TEMPERED FRACTIONAL BROWNIAN MOTION
By
Farzad Sabzikar

A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
Statistics – Doctor of Philosophy
2014

ABSTRACT
TEMPERED FRACTIONAL BROWNIAN MOTION
By
Farzad Sabzikar
Tempered fractional Brownian motion TFBM modifies the power law kernel in the moving average representation of a fractional Brownian motion (FBM), adding an exponential
tempering. It also has a harmonizable representation. The increments of TFBM are stationary, and the autocovariance of the resulting tempered fractional Gaussian noise TFGN
has semi-long range dependence, in which the autocorrelations decay like a power law over a
moderate length scale, but eventually fall off more rapidly. TFBM can be represented as the
linear combination of tempered fractional derivative (or tempered fractional integral) of the
indicator functions. This representation and the classical Itˆo isometry provides to characterize the class of all deterministic functions for which the stochastic integral with respect to
TFBM is well defined. Replacing the Gaussian random measure (Brownian motion) in the
moving average or harmonizable representation of TFBM by a stable random measure, a linear tempered fractional stable motion (LTFSM), or a real harmonizable tempered fractional
stable motion (HTFSM), respectively. Unlike the Gaussian case, LTFSM and HTFSM are
two completely different processes. Existence, basic properties, sample path behavior, and
dependence structure of both processes will be described.

Keywords: Fractional Brownian motion, tempered fractional derivative, harmonizable
representation, long range dependence, reproducing kernel Hilbert space.

I dedicate this dissertation to my lovely wife, Sara Hazinia.

iv

ACKNOWLEDGMENTS

I wish to express my deepest gratitude to my advisor Dr. Mark M. Meerschaert. His
guidance, patience and thoughtfulness made this work possible. It was a privilege to work
with him.
I would like to thank my thesis committee members Drs. Hira Koul, V. S. Mandrekar,
Yimin Xiao and Mantha S. Phanikumar, for their time and interest. I appreciate the help
and support of the Department of Statistics and Probability in last five years. Finally, I
thank my wife, Sara Hazinia for being there for me during this journey and my parents for
their unconditional support.

v

TABLE OF CONTENTS

LIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

vii

Chapter 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

Chapter 2 Tempered Fractional Brownian Motion
2.1 Moving average representation . . . . . . . . . .
2.2 Harmonizable representation . . . . . . . . . . . .
2.3 Tempered fractional Gaussian noise . . . . . . . .
2.4 Sample path properties . . . . . . . . . . . . . .
2.5 Discussion . . . . . . . . . . . . . . . . . . . . . .

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4
4
8
9
13
15

Chapter 3 Tempered Fractional Calculus . . .
3.1 Tempered fractional calculus . . . . . . . . .
3.2 Stochastic Integrals . . . . . . . . . . . . . .
3.2.1 Semi-long range dependence . . . . .
3.2.2 Anti-persistence . . . . . . . . . . . .
3.2.3 Harmonizable representation . . . . .
3.3 Discussion . . . . . . . . . . . . . . . . . . .
3.3.1 White noise approach . . . . . . . . .
3.3.2 Reproducing kernel Hilbert space . .
3.3.3 Tempered distributions as integrands

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17
17
29
35
40
46
49
50
50
52

Chapter 4 Tempered fractional stable motion . .
4.1 Moving average representation . . . . . . . . . .
4.2 Tempered fractional harmonizable stable motion
4.3 Sample path properties . . . . . . . . . . . . . .
4.4 Local Times and Local nondeterminism . . . . .

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54
54
72
76
81

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

87

vi

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LIST OF FIGURES

Figure 2.1

Figure 2.2

Figure 2.3

Figure 4.1

The autocovariance function (2.16) for TFGN with σ = 1, λ = 0.001
and H = 0.7 (solid line) and for the corresponding FGN with σ = 1,
λ = 0 and H = 0.7 (dotted line). . . . . . . . . . . . . . . . . . . .

11

Left panel: Sample paths of TFBM (thick black line) with λ = 0.03
and H = 0.3, and FBM (thin black line) with H = 0.3. Both graphs
use the same noise realization B(t). The right panel shows the same
plots for λ = 0.01 and H = 0.7. . . . . . . . . . . . . . . . . . . . . .

15

The spectral density (2.21) for TFGN with σ = 1, λ = 0.06 and
H = 0.7 (solid line) and FGN with σ = 1, λ = 0 and H = 0.7 (dotted
line). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

16

Left panel: Sample paths of LTFSM (thick black line) with λ = 0.03
and H = 0.3, and LFSM (thin black line) with H = 0.3. Both graphs
use the same noise realization Zα (t). The right panel shows the same
plots for λ = 0.001, H = 0.7 and α = 1.5. . . . . . . . . . . . . . . .

77

vii

Chapter 1
Introduction
Fractional Brownian motion (FBM) is a Gaussian stochastic process whose increments,
termed fractional Gaussian noise (FGN), can exhibit long range dependence [7, 22]. FBM has
become popular in applications to science and engineering, since it yields a simple tractable
model that captures the correlation structure seen in many natural systems [34, 43]. Formed
by convolving Brownian motion with a power law, FBM is essentially the fractional integral or derivative of that Brownian motion [48, 56]. This is also reflected in the correlation
function of FGN, which falls off like a power law with lag, and the corresponding spectral
density, which behaves like a power law near the origin. The increments of long range dependent FBM have a spectral density that blows up like a power law at the origin. This
diverging spectral density is one of the hallmarks of long range dependence. For wind speed
measurements, the spectral density follows this power law model for moderate frequencies,
but the data deviates from that model at low frequencies, and the measured spectal density
remains bounded [55, 28, 53].
In Chapter 2 we define a closely related process, which we call tempered fractional Brownian motion (TFBM). It is defined by exponentially tempering the power law kernel in the
moving average representation of a fractional Brownian motion. TFBM is a Gaussian process
with stationary increments, and we call those increments tempered fractional Gaussian noise
(TFGN). When FGN is long range dependent, TFGN exhibits semi-long range dependence.
Its autocovariance function closely resembles that of FGN on an intermediate scale, but then
1

it eventually falls off more rapidly. Its spectral density resembles a negative power law for
low frequencies, but eventually converges to zero at very low frequencies. TFBM can be a
useful stochastic model for applications where the data follows FBM at some intermediate
scale, but then deviates from FBM at longer scales. For example, wind speed measurements
typically resemble long range dependent FBM over a range of frequencies, but deviate significantly at very low frequencies (corresponding to very long spatial scales). Since the spectal
density of semi-long range dependent TFGN follows the same pattern, it can provide a useful
model for such data.
In Chapter 3 we develop the theory of stochastic integration for TFBM. Our approach follows the seminal work of Pipiras and Taqqu [56] for FBM. A FBM is the fractional derivative
(or integral) of a Brownian motion, in a sense made precise in [56]. A fractional derivative
(or integral) is a (distributional) convolution with a power law [48, 54, 60]. Multiplying that
power law by an exponential factor leads to tempered fractional derivatives and integrals.
TFBM can be written in terms of tempered fractional derivatives (or integrals) of a Brownian motion. This representation and the classical Itˆo Isometry allows us to characterize the
class of all deterministic functions for which the stochastic integral with respect to TFBM
is well defined. We also apply the harmonizable representation of TFBM to define another
type of stochastic integral of deterministic functions with respect to TFBM.
In chapter 4 we consider heavy tailed analogues to TFBM. Replacing the Gaussian random measure (Brownian motion) in the moving average or harmonizable representation of
TFBM by a stable random measure, we obtain a linear tempered fractional stable motion
(LTFSM), or a real harmonizable tempered fractional stable motion (HTFSM), respectively.
Unlike the Gaussian case, LTFSM and HTFSM are two completely different processes. Existence, basic properties, sample path behavior, and dependence structure of both processes
2

are described in this thesis. We also prove that LTFSM and HTFSM are locally nondeterministic on every compact interval.

3

Chapter 2
Tempered Fractional Brownian
Motion
This Chapter has five sections. In Section 2.1, we define tempered fractional Brownian
motion (TFBM) using a moving average representation, and we establish some of its basic
properties. Section 2.2 develops the harmonizable representation of TFBM, and Section 2.3
discusses tempered fractional Gaussian noise (TFGN). Sample path properties of TFBM are
proven in Section 2.4, and an application to wind speed is discussed in Section 2.5.

2.1

Moving average representation

Let {B(t)}t∈R be a real-valued Brownian motion on the real line, a process with stationary
independent increments such that B(t) has a Gaussian distribution with mean zero and
variance σ 2 |t| for all t ∈ R, for some σ > 0. Define an independently scattered Gaussian
random measure B(dx) with control measure m(dx) = σ 2 dx by setting B[a, b] = B(b)−B(a)
for any real numbers a < b, and then extending to all Borel sets. Then the stochastic integrals
∫
∫
I(f ) := f (x)B(dx) are defined for all functions f : R → R such that f (x)2 dx < ∞, as
∫
Gaussian random variables with mean zero and covariance E[I(f )I(g)] = σ 2 f (x)g(x)dx,
see for example Chapter 3 in [61].
Definition 2.1.1. Given an independently scattered Gaussian random measure B(dx) on R
4

with control measure σ 2 dx, for any α < 12 and λ ≥ 0, the stochastic integral
∫ +∞ [
]
−λ(−x)+ (−x)−α B(dx),
Bα,λ (t) :=
e−λ(t−x)+ (t − x)−α
−
e
+
+
−∞

(2.1)

where (x)+ = xI(x > 0), and 00 = 0, will be called a tempered fractional Brownian motion
(TFBM).
It is easy to check that the function

−λ(−x)+ (−x)−α
gα,λ,t (x) := e−λ(t−x)+ (t − x)−α
+ −e
+

(2.2)

is square integrable over the entire real line for any α < 21 , so that TFBM is well-defined,
and of course FBM is a special case of TFBM with λ = 0. Note also that

gα,λ,ct (cx) = c−α gα,cλ,t (x),

(2.3)

for all t, x ∈ R and all c > 0. The next result shows that TFBM has a nice scaling property,
involving both the time scale and the tempering. Here the symbol

indicates equality of

finite dimensional distributions.
Proposition 2.1.2. TFBM (2.1) is Gaussian stochastic process with stationary increments,
such that
{

}

Bα,λ (ct) t∈R

{
}
H
c Bα,cλ (t)

t∈R

(2.4)

for any scale factor c > 0, where the Hurst index H = 1/2 − α.
Proof. Since B(dx) has control measure m(dx) = σ 2 dx, the random measure B(c dx) has

5

control measure c1/2 σ 2 dx. Given t1 < t2 < · · · < tn , a change of variable x = cx′ then yields
(

)
Bα,λ (cti ) : i = 1, . . . , n =

(∫
(∫

)
gα,λ,cti (x)B(dx) : i = 1, . . . , n
c−α gα,cλ,ti (x′ )c1/2 B(dx′ ) : i = 1, . . . , n

)

so that (2.4) holds with H = 1/2 − α. For any s, t ∈ R, the integrand (2.2) satisfies
gα,λ,s+t (s + x) − gα,λ,s (s + x) = gα,λ,t (x), and hence a change of variable x = s + x′ in the
moving average representation yields
(

Bα,λ (s + ti ) − Bα,λ (s) : i = 1, . . . , n

(∫

)

gα,λ,ti (x′ )B(dx′ ) : i = 1, . . . , n

)

which shows that TFBM has stationary increments.
Proposition 2.1.3. The covariance function of TFBM (2.1) has the form
]
[
] σ 2 [ 2 2H
2 |t − s|2H ,
Ct |t| + Cs2 |s|2H − Ct−s
Cov Bα,λ (t), Bα,λ (s) =
2

(2.5)

for any s, t ∈ R, where H = 1/2 − α. Here

Ct2 =

2Γ(H + 12 )
2Γ(2H)
1
√
−
KH (λ|t|),
2H
π
(2λ|t|)
(2λ|t|)H

(2.6)

for t ̸= 0, Ct2 = 0 when t = 0, where Kν (z) is the modified Bessel function of the second
kind.

6

Proof. Use the moving average representation (2.1) with σ = 1 to define

Ct2 : = E[Bα,λ|t| (1)2 ] =
∫ +∞
=

−∞

−2

∫ +∞ [
]
−λt(−x)+ (−x)−α 2 dx
e−λt(1−x)+ (1 − x)−α
−
e
+
+
−∞

e−2λt(1−x)+ (1 − x)−2α
+ dx +

∫ +∞
−∞

∫ +∞
−∞

e−2λt(−x)+ (−x)−2α
+ dx

(2.7)

−λt(−x)+ (−x)−α dx.
e−λt(1−x)+ (1 − x)−α
+ e
+

Apply the definition of the gamma function, along with a standard integral formula from
Page 344 in [24], to see that (2.6) holds. Since TFBM has stationary increments, it follows
from (2.4) that E[Bα,λ (t)2 ] = |t|2H Ct2 for all t real. Recall the elementary formula ab =
1 2
2
2
α,λ (t) and b = B α,λ (s), take expectations, and use the
2 [a + b − (a − b) ], set a = B

stationary increments property again, to see that (2.5) holds.
Remark 2.1.4. The integral representation (2.1) is causal, i.e., Bα,λ (t) depends only on the
values of B(s) for s ≤ t. For applications to spatial statistics, consider
p,q
Bα,λ (t) = p

∫ +∞ [

−∞
∫ +∞ [

+q

−∞

−λ(−x)+ (−x)−α
e−λ(t−x)+ (t − x)−α
+ −e
+

−λ(x)+ (x)−α
e−λ(x−t)+ (x − t)−α
+ −e
+

]
B(dx)
(2.8)

]
B(dx)

for p, q ≥ 0. It is not hard to check, by mimicking the proof of Proposition 2.1.2, that this
process also has stationary increments, and satisfies the scaling property
{
}
p,q
Bα,λ (ct)

t∈R

{
}
p,q
cH Bα,cλ (t)

t∈R

(2.9)

for any scale factor c > 0, where the Hurst index H = 1/2 − α. When p = q > 1, (2.8) is a
well-balanced TFBM.

