J Theor ProbabDOI 10.1007/s10959-014-0542-3

Cramér-Type Moderate Deviation for StudentizedCompound Poisson Sum

Bing-Yi Jing · Qiying Wang · Wang Zhou

Received: 4 September 2013 / Revised: 22 January 2014© Springer Science+Business Media New York 2014

Abstract Let N be a Poisson distributed random variable (r.v.) with parameter λ.Let {X, Xi , i ≥ 1} be a sequence of i.i.d. r.v’s that are independent of N . Set SN =∑N

j=1 X j and V 2N = ∑N

j=1 X2j . Assume that 0 < μ = E X < ∞ and E X4 < ∞.

In this paper, it is proved that P(SN − λμ ≥ xVN )/{1 − �(x)} → 1 uniformly inx ∈ [0, o(λ1/6)), as λ → ∞.

Keywords Cramér large deviation · Self-normalized compound Poisson sum ·Studentized compound Poisson sum

Mathematics Subject Classification (2010) 62E20 · 60G50

Bing-Yi Jing was partially supported by Hong Kong RGC Grants HKUST6019/10P, HKUST6019/12P,HKUST6022/13P, and Cheung Kong Chair Professorship under the Education Ministry of China and LiKa Shing Foundation. Qiying Wang was partially supported by the Australia Research Council. WangZhou was partially supported by a Grant R-155-000-139-112 at the National University of Singapore.

B.-Y. JingDepartment of Mathematics, Hong Kong University of Science and Technology, Clear Water Bay,Hong Konge-mail: majing@ust.hk

Q. WangSchool of Mathematics and Statistics, University of Sydney, Sydney, Australiae-mail: qiying.wang@sydney.edu.au

W. Zhou (B)Department of Statistics and Applied Probability, National University of Singapore,Singapore, Singaporee-mail: stazw@nus.edu.sg

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1 Introduction and Main Results

During the past century, the classical limit theorems for sums of independent andidentically distributed (i.i.d.) random variables (r.v.’s) have been extended in manydirections. One such extension is to the case where the number of summands is randomitself, i.e. to compound sums. The need to develop these constructions originates fromnumerous practical problems where they serve as mathematical models.

For example, in actuarial mathematics, consider the compound Poisson process

Zt =Nt∑

i=1

Xi ,

where Nt is the number of insurance payments during a time period (0, t] and assumedto be a homogeneous Poisson process with some constant intensity α; Xi are claimsthat are i.i.d. r.v.’s. Clearly, Zt is nothing but the total claim size for an insurancecompany during time period (0, t]. Therefore, the “classical risk model,” namely theCramér–Lundberg model, describes the risk process as

R(t) =: u + pt −Nt∑

i=1

Xi = u + pt − Zt , (1.1)

where u is the initial reserve, and p is the gross risk premium rate. There are manyquantities of interest that are related to R(t), or equivalently Zt . The ruin probability issuch an example. As another example, for economic planning, we are often interestedin the expected total reserve, E R(t), or equivalently the expected total claim size,E Zt , at time t before ruin.

There has been a lot of work on the compound Poisson sum. The studies can beloosely categorized into the following two groups:

• The claim amount has a light-tailed distributionHere, the light-tailed distribution means that the moment generating function ofthe claim distribution exists in a neighborhood of zero. In this case, it is naturalto use saddlepoint approximations. Historically, this application actually seemsto be the origin of saddlepoint approximations of distribution functions. Esscher[5] introduced saddlepoint approximations in the context of the compound Poisonsum. Esscher did not give precise mathematical results, and it seems that such resultswere first given in [4], and later by Jensen [6,7]. Saddlepoint approximations of ruinprobabilities in other contexts have also been well studied; see Jensen [8, Chapter7] for references within.

