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El e c t ro n ic

Jo ur

n a l o

f P

r o

b a b il i t y Vol. 15 (2010), Paper no. 2, pages 44–74.

Journal URL

http://www.math.washington.edu/~ejpecp/

Ratio of The Tail of An Infinitely

Divisible Distribution on The Line

to That of Its Lévy Measure

Toshiro Watanabe

Kouji Yamamuro

Center for Mathematical Science,

Faculty of Engineering,

University of Aizu,

Gifu University,

Aizu-Wakamtsu 965-8580, Japan

Gifu 501-1193, Japan

[email protected]

[email protected]

Abstract

A necessary and sufficient condition for the tail of an infinitely divisible distribution on the real line to be estimated by the tail of its Lévy measure is found. The lower limit and the upper limit of the ratio of the right tailµ(r)of an infinitely divisible distributionµto the right tailν(r)of its Lévy measureνasr→ ∞are estimated from above and below by reviving Teugels’s classical method. The exponential class and the dominated varying class are studied in detail.

Key words: infinite divisibility, Lévy measure,O-subexponentiality, dominated variation, expo-nential class.

AMS 2000 Subject Classification:Primary 60E07; Secondary: 60F99.

Submitted to EJP on September 1, 2009, final version accepted December 23, 2009.


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1

Introduction

The study of tails of distributions is important in theoretical and applied probability. In one direction, it has been developed in relation to infinitely divisible distributions and their Lévy measures since early works such as Feller [16; 17], Cohen [8], and Embrechts et al. [13]. It is known that the two-sided case is harder to analyze than the one-sided case (distributions on

R+ = [0,∞)). Letη(r)be the right tail of a measureη, that is, η(r):= η(r,∞) =η((r,∞)). For positive functions f(r) and g(r), the relation f(r)∼ g(r) means that limr→∞f(r)/g(r) =1. We defineρb(s):=RRes xρ(d x)fors∈R. Denote byρηthe convolution of distributionsρandη.

Definition 1.1. Letρbe a distribution onR. Suppose thatρ(r)>0 for allrR. Letγ0.

(1) We sayρ∈ L(γ)ifρ(r+a)∼eaγρ(r)for alla∈R.

(2) We sayρ∈ S(γ)ifρ∈ L(γ),ρb(γ)<∞, andρρ(r)∼2ρb(γ)ρ(r).

(3) We say ρ ∈ SD(γ) if, for some a > 0, ρ is a distribution on aZ := {0,±a,±2a, . . .} with ρ({na})>0 for all sufficiently largen∈Z, and

lim

n→∞

ρ({(n+1)a}) ρ({na}) =e

−γa

and ifρb(γ)<∞and

lim

n→∞

ρρ({na})

ρ({na}) =2ρb(γ).

LetL andS denoteL(0)andS(0), respectively. The distributions inL andS, respectively, are called long-tailed andsubexponential. Those in S(γ) are called convolution equivalent. The class

L(γ)is often calledexponential class.

We say µID+ if µ is an infinitely divisible distribution on R with Lévy measure ν satisfying

ν(r)>0 for allr∈R. LetµID+. We defineC∗andC∗as

C∗:=lim inf

r→∞ µ(r)

ν(r) and C

:=lim sup

r→∞ µ(r) ν(r).

If 0<C∗≤C<∞, then we can estimate the tailµ(r)by the tailν(r)in a weak sense. That is, for anyε∈(0, 1), there isR>0 such that

(1−ε)Cν(r)≤µ(r)≤(1+ε)Cν(r) (1.1) whenever r >R. To make the estimate (1.1) more meaningful, we should give the expression of

C andC∗. But the expression is known only in some special cases. See Theorem 1.3 of[27] for one-sided strictly semistable distributions. However, there is an important case where the tailµ(r) is estimated by the tailν(r)in a stronger sense. Namely, it is proved in Theorem 1.1 of[31](Lemma 5.3 of Section 5 below) that ifµ∈ S(γ)withγ≥0, then there is an explicitC∈(0,∞)such that

µ(r)∼(r). (1.2)

Prior to[31], the caseγ=0 is already treated in the one-sided case by[13]and in the two-sided case by[24]. There is a lattice-version of the statement above. That is, ifµ ∈ SD(γ) on R+ with


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γ≥0, then (1.2) holds with some C ∈(0,∞). See Theorem 2 of[14]. We do not know whether the converse is true. Namely, there might beµ6∈Sγ≥0(S(γ)∪ SD(γ))such that (1.2) holds with

C ∈(0,∞). Thus we are led to the following two problems.

Problem 1. What is a necessary and sufficient condition in order that 0< C∗≤ C<∞ ? In the case where 0<CC<∞, what are the expressions ofCandC∗or the lower and upper bounds of them ?

Problem 2. What is a necessary and sufficient condition in order that (1.2) holds with 0<C <∞? In the case where (1.2) holds with 0<C <∞, what is the expression ofC ?

We will give answers to Problem 1 and to the second question of Problem 2. A partial answer to the first question of Problem 2 will be given too. Our results are important from the viewpoint of the asymptotic estimates of the transition probabilities of Lévy processes.

In Section 2 we give definitions of classes such as O S, O L, H, and D and formulate our main results in Theorems 2.1, 2.2, and 2.3 and two corollaries. Relations with other works are given in detail. Section 3 discusses the meaning of O S in infinite divisibility. Section 4 studies a bound separatingC∗andC∗forµ∈ H. In Section 5 we prove Theorem 2.1 and its corollaries. In Sections 6 and 7, we prove Theorems 2.2 and 2.3, respectively.

2

Main results

For positive functions f(x)andg(x), the relation f(r)≍g(r)means that lim infr→∞f(r)/g(r)>0 and lim supr→∞f(r)/g(r)<∞. We introduce some basic classes of distributions onR in addition toL(γ),S(γ), andSD(γ)in Definition 1.1.

Definition 2.1. Letρbe a distribution onRsatisfyingρ(r)>0 for allrR.

(1) We sayρ∈ O S ifρρ(r)≍ρ(r). (2) We sayρ∈ H ifρb(s) =∞for alls>0. (3) We sayρ∈ Difρ(2r)≍ρ(r).

(4) We sayρ∈ O L ifρ(r+a)≍ρ(r)for alla∈R.

The distributions inO S are calledO-subexponentialand those inH are sometimes called heavy-tailed. Those inDare calleddominatedly varying.

Remark 2.1. Let γ ≥ 0. The classes in Definitions 1.1 and 2.1 satisfy the following inclusion

relations:

(i)D ∩ L ⊂ S ⊂ O S ∩ L. (ii)D ∪ L ⊂ H.

(iii)D ∪Sγ≥0(S(γ)∪ SD(γ))⊂ O S ⊂ O L. (iv)Sγ≥0L(γ)⊂ O L.


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LetµID+with Lévy measureν. Denote byµtt-th convolution power ofµfor t>0. We define the normalized Lévy measureν(1)on(1,∞)by

ν(1)(d x):=c0−11{x>1}(x)ν(d x), wherec0:=ν(1,∞). Define quantitiesd∗andγ∗as

d∗:=lim sup

r→∞

ν(1)∗ν(1)(r)

ν(1)(r)

, and

γ∗:=sup{γ≥0 :µb(γ)<∞}.

Under the assumption thatν(1)∈ O L, we define the following. LetΛbe the totality of increasing

sequences{λn}∞n=1 with limn→∞λn=∞such that, for every x ∈R, the followingm(x;{λn})exists

and is finite:

m(x;{λn}):=nlim→∞

ν(λnx)

ν(λn)

.

The idea of the use of the functionm(x;{λn})goes back to Teugels[29]. See Remark 5.1 in Section

5 for richness of the setΛ. Given a distributionρ, define

I(ρ):= inf {λn}∈Λ

Z ∞

−∞

m(x;{λn})ρ(d x),

I∗(ρ):= sup {λn}∈Λ

Z ∞

−∞

m(x;{λn})ρ(d x).

