Tempered infinitely divisible distributions and processes

by Michele Leonardo Bianchi, Svetlozar T. Rachev, Young Shin Kim, Frank J. Fabozzi

No. 26 | MAY 2011 WORKING PAPER SERIES IN ECONOMICS

KIT – University of the State of Baden-Wuerttemberg and National Laboratory of the Helmholtz Association

econpapers.wiwi.kit.edu

Impressum Karlsruher Institut für Technologie (KIT) Fakultät für Wirtschaftswissenschaften Institut für Wirtschaftspolitik und Wirtschaftsforschung (IWW) Institut für Wirtschaftstheorie und Statistik (ETS) Schlossbezirk 12 76131 Karlsruhe

KIT – Universität des Landes Baden-Württemberg und nationales Forschungszentrum in der Helmholtz-Gemeinschaft

Working Paper Series in Economics No. 26, May 2011 ISSN 2190-9806

econpapers.wiwi.kit.edu

Tempered infinitely divisible distributions and processes Michele Leonardo Bianchi∗ Department of Mathematics, Statistics, Computer Science and Applications, University of Bergamo E-mail: [email protected]

Svetlozar T. Rachev

†

Chair-Professor, Chair of Econometrics, Statistics and Mathematical Finance, School of Economics and Business Engineering, University of Karlsruhe and Karlsruhe Institute of Technology (KIT) and Department of Statistics and Applied Probability, University of California, Santa Barbara E-mail: [email protected]

Young Shin Kim Department of Econometrics, Statistics and Mathematical Finance, School of Economics and Business Engineering, University of Karlsruhe and Karlsruhe Institute of Technology (KIT) E-mail: [email protected]

Frank J. Fabozzi School of Management, Yale University E-mail: [email protected]

July 29, 2008

∗

Michele Leonardo Bianchi’s research was supported by a Ph.D. scholarship of the University of Bergamo. † Svetlozar T. Rachev’s research was supported by grants from Division of Mathematical, Life and Physical Science, College of Letters and Science, University of California, Santa Barbara, and the Deutschen Forschungsgemeinschaft.

1

Abstract. In this paper, we construct the new class of tempered infinitely divisible (TID) distributions. Taking into account the tempered stable distribution class, as introduced by in the seminal work of Rosi´ nsky [10], a modification of the tempering function allows one to obtain suitable properties. In particular, TID distributions may have exponential moments of any order and conserve all proper properties of the Rosi´ nski setting. Furthermore, we prove that the modified tempered stable distribution is TID and give some further parametric example. M.S.C. classification: 60E07, 60G52. Key words: stable distributions, tempered stable distributions, tempered infinitely divisible distributions, modified tempered stable distributions.

2

1

Introduction

The formal and elegant definition of tempered stable distributions and processes has been proposed in the work of Rosi´ nski [10] where a completely monotone function is chosen to transform the L´evy measure of a stable distribution. Tempered stable distributions may have all moments finite and exponential moments of some order. The idea of selecting a different tempering function has been already considnski ered in the literature, see [4]. In this paper, by following the approach of Rosi´ [10] and considering a particular family of tempering functions, a new class of distributions is introduced with the same suitable properties of the tempered stable class, but with the advantage that it may admit exponential moments of any order. By multiplying the L´evy measure of a stable distribution with a positive definite radial function, see [13], instead of with a completely monotone function as in [10], we obtain the class of tempered infinitely divisible (TID) distributions. In some cases, the characteristic function of a TID random variable is extendible to an entire function on C, that is, it admits any exponential moment. Some practical problems in the field of mathematical finance have motivated our studies. Furthermore, we want to fill a gap in the literature. The modified tempered stable (MTS) distribution is not a tempered stable distribution of the Rosi´ nski type [8] even though its properties are very close to that class. We will prove that the MTS distribution is in the TID class. Although this distributional family is constructed by tempering the L´evy measure of a stable distribution, any stability property is lost. We will proceed as following. In Section 2, basic definitions and distributional properties are given. Working with the L´evy measure may be a difficult task, therefore a spectral measure R is needed to figure out all characteristics of this class. This measure describes all distributional properties and allows one to obtain a close formula for the characteristic function. Since TID distributions are by construction infinitely divisible, a TID L´evy process can be considered. In Section 3, TID processes are analyzed. If the time scale increases, the TID process looks like a Gaussian process; conversely, if the time scale decreases, it looks like a stable process. Furthermore, under some condition on the tempering function, the change of measure problem between stable and TID processes can be solved. In Section 4, a view toward simulation is given. Taking into consideration [9, 10], a series representation is derived in terms of a measure Q, as already proved for the tempered stable class. Similar to the tempered stable framework, this class of distribution has an infinite dimensional parametrization by a family of measures [14], making it difficult to use. For this reason, in Section 5 some parametric examples in one dimension are proposed, and characteristic functions are derived.

2

Tempered infinitely divisible distribution

It is well known that the L´evy measure M0 of an α-stable distribution on Rd can be written in polar coordinates in the form M0 (dr, du) = r−α−1 dr σ(du),

3

(2.1)

where α ∈ (0, 2) and σ is a finite measure on the unit sphere S d−1 . Theorem 2.1. If µ0 is an α-stable distribution, then its characteristic function has the form  R  sgnhy, ui)σ(du) + ihy, ai exp −cα R S d−1 |hy, ui|α (1 − i tan πα 2 , α 6= 1, µ ˆ0 (y) = exp −c1 S d−1 (|hy, ui| + i π2 hy, ui log |hy, ui|)σ(du) + ihy, ai , α = 1, (2.2) where a ∈ Rd and  |Γ(−α) cos( πα )|, α 6= 1, 2 cα = π , α = 1. 2 Proof. See [12, Theorem 14.10]. Definition 2.2. If Y is an α-stable random vector with characteristic function (2.2), we will write Y ∼ Sα (σ, a). Taking into account the approach of [10], we want to modify the radial component of M0 and obtain a probability distribution with lighter tails than stable ones. A TID distribution is defined by tempering the radial term of M0 as follows. Definition 2.3. Let µ be a infinitely divisible probability measure on Rd without gaussian part. We call µ tempered infinitely divisible (TID) if its L´evy measure M can be written in polar coordinates as M (dr, du) = r−α−1 q(r, u)dr σ(du)

(2.3)

where α is a real number α ∈ [0, 2), σ a finite measure on the unit sphere S d−1 and q : (0, ∞) × S d−1 7→ (0, ∞) is a Borel function defined by Z ∞ 2 2 q(r, u) := e−r s /2 Q(ds|u), (2.4) 0

with {Q(·|u)}u∈S d−1 a measurable family of Borel measure on (0, ∞). If q(0+, u) = 1 for each u ∈ S d−1 , µ is referred to as a proper TID. The function q is called a tempering function. In the case where {Q(·|u)}u∈S d−1 are finite non-negative Borel measures on (0, ∞), q(·, u) are positive definite radial functions on Rd . By [13], the following results holds. Theorem 2.4. A continuous function ϕ : (0, ∞) → R is positive definite and radial on Rd for all d if and only if it is of the form Z ∞ 2 2 ϕ(r) = e−r s µ(ds), 0

where µ is a finite non-negative Borel measure on (0, ∞).

4

Define a measure Q on Rd by Z Z ∞ Q(A) := IA (ru)Q(dr|u)σ(du), S d−1

A ∈ B(Rd ).

(2.5)

0

It is easy to check that Q({0}) = 0. We also define a measure R on Rd by Z x IA ( R(A) := )kxkα Q(dx), A ∈ B(Rd ). 2 kxk d R

(2.6)

The measure R is equivalent to the measure Q and clearly R({0}) = 0. By definition of R, for each Borel function F , the following equality is satisfied Z Z x F (x)R(dx) = F( )kxkα Q(dx), (2.7) kxk2 Rd Rd in the sense that when one sides exists then the other exists and are equal. By choosing x F (x) = IA ( )kxkα , 2 kxk then Q can be written as Z Q(A) := Rd

IA (

x )kxkα R(dx), kxk2

A ∈ B(Rd ).