• The claim amount has a heavy-tailed distributionHere, the heavy-tailed distribution means that the moment generating function of theclaim distribution does not exist in a neighborhood of zero. Naturally, the saddle-point approximations for compound sums for the light-tailed distributions cannotbe used here. Typically, the following approach is adopted:

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– At the center of the distribution, one can use the normal approximations or Edge-worth expansions (which hold under low order moment conditions). In this regard,we refer to [1] and [2] for more details.

– At the tail of the distribution, one could use large deviation-type approximations.For example, for some special classes of heavy-tailed distributions, such as the sub-

exponential family, one could obtain the result P(∑Nt

i=1 Xi ≥ y)

∼ λP(X1 ≥ y),

see, for example, [3,10–13]. In applications, however, y usually has to be very largefor this to be a good approximation. Sometimes, the sub-exponential requirementis also too restrictive.

From the above discussion, we see that when the claim amount has a light-taileddistribution, and the problem of approximating the df of Zt has been solved com-pletely. However, it is commonly believed that the distributions of the claim amountsare typically heavy tailed. Unfortunately, accurate approximations to the distributionof a compound Poisson sum Zt under heavy-tailed distributions are less than satis-factory, and much work in this area remains to be done. For instance, one needs tofind accurate approximations of the distribution of Zt for intermediate values of y,where neither normal-type approximations nor large deviation-type approximationswork well.

In this paper, we will show that the situation will be totally different if self-normalization is applied. Indeed, the main result (see Theorem 1.1 below) inthis paper shows that the distribution of the studentized compound Poisson sum

(Zt − E Zt )/

√∑Nti=1 X2

i can be precisely approximated by a normal distribution in awide range under only finite fourth moment. Note that E Zt denotes the expected totalclaim size over the period [0, t]. Our result provides a feasible method in the inferencein relation to the expected value of the total claim size when the claim amount has aheavy-tailed distribution.

To see why self-normalization can be used to reduce moment conditions for thecompound Poisson sum Zt , let us give a brief review on the related result for theso-called self-normalized sum Sn/Vn , where

Sn =n∑

j=1

X j and V 2n =

n∑

j=1

X2j .

In this regard, past decades have witnessed a significant development. It is nowwell-known that the limit theorems for Sn/Vn usually require much weaker momentassumptions than those for their classical analogs. For instance, several papers haveshown: if E X = 0 and E |X |3 < ∞, then

limn→∞

P(Sn ≥ x Vn

)

1 − �(x)= 1, (1.2)

uniformly for x ∈ [0, o(n1/6)), while a finite moment generating condition is neces-sary for a similar result in relation to the standard sum Sn/

√var Sn . For more related

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results on the Cramér-type moderate deviation for the self-normalized sum Sn/Vn , werefer to Jing et al. [9] and Shao [14], Wang [15,16] and Wang and Hall [17].

In view of (1.2), it is natural to expect that a similar result should hold true for

the studentized compound Poisson sum (Zt − E Zt )/

√∑Nti=1 X2

i . This is confirmedin Theorem 1.1. We now state our main result. To make our presentation easier, fromnow on, we will write

N = Nt , TN = (SN − λμ)/VN ,

where μ = E X , N is a Poisson distributed r.v. with parameter λ which is independentof X, X1, X2, · · · , and similarly to the definitions of Sn and Vn , SN = ∑N

j=1 X j and

V 2N = ∑N

j=1 X2j with

∑0j=1 ≡ 0. Our main result is as follows.

Theorem 1.1 Assume that 0 < E X < ∞ and E X4 < ∞. Then, uniformly on0 ≤ x ≤ o(λ1/6), we have

limλ→∞

P(

SN − λ μ ≥ xVN

)

1 − �(x)= 1. (1.3)

Remark 1.1 The technique in establishing (1.3) is similar to that of (1.2), but requir-ing a different truncation on Xi to offset the accumulated effect from the Poissondistributed r.v. N . This explains why a higher moment condition is required to prove(1.3). We conjecture that (1.3) is still true under E |X |3 < ∞, but it may need a quitedifferent technique. We leave this topic for future work.

