Short and Long Memory Fractional Ornstein–Uhlenbeck α-Stable Processes

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Short and Long Memory Fractional Ornstein–Uhlenbeckα-Stable ProcessesMarcin Magdziarz aa Institute of Mathematics and Computer Science , Wroclaw University of Technology ,Wroclaw, PolandPublished online: 16 Jul 2007.

To cite this article: Marcin Magdziarz (2007) Short and Long Memory Fractional Ornstein–Uhlenbeck α-Stable Processes,Stochastic Models, 23:3, 451-473, DOI: 10.1080/15326340701471091

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Stochastic Models, 23:451–473, 2007Copyright © Taylor & Francis Group, LLCISSN: 1532-6349 print/1532-4214 onlineDOI: 10.1080/15326340701471091

SHORT AND LONG MEMORY FRACTIONALORNSTEIN–UHLENBECK �-STABLE PROCESSES

Marcin Magdziarz � Institute of Mathematics and Computer Science,Wroclaw University of Technology, Wroclaw, Poland

� We show that, similar to the Gaussian case, the fractional Ornstein–Uhlenbeck �-stableprocess obtained via the Lamperti transformation of the linear fractional stable motion is adifferent stationary process than the one defined as the solution of the Langevin equation drivenby a linear fractional stable noise. We investigate the asymptotic dependence structure of thefirst process and prove that, in contrast to the second case, it is a short-memory process in thesense of the measure of dependence appropriate for processes with infinite second moment.

Keywords Lamperti transformation; Langevin equation; Long memory; Ornstein–Uhlenbeck process; Stable processes.

Mathematics Subject Classification Primary 60H10; Secondary 60G52.

1. INTRODUCTION

Linear fractional stable motion (LFSM) is an extension of the well-known fractional Brownian motion to the �-stable case. It is defined inthe following way[11]: Let 0 < � ≤ 2, 0 < H < 1, H �= 1/� and a, b ∈ �,|a| + |b|> 0. Then the process

L�,H (t) =∫ ∞

−∞

(a[(t − s)H−1/�

+ − (−s)H−1/�+

]+ b

[(t − s)H−1/�

− − (−s)H−1/�−

])L�(ds), t ∈ �, (1)

is called LFSM. Here x+ = max�x , 0�, x− = max�−x , 0� and L�(ds) is thestandard symmetric �-stable random measure on � with control measureas Lebesque measure (Refs.[5,11]). L�,H (t) is a self-similar, stationary-increment process. For � = 2 it reduces to the fractional Brownian motion.

Received July 2006; Accepted March 2007Address correspondence to Marcin Magdziarz, Institute of Mathematics and Computer Science,

Wroclaw University of Technology, Wroclaw 50-370, Poland; E-mail: [email protected]

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452 Magdziarz

In the Gaussian case, the prominent example of a stationary stochasticprocess is the celebrated Ornstein–Uhlenbeck (O–U) process. It can bederived either as a stationary solution of the Langevin equation

dY (t) = −�Y (t)dt + dB(t), � > 0, (2)

where B(t) is the Brownian motion, or as the Lamperti transformation(Refs.[2,6]) of B(t). In Ref.[3], the authors introduce two types of fractionalO–U processes using a similar analogy. The first one is defined as thesolution of (2) with the noise process B(t) replaced by the fractionalBrownian motion BH (t), i.e.,

dY1(t) = −�Y1(t)dt + dBH (t), � > 0� (3)

The second one is defined as the Lamperti transformation of BH (t)

Y2(t) = e−tH BH (e t)� (4)

As it is shown in Ref.[3], these are two different stationary Gaussianprocesses, and the essential difference lies in the dependence structure.Let us recall that in the Gaussian case a stationary process Y (t) has long-range dependence (LRD) (Ref.[1]), if its covariance function satisfies

∞∑t=1

|Cov(Y (t),Y (0)) | = ∞� (5)

For H �= 1/2 the covariance function of Y1(t) decays as the power function,i.e.,

Cov(Y1(t),Y1(0)) ∼ ct 2H−2

as t → ∞. Here c is the appropriate constant. This implies that forH ∈ (1/2, 1) we have

∑∞n=0 |Cov(Y1(n),Y1(0)) | = ∞, thus the process

exhibits LRD. The covariance function of Y2(t) decays exponentially andin this case the process does not have LRD property.

In what follows, we extend the above considerations to the �-stable caseby replacing BH (t) with the linear fractional stable motion L�,H (t) definedin (1). However, for a stationary stable process with index of stability� < 2 the covariance function is not defined. Instead, another measure ofdependence must be used. Basing on the results in Refs.[7–9,11], we adoptthe following measure of dependence for a stationary �-stable process Y (t)

r (�1; �2; t) = E[exp�i(�1Y (t) + �2Y (0))�

]− E

[exp�i�1Y (t)�

]E

[exp�i�2Y (0)�

], �1, �2 ∈ �� (6)

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Fractional Ornstein–Uhlenbeck �-Stable Processes 453

Unlike the covariance function, r (�1; �2; t) is always well defined andis equal to zero for independent Y (t) and Y (0). Additionally, if theprocess has finite second moment, it is asymptotically equivalent to thecovariance function. The function r (�1; �2; t) is closely related to the so-called “generalized codifference” (Ref.[11])

I (�1; �2; t) = − lnE[exp�i(�1Y (t) + �2Y (0))�

]+ lnE

[exp�i�1Y (t)�

] + lnE[exp�i�2Y (0)�

], (7)

since r (�1; �2; t)=C(�1; �2) · (e−I (�1;�2;t) − 1) with C(�1; �2)=E [exp�i�1Y (0)�]E [exp�i�2Y (0)�]. Thus, if I (�1; �2; t) → 0 as t → ∞, then r (�1;�2;t)

I (�1;�2;t)→

−C(�1; �2) as t → ∞, which implies that r (·) and I (·) are asymptoticallyequivalent. In the special case, when �1 = 1, �2 = −1, the generalizedcodifference I (1,−1, t) becomes (up to a constant) the standardcodifference (Ref.[11]). As shown in Ref.[10], the asymptotic behaviourof the codifference determines the chaotic properties (ergodicity, weakmixing, mixing) of the stationary �-stable processes. Thus, the asymptoticequivalence of r (·) and I (·) indicates that r (·) can also be helpful indetecting the ergodic properties of the �-stable processes. The presenceof the parameters �1 and �2 in the definition of r (·) has the followingadvantage: consider two stationary �-stable stochastic processes Y andY ′. In order to show that the two processes are different, we examinethe asymptotic behaviour of the corresponding measures of dependencerY (�1; �2; t) and rY ′(�1; �2; t). If the measures are not asymptoticallyequivalent at least for one specific choice of �1 and �2, then the processesY and Y ′ must be different.

