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On weak laws of large numbers for maximal partial sums of pairwise independent random variables

Lê Vǎn Thành Department of Mathematics, Vinh University, 182 Le Duan, Vinh, Nghe An, Vietnam [email protected]
Abstract.

This paper develops Rio’s method [C. R. Acad. Sci. Paris Sér. I Math., 1995] to prove the weak law of large numbers for maximal partial sums of pairwise independent random variables. The method allows us to avoid using the Kolmogorov maximal inequality. As an application, a weak law of large numbers for maximal partial sums of pairwise independent random variables under a uniform integrability condition is also established. The sharpness of the result is illustrated by an example.

2020 Mathematics Subject Classification:
Primary 60F05

1. Introduction and results

A real-valued function L()L(\cdot) is said to be slowly varying (at infinity) if it is a positive and measurable function on [A,)[A,\infty) for some A0A\geq 0, and for each λ>0\lambda>0,

limxL(λx)L(x)=1.\lim_{x\to\infty}\dfrac{L(\lambda x)}{L(x)}=1.

In [6], de Bruijn proved that if L()L(\cdot) is a slowly varying function, then there exists a slowly varying function L~()\tilde{L}(\cdot), unique up to asymptotic equivalence, satisfying

limxL(x)L~(xL(x))=1 and limxL~(x)L(xL~(x))=1.\lim_{x\to\infty}L(x)\tilde{L}\left(xL(x)\right)=1\ \text{ and }\lim_{x\to\infty}\tilde{L}(x)L\left(x\tilde{L}(x)\right)=1.

The function L~()\tilde{L}(\cdot) is called the de Bruijn conjugate of L()L(\cdot) ([2, p. 29]). Bojanić and Seneta [3] showed that for most of “nice” slowly varying functions, we can choose L~(x)=1/L(x)\tilde{L}(x)=1/L(x). Especially, if L(x)=logγ(x)L(x)=\log^{\gamma}(x) or L(x)=logγ(log(x))L(x)=\log^{\gamma}(\log(x)) for some γ\gamma\in\mathbb{R}, then L~(x)=1/L(x)\tilde{L}(x)=1/L(x). Here and thereafter, for a real number xx, log(x)\log(x) denotes the natural logarithm (base e\mathrm{e}) of max{x,e}\max\{x,\mathrm{e}\}.

Let L()L(\cdot) be a slowly varying function and let r>0r>0. By using a suitable asymptotic equivalence version (see Lemma 2.2 and Lemma 2.3 (i) in Anh et al. [1]), we can assume that L()L(\cdot) is positive and differentiable on [a,)[a,\infty), and xrL(x)x^{r}L(x) is strictly increasing on [a,)[a,\infty) for some large aa. Next, let L1()L_{1}(\cdot) be a slowly varying function with L1(0)=0L_{1}(0)=0 with a linear growth to L(a)L(a) over [0,a)[0,a), and L1(x)L(x)L_{1}(x)\equiv L(x) on [a,)[a,\infty). Then (i) L1(x)L_{1}(x) is continuous on [0,)[0,\infty) and differentiable on [a,)[a,\infty), and (ii) xrL1(x)x^{r}L_{1}(x) is strictly increasing on [0,)[0,\infty). In this paper, we will assume, without loss of generality, that properties (i) and (ii) are fulfilled for all slowly varying functions.

The starting point of the current research is the following weak law of large numbers (WLLN) which was proved by Gut [8]. Hereafter, 𝟏(A)\mathbf{1}(A) denotes the indicator function of a set AA.

Theorem 1 (Gut [8]).

Let 0<p10<p\leq 1 and let {X,Xn,n1}\{X,X_{n},\,n\geq 1\} be a sequence of independent identically distributed (i.i.d.) random variables. Let L()L(\cdot) be a slowly varying function and let bx=x1/pL(x),x0b_{x}=x^{1/p}L(x),\ x\geq 0. Then

(1.1) i=1nXin𝔼(X𝟏(|X|bn))bn0 as n\begin{split}\dfrac{\sum_{i=1}^{n}X_{i}-n\mathbb{E}\left(X\mathbf{1}(|X|\leq b_{n})\right)}{b_{n}}\overset{\mathbb{P}}{\to}0\ \text{ as }n\to\infty\end{split}

if and only if

(1.2) limnn(|X|>bn)=0.\lim_{n\to\infty}n\mathbb{P}(|X|>b_{n})=0.

The above WLLN has been extended in several directions, see [12, 13] for WLLNs with random indices for arrays of independent random variables taking values in Banach spaces, and see [4, 5, 9, 10] and the references therein for WLLNs for dependent random variables and dependent random vectors. Boukhari [4, Theorem 1.2] showed that for 0<p<10<p<1, condition (1.2) implies

(1.3) max1jn|i=1jXi|bn0 as n,\begin{split}\dfrac{\max_{1\leq j\leq n}\left|\sum_{i=1}^{j}X_{i}\right|}{b_{n}}\overset{\mathbb{P}}{\to}0\ \text{ as }n\to\infty,\end{split}

irrespective of the joint distribution of the XnX_{n}’s. Boukhari [4] presented an example showing that his result does not hold when p=1p=1. The proof of the sufficient part of Theorem 1 in [8] works well with pairwise independent random variables since we do not involve the maximal partial sums. Krulov [10] and Chandra [5] established WLLNs for maximal partial sums for the case where the summands are negatively associated and asymptotically almost negatively associated, respectively. The authors in [5, 10] considered general normalizing constants, which showed that the sufficient part of Theorem 1 also holds for 1p<21\leq p<2. However, the method used in [5, 10] requires a Kolmogorov-type maximal inequality (see Lemma 1.2 in [5]) which does not hold for pairwise independent random variables.