The proof of this theorem is provided in the next section. In Sect. 3, we will proveseveral propositions which are needed in the proof of Theorem 1.1.

2 Proof of Theorem 1.1

By virtue of the fact that TN →D N (0, 1) as λ → ∞, the result (1.3) is trivial for0 ≤ x ≤ 4. In order to prove (1.3) uniformly on 4 ≤ x ≤ o(λ1/6), write,

Xi = Xi I[|Xi | ≤ (

√λ/x)α

], Sn =

n∑

i=1

Xi , V 2n =

n∑

i=1

X2i , (2.1)

where 0 < α < 1 will be chosen later. The proof is based on the following threepropositions. The proofs of these propositions will be given in the next section.

Proposition 2.1 If E |X |3 < ∞ and 0 < μ = E X < ∞, then for 0 ≤ x ≤ λ1/6,

P(SN ≥ xVN + λ μ) ≤ C e−x2/2, (2.2)

where C is a constant not depending on x and n.

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Proposition 2.2 If E |X |(α+2)/α < ∞ and E X2 = 1, then

P(SN ≥ x√

λ + λ μ)

1 − �(x)= 1 + O(x3/

√λ), (2.3)

uniformly on 4 ≤ x ≤ ελ λ1/6, where ελ → 0 as λ → ∞.

Proposition 2.3 If E |X |max{(α+2)/α,4} < ∞, then

limλ→∞

P(SN ≥ x VN + λ μ)

1 − �(x)= 1, (2.4)

uniformly on 4 ≤ x ≤ ελ λ1/6, where ελ → 0 as λ → ∞.

We are now ready to prove (1.3) uniformly on 4 ≤ x ≤ o(λ1/6). We have

P(

TN ≥ x)

≤ P(

SN ≥ x VN + λ μ)

+P(

SN ≥ x VN + λ μ, Xi = Xi , some i = 1, 2, ..., N)

= p1N + p2N , say. (2.5)

Put S( j)n = Sn − Xi and V ( j)

n =√

V 2n − X2

j . Noting that for any real numbers s and t

and nonnegative number c and x ≥ 1

{s + t ≥ x√

c + t2} ⊂ {s ≥√

x2 − 1√

c}

(see p. 2181 in [9])), we have

{Sn ≥ xVn + λ μ} ⊂ {S( j)n ≥

√x2 − 1 V ( j)

n + λ μ}

for each 1 ≤ j ≤ n. Let α = 7/8 in (2.1). It is readily seen by the independence ofX j and S( j)

n /V ( j)n , the iid properties of Xi and Proposition 2.1 that, for each 1 ≤ j ≤

n, n ≥ 1,

P(Sn ≥ xVn + λ μ, X j = X j

)

= P(|X j | > (√

λ/x)7/8) P(S( j)

n ≥√

x2 − 1 V ( j)n + λμ

)

= O(1)(x/√

λ)7/2 P(Sn−1 ≥

√x2 − 1 Vn−1 + λμ

)

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uniformly in x ≥ 1, provided E X4 < ∞. It follows from Proposition 2.1 that

p2N = P(N = 0) +∞∑

n=1

P(N = n)

n∑

j=1

P(Sn ≥ xVn + λ μ, X j = X j

)

≤ e−λ + O(1)(x/√

λ)7/2∞∑

n=1

n P(N = n) P(Sn−1 ≥

√x2 − 1 Vn−1 + λμ

)

≤ e−λ+O(1)(x/√

λ)7/2 λ

∞∑

n=1

P(N = n − 1) P(Sn−1 ≥

√x2 − 1 Vn−1+λμ

)

= e−λ + O(1)(x/√

λ)7/2 λ P(SN ≥

√x2 − 1 VN + λμ

)

≤ e−λ + O(1) x7/2 λ−3/4 e−x2/2

= o(1)[1 − �(x)

], (2.6)

uniformly on 4 ≤ x ≤ o(λ1/6), as λ → ∞, where we have used the well-known factthat, whenever y > 0,