The above mentioned properties of r (�1; �2; t) indicate that it can beregarded as an appropriate tool to examine the dependence structure ofthe �-stable processes.

Now, let us introduce the following definition of LRD in the �-stablecase. We say that a stationary �-stable process has long memory, if thecorresponding measure of dependence r (�1; �2; t) satisfies

∞∑n=1

| r (�1; �2; t) | = ∞� (8)

The above definition of long memory is a straightforward extension of theone used in the Gaussian case, (5), and has found a widespread acceptancein the �-stable case (Refs.[7–9]). For � = 2 the definitions (5) and (8) areequivalent. The terms “long memory” and “long-range dependence” areused interchangeably (Refs.[1,4]), and define the phenomena in which themeasure of dependence decays slowly. Let us emphasize that the conceptof long memory in the non-Gaussian world is still not well-formulated and

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454 Magdziarz

is a subject of extensive research. Therefore, the presented definition ofLRD and the obtained results should be viewed as one of the possibleapproaches to long memory for �-stable processes.

The fractional Langevin equation with linear fractional stable noise isdefined as

dZ1(t) = −�Z1(t)dt + dL�,H (t), � > 0� (9)

As it is shown in a recent paper by Maejima and Yamamoto[8], for � ∈ (1, 2)and H > 1/�, the unique stationary solution of (9) has the form

Z1(t) = L�,H (t) − �

∫ t

−∞e−�(t−s)L�,H (s)ds� (10)

Moreover, the authors prove that for the process Z1(t) the measure ofdependence satisfies r (�1; �2; t) ∼ c | t |�(H−1) as t → ∞, where c is theappropriate constant, thus for H > 1/� the process has long memory,which is the analogous result to the Gaussian case. Z1(t) is called the longmemory �-stable fractional O–U Process.

In the next section, we investigate the dependence structure of anotherstationary �-stable process, being the extension of (4) to the �-stable case,and defined as the following Lamperti transformation

Z2(t) = e−tH L�,H (e t)� (11)

We show that, similarly to the Gaussian counterpart (4), the measure ofdependence r (·) of Z2(t) decays exponentially and therefore we can callZ2(t) the short memory �-stable fractional O–U process.

2. ASYMPTOTIC DEPENDENCE STRUCTURE

In the next three theorems we give the precise formulas for theasymptotic behaviour of the measure of dependence r (·) for Z2(t). Weexclude the two cases �1�2 = 0 and a = b = 0, since then r (�1; �2; t) = 0.

In the proofs we frequently use the following property (Ref.[11], p. 122)

E[exp

{i�

∫Bf (x)L�(dx)

}]= exp

{− |�|�

∫B| f (x) |�dx

}(12)

with B ⊂ � and f ∈ L�((B), dx). We also take advantage of the two keyinequalities (Ref.[8]): For r , s ∈ �

‖r + s|� − |r |� − |s|�| ≤{2|r |� if 0 < � ≤ 1(� + 1)|r |�+�|r‖s|�−1 if 1 < � ≤ 2�

(13)

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Theorem 2.1. Let 0 < � < 1 and 0 < H < 1. Then the measure of dependenceof Z2(t) satisfies

r (�1; �2; t) ∼ −C�(�1; �2)A�(�1; �2)e−t�H (1−H )

as t → ∞, where

C�(�1; �2) = exp{−

2∑i=1

|�i |�∫ ∞

−∞

∣∣a[(1 − s)H−1/�

+ − (−s)H−1/�+

]

+ b[(1 − s)H−1/�

− − (−s)H−1/�−

]∣∣�ds} (14)

and

A�(�1; �2) =∫ ∞

0

{∣∣−�1asH−1/� + �2a(H − 1/�)sH−1/�−1∣∣�

− ∣∣�1asH−1/�∣∣� − ∣∣�2a(H − 1/�)sH−1/�−1

∣∣�}ds+

∫ ∞

0

{∣∣−�1bsH−1/� + �2b(1/� − H )sH−1/�−1∣∣�

− ∣∣�1bsH−1/�∣∣� − ∣∣�2b(1/� − H )sH−1/�−1

∣∣�}ds� (15)

Proof. We have Z2(t) = e−tH L�,H (e t) = ∫ ∞−∞ f (s, t)L�(ds) with

f (s, t) = e−tH a[(e t − s)H−1/�

+ − (−s)H−1/�+

] + e−tH b[(e t − s)H−1/�

− − (−s)H−1/�−

]�

For the generalized codifference I (�1; �2; t), the following relationholds true r (�1; �2; t) = C�(�1; �2) · (e−I (�1;�2;t) − 1), where C�(�1; �2) =E [exp�i�1Y (0)�]E [exp�i�2Y (0)�]. Note that by formula (12), the constantC�(·) is given by (14). Therefore, it is enough to investigate the asymptoticbehaviour of I (·). Taking advantage of (7) and (12) we obtain

I (�1; �2; t) =∫ ∞

−∞�|�1 f (s, t) + �2 f (s, 0)|� − |�1 f (s, t)|�−|�2 f (s, 0)|��ds

=∫ 0

−∞� � � ds +

∫ 1

0� � � ds +

∫ e t

1� � � ds +

∫ ∞

e t� � � ds

=: I1(t) + I2(t) + I3(t) + I4(t)� (16)

In what follows, we estimate every Ij(t), j = 1, � � � , 4, separately.