The aim of this paper is to establish WLLNs for maximal partial sums of pairwise independent random variables thereby extending the sufficient part of Theorem 1 for the case p=1p=1 to WLLN for maximal partial sums from sequences of pairwise independent random variables. We use a technique developed by Rio [11] to avoid using the Kolmogorov maximal inequality. In addition, we also establish a WLLN for maximal partial sums of pairwise independent random variables under a uniform integrability condition, and present an example to show that this result does not hold in general if the uniform integrability assumption is weakened to the uniform boundedness of the moments.

Let Λ\Lambda be a nonempty index set. A family of random variables {Xλ,λΛ}\{X_{\lambda},\lambda\in\Lambda\} is said to be stochastically dominated by a random variable XX if

(1.4) supλΛ(|Xλ|>t)(|X|>t) for all t0.\sup_{\lambda\in\Lambda}\mathbb{P}(|X_{\lambda}|>t)\leq\mathbb{P}(|X|>t)\mbox{ for all }t\geq 0.

Some authors use an apparently weaker definition of {Xλ,λΛ}\{X_{\lambda},\lambda\in\Lambda\} being stochastically dominated by a random variable YY, namely that

(1.5) supλΛ(|Xλ|>t)C1(C2|Y|>t) for all t0\sup_{\lambda\in\Lambda}\mathbb{P}(|X_{\lambda}|>t)\leq C_{1}\mathbb{P}(C_{2}|Y|>t)\mbox{ for all }t\geq 0

for some constants C1,C2(0,)C_{1},C_{2}\in(0,\infty). It is shown recently by Rosalsky and Thành [14] that (1.4) and (1.5) are indeed equivalent. We note that if (1.4) is satisfied, then for all t>0t>0 and r>0r>0

supλΛ𝔼(|Xλ|r𝟏(|Xλ|t))𝔼(|X|r𝟏(|X|t))+tr{|X|>t},\sup_{\lambda\in\Lambda}\mathbb{E}(|X_{\lambda}|^{r}\mathbf{1}(|X_{\lambda}|\leq t))\leq\mathbb{E}(|X|^{r}\mathbf{1}(|X|\leq t))+t^{r}\mathbb{P}\{|X|>t\},

and

supλΛ𝔼(|Xλ|r𝟏(|Xλ|>t))𝔼(|X|r𝟏(|X|>t)).\sup_{\lambda\in\Lambda}\mathbb{E}(|X_{\lambda}|^{r}\mathbf{1}(|X_{\lambda}|>t))\leq\mathbb{E}(|X|^{r}\mathbf{1}(|X|>t)).

The following theorem is the main result of this paper.

Theorem 2.

Let 1p<21\leq p<2 and let {Xn,n1}\{X_{n},\,n\geq 1\} be a sequence of pairwise independent random variables which is stochastically dominated by a random variable XX. Let L()L(\cdot) be a slowly varying function and let bn=n1/pL(n)b_{n}=n^{1/p}L(n), n1n\geq 1. If

(1.6) limnn(|X|>bn)=0,\lim_{n\to\infty}n\mathbb{P}(|X|>b_{n})=0,

then

(1.7) max1jn|i=1j(Xi𝔼(Xi𝟏(|Xi|bn)))|bn0 as n.\begin{split}\dfrac{\max_{1\leq j\leq n}\left|\sum_{i=1}^{j}\left(X_{i}-\mathbb{E}\left(X_{i}\mathbf{1}(|X_{i}|\leq b_{n})\right)\right)\right|}{b_{n}}\overset{\mathbb{P}}{\to}0\ \text{ as }n\to\infty.\end{split}

We postpone the proof of Theorem 2 to Section 2. From Theorem 3.2 of Boukhari [4], we have that if {Xn,n1}\{X_{n},n\geq 1\} is a sequence of pairwise independent random variables, and {bn,n1}\{b_{n},n\geq 1\} is a sequence of positive constants, then

(1.8) max1in|Xi|bn0 if and only if i=1n(|Xi|>bnε)0 for all ε>0.\frac{\max_{1\leq i\leq n}|X_{i}|}{b_{n}}\overset{\mathbb{P}}{\to}0\text{ if and only if }\sum_{i=1}^{n}\mathbb{P}(|X_{i}|>b_{n}\varepsilon)\to 0\text{ for all }\varepsilon>0.

By using Theorem 2 and (1.8), we obtain the following corollary.

Corollary 3.

Let {X,Xn,n1}\{X,X_{n},\,n\geq 1\} be a sequence of pairwise i.i.d. random variables. Let pp, L()L(\cdot), bnb_{n} be as in Theorem 2. Then (1.6) and (1.7) are equivalent.

Proof.