(2.8)

Sometimes the only knowledge of the L´evy measure cannot be enough to obtain analytical properties of tempered infinitely divisible distributions. Therefore, the definitions of measures Q and R allow one to overcome this problem and to obtain explicit analytic formulas and more explicit calculations. The following result allows one to figure out relations between the L´evy measure M and the measure R above defined. Proposition 2.5. Let µ be a TID distribution, the corresponding L´evy measure M be defined as in (2.3) and R as in (2.6). Then M can be written in the form Z Z ∞ 2 IA (tx)t−α−1 e−t /2 dtR(dx), A ∈ B(Rd ), (2.9) M (A) = Rd

0

if and only if the measure R on Rd satisfies the following conditions, R({0}) = 0 and  R 2 α 0 < α < 2, RRd (kxk ∧ kxk )R(dx) < ∞, (2.10) (log(1 + kxk) + 1)R(dx) < ∞, α = 0. Rd Proof. Let µ be a TID distribution with L´evy measure M . First we will prove that there exists at least one measure R, defined in (2.6), such that M , defined in (2.3), can be written in the form (2.9). To show this result, we take the measure R as in (2.6) and for each A ∈ B(Rd ), by considering (2.4), (2.5), (2.6), (2.7) and Fubini

5

theorem, we have Z Z ∞ IA (ru)r−α−1 q(r, u)drσ(du) M (A) = d−1 ZS Z0 ∞ Z ∞ 2 2 = IA (ru)r−α−1 e−r s /2 Q(ds|u)drσ(du) d−1  ZS Z0 ∞ 0Z ∞ −α−1 −r 2 s2 /2 IA (ru)r e dr Q(ds|u)σ(du) = 0 S d−1 0  Z Z ∞Z ∞ t −α−1 −t2 /2 = IA ( u)t e dt sα Q(ds|u)σ(du) s d−1  Z0 ∞ ZS∞  Z0 u α 2 IA (t )s Q(ds|u)σ(du) t−α−1 e−t /2 dt = s d−1 0  Z0 ∞  ZS x 2 α = IA (t )kxk Q(dx) t−α−1 e−t /2 dt 2 kxk d  Z0 ∞  ZR 2 IA (ty)R(dy) t−α−1 e−t /2 dt = d Z0 Z ∞ R 2 = IA (ty)t−α−1 e−t /2 dtR(dy) Rd

(2.11)

0

Conversely, given a measure R, let Q be the measure defined by (2.8) and let us consider the decomposition Q(dr, du) = Q(dr|u)σ(du), where σ is a finite measure on S d−1 . Thus, we can define q(r, u) by (2.4) and the computation (2.11) proves that M can be written in the form (2.3). Now we want to prove that M is a L´evy measure if and only if (2.10) holds. Suppose M a L´evy measure, then we have Z kxk2 M (dx) < ∞ kxk≤1

and by considering (2.9) and α 6= 0, we obtain Z

Z

2

kxk M (dx) =

∞>

Rd 2

≥

Z

1

kxk kxk≤1 −1/2

≥ e

0 −1

1−α −t2 /2

t Z

Z

1/kxk

e

2 /2

t1−α e−t

kxk

kxk≤1

Z

2

dtR(dx)

0

Z kxk

dtR(dx) +

Z

2

kxk>1 −1/2

2

(2 − α)

kxk R(dx) + e

1/kxk

2 /2

t1−α e−t

dtR(dx)

0 −1

Z

(2 − α)

kxk≤1

kxkα R(dx)

kxk>1

thus, we obtain the desired inequality Z (kxk2 ∧ kxkα )R(dx) < ∞. Rd

Now, let us consider the case α = 0. By definition of the L´evy measure we have Z M (dx) < ∞ kxk≥1

6

and by considering (2.9) with α = 0, the following inequalities are satisfied Z Z Z ∞ 2 ∞> M (dx) = t−1 e−t /2 dtR(dx) 1 kxk

Rd

kxk>1

Z

Z

∞

−1 −t2 /2

t e

=

Z

−1/2

≥

dtR(dx)

1 kxk

kxk>1

Z

2 /2

t−1 e−t

dtR(dx) +

1 kxk

kxk≤1

∞

Z

Z

KR(dx) + e

log(kxk)R(dx)

kxk≤1

kxk>1

where K is a finite constant. Then, also when α = 0, condition (2.10) is a necessary condition. Conversely, now we prove that (2.10) is also sufficient. Suppose that there is a measure R satisfying (2.10). Then the measure M can be written in the form (2.9). If α 6= 0, we can write Z kxk2 M (dx) = kxk≤1

Z

2

1 kxk

Z

kxk

= Rd

0

Z

2 /2

t1−α e−t ∞

Z

dtR(dx)

Z 1 ≤ kxk t e dtR(dx) + kxkα R(dx) 2 − α kxk≤1 0 Z Z kxk>1 α α 1 = 2− 2 Γ(1 − ) kxk2 R(dx) + kxkα R(dx) < ∞ 2 kxk≤1 2 − α kxk>1 2

1−α −t2 /2

and Z

Z

Z

∞

Z

∞

Z

Z

∞

t−α−1 dtR(dx)

t dtR(dx) + kxk≤1

C 2

Z

−3

≤ C =

dtR(dx)

1 kxk

Rd

kxk>1

2 /2

t−1−α e−t

M (dx) =

Z

1 kxk

kxk>1

kxk2 R(dx) +

kxk≤1

1 α

Z

1 kxk

kxkα R(dx),

kxk>1

2

where C := supt≥1 t2−α e−t /2 . Thus M is a L´evy measure. Considering the case where α = 0, we obtain Z

Z

2

2

kxk M (dx) =

kxk Rd

kxk≤1

Z

Z ≤

1 kxk

2 /2

te−t

dtR(dx)

0 −

kxk2 (1 − e

1 2kxk2

)R(dx) Z 2 kxk R(dx) + R(dx)

d ZR

≤ kxk≤1

7

kxk>1

and Z kxk>1

M (dx) = Z Z ∞ 2 = t−1 e−t /2 dtR(dx) Rd

1 kxk

Z ∞ Z 1 2 dtR(dx) + ≤ t−1 e−t /2 dtR(dx) 2 1 1 t(1 + t /2) kxk>1 kxk kxk≤1 kxk Z Z ≤ kxk2 R(dx) + (log(kxk) + e−1/2 )R(dx). Z

Z

kxk≤1

∞

kxk>1

Thus, M is a L´evy measure. Now, in order to show that (2.9) is well defined, we want to show that R is uniquely determined. We will prove it by contradiction. Let R1 and R2 be two measures on Rd satisfying (2.9). Then, by previous argument, (2.10) has to be satisified also. By contradiction, we suppose that there exists a Borel set A such that R1 (A) 6= R2 (A). By equation (2.8), we can define Q1 and Q2 from R1 and R2 and consider the polar representation Qi (dr, du) = Qi (dr|u)σ(du) where σ is a probability measure on S d−1 and {Qi (·|u)}u∈S d−1 are measurable families of Borel measure on (0, ∞). Without any loss of generality, we assume that σ is not the null measure on S d−1 . If α 6= 0, by definition (2.8) and conditions (2.10), the inequality Z Z 2 α ∞> (kxk ∧ kxk )Ri (dx) = (kxk−2 ∧ kxk−α )kxkα Qi (dx) d d R ZR Z ∞ = (sα−2 ∧ 1)Qi (ds|u). S d−1

0

holds. Therefore, the tempering function Z ∞ 2 2 qi (r, u) = e−r s /2 Qi (ds|u) 0

is well defined. Since R1 (A) 6= R2 (A) also Q1 (A) 6= Q2 (A). By assumption, Ri verifies (2.10). Then, by using the same calculus done to obtain (2.11), we can find M (A) and write Z Z ∞ IA (ru)r−α−1 (q1 (r, u) − q2 (r, u))drσ(du) = 0 S d−1

0

and we find the contradiction. A similar argument shows the uniqueness of the measure R also in the case α = 0. Remark 2.6. The case α = 0 is consider only for completeness and the theory will be not completely extended to this limiting case. It may be an interesting case in some applications. 8