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Let us begin with I1(t). After some standard calculations and by thechange of variables s → −e tH s, we get

I1(t) = e t�H2∫ ∞

0�|p(s, t) + q(s, t)|� − |p(s, t)|� − |q(s, t)|��ds,

where

p(s, t) = e−tH�1a[(e t(1−H ) + s)H−1/� − sH−1/�

]and (17)

q(s, t) = �2a[(e−tH + s)H−1/� − sH−1/�

]� (18)

For fixed s ∈ (0,∞) we see that

e tH p(s, t) −→ −�1asH−1/� =: p∞(s)

as t → ∞. Using the mean-value theorem

f (r + s) − f (r ) = s∫ 1

0f ′(r + us)du, (19)

where f is accordingly smooth, and the dominated convergence theorem(DCT), we obtain

e tH q(s, t) = �2a(H − 1/�)∫ 1

0(s +ue−tH )H−1/�−1du −→ �2a(H − 1/�)sH−1/�−1

=: q∞(s)

as t → ∞. Consequently, for fixed s ∈ (0,∞)

e t�H �|p(s, t) + q(s, t)|� − |p(s, t)|� − |q(s, t)|��−→ �|p∞(s) + q∞(s)|� − |p∞(s)|� − |q∞(s)|�� (20)

as t → ∞. To apply DCT, we use inequality (13) together with the mean-value theorem and get

supt>1

e t�H‖p(s, t) + q(s, t)|� − |p(s, t)|� − |q(s, t)|�|

≤ supt>1

1(0,1)(s)e t�H‖p(s, t) + q(s, t)|� − |p(s, t)|� − |q(s, t)|�|

+ supt>1

1[1,∞)(s)e t�H‖p(s, t) + q(s, t)|� − |p(s, t)|� − |q(s, t)|�|

≤ supt>1

1(0,1)(s)e t�H2|p(s, t)|� + supt>1

1[1,∞)(s)e t�H2|q(s, t)|�

≤ 1(0,1)(s)c1sH �−1 + 1[1,∞)(s)c2sH �−1−�,

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which is integrable on (0,∞). Here c1 and c2 are the appropriate constantsindependent of s and t . Thus, from DCT we get

I1(t) ∼ e t�H2e−t�H

∫ ∞

0

{|p∞(s) + q∞(s)|� − |p∞(s)|� − |q∞(s)|�}ds (21)

as t → ∞.For the next term, we use inequality (13) together with DCT and show

in a similar manner that I2(t) = O(e−t�H ). Thus, the contribution of I2 isnegligible. The same statement applies to I4(t) = O(e−t�(1−H )).

We continue our estimations for I3(t). After the change of variabless → e tH s, we have

I3(t) = e t�H2∫ ∞

0�|w(s, t)+ z(s, t)|� − |w(s, t)|�−|z(s, t)|��ds,

where

w(s, t) = e−tH�1[a(e t(1−H ) − s)H−1/� − bsH−1/�

] · 1(e−tH ,e t(1−H ))(s) (22)

and

z(s, t) = �2b[(s − e−tH )H−1/� − sH−1/�

] · 1(e−tH ,e t(1−H ))(s)� (23)

In a similar manner as for I1(t), we get that for fixed s ∈ (0,∞)

e tHw(s, t) −→ −�1bsH−1/� =: w∞(s)

and also

e tH z(s, t) −→ �2b(1/� − H )sH−1/�−1 =: z∞(s)

as t → ∞. Consequently, DCT yields


∫ ∞

0

{|w∞(s) + z∞(s)|� − |w∞(s)|� − |z∞(s)|�}ds (24)

as t → ∞. Finally, putting together formulas (21) and (24), we get thedesired result.

We pass on to the case � = 1. We recall the fact that the two cases�1�2 = 0 and a = b = 0 are excluded, since then r (�1; �2; t) = 0.

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458 Magdziarz

Theorem 2.2. Let � = 1 and 0 < H < 1. Then the measure of dependence ofZ2(t) satisfies

(i) If b = 0 and �1�2 > 0 then r (�1; �2; t) = 0,(ii) If a = 0 and �1�2 < 0 then

r (�1; �2; t) ∼ 2|b|C1(�1; �2)( |�1|

He−tH · 1(0,1/2](H ) + |�2|e−t(1−H ) · 1[1/2,1)(H )

)

as t → ∞.Otherwise

r (�1; �2; t) ∼ −C1(�1; �2)A1(�1; �2)e−tH (1−H )

as t → ∞, where C1 and A1 are given in (14) and (15), respectively.

Proof. First, we determine, in which case the constant A1(�1; �2) = 0.From (15) we get

A1(�1; �2) =∫ ∞

0� � � ds +

∫ ∞

0� � � ds =: A11(�1; �2) + A12(�1; �2)�

From the triangle inequality we see that A1(�1; �2) = 0 ⇔ �A11(�1; �2)= 0and A12(�1; �2) = 0�. Additionally, we have A11(�1; �2) = 0 ⇔ �a = 0 or�1�2 > 0� as well as A12(�1; �2) = 0 ⇔ �b = 0 or �1�2 < 0�. Since the cases�1�2 = 0 or a = b = 0 are excluded, we obtain

A1(�1; �2) = 0 ⇔ �a = 0 and �1�2 < 0� or �b = 0 and �1�2 > 0��

The case �b = 0 and �1�2 > 0� is trivial, since then it is easy to verify thatfor every term in formula (16) we have Ij(t) = 0, i = 1, � � � , 4. Thus, weobtain part (i) of the theorem.

We pass on to the second possibility �a = 0 and �1�2 < 0�. In this caseonly I1(t) and I3(t) from (16) disappear, therefore we need to find theasymptotic behaviour of I2(t) and I4(t). Using some standard argumentsand DCT we show that

I2(t) ∼ −2|�1b|H

e−tH (25)

and

I4(t) ∼ −2|�2b|e−t(1−H ) (26)

as t → ∞.

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Now, from (25) and (26) we get that for H < 1/2 we obtainI (�1; �2; t) ∼ I2(t), for H > 1/2 we obtain I (�1; �2; t) ∼ I4(t) and forH = 1/2 we get I (�1; �2; t) ∼ I2(t) + I4(t) as t → ∞. Thus, we have provedpart (ii) of the theorem.

In any other case, i.e., when A1(�1; �2) is a non-zero constant,the proof of Theorem 2.1 applies and we get r (�1; �2; t) ∼−C1(�1; �2)A1(�1; �2)e−tH (1−H ) as t → ∞.

The next theorem determines the asymptotic dependence structure ofZ2(t) when the index of stability is such that 1 < � < 2.

