# A Kind of Complete Moment Convergence for Sums of Independent and Nonidentically Distributed Random Variables

Authors: Bao Wang; Quanxin Zhu
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

Let{X,Xn,n≥1} be a sequence of independent and nonidentically distributed random variables. We obtain a new kind of complete moment convergence for their sums under the Lyapunov condition. Moreover, our result extends and improves the corresponding result of the independent and identically distributed (i.i.d.) cases.


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## Body

## 1. Introduction and Main Result

Let{X,Xn,n≥1} be a sequence of random variables, and Sn=∑k=1n‍Xk. If for every ɛ>0, ∑n=1+∞‍P{|Xn|>ɛ}<∞, then {X,Xn,n≥1} is said to converge to 0 completely.Hsu and Robbins [1] proved that if {X,Xn,n≥1} is a sequence of independent and identically distributed (i.i.d.) random variables with EX1=μ, and EX12<∞, then Sn/n→μ completely.Erdos [2, 3] proved that if {X,Xn,n≥1} is a sequence of i.i.d. random variables, then for every ɛ>0, ∑n=1+∞‍P{|Sn|/n>ɛ}<∞ holds if and only if EX1=μ and EX12<∞.Obviously the sum tends to infinity asɛ↓0, and it is necessary to study the convergence rate in which this occurs; Heyde [4] proved that
(1)lim⁡ɛ↓0ɛ2∑n=1+∞‍P(|Sn|≥ɛn)=EX2,
when EX=0 and EX2<∞. This research direction is known as the precise asymptotics. For analogous results in more general case, we refer the reader to [5–14] and the references therein.Recently, Liu and Lin [15] have introduced a new kind of complete moment convergence and obtained the following result.Theorem A (see [15]).
Suppose that{X,Xn,n≥1} is a sequence of independent and identically distributed (i.i.d.) random variables. Then
(2)lim⁡ɛ↓01-log⁡ɛ∑n=1∞‍1n2E[Sn2I{|Sn|≥ɛn}]=2σ2
holds if and only if EXk=0, E[Xk2]=σ2, and E[Xk2log⁡+|Xk|]<∞, where k∈Z and k≥1.However, the condition of identical distribution is very strong and rather difficult to verify in some real cases. The following theorem gives a sufficient condition of complete moment convergence for independent nonidentically distributed random variables.Theorem 1.
Let{X,Xn,n≥1} be independent random variables such that EXn=0 and E[Xn2]=σn2<∞, n∈N. Assume that there exists a constant c such that |Xk|≤cBn, a.s., where Bn2=∑k=1n‍σk2. Moreover, one also assumes that the following Lyapunov condition [16, page 298] is satisfied:
(3)lim⁡n→+∞Bn-2-δ∑j=1n‍E|Xj|2+δ=0,
where 0<δ≤1. Then, one has
(4)lim⁡ɛ↓01-log⁡ɛ∑n=1+∞‍1nE[Sn2Bn2I{|Sn|/Bn≥ɛn}]=2.Remark 2.
Suppose that{X,Xn,n≥1} is a sequence of independent and identically distributed (i.i.d.) random variables with E[|Xn2+δ|]<∞, where 0<δ≤1 is a constant. It is easy to verify the Lyapunov condition (3) in real applications, and so the Lyapunov condition (3) is much weaker than the identically distributed condition. Moreover, the Lyapunov condition (3) constrains the growth rate of moment.
Many sequences of independent random variables satisfy Lyapunov’s condition; here we give some examples.Example 1.
Let{X,Xn,n≥1} be a sequence of independent random variables satisfying EXk=μk, Var⁡Xk=σk2, k≥1, and Bn2→∞(n→∞). Suppose that Xn, n≥1 are uniformly bounded; that is, there exists a constant M such that |Xn|≤M for all n≥1, and then we have
(5)∑k=1n‍E|Xk-μk|2+1≤2M∑k=1n‍E|Xk-μk|2=2MBn2,lim⁡n→+∞Bn-2-1∑j=1n‍E|Xj|2+1≤lim⁡n→+∞2MBn2Bn2+1=2Mlim⁡n→+∞1Bn=0,
which verifies that {X,Xn,n≥1} satisfies the Lyapunov condition (3).Example 2.
Let{X,Xn,n≥1} be a sequence of independent random variables, which satisfies P(Xn=1)=pn, P(Xn=0)=1-pn, and
(6)pn={12,n=2k,k∈N,13,n=2k+1,k∈N.
By Example 1, we know that {X,Xn,n≥1} satisfies the Lyapunov condition (3).Remark 3.
Suppose that{X,Xn,n≥1} is a sequence of independent and identically distributed (i.i.d.) random variables such that EXn=0, E[Xn2]=σ2 for all n≥1, E[|Xn2+δ|]<∞, where δ>0 is a constant. Then, from Remark 2, we know that it satisfies Lyapunov’s condition. Therefore, by Theorem 1, we have
(7)lim⁡ɛ↓01-log⁡ɛ∑n=1∞‍1n2E[Sn2I{|Sn|≥ɛn}]=2σ2.
Obviously, this case is the result of Liu and Lin [15]. Therefore, our condition of Theorem 1 is different from the conditions of Theorem A, and our result partly extends and improves those given in Liu and Lin [15].