1√2π

(1

y− 1

y3 )e−y2/2 ≤ 1 − �(y) ≤ 1√2π

1

ye−y2/2. (2.7)

The results (2.5) and (2.6), together with Proposition 2.3, imply that

limλ→∞

P(

TN ≥ x)

1 − �(x)≤ 1, (2.8)

uniformly on 4 ≤ x ≤ o(λ1/6), as λ → ∞.On the other hand, we have

P(

SN ≥ x VN + λ μ)

≤ P(

TN ≥ x)

+ P(

SN ≥ x VN + λ μ, Xi = Xi , some i = 1, 2, ..., N).

(2.9)

Similar arguments as in the proof of (2.6), by taking α = 7/8, yield that the last termof (2.9) is less than

e−λ + O(1)x7/2 λ−3/4 P(SN ≥

√x2 − 1 VN + λμ

) = o(1)[1 − �(x)

],

uniformly on 4 ≤ x ≤ o(λ1/6), as λ → ∞, by Proposition 2.3. Taking this fact into(2.9) and using Proposition 2.3 again, we obtain

limλ→∞

P(

TN ≥ x)

1 − �(x)≥ 1, (2.10)

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uniformly on 4 ≤ x ≤ o(λ1/6), as λ → ∞. Combining (2.8) and (2.10), we completethe proof of Theorem 1.1.

3 Proofs of Propositions

Throughout this section, we let b = x/√

λ and denote by C, C1, . . . , the constantsnot depending on x and λ. The constants C, C1, . . . , may be different at each of theirappearance.

Proof of Proposition 2.1 We start with a lemma. Its proof is similar to (2.7) of [14].We omit the details.

Lemma 3.1 Let X be a random variable with E X = μ, E X2 = 1 and E |X |3 < ∞.Then, for any η > 0, θ > 0 and 0 < x ≤ √

λ,

Eeη b(X−μ)−θ(bX)2 = 1 + [η2(1 − μ2)/2 − θ ] b2 + O(1) b3 E |X |3, (3.1)

where O(1) is bounded by a constant depending only on η and θ . Consequently,

log Eeη bX−θ(bX)2 = η bμ + [η2(1 − μ2)/2 − θ ] b2 + O(1) b3 E |X |3, (3.2)

whenever 0 < x ≤ o(√

λ).

We now turn back to the proof of Proposition 2.1. Since (2.2) is trivial for 0 ≤ x < 1,we assume x ≥ 1. Furthermore, we assume E X2 = 1. Otherwise, one only needs toreplace X j by X j/

√E X2. Observe that

P(SN ≥ xVN + λ μ) ≤ P(2bSN ≥ b2V 2

N + x2 − 1 + 2bλ μ)

+ P(SN ≥ xVN + λ μ, 2bxVN ≤ b2V 2

N + x2 − 1)

= P(2bSN ≥ b2V 2

N + x2 − 1 + 2bλ μ)

+P(SN ≥ xVN + λ μ, (bVN − x)2 ≥ 1)

≤ P(2bSN − b2V 2

N ≥ x2 − 1 + 2bλ μ)

+P(SN ≥ xVN + λ μ, b2V 2

N ≥ x2 + x)

+P(Sn ≥ xVN + λ μ, b2V 2

N ≤ x2 − x)

:= 1,N (x) + 2,N (x) + 3,N (x), say. (3.3)

By (3.2) with η = 1 and θ = 1/2 and noting μ > 0, it follows from Markov’sinequality that

1,N (x) ≤ √e e−x2/2−bλμ E exp

{

bSN − 1

2b2V 2

N

}

= √e e−x2/2−bλμ

∞∑

n=0

(EebX−b2 X2/2)n

P(N = n)

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≤ √e e−x2/2−bλμ exp{λ(ebμ−μ2b2/2+O(1) b3 E |X |3 − 1)}

≤ C e−x2/2, (3.4)

whenever 0 ≤ x ≤ λ1/6, where C is a constant not depending on x and λ and we haveused the fact that

ebμ−μ2b2/2+O(1) b3 E |X |3 − 1 ≤ bμ − μ2b2/2 + (bμ − μ2b2/2)2/2 + C b3

≤ b μ + C1b3.

