ n , |X| > n nk nk k=1 –1/p 1/p 1/β + n E|a X|I |a X|≤ n , |X|≤ n nk nk k=1 n n –1/p (1–p)/p p 1/β –1/p 1/β ≤ n n E|a X| I |X| > n + n E|a X|I |X|≤ n nk nk k=1 k=1 p 1/β –1/p+1/(α∧1) 1/β ≤ CE|X| I |X| > n + Cn E|X|I |X|≤ n p 1/β –1/p+1/(α∧1)+(1–p)/β p ≤ CE|X| I |X| > n + Cn E|X| → 0 as n →∞,since –1/p+1/(α ∧1)+(1–p)/β =–p/β if α ≤ 1and –1/p+1/(α ∧1)+(1–p)/β = –(1 – p)/α if α >1. Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 11 of 16 3 Main results We ﬁrst present complete convergence for weighted sums of ρ -mixing random variables. Theorem 3.1 Let r ≥ 1, 1 ≤ p <2, α >0, β >0 with 1/α +1/β =1/p. Let {a ,1 ≤ k ≤ nk n, n ≥ 1} be an array of constants satisfying (1.1), and let {X, X , n ≥ 1} be a sequence of identically distributed ρ -mixing random variables. If (r–1)β E|X| < ∞ if α < rp, (r–1)β EX =0, (3.1) E|X| log(1 + |X|)< ∞ if α = rp, rp E|X| < ∞ if α > rp, then (1.4) holds. Conversely, if (1.4) holds for any array {a ,1 ≤ k ≤ n, n ≥ 1} satisfying (1.1) for some nk rp (r–1)β α > p, then EX =0, E|X| < ∞ and E|X| < ∞. Remark 3.1 When 0 < p <1,(3.1) without the condition EX = 0 implies (1.4). The proof is the same as that of Theorem 3.1 except that Lemma 2.5 is replaced by Lemma 2.6. Remark 3.2 The case α > rp (r >1) of Theorem 3.1 corresponds to Theorem 2.2 of Sung [1], and the proof is much simpler than that of Sung [1]. Hence Theorem 3.1 generalizes the result of Sung [1]. Remark 3.3 Suppose that r ≥ 1, 1 ≤ p <2, α >0, β >0 with 1/α +1/β =1/p. Then the case α < rp is equivalent to the case rp <(r –1)β,and in this case, α < rp <(r –1)β.The case α = rp is equivalent to the case rp =(r –1)β,and in this case, α = rp =(r –1)β.The case α > rp is equivalent to the case rp >(r –1)β,and in this case, α > rp >(r –1)β. Remark 3.4 In two cases α < rp and α > rp, the moment conditions are necessary and (r–1)β suﬃcient conditions, but in the case α = rp, the moment condition E|X| log(1 + |X|)= rp E|X| log(1 + |X|)< ∞ is only suﬃcient for (1.4). It may be diﬃcult to prove (1.4)under rp the necessary moment condition E|X| < ∞.Anand Yuan [17]proved(1.4)under the rp moment condition E|X| < ∞ and the condition –δ rp sup n |a | < ∞ nk n≥1 k=1 for some δ ∈ (0, 1). However, their result is not an extension of the classical one and is a particular case of Sung [1]. In fact, if we set α = rp/δ,then α > rp,and (1.1)holds. Proof of Theorem 3.1 Suﬃciency. For any 1 ≤ k ≤ n and n ≥ 1, set 1/p X = a X I |a X |≤ n . nk nk k nk k Note that m m 1/p n 1/p 1/p max a X > εn ⊂∪ |a X | > n ∪ max X > εn . nk k nk k nk k=1 1≤m≤n 1≤m≤n k=1 k=1 Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 12 of 16 Then by Lemmas 2.2 and 2.5,toprove (1.4), it suﬃces to prove that ∞ m r–2 1/p n P max (X – EX ) > εn < ∞, ∀ε > 0. (3.2) nk nk 1≤m≤n n=1 k=1 When r >1, set s ∈ (p, min{2, α})if α ≤ rp and s ∈ (p, min{2, rp})if α > rp.Notethat, when r = 1,wecannotchoosesuch s,since α > p = rp.Then p < s < min{2, α},and E|X| < ∞ by Remark 3.3.Taking q > max{2, α,(r –1)β,2p(r –1)/(s – p)},wehavebythe Markov inequality and Lemma 2.1 that 1/p P max (X – EX ) > εn nk nk 1≤m≤n k=1 q/2 n n –q/p 2 –q/p q ≤ Cn E(X – EX ) + Cn E|X – EX | . (3.3) nk nk nk nk k=1 k=1 Since q >2p(r –1)/(s – p), we have that r –2+ q(1 – s/p)/2 < –1. It follows that q/2 ∞ n r–2 –q/p 2 n · n