7

2.2

Harmonizable representation

ˆ1 and B
ˆ2 be independent Gaussian random measures with B
ˆ1 (A) = B
ˆ1 (−A), B
ˆ2 (A) =
Let B
ˆ2 (−A) and E[(B
ˆi (A))2 ] = m(A)/2, where m(dx) = σ 2 dx, and define the complex-valued
−B
ˆ=B
ˆ1 + iB
ˆ2 . If f (x) is a complex-valued function of x real such
Gaussian random measure B
∫
∫
that its Fourier transform fˆ(k) := (2π)−1/2 e−ikx f (x) dx exists and |fˆ(k)|2 dk < ∞,
∫
∫
∫
ˆ fˆ) = fˆ(k)B(dk)
ˆ
ˆ1 (dk) − fˆ2 (k)B
ˆ2 (dk),
we define the stochastic integral I(
:= fˆ1 (k)B
ˆ fˆ) is a Gauswhere fˆ = fˆ1 + ifˆ2 is separated into real and imaginary parts. Then I(
∫
ˆ fˆ)I(ˆ
ˆ g )] = fˆ(k)ˆ
sian random variable with mean zero, such that E[I(
g (k) dk. The Par∫
∫
∫
∫
seval identity f (x)g(x) dx = fˆ(k)ˆ
g (k) dk implies that ( f (x)B(dx), g(x)B(dx))
∫
∫
ˆ
ˆ
( fˆ(k)B(dk),
gˆ(k)B(dk)),
see Proposition 7.2.7 in [61].
Proposition 2.2.1. The TFBM (2.1) has the harmonizable representation
Γ(1 − α)
B α,λ (t) = √
2π

∫ +∞ −itk
e
−1 ˆ
B(dk).
1−α
−∞ (λ − ik)

(2.10)

Proof. To show that the stochastic integral (2.10) exists, note that
∫ +∞
−∞

2
∫ +∞
e−itx − 1
4
dx < ∞,
dx ≤
1−α
2
2 1−α
(λ − ix)
−∞ (λ + x )

since the last integrand is bounded and O(x2α−2 ) as |x| → ∞, for α < 21 . Observe that the

8

function gα,λ,t , given by (2.2), has the Fourier transform
∫ +∞
[
]
1
−λ(−x)+ (−x)−α dx
gα,λ,t (k) = √
e−ikx e−λ(t−x)+ (t − x)−α
−
e
+
+
2π −∞
[∫ t
]
∫ 0
1
−ikx
−λ(t−x)
−α
−ikx
λx
−α
=√
e
e
(t − x) dx −
e
e (−x) dx
2π −∞
−∞
[
]
∫ +∞
∫ +∞
1
−ikt
−u(λ−ik)
−α
−u(λ−ik)
−α
=√
e
e
u du −
e
u du
2π
0
0
Γ(1 − α) e−ikt − 1
.
= √
2π (λ − ik)1−α

(2.11)

Hence by (2.1),
∫ +∞
Bα,λ (t) =

−∞
∫ +∞

gα,λ,t (x)B(dx)

∫
Γ(1 − α) +∞ e−ikt − 1 ˆ
ˆ
gα,λ,t (k)B(dk) = √
B(dk),
1−α
2π
−∞
−∞ (λ − ik)

which is equivalent to (2.10).
Remark 2.2.2. The spectral representation (2.10) reduces to that of causal FBM in the
special case λ = 0, see for example Equation 7.2.17 in [61]. The general TFBM (2.8) has
spectral representation
[
]
∫ −itk
Γ(1 − α)
e
−1
p ik
q ik
p,q
Bα,λ (t) = √
−
B(dk).
ik
(λ − ik)1−α (λ + ik)1−α
2π
R

2.3

(2.12)

Tempered fractional Gaussian noise

Given a TFBM (2.1), we define tempered fractional Gaussian noise (TFGN)

Xj = Bα,λ (j + 1) − Bα,λ (j),

9

for integers −∞ < j < ∞.

(2.13)

It follows easily from (2.1) that TFGN has the moving average representation
∫ +∞ [
]
−λ(j−x)+ (j − x)−α B(dx).
Xj =
e−λ(j+1−x)+ (j + 1 − x)−α
−
e
+
+
−∞

(2.14)

Using (2.10), it also follows that the harmonizable representation of TFGN is
∫
Γ(1 − α) +∞ −ikj e−ik − 1 ˆ
Xj = √
e
B(dk).
(λ − ik)1−α
2π
−∞

(2.15)

It follows from (2.5) that TFGN is a stationary Gaussian time series with mean zero and
covariance function

r(j) := E[X0 Xj ] =

]
σ2 [
2 − 2 |j|2H C 2 + |j − 1|2H C 2
|j + 1|2H Cj+1
j
j−1 ,
2

(2.16)

where H = 1/2 − α, and Cj is given by (2.6).
Remark 2.3.1. Using the well-known fact that Kν (x) ∼

√
π(2x)−1/2 e−x as x → ∞, it

follows easily from (2.6) that

t2H Ct2 → 2Γ(2H)(2λ)−2H ,

as t → ∞.

(2.17)

Hence Cj ∼ Cj+1 as j → ∞. Then, (2.16) along with a Taylor series expansion, shows that

r(j) ∼ σ 2 Cj2 H(2H − 1)|j|2H−2

as j → ∞.

Compare this with Proposition 7.2.10 in [61]. For λ > 0, sufficiently small, the power
law terms in (2.7) dominate, Cj2 remains almost constant, and r(j) falls off like |j|2H−2
10

for moderate values of j > 0. For larger j, the exponential terms in (2.7) dominate, and
(2.17) implies that r(j) ∼ j −2 2H(2H − 1)Γ(2H)(2λ)−2H , as j → ∞. Hence TFGN is
short range dependent, but its covariance function is arbitrarily close to that of long range
dependent FGN for small values of λ, and moderate lags. We call this property semi-long
range dependence, since it is analogous to the semi-heavy tails of Barndorff and Nielsen [21].
Figure 2.1 shows a log-log plot of r(j) in the case H = 0.7 and λ = 0.001, where FGN
exhibits long range dependence.

1

0.1

0.01

0.001
1

10

100

lag

Figure 2.1: The autocovariance function (2.16) for TFGN with σ = 1, λ = 0.001 and H = 0.7
(solid line) and for the corresponding FGN with σ = 1, λ = 0 and H = 0.7 (dotted line).

Proposition 2.3.2. TFGN (2.13) has the spectral density

h(k) =

+∞
2 ∑
Γ(1 − α)2 −ik
σ2
.
e
−1
2π
2 + (k + 2πℓ)2 ]H+1/2
[λ
ℓ=−∞

(2.18)

Proof. Recall that the spectral density
∫ π
+∞
1 ∑ ikj
h(k) =
e r(j) and r(j) =
e−ikj h(k)dk.
2π
−π
j=−∞

11

(2.19)

Define C =

√

2π/Γ(1 − α) and apply (2.15) to write
2

−ik − 1
∫
σ 2 +∞ −ikj e
r(j) = 2
e
dk
C −∞
(λ2 + k 2 )(1−α)
∫ +π
+∞
2 ∑
1
σ2
−ikj
−ik
= 2
e
e
−1
dk
(1−α)
C −π
[λ2 + (k + 2πℓ)2 ]

(2.20)

ℓ=−∞

and then it follows from (2.19) that the spectral density of TFGN is given by (2.18).
Remark 2.3.3. Extending (2.13) to all j real, we obtain the continuous parameter TFGN

Xt = Bα,λ (t + 1) − Bα,λ (t).

The harmonizable representation of this process is given by (2.15) with j replaced by t, and
the proof of Proposition 2.3.2 implies that Xt has spectral density

h(ω) =

2
Γ(1 − α)2 −iω
σ2
e
−1
H+1/2
2π
[λ2 + ω 2 ]

(2.21)

for all real ω. The fact that e−iω −1 ∼ −iω as ω → 0 yields the low frequency approximation
σ 2 Γ(1 − α)2
ω2
h(ω) ≈
.
H+1/2
2π
(λ2 + ω 2 )
See Section 2.5 for an application to wind speed data.

12

2.4

Sample path properties

We say that the sample paths of a stochastic process X(t) satisfy a uniform H¨older condition
of order β on the compact set K ⊂ R if there exists a positive random variable A such that

|X(x) − X(y)| ≤ A|x − y|β ,

almost surely for all x, y ∈ K. We say that the process has H¨older critical exponent γ ∈ (0, 1)
if the process satisfies a uniform H¨older condition of any order β ∈ (0, γ) on any compact
set K ⊂ R, and fails to satisfy this condition for β ∈ (γ, 1).
Theorem 2.4.1. The sample paths of the TFBM (2.1) have H¨older critical exponent H =
1/2 − α, for any α ∈ (−1/2, 1/2), and for any λ ≥ 0.
Proof. Since Bα,λ (0) = 0, it follows from Proposition 4 in [10] that if
{
[
]
(
)
γ = sup β > 0 : E Bα,λ (t)2 = o |t|2β

}
as |t| → 0 ,

(2.22)

then the TFBM Bα,λ (t) satisfies a uniform H¨older condition of order β on any compact set
for any β ∈ (0, γ). Moreover, if we also have
{
( [
])
γ = inf β > 0 : |t|2β = o E Bα,λ (t)2

}
as |t| → 0 ,

(2.23)

then this TFBM has H¨older critical exponent γ. Use the harmonizable representation (2.10)

13

to write
[ −itk
]
∫ +∞ −itk
[
]
1
e
−1
e
−1
2
dk
E Bα,λ (t) = 2
C −∞ (λ − ik)1−α (λ − ik)1−α
∫ +∞
2
= 2
[1 − cos(tk)] (λ2 + k 2 )α−1 dk,
C −∞
√
2π/Γ(1−α), and apply the Tauberian theorem for Fourier transforms, Theorem
[
]
1 in [57], to see that E Bα,λ (t)2 ∼ H(1/t) as t → 0, where
where C =

∫
2
H(x) = 2
(λ2 + k 2 )α−1 dk.
C |k|>x
Since λ2 +k 2 ∼ k 2 as k → ∞, for any ε > 0, there exists some M > 0 such that (1−ε)k 2α−2 <
(λ2 + k 2 )α−1 < (1 + ε)k 2α−2 for all k > M , and hence we have
4(1 − ε)
2α−1 < H(x) < 4(1 + ε) x2α−1
x
(1 − 2α)C 2
(1 − 2α)C 2
, for all x > M . Substitute t = 1/x to see that both (2.22) and (2.23) hold with γ = 1−2α =
2H, which completes the proof.
Remark 2.4.2. The harmonizable representation
∫ +∞ (
X(t) =

−∞

)

ˆ
e−itk − 1 fˆ(k)B(dk)

defines a mean zero Gaussian processes with stationary increments for any Fourier filter fˆ(k)
∫
such that [1 − cos(tk)] |fˆ(k)|2 dk < ∞. If |fˆ(k)|2 is regularly varying at infinity with index
2α − 2 for some −1/2 < α < 1/2, the Karamata Theorem (e.g., see Lemma 5.3.8 (d) in
[47]) implies that H(x) varies regularly at infinity with index 2α − 1, and then the proof
14

X(t)

0

10 20

10
5
−5

−20

0

X(t)

0

100

200

300

400

500

0

t

100

200

300

400

500

t

Figure 2.2: Left panel: Sample paths of TFBM (thick black line) with λ = 0.03 and H = 0.3,
and FBM (thin black line) with H = 0.3. Both graphs use the same noise realization B(t).
The right panel shows the same plots for λ = 0.01 and H = 0.7.
of Theorem 2.4.1 extends to show that X(t) has H¨older critical exponent 1 − 2α. Several
examples of such processes are given in [10].
The sample paths of TFBM closely resemble that of FBM for small values of the tempering parameter λ > 0. The left panel in Figure 2.2 compares a typical sample path of both
processes, simulated using the same white noise B(dx), in a case where FBM is negative
dependent. The right panel shows the corresponding sample paths in a case where FBM
is long range dependent. These simulations use a discretized version of the moving average
representation (2.1). It would also be interesting to develop a simulation method based on
the harmonizable representation (2.10).

2.5

Discussion

Wind speed data are important for electrical power generation and structural engineering.
The most popular model for wind speed near the earth surface, due to Davenport [16], see
also [38], can be written in the form st = µ + Xt , where µ = E[st ] is the average wind speed,

15

0.3
0.25
0.2
0.15
0.1
0.05
0
0

0.5

1

1.5

2

frequency

Figure 2.3: The spectral density (2.21) for TFGN with σ = 1, λ = 0.06 and H = 0.7 (solid
line) and FGN with σ = 1, λ = 0 and H = 0.7 (dotted line).
and Xt has normalized spectral density

4800DV10

x2
4
(1 + x2 ) 3

,

(2.24)

where V10 is the mean velocity (m/sec) at an altitude of 10 meters, D is the corresponding
drag coefficient, and x = 1200ω/V10 . In view of Remark 2.3.3, it is not hard to check
that (2.24) corresponds to the the spectral density of a continuous parameter TFGN with
λ = V10 /1200 and H = 5/6. Hence TFGN can provide a useful stochastic process model for
wind speed data. Figure 2.3 compares the spectral density of TFGN and FGN in the case
where FGN is long range dependent. The spectral density of FGN blows up at the origin
like a power law. The spectral density of TFGN follows the same power law at moderate
frequencies, but remains bounded at very low frequencies, a behavior typically seen in wind
speed data for example in [55, 16, 53, 28].