Remark 2.7. If alpha ∈ (0, 2) and R satisfies the following additional inequality Z kxkα R(dx) < ∞, 0 < α < 2, (2.12) Rd

we will call µ a proper TID distribution. The measure Q has the form Z d kxkα R(dx), 0 < α < 2, Q(R ) =

(2.13)

Rd

In this case Q is a finite measure and it can be represented in polar coordinates as Q(dr, du) = Q(dr|u)σ(du), where Q(·|u) are finite measures and σ is a finite measure on S d−1 . Definition 2.8. The unique measure R in (2.9) is called a spectral measure of the corresponding TID distribution. We will call R the Rosi´ nski measure [14]. We focus on the following result. Remark 2.9. If µ is a proper TID distribution, then Q is a finite measure and {Q(·|u)}u∈S d−1 is a measurable family of finite Borel measures on (0, ∞). Since the equation q(0+, u) = 1 holds, they are probability measures. Furthermore, for any fixed u ∈ S d−1 , function q(·, u) are positive definite radial functions. Taking into consideration Lemma 2.14 of [10], we want to figure out the relation between parameters of the proper TID distribution and stable ones. Proposition 2.10. Let M be a L´evy measure of a proper TID distribution, as in (2.3), with corresponding spectral measure R. Then, the L´evy measure M0 of an α-stable distribution, given in (2.1), can be written in the following form Z Z ∞ M0 (A) = IA (tx)t−α−1 dtR(dx), A ∈ B(Rd ) (2.14) Rd

0

and additionally 

Z σ(B) =

IB Rd

x kxk



kxkα R(dx),

B ∈ B(S d−1 ).

(2.15)

Proof. By definitions (2.6) and (2.5) we obtain for A ∈ B(Rd ) Z Z ∞ Z Z ∞ x −α−1 IA (t IA (tx)t dtR(dx) = )t−α−1 kxkα dtQ(dx) 2 kxk d d R 0 ZR Z0 ∞ x IA (s = )s−α−1 kxkα dsQ(dx) 2 kxk d ZR 0Z ∞ = IA (su)s−α−1 dsσ(du) S d−1

0

= M0 (A). Since we are considering a proper TID distribution, then, by remark 2.9, Q(·|u) are finite measures on (0, ∞). Therefore we can write     Z Z x x α IB kxk R(dx) = IB Q(dx) kxk kxk Rd Rd Z Z ∞ = Q(ds|u)σ(du) = cσ(B). B

0

Since Q(ds|u) are probability measures, then c = 1 and (2.15) holds. 9

2.1

Distributional properties

A TID distribution may have moments and also exponential moments of any order. The behavior of the tails depends on the measure R. Proposition 2.11. Let µ be a TID distribution with L´evy measure M given by (2.9) and α ∈ (0, 2). Then (a) For p ∈ (0, α) Z

kxkp µ(dx) < ∞;

Rd

(b)

R Rd

kxkα µ(dx) < ∞ ⇐⇒

R kxk>1

kxkα log(kxk)R(dx) < ∞;

(c) If p > α, then Z

Z

p

kxkp R(dx) < ∞;

kxk µ(dx) < ∞ ⇐⇒ Rd

kxk>1

(d) For each θ > 0, we have Z Z θkxk e µ(dx) < ∞ ⇐⇒ Rd

kxk−(α+1) e

θ 2 kxk2 2

R(dx) < ∞.

kxk>1

(e) If α = 0 and p > 0, then Z Z p kxk µ(dx) < ∞ ⇐⇒ Rd

kxkp R(dx) < ∞;

kxk>1

Proof. It is well known that moments conditions for µ are related to the corresponding conditions for M|{kxk>1} , see [12]. Let we consider p > 0. Then we obtain Z Z Z ∞ 2 p p tp−α−1 e−t /2 dtR(dx) kxk M (dx) = kxk kxk>1

1 kxk

kxk≤1

Z kxk

+ kxk>1

p

Z

∞

2 /2

tp−α−1 e−t

dtR(dx)

1 kxk

= I (1) (x) + I (2) (x). By (2.10), the following inequality holds Z Z ∞ Z C (1) −3 I (x) ≤ C t dtR(dx) ≤ kxk2 R(dx) < ∞ 1 2 kxk≤1 kxk kxk≤1 2

(2.16)

where C := supt≥1 tp+2−α e−t /2 . The inequality (2.16) shows that the integral I (1) (x) is always finite. If p < α, then Z Z ∞ Z 1 (2) p p−α−1 I (x) ≤ kxk t dtR(dx) = kxkα R(dx) < ∞, 1 α − p kxk>1 kxk>1 kxk 10

by inequality (2.10), condition (a) is fulfilled. If p = α, we have Z Z 1 Z Z ∞ 2 (2) α −1 α e−t /2 dtR(dx) t dtR(dx) + kxk I (x) ≤ kxk 1 kxk

kxk>1

Z

kxk>1

1

kxkα (log(kxk) + C1 )R(dx)

= kxk>1

where C1 is a finite constant, and from the other side, Z (2) −1/2 I (x) ≥ e kxkα log(kxk)R(dx). kxk>1

Therefore condition (b) is satisfied. Now, we suppose p > α. Let us define Z ∞ 2 ¯ tp−α−1 e−t /2 dt. C= 1

Then, the following inequality holds I

(2)

(x) ≥ C¯

Z

kxkp R(dx)

kxk>1

and furthermore, I

(2)

Z (x) ≤

kxk

p

Z

kxk>1

∞

2 /2

tp−α−1 e−t

dtR(dx).

0

By changing variables in the integral, we have Z ∞ Z ∞ p − α p−α−1 −t2 /2 (p−α)/2−1 t e dt = 2 z (p−α)/2−1 e−z dz = 2(p−α)/2−1 Γ , 2 0 0 thus I

(2)

(p−α)/2−1

(x) ≤ 2

p − α Z Γ kxkp R(dx). 2 kxk>1

This proves (c). In order to prove (d), we consider the integral Z Z Z ∞ 2 θkxk e M (dx) = eθtkxk t−(α+1) e−t /2 dtR(dx) Rd

kxk>1

and we define

Z

1 kxk

∞

2 /2

eθtkxk t−(α+1) e−t

Iθ (x) :=

dt.

1 kxk

It is easy to check that as kxk → 0, then Iθ (x) goes to 0 exponentially fast. Now, let us consider the case kxk → ∞. We have Z ∞ 2 θ2 kxk2 /2 Iθ (x) = e t−(α+1) e−(t−θkxk) /2 dt. 1 kxk

11

Define Kθ (x) by Z

∞

t−(α+1) e−(t−θkxk)

Kθ (x) :=

2 /2

dt,

1 kxk

by changing variables in the integral we obtain Z ∞ 2 Kθ (x) = (t + θkxk)−(α+1) e−t /2 dt 1 −θkxk kxk

Z

−

θkxk 2

−(α+1) −t2 /2

(t + θkxk)

=

e

Z

∞

2 /2

(t + θkxk)−(α+1) e−t

dt +

dt

θkxk − 2

1 −θkxk kxk

(1)

(2)

= Kθ (x) + Kθ (x). Furthermore, the following inequality is satisfied −

Z

−θ2 kxk2 /8

(1) Kθ (x)

≤ e

θkxk 2

(t + θkxk)−(α+1) dt

1 −θkxk kxk

≤ Ckxkα+2 e−θ

2 kxk2 /8

.

and for kxk → ∞, (2) Kθ (x)

Z

∞

≤

2 /2

(t + θkxk)−(α+1) e−t

dt

∞

√

θkxk − 2

−(α+1)

Z

2 /2

e−t

∼ (θkxk)

dt =

2π(θkxk)−(α+1) ,

−∞

It follows that, for kxk → ∞, Kθ (x) ∼ and Iθ (x) ∼

√

√

2π(θkxk)−(α+1) ,

2π(θkxk)−(α+1) eθ

2 kxk2 /2

,

therefore condition (d) holds. Part (e) can be proved with a similar argument to (c). Remark 2.12. If the measure R has a bounded support, then E(eθkXk ) < ∞ for all θ > 0. We have exponential moments of any order. As we said before, sometimes it is more convenient to work with the measure R; in order to find some distributional property of a TID distribution. Taking into account Proposition 2.8 of [10], we will show a result about finite variation. Proposition 2.13. Let M be the L´evy measure of a TID distribution, and R a measure as in (2.9) and (2.6). These conditions are equivalent R (i) kxk≤1 kxkM (dx) < ∞ R (ii) kxk≤1 kxkR(dx) < ∞ and α ∈ (0, 1) 12

