Research Article | Open Access

Wei Li, Pingyan Chen, Soo Hak Sung, "Complete Moment Convergence for Sung’s Type Weighted Sums of -Valued Random Elements", *Discrete Dynamics in Nature and Society*, vol. 2016, Article ID 1484160, 8 pages, 2016. https://doi.org/10.1155/2016/1484160

# Complete Moment Convergence for Sung’s Type Weighted Sums of -Valued Random Elements

**Academic Editor:**Ivan Area

#### Abstract

Let and Let be a sequence of independent and identically distributed -valued random elements and let be an array of real numbers satisfying for some We give necessary and sufficient conditions for complete moment convergence of the form , where A strong law of large numbers for weighted sums of independent -valued random elements is also obtained.

#### 1. Introduction

Let be a sequence of random variables (or random elements) and let be an array of real numbers. The weighted sums include many useful linear statistical estimators, such as least squares estimators, nonparametric regression function estimators, and jackknife estimators. So it is interesting and meaningful to study the limiting behavior for them. In fact, many authors have studied some limiting properties. We refer to Bai and Cheng [1], Chen et al. [2], Cuzick [3], Sung [4, 5], Wang et al. [6], Wu [7], and Zhang [8].

Recently, Sung [5] obtained a complete convergence result for weighted sums of identically distributed -mixing random variables (we call them Sung’s type weighted sums).

Theorem A. *Let and . Let be a sequence of identically distributed -mixing random variables and let be an array of real numbers with for some . Then, and imply that Conversely, if (2) holds for any array with (1) for some , then *

The weights satisfying (1) are very general. For example, set for all and Then, (1) holds for any and therefore the weighted sums include the partial sums. Set if and for some Then, (1) holds; meanwhile, (1) does not hold for any , and obviously the weights are unbounded in this case. So Sung’s type weights are very rich and interesting, but very few authors continue to study the kind of weighted sums except Zhang [8] who obtained Theorem A for END random variables.

Chow [9] first investigated the complete moment convergence as follows.

Theorem B. *Let be a sequence of independent and identically distributed random variables with and , where . Then, where means for any real number .*

Chen and Wang [10] pointed out that (3) is equivalent to or Li and Sptaru [11] called (4) the refined result of complete convergence. For some applications in the theory of branching processes, Sptaru [12] obtained (4) for the special case and

Obviously, (4) or (5) implies that Formula (6) is called complete convergence which was introduced by Hsu and Robbins [13]. They first obtained (6) for the special case and . Therefore, the complete moment convergence and the refined result of complete convergence are more exact than the complete convergence.

The complete convergence, the complete moment convergence, and the refined result of complete convergence have attracted many authors. We refer to Bai and Su [14], Baum and Katz [15], Chen [16], Chen et al. [17, 18], Chen and Wang [10], Katz [19], Li and Sptaru [11, 20], Qiu et al. [21], Rosalsky et al. [22], Sung [23], Wang and Su [24], and Wu et al. [25].

The purpose of this paper is to extend Theorem A to complete moment convergence for independent and identically distributed random elements taking values in a Banach space . We also consider the case No geometric conditions are imposed on the Banach space. Our results also partially extend the results of Chen [16] and Li and Sptaru [20] from the partial sums to the weighted sums.

#### 2. Preliminaries

Let be a real separable Banach space with norm and let be a probability space. A random element taking values in is defined as a Borel measurable function from into with the Borel sigma-algebra. The expected value of a -valued random element is defined as the Bochner integral and denoted by .

A Banach space is said to be of Rademacher type , , if there exists a constant such that for all and each sequence of independent random elements taking values in with mean zero and finite th moments. It is well known that if is of Rademacher type , , then is of Rademacher type , . Set for some and . It is well known that is a Banach space with norm . Then, is of Rademacher type (see, e.g., Ledoux and Talagrand [26]).

The following assertion gives us a useful contraction principle and can be found in Lemma 6.5 of Ledoux and Talagrand [26].

Lemma 1. *Let be a sequence of symmetric -valued random elements. Let and be real random variables such that , where is symmetric (even), similarly for . If almost surely for every , then for any *

Checking carefully the arguments of ()–() and ()–() in Sung [5], we have the following lemma.

Lemma 2. *Let and . Let be a nonnegative random variable with . Assume that is an array of real numbers with for some and or . Then, there exist two positive constants and not depending on such that where if and if .*

Lemma 3. *Let . Let be a sequence of independent and identically distributed -valued random elements with and let be an array of real numbers with (10) for some . Assume that is symmetric and in probability. Then, *

*Proof. *Note that From (10), for all and . Then, by (10) and Hölder’s inequality, as since and . By Lemma 7.2 in Ledoux and Talagrand [26], as . Hence, combining (13)–(15) gives (12).