16

Chapter 3
Tempered Fractional Calculus
This chapter has three sections. In Section 3.1 we prove some basic results on tempered
fractional calculus, which will be needed in the sequel. In Section 3.2 we apply the methods
of Section 3.1 to construct a suitable theory of stochastic integration for tempered fractional
Brownian motion. Finally, in Section 3.3 we discuss model extensions, related results, and
some open questions.

3.1

Tempered fractional calculus

In this section, we define tempered fractional integrals and derivatives, and establish their
essential properties. These results will form the foundation of the stochastic integration
theory developed in Section 3.2. We begin with the definition of a tempered fractional
integral.
Definition 3.1.1. For any f ∈ Lp (R) (where 1 ≤ p < ∞), the positive and negative tempered
fractional integrals are defined by

and

∫ +∞
1
α,λ
−λ(t−u)+ du,
I+ f (t) =
f (u)(t − u)α−1
+ e
Γ(α) −∞

(3.1)

∫ ∞
1
α,λ
I− f (t) =
e−λ(u−t)+ du,
f (u)(u − t)α−1
+
Γ(α) −∞

(3.2)

17

∫
respectively, for any α > 0 and λ > 0, where Γ(α) = 0+∞ e−x xα−1 dx is the Euler gamma
function, and (x)+ = xI(x > 0).
When λ = 0 these definitions reduce to the (positive and negative) Riemann-Liouville
fractional integral [48, 54, 60], which extends the usual operation of iterated integration to a
fractional order. When λ = 1, the operator (3.1) is called the Bessel fractional integral [60,
Section 18.4].
α,λ

Lemma 3.1.2. For any α > 0, λ > 0, and p ≥ 1, I±

is a bounded linear operator on

Lp (R) such that
∥I± f ∥p ≤ λ−α ∥f ∥p ,
α,λ

(3.3)

for all f ∈ Lp (R).
Proof. Young’s Theorem [60, p. 12] states that if ϕ ∈ L1 (R) and f ∈ Lp (R) then ϕ∗f ∈ Lp (R)
and the inequality
∥ϕ ∗ f ∥p ≤ ∥ϕ∥1 ∥f ∥p ,

(3.4)

where ∗ denotes the convolution

[f ∗ ϕ](t) =

∫ +∞
−∞

f (u)ϕ(t − u)du = [ϕ ∗ f ](t).

Obviously I± is linear, and I± f (t) = [f ∗ ϕ±
α ](t) where
α,λ

α,λ

ϕ+
α (t) =

1 α−1 −λt
t
e 1(0,∞) (t),
Γ(α)

1
ϕ−
(−t)α−1 e−λ(−t) 1(−∞,0) (t)
α (t) =
Γ(α)

18

(3.5)

for any α, λ > 0. But

∥ϕ±
α ∥1 =

∫ +∞
]
1
1 [ −α
λ Γ(α) = λ−α ,
tα−1 e−λt dt =
Γ(α) 0
Γ(α)

using the formula for the Laplace transform (moment generating function) of the gamma
probability density, and then (3.3) follows from Young’s Inequality (3.4).
Next we prove a semigroup property for tempered fractional integrals, which follows easily
from the following property of the convolution kernels in the definition (3.1.1).
Lemma 3.1.3. For any λ > 0 the functions (3.5) satisfy

±
±
ϕ±
α ∗ ϕβ = ϕα+β ,

(3.6)

for any α > 0 and β > 0.
Proof. For t > 0 we have
∫ t
1
+
+
(t − s)α−1 e−λ(t−s) sβ−1 e−λs ds
ϕα ∗ ϕβ (t) =
Γ(α)Γ(β) 0
=

1
tα+β−1 e−λt = ϕ+
α+β (t)
Γ(α + β)

using the formula for the beta probability density. The proof for ϕ−
α is similar.
The following lemma establishes the semigroup property for tempered fractional integrals
on Lp (R). In the case λ = 0, the semigroup property for fractional integrals is well known
(e.g., see Samko et al. [60, Theorem 2.5]).

19

Lemma 3.1.4. For any λ > 0 we have

α,λ β,λ

α+β,λ

I± I± f = I±

f

(3.7)

for all α, β > 0 and all f ∈ Lp (R).
Proof. Lemma 3.1.2 shows that both sides of (3.7) belong to Lp (R) for any f ∈ Lp (R), and
α,λ

then the result follows immediately from Lemma 3.1.3 along with the fact that I± f (t) =
[f ∗ ϕ±
α ](t).
The next result shows that positive and negative tempered fractional integrals are adjoint
∫
operators with respect to the inner product ⟨f, g⟩2 = f (x)g(x) dx on L2 (R).
Lemma 3.1.5 (Integration by parts). Suppose f, g ∈ L2 (R). Then

α,λ

α,λ

⟨f, I+ g⟩2 = ⟨I− f, g⟩2

(3.8)

for any α > 0 and any λ > 0.
Proof. Write
∫ +∞
−∞

∫ x
1
g(u)(x − u)α−1 e−λ(x−u) du dx
f (x)
Γ(α) −∞
−∞
∫
∫ +∞
g(u) +∞
=
f (x)(x − u)α−1 e−λ(x−u) dx du
Γ(α)
u
−∞
∫ +∞
α,λ
=
I− f (x)g(x) dx

α,λ
f (x)I+ g(x) dx =

∫ +∞

−∞

and this completes the proof.
Next we discuss the relationship between tempered fractional integrals and Fourier trans20

forms. Recall that the Fourier transform
∫ +∞
1
ˆ
F[f ](k) = f (k) = √
e−ikx f (x)dx
2π −∞
, for functions f ∈ L1 (R) ∩ L2 (R) can be extended to an isometry (a linear onto map that
preserves the inner product) on L2 (R) such that
∫ n
1
f (k) = lim √
e−ikx f (x) dx,
n→∞ 2π −n

(3.9)

for any f ∈ L2 (R), see for example [31, Theorem 6.6.4].
Lemma 3.1.6. For any α > 0 and λ > 0 we have

F[I± f ](k) = fˆ(k)(λ ± ik)−α ,
α,λ

(3.10)

for all f ∈ L1 (R) and all f ∈ L2 (R).
Proof. The function ϕ+
α in (3.5) has Fourier transform

F[ϕ+
α ](k) =

∫ ∞
1
1
√
e−ikt tα−1 e−λt dt = √ (λ + ik)−α
Γ(α) 2π 0
2π

(3.11)

by the formula for the Fourier transform of a gamma density. For any two functions f, g ∈
√
L1 (R), the convolution f ∗ g ∈ L1 (R) has Fourier transform 2π fˆ(k)ˆ
g (k) (e.g., see [48,
α,λ

p. 65]), and then (3.10) follows. The argument for I−

is quite similar. If f ∈ L2 (R),

approximate by the L1 function f (x)1[−n,n] (x) and let n → ∞.
Remark 3.1.7. Recall that the space of rapidly decreasing functions S(R) consists of the
21

infinitely differentiable functions g : R → R such that

sup xn g (m) (x) < ∞,
x∈R

where n, m are non-negative integers, and g (m) is the derivative of order m. The space S ′ (R)
of continuous linear functionals on S(R) is called the space of tempered distributions. The
Fourier transform, and inverse Fourier transform, can then be extended to linear continuous
mappings of S ′ (R) into itself. If f : R → R is a measurable function with polynomial growth,
∫
∫
so that |f (x)|(1 + |x|)−p dx < ∞ for some p > 0, then Tf (φ) = f (x)φ(x) dx := ⟨f, φ⟩1
is a tempered distribution, also called a generalized function. The Fourier transform of this
generalized function is defined as Tˆf (φ) = ⟨fˆ, φ⟩1 = ⟨f, φ⟩
ˆ 1 = Tf (φ)
ˆ for φ ∈ S(R). See
Yosida [68, Ch.VI] for more details. If f is a tempered distribution, then the tempered
fractional integrals I± f (x) exist as convolutions with the tempered distributions ϕ±
α in
α,λ

(3.5). The same holds for Riemann-Liouville fractional integrals (the case λ = 0), but that
1
case is more delicate, because the power law kernels ϕ±
α of (3.5) with λ = 0 are not in L (R).

Next we consider the inverse operator of the tempered fractional integral, which is called
a tempered fractional derivative. For our purposes, we only require derivatives of order
0 < α < 1, and this simplifies the presentation.
Definition 3.1.8. The positive and negative tempered fractional derivatives of a function
f : R → R are defined as
∫ t
α
f (t) − f (u) −λ(t−u)
α,λ
α
e
du,
D+ f (t) = λ f (t) +
Γ(1 − α) −∞ (t − u)α+1

22

(3.12)

and

∫ +∞
f (t) − f (u) −λ(u−t)
e
du,
(u − t)α+1

α
α,λ
D− f (t) = λα f (t) +
Γ(1 − α) t

(3.13)

respectively, for any 0 < α < 1 and any λ > 0.
If λ = 0, the definitions (3.12) and (3.13) reduce to the positive and negative Marchaud
fractional derivatives [60, Section 5.4].
Note that tempered fractional derivatives cannot be defined pointwise for all functions
f ∈ Lp (R), since we need |f (t) − f (u)| → 0 fast enough to counter the singularity of the
denominator (t − u)α+1 , as u → t.
Next we establish the existence and compute the Fourier transform of tempered fractional
derivatives on a natural domain.
Theorem 3.1.9. Assume f and f ′ are in L1 (R). Then the tempered fractional derivative
α,λ

D+ f (t) exists and
α,λ

F[D± f ](k) = f (k)(λ ± ik)α ,

(3.14)

for any 0 < α < 1 and any λ > 0.
Proof. A standard argument from functional analysis (e.g., see [50, Proposition 2.2]) shows
that if f, f ′ ∈ L1 (R), then
∫ ∫
I :=

|f (t) − f (u)|
dt du < ∞,
1+α
R R |t − u|

23

(3.15)

for any 0 < α < 1. To see this, write I = I1 + I2 where
∫ ∫

|f (t) − f (u)|
dt du
1+α
R R∩{|t−u|<1} |t − u|
∫ ∫
|f (t) − f (z + t)|
=
dz dt
|z|1+α
R {|z|<1}
∫ ∫
∫ 1
−α
≤
|z|
|f ′ (t + uz)| du dz dt =

I1 : =

R {|z|<1}

0

2
∥f ′ ∥L1 (R) < ∞
1−α

and
∫ ∫

|f (t) − f (u)|
dt du
1+α
R R∩{|t−u|≥1} |t − u|
∫ ∫
|f (t)| + |f (z + t)|
2
≤
dt dz = ∥f ∥L1 (R) < ∞.
1+α
α
|z|
R {|z|≥1}

I2 : =

Now it follows easily from (3.15) that D± f exists for all f, f ′ ∈ L1 (R). Define
α,λ

∫ t
f (t) − f (u) −λ(t−u)
α
e
du,
F (t) :=
Γ(1 − α) −∞ (t − u)α+1
and apply the Fubini Theorem, along with the shift property F[f (t − y)](k) = e−iky fˆ(k) of
the Fourier transform, to see that
∫ +∞
∫ ∞
α
f (t) − f (t − y) −λy
−ikt
√
e
e
dy dt
F (k) =
y α+1
Γ(1 − α) 2π −∞
0
∫ +∞
(
)
α
I (α)
=
f (k),
y −α−1 e−λy 1 − e−iky f (k) dy = λ
Γ(1 − α) 0
Γ(1 − α)
where
∫ +∞ (
Iλ (α) =

0

)
e−λy − e−(λ+ik)y αy −α−1 dy.

24

(3.16)

Integrate by parts with u = e−λy − e−(λ+ik)y and use the fact that e−λy − e−(λ+ik)y = O(y)
as y → 0, to obtain
[(
Iλ (α) =

e−λy − e−(λ+ik)y

)(

)
−y −α

∫ ∞

] ∞
+

0

0

[

y −α −λe−λy + (λ + ik)e−(λ+ik)y

]
dy.