Let B := d∗−I(ν(1))−I∗(ν(1)). Note that 0≤ B <∞whenever ν(1)∈ O S, as will be shown in

Lemma 5.2 (iv) of Section 5. Defineµ1 as the compound Poisson distribution with Lévy measure 1{x>1}ν(d x)and letµ2be the infinitely divisible distribution satisfyingµ=µ1∗µ2. Then, under the assumption thatν(1)∈ O S, define

J(µ):=

  

I∗(µ2)exp(c0(I∗(ν(1))−1)) ifB=0,

I∗(µ2)exp(c0(I∗(ν(1))−1))

exp(c0B)−1

c0B

ifB>0.

We answer Problem 1 in the main theorem (Theorem 2.1 below). Shimura and Watanabe [27]

solved the first question of Problem 1 in the one-sided case. We will give a method to reduce the two-sided case to the one-sided case. Concerning the second question, the existing knowledge is that C∗ ≥1 in the one-sided case and that if moreoverµ∈ H, then C∗ =1. See Proposition 2 of

[11]and the proof of Theorem 7 of[9]. We can give general lower and upper bounds ofCandC

by evolving the theory ofO-subexponentiality with the help of Teugels’s idea. The most involved part of our discussion is in finding the upper bound ofC∗.

Theorem 2.1. Letµbe a distribution inID+ with Lévy measureν.


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(ii)Letν(1)∈ O S. Then0≤γ<and0<I∗(µ)≤µb(γ∗)≤J(µ)<. Moreover

I∗(µ)≤C∗≤µb(γ∗), (2.1)

and

b

µ(γ∗)≤C∗≤J(µ). (2.2)

Our results concerning Problem 2 are as follows. The first question in Problem 2 is hard to solve and our answer (Corollary 2.2 below) is only partial, which is an improvement of Corollary 1.3 of

[27]. It is a long-standing problem for thirty years since[13; 14]and is still open. Moreover, we do not know yet whetherµ∈ SD(γ) impliesν(1)∈ SD(γ) in the two-sided case. The second question

is solved by Theorem 1.1 of [31]under the assumption thatν(1) ∈ L(γ) withγ≥ 0. We show in

Corollary 2.1 below that such an additional assumption is not needed.

Corollary 2.1. Letµbe a distribution inID+ with Lévy measureν. If(1.2)holds with0< C <,

then0≤γ<b(γ∗)<, and C =µb(γ∗).

Corollary 2.2. Letµbe a distribution inID+with Lévy measureν. Suppose that there are real numbers

a1,a2 with a2 6= 0 and a1/a2 being irrational such that, for a = 0,a1,a2, there is C(a) ∈(0,∞)

satisfying

µ(r+a)∼C(a)ν(r).

Then0≤γ<andµ∈ S(γ∗).

Remark 2.2. Suppose that all assumptions of Corollary 2.2 are satisfied except the irrationality of

a1/a2. Then 0≤γ<∞andµb(γ∗)<∞. In caseγ∗=0 we haveµ∈ S, but in caseγ>0 we may haveµ∈ SD(γ∗)andµ /∈ S(γ∗).

We present explicit lower and upper bounds ofC and C∗ forµin (L(γ)∪ D)∩ID+ in Theorems 2.2 and 2.3 below. The classL(γ)is extensively studied by [7; 10; 12]in the one-sided case and by[1; 5; 24]also in the two-sided case. Concerning Theorem 2.2 (i), a recent paper[1]of Albin contains an assertion that ifν(1)∈ L(γ), thenµ∈ L(γ). However, his proof forγ >0 depends on

an incorrect lemma, as will be explained in Remark 6.1 of Section 6. Braverman [5]also proved that if ν(1)∈ L(γ) andνb(1)(γ) =∞, thenµ∈ L(γ) forγ >0. Applications of the class L(γ)to

Lévy processes are found in[2; 5].

Theorem 2.2. Letµbe a distribution inID+ with Lévy measureν. Letγ≥0.

(i)Ifν(1)∈ L(γ), thenµt∗∈ L(γ)for every t>0. In the converse direction, ifµt∗∈ L(γ)for every

t>0andν(1)∈ O S, thenν(1)∈ L(γ).

(ii)Suppose thatν(1)∈ L(γ). Then0<C∗=µb(γ)≤ ∞and the following are true: (1)If d∗=2νb(1)(γ)<, thenν(1)∈ S(γ)and C∗=C∗=µb(γ)<.


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(2)If 2νb(1)(γ)<d<, thenν(1)∈(L(γ)∩ O S)\S(γ)and

b

µ(γ)<C∗≤µb(γ)exp(c0(d2

b

ν(1)(γ)))−1

c0(d2νb

(1)(γ))

<∞. (2.3)

(3)If d∗=∞, thenν(1)∈ L(γ)\O S and C∗=∞.

Remark 2.3. Letγ≥0. There existsµin ID+ such that ν(1)∈ L/ (γ)butµt∗∈ L(γ)for all t >0.

We shall show this in a forthcoming paper. Embrechts and Goldie[10; 12] conjectured that the class L(γ) is closed under convolution roots. However, Shimura and Watanabe [28] disproved their conjecture for all γ≥0. We do not know yet whether the classL(γ)∩ID+ is closed under convolution roots.

Remark 2.4. Letγ≥0. We see from Lemma 5.4 of[24]thatν(1)∈ L(γ)impliesd∗≥2νb(1)(γ).

The class(O S ∩ L(γ))\S(γ)is not empty. We know that there areρ andρ′ both inS(γ)such thatρρ′is not inS(γ). As the classesO S andL(γ) are closed under convolution, thisρρ′ is inO S ∩ L(γ). See Lemma 3.1 (iii) of Section 3 and Lemma 2.5 of[31]. Hence thisρρ′is in (O S ∩ L(γ))\S(γ). For example, in the case ofγ=0, we can take the distributions in Section 6 of[23]asρandρ′. In the case ofγ >0, we can take the distributions in the proof of Theorem 2 of

[22]asρandρ′. The classL(γ)\O S is not empty. For example, in the case ofγ=0, a distribution onR+in Section 3 of[10]belongs toL \O S. In the case ofγ >0, any distributionρ∈ L(γ)with

b

ρ(γ) =∞can be taken asρ∈ L(γ)\O S, becauseρ∈ O S ∩ L(γ) impliesρb(γ)<∞by Lemma 6.4 of[31]. Thus none of the cases (1)–(3) in Theorem 2.2 (ii) is empty.

Feller [16] started the study of dominated variation of infinitely divisible distributions. But his assertion thatν(1)∈ D impliesµ(r)∼ν(r)is not true. In the following Theorem 2.3 we clarify the

role of dominated variation in our problems. In the one-sided case, Watanabe[30]proved assertion (i) by preparing a Tauberian theorem similar to Theorem 1 of [20] and Shimura and Watanabe

[27]gave an alternative proof by employingO-subexponentiality. However, they did not discuss the values ofC∗ andC∗ forµin D ∩ID+. A result weaker than assertion (ii) is given in Theorem 1 of

Yakimiv[33]. An application of the classDto selfsimilar processes with independent increments is found in[30].

Define constantsQ∗andQ∗as

Q:= lim

N→∞lim infr→∞

ν(r+N) ν(r) ,

Q∗:= lim

N→∞lim supr→∞

ν(rN)

ν(r) = (Q∗) −1.

Theorem 2.3. Letµbe a distribution inID+ with Lévy measureν.

(i)µ∈ Dif and only ifν(1)∈ D.

(ii)Ifν(1)∈ D, then0<Q∗≤1≤Q<,

1−(1−Q)µ(−∞, 0)≤C≤1, (2.4)

and


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Remark 2.5. The classD ∩S contains all distributions with regularly varying tails. The lognormal distributions are infinitely divisible and belong to the classS \ D. On the other hand, the Peter and Paul distribution belongs to the classD \ S. One-sided strictly semistable distributions with discrete Lévy measures are infinitely divisible distributions in the classD \ S withC∗=1 andC∗=Q>1. See Theorem 1.3 of[27]. We show in Example 7.1 of Section 7 that there existsµ∈ D ∩ID+such thatQ<C<1<C<Q∗.