Proof. Suppose condition (i) is fulfilled. Choose r ≥ 1 such that R({kxk ≤ r}) 6= 0. We can write the following relations Z Z 1/kxk Z 2 kxkM (dx) = kxk t−α e−t /2 dtR(dx) Rd

kxk≤1

0

Z

Z

≥

1/kxk

kxk≤r

2 /2

t−α e−t

kxk

dtR(dx)

0

− 21 (α+1)

Z

Z

kxk≤r

1

z − 2 (α+1) e−z dzR(dx)

kxk

= 2

− 12 (α+1)

1/(2kxk2 )

0

Z

1/(2r 2 )

Z

kxk≤r

1

z − 2 (α+1) e−z dz.

kxkR(dx)

≥ 2

0

By condition (i), we obtain α < 1 and Z kxkR(dx) < ∞. kxk≤1

Conversely, if (ii) holds, then Z kxkM (dx) = kxk≤1

Z

Z

1/kxk

kxk

= kxk≤1

Z

Z

−α −t2 /2

t

e

kxk>1 ∞

1/kxk

kxk

dtR(dx) +

0

Z

Z

2 /2

t−α e−t

dtR(dx)

0

Z 1 ≤ kxk t e dtR(dx) + kxkα R(dx) 1 − α kxk≤1 0 kxk>1 Z Z α 1 1 1 α = 2− 2 − 2 Γ( − ) kxkR(dx) + kxkα R(dx) < ∞, 2 2 kxk≤1 1 − α kxk>1 −α −t2 /2

which proves the converse.

2.2

Characteristic function of a TID distribution

It is well known that given a L´evy measure of a infinitely divisible distribution, we have an explicit formula for the characteristic function, see [12]. Sometimes, working with a L´evy measure of the form (2.3) may be difficult and, as a consequence, we will provide an expression for the characteristic function of a TID distribution with respect to the measure R. The measure R allows one to obtain explicit analytic formulas and more explicit calculations. Let us now define functions Z ∞ 2 ψα (s) = (eist − 1 − ist)t−α−1 e−t /2 dt, (2.17) 0

and ψα0 (s)

Z =

∞

2 /2

(eist − 1)t−α−1 e−t

0

13

dt.

(2.18)

In order to find a more useful form for the characteristic function of a TID distribution, we will need the following results. Lemma 2.14. The following limits are verified α

lims→0 s−2 ψα (s) = −2−p2 −1 Γ(1 − α2 ), lims→∞ s−1 ψ0 (s) = −i π2 , 1 α lims→∞ s−1 ψα (s) = −2− 2 − 2 Γ( 21 − α2 )i, lims→∞ (s−1 ψ1 (s) + i log s) = − π2 + i, π lims→∞ s−α ψα (s) = Γ(−α)e−iα 2 ,

α ∈ (0, 2) α=0 α ∈ (0, 1) α=1 α ∈ (1, 2)

(2.19)

Furthermore, if α ∈ (0, 1) we have α

1

lims→∞ s−1 ψα0 (s) = 2− 2 − 2 Γ( 12 − α2 )i, π lims→∞ s−α ψα0 (s) = Γ(−α)e−iα 2 ,

(2.20)

Then, there exists for each α a finite positive constant Cα such that for all s ∈ R the following inequalities are fulfilled Cα−1 (s2 ∧ |s|α∨1 ) ∧ |s|(1 + log+ |s|)] Cα−1 (s2 ∧ |s|α ) −1 C0 [1 + log(1 + s)]

C1−1 [s2

≤ |ψα (s)| ≤ ≤ |ψ1 (s)| ≤ ≤ |ψα0 (s)| ≤ ≤ |ψ0 (s)| ≤

Cα (s2 ∧ |s|α∨0,1 ), α 6= 1, C1 [s2 ∧ |s|(1 + log+ |s|)], α = 1, 2 α Cα (s ∧ |s| ), α ∈ (0, 1). C0 [1 + log(1 + s)], α = 0. (2.21)

Proof. By solving the limit and using [12, Lemma 14.11], (2.19) and (2.20) are verified. Lemma 2.15. Let us consider the confluent equation x

d2 y dy + (c − x) − ax = 0. 2 dx dx

(2.22)

Then the solution of this differential equation is y = AM (a, c; z) + BU (a, c; z) where A and B are constant and M (a, c; z) is the Kummer’s or confluent hypergeometric function of first kind [1, 13.1.2] and U (a, c; z) is the confluent hypergeometric function of second kind [1, 13.1.3]. Proof. For a complete overview on confluent hypergeometric function see [15] or [1]. Lemma 2.16. Let α ∈ (0, 2), α 6= 1. Then we have       ∞ X α α 1  z 2 zn 1 Γ (n − α) = Γ − M − , ; n! 2 2 2 2 2 n=0     1−α 1 α 3  z 2 M − , ; + zΓ 2 2 2 2 2 14

(2.23)

Proof. Since the series converges, we can split it in two parts   X ∞ ∞ X zn 1 z 2n  α Γ (n − α) = Γ n− n! 2 (2n)! 2 n=0 n=0   ∞ X z 2n+1 1−α + Γ n+ . (2n + 1)! 2 n=0 By the Legendre duplication formula [1, 6.1.18], we obtain the following equalities   1 2n , (2n)! = n!2 2 n   3 2n , (2n + 1)! = n!2 2 n and we define the Pochhammer’s symbols as (a)n = Γ(n + a)/Γ(a) [1, 6.1.22]. Thus, we obtain   ∞ X n−α zn Γ = n! 2 n=0

  ∞ ∞  α X z 2n − α2 n 1 − α X z 2n 1−α 2
+ zΓ
n =Γ − 3 2n 2 n=0 n!22n 12 n 2 n!2 2 n n=0      α   α 1  z 2  1−α 1 − α 3  z 2 M − , ; , ; =Γ − + zΓ M . 2 2 2 2 2 2 2 2

Theorem 2.17. (Characteristic function) Let µ be a TID distribution with L´evy measure given by (2.9), α ∈ [0, 2) and α 6= 1. If the distribution has finite mean, R i.e. Rd kxkµ(dx) < ∞, then nZ o µ ˆ(y) = exp ψα (hy, xi)R(dx) + ihy, mi (2.24) Rd

where √     α α 1  i 2s 2 ψα (s) = 2 Γ − M − , ; 2 2 2 2 √     √ 1−α 1 α 3  i 2s 2 M − , ; + i 2sΓ 2 2 2 2 2    ! √ 1 α α − i 2sΓ − −Γ − . 2 2 2 −α −1 2

(2.25)

R and m = Rd xµ(dx). Furthermore, if 0 < α < 1, the characteristic function can be written in an alternative form nZ o 0 µ ˆ(y) = exp ψα (hy, xi)R(dx) + ihy, m0 i (2.26) Rd

15

where

√     α 1  i 2s 2 α ψα (s) = 2 M − , ; Γ − 2 2 2 2 √     !   2 √ 1 α 3 i 2s α 1−α M − , ; −Γ − . + i 2sΓ 2 2 2 2 2 2 −α −1 2

(2.27)

Proof. First, integrals (2.24) and (2.26) are well defined due to conditions (2.10) and (2.21) of Lemma 2.14. It is well R known that if the mean is finite, that is if the first absolute moment exists, i.e. Rd kxkµ(dx) < ∞, then µ ˆ can be written as Z  ihy,xi µ ˆ = exp (e − 1 − ihy, xi)ν(dx) + ihy, mi Rd

R

xµ(dx). By (2.9), we obtain the equality (2.24), where, if α ∈ [0, 2) Z ∞ 2 (eist − 1 − ist)t−α−1 e−t /2 dt, (2.28) ψα (s) = 0 R R If α ∈ [0, 1) and kxk≤1 kxkR(dx) < ∞, by Proposition 2.13 kxk≤1 kxkν(dx) < ∞, in which case µ ˆ can be written as  Z ihy,xi (e − 1)ν(dx) + ihy, m0 i , exp where m =

Rd

Rd

where m0 is the drift as defined in [12]. By (2.9), we obtain the equality (2.26), where Z ∞ 2 0 (eist − 1)t−α−1 e−t /2 dt, (2.29) ψα (s) = 0

and, furthermore, the equality ψα (s) =

ψα0 (s)

Z − is

∞

2 /2

t−α e−t

dt

(2.30)

0

holds. Now we will prove (2.25) and (2.27). If α ∈ (0, 1), we obtain by equality (2.23) Z ∞ 2 (eist − 1)t−α−1 e−t /2 dt = 0

=

Z ∞ X (is)n n=1 ∞ X

n!