To estimate 3,N , write B1 = {(s, t) : s ≥ x√

t + λ bμ, 0 ≤ t ≤ x2 − x}. Bynoting that

√1 − x−1/2 ≥ 1 − x−1/2 − x−2 since x ≥ 4, it follows easily from

(3.2) with η = 1 and θ = 2, and similar arguments as in the proof of (3.4) that, for4 ≤ x ≤ λ1/6,

3,N = P{(b SN , b2 V 2

N ) ∈ B1}

≤ E exp(

b SN − 2 b2 V 2N

)exp

{ − inf(s,t)∈B1

(s − 2 t)}

≤ E exp(b SN − 2 b2 V 2

N

)exp

{ − λ bμ − x√

x2 − x + 2(x2 − x

)}

≤ e∞∑

n=0

(EebX−2b2 X2)n

P(N = n) exp{ − λ bμ + x2 − 3x/2

}

≤ C exp{λ(ebμ−(μ2+3)b2/2+O(1)b3 E |X |3 − 1)

}exp

{ − λ bμ + x2 − 3x/2}

≤ C e−x2/2−3x/2. (3.5)

As for 2,N , we have

2,N ≤ P{

SN ≥ x VN + λ μ,(1 + x−1)λ ≤ V 2

N ≤ 9λ}

+ P( SN ≥ x VN + λ μ, V 2N ≥ 9λ)

= (1)2,N +

(2)2,N , say. (3.6)

Let B2 = {(s, t) : s ≥ x√

t + λ b μ, x2 + x ≤ t ≤ 9x2}. Similar to the proof of (3.5)but choosing η = 1 and θ = 1/6 in (3.2), we get, for 4 ≤ x ≤ λ1/6,

(1)2,N = P

{(b SN , b2 V 2

N ) ∈ B2}

≤ E exp(b SN − b2 V 2

N /6)

exp{ − inf

(s,t)∈B2(s − t/6)

}

≤ E exp(b SN − b2 V 2

N /6)

exp{ − λ bμ − x

√x2 + x + (

x2 + x)/6

}

≤ C∞∑

n=0

(EebX−b2 X2/6)n

P(N = n) exp{ − λ bμ − 5x2/6 − x/3

}

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≤ C exp{λ(ebμ+(2−3u2)b2/6+O(1)b3 E |X |3 − 1)

}exp

{ − λ bμ − 5x2/6 − x/3}

≤ C e−x2/2−x/3, (3.7)

where we have used the fact that√

1 + x−1 ≥ 1 + x−1/2 − x−2. On the other hand,by letting SN = ∑N

j=1 X j I (|X j | ≤ A0b−1), where A0 > 0 is a constant chosen later,we have,

(2)2,N ≤ P(SN ≥ x VN /2+λμ, V 2

N ≥ 9λ)+P

⎛

⎝N∑

j=1

X j I (|X j |≥ A0b−1)≥ x VN /2

⎞

⎠

≤ P(SN ≥ 3 x√

λ/2 + λμ) + P

⎛

⎝N∑

j=1

X j I (|X j | ≥ A0b−1) ≥ x VN /2

⎞

⎠ .

(3.8)

Note that es ≤ 1 + s + s2/2 + |s|3es/6. For each n ≥ 1,

P(Sn ≥ 3 x√

λ/2 + λμ) ≤ e−λ b μ−3x2/2+bE Sn Eeb(Sn−E Sn)

≤ e−λ b μ−3x2/2+nbE X I (|X |≥A0b−1)

n∏

j=1

{1 + b2var

[X2

j I (|X j | ≤ A0b−1)]/2 + 8b3 E |X j |3eA0/6

}

≤ e−λ b μ−3x2/2enbμ+nb2(1−μ2)/2+Cnb3.