3

Class

O S

and Infinite Divisibility

The classO S was introduced by Shimura and Watanabe[27]. They studied the asymptotic relation between an infinitely divisible distribution onR+and its Lévy measure by usingO-subexponentiality.

In this section, we extend their results to the two-sided case. Letδa(d x) be the delta measure at

a∈R. For a distributionρonR, we define

ρ+(d x):=ρ(−∞, 0]δ0(d x) +1(0,∞)(x)ρ(d x).

Denote byρnn-th convolution power of ρ with the understanding thatρ0∗(d x) :=δ0(d x). Let µID+with Lévy measureν. In what follows, defineµ3as the compound Poisson distribution with Lévy measure 1{x>c}ν(d x) withc > 1 and let µ4 be the infinitely divisible distribution satisfying µ=µ3∗µ4. We define the distributionν(c)by

ν(c)(d x):= (¯ν(c))−11{x>c}(x)ν(d x). We choose sufficiently largec>1 such thatµ4(0,∞)>0.

Lemma 3.1. Letρandηbe distributions onR.

(i)Ifρ(r)≍η(r)for someη∈ O S , thenρ∈ O S.

(ii)ρ∈ O S if and only ifρ+∈ O S.

(iii)Ifρ,η∈ O S, thenρη∈ O S. In particular, ifρ∈ O S, thenρn∗∈ O S for all n≥1.

Proof. First we prove (i). Suppose thatρ(r)≍η(r)for someη∈ O S. Then we have

ρρ(r) =

Z R

ρ(ry)ρ(d y)

Z R

η(ry)ρ(d y) =

Z R

ρ(ry)η(d y)

Z R

η(ry)η(d y) =ηη(r)

η(r)≍ρ(r). (3.1)

Thus we haveρ∈ O S . We haveρ(r) =ρ+(r)forr>0. It follows from (i) that (ii) holds. Suppose

thatρ,η∈ O S . We see as in (3.1) that


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Thus we obtain from (i), (ii), and Proposition 2.5 of [27] that ρ+∗η+ ∈ O S and thereby

ρη∈ O S. The second assertion of (iii) is clear. ƒ

Lemma 3.2. Letρbe a distribution onR.

(i)Ifρ∈ O S, thenρ∈ O L.

(ii)Ifρ∈ O L, then there isγ1∈(0,∞)such thatρb(γ1) =∞. (iii)Ifρ∈ O S, there is K>0such that, for all n≥1and r∈R,

ρn(r)Knρ(r).

Proof. Suppose thatρ∈ O S. Thenρ+∈ O S. Letr>0. By Proposition 2.1 (ii) of[27], we have

ρ+∈ O L andρ(r+a) =ρ+(r+a)≍ρ+(r) =ρ(r)for anya≥0 and therebyρ∈ O L. Thus we

have proved (i). We see from Proposition 2.2 of[27]that ifρ∈ O L, then there isγ1∈(0,∞)such that ρb+(γ1) =∞, that is, ρb(γ1) = ∞. Thus assertion (ii) is true. Finally assertion (iii) is due to

Lemma 6.3 (ii) of[31]. ƒ

Lemma 3.3. Letµbe a distribution inID+. Thenµ∈ O S if and only ifµ1 ∈ O S. Furthermore, if µ∈ O S, thenµ(r)≍µ1(r)andµ2(r) =o(µ1(r)) =o(µ(r)).

Proof. Suppose that µ1 ∈ O S. We see from Lemma 3.2 (i) that µ1(logr) is in OR. As regards

the definition of the class OR, see [3]. By virtue of Theorem 2.2.7 of [3], there is c1 > 0 such that ec1rµ

1(r) → ∞as r → ∞. Furthermore, by Theorem 26.8 of [26], there is c2 > 0 such that µ2(r) = o(ec2rlogr). Thus µ2(r) = o(µ1(r)) and there isc3 > 0 such that µ2(r)≤ c3µ1(r). We obtain that

µ(r) =

Z R

µ2(ry)µ1(d y) ≤ c3

Z R

µ1(ry)µ1(d y) =c3µ1∗µ1(r)≍µ1(r). Chooseb∈Rsuch thatc4:=µ2(b,)>0. Then we have

µ(r)≥

Z ∞

b+

µ1(ry)µ2(d y)≥c4µ1(rb). (3.3) Hence by Lemma 3.2 (i) we haveµ(r)≍µ1(r). It follows from Lemma 3.1 (i) thatµ∈ O S. Conversely, suppose that µ∈ O S. It follows thatµ2(r) = o(µ(rb)) in the same way as above. Letε >0. There isa>0 such thatµ2(r)≤εµ(rb)forra. Hence we have

µ(r+a) ≤

Z r+

−∞

µ2(r+ay)µ1(d y) +µ1(r)

ε

Z r+

−∞


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εµµ1(r+ab) +µ1(r) = ε

Z R

µ1(r+aby)µ(d y) +µ1(r) ≤ εc−14

Z R

µ(r+ay)µ(d y) +µ1(r) = εc−14 µµ(r+a) +µ1(r).

Here we used (3.3) in the last inequality. Sinceµ∈ O S, there isc5>0 such thatµµ(r)≤c5µ(r). Hence we have(1−εc4−1c5)µ(r+a)≤µ1(r). Here we can takeεsatisfyingεc4−1c5<1. Therefore we see from Lemma 3.2 (i) that there is c6 > 0 such that µ(r + b) ≤ c6µ1(r). Combining this inequality with (3.3), we haveµ(r)≍µ(r+b)≍µ1(r). Furthermore, we haveµ1∈ O S by Lemma 3.1 (i). Finally, we see thatµ(r)≍µ1(r)andµ2(r) =o(µ1(r)) =o(µ(r)). ƒ

Lemma 3.4.(Theorem 1.1 of[27]) Letµbe a distribution onR+inID+with Lévy measureν.

(i)µ(r)≍ν(r)if and only ifν(1)∈ O S.

(ii)The following statements are equivalent:

(1)µ∈ O S;

(2)(ν(1))n∗∈ O S for some n≥1;

(3)µ(r)≍(ν(1))n∗(r)for some n≥1.

Next we present the main result of this section.

Propsition 3.1. Letµbe a distribution inID+with Lévy measureν.

(i)We have0<C.

(ii)C<if and only ifν(1)∈ O S.

(iii)The following statements are equivalent:

(1)µ∈ O S;

(2)(ν(1))n∗∈ O S for some n≥1;

(3)µ(r)≍(ν(1))n∗(r)for some n≥1.

Remark 3.1. In Proposition 3.1 (iii), we cannot replace statement (2) by the statementν(1)∈ O S.

The classO S is not closed under convolution roots. See Remark 1.3 and Proposition 1.1 of[27]

for further details. We see from Proposition 3.1 (iii) thatO S ∩ID+is closed under convolution roots.

Proof of Proposition 3.1. We prove (i). Notice that, forr>0, µ3(r) =ea

X

n=1

an(n!)−1(ν(c))n∗(r),

witha:=ν(c,∞). Then we have µ(r) ν(r) ≥

Z R

e(c)(ry)

ν(c)(r)


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e4(0)>0. HenceC>0.

Next we show (ii). We see from Lemma 3.4 that ν(1)∈ O S if and only if µ1(r) ≍ν(r). We find from (i) that C<∞if and only ifµ(r)≍ν(r). Suppose thatν(1)∈ O S. Then µ1(r)≍ν(r)and µ1∈ O S. We see from Lemma 3.3 thatµ∈ O S andµ(r)≍µ1(r). It follows thatµ(r)≍ν(r), that is,C<∞. Conversely, suppose thatC<∞, that is,µ(r)≍ν(r). Letr>0 andc7:=µ4(0,∞)>0. We have

µ(r)≥

Z

(0,∞)

µ3(ry)µ4(d y)≥c7µ3(r). (3.4) Asµ3 is compound Poisson, there isc8 >0 such thatµ3(r)≥ c8ν(r). Hence we obtain from (3.4) thatc7µ3(r)≤µ(r)≍ν(r)≤c8−1µ3(r). This implies thatµ3(r)≍ν(r). Hence we see from Lemmas 3.1 (i) and 3.4 (i) that(ν(c))(1)=ν(c)∈ O S and therebyν(1)∈ O S.