∞

2 /2

tn−α−1 e−t

dt

0

  (is)n 1 (n−α−2) 1 = 22 Γ (n − α) n! 2 n=1 √ n   ∞ X (i 2s) 1 −α −1 =2 2 Γ (n − α) n! 2 n=1 √     α α 1  i 2s 2 −α −1 =2 2 Γ − M − , ; 2 2 2 2 √      ! √ 1−α 1 α 3  i 2s 2 α + i 2sΓ M − , ; −Γ − . 2 2 2 2 2 2 16

With a similar calculus, if α ∈ (0, 2) and α 6= 1, we obtain Z ∞ 2 (eist − 1 − ist)t−α−1 e−t /2 dt = 0

=

Z ∞ X (is)n n=2

n!

∞ X (is)n

∞

2 /2

tn−α−1 e−t

dt

0

  1 = 2 Γ (n − α) n! 2 n=2 √   ∞ X (i 2s)n 1 −α −1 =2 2 Γ (n − α) n! 2 n=2 √     α 1  i 2s 2 α −α −1 =2 2 M − , ; Γ − 2 2 2 2 √     √ 1−α 1 α 3  i 2s 2 + i 2sΓ M − , ; 2 2 2 2 2    ! √ 1 α α − 2isΓ − −Γ − 2 2 2 1 (n−α−2) 2

Remark 2.18. With a similar technique, the characteristic exponent also can be calculated also for both cases α = 0 and α = 1. Definition 2.19. We will write X ∼ T IDα (R, m) to indicate that X is a TID random variable with characteristic function (2.24) and X ∼ T IDα0 (R, m0 ) to indicate that X is a TID random variable with characteristic function (2.26). The constant m is exactly the mean m = E[X].

3

TID processes

In this section, we will introduce TID processes. By Definition 2.3, if µ is a TID distribution, it is infinitely divisible and therefore there exists a L´evy process {X(t) : t ≥ 0} such that µ is the distribution of X(1).

3.1

Short and long time behavior

The following theorems will show the different behavior of a TID process for different time scale. If one decreases the time scale, a TID process looks like a stable process; otherwise, if one increases the time scale, it looks like a Brownian motion. To figure out this different time behavior, we consider the time rescaled process {Xh (t) : t ≥ 0} = {X(ht) : t ≥ 0}, (3.1) where h > 0.

17

Theorem 3.1. (Short time behavior) Let {X(t) : t ≥ 0} be a TID L´evy process in Rd such that the distribution of X(1) has spectral measure R. (a) Let us consider a TID process with X(1) ∼ T IDα0 (R, 0), if α ∈ (0, 1) and with X(1) ∼ T IDα (R, 0), if α ∈ (1, 2). Assume that Z kxkα R(dx) < ∞ (3.2) Rd

and let σ be the finite measure on S d−1 defined in (2.15). Then d

h−1/α Xh → Y,

(3.3)

as h → 0, where {Y (t) : t ≥ 0} is a strictly α-stable L´evy process with Y (1) = Sα (σ, 0). (b) Let us consider a TID process with X(1) ∼ T IDα (R, 0), if α = 1. Assume that Z kxk |log kxk| R(dx) < ∞. (3.4) Rd

Then

d

h−1/α Xh − ah → Y, where

Z ah (t) = t log h

xR(dx), Rd

and {Y (t) : t ≥ 0} is an α-stable L´evy process with Y (1) ∼ S1 (σ, b) with Z b= x(1 − log kxk)R(dx). Rd

Proof. Since {h−1/α Xh (t) : t ≥ 0} is a L´evy process, by [5, Theorem 13.17], it is enough to show the convergence in distribution of h−1/α Xh (1) to Y (1). By a Paul L´evy theorem (also called the continuity theorem) [2, Theorem 2, p.508], the convergence in distribution can be proved by considering the pointwise convergence of the respective characteristic functions. First, we want to prove (a). If α(0, 1), we obtain E[eihy,h

−1/α X (1) h

−1/α y,X(h)

] = E[eihh Z = exp

Rd

] 

hψα0 (h−1/α hy, xi)R(dx)

,

(3.5)

The upper bounds (2.21) of Lemma 2.14 and condition (3.2) allow one to apply the dominated convergence theorem to the above integral. By definitions (2.25) and (2.27) it easy to check that ψα0 (−s) = ψα0 (s) and ψα (−s) = ψα (s). Now, by (2.20) and [12, Theorem 14.10], we calculate the limit h → 0 under the integral (3.5) n απ o 0 −1/α α lim hψα (h hy, xi) = Γ(−α)|hy, xi| exp −i sgnhy, xi h→0 2  απ απ α = Γ(−α) cos |hy, xi| 1 − i tan sgnhy, xi . 2 2 18

Therefore, under the assumption α ∈ (0, 1), (3.3) holds. A similar argument proves (a) also in the case α ∈ (1, 2). Let us consider the case α = 1. By definition of ah , the equality Z  −1 −1 E[exp{ihy, h Xh (1) − ah (1)i}] = exp (hψ1 (h hy, xi) − ihy, xi log h)R(dx) . Rd

(3.6)

is fulfilled, then, by assumption (3.4) and [10, Theorem 3.1], (b) holds. Now, we will prove that if one increases the time scale, a TID process looks like a Brownian motion. Theorem 3.2. (Long time behavior) Let {X(t) : t ≥ 0} be a TID L´evy process in Rd such that the distribution of X(1) ∼ T IDα (R, 0) and α ∈ (0, 2). Assume that Z kxk2 R(dx) < ∞. (3.7) Rd

Then

1

d

h− 2 Xh → B, as h → ∞, where {B(t) : t ≥ 0} is a Brownian motion with characteristic function   Z α −1 2 ihy,B(t)i −α hy, xi R(dx) . (3.8) E[e ] = exp −t2 2 Γ(1 − ) 2 Rd 1

Proof. Since {h− 2 Xh (t) : t ≥ 0} is a L´evy process, by [5, Theorem 13.17], it is 1 enough to show the convergence in distribution of h− 2 Xh (1) to B(1). By the continuity theorem [2, Theorem 2, p.508], the convergence in distribution can be proved by considering the pointwise convergence of the respective characteristic functions. By considering equality (2.24), we can write E[eihy,h

1 −2

Xh (1)

−1

] = E[eihh 2 y,X(h) ]  Z − 12 hψ(h hy, xi)R(dx) . = exp Rd

The upper bounds (2.21) and condition (3.7) allow one to apply the dominated convergence theorem to the above integral and by considering (2.19) we obtain 1 α α lim hψ(h− 2 hy, xi) = −2− 2 −1 Γ(1 − )hy, xi2 , h→∞ 2 which verifies (3.8).