Theorem 2.3. Let 1 < � < 2, 0 < H < 1 and H �= 1/�. Then the measure ofdependence of Z2(t) satisfies

(i) If 1 − 1�< H < 1

�then

r (�1; �2; t) ∼ −C�(�1; �2)A�(�1; �2)e−t�H (1−H )

(ii) If H < 1 − 1�then

r (�1; �2; t) ∼ −C�(�1; �2)B�(�1; �2)e−tH

(iii) If H > 1�then

r (�1; �2; t) ∼ −C�(�1; �2)D�(�1; �2)e−t(1−H )

as t → ∞. The constants C� and A� are given in (14) and (15), respectively,whereas

B�(�1; �2) =∫ ∞

0��1sgn��2�|�2|�−1|a|�∣∣(1 + s)H−1/� − sH−1/�

∣∣�−1sH−1/�ds

−∫ 1

0��1bsH−1/�

∣∣�2[a(1 − s)H−1/� − bsH−1/�]∣∣�−1

× sgn{�2[a(1 − s)H−1/� − bsH−1/�]}ds

−∫ ∞

1��1sgn��2b�bsH−1/�

∣∣�2b[(s − 1)H−1/� − sH−1/�]∣∣�−1ds,

and

D�(�1; �2) = |a|�∫ ∞

0�(H − 1/�)�2|�1|�−1sgn��1�sH−1/�−1

× |(s + 1)H−1/� − sH−1/�|�−1ds

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460 Magdziarz

+∫ 1

0�|�1|�−1�2b(1/� − H )sH−1/�−1|a(1 − s)H−1/� − bsH−1/�|�−1

× sgn��1[a(1 − s)H−1/� − bsH−1/�]�ds+

∫ ∞

1��2b(H − 1/�)sgn��1b�|�1b|�−1

× |(s − 1)H−1/� − sH−1/�|�−1sH−1/�−1ds�

Proof. (i) Recall the decomposition in (16). We have I1(t) =e t�H

2 ∫ ∞0 �|p(s, t) + q(s, t)|� − |p(s, t)|� − |q(s, t)|��ds, where p(s, t) and q(s, t)

are given in (17) and (18), respectively. For fixed s ∈ (0,∞) we get that(20) holds, since H − 1/� < 0. Next, we apply DCT and the second partof inequality (13) in combination with the mean-value theorem (19).We obtain


∫ ∞

0

{|p∞(s) + q∞(s)|� − |p∞(s)|� − |q∞(s)|�}ds (27)

as t → ∞.For the next term I2(t) we use inequality (13) together with DCT

and show in a standard way that I2(t) = O(e−tH ), t → ∞. Thus, since�(1 − H ) < 1, the integral I2(t) decays faster than I1(t).

In case of I3(t) we cannot use DCT directly, we need more delicateestimations. We have I3(t) = e t�H

2 ∫ ∞0 �|w(s, t)+ z(s, t)|� − |w(s, t)|�−

|z(s, t)|��ds, where w(s, t) and z(s, t) are given in (22) and (23),respectively. Set

G(s, t) := |w(s, t)+ z(s, t)|� − |w(s, t)|� − |z(s, t)|�,G∞(s) := |w∞(s)+ z∞(s)|� − |w∞(s)|� − |z∞(s)|�

with w∞(s) = −�1bsH−1/� and z∞(s) = �2b(1/� − H )sH−1/�−1. We will showthat ∣∣∣∣

∫ ∞

0

[e t�HG(s, t) − G∞(s)

]ds

∣∣∣∣ −→ 0

as t → ∞. Fix � > 0 and put

∫ ∞

0

[e t�HG(s, t) − G∞(s)

]ds =

∫ e−tH

0� � � ds +

∫ e−tH +�

e−tH� � � ds +

∫ e t(1−H )−�

e−tH +�

� � � ds

+∫ e t(1−H )

e t(1−H )−�

� � � ds +∫ ∞

e t(1−H )

� � � ds

= : J1(t)+ J2(t)+ J3(t)+ J4(t)+ J5(t)�

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Since G∞(s) is integrable on (0,∞), we obtain | J1(t)| −→ 0 and| J5(t)| −→ 0 as t → ∞. For the second term we have from (13) that

|J2(t)| ≤∫ e−tH +�

e−tH|e t�HG(s, t) − G∞(s)|ds ≤ (� + 1)

∫ e−tH +�

e−tHe t�H |w(s, t)|�ds

+ �

∫ e−tH +�

e−tHe t�H |w(s, t)‖z(s, t)|�−1ds + (� + 1)

∫ e−tH +�

e−tH|w∞(s)|�ds

+ �

∫ e−tH +�

e−tH|w∞(s)‖z∞(s)|�−1ds =: J21(t) + J22(t) + J23(t) + J24(t)�

Additionally, since H − 1/� < 0, we have

J21(t) ≤ d1

∫ e−tH +�

e−tH

[(1 − s)H �−1 + sH �−1

]ds

= − d1H �

[(1 − e−tH − �)H � − (1 − e−tH )H �

]+ d1

H �

[(e−tH + �)H � − (e−tH )H �

]�

Next, since H − 1/� < 0, we get for s ∈ (e−tH , e−tH+�)

|(s − e−tH )H−1/� − sH−1/�| ≤ (1/� − H )e−tH (s − e−tH )H−1/�−1,

and consequently

J22(t) ≤ d2

∫ e−tH +�

e−tH

[(1 − s)H−1/� + sH−1/�

](s − e−tH

)(H−1/�−1)(�−1)ds

≤ d3(1 − e−tH − �

)H−1/��H �−�+1/�−H+1 + d4�H �−�+1�

Moreover,

J23(t) = d5[(e−tH + �)H � − (e−tH )H �

],

J24(t) = d6[(e−tH + �)H �−�+1 − (e−tH )H �−�+1

],

where di , i = 1, � � � , 6 are the appropriate constants independent of t , s,and �. Using the fact that 1 − 1

�< H < 1

�, we obtain lim�↘0 limt→∞ J2i(t) = 0

for every i = 1, � � � , 4, which implies lim�↘0 limt→∞ J2(t) = 0.We pass on to J3(t). For fixed s ∈ (0,∞) we get from the proof of

Theorem 2.1

e t�HG(s, t)1(e−tH +�,e t(1−H )−�)(s) −→ G∞(s)1(�,∞)(s)�

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as t → ∞. Additionally,

supt> 2

H

e t�H |G(s, t)|1(e−tH +�,e t(1−H )−�)(s)