If (1.6) holds, then (1.7) follows immediately from Theorem 2. Now, assume that (1.7) holds. By the symmetrization procedure, it suffices to check the case where the random variables XnX_{n}, n1n\geq 1 are symmetric. In this case, (1.7) becomes

(1.9) max1jn|Sj|bn0 as n,\dfrac{\max_{1\leq j\leq n}\left|S_{j}\right|}{b_{n}}\overset{\mathbb{P}}{\to}0\ \text{ as }n\to\infty,

where Sj=i=1jXi,j1S_{j}=\sum_{i=1}^{j}X_{i},j\geq 1. Putting S0=0S_{0}=0, and applying (1.9) and inequality

max1jn|Xj|max1jn|Sj|+max1jn|Sj1|,\max_{1\leq j\leq n}|X_{j}|\leq\max_{1\leq j\leq n}|S_{j}|+\max_{1\leq j\leq n}|S_{j-1}|,

we obtain

(1.10) max1jn|Xj|bn0 as n.\dfrac{\max_{1\leq j\leq n}\left|X_{j}\right|}{b_{n}}\overset{\mathbb{P}}{\to}0\ \text{ as }n\to\infty.

By combining (1.8) and (1.10), and using the identical distribution assumption, we obtain (1.6). ∎

Theorem 2 also enables us to establish a WLLN for maximal partial sums of pairwise independent random variables under a uniform integrability condition. After this paper was submitted, Thành [16, Corollary 4.10] established a similar WLLN for triangular arrays of random variables satisfying a Kolmogorov-type maximal inequality. Theorem 4 and Corollary 4.10 of Thành [16] do not imply each other.

Hereafter, we denote the de Bruijn conjugate of a slowly varying function L()L(\cdot) by L~()\tilde{L}(\cdot).

Theorem 4.

Let 1p<21\leq p<2 and let {Xn,n1}\{X_{n},n\geq 1\} be a sequence of pairwise independent random variables. Let L()L(\cdot) be a slowly varying function. If {|Xn|pL(|Xn|p),n1}\{|X_{n}|^{p}L(|X_{n}|^{p}),n\geq 1\} is uniformly integrable, then

(1.11) max1jn|i=1j(Xi𝔼(Xi))|bn0 as n,\begin{split}\dfrac{\max_{1\leq j\leq n}\left|\sum_{i=1}^{j}\left(X_{i}-\mathbb{E}(X_{i})\right)\right|}{b_{n}}\overset{\mathbb{P}}{\to}0\ \text{ as }n\to\infty,\end{split}

where bn=n1/pL~1/p(n)b_{n}=n^{1/p}\tilde{L}^{1/p}(n), n1n\geq 1.

Proof.

Let f(x)=xpL(xp),g(x)=x1/pL~1/p(x),x0f(x)=x^{p}L(x^{p}),\ g(x)=x^{1/p}\tilde{L}^{1/p}(x),\ x\geq 0. Recalling that we have assumed, without loss of generality, that ff and gg are strictly increasing on [0,)[0,\infty). From Lemma 2.5 in Anh et al. [1], we have

limxf(g(x))x=1,\lim_{x\to\infty}\frac{f(g(x))}{x}=1,

and therefore

(1.12) f(g(n))>n/2 for all large n.f(g(n))>n/2\text{ for all large }n.

By the de La Vallée Poussin criterion for uniform integrability, there exists a nondecreasing function hh defined on [0,)[0,\infty) with h(0)=0h(0)=0 such that

(1.13) limxh(x)x=,\lim_{x\to\infty}\dfrac{h(x)}{x}=\infty,

and

(1.14) supi1𝔼(h(f(|Xi|)))=supi1𝔼(h(|Xi|pL(|Xi|p)))<.\sup_{i\geq 1}\mathbb{E}(h(f(|X_{i}|)))=\sup_{i\geq 1}\mathbb{E}(h(|X_{i}|^{p}L(|X_{i}|^{p})))<\infty.

By using Theorem 2.5 (i) of Rosalsky and Thành [14], (1.14) implies that the sequence {Xn,n1}\{X_{n},n\geq 1\} is stochastically dominated by a nonnegative random variable XX with distribution function

F(x)=1supi1(|Xi|>x),x.F(x)=1-\sup_{i\geq 1}\mathbb{P}(|X_{i}|>x),\ x\in\mathbb{R}.

We thus have by (1.12), (1.13), (1.14) and the Markov inequality that

limnn(X>bn)=limnnsupi1(|Xi|>g(n))=limnnsupi1(f(|Xi|)>f(g(n)))limnnsupi1(f(|Xi|)n/2)limnnsupi1(h(f(|Xi|))h(n/2))limnnsupi1𝔼(h(f(|Xi|)))h(n/2))=2supi1𝔼(h(f(|Xi|)))limnn/2h(n/2)=0.\begin{split}\lim_{n\to\infty}n\mathbb{P}\left(X>b_{n}\right)&=\lim_{n\to\infty}n\sup_{i\geq 1}\mathbb{P}(|X_{i}|>g(n))\\ &=\lim_{n\to\infty}n\sup_{i\geq 1}\mathbb{P}(f(|X_{i}|)>f(g(n)))\\ &\leq\lim_{n\to\infty}n\sup_{i\geq 1}\mathbb{P}(f(|X_{i}|)\geq n/2)\\ &\leq\lim_{n\to\infty}n\sup_{i\geq 1}\mathbb{P}(h(f(|X_{i}|))\geq h(n/2))\\ &\leq\lim_{n\to\infty}n\sup_{i\geq 1}\dfrac{\mathbb{E}(h(f(|X_{i}|)))}{h(n/2))}\\ &=2\sup_{i\geq 1}\mathbb{E}(h(f(|X_{i}|)))\lim_{n\to\infty}\dfrac{n/2}{h(n/2)}=0.\end{split}