This yields that, for 4 ≤ x ≤ λ1/6,

P(SN ≥ 3 x√

λ/2 + λμ) ≤∞∑

n=0

P(N = n) P(Sn ≥ 3 x√

λ/2 + λμ)

≤ e−λ b μ−3x2/2 exp{λ (ebμ+b2(1−μ2)/2+Cb3 − 1)}≤ e−x2+C λb3 ≤ Ce−x2

.

Similarly, it follows from es ≤ 1 + |s|es that, for each n ≥ 1,

P

⎛

⎝n∑

j=1

I (|X j | ≥ A0b−1) ≥ x2 /4

⎞

⎠ ≤ e−t x2/4n∏

j=1

Eet I (|X j |≥A0b−1)

≤ e−t x2/4n∏

j=0

(1 + et P(|X j | ≥ A0b−1))

≤ exp{−t x2/4 + nb2et/A20}

≤ e−2x2+nb2/2,

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if we choose t = 8 and A0 = 2e4. This yields that, for 4 ≤ x ≤ λ1/6,

P

⎛

⎝N∑

j=1

X j I (|X j | ≥ A0b−1) ≥ x VN /2

⎞

⎠

≤∞∑

n=0

P(N = n) P( n∑

j=1

I (|X j | ≥ A0b−1) ≥ x2 /4)

≤ e−2x2exp{λ(eb2/2 − 1)} ≤ C e−x2

.

Taking these estimates into (3.8), we derive

(2)2,N ≤ C e−x2

, (3.9)

for 4 ≤ x ≤ λ1/6. Combining (3.3)–(3.9), we obtain (2.2). The proof of Proposition2.1 is now complete.

Proof of Proposition 2.2 We start with some preliminaries. Write Y = X I (|X | ≤b−α). Recall b = x/

√λ and 4 ≤ x ≤ λ1/6. Since E X2 = 1 and E |X |(α+2)/α < ∞,

it is readily seen that

EY = μ + o(b2) and EY 2 = 1 + o(b2−α). (3.10)

Consequently, it follows from b|Y | ≤ b1−α ≤ 1, since 0 < α < 1, and Taylor’sexpansion that

Eeb Y = 1 + b μ + b2/2 + O(b3), (3.11)

EY eb Y = μ + b + O(b2), (3.12)

EY 2eb Y = 1 + O(b), (3.13)

and E |Y |3eb Y ≤ C . Next let

g(t) = EetY/√

λ, S = (SN − λμ)/√

λ and F(u) = Eex S I (S ≤ u)/Eex S .

We have that, for t ∈ R,

Ee(i t+x)S = e−(i t+x)√

λμ Ee(i t+x)SN /√

λ

= e−(i t+x)√

λμ∞∑

n=0

Ee(i t+x)Sn/√

λ P(N = n)

= exp{λ[g(i t + x) − 1 − (i t + x)μ/

√λ]}

. (3.14)

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This implies that, whenever |t | ≤ ε√

λ, where ε is small enough,

log Ee(i t+x)S − log Eex S = λ[g(i t + x) − g(x) − i t μ/

√λ]

= −i t x − t2/2 + O(1)(|t |3 + |t |x2)/

√λ,

since Taylor’s expansion, together with (3.12)–(3.13), yields that

g(i t + x) = g(x) + i t√λ

EY ebY − t2

2λEY 2ebY + O(1)

|t |3λ3/2 E |Y |3ebY

= g(x) + i t√λ

(μ + b) − t2

2λ+ O(1)

( |t |3λ3/2 + |t |√

λb2 + |t |2

λb).