≤ supt> 2

H

1(e−tH +�,1)(s)e t�H (� + 1)|w(s, t)|�

+ supt> 2

H

1(e−tH +�,1)(s)e t�H�|w(s, t)‖z(s, t)|�−1

+ supt> 2

H

1[1,e t(1−H )−�)(s)et�H (� + 1)|z(s, t)|�

+ supt> 2

H

1[1,e t(1−H )−�)(s)et�H�|z(s, t)‖w(s, t)|�−1

≤ 1(�,1)(s)k1[(1 − s)H �−1 + sH �−1

]+ 1(�,1)(s)k2

[(1 − s)H−1/� + sH−1/�

]�(H−1/�−1)(�−1)

+ 1[1,∞)(s)k3(s − 1/2)H �−1−�

+ 1[1,∞)(s)k4(s − 1/2)H−1/�−1[�(H−1/�)(�−1) + s(H−1/�)(�−1)

],

which is integrable on (0,∞). Thus, from DCT we get J3(t) −→ 0 ast → ∞. For J4(t) we have

|J4(t)| ≤∫ e t(1−H )

e t(1−H )−�

|e t�HG(s, t) − G∞(s)|ds ≤ (� + 1)∫ e t(1−H )

e t(1−H )−�

e t�H |z(s, t)|�ds

+ �

∫ e t(1−H )

e t(1−H )−�

e t�H |z(s, t)‖w(s, t)|�−1ds + (� + 1)∫ e t(1−H )

e t(1−H )−�

|z∞(s)|�ds

+ �

∫ e t(1−H )

e t(1−H )−�

|z∞(s)‖w∞(s)|�−1ds =: J41(t)+ J42(t)+ J43(t)+ J44(t),

and one shows similarly, as for J2(t) that lim�↘0 limt→∞ J4(t) = 0. Finally,we have proved

lim�↘0

limt→∞

Ji(t) = 0

for i = 1, � � � , 5. Thus,

∣∣∣∣∫ ∞

0

[e t�HG(s, t) − G∞(s)

]ds

∣∣∣∣ −→ 0,

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which implies for the term I3(t) in (16) that

I3(t) ∼ e−t�H (1−H )

∫ ∞

0G∞(s)ds

as t → ∞.For the last term we show in a standard manner that I4(t) = O(e−t(1−H ))

and, since �H < 1, its contribution is negligible.Finally, we get that I2(t) and I4(t) decay faster than I1(t) and I3(t).

Therefore, I (�1; �2; t) ∼ I1(t) + I3(t) as t → ∞, which completes the proofof part (i).

(ii) Recall the decomposition (16). For the first term we have

I1(t) = e t�H2∫ ∞

0

{|p(s, t)+ q(s, t)|� − |p(s, t)|� − |q(s, t)|�}ds,= e t�H

2∫ 1

0� � � ds + e t�H

2∫ ∞

1� � � ds =: I11(t) + I12(t),

where p(s, t) and q(s, t) are given in (17) and (18), respectively. For I12(t)one shows similarly as in the part (i) of the proof that

I12(t) ∼ e−t�H (1−H )

∫ ∞

1|p∞(s)+ q∞(s)|� − |p∞(s)|� − |q∞(s)|�ds,

as t → ∞. Here p∞(s) = −�1asH−1/� and q∞(s) = �2a(H − 1/�)sH−1/�−1.For I11(t), after the change of variables s → e−tH s, we get

I11(t) =∫ e tH

0

{|p̃(s, t)+ q̃(s)|� − |p̃(s, t)|� − |q̃(s)|�}ds,where p̃(s, t) = e−tH�1a

[(e t + s)H−1/� − sH−1/�

]and q̃(s) = �2a

[(1 +

s)H−1/� − sH−1/�]. For fixed s ∈ (0,∞) we have that e tH p̃(s, t)1(0,e tH )(s) −→

−�1asH−1/� as t → ∞, and from the mean-value theorem we obtain

e tH{|p̃(s, t)+ q̃(s)|� − |q̃(s)|�}1(0,e tH )(s)

−→t→∞ ��1sgn��2�|�2|�−1|a|�∣∣(1 + s)H−1/� − sH−1/�∣∣�−1

sH−1/� =: H∞(s),

since ddx |x |�= �|x |�−1sgn�x� for x �= 0. Using inequality (13) and DCT we

show in a standard way that I11(t) ∼ e−tH∫ ∞0 H∞(s)ds as t → ∞. Since �(1 −

H ) > 1, we see that I12(t) decays faster than I11(t), and we finally obtain

I1(t) ∼ e−tH

∫ ∞

0H∞(s)ds

as t → ∞.

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Next, we have I2(t) = ∫ 10 �|v(s, t) + u(s)|� − |v(s, t)|� − |u(s)|��ds, with

v(s, t) = e−tH�1[a(e t − s)H−1/� − bsH−1/�

](28)

and

u(s) = �2[a(1 − s)H−1/� − bsH−1/�

]� (29)

For fixed s ∈ (0, 1) we obtain e tH v(s, t) −→ −�1bsH−1/� as t → ∞, and fromthe mean value theorem we get

e tH{|u(s)+ v(s, t)|� − |u(s)|�}−→t→∞ − ��1bsH−1/�

∣∣�2[a(1 − s)H−1/� − bsH−1/�]∣∣�−1

× sgn{�2

[a(1 − s)H−1/� − bsH−1/�

]}=: M∞(s)�

Since for the appropriate constants m1 and m2 we get

supt>1

e tH‖v(s, t) + u(s)|� − |v(s, t)|� − |u(s)|�|

≤ supt>1

e tH (� + 1)|v(s, t)|�+ supt>1

e tH�|v(s, t)‖u(s)|�−1

≤ m1(1 − s)H �−1 + m2sH �−1,

which is integrable on (0, 1), DCT yields

I2(t) ∼ e−tH

∫ 1

0M∞(s)ds�

For the next component we need some more delicate estimations.We have I3(t) = ∫ e t