3.2

Change of measure

In this section, a result on density transformations between stable and TID processes is considered. Theorem 3.3. Let P0 and P be probability measures on (Ω, F) such that the canonical process {X(t) : t ≥ 0} is a L´evy α-stable process Sα (σ, a) under P0 , while it is a proper TID process T IDα (R, b) under P , where σ is given by equation (2.15). Then 19

(i) P0|Ft and P|Ft are mutually absolutely continuous for every t > 0 if and only if Z

Z

S d−1

1



2 1 − q(r, u) r−α−1 drσ(du) < ∞,

(3.9)

0

and  0 0 if and only if Z  2 φ(x) −1 2 e − 1 M0 (dx) < ∞ (3.13) Rd

20

and Bα = 0,

(3.14)

where Bα is defined by  R R R  xM0 (dx)) − kxk≤1 x(M − M0 )(dx), 0 < α < 1,  b + Rkxk≤1 xM (dx) − (a + kxk≤1 R R b − kxk>1 xM (dx) − (a − c S d−1 uσ(du)) − kxk≤1 x(M − M0 )(dx), α = 1, Bα = R R   b−R xM (dx) − (a − kxk>1 xM0 (dx)) − kxk≤1 x(M − M0 )(dx), 1 < α < 2. kxk>1 In the case α = 1, c = 1 − γ, where γ is the Euler constant. The inequality (3.13) can be written as 2 Z   x  1/2 1−q kxk, M0 (dx) < ∞ (3.15) kxk Rd Since the integrand is bounded and M0 is a L´evy measure, we may focus our attention only on integration over {kxk ≤ 1}. Applying elementary inequalities 1 √ (1 − y)2 ≤ (1 − y)2 ≤ (1 − y)2 4 for y ∈ [0, 1], inequality (3.15) becomes Z {kxk≤1}



x  1 − q kxk, kxk 

2 M0 (dx),

and writing the above integral in polar coordinates, we obtain (3.9). Now, we will prove the equivalence between conditions (3.10) and (3.14). By finiteness of the integral above and H¨older inequality, we have Z Z   x  M0 (dx) kxk(M0 − M )(dx) = kxk 1 − q kxk, kxk kxk≤1 kxk≤1 1/2  Z Z 1/2   x 2 2 1 − q kxk, ≤ kxk M0 (dx) M0 (dx) 1 kxk>1 and, since we are considering a proper TID process  x  q kxk, ≤ 1, kxk and M0 is the L´evy measure of an α-stable distribution with 1 < α < 2, we obtain Z Z kxkM (dx) ≤ kxkM0 (dx) < ∞. kxk>1

kxk>1

21

Furthermore, by (3.16) Z kxk(M0 − M )(dx) < ∞. Rd

By using (2.9) and (2.14) and integrating by parts, the following result is obtained Z Z Z ∞ 2 kxkt−α (1 − e−t /2 )dtR(dx) kxk(M0 − M )(dx) = Rd Rd 0 1 α Z −(1+α)/2 = −2 Γ − kxkR(dx) < ∞. 2 2 Rd With a similar calculus, we can write Z Z 1 α −(1+α)/2 xR(dx) + b − a = 0, Bα = x(M0 − M )(dx) + b − a = −2 Γ( − ) 2 2 Rd Rd where the last equality follows by (3.10), proving (3.14). It remains to verify (3.14) in the case α = 1. By taking into account (3.16), Z Z Z 1/kxk 2 ∞> kxk(M0 − M )(dx) = kxk t−1 (1 − e−t /2 )dtR(dx) Rd

kxk≤1

≥

1 4

Z

Z kxk kxk≤1

1 kxk

0

1 4

t−1 dtR(dx) =

1

Z kxk| log kxkR(dx). kxk≤1

Since the last integral is finite, the integral in (3.10) is well defined and we can calculate Z Z Z 1/kxk 2 x(M0 − M )(dx) = x t−1 (1 − e−t /2 )dtR(dx) kxk≤1 Rd 0 Z   1   1 = x E1 − 2 log kxk − log 2 + γ R(dx) 2 Rd 2kxk2 by changing variable and equation 5.1.39 in [1], where the function E1 (x) is the exponential integral function defines by Z ∞ E1 (x) = t−1 e−t . x

By part (b) of Proposition 2.11, the first moment is finite, thus Z xM (dx) < ∞ kxk>1

and Z

Z

Z

2 /2

t−1 e−t

x

xM (dx) = kxk>1

∞

Rd

1 = 2

dtR(dx)

1/kxk

Z xE1 Rd

22



1  R(dx). 2kxk2

Adding together the above results, we have Z Z  1  1 B1 = b − xE1 R(dx) − a + (1 − γ) uσ(du) 2 Rd 2kxk2 S d−1 Z   1   1 + x E1 − 2 log kxk − log 2 + γ R(dx) 2 Rd 2kxk2 Z Z Z  γ log 2  R(dx) + = b − a + (1 − γ) xR(dx) − x log kxk + xR(dx) = 0 2 2 Rd Rd Rd By considering the remark in [12, Notes page 236], we can complete the proof of part (i). Indeed, since M and M0 are mutually absolutely continuous by (3.12), P0|Ft and P|Ft are mutually absolutely continuous or singular for all t > 0. Part (ii) is an application of Theorem 33.2 of [12], where the form of RadonNikodym derivative is specified for two mutually absolutely continuous L´evy processes.

4

Simulation of proper TID laws and processes

There are different methods to simulate L´evy processes, but most of these methods are not suitable for the simulation of TID processes due to the complicated structure of their L´evy measure. The usual method of the inverse of the L´evy measure is difficult to implement, even if the spectral measure R has a simple form, readers are referred to [10]. To overcome this problem, we will find a shot noise representation for proper TID distributions, and consequently also TID processes, without constructing any inverse. The representation, we will show, is based on results in [9] and [10]. Let M be the L´evy measure of a proper TID distribution on Rd , given by (2.3), and Q and R corresponding measures defined in (2.5) and (2.6). Let us define kσk as kσk := σ(S d−1 ), (4.1) and by equalities (2.13) and (2.15), we obtain Z d kσk = Q(R ) = kxkα R(dx) < ∞. Rd

Let {vj } be an i.i.d. sequence of random vector in Rd with distribution Q/kσk. Let {uj } be an i.i.d. sequence of uniform random variables on (0, 1) and let {ej } and {e0j } be i.i.d. sequences of exponential random variables with parameters 1. Furthermore, we assume that {vj }, {uj }, {ej } and {e0j } are independent. We consider γj = e01 + . . . + e0j and, by definition of {e0j }, {γj } is a Poisson point process on (0, ∞) with Lebesgue intensity measure. Now, we will prove a useful lemma. Lemma 4.1. Under the above definitions, let us define the function  H(γj , (vj , ej , uj )) :=

αγj kσk

−1/α ∧

23

√

1/2 1/α 2ej uj kvj k−1



vj . kvj k

(4.2)

Then, for every non-empty set A ∈ B(Rd ), the equality Z ∞ P (H(s, (v1 , e1 , u1 )) ∈ A)ds = M (A) 0

is verified. Proof. Let A be a set of the form   x d A = x ∈ R : kxk > a, ∈B , kxk where a > 0 and B ∈ B(S d−1 ). Then, we can write Z ∞ P (H(s, (v1 , e1 , u1 )) ∈ A)ds = 0

Z

∞

 P

= 0

αs kσk

−1/α ∧

√



 v1 ∈ A ds, kv1 k  v1 > akv1 k, ∈ B ds kv1 k  ∈B

1/2 1/α 2e1 u1 kv1 k−1

−1/α √ 1/2 1/α αs I =E > a, 2e1 u1 kσk 0  −α √ 1/2 1/α kσka v1 = EI 2e1 u1 > akv1 k, α kv1 k  −α Z Z ∞ √ a 1/2 1/α = P 2e1 u1 > as Q(ds|u)σ(du). α B 0 Z

∞



By conditioning, the probability in the integral can be calculated Z 1Z ∞  √ 1/2 1/α e−x dxdu 2e1 u1 > as = P 2 2 a s 2u2/α

0

1

Z =

−

e 0 Z α =a α

a2 s2 2u2/α

du

∞

e−r

2 s2 /2

r−α−1 dr,

a

therefore we obtain Z ∞ Z Z ∞Z ∞ 2 2 P (H(s, (v1 , e1 , u1 )) ∈ A)ds = e−r s /2 r−α−1 drQ(ds|u)σ(du) 0 ZB Z0 ∞ a = q(r, u)r−α−1 drσ(du) = M (A). B

a

First, we consider a simple case. Theorem 4.2. (α ∈ (0, 1) and symmetric case) Suppose that all the above assumption are fulfilled. If α ∈ (0, 1), or if α ∈ [1, 2) and Q is symmetric, the series  −1/α ∞  X √ 1/2 1/α vj αγj −1 S0 = ∧ 2ej uj kvj k . (4.3) kσk kv k j j=1 converges a.s.. Furthemore, we have that S0 ∼ T IDα0 (R, 0) for α ∈ (0, 1) and S0 ∼ T IDα (R, 0) for α ∈ [1, 2). 24