The following moment inequality is due to de Acosta [27].

Lemma 4. *For every , there exists a positive constant such that, for any separable Banach space and any finite sequence of independent -valued random elements with for every , the following inequalities hold:*(i)*For , *(ii)*For , *

In the following, will be used to denote various positive constants whose exact value is immaterial.

#### 3. Main Results

We now state the main results and give the proofs.

Theorem 5. *Let , , and . Let be a sequence of independent and identically distributed -valued random elements and let be an array of real numbers with (1) for some . Then, and in probability imply that and henceConversely, if (18) or (19) holds for any array with (1) for some , then *

*Proof. *
*Sufficiency*. By in probability, for any fixed , there exists an integer , such that for all Let be an independent copy of By formula (6.1) of Ledoux and Talagrand [26], for all and , By Proposition 1.1 in Chen and Wang [10], (18) is equivalent to Hence, to prove (22), by (21), it is enough to prove that Therefore, we can assume that is symmetric. Without loss of generality, we can assume that for all . Set and if and and if . Then, . Hence, to prove (22), it is enough to prove thatBy Lemma 1, for any It follows from the assumption that in probability. By the same argument as in Chen [16] or Li and Sptaru [20], it is not hard to prove (24). Here, we omit the details.

Set for all , , and . Note that Therefore, in order to prove (25), it is enough to prove that We first prove that . Taking in Lemma 2, we have Now, we prove that . By Lemma 1, for any It follows that in probability, and hence by Lemma 3. By Lemma 1 again, and so Hence, to prove , it is enough to prove that By the Markov inequality and Lemma 4, we have, for any ,If , then Choose such that . Then, and, by taking in Lemma 2, If , then we choose . In this case, . By Lemma 2 again, .

Formula (22) implies that, for every and ,Hence, (19) holds. *Necessity*. Set for all and . Then, (1) holds for all . In this case, (19) reduces to which implies by Yang and Wang [28]. So we complete the proof.

Now, we consider the very interesting case in Theorem 5.

Theorem 6. *Let and . Let be a sequence of independent and identically distributed -valued random elements and let be an array of real numbers with (1) for some . Then, and in probability imply that and hence Conversely, if (37) or (38) holds for any array with (1) for some , then *

*Proof. *By the same argument as in the proof of Theorem 5, we can assume that is symmetric. By the Hölder inequality, we can also assume that for . By Proposition 1.1 in Chen and Wang [10], (37) is equivalent to Hence, it is enough to prove that If , then we take such that Then, we have by the Markov inequality, the -inequality, the Hölder inequality, and a standard computationIf , then we take such that . Note that, by Lemmas 1 and 3, we have as . Then, by the Markov inequality, Lemma 4, the Hölder inequality, and a standard computation, Therefore, (40) holds. By Lemmas 1 and 3 again, as . Then, by the Markov inequality, Lemma 4, and a standard computation, Therefore, (41) holds. The rest of the proof is similar to that of Theorem 5. So we complete the proof.

*Remark 7. *The condition cannot be weakened to . For example, when , set if and . Then, (19) reduces to which is equivalent to . Note that , so the moment condition is stronger than the moment condition When , set if and , and set Then, for ,that is, (19) does not hold. When and , Sung [29] studied the complete convergence under NA setup by taking instead of , where . But, as far as we know, there are no results when and .

*Remark 8. *In the proof of Theorem 5, our method uses not only the truncation of random elements but also the truncation of weights. But in the proof of Theorem 6 we only truncate the random elements. Since the proof of Lemma 2 depends on the condition , the method of the proof of Theorem 5 cannot be applied to that of Theorem 6. If we only truncate the random elements in the proof of Theorem 5, then it is hard to estimate when and is not large enough. So we need two different methods to prove Theorems 5 and 6.

By Theorem 6, we have a strong law of large numbers.

Theorem 9. *Let . Let be a sequence of independent and identically distributed -valued random elements and let be a sequence of real numbers with for some . Then, and in probability imply that Conversely, if (50) holds for any sequence with (49) for some , then *

*Proof. *
*Sufficiency*. Since in probability, by a standard symmetric argument, we can assume that is symmetric. Set for and Then, (1) holds by (49). Therefore, by Theorem 6, By the Lévy inequality (see Proposition 2.3 in Ledoux and Talagrand [26]), which, together with (51), implies that Then, (50) holds by a standard argument (see, e.g., the proof of Theorem 2 in Chen et al. [18]).*Necessity*. Set for all Then, (50) reduces to Hence,