## 2. Proof of Theorem1

In this section, we will prove Theorem1. We first present the following two lemmas, which play a key role in the proof of Theorem 1.Lemma 4 (see [17]).
Suppose that{X,Xn,n≥1} are independent random variables with EXn=0 and E[Xn2]=σn2<∞, where n∈N. Let Bn2=∑j=1n‍σj2, Fn(x)=P((∑j=1n‍Xj)/Bn≤x), and Δn(x)=|Fn(x)-Φ(x)|, where Φ(x) is the standard normal distribution function. If E|Xj|2+δ<∞, j=1,2,…,n, for some 0≤δ≤1, then for every x,
(8)Δn(x)≤ABn-2-δ∑j=1n‍E|Xj|2+δ(1+|x|2+δ)-1
holds.Lemma 5 (see page 73 of [18]).
Under the conditions of Lemma4, if |Xj|≤cBn a.s., j=1,2,…,n, where c>0, then for every x>0,
(9)P(|Sn|Bn≥x)≤2exp⁡(-x2csinh⁡(xc2)).Proof of Theorem1.
Similar to [15], we have
(10)∑n=1+∞‍1nE[Sn2Bn2I{|Sn|Bn≥ɛn}]=ɛ2∑n=1+∞‍P(|Sn|Bn≥ɛn)+∑n=1+∞‍1n∫ɛn+∞‍2tP(|Sn|Bn≥t)dt:=I1+I2.
To prove Theorem1, we only need to study I1 and I2. We will divide the proof into two steps.
Step
1. We first prove the equality as follows:
(11)lim⁡ɛ↓01-log⁡ɛɛ2∑n=1+∞‍P(|Sn|Bn≥ɛn)=0.
In fact, it follows from Proposition 2.1.1 of [19] that
(12)lim⁡ɛ↓0ɛ2∑n=1+∞‍P(|N|≥ɛn)=1.
By (12), we obtain
(13)lim⁡ɛ↓01-log⁡ɛɛ2∑n=1+∞‍P(|N|≥ɛn)=0.
To establish the equality (11), from (13) we only need to prove that
(14)lim⁡ɛ↓01-log⁡ɛɛ2∑n=1+∞‍|P(|Sn|Bn≥ɛn)-P(|N|≥ɛn)|=0.
Obviously, it follows from Lemma4 that
(15)|P(|Sn|Bn≥ɛn)-P(|N|≥ɛn)|=2Δn(ɛn)≤2ABn-2-δ(1+|ɛn|2+δ)-1∑j=1n‍E|Xj|2+δ.
Combining (3) and (15), we get
(16)lim⁡n→+∞|P(|Sn|Bn≥ɛn)-P(|N|≥ɛn)|=0.
Since
(17)lim⁡ɛ↓0∑n=1[1/ɛ2]‍ɛ2≤1,
it follows from Toeplitz’s lemma (page 120 of [20]) that
(18)lim⁡ɛ↓0ɛ2∑n=1[1/ɛ2]‍|P(|Sn|Bn≥ɛn)-P(|N|≥ɛn]|=0.
By (18), we have
(19)lim⁡ɛ↓01-log⁡ɛɛ2∑n=1[1/ɛ2]‍|P(|Sn|Bn≥ɛn)-P(|N|≥ɛn]|=0.
On the other hand, it follows from Lemma5 that
(20)ɛ2∑n≥[1/ɛ2]‍P(|Sn|Bn≥ɛn)≤2ɛ2∫1/ɛ2+∞‍exp⁡(-ɛt2csinh⁡(ɛtc2))dt=2ɛ2∫1+∞‍exp⁡(-u2csinh⁡(uc2))2uɛ2dt=4∫1+∞‍uexp⁡(-u2csinh⁡(uc2))du.
Noting that
(21)∫1+∞‍uexp⁡(-u2csinh⁡(uc2))du<∞,
the inequality (20) yields