Use the definition of the gamma function, and the formula for the Fourier tranform of the
gamma probability density, to obtain
∫ ∞

∫ ∞

y −α e−(λ+ik)y dy
0
0
(
)
Γ(1 − α)
ik α−1
α
= −λ Γ(1 − α) + (λ + ik) 1−α
1+
λ
λ

Iλ (α) = −λ

y −α e−λy dy + (λ + ik)

= Γ(1 − α) [(λ + ik)α − λα ] .
α,λ

Then F (k) = f (k) [(λ + ik)α − λα ], and hence F[D+ f ](k) = (λ + ik)α fˆ(k). The proof for
α,λ

F[D− f ](k) is similar.
Remark 3.1.10. Theorem 3.1.9 can also be proven, under somewhat stronger conditions,
using the generator formula for infinitely divisible semigroups [48, Theorem 3.17 and Theorem 3.23 (b)].
Next we extend the definition of tempered fractional derivatives to a suitable class of
functions in L2 (R). For any α > 0 and λ > 0, we define the fractional Sobolev space
∫
W α,2 (R) := {f ∈ L2 (R) :

R

(λ2 + k 2 )α |fˆ(k)|2 dk < ∞},

(3.17)

which is a Banach space with norm ∥f ∥α,λ = ∥(λ2 + k 2 )α/2 fˆ(k)∥2 . The space W α,2 (R) is
the same for any λ > 0 (typically we take λ = 1) and all the norms ∥f ∥α,λ are equivalent,
25

since 1 + k 2 ≤ λ2 + k 2 ≤ λ2 (1 + k 2 ) for all λ ≥ 1, and λ2 + k 2 ≤ 1 + k 2 ≤ λ−2 (1 + k 2 ) for
all 0 < λ < 1.
α,λ

Definition 3.1.11. The positive (resp., negative) tempered fractional derivative D± f (t) of
a function f ∈ W α,2 (R) is defined as the unique element of L2 (R) with Fourier transform
f (k)(λ ± ik)α , for any α > 0 and any λ > 0.
Remark 3.1.12. The pointwise definition of the tempered fractional derivative in real space
is more complicated when α > 1. For example, when 1 < α < 2 we have
∫ t
α
f (u) − f (t) + (t − u)f ′ (t) −λ(t−u)
α,λ
α
α−1
′
f (x) +
D+ f (t) = λ f (t) + αλ
e
du,
Γ(1 − α) −∞
(t − u)α+1
for all f ∈ W 1,2 (R), compare [48, Remark 7.11].
Lemma 3.1.13. For any α > 0, β > 0 and λ > 0 we have

α,λ β,λ

α+β,λ

D± D± f (t) = D±

f (t),

for any f ∈ W α+β,2 (R).
Proof. It is obvious from (3.17) that W α,2 (R) ⊂ W β,2 (R) for α > β. It is clear from
β,λ

Definition 3.1.11 that D± f (t) exists and belongs to W α,2 (R) for any f ∈ W α+β,2 (R), and
α,λ
likewise, D± f (t) exists and belongs to L2 (R) for any f ∈ W α,2 (R).

Lemma 3.1.14. For any α > 0 and λ > 0, we have

α,λ α,λ

D± I± f (t) = f (t)

26

(3.18)

for any function f ∈ L2 (R), and

α,λ α,λ

I± D± f (t) = f (t)

(3.19)

for any f ∈ W α,2 (R).
α,λ
Proof. Given f ∈ L2 (R), note that g(t) = I± f (t) satisfies gˆ(k) = fˆ(k)(λ±ik)−α by Lemma

3.1.6, and then it follows easily that g ∈ W α,2 (R). Definition 3.1.11 implies that

α,λ

α,λ α,λ

F[D± I± f ](k) = F[D± g](k) = g(k)(λ ± ik)α = fˆ(k),

(3.20)

and then (3.18) follows using the uniqueness of the Fourier transform. The proof of (3.19)
is similar.
Lemma 3.1.15. Suppose f, g ∈ W α,2 (R). Then

α,λ

α,λ

⟨f, D+ g⟩2 = ⟨D− f, g⟩2 ,

(3.21)

for any α > 0 and any λ > 0.
Proof. Apply the Plancherel Theorem along with Definition 3.1.11 to see that

α,λ
⟨f, D+ g⟩2 =

∫

α,λ
α,λ
f (x)D+ g(x) dx = ⟨fˆ, (λ + ik)α gˆ⟩2 = ⟨(λ − ik)α fˆ, gˆ⟩2 = ⟨D− f, g⟩2

and this completes the proof.
Remark 3.1.16. One can also prove (3.21) for f, f ′ , g, g ′ ∈ L1 (R) ∩ L2 (R) using integration
by parts, compare [69, Appendix A.1].
27

A slightly different tempered fractional derivative
∫ t
α
f (t) − f (u) −λ(t−u)
α,λ
D+ f (t) =
e
du,
Γ(1 − α) −∞ (t − u)α+1
∫ +∞
α
f (t) − f (u) −λ(u−t)
α,λ
e
du,
D− f (t) =
Γ(1 − α) t
(u − t)α+1

(3.22)

was proposed by Cartea and del-Castillo-Negrete [12] for a problem in physics, and studied
further by Baeumer and Meerschaert [4, 48] using tools from probability theory and semiα,λ
groups. When 0 < α < 1, F[D± f ](k) = f (k)[(λ ± ik)α − λα ]fˆ(k) for suitable functions f .

The additional λα term makes the evolution equation

∂
α,λ
α,λ
u(x, t) = [pD+ + qD− ]u(x, t),
∂t

(3.23)

for p, q ≥ 0 mass preserving, which can easily be seen by considering the Fourier transform
uˆ(k, t) = exp(t[(λ ± ik)α − λα ]) of point source solutions to the tempered fractional diffusion
equation (3.23). Now x → u(x, t) are the probability density functions of a tempered stable
L´evy process, as in Rosi´
nski [59]. That process arises as the long-time scaling limit of a random walk with exponentially tempered power law jumps, see Chakrabarty and Meerschaert
[13]. The tempered fractional diffusion equation (3.23) has been applied to contaminant
plumes in underground aquifers, and sediment transport in rivers [49, 70, 71].
Remark 3.1.17. Tempered fractional derivatives are a natural analogues of integer (and
fractional) order derivatives. For suitable functions f (x), the Fourier transform of the derivative f ′ (x) is (ik)fˆ(k) (e.g., see [48, p. 8]), and one can define the fractional derivative Dα
± f (t)
as the function with Fourier transform (ik)α fˆ(k). Definition 3.1.11 extends to tempered fractional derivatives.

28

3.2

Stochastic Integrals

In this section, we apply tempered fractional calculus to define stochastic integrals with
respect to TFBM. First we recall the moving average representation of TFBM as a stochastic
integral with respect to Brownian motion.
Recall from Definition 2.1.1, the stochastic integral
∫ +∞ [
Bα,λ (t) =

−∞

−λ(−x)+ (−x)−α
e−λ(t−x)+ (t − x)−α
+ −e
+

]
B(dx).

(3.24)

When λ = 0 and α < −1/2, the right-hand side of (3.24) does not exist, since the integrand
is not in L2 (R). However, TFBM with λ > 0 and α < −1/2 is well-defined, because the
exponential tempering keeps the integrand in L2 (R). In fact, if α < −1/2 and λ > 0, or
if α = 0 and λ > 0, we will now show that TFBM is a semimartingale, and hence one
∫
can define stochastic integrals I(f ) := f (x)Bα,λ (dx) in the standard manner, via the Itˆo
stochastic calculus (e.g., see Kallenberg [30, Chapter 15]).
Theorem 3.2.1. A tempered fractional Brownian motion {Bα,λ (t)}t≥0 with α < −1/2 and
λ > 0 is a continuous semimartingale with the canonical decomposition

Bα,λ (t) = −λ

where

∫ t
0

Mα,λ (s) ds − α

∫ +∞
Mα,λ (t) :=

−∞

∫ t
0

Mα+1,λ (s) ds

e−λ(t−x)+ (t − x)−α
+ B(dx).

(3.25)

(3.26)

Moreover, {Bα,λ (t)}t≥0 is a finite variation process. The same is true if α = 0 and λ > 0.
Proof. Let {FtB }t≥0 be the σ-algebra generated by {Bs : 0 ≤ s ≤ t}. Given a function
29

g : R → R such that g(t) = 0 for all t < 0, and
∫ t
g(t) = C +

h(s), ds for all t > 0,

(3.27)

0

for some C ∈ R and some h ∈ L2 (R), a result of Cheridito [14, Theorem 3.9] shows that the
Gaussian stationary increment process

g
Yt :=

∫
R

[g(t − u) − g(−u)] B(du), t ≥ 0

(3.28)

is a continuous {FtB }t≥0 semimartingale with canonical decomposition

g
Yt = g(0)Bt +

∫ t∫ s
0

−∞

h(s − u)B(du)ds,

(3.29)

and conversely, that if (3.28) defines a semimartingale on [0, T ] for some T > 0, then g
satisfies these properties. Define g(t) = 0 for t ≤ 0 and

g(t) := e−λt t−α

for t > 0.

(3.30)

It is easy to check that the function g(t − u) − g(−u), which is the integrand in (3.24), is
square integrable over the entire real line for any α < 1/2 and λ > 0. Next observe that
(3.27) holds with C = 0, h(s) = 0 for s < 0 and

h(s) :=

d −λs −α
[e
s ] = −λe−λs s−α − αe−λs s−α−1 ∈ L2 (R)
ds

(3.31)

for any α < −1/2 and λ > 0. Then it follows from [14, Theorem 3.9] that TFBM is a

30

continuous semimartingale with canonical decomposition
∫ t∫ s
Bα,λ =

0

−∞

−λe−λ(s−u) (s − u)−α − αe−λ(s−u) (s − u)−α−1 B(du) ds

(3.32)

which reduces to (3.25). Since C = 0, Theorem 3.9 in [14] implies that {Bα,λ (t)} is a finite
variation process. The proof for α = 0 is similar, using g(t) = e−λt for t > 0.
Remark 3.2.2. When α = 0 and λ > 0, the Gaussian stochastic process (3.26) is an
Ornstein-Uhlenbeck process. When α < −1/2 and λ > 0, it is a one dimensional Mat´ern
stochastic process [5, 23, 26], also called a “fractional Ornstein-Uhlenbeck process” in the
physics literature [40]. It follows from Knight [32, Theorem 6.5] that Mα,λ (t) is a semimartingale in both cases.
Cheridito [14, Theorem 3.9] provides a necessary and sufficient condition for the process (3.28) to be a semimartingale, and then it is not hard to check that TFBM is not a
semimartingale in the remaining cases when −1/2 < α < 0 or 0 < α < 1/2. Next we will
investigate the problem of stochastic integration with deterministic integrands in these two
cases. Our approach follows that of Pipiras and Taqqu [56].
Next we establish a link between TFBM and tempered fractional calculus.
Lemma 3.2.3. For a tempered fractional Brownian motion (3.24) with λ > 0, we have:
(i) When −1/2 < α < 0, we can write
∫ +∞ [
]
κ,λ
κ+1,λ
Bα,λ (t) = Γ(κ + 1)
I− 1[0,t] (x) − λI−
1[0,t] (x) B(dx)
−∞

where κ = −α.
31

(3.33)

(ii) When 0 < α < 1/2, we can write
∫ +∞ [
]
1−α,λ
α,λ
1[0,t] (x) B(dx).
Bα,λ (t) = Γ(1 − α)
D− 1[0,t] (x) − λI−
−∞

(3.34)

Proof. To prove part (i), write the kernel function from (3.24) in the form

−λ(−x)+ (−x)−α
gt,λ (x) : = e−λ(t−x)+ (t − x)−α
+ −e
+
[
]
∫ t d e−λ(u−x)+ (u − x)κ
+
=
du
du
0
∫ +∞
(κ+1)−1
= −λ
1[0,t] (u)e−λ(u−x)+ (u − x)+
du
−∞
∫ +∞
+κ
1[0,t] (u)e−λ(u−x)+ (u − x)κ−1
+ du
−∞

and apply the definition (3.2) of the tempered fractional integral.
To prove part (ii), it suffices to show that the integrand

−λ(0−x)+ (0 − x)−α =: ϕ (x) − ϕ (x)
gt,λ (x) = e−λ(t−x)+ (t − x)−α
t
0
+ −e
+

in (3.24) equals the integrand in (3.34). We will prove this using Fourier transforms. The
substitution u = t − x shows that
∫ t
1
e−ikt Γ(1 − α)
,
ϕt (k) = √
e−ikx e−λ(t−x) (t − x)−α dx = √
2π −∞
2π(λ − ik)1−α
using the formula for the Fourier transform of the gamma density, and hence
e−ikt − 1
gt,λ (k) = ϕt (k) − ϕ0 (k) = Γ(1 − α) √
.
2π(λ − ik)1−α

32

(3.35)

On the other hand, from Lemma 3.1.6 and Theorem 3.1.9 we obtain

α,λ

1−α,λ

F[D− 1[0,t] − λI−

1[0,t] ](k) = [(λ − ik)α − λ(λ − ik)α−1 ] ·

e−ikt − 1
√
(−ik) 2π

e−ikt − 1
= (λ − ik)α−1 · √
,
2π

(3.36)

where we have used the formula (which is easy to verify)
e−ikb − e−ika
ˆ
√
h(k)
= F[1[a,b) ](k) =
.
(−ik) 2π

(3.37)

The desired result now follows by the uniqueness of the Fourier transform.
Next, we explain the connection between the fractional calculus representations (3.33) and
(3.34). Substitute κ = −α into (3.33) and note that the resulting formula differs from (3.34)
−α,λ

only in that the tempered fractional integral I−
α,λ

α,λ

is replaced by the tempered fractional
α,λ

derivative D− . Lemma 3.1.14 shows that I− and D− are inverse operators, and hence it
−α,λ

makes sense to define I±

α,λ

:= D± when 0 < α < 1. Now equations (3.33) and (3.34) are

equivalent.
Next, we discuss a general construction for stochastic integrals with respect to TFBM.
For a standard Brownian motion {B(t)}t∈R on (Ω, F, P ), the stochastic integral I(f ) :=
∫
f (x)B(dx) is defined for any f ∈ L2 (R), and the mapping f → I(f ) defines an isometry
from L2 (R) into L2 (Ω), called the Itˆ
o isometry:
∫
⟨I(f ), I(g)⟩L2 (Ω) = Cov[I(f ), I(g)] =

f (x)g(x) dx = ⟨f, g⟩L2 (R) .