Applying Theorem 2, we obtain

(1.15) max1jn|i=1j(Xi𝔼(Xi𝟏(|Xi|bn)))|bn0 as n.\begin{split}\dfrac{\max_{1\leq j\leq n}\left|\sum_{i=1}^{j}\left(X_{i}-\mathbb{E}\left(X_{i}\mathbf{1}(|X_{i}|\leq b_{n})\right)\right)\right|}{b_{n}}\overset{\mathbb{P}}{\to}0\ \text{ as }n\to\infty.\end{split}

By the same argument as in the proof of Corollary 4.10 of Thành [16] (see Equation (4.45) in [16]), we have

(1.16) max1jn|i=1j𝔼(Xi𝟏(|Xi|>bn))|bn0 as n.\begin{split}\dfrac{\max_{1\leq j\leq n}\left|\sum_{i=1}^{j}\mathbb{E}\left(X_{i}\mathbf{1}(|X_{i}|>b_{n})\right)\right|}{b_{n}}\to 0\ \text{ as }n\to\infty.\end{split}

Combining (1.15) and (1.16) yields (1.11). ∎

The following example shows that in Theorem 4, the assumption that {|Xn|pL(|Xn|p),n1}\{|X_{n}|^{p}L(|X_{n}|^{p}),n\geq 1\} is uniformly integrable cannot be weakened to

(1.17) supn1𝔼(|Xn|pL(|Xn|p))<.\sup_{n\geq 1}\mathbb{E}(|X_{n}|^{p}L(|X_{n}|^{p}))<\infty.
Example 5.

Let 1p<21\leq p<2, and let {Xn,n1}\{X_{n},n\geq 1\} be a sequence of independent symmetric random variables with

(Xn=0)=11n,(Xn=n1/p)=(Xn=n1/p)=12n,n1.\mathbb{P}(X_{n}=0)=1-\frac{1}{n},\ \mathbb{P}(X_{n}=n^{1/p})=\mathbb{P}(X_{n}=-n^{1/p})=\frac{1}{2n},\ n\geq 1.

Consider the case where the slowly varying function L(x)1L(x)\equiv 1. Then it is clear that

supn1𝔼(|Xn|pL(|Xn|p))=supn1𝔼(|Xn|p)=1<\sup_{n\geq 1}\mathbb{E}(|X_{n}|^{p}L(|X_{n}|^{p}))=\sup_{n\geq 1}\mathbb{E}(|X_{n}|^{p})=1<\infty

and

supn1𝔼(|Xn|pL(|Xn|p)𝟏(|Xn|pL(|Xn|p)>a))=supn1𝔼(|Xn|p𝟏(|Xn|>a1/p))=1\sup_{n\geq 1}\mathbb{E}(|X_{n}|^{p}L(|X_{n}|^{p})\mathbf{1}(|X_{n}|^{p}L(|X_{n}|^{p})>a))=\sup_{n\geq 1}\mathbb{E}(|X_{n}|^{p}\mathbf{1}(|X_{n}|>a^{1/p}))=1

for all a>0a>0. Therefore (1.17) is satisfied but {|Xn|pL(|Xn|p),n1}\{|X_{n}|^{p}L(|X_{n}|^{p}),n\geq 1\} is not uniformly integrable. For a real number xx, let x\lfloor x\rfloor denote the greatest integer that is smaller than or equal to xx. Then for 0<ε<1/40<\varepsilon<1/4 and for n2n\geq 2, we have

(1.18) i=1n(|Xi|>εn1/p)i=n/2n(|Xi|>εn1/p)i=n/2n1n12.\begin{split}\sum_{i=1}^{n}\mathbb{P}(|X_{i}|>\varepsilon n^{1/p})&\geq\sum_{i=\lfloor n/2\rfloor}^{n}\mathbb{P}(|X_{i}|>\varepsilon n^{1/p})\\ &\geq\sum_{i=\lfloor n/2\rfloor}^{n}\frac{1}{n}\geq\frac{1}{2}.\end{split}

Combining (1.8) and (1.18) yields

max1in|Xi|n1/p0.\frac{\max_{1\leq i\leq n}|X_{i}|}{n^{1/p}}\overset{\mathbb{P}}{\nrightarrow}0.

This implies that (1.11) (with bnn1/pb_{n}\equiv n^{1/p}) fails.

2. Proof of Theorem 2

The following lemma plays an important role in the proof of Theorem 2. It gives a general approach to the WLLN. In this lemma, we do not require any dependence structure. Throughout this section, we use the symbol CC to denote a universal positive constant which is not necessarily the same in each appearance.

Lemma 6.

Let {bn,n1}\{b_{n},n\geq 1\} be a nondecreasing sequence of positive numbers satisfying

(2.1) supm1b2mb2m1<.\sup_{m\geq 1}\frac{b_{2^{m}}}{b_{2^{m-1}}}<\infty.

Let {Xn,n1}\{X_{n},n\geq 1\} be a sequence of random variables which is stochastically dominated by a random variable XX and let Xi,n=Xi𝟏(|Xi|bn),n1,i1X_{i,n}=X_{i}\mathbf{1}(|X_{i}|\leq b_{n}),\ n\geq 1,i\geq 1. Assume that

(2.2) limnn(|X|>bn)=0.\lim_{n\to\infty}n\mathbb{P}(|X|>b_{n})=0.