(22)ɛ2∑n≥[1/ɛ2]‍P(|Sn|Bn≥ɛn)<∞.
By (22), we obtain
(23)lim⁡ɛ↓01-log⁡ɛɛ2∑[1/ɛ2]+∞‍|P(|Sn|Bn≥ɛn)-P(|N|≥ɛn)|=0.
Combining (19) and (23), we see that the equality (11) is satisfied.
Step
2. Next, we need to prove the following equality:
(24)lim⁡ɛ↓01-log⁡ɛ∑n=1+∞‍1n∫ɛn+∞‍2tP(|Sn|Bn≥t)dt=2.
Obviously, it follows from Proposition 3.1 of [15] that
(25)lim⁡ɛ↓01-log⁡ɛ∑n=1+∞‍1n∫ɛn+∞‍2tP(|N|≥t)dt=2.
To establish (24), from (25) we only need to prove
(26)lim⁡ɛ↓01-log⁡ɛ∑n=1+∞‍1n∫ɛn+∞‍2t|P(|Sn|Bn≥t)-P(|N|≥t)|dt=0.
Lettingx=t/n-ɛ and Ln=(A∑j=1n‍E|Xj|2+δ)/Bn2+δ, we apply Lemma 4 to obtain
(27)∫ɛn+∞‍2t|P(|Sn|Bn≥t)-P(|N|≥t)|dt=∫0+∞‍2n(x+ɛ)|P(|Sn|Bn≥n(x+ɛ))-P(|N|≥n(x+ɛ))P(|Sn|Bn≥n(x+ɛ))|ndx=2∫0+∞‍n(x+ɛ)2Δn(n(x+ɛ))dx≤2∫0+∞‍n(x+ɛ)Ln11+[n(x+ɛ)]2+δdx.
Ifn≤[1/ɛ2], then it follows from (3) that
(28)∫01/n‍n(x+ɛ)Ln11+[n(x+ɛ)]2+δdx≤∫01/n‍n(x+ɛ)Lndx=12nLn(x+ɛ)2|01/n≤32Ln⟶0(n⟶∞),∫1/n1‍n(x+ɛ)Ln11+[n(x+ɛ)]2+δdx≤∫1/n1‍n-δ/2(x+ɛ)-1-δLndx≤∫1/n1‍n-δ/2x-1-δLndx=Lnn-δ/2[-1δx-δ]|1/n1≤1δLn⟶0(n⟶∞),∫1+∞‍n(x+ɛ)1n1+δ/2(x+ɛ)2+δdx=n-δ/2∫1+∞‍(x+ɛ)-1-δdx⟶0(n⟶∞).
Hence, by (27) and (28), we have that for n≤[1/ɛ2], the following holds:
(29)∫ɛn+∞‍2t|P(|Sn|Bn≥t)-P(|N|≥t)|dt⟶0(n⟶∞).
Noting the fact that the weighted average of a sequence that converge to 0 also converges to 0, we have(30)∑n=1[1/ɛ2]‍1n∫ɛn+∞‍2t|P(|Sn|Bn≥t)-P(|N|≥t)|dt⟶0,asɛ⟶0,
and so
(31)lim⁡ɛ↓01-log⁡ɛ∑n=1[1/ɛ2]‍1n∫ɛn+∞‍2t|P(|Sn|Bn≥t)-P(|N|≥t)|dt=0.
Ifn≥[1/ɛ2], it follows from (8) and (3) that
(32)∫ɛn+∞‍2t|P(|Sn|Bn≥t)-P(|N|≥t)|dt≤∫1+∞‍4tABn-2-δ(1+|t|2+δ)-1∑j=1n‍E|Xj|2+δdt≤4Ln∫1+∞‍1t1+δdt⟶0(n⟶∞).
Obviously, by (32), we get
(33)lim⁡ɛ↓01-log⁡ɛ∑n=[1/ɛ2]+∞‍1n∫ɛn+∞‍2t|P(|Sn|Bn≥t)-P(|N|≥t)|dt=0.
Combining (31) and (33), we have
(34)lim⁡ɛ↓01-log⁡ɛ∑n=1+∞‍1n∫ɛn+∞‍2t|P(|Sn|Bn≥t)-P(|N|≥t)|dt=0,
which implies that (24) is satisfied.
Therefore, from (11) and (34), we see that (4) is true. This completes the proof of Theorem 1.



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*Source: 379417-2014-10-14.xml*