(3.38)

Since this isometry maps L2 (R) onto the space Sp(B) = {I(f ) : f ∈ L2 (R)}, we say that

33

these two spaces are isometric. For any elementary function (step function)

f (u) =

n
∑
i=1

ai 1[t ,t ) (u),
i i+1

(3.39)

where ai , ti are real numbers such that ti < tj for i < j, it is natural to define the stochastic
integral

∫
I α,λ (f ) =

R

f (x)Bα,λ (dx) =

n
∑

[
]
ai Bα,λ (ti+1 ) − Bα,λ (ti ) ,

(3.40)

i=1

and then it follows immediately from (3.33) that for f ∈ E, the space of elementary functions,
the stochastic integral
∫
I α,λ (f ) =

R

∫ [
]
κ,λ
κ+1,λ
f (x)Bα,λ (dx) = Γ(κ + 1)
I− f (x) − λI−
f (x) B(dx)
R

is a Gaussian random variable with mean zero, such that for any f, g ∈ E we have
(∫
⟨I α,λ (f ), I α,λ (g)⟩
= Γ(κ + 1)2

∫ [
R

=E
L2 (Ω)

R

∫
f (x)Bα,λ (dx)

R

)
g(x)Bα,λ (dx)

][
]
κ,λ
κ+1,λ
κ,λ
κ+1,λ
I− f (x) − λI−
f (x) I− g(x) − λI−
g(x) dx,

(3.41)

in view of (3.33) and the Itˆo isometry (3.38). The linear space of Gaussian random variables
{
}
I α,λ (f ), f ∈ E is contained in the larger linear space
{
}
Sp(Bα,λ ) = X : I α,λ (fn ) → X in L2 (Ω) for some sequence (fn ) in E .

An element X ∈ Sp(Bα,λ ) is mean zero Gaussian with variance

Var(X) = lim Var[I α,λ (fn )],
n→∞

34

(3.42)

and X can be associated with an equivalence class of sequences of elementary functions (fn )
such that I α,λ (fn ) → X in L2 (R). If [fX ] denotes this class, then X can be written in an
integral form as

∫
X=

R

[fX ]dBα,λ

(3.43)

and the right hand side of (3.43) is called the stochastic integral with respect to TFBM
on the real line (see, for example, Huang and Cambanis [27], page 587). In the special
case of a Brownian motion λ = α = 0, I α,λ (fn ) → X along with the Itˆo isometry (3.38)
implies that (fn ) is a Cauchy sequence, and then since L2 (R) is a (complete) Hilbert space,
there exists a unique f ∈ L2 (R) such that fn → f in L2 (R), and we can write X =
∫
R f (x)B(dx). However, if the space of integrands is not complete, then the situation is
more complicated. We begin with the case −1/2 < α < 0, where the corresponding FBM is
long range dependent.

3.2.1

Semi-long range dependence

Here we investigate stochastic integrals with respect to TFBM in the case −1/2 < α < 0, so
that 1/2 < H < 1 in (2.4). Equation (3.41) suggests the appropriate space of integrands for
TFBM, in order to obtain a nice isometry that maps into the space Sp(Bα,λ ) of stochastic
integrals.
Theorem 3.2.4. Given −1/2 < α < 0 and λ > 0, let κ = −α. Then the class of functions
{
∫
2
A1 := f ∈ L (R) :

R

}

2
κ,λ
κ+1,λ
I− f (x) − λI−
f (x) dx < ∞

35

,

(3.44)

is a linear space with inner product

⟨f, g⟩A := ⟨F, G⟩L2 (R)

(3.45)

1

where
κ+1,λ

κ,λ

F (x) = Γ(κ + 1)[I− f (x) − λI−

f (x)],
(3.46)

κ,λ
κ+1,λ
G(x) = Γ(κ + 1)[I− g(x) − λI−
g(x)].

The set of elementary functions E is dense in the space A1 . The space A1 is not complete.
κ,λ

κ+1,λ

The proof of Theorem 3.2.4 requires one simple lemma, which shows that I− − λI−
is a bounded linear operator on Lp (R) for any 1 ≤ p < ∞.

Lemma 3.2.5. Under the assumptions of Theorem 3.2.4, suppose 1 ≤ p < ∞. Then for
any f ∈ Lp (R) we have

κ,λ

κ+1,λ

∥I− f (x) − λI−

f (x)∥p ≤ C∥f ∥p

(3.47)

where C is a constant depending only on α and λ.
κ,λ

κ+1,λ

Proof. It follows from Lemma 3.1.2 that I− f (x) − λI−

κ,λ

κ+1,λ

∥I− f (x) − λI−

κ,λ

f (x) ∈ Lp (R) and that

κ+1,λ

f (x)∥p ≤ ∥I− f (x)∥p + λ∥I−

f (x)∥p ≤ 2λ−κ ∥f ∥p

for any f ∈ Lp (R).
Remark 3.2.6. It follows from Lemma 3.2.5 that A1 contains every function in L2 (R), and
hence they are the same set, but endowed with a different inner product. The inner product
36

on the space A1 is required to obtain a nice isometry.
Proof of Theorem 3.2.4. The proof is similar to [56, Theorem 3.2]. To show that A1 is an
inner product space, we will check that ⟨f, f ⟩A = 0 implies f = 0 almost everywhere. If
1

⟨f, f ⟩A = 0, then in view of (3.45) and (3.46) we have ⟨F, F ⟩2 = 0, so F (x) = Γ(1 +
1

κ,λ

κ+1,λ

κ)[I− f (x) − λI−

f (x)] = 0 for almost every x ∈ R. Then

κ,λ

κ+1,λ

I− f (x) = λI−

f (x) for almost every x ∈ R.

(3.48)

κ,λ

Apply D− to both sides of equation (3.48) and use Lemma 3.1.4 along with Lemma 3.1.14
to get

κ,λ κ,λ

κ,λ

κ+1,λ

f (x) = D− I− f (x) = D− λI−

[
]
κ,λ κ,λ 1,λ
1,λ
f (x) = λ D− I− I− f (x) = λI− f (x)

for almost every x ∈ R, and in view of the definition (3.1) this is equivalent to
∫ +∞
f (x) = λ

f (u)e−λ(u−x) du = λeλx

x

∫ +∞

f (u)e−λu du

(3.49)

x

for almost every x ∈ R. Observe that the functions f (u) and e−λu are in L2 [x, ∞) for any
x ∈ R and then, by the Cauchy-Schwartz inequality, the function f (u)e−λu is in L1 [x, ∞). It
∫
follows that x+∞ f (u)e−λu du is absolutely continuous, and so the function f (x) in (3.49) is
also absolutely continuous. Taking the derivative on both sides of (3.49) using the Lebesgue
Differentiation Theorem (e.g., see [66, Theorem 7.16]) we get

f ′ (x) = λf (x) − λeλx f (x)e−λx = 0 for almost every x ∈ R.

37

Then for any a, b ∈ R we have
∫ b
f (b) = f (a) +

f ′ (x) dx = f (a).

a

and so f (x) is a constant function. Since f ∈ L2 (R), it follows that f (x) = 0 for all x ∈ R,
and hence A1 is an inner product space.
Next, we want to show that the set of elementary functions E is dense in A1 . For any
f ∈ A1 , we also have f ∈ L2 (R), and hence there exists a sequence of elementary functions
(fn ) in L2 (R) such that ∥f − fn ∥2 → 0. But

∥f − fn ∥A1 = ⟨f − fn , f − fn ⟩A = ⟨F − Fn , F − Fn ⟩2 = ∥F − Fn ∥2 ,
1

κ,λ

κ+1,λ

where Fn (x) = I− fn (x) − λI−

fn (x) and F (x) is given by (3.46). Lemma 3.2.5 implies

that

κ,λ

κ+1,λ

∥f − fn ∥A1 = ∥F − Fn ∥2 = ∥I− (f − fn ) − λI−

(f − fn )∥2 ≤ C∥f − fn ∥2

for some C > 0, and since ∥f − fn ∥2 → 0, it follows that the set of elementary functions is
dense in A1 .
Finally, we provide an example to show that A1 is not complete. The functions

fn (k) = |k|−p 1{1<|k|<n} (k), p > 0,

are in L2 (R), fn (k) = fn (−k), and hence they are the Fourier transforms of functions
fn ∈ L2 (R). Apply Lemma 3.1.6 to see that the corresponding functions Fn (x) = Γ(κ +

38

κ,λ

κ+1,λ

1)[I− fn (x) − λI−

fn (x)] from (3.46) have Fourier transform

−ikΓ(1 − α) ˆ
F[Fn ](k) = Γ(1 − α)[(λ − ik)α − λ(λ − ik)α−1 ]fˆn (k) =
fn (k).
(λ − ik)1−α

(3.50)

Since α < 0, it follows that

∥Fn ∥22 = ∥Fˆn ∥22 = Γ(1 − α)2

∫ ∞
−∞

fn (k)

2

k2
dk < ∞
(λ2 + k 2 )1−α

for each n, which shows that fn ∈ A1 . Now it is easy to check that fn − fm → 0 in
A1 , as n, m → ∞, whenever p > 1/2 + α, so that (fn ) is a Cauchy sequence. Choose
p ∈ (1/2 + α, 1/2) and suppose that there exists some f ∈ A1 such that ∥fn − f ∥A1 → 0 as
n → ∞. Then

∫ ∞
−∞

fn (k) − f (k)

2

k2
dk → 0
(λ2 + k 2 )1−α

(3.51)

as n → ∞, and since, for any given m ≥ 1, the value of fn (k) does not vary with n > m
whenever k ∈ [−m, m], it follows that fˆ(k) = |k|−p 1{|k|>1} on any such interval. Since
m is arbitrary, it follows that fˆ(k) = |k|−p 1{|k|>1} , but this function is not in L2 (R), so
fˆ(k) ∈
/ A1 , which is a contradiction. Hence A1 is not complete, and this completes the
proof.
We now define the stochastic integral with respect to TFBM for any function in A1 in
the case where 1/2 < H < 1 in (2.4).
Definition 3.2.7. For any −1/2 < α < 0 and λ > 0, we define
∫
R

∫ [
f (x)Bα,λ (dx) := Γ(κ + 1)

R

κ,λ

κ+1,λ

I− f (x) − λI−

39

]
f (x) B(dx)

(3.52)

for any f ∈ A1 , where κ = −α.
Theorem 3.2.8. For any −1/2 < α < 0 and λ > 0, the stochastic integral I α,λ in (3.52)
is an isometry from A1 into Sp(Bα,λ ). Since A1 is not complete, these two spaces are not
isometric.
Proof. It follows from Lemma 3.2.5 that the stochastic integral (3.52) is well-defined for any
f ∈ A1 . Proposition 2.1 in Pipiras and Taqqu [56] implies that, if D is an inner product
space such that (f, g)D = ⟨I α,λ (f ), I α,λ (g)⟩L2 (Ω) for all f, g ∈ E, and if E is dense D, then
there is an isometry between D and a linear subspace of Sp(Bα,λ ) that extends the map
f → I α,λ (f ) for f ∈ E, and furthermore, D is isometric to Sp(Bα,λ ) itself if and only if D
is complete. Using the Itˆo isometry and the definition (3.52), it follows from (3.45) that for
any f, g ∈ A1 we have

⟨f, g⟩A = ⟨F, G⟩L2 (R) = ⟨I α,λ (f ), I α,λ (g)⟩L2 (Ω) ,
1

and then the result follows from Theorem 3.2.4.

3.2.2

Anti-persistence

Next we investigate stochastic integrals with respect to TFBM in the case 0 < α < 1/2, so
that 0 < H < 1/2 in (2.4). It follows from (3.34) that the stochastic integral (3.40) can be
written in the form
∫
I α,λ (f ) =

R

f (x)Bα,λ (dx) = Γ(1 − α)

∫ ∞ [

40

−∞

α,λ

1−α,λ

D− f (x) − λI−

]
f (x) B(dx)

for any f ∈ E, the space of elementary functions. Then I α,λ (f ) is a Gaussian random
variable with mean zero, such that
(∫
⟨I α,λ (f ), I α,λ (g)⟩

=E
L2 (Ω)

R

∫
f (x)Bα,λ (dx)

R

)
g(x)Bα,λ (dx)

∫ [
][
]
α,λ
1−α,λ
α,λ
1−α,λ
2
D− f (x) − λI−
f (x) D− g(x) − λI−
g(x) dx.
= Γ(1 − α)

(3.53)

R

for any f, g ∈ E, using (3.34) and the Itˆo isometry (3.38). Equation (3.53) suggests the
following space of integrands for TFBM in the case 0 < H < 1/2. Recall that W α,2 (R) is
the fractional Sobolev space (3.17).
Theorem 3.2.9. For any 0 < α < 1/2 and λ > 0, the class of functions
{
}
α,λ
1−α,λ
α,2
2
A2 := f ∈ W (R) : φf = D− f − λI−
f for some φf ∈ L (R). .

(3.54)

is a linear space with inner product

⟨f, g⟩A := ⟨F, G⟩L2 (R)

(3.55)

[
]
α,λ
1−α,λ
F (x) = Γ(1 − α) D− f (x) − λI−
f (x)
[
]
α,λ
1−α,λ
G(x) = Γ(1 − α) D− g(x) − λI−
g(x) .

(3.56)

2

where

The set of elementary functions E is dense in the space A2 . The space A2 is not complete.
We begin with the two lemmas. The first lemma shows that the set A2 contains every
function in W α,2 (R), and hence they are the same set, but different spaces, since they have

41

different inner products.
Lemma 3.2.10. Under the assumptions of Theorem 3.2.9, every f ∈ W α,2 (R) is an element
of A2 .
Proof. Given f ∈ W α,2 (R), we need to show that

1−α,λ

α,λ

φf = D− f − λI−

f,

for some φf ∈ L2 (R). From the definition (3.17) we see that

(3.57)
∫

(λ2 + k 2 )α |fˆ(k)|2 dk < ∞.