Then

(2.3) max1jn|i=1j(Xi𝔼(Xi,n))|bn0 as n\dfrac{\max_{1\leq j\leq n}\left|\sum_{i=1}^{j}\left(X_{i}-\mathbb{E}\left(X_{i,n}\right)\right)\right|}{b_{n}}\overset{\mathbb{P}}{\to}0\ \text{ as }n\to\infty

if and only if

(2.4) max1j<2n|i=1j(Xi,2n𝔼(Xi,2n))|b2n0 as n.\dfrac{\max_{1\leq j<2^{n}}\left|\sum_{i=1}^{j}\left(X_{i,2^{n}}-\mathbb{E}\left(X_{i,2^{n}}\right)\right)\right|}{b_{2^{n}}}\overset{\mathbb{P}}{\to}0\ \text{ as }n\to\infty.
Proof.

We firstly prove under (2.2) that

(2.5) max1j<2n|i=1j(XiXi,2n)|b2n0 as n.\begin{split}\dfrac{\max_{1\leq j<2^{n}}\left|\sum_{i=1}^{j}\left(X_{i}-X_{i,2^{n}}\right)\right|}{b_{2^{n}}}\overset{\mathbb{P}}{\to}0\ \text{ as }n\to\infty.\end{split}

To see this, let ε>0\varepsilon>0 be arbitrary. Then

(max1j<2n|i=1j(XiXi,2n)|b2n>ε)(i=12n1(XiXi,2n))i=12n1(XiXi,2n)2n(|X|>b2n)0 as n (by (2.2))\begin{split}\mathbb{P}\left(\dfrac{\max_{1\leq j<2^{n}}\left|\sum_{i=1}^{j}\left(X_{i}-X_{i,2^{n}}\right)\right|}{b_{2^{n}}}>\varepsilon\right)&\leq\mathbb{P}\left(\bigcup_{i=1}^{2^{n}-1}\left(X_{i}\not=X_{i,2^{n}}\right)\right)\\ &\leq\sum_{i=1}^{2^{n}-1}\mathbb{P}\left(X_{i}\not=X_{i,2^{n}}\right)\\ &\leq 2^{n}\mathbb{P}\left(|X|>b_{2^{n}}\right)\to 0\text{ as }n\to\infty\text{ (by \eqref{lem.eq03})}\end{split}

thereby proving (2.5) since ε>0\varepsilon>0 is arbitrary.

Next, assume that (2.3) holds. Then

(2.6) max1j<2n|i=1j(Xi𝔼(Xi,2n))|b2n0 as n.\dfrac{\max_{1\leq j<2^{n}}\left|\sum_{i=1}^{j}\left(X_{i}-\mathbb{E}\left(X_{i,2^{n}}\right)\right)\right|}{b_{2^{n}}}\overset{\mathbb{P}}{\to}0\ \text{ as }n\to\infty.

Combining (2.5) and (2.6), we obtain (2.4).

Finally, assume that (2.4) holds. It follows from (2.4) and (2.5) that

(2.7) max1j<2n|i=1j(Xi𝔼(Xi,2n))|b2n0 as n.\begin{split}\dfrac{\max_{1\leq j<2^{n}}\left|\sum_{i=1}^{j}\left(X_{i}-\mathbb{E}\left(X_{i,2^{n}}\right)\right)\right|}{b_{2^{n}}}\overset{\mathbb{P}}{\to}0\ \text{ as }n\to\infty.\end{split}

Now, for m1m\geq 1, set

Km=max2m1n<2mmax1j<2m|i=1j𝔼(Xi,2mXi,n)|b2m1.K_{m}=\max_{2^{m-1}\leq n<2^{m}}\frac{\max_{1\leq j<2^{m}}\left|\sum_{i=1}^{j}\mathbb{E}(X_{i,2^{m}}-X_{i,n})\right|}{b_{2^{m-1}}}.

Then by using (2.1), (2.2) and the stochastic domination assumption, we have

(2.8) Kmi=12m𝔼(|Xi|𝟏(b2m1<|Xi|b2m))b2m1i=12mb2m(|Xi|>b2m1)b2m1C2m(|X|>b2m1)0 as m.\begin{split}K_{m}&\leq\dfrac{\sum_{i=1}^{2^{m}}\mathbb{E}\left(|X_{i}|\mathbf{1}(b_{2^{m-1}}<|X_{i}|\leq b_{2^{m}})\right)}{b_{2^{m-1}}}\\ &\leq\dfrac{\sum_{i=1}^{2^{m}}b_{2^{m}}\mathbb{P}(|X_{i}|>b_{2^{m-1}})}{b_{2^{m-1}}}\\ &\leq C2^{m}\mathbb{P}(|X|>b_{2^{m-1}})\to 0\text{ as }m\to\infty.\end{split}

For n1n\geq 1, let m1m\geq 1 be such that 2m1n<2m2^{m-1}\leq n<2^{m}. Then by (2.1), (2.7) and (2.8)), we have

max1jn|i=1j(Xi𝔼(Xi,n))|bnmax1j<2m|i=1j(Xi𝔼(Xi,2m))|b2m1+max1j<2m|i=1j𝔼(Xi,2mXi,n)|b2m1Cmax1j<2m|i=1j(Xi𝔼(Xi,2m))|b2m+Km0 as m\begin{split}\frac{\max_{1\leq j\leq n}\left|\sum_{i=1}^{j}(X_{i}-\mathbb{E}(X_{i,n}))\right|}{b_{n}}&\leq\dfrac{\max_{1\leq j<2^{m}}\left|\sum_{i=1}^{j}(X_{i}-\mathbb{E}(X_{i,2^{m}}))\right|}{b_{2^{m-1}}}\\ &\qquad+\dfrac{\max_{1\leq j<2^{m}}\left|\sum_{i=1}^{j}\mathbb{E}(X_{i,2^{m}}-X_{i,n})\right|}{b_{2^{m-1}}}\\ &\leq\dfrac{C\max_{1\leq j<2^{m}}\left|\sum_{i=1}^{j}(X_{i}-\mathbb{E}(X_{i,2^{m}}))\right|}{b_{2^{m}}}+K_{m}\\ &\to 0\text{ as }m\to\infty\end{split}

thereby establishing (2.3).