E(X – EX ) nk nk n=1 k=1 q/2 ∞ n r–2 –q/p 2 1/p ≤ n · n E|a X | I |a X |≤ n nk k nk k n=1 k=1 q/2 ∞ n r–2 –q/p (2–s)/p s 1/p ≤ n · n n E|a X | I |a X |≤ n nk k nk k n=1 k=1 q/2 ∞ n r–2 –q/p (2–s)/p s s ≤ n · n n |a | E|X| nk n=1 k=1 r–2+q(1–s/p)/2 ≤ C n < ∞. (3.4) n=1 By Lemma 2.3 we have ∞ n r–2 –q/p q n · n E|X – EX | nk nk n=1 k=1 ∞ n r–2 –q/p q 1/p ≤ C n · n E|a X | I |a X |≤ n nk k nk k n=1 k=1 < ∞. (3.5) Hence (3.2)holds by (3.3)–(3.5). When r = 1, we always have that α > p = rp.If(1.1) holds for some α >0, then (1.1)also holds for any α (0 < α ≤ α)byRemark 2.1.Thuswemay assume that p < α <2. Taking Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 13 of 16 q = 2,wehavebythe Markov inequality andLemmas 2.1 and 2.3 that ∞ m r–2 1/p n P max (X – EX ) > εn nk nk 1≤m≤n n=1 k=1 ∞ n r–2 –2/p 2 1/p ≤ C n · n E|a X | I |a X |≤ n nk k nk k n=1 k=1 < ∞. Necessity. Set a =1 for all 1 ≤ k ≤ n and n ≥ 1. Then (1.4)can be rewrittenas nk ∞ m r–2 1/p n P max X > εn < ∞, ∀ε >0, 1≤m≤n n=1 k=1 rp which implies that EX =0 and E|X| < ∞ (see Theorem 2 in Peligrad and Gut [5]). Set 1/α a =0 if 1 ≤ k ≤ n –1 and a = n .Then(1.4)can be rewrittenas nk nn r–2 1/α 1/p n P n |X | > εn < ∞, ∀ε >0, n=1 (r–1)β which is equivalent to E|X| < ∞. The proof is completed. Now we extend Theorem 3.1 to complete moment convergence. Theorem 3.2 Let q >0, r ≥ 1, 1 ≤ p <2, α >0, β >0 with 1/α +1/β =1/p. Let {a ,1 ≤ nk k ≤ n, n ≥ 1} be an array of constants satisfying (1.1), and let {X, X , n ≥ 1} be a sequence of identically distributed ρ -mixing random variables. Assume that one of the following conditions holds. (1) If α < rp, then (r–1)β E|X| < ∞ if q <(r –1)β, (r–1)β EX =0, E|X| log(1 + |X|)< ∞ if q =(r –1)β, (3.6) E|X| < ∞ if q >(r –1)β. (2) If α = rp, then ⎨ (r–1)β E|X| log(1 + |X|)< ∞ if q ≤ α = rp, EX =0, (3.7) ⎩ q E|X| < ∞ if q > α = rp. (3) If α > rp, then rp E|X| < ∞ if q < rp, rp EX =0, (3.8) E|X| log(1 + |X|)< ∞ if q = rp, E|X| < ∞ if q > rp. Then (1.5) holds. Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 14 of 16 Remark 3.5 As stated in the Introduction,if(1.5) holds for some q >0, then (1.4)also (r–1)β holds. If α < rp, EX =0, and E|X| < ∞,then(3.6) holds for some 0 < q <(r –1)β.If (r–1)β α = rp, EX =0,and E|X| log(1+ |X|)< ∞,then(3.7) holds for some 0 < q ≤ α.If α > rp, rp EX =0, and E|X| < ∞,then(3.8) holds for some 0 < q < rp. Therefore the suﬃciency of Theorem 3.1 holds by Theorem 3.2. Remark 3.6 The case α > rp of Theorem 3.2 corresponds to combining Theorems 3.1 and 3.2 in Wu et al. [2]. The condition on weights {a } in Wu et al. [2]is nk –1 t sup n |a | < ∞ for some t > max{rp, q}, nk n≥1 k=1 which is stronger than (1.1)with α > rp.Hence Theorem 3.2 generalizes and improves the results of Wu et al. [2]. Remark 3.7 In this paper, the ρ -mixing condition is only used in Lemma 2.1.There- fore our main results (Theorems 3.1 and 3.2) also hold for random variables satisfying Lemma 2.1. Proof of Theorem 3.2 We apply Theorems 2.1 and 2.2 in Sung [18]with X = a X , b = nk nk k n r–2 1/p n , a = n . When the second moment of X does not exist, we apply Theorem 2.1 in Sung [18]. We can easily prove that Theorem 2.1 in Sung [18] still holds for 0 < q <1.When the second moment of X exists, we apply Theorem 2.2 in Sung [18]. (1) If