Lastly, we prove (iii). Lemma 3.3 states thatµ∈ O S if and only ifµ1∈ O S. We see from Lemma 3.4 thatµ1 ∈ O S if and only if(ν(1))n∗∈ O S for some n≥ 1. Thus we have proved that (1) is

equivalent to (2). Suppose that µ(r)≍(ν(1))n∗(r). Since µ3 is compound Poisson, there isc9 >0 such thatµ3(r)≥c9(ν(c))n∗(r). Hence we obtain from (3.4) that

c7µ3(r)≤µ(r)≍(ν(1))n∗(r)≍(ν(c))n∗(r)≤c9−1µ3(r).

This implies that µ3(r) ≍ (ν(c))n∗(r) and thereby we see from Lemma 3.4 that µ3 ∈ O S and (ν(c))n∗∈ O S. It follows that(ν(1))n∗∈ O S and therebyµ1∈ O S. Thus we find from Lemma 3.3 thatµ∈ O S . Conversely, suppose thatµ∈ O S. We obtain from Lemmas 3.3 and 3.4 (ii) that

µ(r)≍µ1(r)≍(ν(1))n∗(r)

for somen≥1. We have proved that (1) is equivalent to (3). ƒ

4

Bound separating

C

and

C

for

µ

in

H

The study of the lower and upper limits of ratios of tails of distributions on R+ was initiated by

[11; 25]and progressed by[9; 18]. Watanabe and Yamamuro[32]discussed them for distributions onRand showed basic diferrences between the one-sided and two-sided cases. In this section, we

study a bound separatingC∗andC∗forµ∈ H. We need the following lemma of Denisov et al.[9].

Lemma 4.1. (Lemma 2 of [9]) If a distributionρ onR+ is in H, then there exists an increasing

concave function h:R+R+such that

Z R+

eh(x)ρ(d x)<and Z

R+


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Now we present the main result of this section.

Proposition 4.1. Letµbe a distribution inID+ with Lévy measureν.

(i)µ∈ H if and only ifν(1)∈ H.

(ii)Ifν(1)∈ H, then0<C∗≤1≤C∗≤ ∞.

Proof. Assertion (i) is obvious by virtue of Theorem 25.17 of[26]. We prove only (ii). Suppose

thatν(1)∈ H. First we show thatC∗≥1. We have

µ(r) ν(r)=

Z R

µ3(ry)

ν(r) µ4(d y)≥

µ3(r+a)

ν(r) µ4(−a) (4.1)

for anya>0. Letr>0. We have for anyk≥1

(ν(c))k∗(r)≥ν(c)(r).

Notice thatµ3(r) =e−ν(c)

P∞

n=1ν(c)

n(n!)−1(ν

(c))n∗(r). It follows that

µ3(r+a) = e−ν(c) ∞

X

n=1

ν(c)n(n!)−1(ν(c))n∗(r+a)

e−ν(c)

X

n=1

ν(c)n(n!)−1ν(c)(r+a). (4.2)

Asν(1)∈ H, we haveν(c)∈ H. Suppose that, for somea>0 andδ >0,

lim sup

r→∞

ν(c)(r+a)

ν(c)(r)

e−δ.

Then there is sufficiently larger0>0 such thatν(c)(r0+ak)≤ekδ/2ν(c)(r0)for all integersk≥1. Thus we haveνb(c)(γ1) =γ1

R∞ −∞e

γ1xν

(c)(x)d x <∞ forγ1 :=δ/(4a) >0. This is a contradiction. Thus we have, for anya>0,

lim sup

r→∞

ν(c)(r+a)

ν(c)(r)

=1. Hence we see from (4.1) and (4.2) that

C∗ ≥ lim sup

r→∞

µ3(r+a) ν(c)ν(c)(r)

µ4(−a) ≥ e−ν(c)

X

n=1

ν(c)n(n!)−1µ4(−a) ν(c) =

1−e−ν(c)

ν(c) µ4(−a). Asa→ ∞andc→ ∞, the right-hand side goes to 1. We have proved thatC∗≥1.

Next we show that C∗ ≤ 1. The proof is suggested by that of Theorem 4 of [9]. Lemma 4.1 is applied to ν(c). Then we can take a function h(x) of Lemma 4.1. For any b > 0, we consider a


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concave function hb(x) := min{h(x),b x}. Hence there is x1 > 0 such that hb(x) = h(x) for all

xx1. Thus we have

Z R+

ehb(x)ν

(c)(d x)<∞ and

Z R+

x ehb(x)ν

(c)(d x) =∞. (4.3)

Furthermore, we see from Theorem 25.17 of[26]thatµb4(γ)<∞for allγ >0. Hence

Z R

(y∨0)ehb(y∨0)µ

4(d y)≤

Z R+

y eb yµ4(d y)<∞. (4.4)

For reala and t, we use the notation of [9]and put a[t] = min{a,t}and at = max{a,t}. Let

t>0. We note that(x+y)[t]≤x[t]+y[t]forx,y ≥0. Hence we obtain that

Z R

(x∨0)[t]ehb(x∨0)µ(d x) =

ZZ

((x+ y)∨0)[t]ehb((x+y)∨0)µ

3(d x)µ4(d y)

ZZ

(x∨0)[t]ehb((x+y)∨0)µ

3(d x)µ4(d y)

+

ZZ

(y∨0)[t]ehb((x+y)∨0)µ

3(d x)µ4(d y) =J1+J2.

By concavity of the functionhb(x), we havehb(x+y)≤hb(x) +hb(y)forx,y≥0. Hence

J1

Z R+

x[t]ehb(x)µ 3(d x)

Z R

ehb(y∨0)µ 4(d y).

Here we see that, for any positive integern,

Z R+

x[t]ehb(x)(ν

(c))n∗(d x)

n Z

· · ·

Z

R+×···×R+

x[1t]ehb(x1+···+xn)ν

(c)(d x1)· · ·ν(c)(d xn)

n Z

R+

x[1t]ehb(x1)ν

(c)(d x1)

Z R+

ehb(x)ν

(c)(d x)

!n−1 .

Thus it follows that

Z R+

x[t]ehb(x)µ 3(d x)

=e−ν(c)

X

n=1 ν(c)n

n!

Z R+

x[t]ehb(x)(ν

(c))n∗(d x)

ν(c)exp

 ν(c)

Z R+

ehb(x)ν

(c)(d x)−1

!  Z

R+

x[t]ehb(x)ν


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Therefore we obtain from (4.3) and (4.4) that lim sup b→0 J1 Z R+

x[t]ehb(x)ν

(c)(d x)

ν(c). (4.6)

Furthermore, we have

Z R+

ehb(x)µ

3(d x)≤exp

 ν(c)

Z R+

ehb(x)ν

(c)(d x)−1

! . Thus it follows from (4.3) and (4.4) that

lim sup

t→∞

J2 Z

R+

x[t]ehb(x)ν

(c)(d x)

≤ lim

t→∞

Z R

ehb(x∨0)µ 3(d x)

Z R+

y ehb(y)µ 4(d y)

Z R+

x[t]ehb(x)ν

(c)(d x)

=0. (4.7)

Let δ1 > 0. Consequently, we obtain from (4.6) and (4.7) that, for sufficiently small b and suffi-ciently larget,

Z R+

x[t]ehb(x)µ(d x)

Z R+

x[t]ehb(x)ν

(c)(d x)

< ν(c) +δ1. (4.8)

Let x0>0. Suppose thatµ(x)≥ν(c)(x)(ν(c) +δ1)for all xx0. Here we notice that

Z R+

x ehb(x)µ(d x)

ZZ R+×R+

(x+y)ehb(x+y)µ

3(d x)µ4(d y)

µ4(R+)

Z R+

x ehb(x)µ 3(d x)

µ4(R+)e−ν(c)ν(c)

Z R+

x ehb(x)ν

(c)(d x) =∞. (4.9)

We see from (4.3) and (4.9) that both the numerator and the denominator of (4.8) goes to infinity ast → ∞. Hence we have

lim inf

t→∞

Z R+

x[t]ehb(x)µ(d x)

Z R+

x[t]ehb(x)ν

(c)(d x)

= lim inf

t→∞

Z ∞

x0+

x[t]ehb(x)µ(d x)

Z ∞

x0+

x[t]ehb(x)ν


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= lim inf

t→∞

Z ∞

x0+

µ(x)d(x[t]ehb(x))

Z ∞

x0+

ν(c)(x)d(x[t]ehb(x))

ν(c) +δ1.