Proof of Theorem 2.

Let

Xi,n=Xi𝟏(|Xi|bn),n1,i1.X_{i,n}=X_{i}\mathbf{1}(|X_{i}|\leq b_{n}),\ n\geq 1,i\geq 1.

It is clear that the sequence {bn,n1}\{b_{n},n\geq 1\} satisfies (2.1). By Lemma 6, it suffices to show that

(2.9) max1j<2n|i=1j(Xi,2n𝔼(Xi,2n))|b2n0 as n.\begin{split}\dfrac{\max_{1\leq j<2^{n}}\left|\sum_{i=1}^{j}\left(X_{i,2^{n}}-\mathbb{E}\left(X_{i,2^{n}}\right)\right)\right|}{b_{2^{n}}}\overset{\mathbb{P}}{\to}0\ \text{ as }n\to\infty.\end{split}

For m0,m\geq 0, set S0,m=0S_{0,m}=0 and

Sj,m=i=1j(Xi,2m𝔼Xi,2m),j1.S_{j,m}=\sum_{i=1}^{j}(X_{i,2^{m}}-\mathbb{E}X_{i,2^{m}}),\ j\geq 1.

We will use techniques developed by Rio [11] (see also [15] for the case of regular varying normalizing sequences) as follows. For n1n\geq 1, 1j<2n1\leq j<2^{n} and for 0mn0\leq m\leq n, let k=j/2mk=\lfloor j/2^{m}\rfloor be the greatest integer which is less than or equal to j/2mj/2^{m}. Then 0k<2nm0\leq k<2^{n-m} and k2mj<(k+1)2mk2^{m}\leq j<(k+1)2^{m}. Let jm=k2mj_{m}=k2^{m}, and

Yi,m=|Xi,2mXi,2m1|𝔼(|Xi,2mXi,2m1|).Y_{i,m}=\left|X_{i,2^{m}}-X_{i,2^{m-1}}\right|-\mathbb{E}\left(\left|X_{i,2^{m}}-X_{i,2^{m-1}}\right|\right).

Then we can show that (see [15, Page 1236])

(2.10) max1j<2n|Sj,n|m=1nmax0k<2nm|i=k2m+1k2m+2m1(Xi,2m1𝔼(Xi,2m1))|+m=1nmax0k<2nm|i=k2m+1(k+1)2mYi,m|+m=1n2m+1b2m(|X|>b2m1).\begin{split}\max_{1\leq j<2^{n}}\left|S_{j,n}\right|&\leq\sum_{m=1}^{n}\max_{0\leq k<2^{n-m}}\left|\sum_{i=k2^{m}+1}^{k2^{m}+2^{m-1}}\left(X_{i,2^{m-1}}-\mathbb{E}(X_{i,2^{m-1}})\right)\right|+\sum_{m=1}^{n}\max_{0\leq k<2^{n-m}}\left|\sum_{i=k2^{m}+1}^{(k+1)2^{m}}Y_{i,m}\right|\\ &\quad+\sum_{m=1}^{n}2^{m+1}b_{2^{m}}\mathbb{P}\left(|X|>b_{2^{m-1}}\right).\end{split}

By (1.6), we have

(2.11) 2m+1(|X|>b2m1)0 as m.2^{m+1}\mathbb{P}\left(|X|>b_{2^{m-1}}\right)\to 0\text{ as }m\to\infty.

It follows from Karamata’s theorem (see, e.g., [2, p. 30]) that

(2.12) k=1nαkL(βk)CαnL(βn) for all α>1,β1.\sum_{k=1}^{n}\alpha^{k}L\left(\beta^{k}\right)\leq C\alpha^{n}L(\beta^{n})\text{ for all }\alpha>1,\beta\geq 1.

By using (2.11), (2.12) and Toeplitz’s lemma, we have

(2.13) limnm=1nb2m2m+1(|X|>b2m1)b2n=limnm=1n2m/pL(2m)2m+1(|X|>b2m1)2n/pL(2n)=0.\begin{split}&\lim_{n\to\infty}\dfrac{\sum_{m=1}^{n}b_{2^{m}}2^{m+1}\mathbb{P}\left(|X|>b_{2^{m-1}}\right)}{b_{2^{n}}}\\ &=\lim_{n\to\infty}\dfrac{\sum_{m=1}^{n}2^{m/p}L(2^{m})2^{m+1}\mathbb{P}\left(|X|>b_{2^{m-1}}\right)}{2^{n/p}L(2^{n})}=0.\end{split}

Let ε1>0\varepsilon_{1}>0 be arbitrary, and let aa and bb be positive constants satisfying

1/2<a<1/p,a+b=1/p.1/2<a<1/p,\ a+b=1/p.

For n1n\geq 1, 0mn0\leq m\leq n, let

(2.14) λm,n=ε12bm2anL(2n).\lambda_{m,n}=\varepsilon_{1}2^{bm}2^{an}L(2^{n}).