α < rp,then α < rp <(r –1)β by Remark 3.3. We ﬁrst consider the case q <(r –1)β. (r–1)β In this case, the moment conditions are EX =0 and E|X| < ∞.When q <(r –1)β <2, we prove (1.5) by using Theorem 2.1 in Sung [18]. To apply Theorem 2.1 in Sung [18], we take s = 2. By Lemma 2.1, m n E max X (x)– EX (x) ≤ C E X (x) , ∀n ≥ 1, ∀x >0, nk nk nk 1≤m≤n k=1 k=1 1/q 1/q 1/q 1/q 1/q where X (x)= a X I(|a X |≤ x )+ x I(a X > x )– x I(a X <–x ). By nk k nk k nk k nk k nk Lemma 2.3, ∞ n r–2–s/p s 1/p (r–1)β n E|a X | I |a X |≤ n ≤ CE|X| < ∞. (3.9) nk k nk k n=1 k=1 By Lemma 2.4, ∞ n r–2–q/p q 1/p (r–1)β n E|a X | I |a X | > n ≤ CE|X| < ∞. (3.10) nk k nk k n=1 k=1 By Lemma 2.5 (note that E|X| < ∞,since p ≤ rp <(r –1)β), –1/p 1/p n E|a X |I |a X | > n → 0. (3.11) nk k nk k k=1 Chen and Sung Journal of Inequalities and Applications (2018) 2018:121 Page 15 of 16 Hence all conditions of Theorem 2.1 in Sung [18] are satisﬁed. Therefore (1.5)holds by Theorem 2.1 in Sung [18]. When q <(r –1)β and (r –1)β ≥ 2, we prove (1.5) by using Theorem 2.2 in Sung [18]. To apply Theorem 2.2 in Sung [18], we take s >0 such that s > max{2, q, α,(r –1)β,(r – 1)p(α ∧ 2)/((α ∧ 2) – p)}. By Lemma 2.1, E max X (x)– EX (x) nk nk 1≤m≤n k=1 s/2 n n s 2 ≤ C E X (x) + E X (x) , ∀n ≥ 1, ∀x >0. nk nk k=1 k=1 Since s > max{α,(r –1)β},(3.9) holds. Also, (3.10)and (3.11)hold. Since E|X| < ∞ and s >(r –1)p(α ∧ 2)/((α ∧ 2) – p), we have that s/2 ∞ n ∞ s/2 r–2 –2/p 2 r–2 –2/p 2/(α∧2) n n E|a X | ≤ C n n n < ∞. nk k n=1 k=1 n=1 Hence all conditions of Theorem 2.2 in Sung [18] are satisﬁed. Therefore (1.5)holds by Theorem 2.2 in Sung [18]. For the cases q =(r –1)β and q >(r –1)β, the proofs are similar to that of the previous case and are omitted. The proofs of (2) and (3) are similar to that of (1) and are omitted. Acknowledgements The research of Pingyan Chen is supported by the National Natural Science Foundation of China [grant number 71471075]. The research of Soo Hak Sung is supported by the research grant of Pai Chai University in 2018. Competing interests The authors declare that they have no competing interests. Authors’ contributions Both authors read and approved the manuscript. Author details 1 2 Department of Mathematics, Jinan University, Guangzhou, China. Department of Applied Mathematics, Pai Chai University, Daejeon, South Korea. Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations. Received: 1 April 2018 Accepted: 8 May 2018 References 1. Sung, S.H.: Complete convergence for weighted sums of ρ -mixing random variables. Discrete Dyn. Nat. Soc. 2010, Article ID 630608 (2010) 2. Wu, Y., Wang, X., Hu, S.: Complete moment convergence for weighted sums of weakly dependent random variables and its application in nonparametric regression model. Stat. Probab. Lett. 127, 56–66 (2017) 3. Bradley, R.C.: On the spectral density and asymptotic normality of weakly dependent random ﬁelds. J. Theor. Probab. 5, 355–373 (1992) 4. Bryc, W., Smolenski, W.: Moment conditions for almost sure convergence of weakly correlated random variables. Proc. Am. Math. Soc. 119, 629–635 (1993) 5. Peligrad, M., Gut, A.: Almost-sure results for a class of dependent random variables. J. Theor. Probab. 12, 87–104 (1999) 6. 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