Now, by letting Z be a r.v. with the df F(u) and noting that

log Eeit Z = log Ee(i t+x)S − log Eex S,

simple calculations show that

|Eeit Z − eitx−t2/2| ≤ e−t2/2∣∣ exp{O(|t |)(t2 + x2)/

√λ} − 1

∣∣

≤ C |t | (x2 + t2)e−t2/4+|t |/8/√

λ,

whenever |t | ≤ ε√

λ and 4 ≤ x ≤ ελ λ1/6, where ε is small enough. This, togetherwith the Esseen’s smoothing inequality (see, Petrov (1975), for example), implies that

supu∈R

|P(Z ≤ u) − �(u − x)| ≤ 1

π

∫

|t |≤ε√

λ

|Eeit Z − eitx−t2/2||t | dt + C λ−1/2

≤ C x2/√

λ, (3.15)

whenever 4 ≤ x ≤ ελ λ1/6.We are now ready to prove Proposition 2.2.It follows from the conjugate method that

P(SN ≥ x√

λ + λμ) = P(S ≥ x)

= Eex S

∞∫

x

e−xud F(u)

= Eex S

⎡

⎣

∞∫

x

e−xud�(u − x) +∞∫

x

e−xud{

F(u) − �(u − x)}⎤

⎦

:= Eex S (Lx + Rx). (3.16)

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By virtue of (3.14) with t = 0 and (3.11), we have

Eex S = eλ(g(x)−1−bμ) = ex2/2+O(x3/√

λ).

By (3.15) and integration by parts,

|Rx | = e−x2 ∣∣

∞∫

0

e−xud{

F(u + x) − �(u)}∣∣

≤ 2e−x2sup

u|P(Z ≤ u + x) − �(u)|

≤ C |x |3e−x2/2{1 − �(x)}/√

λ,

where we have used (2.7). On the other hand, simple calculations show that

Lx = e−x2

∞∫

0

e−ux d�(u) = e−x2/2 {1 − �(x)

}.

Taking these estimates into (3.16), we obtain

P(SN ≥ x√

λ + λμ)

1 − �(x)= eO(x3/

√λ)(1 + O(1)x3/

√λ) = 1 + O(x3/

√λ),

whenever 4 ≤ x ≤ ελ λ1/6, where ελ → 0, as λ → ∞. The proof of Proposition 2.2is now complete.

Proof of Proposition 2.3 Without loss of generality, assume E X2 = 1. Otherwise, weonly need to replace X j by X j/

√E X2. First note that, by Proposition 2.2, we have

limλ→∞

P[SN ≥ x

√λ(1 + δλ) + λμ

]

1 − �(x)= 1, (3.17)

uniformly in the range 2 ≤ x ≤ ελλ1/6, where |δλ| ≤ ελ/x2. Also note that, it follows

from

Eet (1−X2j ) ≤ 1 + t E X2 I{|X |≥(

√λ/x)α} + t2 E X4et

and E |X |max{(α+2)/α,4} < ∞ that (by letting t = b = x/√

λ)

Ee−bV 2N =

∞∑

n=0

P(N = n) e−n b (Eeb(1−X2

j ))n

≤ exp{λ(e−b+bE X2 I (|X |≥b−α)+b2 E X4eb − 1)

}

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J Theor Probab

≤ exp{ − λ b + o(1)λ b3−α + C λb2}

≤ exp{ − λ b + C1 λb2},

whenever 0 < b ≤ 1 and 0 < α < 1. Hence, uniformly in the range 2 ≤ x ≤ ελλ1/6,

P(λ − V 2

N ≥ ελλ/x2)

≤ ebλ(1−ελ/x2) Ee−bV 2N

= exp{

− ελb−1 + O(x2)}

≤ o(1)[1 − �(x)

],

as λ → ∞, where we have used the fact that

ελb−1 ≥ λ1/3 ≤ ε−1λ x2.