An elementary calculation shows (see [15, p. 1236])

(2.15) m=1nλm,n2bε1b2n2b1:=C1(b)ε1b2n.\begin{split}\sum_{m=1}^{n}\lambda_{m,n}\leq\dfrac{2^{b}\varepsilon_{1}b_{2^{n}}}{2^{b}-1}:=C_{1}(b)\varepsilon_{1}b_{2^{n}}.\end{split}

By (2.10), (2.13), the proof of (2.9) is completed if we show that

(2.16) In:=(m=1nmax0k<2nm|i=k2m+1(k+1)2mYi,m|C1(b)ε1b2n)0 as n,I_{n}:=\mathbb{P}\left(\sum_{m=1}^{n}\max_{0\leq k<2^{n-m}}\left|\sum_{i=k2^{m}+1}^{(k+1)2^{m}}Y_{i,m}\right|\geq C_{1}(b)\varepsilon_{1}b_{2^{n}}\right)\to 0\text{ as }n\to\infty,

and

(2.17) Jn:=(m=1nmax0k<2nm|i=k2m+1k2m+2m1(Xi,2m1𝔼(Xi,2m1))|C1(b)ε1b2n)0 as n.J_{n}:=\mathbb{P}\left(\sum_{m=1}^{n}\max_{0\leq k<2^{n-m}}\left|\sum_{i=k2^{m}+1}^{k2^{m}+2^{m-1}}\left(X_{i,2^{m-1}}-\mathbb{E}(X_{i,2^{m-1}})\right)\right|\geq C_{1}(b)\varepsilon_{1}b_{2^{n}}\right)\to 0\text{ as }n\to\infty.

We note that for each m1m\geq 1, Yi,m,i1Y_{i,m},i\geq 1 are mean 0 and pairwise independent random variables. Therefore

(2.18) Inm=1n(max0k<2nm|i=k2m+1(k+1)2mYi,m|λm,n) (by (2.15))m=1nλm,n2𝔼(max0k<2nm|i=k2m+1(k+1)2mYi,m|)2 (by Markov’s inequality)m=1nλm,n2k=02nm1𝔼(i=k2m+1(k+1)2mYi,m)2=m=1nλm,n2k=02nm1i=k2m+1(k+1)2m𝔼(Yi,m)2m=1nλm,n2k=02nm1i=k2m+1(k+1)2m𝔼(Xi,2mXi,2m1)2m=1n2nλm,n2b2m2(|X|>b2m1) (by the stochastic domination assumption)=ε1212n(2a1)L2(2n)(m=1n2m(2a1)L2(2m)2m(|X|>b2m1)) (by (2.14))0 as n (by noting 2a1>0 and using (2.11), (2.12), and Toeplitz’s lemma).\begin{split}I_{n}&\leq\sum_{m=1}^{n}\mathbb{P}\left(\max_{0\leq k<2^{n-m}}\left|\sum_{i=k2^{m}+1}^{(k+1)2^{m}}Y_{i,m}\right|\geq\lambda_{m,n}\right)\text{ (by \eqref{30})}\\ &\leq\sum_{m=1}^{n}\lambda_{m,n}^{-2}\mathbb{E}\left(\max_{0\leq k<2^{n-m}}\left|\sum_{i=k2^{m}+1}^{(k+1)2^{m}}Y_{i,m}\right|\right)^{2}\text{ (by Markov's inequality)}\\ &\leq\sum_{m=1}^{n}\lambda_{m,n}^{-2}\sum_{k=0}^{2^{n-m}-1}\mathbb{E}\left(\sum_{i=k2^{m}+1}^{(k+1)2^{m}}Y_{i,m}\right)^{2}\\ &=\sum_{m=1}^{n}\lambda_{m,n}^{-2}\sum_{k=0}^{2^{n-m}-1}\sum_{i=k2^{m}+1}^{(k+1)2^{m}}\mathbb{E}\left(Y_{i,m}\right)^{2}\\ &\leq\sum_{m=1}^{n}\lambda_{m,n}^{-2}\sum_{k=0}^{2^{n-m}-1}\sum_{i=k2^{m}+1}^{(k+1)2^{m}}\mathbb{E}\left(X_{i,2^{m}}-X_{i,2^{m-1}}\right)^{2}\\ &\leq\sum_{m=1}^{n}2^{n}\lambda_{m,n}^{-2}b_{2^{m}}^{2}\mathbb{P}(|X|>b_{2^{m-1}})\text{ (by the stochastic domination assumption)}\\ &=\varepsilon_{1}^{-2}\frac{1}{2^{n(2a-1)}L^{2}(2^{n})}\left(\sum_{m=1}^{n}2^{m(2a-1)}L^{2}(2^{m})2^{m}\mathbb{P}(|X|>b_{2^{m-1}})\right)\text{ (by \eqref{eq.main24})}\\ &\to 0\text{ as }n\to\infty\text{ (by noting $2a-1>0$ and using \eqref{29}, \eqref{rem08}, and Toeplitz's lemma)}.\end{split}