1

{|d(s, t) + f (s)|� − |d(s, t)|� − |f (s)|�}ds with d(s, t) =e−tH�1

[a(e t − s)H−1/� − bsH−1/�

]and f (s) = �2b

[(s − 1)H−1/� − sH−1/�

]. For

fixed s ∈ (1,∞) we get

e tH d(s, t)−→t→∞ − �1bsH−1/�

and from the mean-value theorem

e tH{|d(s, t) + f (s)|� − |f (s)|�}−→t→∞ − ��1bsH−1/�

∣∣�2b[(s − 1)H−1/� − sH−1/�]∣∣�−1

sgn��2b�

=: N∞(s)�

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Note that N∞(s) is integrable on (1,∞). Set

N (s, t) := |d(s, t)+ f (s)|� − |d(s, t)|� − |f (s)|��

Then, we have

I3(t) =∫ 2

1N (s, t)ds +

∫ e t

2N (s, t)ds =: I31(t) + I32(t)�

We will find the rate of convergence for every I3i , i = 1, 2, separately.For I31(t), fix s ∈ (1, 2), then we get e tHN (s, t) −→ N∞(s) as t → ∞.Additionally,

supt>2

e tH |N (s, t)| ≤ supt>2

e tH (� + 1)|d(s, t)|�+ supt>2

e tH�|d(s, t)‖f (s)|�−1

≤ p1[(3 − s)H �−1 + sH �−1

]+ p2

[(3 − s)H−1/� + sH−1/�

][(s − 1)(H−1/�)(�−1) + s(H−1/�)(�−1)

],

which is integrable on (1, 2). Here p1 and p2 are the appropriate constantsindependent of s and t . Hence,

I31(t) ∼ e−tH

∫ 2

1N∞(s)ds�

For I32(t) we need more subtle estimations. In what follows we show that∣∣∣∣∫ ∞

2

[e tHN (s, t)1(2,e t )(s) − N∞(s)

]ds

∣∣∣∣ −→t→∞0

Fix � > 0 appropriately small and put∫ ∞

2

[e tHN (s, t)1(2,e t )(s) − N∞(s)

]ds

=∫ e tH

2� � � ds +

∫ e t−e tH −�

e tH� � � ds +

∫ e t−�

e t−e tH −�

� � � ds +∫ e t

e t−�

� � � ds +∫ ∞

e t� � � ds

=: J1(t)+ J2(t)+ J3(t)+ J4(t)+ J5(t)�

Let us begin with J1(t). For fixed s ∈ (2,∞) we obtain

e tHN (s, t)1(2,e tH )(s)−→t→∞ N∞(s)�

Additionally, we use the fact that s ∈ (2, e tH ), which implies that thereexist t0 such that for every t > t0 and every s ∈ (2, e tH ) we have e t − s > s.

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Hence,

supt>t0

e tH |N (s, t)|1(2,e tH )(s)

≤ supt>t0

e tH (� + 1)|d(s, t)|�1(2,e tH )(s) + supt>t0

e tH�|d(s, t)‖f (s)|�−11(2,e tH )(s)

≤ supt>t0

q1e tH e−t�H sH �−11(2,e tH )(s) + supt>t0

q2sH−1/�(s − 1)(H−1/�−1)(�−1)1(2,e tH )(s)

≤ q1sH �−� + q2sH−1/�(s − 1)(H−1/�−1)(�−1),

which is integrable on (2,∞). Here q1 and q2 are the appropriate constantsindependent of s and t . Hence, DCT yields J1(t) −→ 0 as t → ∞.

For the next component we have

|J2(t)| ≤∫ e t−e tH −�

e tHe tH |N (s, t)|ds +

∫ e t−e tH −�

e tH|N∞(s)|ds =: J21(t)+ J22(t)�

Since N∞(s) is integrable on (1,∞), we get J22(t) −→ 0 as t → ∞.Furthermore,

J21(t) ≤∫ e t−e tH −�

e tHe tH (� + 1)|f (s)|�ds +

∫ e t−e tH −�

e tHe tH�|f (s)‖d(s, t)|�−1ds

≤ u1e tH∫ e t−e tH −�

e tH(s − 1)H �−1−�ds

+ u2e tH e−tH (�−1)

∫ e t−e tH −�

e tH(s − 1)H−1/�−1(e t − s)(H−1/�)(�−1)ds

+ u3e tH e−tH (�−1)

∫ e t−e tH −�

e tH(s − 1)H−1/�−1s(H−1/�)(�−1)ds −→t→∞ 0,

where ui , i = 1, 2, 3, are the appropriate constants independent of s, t , and�. Hence, |J2(t)| −→ 0 as t → ∞.

Next, for the third term we get

|J3(t)| ≤∫ e t−�

e t−e tH −�

e tH |N (s, t)|ds +∫ e t−�

e t−e tH −�

|N∞(s)|ds =: J31(t)+ J32(t)�

Since N∞(s) is integrable on (1,∞), we obtain J32(t) −→ 0 as t → ∞.Additionally,

J31(t) ≤∫ e t−�

e t−e tH −�

e tH (� + 1)|f (s)|�ds +∫ e t−�

e t−e tH −�

e tH�|f (s)‖d(s, t)|�−1ds

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≤ w1e tH∫ e t−�

e t−e tH −�

(s − 1)H �−1−�ds

+ w2e tH e−tH (�−1)

∫ e t−�

e t−e tH −�

(s − 1)H−1/�−1�(H−1/�)(�−1)ds

+ w3e tH e−tH (�−1)

∫ e t−�

e t−e tH −�

(s − 1)H−1/�−1s(H−1/�)(�−1)ds −→t→∞ 0,

where wi , i = 1, 2, 3, are the appropriate constants independent of s, t , and�. Therefore, |J3(t)|−→ 0 as t → ∞.