Define h1 (k) = (λ − ik)α fˆ(k) and note that h1 is the Fourier transform of some function
φ1 ∈ L2 (R). Define h2 (k) := (λ − ik)α−1 f (k), and observe that
∫

∫
|h2 (k)|2 dk =

|f (k)|2 (λ2 + k 2 )α−1 dk
∫

=

|h1 (k)|2
dk < ∞,
λ2 + k 2

since h1 ∈ L2 (R) and 1/(λ2 + k 2 ) is bounded. Hence there is another function φ2 ∈ L2 (R)
such that h2 = φˆ2 . Define φf := φ1 − λφ2 so that

φf (k) = φ1 (k) − λφ2 (k) = f (k)(λ − ik)α − f (k)λ(λ − ik)α−1 .

(3.58)

Since f ∈ W α,2 (R) ⊂ L2 (R), we can apply Definition 3.1.11 and Lemma 3.1.6 to see that
(3.57) holds.
Lemma 3.2.11. Under the assumptions of Theorem 3.2.9, if f ∈ W α,2 (R), then there exists

42

a sequence of elementary functions (fn ) such that fn → f in L2 (R), and also
∫ +∞
−∞

|fn (k) − f (k)|2 |k|2α dk → 0

as n → ∞.

(3.59)

Proof. Equation (3.59) is proven in [56, Lemma 5.1]. For any L > 0, that proof constructs
a sequence of elementary functions fn such that fˆn (k) → 1[−1,1] (k) almost everywhere on
−L ≤ x ≤ L, and shows that |fˆn (k)| ≤ C min{1, |k|−1 } for all k ∈ R and all n ≥ 1. In
the notation of that paper, we have fˆn (k) = k −1 Un (k). Apply the dominated convergence
theorem to see that

∫ +L
−L

and note that

|fˆn (k) − 1[−1,1] (k)|2 dk → 0

∫
|k|>L

|fˆn (k) − 1[−1,1] (k)|2 dk ≤ 2C 2

∫ ∞
dk
2C 2
≤
.
2
L
L k

Since L is arbitrary, it follows that fˆn (k) → 1[−1,1] (k) in L2 (R), and then the result follows
as in [56, Lemma 5.1].
Proof of Theorem 3.2.9. For f ∈ A2 we define

∥f ∥A2 =

√

⟨f, f ⟩A =

√

2

⟨φf , φf ⟩2 = ∥φf ∥2 .

(3.60)

where φf is given by (3.57). Next, use (3.58) to see that

φf (k) = (−ik)(λ − ik)α−1 f (k).

43

(3.61)

To verify that (3.55) is an inner product, note that if ⟨f, f ⟩A = 0 then
2

∥f ∥2A = ∥φf ∥22 = ∥φf ∥22 =
2

∫ ∞
−∞

|f (k)|2

k2
dk
(λ2 + k 2 )1−α

(3.62)

equals zero, which implies that |f (k)| = 0 almost everywhere, and then f = 0 almost
everywhere. This proves that (3.57) is an inner product.
Next we show that E is dense in A2 . Apply Lemma 3.2.11 to obtain a sequence (fn )
in E such that ∥fn − f ∥2 → 0 and (3.59) holds. It is easy to check using (3.37) that any
elementary function is an element of W α,2 (R), and then Lemma 3.2.10 implies that it is also
an element of A2 . Now use (3.62) to write
∥fn − f ∥2A =
2

∫ +∞
−∞

− λ2

2

fn (k) − f (k) (k 2 + λ2 )α dk

∫ +∞
−∞

fn (k) − f (k)

2

1
(λ2 + k 2 )1−α

dk.

Since 1/(λ2 + k 2 )1−α is bounded, it follows easily using (3.59) and ∥fn − f ∥2 → 0 that
∥fn − f ∥A2 → 0, and hence E is dense in A2 .
Finally, we want to show that A2 is not complete. The proof is similar to that of Theorem
3.2.4. The functions
fn (k) = |k|−p 1{1/n<|k|<1} (k).
are the Fourier transforms of some functions fn ∈ L2 (R). Clearly fn ∈ W α,2 (R), and
then it follows from Lemmas 3.1.6 and 3.1.9 that the corresponding functions Fn (x) =
α,λ

1−α,λ

Γ(1 − α)[D− fn (x) − λI−

fn (x)] from (3.56) have Fourier transform (3.50), that is,

F[Fn ](k) = Γ(1 − α)

44

−ik
fˆn (k).
(λ − ik)1−α

Then
∫ ∞
2
k2
2
2
2
2
∥fn ∥A = ∥Fn ∥2 = ∥Fn ∥2 = Γ(1 − α)
fn (k)
dk < ∞
2
(λ2 + k 2 )1−α
−∞
for any p < 3/2, so that fn ∈ A2 . Now it is easy to check that fn − fm → 0 in A2 ,
as n, m → ∞, so that (fn ) is a Cauchy sequence. Suppose 1/2 < p < 3/2 and that
∥fn − f ∥A2 → 0 for some f ∈ A2 . Then fˆ(k) = |k|−p 1{0<|k|<1} , but this fˆ is not in L2 (R),
so fˆ ∈
/ A2 , and hence A2 is not complete.
We now define the stochastic integral with respect to TFBM for any function in A2 in
the case where 0 < H < 1/2 in (2.4).
Definition 3.2.12. For any 0 < α < 1/2 and λ > 0, we define
∫
I α,λ (f ) =

R

∫ [
]
α,λ
1−α,λ
f (x)Bα,λ (dx) : = Γ(1 − α)
D− f (x) − λI−
f (x) B(dx)

(3.63)

R

for any f ∈ A2 .
Theorem 3.2.13. For any 0 < α < 1/2 and λ > 0, the stochastic integral I α,λ is an isometry from A2 into Sp(Bα,λ ). Since A2 is not complete, these two spaces are not isometric.
Proof. The proof is similar to that of Theorem 3.2.8. It follows from Lemma 3.2.10 that the
stochastic integral (3.63) is well-defined for any f ∈ A2 . Use Proposition 2.1 in Pipiras and
Taqqu [56], and note that the Itˆo isometry, the definition (3.63), and equation (3.55) imply
that for any f, g ∈ A2 we have

⟨f, g⟩A = ⟨F, G⟩L2 (R) = ⟨I α,λ (f ), I α,λ (g)⟩L2 (Ω) .
2

45

Then the result follows from Theorem 3.2.9.

3.2.3

Harmonizable representation

By now it should be clear that the Fourier transform plays an important role in the theory of
stochastic integration for TFBM. Here we apply the harmonizable representation of TFBM
to unify the two cases −1/2 < α < 0 and 0 < α < 1/2.
For any −1/2 < α < 1/2 and any λ > 0, Proposition 3.1 in [46] shows that TFBM has
the harmonizable representation
∫
Γ(1 − α) ∞ e−itk − 1 ˆ
B(dk)
Bα,λ (t) = √
1−α
2π
−∞ (λ − ik)
ˆ = B
ˆ1 + iB
ˆ2 is a complex-valued Gaussian random measure constructed as folwhere B
ˆ1 and B
ˆ2 be two independent Brownian motions on the positive real line with
lows. Let B
ˆi (t))2 ] = t/2 for i = 1, 2, and define two independently scattered Gaussian random
E[(B
ˆi [a, b] = B
ˆi (b) − B
ˆi (a), extend to Borel subsets of the positive real line,
measures by setting B
ˆ1 (A) = B
ˆ1 (−A), B
ˆ2 (A) = −B
ˆ2 (−A).
and then extend to the entire real line by setting B
Apply the formula (3.37) for the Fourier transform of an indicator function to write this
harmonizable representation in the form

Bα,λ (t) = Γ(1 − α)

∫ +∞
−∞

1[0,t] (k)

(−ik)
ˆ
B(dk).
(λ − ik)1−α

It follows easily that for any elementary function (3.39) we may write

I α,λ (f ) = Γ(1 − α)

∫ ∞
−∞

f (k)

46

(−ik)
ˆ
B(dk),
(λ − ik)1−α

(3.64)

and then for any elementary functions f and g we have

⟨I α,λ (f ), I α,λ (g)⟩

L2 (Ω)

= Γ(1 − α)2

∫ ∞
−∞

f (k)g(k)

k2
dk.
(λ2 + k 2 )1−α

(3.65)

Theorem 3.2.14. For any α ∈ (−1/2, 0) ∪ (0, 1/2) and λ > 0, the class of functions
{
∫
2
2
A3 := f ∈ L (R) :
f (k)

}
k2
dk < ∞ .
(λ2 + k 2 )1−α

(3.66)

is a linear space with the inner product
∫ +∞
k2
2
f (k)g(k) 2
⟨f, g⟩A = Γ(1 − α)
dk.
3
(λ + k 2 )1−α
−∞

(3.67)

The set of elementary functions E is dense in the space A3 . The space A3 is not complete.
Proof. The proof combines Theorems 3.2.4 and 3.2.9 using the Plancherel Theorem. First
α,λ

1−α,λ

suppose that 0 < α < 1/2 and recall that φf = D− f − λI−

f is a function with Fourier

transform

φˆf = [(λ − ik)α − λ(λ − ik)α−1 ]fˆ = [λ − ik − λ](λ − ik)α−1 fˆ = (−ik)(λ − ik)α−1 fˆ.

Then it follows from the Plancherel Theorem that
⟨f, g⟩A = Γ(1 − α)2 ⟨φf , φg ⟩2 = Γ(1 − α)2 ⟨φˆf , φˆg ⟩2
2
∫ +∞
2
= Γ(1 − α)
f (k)g(k)
−∞

47

k2
dk = ⟨f, g⟩A
3
(λ2 + k 2 )1−α

and hence the two inner products are identical. If f ∈ A3 , then
∫ +∞
−∞

|fˆ(k)|2 (λ2 + k 2 )α dk =

∫ +∞
−∞

+ λ2

|fˆ(k)|2

∫ +∞
−∞

k2
dk
(λ2 + k 2 )1−α

|fˆ(k)|2

1
dk.
(λ2 + k 2 )1−α

(3.68)

The first integral on the right-hand side is finite by (3.66), and the second is finite since
1/(λ2 + k 2 )1−α is bounded. Then it follows from the definition (3.17) that f ∈ W α,2 (R).
Conversely, if f ∈ W α,2 (R) then since
k2
k2
=
(λ2 + k 2 )α ≤ (λ2 + k 2 )α
(λ2 + k 2 )1−α
λ2 + k 2
it follows immediately that f ∈ A3 , and hence W α,2 (R) and A3 are the same set of functions.
Then it follows from Lemma 3.2.10 that A2 and A3 are identical when 0 < α < 1/2, and
the conclusions of Theorem 3.2.14 follow from Theorem 3.2.9 in this case.
If −1/2 < α < 0, then the function k 2 /(λ2 + k 2 )1−α is bounded by a constant C(α, λ)
that depends only on α and λ, so for any f ∈ L2 (R) we have
∫
R

f (k)

2

∫
2
k2
dk
≤
C(α,
λ)
f
(k)
dk < ∞
(λ2 + k 2 )1−α
R

(3.69)

and hence f ∈ A3 . Since A3 ⊂ L2 (R) by definition, this proves that L2 (R) and A3 are the
same set of functions, and then it follows from Lemma 3.2.5 that A1 and A3 are the same
κ,λ

κ+1,λ

set of functions in this case. Let κ = −α and note that φf = I− f − λI−
function with Fourier transform

φˆf = [(λ − ik)α − λ(λ − ik)α−1 ]fˆ = (−ik)(λ − ik)α−1 fˆ.

48

f is again a

Then it follows from the Plancherel Theorem that
⟨f, g⟩A = Γ(κ + 1)2 ⟨φf , φg ⟩2 = Γ(1 − α)2 ⟨φˆf , φˆg ⟩2
1
∫ +∞
2
= Γ(1 − α)
f (k)g(k)
−∞

k2
dk = ⟨f, g⟩A
3
(λ2 + k 2 )1−α

and hence the two inner products are identical. Then the conclusions of Theorem 3.2.14
follow from Theorem 3.2.4 in this case as well.
Definition 3.2.15. For any α ∈ (−1/2, 0) ∪ (0, 1/2) and λ > 0, we define

I α,λ (f ) = Γ(1 − α)

∫ ∞
−∞

f (k)

(−ik)
ˆ
B(dk)
(λ − ik)1−α

(3.70)

for any f ∈ A3 .
Theorem 3.2.16. For any α ∈ (−1/2, 0) ∪ (0, 1/2) and λ > 0, the stochastic integral I α,λ
in (3.70) is an isometry from A3 into Sp(Bα,λ ). Since A3 is not complete, these two spaces
are not isometric.
Proof. The proof of Theorem 3.2.14 shows that A1 and A3 are identical when −1/2 < α < 0,
and A2 and A3 are identical when 0 < α < 1/2. Then the result follows immediately from
Theorems 3.2.8 and 3.2.13.

3.3

Discussion

In this section, we collect some remarks and extensions.

49

3.3.1

White noise approach

Heuristically, the TFBM (3.33) with 1/2 < H < 1 in (2.4) can be written in terms of
tempered fractional integrals of the white noise W (x)dx = B(dx), since in view of (3.8) we
can write
∫ +∞ [
]
κ+1,λ
κ,λ
W (x) 1[0,t] (x) dx.
Bα,λ (t) = Γ(κ + 1)
I+ W (x) − λI+
−∞

In the same way, when 0 < H < 1/2 we can write

Bα,λ (t) = Γ(1 − α)

∫ +∞ [
−∞

]

α,λ
1−α,λ
W (x) 1[0,t] (x) dx,
D+ W (x) − λI+

using Lemma 3.1.15. These ideas could be made rigorous using white noise theory [36]. Setting λ = 0, we recover the fact that FBM is the fractional integral or derivative of a Brownian
motion [56, p. 261]. The white noise approach is preferred in engineering applications (e.g.,
see [6]).