Similarly,

(2.19) Jnm=1n(max0k<2nm|i=k2m+1k2m+2m1(Xi,2m1𝔼(Xi,2m1))|λm,n)m=1n2nλm,n2(𝔼X2𝟏(|X|b2m1)+b2m12(|X|>b2m1))=m=1n2nλm,n2𝔼X2𝟏(|X|b2m1)+o(1),\begin{split}J_{n}&\leq\sum_{m=1}^{n}\mathbb{P}\left(\max_{0\leq k<2^{n-m}}\left|\sum_{i=k2^{m}+1}^{k2^{m}+2^{m-1}}\left(X_{i,2^{m-1}}-\mathbb{E}(X_{i,2^{m-1}})\right)\right|\geq\lambda_{m,n}\right)\\ &\leq\sum_{m=1}^{n}2^{n}\lambda_{m,n}^{-2}\left(\mathbb{E}X^{2}\mathbf{1}(|X|\leq b_{2^{m-1}})+b_{2^{m-1}}^{2}\mathbb{P}(|X|>b_{2^{m-1}})\right)\\ &=\sum_{m=1}^{n}2^{n}\lambda_{m,n}^{-2}\mathbb{E}X^{2}\mathbf{1}(|X|\leq b_{2^{m-1}})+o(1),\end{split}

where we have applied (2.18) in the final step. By using integration by parts, and proceeding in a similar manner as the last two lines of (2.18), we have

(2.20) m=1n2nλm,n2𝔼X2𝟏(|X|b2m1)m=1n2nλm,n20b2m12x(|X|>x)dxε122n(12a)L2(2n)m=1n(22mbk=1mb2k1b2k2x(|X|>x)dx+22mb0b12xdx)ε122n(12a)L2(2n)(k=1n(m=kn22bm)b2k2(|X|>b2k1)+m=1n22mbb12)C2n(2a1)L2(2n)(k=1n2k(2a1)L2(2k)2k(|X|>b2k1)+1)0 as n.\begin{split}&\sum_{m=1}^{n}2^{n}\lambda_{m,n}^{-2}\mathbb{E}X^{2}\mathbf{1}(|X|\leq b_{2^{m-1}})\leq\sum_{m=1}^{n}2^{n}\lambda_{m,n}^{-2}\int_{0}^{b_{2^{m-1}}}2x\mathbb{P}(|X|>x)\operatorname{d}x\\ &\leq\varepsilon_{1}^{-2}2^{n(1-2a)}L^{-2}(2^{n})\sum_{m=1}^{n}\left(2^{-2mb}\sum_{k=1}^{m}\int_{b_{2^{k-1}}}^{b_{2^{k}}}2x\mathbb{P}(|X|>x)\operatorname{d}x+2^{-2mb}\int_{0}^{b_{1}}2x\operatorname{d}x\right)\\ &\leq\varepsilon_{1}^{-2}2^{n(1-2a)}L^{-2}(2^{n})\left(\sum_{k=1}^{n}\left(\sum_{m=k}^{n}2^{-2bm}\right)b_{2^{k}}^{2}\mathbb{P}\left(|X|>b_{2^{k-1}}\right)+\sum_{m=1}^{n}2^{-2mb}b_{1}^{2}\right)\\ &\leq\dfrac{C}{2^{n(2a-1)}L^{2}(2^{n})}\left(\sum_{k=1}^{n}2^{k(2a-1)}L^{2}(2^{k})2^{k}\mathbb{P}\left(|X|>b_{2^{k-1}}\right)+1\right)\to 0\text{ as }n\to\infty.\end{split}

Combining (2.18)–(2.20), we obtain (2.16) and (2.17) thereby completing the proof of (2.9).

3. Concluding remarks

This paper establishes WLLNs for maximal partial sums of pairwise independent random variables without using the Kolmogorov maximal inequality. The method can be easily adapted to dependent random variables. We have the following result:

Theorem 7.

Let {Xn,n1}\{X_{n},n\geq 1\} be a sequence of random variables and let pp, L()L(\cdot) and bnb_{n} be as in Theorem 2. Assume that there exists a constant CC such that for all nondecreasing functions fif_{i}, i1i\geq 1 we have

(3.1) Var(i=k+1k+fi(Xi))Ci=k+1k+Var(fi(Xi)),k0,1,\operatorname{Var}\left(\sum_{i=k+1}^{k+\ell}f_{i}(X_{i})\right)\leq C\sum_{i=k+1}^{k+\ell}\operatorname{Var}(f_{i}(X_{i})),\ k\geq 0,\ell\geq 1,

provided the variances exist. If {Xn,n1}\{X_{n},\,n\geq 1\} is stochastically dominated by a random variable XX such that (1.6) is satisfied, then we obtain WLLN (1.7).

Theorem 7 can be proved by assuming that Xn0X_{n}\geq 0, n1n\geq 1 since we can use identity Xn=Xn+XnX_{n}=X_{n}^{+}-X_{n}^{-} in the general case. We then use truncation Xi,n=Xi𝟏(Xibn)+bn𝟏(Xi>bn),n1,i1X_{i,n}=X_{i}\mathbf{1}(X_{i}\leq b_{n})+b_{n}\mathbf{1}(X_{i}>b_{n}),\ n\geq 1,i\geq 1 (to ensure that the truncated sequence {Xn,i,i1}\{X_{n,i},i\geq 1\} satisfies (3.1)), and modify the arguments given in the proofs of Lemma 6 and Theorem 2 accordingly. The details are straightforward and, hence, are omitted.

Many dependence structures satisfy (3.1), including mm-pairwise negative dependence, extended negative dependence, φ\varphi-mixing, etc (see, e.g., [7, 11]).

It is obvious that (1.7) implies (1.1). For the i.i.d case and 1p<21\leq p<2, Kruglov [10, Theorem 2] proved that (1.7) and (1.1) are equivalent. It is an open problem as to whether or not this also holds for the pairwise i.i.d. case.

Acknowledgements

The author would like to thank the Editor for offering useful comments and suggestions which enabled him to improve the paper. The author is also grateful to Professor Fakhreddine Boukhari for his interest in this work.

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