Proof. To prove this theorem, we are going to use [9, Theorem 4.1] and [10, Theorem 5.1]. If H is defined as in (4.2), we can apply Lemma 4.1. Let us consider the case α ∈ (0, 1), then by Proposition 2.13 we can write Z Z ∞ kxkM (dx) < ∞ E(kH(s, (v1 , e1 , u1 ))kI(kH(s, (v1 , e1 , u1 ))k ≤ 1))ds = kxk≤1

0

and [9, Theorem 4.1(A)] proves the theorem in the case α ∈ (0, 1). If α ∈ [1, 2), then by Proposition 2.11 we have Z Z ∞ E(kH(s, (v1 , e1 , u1 ))kI(kH(s, (v1 , e1 , u1 ))k > 1))ds = kxkM (dx) < ∞ kxk>1

0

and by [9, Theorem 4.1(B)], we can consider a series  −1/α  ∞  X √ 1/2 1/α vj αγj −1 ¯ S0 = ∧ 2ej uj kvj k − cj kσk kv k j j=1 which converges a.s. and S¯0 ∼ T IDα (R, 0), where −1/α   Z j  √ 1/2 1/α αγj vj −1 cj = E ∧ 2ej uj kvj k ds. kσk kvj k j−1

(4.4)

If Q is symmetric, cj is equal to zero. It follows that S0 = S¯0 . This completes the proof. Now we consider the non-symmetric case. Theorem 4.3. (Non-symmetric case) Under the above notation, suppose α ∈ [1, 2), Q is non-symmetric and additionally that Z kxk| log kxk|R(dx) < ∞ (4.5) Rd

when α = 1 and that

Z kxkR(dx) < ∞

(4.6)

Rd

when α ∈ (1, 2). Then, the series −1/α   −1/α  ∞  X √ 1/2 1/α αγj vj αj −1 S1 = ∧ 2ej uj kvj k − x0 + b kσk kvj k kσk j=1

(4.7)

where vj x0 = E = kσk−1 kvj k Z x1 = xR(dx),

Z uσ(du), S d−1

Rd

 b=


1 α Γ x1 , 1 < α < 2, ζ α α−1/α kσk1/α x0 − 2−(1+α)/2 − 2 2 R log 2 3 ( 2 γ − 2 + log kσk)x1 − Rd x log kxkR(dx), α = 1,
1

(4.8)

ζ denotes the Riemann zeta function and γ is the Euler constant, converges a.s.. Furthermore, we have that S1 ∼ T IDα (R, 0). 25

Proof. To prove this theorem, we are going to use [9, Theorem 4.1] and [10, Theorem 5.1]. If H is defined as in (4.2), we can apply Lemma 4.1. If α ∈ [1, 2), then by Proposition 2.11 we have Z Z ∞ kxkM (dx) < ∞ E(kH(s, (v1 , e1 , u1 ))kI(kH(s, (v1 , e1 , u1 ))k > 1))ds = kxk>1

0

and [9, Theorem 4.1(B)], we can consider a series   −1/α ∞  X √ 1/2 1/α vj αγj −1 S1 = ∧ 2ej uj kvj k − cj kσk kvj k j=1 which converges a.s. and S1 ∼ T IDα (R, 0), where −1/α   Z j  √ 1/2 1/α αγj vj −1 E cj = ∧ 2ej uj kvj k ds. kσk kvj k j−1 We have to prove that the equality −1/α  ∞  X αj x0 − cj = b kσk j=1

(4.9)

holds, where b is given by (4.8). First consider the case α ∈ (1, 2). Define for j ≥ 1 [10, equation (5.8)] −1/α  Z j  αs α1−1/α kσk1/α 1−1/α v1 0 [j − (j − 1)1−1/α ]x0 . (4.10) cj = E ds = kσk kv1 k α−1 j−1 We have kc0j

Z

j

− cj k ≤

 E

j−1

αs kσk

−1/α

 −

αs kσk

−1/α ∧

√

1/2 1/α 2e1 u1 kv1 k−1

 ds.

Furthermore by [10, equation 5.9], for every θ > 0 the equality −1/α  −1/α  Z ∞  αs αs kσk θ1−α (4.11) − ∧ θ ds = kσk kσk α(α − 1) 0 √ 1/2 1/α holds. Using this identity for θ = 2e1 u1 kv1 k−1 pointwise, we obtain −1/α  −1/α  Z ∞  ∞ X √ 1/2 1/α αs αs 0 −1 kcj − cj k = E − ∧ 2e1 u1 kv1 k ds kσk kσk 0 j=1 1 1  1  kσk (1−α) −1+ α E 2 2 (1−α) e12 u1 kv1 kα−1 α(α − 1)   1 3 α kσk (1−α) = 22 Γ − Ekv1 kα−1 2 2 α−1  Z 1 α −(1+α)/2 = −2 Γ − kxkR(dx) < ∞. 2 2 Rd

=

(4.12) 26

By using (4.11) we obtain   Z ∞  −1/α  −1/α  ∞ X √ 1/2 1/α αs v1 αs −1 0 ds − ∧ 2e1 u1 kv1 k (cj − cj ) = E kσk kσk kv1 k 0 j=1   1 1 1 v1 kσk (1−α) −1+ α =E 2 2 (1−α) e12 u1 kv1 kα−1 α(α − 1) kv1 k   Z 1 3 α 1 = 2 2 (1−α) Γ − xkxkα−2 Q(dx) 2 2 α − 1 Rd Z  1 α −(1+α)/2 − xR(dx) = −2 Γ 2 2 Rd   1 α −(1+α)/2 = −2 Γ − x1 2 2 Then we have −1/α   X n  n X αj α 1−1/α 0 −1/α n α−1/α kσk1/α x0 . x0 − cj = j − kσk α − 1 j=1 j=1 From a classical formula [1, 23.2.9], Z ∞ n X s − [s] n1−z −z = ζ(z) + z ds, j − 1−z sz+1 n j=1 we obtain

Re(z) > 0, Re(z) 6= 1,

(4.13)

−1/α    ∞  X αj 1 0 x0 − cj = ζ α−1/α kσk1/α x0 kσk α j=1

and we can write −1/α −1/α   X ∞  ∞  ∞ X X αj αj 0 x0 − cj = x0 − cj + (c0j − cj ) kσk kσk j=1 j=1 j=1     1 α 1 −1/α 1/α −(1+α)/2 α kσk x0 − 2 Γ − x1 = b =ζ α 2 2 which proves (4.9). Now, we consider the case α = 1. By the same computation above, define for j ≤ 2 −1/α  Z j  s v1 0 cj = E ds = (log j − log(j − 1))kσkx0 , (4.14) kσk kv1 k j−1 and put c01 = 0. For every θ > 0 [10, equation (5.14)], we have −1  −1  Z ∞  s s − ∧ θ ds kσk kσk 1 = {θ − kσk log θ + kσk log kσk − kσk}I(θ ≤ kσk)   kσk + ≤ kσk log . θ 27

(4.15)

By assumption (4.18), we can write −1  −1  Z ∞  ∞ X √ 1/2 s s 0 −1 kcj − cj k = E − ∧ 2e1 u1 kv1 k ds kσk kσk 1 j=1 ! kσkkv k 1 ≤ kσkE log+ √ 1/2 (4.16) 2e1 u1 √ 1/2 ≤ kσk(| log kσk| + E| log kv1 k| + E| log 2e1 u1 |) Z | log kxk|kxkR(dx) + Kkσk < ∞, = kσk| log kσk| + Rd