This contradicts (4.8). Hence we have lim infr→∞µ(r)(c)(r)≤ν(c) +δ1, because x0 is arbitrary. Lettingδ1↓0, we obtain thatC∗≤1. We see from Proposition 3.1 (i) that 0<C∗. ƒ

5

Proofs of Theorem 2.1 and Its Corollaries

Let γ∈ R. We define the class M(γ) as the totality of distributionsρ satisfyingρb(γ) < . For

ρ∈ M(γ), we define the exponential tiltρ〈γ〉 ofρas ρ〈γ〉(d x):= 1

b

ρ(γ)e

γxρ(d x). (5.1)

Note that the exponential tilt conserves convolution. That is, forρ,η∈ M(γ),(ρη)〈γ〉=ρ〈γ〉η〈γ〉. Let {Xj}∞j=0 be i.i.d. random variables with distribution ν(1). Let Y be a random variable with

distributionµ2 independent of{Xj}. Define a random walk{Sn}∞n=0asSn:=

Pn

j=1Xj forn≥1 and

S0:=0. Recall thatc0:=ν(1,∞).

Lemma 5.1. Letµbe a distribution inID+with Lévy measureν.

(i)If0≤γ<andµb(γ∗) =∞, then C∗=∞.

(ii)If0≤γ<andµb(γ∗)<, then C∗≤µb(γ∗)≤C.

Proof. Suppose that 0< γ<∞andµb(γ∗) =∞. Sinceµ∈ M(γ)for 0< γ < γ∗, µ〈γ〉 exists. We haveµ〈γ〉 = (µ3)〈γ〉∗(µ4)〈γ〉 andµb3(γ∗) =∞. Defineν1(d x):=eγxν(d x). Note thatν1 is the Lévy measure ofµ〈γ〉. Since(µ3)〈γ〉 is one-sided, we have

lim inf

r→∞

(µ3)〈γ〉(r) ν1(r)

≥1. By using integration by parts, we have

¯

µ〈γ〉(r) =

eγrµ¯(r)

b

µ(γ) + γ

b

µ(γ)

Z ∞

r

eγtµ¯(t)d t, (5.2) and

ν1(r) =eγrν¯(r) +γ

Z ∞

r

eγtν¯(t)d t. (5.3) Thus we see from Fatou’s lemma that

C∗ b

µ(γ) ≥ lim infr→∞

µ〈γ〉(r) ν1(r)


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Z ∞

0− lim inf

r→∞

(µ3)〈γ〉(ry) ν1(ry)

(µ4)〈γ〉(d y)

≥ 1

b

µ4(γ)

Z ∞

0−

eγyµ4(d y). Hence we have

C∗≥lim inf γ↑γµb3(γ)

Z ∞

0−

eγyµ4(d y) =∞.

Thus assertion (i) is true. It is clear from Proposition 4.1 that assertion (ii) is true for γ∗ = 0. Suppose that 0≤γ<∞andµb(γ∗)<∞. Then we see thatµ〈γ∗∈ H. Defineν2(d x):=

x ν(d x). Thus we obtain from Proposition 4.1 and (5.2) and (5.3) with replacingγbyγ∗that

C∗ b

µ(γ) ≥ lim sup

r→∞

µ〈γ∗(r) ν2(r)

≥1, and

C

b

µ(γ) ≤lim infr→∞

µ〈γ∗(r) ν2(r)

≤1.

Thus we have proved the lemma. ƒ

Proposition 5.1. Letµbe a distribution inID+with Lévy measureν. Suppose thatν(1)∈ O S . Then

we have0≤γ<b(γ∗)<, and0<Cµb(γ∗)≤C<.

Proof. Suppose that ν(1) ∈ O S. Then we find from Proposition 3.1 that µ ∈ O S ⊂ O L and

0<CC<∞. We see from Lemma 3.2 (ii) that 0≤γ<∞. We see from Lemma 5.1 (i) that

b

µ(γ∗)<∞. Thus the proposition follows from Lemma 5.1 (ii). ƒ

Remark 5.1. Suppose thatν(1)∈ O L. As is mentioned in the proof of Theorem 1 of[29], there

exists an increasing subsequence{λn} ∈Λof{xn}for each sequence{xn}∞n=1 with limn→∞xn=∞.

Proof. DefineTn(y)as

Tn(y):= ν(1)(xny) ν(1)(xn)

.

Since{Tn(y)}∞n=1is a sequence of increasing functions, uniformly bounded on all finite intervals, by the selection principle (see Chap. VIII of[17]) there exists an increasing subsequence{λn}of{xn}

with limn→∞λn=∞such that everywhere onR

lim

n→∞

ν(1)(λny)

ν(1)(λn)

= lim

n→∞ ¯

ν(λny)

¯ ν(λn)

=:m(y;{λn}).

The limit functionm(x;{λn})is increasing and is finite. That is,{λn} ∈Λ. ƒ

Proposition 5.2. Letµbe a distribution inID+with Lévy measureν. Suppose thatν(1)∈ O L. Then


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Proof. Suppose thatν(1) ∈ O L. Defineh∗(x):=lim infr→∞ν(rx)(r). Since ν(1)∈ O L, we

have for{λn} ∈Λandx∈R,

0<h∗(x)≤m(x;{λn})<∞.

Thus it follows that 0<R−∞h∗(x)µ(d x)≤I∗(µ). Choose{λn} ∈Λsuch that

C∗= lim

k→∞ ∞

X

n=0

ec0c

n−1

0 P(Y+Sn> λk)

n!P(X0> λk)

. (5.4)

Define the eventsAjfor 1≤ jnandb>0 as

Aj:={Xj> λkbandY+Sn> λk}.

LetBn:={(i,j):j=6 i, 1≤in, 1≤ jn}. We have

P(Y+Sn> λk)≥P( n

[

j=1

Aj)≥ n

X

j=1

P(Aj)−

X

(i,j)∈Bn

P(AiAj). (5.5)

We obtain from Fatou’s lemma that, for 1≤ jn, lim inf

k→∞

P(Aj)

P(X0> λk) ≥

Z b+

−∞ lim inf

k→∞

P(Xj> λku)P(Y+SnXjdu)

P(X0> λk)

=

Z b+

−∞

m(u;{λn})P(Y+Sn−1∈du). Lettingb→ ∞, we have

lim inf

b→∞ lim infk→∞

P(Aj)

P(X0> λk) ≥

Z ∞

−∞

m(u;{λn})P(Y +Sn−1∈du). (5.6) Moreover, we have, fori6= j,

lim sup

k→∞

P(AiAj)

P(X0> λk) ≤ lim sup

k→∞

P(Xi> λkb)P(Xj> λkb)

P(X0> λk)

= m(b;{λn})·0=0. (5.7)

We see from Lemma 3.3 that

lim

k→∞P(Y > λk)/P(X0> λk) =0. (5.8) Thus we established from (5.4)-(5.8) that

C∗ = lim

k→∞ ∞

X

n=1

ec0c

n−1

0 P(Y +Sn> λk)


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≥ ∞

X

n=1

ec0c

n−1 0 n

n!

Z ∞

−∞

m(u;{λk})P(Y+Sn−1∈du)

Z ∞

−∞

m(u;{λk}) ∞

X

n=0

ec0c

n

0

n!P(Y+Sndu) =

Z ∞

−∞

m(u;{λk})µ(du)≥I∗(µ).