3.3.2

Reproducing kernel Hilbert space

The reproducing kernel Hilbert space (RKHS) of TFBM provides another approach to
stochastic integration that produces an isometric space of deterministic integrands. The
RKHS for FBM was computed in [6, 56]. For any mean zero Gaussian process {Xt }t∈R
with covariance function R(s, t) = E[Xs Xt ], the RKHS of X is the unique Hilbert space
H(X) of measurable functions f : R → R such that R(·, t) ∈ H(X) for all t ∈ R, and
⟨f, R(·, t)⟩H(X) = f (t) for all t ∈ R and f ∈ H(X) [25, 65]. As noted in [25], if there exists

50

a measure space (Λ, B, ν) and a set of functions {ft } ⊂ L2 (R, ν) such that
∫
R(s, t) =
Λ

fs (x)ft (x)ν(dx) for all s, t ∈ R,

Then H(X) consists of the functions g(t) =

∫

(3.71)

ft (x)g ∗ (x)ν(dx) for g ∗ ∈ Sp{ft }, the closure

in L2 (R, ν) of the set of linear combinations of functions ft . Then H(X) is a Hilbert space
with the inner product

∫
⟨g, h⟩H(X) =

g ∗ (x)h∗ (x)ν(dx).
Λ

Let Sp(X) denote the closure of the set of linear combinations of random variables {Xt }
in the space L2 (Ω). The mapping J that sends
J
∑

aj R(·, tj ) →

j=1

J
∑

aj Xtj

j=1

is an isometry that maps H(X) onto Sp(X), and hence these two Hilbert spaces are isometric.
Then J (f ) is the stochastic integral of any f ∈ H(X).
∫
For TFBM with −1/2 < α < 0, let κ = −α. Since Bα,λ (t) = R 1[0,t] (x)Bα,λ (dx), it
follows immediately from the definition (3.52) that TFBM has covariance function

R(s, t) = Γ(κ + 1)2

∫ [
][
]
κ,λ
κ+1,λ
κ,λ
κ+1,λ
I− 1[0,s] (x) − λI−
1[0,s] I− 1[0,t] (x) − λI−
1[0,t] dx,
R

and hence the RKHS H(Bα,λ ) consists of functions
∫ [
]
k,λ
k+1,λ
g(t) = Γ(k + 1)
I− − λI−
1[0,t] (x)g ∗ (x)dx
R

51

for g ∗ ∈ L2 (R), with the inner product
∫
⟨g, h⟩H(X) =

R

g ∗ (x)h∗ (x)dx = ⟨g ∗ , h∗ ⟩L2 (R) .

(3.72)

For TFBM with 0 < α < 1/2 and λ > 0, the RKHS H(Bα,λ ) consists of functions

g(t) = Γ(1 − α)2

∫ [
]
1−α,λ
α,λ
1[0,t] (x)g ∗ (x)dx
D− − λI−
R

for g ∗ ∈ L2 (R), with the same inner product (3.72). The proof is similar to [56, Section
6]. Complete details will be provided in the forthcoming paper [45]. Here we take Λ =
L2 (R), with ν the Lebesgue measure on R. The main technical difficulty is to show that
k,λ

k+1,λ

L2 (R) = Sp{ft }, where ft (x) = Γ(k + 1)[I− − λI−
α,λ

1−α,λ

and ft (x) = Γ(1 − α)[D− − λI−

3.3.3

]1[0,t] (x) in the case −1/2 < α < 0,

]1[0,t] (x) for 0 < α < 1/2.

Tempered distributions as integrands

Jolis [29] proved that the exact domain of the Wiener integral for a fractional Brownian
motion BH (t) is given by

ΛH = {f ∈ S ′ (R) =

∫
R

|f (k)|2 |k|1−2H dk < ∞},

where S ′ (R) is the space of tempered distributions. This gives an isometry using the inner
product (for a standard FBM)

Γ(2H + 1) sin(πH)
⟨f, g⟩ =
2π

52

∫
fˆ(k)ˆ
g (k)|k|1−2H dk,

that makes ΛH isometric to Sp(BH ). She also proved that this space contains distributions
that cannot be represented by locally integrable functions in the case of long range dependence (1/2 < H < 1). Tudor [63] extended this result to subfractional Brownian motion.
The distributional approach is useful in the study of partial differential equations with a
Gaussian forcing term [9, 15, 64].
Following along these lines, we conjecture that the exact domain of the Wiener integral
with respect to TFBM is given by the distributional fractional Sobolev space

Λα,λ = {f ∈ S ′ (R) :

∫
R

|f (k)|2 (λ2 + k 2 )α dk < ∞}

with the inner product
∫
⟨f, g⟩ = Cα,λ

fˆ(k)ˆ
g (k)(λ2 + k 2 )α dk.

Proving this using [29, Theorem 3.5] would require computing the second derivative of the
variance function (2.5) and taking the (inverse) Fourier transform of the result. This computation seems difficult, due to the Bessel function term.

53

Chapter 4
Tempered fractional stable motion
This chapter has four sections. In Section 4.1, we define linear tempered fractional stable
motion (LTFSM) using a moving average representation, and we establish the dependence
structure of its increments, which we call tempered fractional stable noise (TFSN). Section 4.2 defines tempered fractional harmonizable stable motion (HTFSM) and shows that
LTFSM is different from HTFSM. Sample path properties of LTFSM and TFHSM are proven
in Section 4.3. Finally, Section 4.4 investigates the local times and local nondeterminism
properties for LTFSM and HTFSM .

4.1

Moving average representation

Let X be a real-valued random variable. We say that X has a symmetric α-stable, SαS,
distribution if its characteristic function has the form

E [exp {i(θX)}] = exp {−c|θ|α } ,

for some constant c > 0 and 0 < α ≤ 2. The parameters α and σ are called the index of
stability and the scale parameter respectively (see Chapter 1 in [61]). We denote the SαS
distribution by Sα (σ, 0, 0) and write X ≃ Sα (σ, 0, 0) to indicate that that X has the stable
distribution Sα (σ, 0, 0). A real-valued stochastic process {X(t)} is called SαS if all linear

54

combinations

∑n

j=1 θj X(tj ) are real-valued SαS random variables. Consider (R, B(R), dx),

where B(R) is the σ-algebra of Borel subsets of R. We say the process {Zα (B), B ∈ B(R)}
is a real-valued SαS random measure with Lebesgue control measure dx if
[

{
}]
E exp iRe(θZα (B)) = exp {−|B||θ|α } ,

for any θ ∈ C where |B| denotes the Lebesgue measure of B (B ∈ B(R)). Now, Let Zα be a
SαS random measure on (R, B(R)) with Lebesgue control measure dx. Then, we define the
stochastic integral

∫ +∞
I(f ) :=

−∞

f (x) Zα (dx)

(4.1)

for all measurable functions f : R → R satisfying the condition
∫ +∞
−∞

f (x)

α

m(dx) < ∞

(4.2)

for any 0 < α < 2, (α ̸= 1). We denote the collection of functions satisfying (4.2) by

Lα (R) =

∫ +∞
{
}
α
f : f : R → R is measurable,
f (x) dx < ∞ .
−∞

(4.3)

Proposition 3.4.1 in [61] shows that I(f ), for any f ∈ Lα (R, B(R), dx), has the characteristic
function

}
[
]
{ ∫ +∞
α
E exp{iθI(f )} = exp −
θf (x) dx .
−∞

55

For the stochastic integral I(f ) in (4.1) we define
α

I(f )

α

∫ +∞
=
(
:=

f (x) Zα (dx)

−∞

α

[
])
− log E exp{i I(f )}

∫ +∞
=

α

−∞

f (x)

α

(4.4)

dx

for any 0 < α < 2.
Definition 4.1.1. A stochastic process {X(t), t ∈ R} is called an SαS L´evy motion if
1. X(0) = 0 a.s.
2. X has independent increments.
1

3. X(t) − X(s) ∼ Sα (σ α |t − s| α , 0, 0) for any −∞ < s < t < ∞ and, for some 0 < α ≤ 2,
and some σ > 0.
Note that the process X has stationary increments. It is Brownian motion when α = 2.
Also it is α1 - self similar , that is, for all c > 0,
{
}
X(ct)

t∈R

where

{
}
1
c α X(t)

,

t∈R

indicates equality in the sense of finite dimensional distributions.

Definition 4.1.2. Given a SαS random measure Zα (dx) on R with control measure m(dx),
for any 0 < α ≤ 2 and H > 0 and λ ≥ 0, the stochastic integral
∫ +∞ [
XH,α,λ (t) :=

−∞

1
1]
H− α
H− α
−λ(t−x)
−λ(−x)
+
+
e
(t − x)+
−e
(−x)+
Zα (dx),

56

(4.5)

where (x)+ = max{x, 0} and 00 = 0 will be called a linear tempered fractional stable motion
(LTFSM).
Remark 4.1.3. When α = 2, {XH,α,λ (t)}t∈R is TFBM defined in (2.1).
It is easy to check that the function
1

1

H−
H−
gα,λ,t (x) := e−λ(t−x)+ (t − x)+ α − e−λ(−x)+ (−x)+ α

(4.6)

belongs to the linear space Lα (R) in (4.3) provided that H > 0, so that LTFSM is well
defined and we have

XH,α,λ (t)

α
α

∫
=
(
:=
∫
=

R

gα,λ,t (x) Zα (dx)

α
α

[
])
− log E exp{i XH,α,λ (t)}

R

gα,λ,t (x)

α

(4.7)

dx,

for any 0 < α < 2. Note also that this function has a scaling property

1

gα,λ,ct (cx) = cH− α gα,cλ,t (x),

(4.8)

for all t, x ∈ R and all c > 0. The next result shows that LTFSM has a nice scaling property,
involving both the time scale and the tempering.
Proposition 4.1.4. LTFSM (4.5) is symmetric α-stable stochastic process with stationary
increments, such that
{

}
XH,α,λ (ct) t∈R

{
}
cH XH,α,cλ (t)

t∈R

57

,

(4.9)

for any scale factor c > 0.
Proof. Since Zα (dx) has control measure m(dx) = σ α dx, the random measure Zα (c dx) has
1

control measure c α σ α dx. Given t1 < t2 < · · · < tn , a change of variable x = cx′ then yields
(

(∫

)

XH,α,λ (cti ) : i = 1, . . . , n =
(∫
=
(∫
≃

)
gα,λ,cti (x)M (dx) : i = 1, . . . , n
gα,λ,cti (cx′ )M (c dx′ ) : i = 1, . . . , n
1
cH− α g

)

′ 1
′
α,cλ,ti (x )c α M (dx ) : i = 1, . . . , n

)

(
)
= cH XH,α,cλ (ti ) : i = 1, . . . , n ,
where ≃ denotes equality in distribution, so that (4.9) holds. For any s, t ∈ R, the integrand
(4.6) satisfies gα,λ,s+t (s + x) − gα,λ,s (s + x) = gα,λ,t (x), and hence a change of variable
x = s + x′ yields
(

)
XH,α,λ (s + ti ) − XH,α,λ (s) : i = 1, . . . , n
)
(∫ [
]
gα,λ,s+ti (x) − gα,λ,s (x) M (dx) : i = 1, . . . , n
=
)
(∫ [
]
′
′
′
gα,λ,s+ti (s + x ) − gα,λ,s (s + x ) M (dx ) : i = 1, . . . , n
≃
(∫
)
′
′
gα,λ,ti (x )M (dx ) : i = 1, . . . , n
=
(
)
= XH,α,λ (ti ) : i = 1, . . . , n

, which shows that LTFSM has stationary increments.
When a stochastic process {Y (t)}t∈R is stationary Gaussian with mean zero, we can
describe its dependence structure by its covariance function E [Y (t)Y (0)]. However, in the
non-Gaussian stable case, the covariance function does not exist. Instead, we use the follow-

58

ing. For a stationary process {Y (t)}, let

r(t) : = r(θ1 , θ2 , t) = E [exp {i(θ1 Y (t) + θ2 Y (0))}]
(4.10)
− E [exp {iθ1 Y (t)}] E [exp {iθ2 Y (0)}] ,

θ1 , θ2 ∈ R,

and

I(t) : = I(θ1 , θ2 , t)
: = − log E [exp {i(θ1 Y (t) + θ2 Y (0))}]
+ log E [exp {iθ1 Y (t)}] + log E [exp {iθ2 Y (0)}] ,

(4.11)
θ1 , θ2 ∈ R.

The following relationship between r(t) and I(t) is valid:
(
)
r(θ1 , θ2 , t) = K(θ1 , θ2 , t) e−I(θ1 ,θ2 ,t) − 1 ,

where
K(θ1 , θ2 , t) = E [exp {iθ1 Y (t)}] E [exp {iθ2 Y (0)}]
= E [exp {iθ1 Y (0)}] E [exp {iθ2 Y (0)}]

(4.12)

:= K(θ1 , θ2 ).
Further, if I(t) → 0 as t → ∞, then r(t) ∼ −K(θ1 , θ2 )I(t) as t → ∞ which means r(t) and
I(t) are asymptotically equivalent. If {Yt }t∈R is Gaussian, −I(1, −1, t) coincides with the
covariance function and thus r(t) is a natural extension. The quantity r(t) was used in [2],
where the authors studied the dependence structure of linear fractional stable motion.