For the fourth part we put

|J4(t)| ≤∫ e t

e t−�

e tH |N (s, t)|ds +∫ e t

e t−�

|N∞(s)|ds =: J41(t)+ J42(t)�

Since N∞(s) is integrable on (1,∞), we get J42(t) −→ 0 as t → ∞. Further,

J41(t) ≤∫ e t

e t−�

e tH (� + 1)|d(s, t)|�ds +∫ e t

e t−�

e tH�)|d(s, t)‖f (s)|�−1ds

≤ z1e tH e−t�H

∫ e t

e t−�

[(e t − s)H �−1 + sH �−1

]ds

+ z2

∫ e t

e t−�

[(e t − s)H−1/� + sH−1/�

](s − 1)(H−1/�−1)(�−1)ds

≤ z3�H � + z4� + z5�H−1/�+1,

where zi , i = 1, � � � , 5, are the appropriate constants independent of s, t ,and �. Thus, we obtain

lim�↘0

limt→∞

|J4(t)| = 0�

The last component J5(t) = ∫ ∞e t N∞(s)ds −→ 0 as t → ∞, since N∞(s) is

integrable on (1,∞). Finally, combining the results for Ji(t), i = 1, � � � , 5,we obtain ∣∣∣∣

∫ ∞

2

[e tHN (s, t)1(2,e t )(s) − N∞(s)

]ds

∣∣∣∣ −→t→∞ 0�

Hence, I32(t) ∼ e−tH∫ ∞2 N∞(s)ds, which implies

I3(t) = I31(t) + I32(t) ∼ e−tH

∫ ∞

1N∞(s)ds

as t → ∞.

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From part (i) of the proof we have I4(t) = O(e−t(1−H )). Additionally, theassumption �(1 − H ) > 1 implies that H < 1/2. Thus, the contribution ofI4(t) is negligible.

Finally, putting together the results for Ij(t), j = 1, � � � , 4, we get

I (�1; �2; t) ∼ e−tH

{ ∫ ∞

0H∞(s)ds +

∫ 1

0M∞(s)ds +

∫ ∞

1N∞(s)ds

}

as t → ∞, which ends the proof of part (ii).

(iii) We begin with showing the following key inequality.

Lemma 2.1.

‖r |� + |s|� − |r − s|�|≤ (� + 1)|r‖s|�−1, ∀r , s ∈ �, 1 < � ≤ 2� (30)

Proof. Inequality (30) follows directly from the two following inequalities

(a) |r � + s� − |r − s|�| = r � + s� − |r − s|�≤ (� + 1)rs�−1,(b) |r � + s� − |r + s|�| = (r + s)� − r � − s� ≤ (� + 1)rs�−1,

where r , s > 0, 1 < � ≤ 2.

Proof of inequality (a) Let r ≥ s. Define fs(r ) := r � + s� −|r − s|�−(� + 1)rs�−1. We have fs(0) = 0 and

f ′s (r ) = �r �−1 − �(r − s)�−1 − (� + 1)s�−1 ≤ �s�−1 − (� + 1)s�−1 ≤ 0�

Thus fs(r ) ≤ 0.Let r < s. Using the mean-value theorem we get

r � + s� − (s − r )� ≤ rs�−1 + �r∫ 1

0[(s − r ) + ru]�−1du

≤ rs�−1 + �rs�−1 = (� + 1)rs�−1,

which proves (a).

Proof of inequality (b) We put hs(r ) = r � + s� + (� + 1)rs�−1 − (r +s)�. Since hs(0) = 0 and h ′

s(r ) = �r �−1 + (� + 1)s�−1 − �(r + s)�−1 ≥ 0, weget hs(r ) ≥ 0.

Now, using the above result we determine the rate of convergence forevery Ij(t), j = 1, � � � , 4, from decomposition (16).

For I1(t), after some standard calculations, we have

I1(t) = e t�H |a|�∫ ∞

0�|p̄(s, t) + q̄(s, t)|� − |p̄(s, t)|� − |q̄(s, t)|��ds

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with p̄(s, t) = e−tH�1[(s + 1)H−1/� − sH−1/�

]and q̄(s, t) = �2

[(s + e−t)H−1/� −

sH−1/�]. For fixed s ∈ (0,∞) we get

e t q̄(s, t)−→t→∞ (H − 1/�)�2sH−1/�−1

and

e t+tH (�−1)�|p̄(s, t) + q̄(s, t)|�−|p̄(s, t)|��−→t→∞ �(H − 1/�)�2|�1|�−1sgn��1�sH−1/�−1|(s + 1)H−1/� − sH−1/�|�−1=: P∞(s)�

Note that P∞(s) is integrable on (0,∞). Now, applying inequality (30) andDCT, we get by some standard arguments

I1(t) ∼ e−t(1−H )|a|�∫ ∞

0P∞(s)ds

as t → ∞.For the next term we have I2(t) = ∫ 1

0 �|v(s, t)+u(s)|� − |v(s, t)|� −|u(s)|��ds, with v(s, t) and u(s) given in (28) and (29), respectively.Additionally, for fixed s ∈ (0, 1), we obtain e t/�v(s, t) −→ a�1 as t → ∞,and by the mean value theorem

e t/��|u(s) + v(s, t)|� − |u(s)|��−→t→∞ �a�1|�2|�−1|a(1 − s)H−1/� − bsH−1/�|�−1sgn

{�2

[a(1 − s)H−1/� − bsH−1/�

]}�

Note that the limit function is integrable on (0, 1). Now, applying DCTin a standard manner, we see that I2(t) = O(e−t/�), thus its contribution isnegligible, since H > 1/� and 1 < � ≤ 2 imply 1/� > 1 − 1/� > 1 − H .

For the next component, after some standard calculations, we haveI3(t) = e t�H

∫ 1e−t �|w̄(s, t)+ z̄(s, t)|� − |w̄(s, t)|� − |z̄(s, t)|��ds, where

w̄(s, t) = �1e−tH[a(1 − s)H−1/� − bsH−1/�

]and

z̄(s, t) = �2b[(s − e−t)H−1/� − sH−1/�

]�

For fixed s ∈ (0, 1), we get from the mean value theorem e t z̄(s, t) −→−�2b(H − 1/�)sH−1/�−1 as t → ∞, and by using the mean value theorem

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again,

e t+tH (�−1){|w̄(s, t) + z̄(s, t)|�−|w̄(s, t)|�}

−→t→∞ − �|�1|�−1�2b(H − 1/�)sH−1/�−1|a(1 − s)H−1/� − bsH−1/�|�−1

× sgn{�1

[a(1 − s)H−1/� − bsH−1/�

]}=: Q∞(s)�

Note that Q∞(s) is integrable on (0, 1). Put Q (s, t) := |w̄(s, t) +z̄(s, t)|� − |w̄(s, t)|� − |z̄(s, t)|�. In what follows, we will show that

∣∣∣∣∫ 1

0

[e t+tH (�−1)Q (s, t)1(e−t ,1)(s) − Q∞(s)