Thus we have proved the proposition. ƒ

Lemma 5.2. Suppose thatν(1)∈ O S. Then we have the following:

(i)There are some b1,b2>0such that m(x;{λn})≤b1eb2(x∨0)for all{λn} ∈Λand for all x∈R.

(ii)I∗(µ2)<.

(iii)For distributionsρandη, I∗(ρη)≤I∗(ρ)I∗(η).

(iv) 2I∗(ν(1))≤d<∞.

Proof. Suppose thatν(1)∈ O S. Let{λn} ∈Λ. Define

h∗(x):=lim sup

r→∞ ¯ ν(rx)

¯ ν(r) .

Since O S ⊂ O L, we have m(x;{λk}) ≤ h∗(x) < ∞ and h∗(x + y) ≤ h∗(x)h∗(y). Thus h∗(x)

is so-called submultiplicative and by Lemma 25.5 of[26]there are b1,b2 > 0 such that h∗(x) ≤

b1eb2(x∨0). Thus (i) is true. Since we see from Theorem 25.17 of[26]thatµb

2(γ)<∞for anyγ >0, we have (ii) by (i). We see that, for x, y ∈R,

lim

n→∞ ¯

ν(λnyx)

¯

ν(λny)

= m(x+y;{λn})

m(y;{λn})

. Hence{λny} ∈Λfor any y ∈R. Thus we have

I∗(ρη) = sup

n}∈Λ

Z ∞

−∞

m(x;{λn})ρη(d x)

= sup {λn}∈Λ

Z ∞

−∞

Z ∞

−∞

m(x+y;{λn})ρ(d x)η(d y)

= sup {λn}∈Λ

Z ∞

−∞

m(x;{λny})ρ(d x)

Z ∞

−∞

m(y;{λn})η(d y)

≤ sup

{λ′ n}∈Λ

Z ∞

−∞

m(x;{λn})ρ(d x) sup {λn}∈Λ

Z ∞

−∞

m(y;{λn})η(d y)

= I∗(ρ)I∗(η). We have, for{λn} ∈Λands>0,

> d∗≥lim sup

k→∞

P(X0+X1> λk)


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≥ 2 lim sup

k→∞

Z s+

−∞

P(X0> λku)

P(X0> λk)

P(X1du)

≥ 2

Z s+

−∞ lim inf

k→∞

P(X0> λku)

P(X0> λk)

P(X1du)

= 2

Z s+

−∞

m(u;{λn})P(X1∈du).

Lettings→ ∞, we see that 2I∗(ν(1))≤d<∞. ƒ

Proposition 5.3. Letµbe a distribution inID+with Lévy measureν. Suppose thatν(1)∈ O S . Then

we have C∗≤J(µ)<∞.

Proof. Suppose thatρ:=ν(1)∈ O S. We find from Lemma 5.2 (ii) and (iv) thatJ(µ)<∞. Define,

forn≥0,

dn:=lim sup

x→∞

P(Sn+Y >x)

P(X0>x) .

Sinceµ∈ O S by Proposition 3.1, we see from Lemma 3.3 that d0=0. Choose{λn} ∈Λsuch that

dn= lim

k→∞

P(Sn+Y > λk)

P(X0> λk)

. Fors>1 andn≥1, define

I1:=

Z (λks)+

−∞

P(Sn−1+Y > λku)P(Xndu),

I2:=

Z s+

−∞

P(Xn> λku)P(Sn−1+Ydu), and

I3:=P(Sn−1+Y >s)P(Xn> λks).

Then we have P(Sn+Y > λk) =

P3

j=1Ij. For any ε >0, we can take sufficiently large s>1 such

that

lim sup

k→∞

I1 P(X0> λk) ≤ (dn−1+ε)lim sup

k→∞

Z (λks)+

−∞

P(X0> λku)

P(X0> λk)

P(Xndu), (5.9)

and

lim sup

k→∞

I3

P(X0> λk) ≤ (dn−1+ε) lim

k→∞

P(Xn> λks)P(X0>s)

P(X0> λk)


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By virtue of the dominated convergence theorem, we see that lim sup

k→∞

I2 P(X0> λk)

=

Z s+

−∞

m(u;{λk})P(Sn−1+Ydu)

Z ∞

−∞

m(u;{λk})P(Sn−1+Ydu)≤I∗(ρ(n−1)∗∗µ2). (5.11)

Note that

lim sup

k→∞

P(X0+Xn> λk)

P(X0> λk) ≥ lim sup

k→∞

Z (λks)+

−∞

P(X0> λku)

P(X0> λk)

P(Xndu) + lim

k→∞

P(Xn> λks)P(X0>s)

P(X0> λk)

+ lim inf

k→∞

Z s+

−∞

P(Xn> λku)

P(X0> λk)

P(X0du). (5.12)

Thus we obtain from (5.9)-(5.12) that, forn≥1,

dn ≤ (dn−1+ε)lim sup

k→∞

P(X0+Xn> λk)

P(X0> λk)

+I∗(ρ(n−1)∗∗µ2)

− (dn−1+ε)lim inf

k→∞

Z s+

−∞

P(Xn> λku)

P(X0> λk) P(X0∈du)

≤ (dn−1+ε)(d∗−

Z s+

−∞

m(u;{λk})ρ(du)) +I∗(ρ(n−1)∗∗µ2).

Lettings→ ∞andε→0, we have by Lemma 5.2 (ii)

dndn−1(d∗−I∗(ρ)) +I∗(ρ)n−1I∗(µ2). (5.13) Noting thatd0=0, we see from (5.13) that

dnI∗(µ2)

n−1

X

k=0

I∗(ρ)k(d∗−I∗(ρ))n−1−k<∞. (5.14)

By Lemma 3.2 (iii), we have forx ≥0

P(Sn+Y >x)

P(X0>x) ≤

Z ∞

−∞

P(Sn> xu)

P(X0>x)

P(Ydu)

Z ∞

−∞

KnP(X0>xu) P(X0>x)


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= KnP(S1+Y >x) P(X0>x)

Kn(d1+δ),

with some positive constantsK andδ. Thus we can use Fatou’s lemma and establish from (5.14) that

C∗ = lim sup

x→∞

ec0

X

n=1

c0n−1 n!

P(Sn+Y >x)

P(X0>x) ≤ ec0

X

n=1

c0n−1 n! dn

ec0

X

n=1

c0n−1 n! I

(µ 2)

n−1

X

k=0

I∗(ρ)k(d∗−I∗(ρ))n−1−k=J(µ).

We have proved the proposition. ƒ

Proof of Theorem 2.1. Assertion (i) is due to Proposition 3.1. Assertion (ii) is due to Propositions

5.1-5.3. ƒ

Proof of Corollary 2.1. We see from Theorem 2.1 (ii) that 0 ≤ γ< ∞, µb(γ∗) < ∞ and

C =C∗=C∗=µb(γ∗). ƒ

The following is due to Theorem 1 of[13]and Theorem 3.1 of[24], and conclusively to Theorem 1.1 of [31]. An interesting history of the establishment of this result is found in [31]. It has an application to the local subexponentiality of an infinitely divisible distribution. See [32]. Applications of the classS(γ)to Lévy processes are found in[4; 6; 21].

Lemma 5.3. Letγ≥0.Letµbe a distribution inID+with Lévy measureν. Then the following are

equivalent:

(1)µ∈ S(γ).

(2)ν(1)∈ S(γ).

(3)ν(1)∈ L(γ)b(γ)<, andµ¯(x)∼µb(γ)ν¯(x).

(4)ν(1)∈ L(γ)and, for some C ∈(0,∞)¯(x)∼¯(x).

Remark 5.2. Letγ≥0. We see from the above lemma thatS(γ)∩ID+is closed under convolution

roots. The classS is closed under convolution roots. Refer to Theorem 2 of[13]in the one-sided case and see Prposition 2.7 of[31]in the two-sided case. However, we do not know whether the classS(γ) is closed under convolution roots for γ >0, so far. We find from Theorem 2.1 of[32]

thatS(γ)onR+is closed under convolution roots for some (equivalently for all)γ >0 if and only

if so is the locally subexponential class onR+.