It is now readily seen that

P(SN ≥ x VN +λμ

) ≤ P(SN ≥ x√

λ(1 − ελ/x2)1/2 + λμ) + P(λ − V 2

N ≥ ελλ/x2)

=[1 + o(1)

][1 − �(x)

],

uniformly in the range 2 ≤ x ≤ ελ λ1/6. Therefore, in order to prove Proposition 2.3,it suffices to show that

limλ→∞

P(SN ≥ x VN + λ μ

)

1 − �(x)≥ 1, (3.18)

In fact, by noting x VN ≤ (x2 + b2V 2N )/2b where b = x/

√λ, we have

P(SN ≥ x VN +λμ

) ≥ P(SN − x V 2

N /(2√

λ)≥ x√

λ/2 + λ μ, V 2N ≤ λ(1 + ελ/x2)

)

≥ P(SN ≥ x

√λ(1 + ελ/x2) + λμ, V 2

N ≤ λ(1 + ελ/x2))

= P(SN ≥ x

√λ(1 + ελ/x2) + λμ

) − Rλ,

where

Rλ = P(SN ≥ x

√λ(1 + ελ/x2) + λμ, V 2

N ≥ λ(1 + ελ/x2)).

This, together with (3.17), implies that (3.18) will follow if we prove

Rλ = o(1)[1 − �(x)

], (3.19)

uniformly in the range 2 ≤ x ≤ ελ λ1/6.In order to prove (3.19), write Y = X I (|X | ≤ b−α) and ξ = Y 2 − EY 2 and

η = Y − EY . Recall b = x/√

λ, |ξ | ≤ b−2α and |η| ≤ 2b−α . For any 0 < θ0 ≤

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J Theor Probab

bα−1, it follows easily from E X = μ, E X2 = 1, E |X |max{(2+α)/α,4} < ∞ and|es − 1 − s − s2/2| ≤ |s|3es∨0 that

f (b) := Eebη+θ0b2ξ

= 1 + 1

2E(bη + θ0b2ξ)2 + O(1)E |bη + θ0b2ξ |3e3

= 1 + 1

2b2 Eη2 + θ0b3 E(ηξ) + 1

2θ2

0 b4 Eξ2 + O(1)e3(b3 E |η|3 + θ30 b6 E |ξ |3)

= 1 + 1

2b2(E X2 − (E X)2) + O(1) (1 + θ0) b3

= 1 + 1

2b2(1 − μ2) + O(1) (1 + θ0) b3. (3.20)

Therefore, for any 0 < θ0 ≤ bα−1, we have

Rλ ≤ P(SN ≥ x√

λ + λμ, V 2N > λ(1 + ελ/x2))

≤ P(bSN + θ0b2V 2N > (1 + θ0) x2 + λ b μ + θ0ελ)

≤ e−(1+θ0) x2−λ b μ−θ0ελ EebSN +θ0b2 V 2N

≤ e−(1+θ0) x2−λ b μ−θ0ελ

∞∑

n=0

P(N = n)(ebEY+θ0b2 EY 2)n[

f (b)]n

≤ e−(1+θ0) x2−λ b μ−θ0ελ exp{λ(ebμ+θ0b2+b2(1−μ2)/2+O(1) (1+θ0) b3 − 1)

}

≤ exp{ − x2/2 − θ0ελ + O(1)(1 + θ0)x3/

√λ}

≤ exp{ − x2/2 − θ0ελ/2}, (3.21)

uniformly in the range 1 ≤ x ≤ ελλ1/6, where we have used the fact that

bEY + θ0b2 EY 2 = bμ + θ0b2 + o(1)b3λ.

Since we may choose ελ → 0 sufficient slow so that ελ ≥ b(1−α)/2, by recalling0 < α < 1 and taking θ0 = bα−1 in (3.21), we obtain

Rλ ≤ exp{−x2/2 − b(α−1)/2} = o(1)[1 − �(x)

],

uniformly in the range 1 ≤ x ≤ ελλ1/6. This proves (3.19) and also completes the

proof of Proposition 2.3. �Acknowledgments The authors would like to thank one referee for his/her helpful suggestions.

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