]ds

∣∣∣∣ −→t→∞ 0�

Fix � > 0 appropriately small and set

∫ 1

0

[e t+tH (�−1)Q (s, t)1(e−t ,1)(s) − Q∞(s)

]ds

=∫ e−t

0� � � ds +

∫ e−t+�

e−t� � � ds +

∫ 1

e−t+�

� � � ds

=: J1(t)+ J2(t)+ J3(t)�

We immediately obtain

|J1(t)| −→t→∞ 0,

since Q∞(s) is integrable on (0, 1). Next, we have

|J2(t)| ≤∫ e−t+�

e−te t+tH (�−1)|Q (s, t)|ds +

∫ e−t+�

e−t|Q∞(s)|ds =: J21(t)+ J22(t)�

Using (30), we obtain

J21(t) =∫ e−t+�

e−te t+tH (�−1)|Q (s, t)|ds

=∫ e−t+�

e−te t+tH (�−1)|Q (s, t)|1(

0, 11+(b/a)1/�−H

](s)ds+

∫ e−t+�

e−te t+tH (�−1)|Q (s, t)|1( 1

1+(b/a)1/�−H ,1)(s)ds

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≤∫ e−t+�

e−te t+tH (�−1)(� + 1)|z̄(s, t)‖w̄(s, t)|�−1

+∫ e−t+�

e−te t+tH (�−1)�|z̄(s, t)‖w̄(s, t)|�−1ds

≤ d1

∫ e−t+�

e−t(s − e−t)H−1/�−1ds ≤ d2�H−1/��

Here d1 and d2 are the appropriate constants independent of �, s, and t .For J22(t), we get

J22(t) ≤ d3

∫ e−t+�

e−tsH−1/�−1ds = d3

H − 1/�

[(e−t + �)H−1/� − (e−t)H−1/�

],

where d3 is the appropriate constant independent of �, s, and t . Finally, weobtain

lim�↘0

limt→∞

J2i(t) = 0,

i = 1, 2, which implies that lim�↘0 limt→∞| J2(t)| = 0.Now, we proceed with J3(t). For fixed s ∈ (0, 1) we have

e t+tH (�−1)Q (s, t)1(e−t+�,1)(s)−→t→∞ Q∞(s)1(�,1)(s)�

Now, using (30) and DCT we show in a standard way that | J3(t)| −→ 0 as→ ∞. Finally, combining the results for every Ji(t), i = 1, 2, 3, we obtain

∣∣∣∣∫ 1

0

[e t+tH (�−1)Q (s, t)1(e−t ,1)(s) − Q∞(s)

]ds

∣∣∣∣ −→t→∞ 0

and consequently

I3(t) ∼ e−t(1−H )

∫ 1

0Q∞(s)ds

as t → ∞.For the last term we show in a standard way that

I4(t) ∼ e−t(1−H )

∫ ∞

1R∞(s)ds�

as t → ∞.

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Finally, combining the results for Ij(t), j = 1, � � � , 4, we obtain

I (�1; �2; t) ∼ e−t(1−H )

{|a|�

∫ ∞

0P∞(s)ds +

∫ 1

0Q∞(s)ds +

∫ ∞

1R∞(s)ds

}

as t → ∞, which ends part (iii) of the theorem.

Corollary 2.1. The fractional O–U �-stable process Z2(t) defined in (11) doesnot have long memory in the sense of (8).

Proof. From Theorems (2.1), (2.2), and (2.3) we see that r (�1; �2; t)decays exponentially as t → ∞, therefore

∞∑t=1

|r (�1; �2; t)|<∞�

Corollary 2.2. The fractional O–U �-stable processes Z1(t) given in (10) andZ2(t) defined in (11) are two different stationary processes.

Proof. Since the rate of convergence of the function r (�1; �2; t) isdifferent for both processes, their finite-dimensional distributions can notbe identical.

The above results show that, similarly to the Gaussian case, theLamperti transformation of the LFSM (1) results in the exponential decayof r (�1; �2; t) for Z2(t), while the long memory property of the noise inthe Langevin equation (9) transfers to its solution Z1(t). The presentedapproach to long memory, basing on the asymptotic properties of themeasure of dependence r (�1; �2, t), should be viewed as one of the existingapproaches. The final formulation of the problem of LRD in the �-stablecase, as well as the behaviour of other dependence measures, is a subjectof our further research.

REFERENCES

1. Beran, J. Statistics for Long-Memory Processes; Chapman & Hall: New York, 1994.2. Burnecki, K.; Maejima, M.; Weron, A. The Lamperti transformation for self-similar processes.

Yokohama Math. J. 1997, 44, 25–42.3. Cheridito, P.; Kawaguchi, H.; Maejima, M. Fractional Ornstein–Uhlenbeck processes. Electron.

J. Probab. 2003, 8, 1–14.4. Doukhan, P.; Oppenheim, G.; Taqqu, M.S. Theory and Applications of Long-Range Dependence;

Birkhäuser: Boston, 2003.5. Janicki, A.; Weron, A. Simulation and Chaotic Behaviour of �-Stable Stochastic Processes; Marcel

Dekker: New York, 1994.6. Lamperti, J.W. Semi-stable stochastic processes. Trans. Amer. Math. Soc. 1962, 104, 62–78.

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7. Lévy, J.B.; Taqqu, M.S. A Characterization of the asymptotic behavior of stationary stableprocesses. In Stable Processes and Related Topics; Cambanis, S.; Samorodnitsky, G.; Taqqu, M.S.,Eds.; Birkhäuser: Boston, 1991; pp. 181–198.

8. Maejima, M.; Yamamoto, K. Long-memory stable Ornstein–Uhlenbeck processes. Electron. J.Probab. 2003, 8, 1–18.

9. Magdziarz, M. The dependence structure of the fractional Ornstein–Uhlenbeck process.Prob. and Math. Stat. 2005, 25, 97–104.

10. Rosinski, J.; Zak, T. Simple conditions for mixing of infinitely divisible processes. StochasticProcess. Appl. 1996, 61, 277–288.

11. Samorodnitsky, G.; Taqqu, M.S. Stable Non-Gaussian Random Processes: Stochastic Models with InfiniteVariance; Chapman and Hall: New York, 1994.

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Short and Long Memory Fractional Ornstein–Uhlenbeck α-Stable Processes

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