Proof of Corollary 2.2. Let a ∈R. Note that µδa is also an infinitely divisible distribution on


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Corollary 2.1 that 0≤γ<∞,µb(γ∗)<∞andC(aj) =µb(γ∗)exp(−γaj)for j=1, 2. Define a set

EasE:={ma1+na2:m, n∈Z}. Then we have, for anyaE,

µ(x+a)∼exp(−γa)µ(x). (5.15) Note thatEis a dense set inRbecausea1/a2is irrational. Thus we have (5.15) for anyaE¯=R. It

follws thatµ∈ L(γ∗)andµ(x)∼µb(γ∗)ν(x). Thus we conclude from Lemma 5.3 thatµ∈ S(γ∗). ƒ

6

Proof of Theorem 2.2

Albin [1] asserted that ifν(1) ∈ L(γ), then µ ∈ L(γ). His proof in the case γ = 0 is complete.

However, his proof in the case γ > 0 depends on an incomplete lemma which is stated without precise proof.

Assertion (Lemma 2.1 of[1]) Letρ∈ L(γ)be supported on[0,∞)for someγ≥0. Given constants

ε >0and t∈R, pick a constant x0Rsuch that

ρ(xt)

ρ(x) ≤(1+ε)e

γt for xx

0. (6.1)

Then

ρn(xt)

ρn(x) ≤(1+ε)e

γt for xn(x

0−t) +t. (6.2)

Remark 6.1. The assertion above is correct forγ=0. But, we can make the following

counterex-ample of this assertion forγ >0. Indeed, letρ(x):=R∞

x e

udu. Thenρ∈ L(1)and the condition

(6.1) is satisfied for x0 =0. Notice thatρn(x) = ((n1)!)−1R∞

x u

n−1eudu. Let t < 0 and put

c:=1−t. We have

lim

n→∞

ρn(cnt)

ρn(cn) = nlim→∞

R∞

0 (cnt+u)

n−1e−(cnt+u)du

R∞

0 (cn+u)

n−1e−(cn+u)du

= et lim

n→∞

R∞ 0 (1+

t+u cn )

n−1eudu

R∞ 0 (1+

u cn)

n−1eudu =e

c−1t

·et.

Takeε satisfying 1+ε < ec−1t. The inequality (6.2) does not hold for sufficiently largen. The example for a generalγ >0 is analogous and is omitted.

Lemma 6.1. Letρandηbe distributions onR. Letγ0.

(i) (Lemma 2.5 of [31]) If ρ,η ∈ L(γ), then ρη ∈ L(γ). In particular, if ρ ∈ L(γ), then


(22)

(ii) (Lemma 1 of[5]) If ρ ∈ L(γ), then, for every a∈R, there is a positive constant M(a) which

depends only on a such that, for every x∈Rand every n≥1,

ρn(xa)M(a)ρn(x). (6.3)

The following lemma is suggested by an argument in[5].

Lemma 6.2. Letγ≥0, a∈R. Suppose thatρ∈ L(γ). For anyε with0< ε <1, there is b >0

such that for all n≥1and for all x∈R,

ρ(n+1)∗(x+a)(1+ε)e−γaρ(n+1)∗(x) +ρn(x+ab), (6.4)

and

ρ(n+1)∗(x+a)(1ε)e−γa(ρ(n+1)∗(x)ρn(xb)). (6.5)

Proof.We have

ρ(n+1)∗(x) =

Z (xb)+

−∞

ρ(xy)ρn∗(d y) +

Z ∞

(xb)+

ρ(xy)ρn∗(d y) = K1(x) +K2(x).

Let 0< ε <1. Sinceρ∈ L(γ), it follows that

(1−ε)e−γaρ(xy+a)

ρ(xy) ≤(1+ε)e −γa

for sufficiently large b>0 and yxb. Notice that

K1(x+a) =

Z (xb)+

−∞

ρ(xy+a)

ρ(xy) ρ(xy)ρ

n(d y).

Hence we obtain that, for sufficiently largeb>0,

(1−ε)e−γaK1(x)≤K1(x+a)≤(1+ε)e−γaK1(x). Here we have K1(x)≤ρ(n+1)∗(x)and K

2(x)≤ρn∗(xb). Hence we get (6.4). Furthermore, we have

K1(x) =ρ(n+1)∗(x)K

2(x)≥ρ(n+1)∗(x)−ρn∗(xb).

Hence we get (6.5). ƒ

Lemma 6.3. Letµ be a distribution inID+ with Lévy measureν. Letγ≥0. Thenµ∈ L(γ)if and only ifµ1∈ L(γ).


(23)

Proof. As in the proof of Theorem 1.1 of[31], we see from Lemma 2.5 of [24]that ifµ∈ L(γ), thenµ1∈ L(γ). Conversely, ifµ1∈ L(γ), then we haveµ∈ L(γ)by Lemma 2.1 of[24]. ƒ

Proposition 6.1. Letµbe a distribution inID+with Lévy measureν. Letγ≥0. Ifν(1)∈ L(γ), then

µt∗ ∈ L(γ)for all t >0. In the converse direction, ifµt∗∈ L(γ) for all t>0andν(1)∈ O S, then

ν(1)∈ L(γ).

Proof. We prove the first assertion. Suppose that ν(1) ∈ L(γ). Without loss of generality, we

can assume that t = 1. Denote ρ := ν(1) and λn := ec0c0n(n!)

−1 with c

0 := ν(1,∞). Note that limn→∞λn+1/λn=0. Let 0< ε <1. LetN be a positive integer satisfyingλn+1≤ελnfor allnN.

Define

IN(x):=

N

X

n=1

λnρn∗(x) and JN(x):=

X

n=N+1

λnρn∗(x).

Thenµ1(x) =IN(x) +JN(x)forx >0. Sinceρ∈ L(γ), we have by Lemma 6.1 (i)

(1−ε)e−γaIN(x)≤IN(x+a)≤(1+ε)e−γaIN(x)

for sufficiently large x. We see from Lemma 6.2 that, for allx ∈R,

JN(x+a)≤

X

n=N+1

λnρn∗(x)(1+ε)e−γa+

X

n=N+1

λnρ(n−1)∗(x+ab).

By Lemma 6.1 (ii), we have ∞

X

n=N+1

λnρ(n−1)∗(x+ab) =

X

n=N

λn+1ρn∗(x+ab)

εM(ba) ∞

X

n=N

λnρn∗(x).

Hence, for all x∈R,

JN(x+a)≤(1+ε)e−γaJN(x) +εM(ba)µ1(x). Furthermore, we obtain from Lemmas 6.1 (ii) and 6.2 that, for all x∈R,

JN(x+a) ≥ (1−ε)e−γa

X

n=N+1

λnρn∗(x)−

X

n=N+1

λnρ(n−1)∗(xb)

≥ (1−ε)e−γa

X

n=N+1

λnρn∗(x)−εM(b)

X

n=N

λnρn∗(x) ≥ (1−ε)e−γa(JN(x)−εM(b)µ1(x)).

In consequence, it follows that, for sufficiently large x,


(1)

(ii)Ifρn∗∈ D for some n≥1, thenρ∈ D.

(iii)Ifρ,η∈ D, thenρη∈ D. In particular, ifρ∈ D, thenρn∗∈ Dfor all n≥1.