√ 1/2 where K = E| log 2e1 u1 | < ∞. P 0 Before computing the series ∞ j=1 (cj − cj ), we recall some useful relations [10]. For every θ > 0 −1 Z 1 s ∧ θds = θI(θ ≤ kσk) + {kσk − kσk log kσk + kσk log θ}I(θ > kσk) kθk 0 and by (4.15) we get −1 −1  −1  Z ∞  Z 1 s s s ∧ θ ds+ − ∧θ = kσk(log kσk−log θ−1). − kθk kθk kθk 1 0 √ 1/2 By using this formula for θ = 2e1 u1 kv1 k−1 we get −1   Z 1  ∞ X √ 1/2 s −1 0 ∧ 2e1 u1 kv1 k ds (cj − cj ) = E − kθk 0 j=1 −1  −1    Z ∞  √ 1/2 s s v1 −1 + − ∧ 2e1 u1 kv1 k ds kθk kθk kv1 k 1   √ 1/2 v1 = kσkE (log kσk + log kv1 k − log( 2e1 u1 ) − 1) . kv1 k The following expectation can be calculated √ 1/2 log 2 1 − 1 − γ, E log( 2e1 u1 ) = 2 2 R∞ where γ = − 0 log(x)e−x dx is the Euler constant, see [1, 6.1.3]. By equation (2.7), the series above can be rewritten as   ∞ X v1 1 log 2 0 (cj − cj ) = kσkE (log kσk + log kv1 k + γ − kv1 k 2 2 j=1 Z x  1 log 2  = log kσk + log kxk + γ − Q(dx) 2 2 Rd kxk Z  1 log 2  = x log kσk − log kxk + γ − R(dx) 2 2 Rd Z 1  log 2 = γ− + log kσk x1 − x log kxkR(dx). 2 2 Rd 28

By [10, Theorem 5.1], the equality −1  ∞  X j 0 x0 − cj = γx1 kσk j=1 holds, where γ is the Euler constant, thus we obtain −1  −1  X ∞  ∞ ∞  X X j j 0 x0 − cj = x0 − cj + (c0j − cj ) kσk kσk j=1 j=1 j=1 Z 3  log 2 x log kxkR(dx) = b, = γ− + log kσk x1 − 2 2 Rd which completes the proof. A series representation for TID processes can be obtained. Theorem 4.4. Under the above notation and assumptions, given a fixed T > 0, let {τj } be a i.i.d. sequence of uniform random variables in [0, T ]. Assume {τj } independent of the random sequences {vj }, {uj }, {ej } and {γj }. (i) If α ∈ (0, 1), or if α ∈ [1, 2) and Q is symmetric, set for every t ∈ [0, T ]   −1/α ∞ X √ 1/2 1/α vj αγj −1 , (4.17) X0 (t) = I(0,t] (τj ) ∧ 2ej uj kvj k T kσk kvj k j=1 then the series converges a.s. uniformly in t ∈ [0, T ] to a L´evy process such that X0 (t) ∼ T IDα0 (tR, 0) if α ∈ (0, 1) and X0 (t) ∼ T IDα0 (tR, 0) if α ∈ [1, 2). (ii) If α ∈ [1, 2), Q is non-symmetric and additionally Z kxk| log kxk|R(dx) < ∞

(4.18)

Rd

when α = 1 and that

Z kxkR(dx) < ∞

(4.19)

Rd

when α ∈ (1, 2), then, the series −1/α   ∞  X √ 1/2 1/α αγj vj −1 ∧ 2ej uj kvj k X1 (t) = I(0,t] (τj ) T kσk kvj k j=1  −1/α  t αj − x0 + tbT . T T kσk

(4.20)

where

ζ α1 α−1/α T −1 (T kσk)1/α x0 R− 2−(1+α)/2 Γ 12 − α2 x1 , 1 < α < 2, bT = ( 23 γ − log2 2 + log T kσk)x1 − Rd x log kxkR(dx), α = 1, (4.21) the series converges a.s. uniformly in t ∈ [0, T ] to a L´evy process such that X1 (t) ∼ T IDα (tR, 0). 

29

Proof. It is enough to show the convergence in distribution of series (4.17) and (4.20) for a fixed t, see [9, 10]. By the same arguments of Lemma 4.1, we obtain Z ∞ P (H(s, (v1 , e1 , u1 , τ1 )) ∈ A)ds = tM (A) 0

where we define   −1/α √ 1/2 1/α αγj vj −1 H(γj , (vj , ej , uj , τj )) := I(0,t] (τj ) ∧ 2ej uj kvj k . (4.22) kσk kvj k By following the proof of TheoremR 4.2, (i) is verified in the case α ∈ (0, 1). By Proposition 2.11 if α ∈ [1, 2), then kxk>1 kxkM (dx) < ∞. By [9, Theorem 4.1(B)] we can consider the series   −1/α  ∞  X √ 1/2 1/α v αγ j j −1 T ¯ 1 (t) = − aj X ∧ 2ej uj kvj k I(0,t] (τj ) kσk kvj k j=1 ¯ 1 (t) ∼ T IDα (tR, 0), where which converges a.s. and X aTj

  −1/α   √ 1/2 1/α αγj vj −1 E I(0,t] (τj ) = ∧ 2ej uj kvj k ds. kσk kvj k j−1 Z

j

If Q is symmetric then aTj = 0 and (i) is proved. To complete the proof, by following [10, Theorem 5.3] and equation (4.4), cj can be viewed as a function of the measure Q, thus we have aTj (t) =

t cj (T Q). T

By Theorem 4.3, where T Q and T R have to be considered instead of Q and R, we have −1/α  ∞  X αγj x0 − cj (T Q) T kσk j=1

 1 α ζ α1 α−1/α (T kσk)1/α x0 − 2−(1+α)/2 Γ − T x1 α − 1, 1 < α < 2, 2 2 R = log 2 3 ( 2 γ − 2 + log T kσk)T x1 − T Rd x log kxkR(dx), α = 1. By definition of aTj (t), we obtain −1/α  ∞   X t αγj T x0 − aj (t) = tbT , T T kσk j=1 which completes the proof. √ 1/2 1/α Remark 4.5. By removing the tempering part 2ej uj kvj k−1 in the shot noise representation, a well-known result for α-stable processes can be found, see [10, Theorem 5.4] or [11]. 30

5

Examples

A real TID law can be defined by fixing a positive definite radial function q with a measure σ on S 1 or alternatively by defining its spectral measure R. We are going to show in the following three parametric examples of TID laws in one dimension. In the first example, the measure R is the sum of two Dirac measures multiplied for opportune constants. The spectral measure R of the second example has a non-trivial bounded support and the derived TID distribution has exponential moments of any order. In the last example, the MTS distribution is considered, see [8, 6], the spectral measure is defined on an unbounded support and there exist exponential moments of some order.

5.1

Example 1: Simple TID distribution

The L´evy measure of the form 2

M (dx) = (C+ e−λ+ x

2 /2

2

1x>0 + C− e−λ− |x|

2 /2

1x0 + C− e−λ− |x| 31

2 /2

1x0 + C− 1x 0, α ∈ (0, 2), and α 6= 1. The tempering function q is of the form ( α+1 (λ+ r) 2 K α+1 (λ+ r), u = 1 2 q(r, u) = (5.8) α+1 (λ− r) 2 K α+1 (λ− r), u = −1, 2

and the measure σ is σ(1) = C+ ,

σ(−1) = C− . 33

Lemma 5.5. Let z > 0 and Kν (x) the modified Bessel function of second kind, then the equality Z ∞ √ ν 1 2z 2 Kν (2 z) = e−zt− t t−ν−1 dt (5.9) 0

is satisfied. Proof. By equality [3, 8.432(7)], we have Z pν ∞ − xt − xp2 −ν−1 e 2 2t t Kν (xp) = dt. 2 0 √ By setting x = 2z and p = 1/ z, then we can write ν Z √ z − 2 ∞ −zt− 1 −ν−1 tt Kν (2 z) = dt e 2 0

(5.10)

hence the equality (5.9) holds. Lemma 5.6. Let µ be a MTS distribution, then 2

2

Q(ds| ± 1) = e−λ± /2s s−α−2 λα+1 ± ds, and R(dx) =



Proof. By setting ν = t = 2s2 /λ2 , we have (λr)

α+1 2

− C+ λα+1 + e α+1 2

2 λ2 +x 2

Ix>0 +

− C− λα+1 − e

2 λ2 −x 2

(5.11) 

Ix0 − x