Proof. Assertion (i) is obvious by the definition of the classD. For (ii), suppose thatρn∗∈ D. Since ρn∗(r)≍(ρ+)n∗(r), we have (ρ+)n∗ ∈ D and, by Proposition 1.1 (iii) of[27], ρ+∈ D and hence ρ∈ D. Ifρ,η∈ D ⊂ O S, thenρ+,η+∈ D ⊂ O S and (3.2) holds. Thus we see from Proposition 2.3 (ii) of[27]that ρ+∗η+∈ D and thus from (3.2) and (i) thatρη∈ D. The second assertion

of (iii) is obvious. ƒ

Lemma 7.2. Letρbe a distribution onR. (i)Ifρ∈ D, then, for all n≥1and a≥0,

lim sup r→∞

ρn∗(ra)

ρ(r) ≤ n·N→∞lim lim supr→∞

ρ(rN)

ρ(r) . (7.1)

(ii)Ifρ∈ D, then there are two positive constants c1and c2 such that, for all n≥1and r>0, ρn∗(r)

ρ(r) ≤cn

c2+1. (7.2)

Proof. First we prove (i). Let{Xj}∞j=1be i.i.d. random variables with a common distributionρ. Let r>a+ (n−1)N1 withN1>0. We have

ρn∗(ra)

=P    n X j=1

Xj>ra, Xj0>n−1(ra)for some j0 with 1≤ j0n    ≤ n X j0=1

P    n X j=1

Xj>ra, Xj0>n−1(ra), XiN1 for anyi6= j0    + n X j0=1

P    n X j=1

Xj>ra, Xj0>n−1(ra), Xi>N1for somei6= j0   

(ra−(n−1)N1) +n(n−1)ρ(N1)ρ(n−1(ra)). (7.3) Here, ifρ∈ D, then there are two positive constantsc1 andc2 such that, for anyn≥1 andr>0,

ρ(n−1r)

ρ(r) ≤c1n

c2. (7.4)

Thus we see from (7.3) and (7.4) that ifρ∈ D, then lim sup

r→∞

ρn∗(ra)

ρ(r) ≤ Nlim1→∞

lim sup r→∞

(ra−(n−1)N1)

ρ(r) = n· lim

N→∞lim supr→∞

ρ(rN)


(2)

As we have

ρn∗(r) =P   

n X

j=1 Xj>r

  ≤

n X

j=1

P(Xj>n−1r) =(n−1r),

it follows from (7.4) that (ii) holds. ƒ

Lemma 7.3. Letµbe a distribution inID+with Lévy measureν. (i)Ifν(1)∈ D, then0<Q∗≤1≤Q<.

(ii)Ifν(1)∈ D andµis a compound Poisson distribution onR+, then C∗≤ lim

N→∞lim supr→∞

µ(rN)

ν(r) ≤Q

. (7.5)

Proof. We prove (i). Suppose thatν(1)∈ D. Then we have 1≥Q∗≥lim infr→∞

ν(2r)

ν(r) >0,

and

1≤Q∗= (Q)−1<∞.

Next we prove (ii). Let r > 0. Notice thatµ(r) = e−aP∞n=1an(n!)−1ρn∗(r), where a:=ν(R +) = ν(0)andρ:=a−1ν. Suppose thatν(1)∈ D. LetN1>0. Now we can use Fatou’s lemma by virtue of Lemma 7.2 (ii). Thus we see from Lemma 7.2 (i) that

C∗ ≤ lim sup r→∞

µ(rN1)

ν(r) ≤ e−a

∞ X n=1

an

n!lim supr→∞

ρn∗(rN 1) (r) ≤Q

.

AsN1→ ∞, we get (7.5). ƒ

Now we prove Theorem 2.3.

Proof of Theorem 2.3. We prove (i). Ifν(1)∈ D ⊂ O S, we see from Theorem 2.1 and Lemma 7.1 (i) thatµ(r)≍ν(r)andµ∈ D. Conversely, suppose thatµ∈ D ⊂ O S. Then we see from Proposition 3.1 (iii) that(ν(1))k∗∈ O S for some k≥1 andµ(r)≍(ν(1))k∗(r). It follows from Lemma 7.1 that

(ν(1))k∗∈ Dandν(1)∈ D.

We show (ii). LetY1 andY2 be independent random variables with distributionµ1 andµ2, respec-tively. Suppose that ν(1) ∈ D. It follows from Lemma 7.3 (i) that 0< Q∗ ≤ 1≤Q< ∞. Since

D ⊂ H, we see from Proposition 4.1 thatC∗≤1≤C∗. Letr>2N >0. Then we have µ(r) =P(Y1+Y2>r)≤

3 X k=1


(3)

whereJ1= P(Y1+Y2>r, |Y2| ≤N),J2=P(Y1+Y2>r, |Y1| ≤N), andJ3=P(Y1+Y2>r, |Y1|> N, |Y2|>N). Now we see from Lemma 7.3 (ii) that

lim

N→∞lim supr→∞ J1

ν(r) ≤ N→∞lim lim supr→∞

P(Y1>rN) ν(r) ≤Q

.

Sinceν(1)∈ D ⊂ O S ⊂ O L, we see from Proposition 3.1 and Lemma 3.3 that lim sup

r→∞ J2

ν(r) ≤ lim supr→∞

P(Y2>rN)

µ1(rN)

µ1(rN)

ν(rN)

ν(rN)

ν(r) =0.

Using Lemma 7.3 (ii) again, we obtain that lim

N→∞lim supr→∞ J3 ν(r) ≤ lim

N→∞lim supr→∞ P(Y

1>2−1r) ν(2−1r)

ν(2−1r)

ν(r) P(|Y2|>N) +P(Y2>2

−1r) µ1(2−1r)

µ1(2−1r)

ν(2−1r)

ν(2−1r)

ν(r) P(|Y1|>N)

=0.

Consequently, we have obtained (2.5). We have, for {λn} ∈ Λ, Q∗ ≤ m(x;{λn}) for x < 0 and 1≤m(x;{λn})for x≥0. Thus we obtain from Theorem 2.1 (ii) that

CI(µ)≥µ[0,∞) +Qµ(−∞, 0) =1−(1−Q)µ(−∞, 0).

Thus we have got (2.4). ƒ

As is mentioned in Section 2, Denisov et al.[9]proved that ifµis a distribution onR+inH ∩ID+, thenC=1. We show in the following example that there existsµ∈ H ∩ID+such thatC<1.

Example 7.1. Letρbe the Peter and Paul distribution, that is, ρ(d x):=

∞ X n=1

2−nδ2n(d x).

Let 0 < λ <1. Let µ1 be a compound Poisson distribution with Lévy measure λρ. Let µ2 be a continuous infinitely divisible distribution on(−∞, 0]. Define µID+ as µ= µ1∗µ2. Then we haveµ∈ DandQ∗=2. Moreover, we can takeλsufficiently small such thatQ<C<1<C<Q∗. Proof. Sinceν(1)=ρ∈ D, we haveµ∈ D by Theorem 2.3 and, obviously,Q∗=2. Note thatµ is continuous, becauseµ2is so. Thus we can take a positive sequence{εn}∞n=1such that limn→∞εn=0 and

lim n→∞

¯

µ(2nεn) ¯

µ(2n) =1. Hence we have

C∗ ≤ lim infn→∞ ¯

µ(2nεn) λρ¯(2nε


(4)

≤ lim n→∞

¯ ρ(2n)

¯ ρ(2nε

n)

lim sup n→∞

¯ µ(2n)

λρ¯(2n)

≤ 2−1C∗. (7.6)

Let{λn} ∈Λ. Note thatm(x;{λn})andI∗(µ2)do not depend on the value ofλ. Sincem(x;{λn})≤ 1 forx ≤0, we see thatI∗(µ2)≤1. Thus we have, forB=0,

lim

λ→0J(µ) =λlim→0I(µ

2)exp(λ(I∗(ρ)−1)) =I∗(µ2)≤1, and, forB>0,

lim

λ→0J(µ) =λlim→0I(µ

2)exp(λ(I∗(ρ)−1))

exp(λB)−1

λB =I

(µ 2)≤1. Hence we obtain from Theorem 2.1 (ii) that

1≤ lim

λ→0C

lim

λ→0J(µ) =I(µ

2)≤1,

that is, I∗(µ2) =1 and limλ→0C∗=1. Thus we conclude from (7.6) and Theorem 2.3 (ii) that, for sufficiently smallλ >0,

2−1<1−2−1µ(−∞, 0)≤C∗≤2−1C<1<2C∗≤C<2.

We have proved all the assertion. ƒ

Acknowledgement.The authors are grateful to Ken-iti Sato for his valuable advice on improvement of the early draft.

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