59

Given a LTFSM (4.5), we define tempered fractional stable noise (TFSN)

YH,α,λ (t) := XH,α,λ (t + 1) − XH,α,λ (t) for integers −∞ < t < ∞.

(4.13)

Remark 4.1.5. For two non-negative functions f (t) and g(t) on R, we will write f (t) ≍ g(t)
f (t)

if C1 ≤ g(t) ≤ C2 for all t sufficiently large, for some 0 < C1 < C2 < ∞.
Theorem 4.1.6. Let 0 < α < 1, 0 < H < 1, λ > 0 and YH,α,λ (t) be the tempered fractional
stable noise (4.13). Then
r(θ1 , θ2 , t) ≍ e−λαt tHα−1 ,
as t → ∞.
Proof. It follows easily from (4.5) that TFSN has the moving average representation
∫ +∞ [
]
H− 1
H− 1
YH,α,λ (t) =
e−λ(t+1−x)+ (t + 1 − x)+ α − e−λ(t−x)+ (t − x)+ α Zα (dx). (4.14)
−∞

1

H−
Define gt (x) = (t − x)+ α e−λ(t−x)+ for t ∈ R and compute I(θ1 , θ2 , t) and K(θ1 , θ2 , t) for

TFSN {YH,α,λ } as follows:
[
]
I(θ1 , θ2 , t) = − log E exp {i(θ1 YH,α,λ (t) + θ2 YH,α,λ (0))}
[
]
[
]
+ log E exp {iθ1 YH,α,λ (t)} + log E exp {iθ2 YH,α,λ (0)}
∫ +∞
α
=
θ1 [gt+1 (x) − gt (x)] + θ2 [g1 (x) − g0 (x)] dx
−

−∞
∫ +∞
−∞

[
]α
dx −
θ1 gt+1 (x) − gt (x)

:= I1 (θ1 , θ2 , t) + I2 (θ1 , θ2 , t),

60

∫ +∞
−∞

[
]α
dx
θ2 g1 (x) − g0 (x)

(4.15)

where
∫ 0
I1 (θ1 , θ2 , t) =
−

−∞
∫ 0
−∞

θ1 [gt+1 (x) − gt (x)] + θ2 [g1 (x) − g0 (x)]
[
]α
θ1 gt+1 (x) − gt (x)
dx −

∫ 0
−∞

α

dx

[
]α
θ2 g1 (x) − g0 (x)
dx

and
∫ 1

α

θ1 [gt+1 (x) − gt (x)] + θ2 g1 (x) dx
0
∫ 1
∫ 1
α
[
]α
−
θ1 gt+1 (x) − gt (x)
dx −
θ2 g1 (x) dx.

I2 (θ1 , θ2 , t) =

0

0

Also,
[
] [
]
K(θ1 , θ2 ) = E eiθ1 Y (t) E eiθ2 Y (0)
[
] [
]
iθ
Y
(0)
iθ
Y
(0)
1
2
=E e
E e
∫ +∞
{
}
α
α
α
= exp − (|θ1 | + |θ2 | )
g1 (x) − g0 (x) dx

(4.16)

−∞

by stationarity. Therefore, I(θ1 , θ2 , t) = K(θ1 , θ2 , t)(I1 (t) + I2 (t)) and to verify the asymptotic dependence of I(t) we just need to verify the asymptotic dependence of I1 (t) and I2 (t)
as t → ∞. We first verify the asymptotic dependence of I1 (t). A change of variable in I1 (t)
for t > 1 gives
∫ ∞
I1 (t) =

0

[
1
1]
θ1 e−λ(t+1+x) (t + 1 + x)H− α − e−λ(t+x) (t + x)H− α

[
1] α
1
+ θ2 e−λ(1+x) (1 + x)H− α − e−λ(x) (x)H− α dx
∫ ∞
[
1
1] α
−
θ1 e−λ(t+1+x) (t + 1 + x)H− α − e−λ(t+x) (t + x)H− α dx
∫0∞
[ −λ(1+x)
1
1] α
H−
−λ(x)
H−
α
α
−
θ2 e
(1 + x)
dx.
−e
(x)
0

61

Let
[
1
1] α
ft+1,t (x) := θ1 e−λ(t+1+x) (t + 1 + x)H− α − e−λ(t+x) (t + x)H− α .

(4.17)

For every t > 1 and x > 0 we get
1

eαλt t−α(H− α ) ft+1,t (x) = θ1

1
( t + x )H− 1 α
α −λ(1+x) ( t + 1 + x )H− α
α
e
− e−λx

t

t

α
α
→ θ1 e−λαx e−λ − 1 as t → ∞

and
α
1
sup eαλt t−α(H− α ) ft+1,t (x) ≤ θ1 (e−λ − 1) e−λαx ,
t>1

which belongs to L1 (0, ∞). Now we can use the Dominated Convergence Theorem to see
that
∫ ∞
0

α
1
ft+1,t (x) dx → θ1 (e−λ − 1) e−λαt tα(H− α )

=

θ1 (e−λ − 1)

α

∫ ∞

e−λαx dx

0

1
e−λαt tα(H− α )

λα

(4.18)

,

as t → ∞. Now consider,

gt,t+1,0,1 (x) :=

[
1]
1
θ1 e−λ(t+1+x) (t + 1 + x)H− α − e−λ(t+x) (t + x)H− α

[
1
1] α
+ θ2 e−λ(1+x) (1 + x)H− α − e−λ(x) (x)H− α
−

θ2

α [ −λ(1+x)
1
1] α
e
(1 + x)H− α − e−λx (x)H− α .

62

(4.19)

Then,

1
eλαt t−α(H− α ) g

[
t,t+1,0,1 (x)

=

θ1

e−λ(1+x)

(

1
1
)
)
(
t + x H− α ]
t + 1 + x H− α
−λx
−e
t
t

[
( 1 + x )H− 1
( x )H− 1 ] α
α
α
−λ(1+x)
λt
−λx
λt
e
+ θ2 e
−e
e
t
t

where

−

( 1 + x )H− 1
( x )H− 1 ] α
α
α
−λ(1+x)
λt
−λx
λt
θ2 e
e
−e
e

=:

at + bt

[
at = θ1 e−λ(1+x)

and

[

t

(

α

− bt

t

α

1
)
)H− 1 ]
(
α
t + 1 + x H− α
t
+
x
− e−λx
t
t

[

]

( 1 + x )H− 1
( x )H− 1
α
α
−λ(1+x)
λt
−λx
λt
bt = θ2 e
e
−e
e
.
t

t

It is obvious that at → Cx := θ1 e−λx (e−λ − 1) and bt → ∞ as t → ∞. Then, |at + bt |α −
|bt |α → 0 as t → ∞ since 0 < α < 1. Therefore

1

eλαt t−α(H− α ) gt,t+1,0,1 → 0,

as t → ∞. Moreover, for any 0 < α < 1, using the inequality |a|α − |b|α ≤ a − b

α

(see

[61], Page 211), we get
gt,t+1,0,1 ≤ ft+1,t ,
where gt,t+1,0,1 and gt,t+1,0,1 are defined in (4.17) and (4.19) respectively, if we let a =

63

θ1 (gt+1 − gt ) + θ2 (g1 − g0 ) and b = θ2 (g1 − g0 ). Consequently
1
1
sup eλαt t−α(H− α ) gt,t+1,0,1 ≤ sup eαλt t−α(H− α ) ft+1,t (x)
t>1

t>1

≤ θ1 (e−λ − 1)

α −λαx
e

which also belongs to L1 (0, ∞). Applying the Dominated Convergence Theorem yields
∫ +∞
−∞

gt,t+1,0,1 (x)dx → 0 as t → ∞.

(4.20)

Therefore from (4.18) and (4.20)
1

|θ (e−λ − 1)|α e−λαt tα(H− α )
,
I1 (t) ∼ − 1
λα

(4.21)

as t → ∞. Consider now,
∫ 1

α

θ1 [gt+1 (x) − gt (x)] + θ2 g1 (x) dx
0
∫ 1
∫ 1
α
α
−
θ1 [gt+1 (x) − gt (x)] dx −
θ2 g1 (x) dx,

I2 (t) =

0

0

Define,
[
1
1]
ut (x) := θ1 e−λ(t+1−x) (t + 1 − x)H− α − e−λ(t−x) (t − x)H− α

(4.22)

and
1

v(x) := θ2 e−λ(1−x) (1 − x)H− α .
We can rewrite

∫ 1
I2 (t) =

0

ξ(ut (x) + v(x)) − ξ(ut (x)) − ξ(v(x)) dx,

64

(4.23)

where
α

ξ(g(x)) = g(x) .

(4.24)

Lemma 3.1 in [2] implies that
∫ 1

ξ(ut (x) + v(x)) − ξ(ut (x)) − ξ(v(x)) dx
∫ 1
α
ut (x) dx,
≤ Pα

I2 (t) =

0

(4.25)

0

where Pα = 2 + 4 tan( πα
2 )(see [2], Page 11). On the other hand

ut (x)

1
1
1
− H)(t − x)H− α −1 e−λ(t−x) + λ(t − x)H− α e−λ(t−x)
α
[
]
1 −1
1
1
H−
α
≤ θ1 e−λ(t−1) ( − H) t − 1
+ λ |t − 1|H− α
α
1
[1
]
H− α
−λ(t−1)
≤ θ1 e
−H +λ t−1
,
α

θ1 (

(4.26)

since 0 < x < 1. From (4.25) and (4.26) we get

I2 (t) ≤ Pα

∫ 1
0

ut (x)

α

dx

]α
[1
Hα−1
α
−H +λ t−1
.
= Pα θ1 e−λα(t−1)

(4.27)

α

Hence from (4.21) and (4.27) we get

I(t) ≍ e−λαt tHα−1

as t → ∞ and this completes the proof.
Theorem 4.1.7. Let 1 < α < 2, α1 < H < 1, λ > 0 and let YH,α,λ (t) be the tempered

65

fractional stable noise (4.13). Then

1

I(t) ≍ e−λt t(H− α ) ,

as t → ∞.
Proof. The proof is similar to that of Theorem 4.1.6. Let ft+1,t (x) be the function which is
defined by (4.17). Then
α
1
1
eλt t−(H− α ) ft+1,t (x) = θ1 eλt t−(H− α )
1 α

1

e−λ(t+1+x) (t + 1 + x)H− α − e−λ(t+x) (t + x)H− α
= at · bt ,
where
α

1

at := θ1 eλt(α−1) t(H− α )(α−1)
and
(
(
x )H− α1 α
1 x )H− α1
− e−λx 1 +
.
bt := e−λ(1+x) 1 + +
t
t
t
Note that at → 0 (since 1 < α < 2) and bt → e−λ(1+x) − e−λx

α

as t → ∞. Now, let

1
h(t) = e−λt(α−1) t(α−1)(H− α ) . Observe that h(t) attains its maximum at t = λ1 (H − α1 ).

Then
α
α
1
sup eλt t−(H− α ) ft+1,t (x) = e−λ − 1 θ1 e−λαx sup h(t)
t>1

t>1

1)
α −λαx −(H− 1 )(α−1) [ H − 1 ](α−1)(H− α
α
−λ
α
α
e
= e − 1 θ1 e
,

λ

and so ft+1,t (x) is bounded by an L1 (0, ∞) function. Therefore the Dominated Convergence
66

Theorem implies that

∫ ∞
0

ft+1,t (x) → 0

(4.28)

1
as t → ∞. Consider now, eλt t−(H− α ) gt,t+1,0,1 where gt,t+1,0,1 is given by (4.19). Then

1

eλt t−(H− α ) gt,t+1,0,1 = at + bt

α

− bt

α

where
[

1)
1)
1 ) −λ(1+x) ( t + 1 + x )(H− α
1 ) −λx ( t + x )(H− α
−λt(1−
−λt(1−
α e
α e
at := θ1 e
−e
1
1
tα
tα

and

[

]

]

1
λt −(H− α ) [ −λ(1+x)
1
1 ]
α
bt := θ2 e α t
e
(1 + x)(H− α ) − e−λx x(H− α ) .

Observe that limt→∞ bt = ∞ and limt→∞ at = 0. Since |at + bt |α − |bt |α ∼ α|at ||bt |α−1 , as
t → ∞, we get,
1

eλt t−(H− α ) gt,t+1,0,1 ∼ α θ1
1)
1)
1 ) −λx ( t + x )(H− α
1 ) −λ(1+x) ( t + 1 + x )(H− α
−λt(1−
−λt(1−
α
α
−e
e
e
e
1
1
tα
tα
α−1 λt(1− 1 ) −(H− 1 )(1− 1 ) −λ(1+x)
1
1 α−1
α t
α
α e
θ2
e
(1 + x)(H− α ) − e−λx x(H− α )

and consequently
1

eλt t−(H− α ) gt,t+1,0,1 → α θ1 e−λ(1+x) − e−λx
θ2

α−1 −λ(1+x)
1
1 α−1
e
(1 + x)H− α − e−λx xH− α
.

67

Moreover,
α
α
1
sup eλt t−(H− α ) gt,t+1,0,1 = sup at + bt − bt
t≥1

t≥1

≤ sup at
t≥1

α

+ α sup at bt

α−1

(4.29)
,

t≥1

where we have used the following inequalities (Lemma 2 in [41]): |a − b|α ≤ aα + bα and
|a + b|α − |b|α ≤ a

α

+α a b

α−1

valid for a ≥ 0 and b ≥ 0 and α ∈ (1, 2). In order to

find an upper bound for supt≥1 |at |α , write

at

α

1)
1) α
α −λt(1− 1 ) −λ(1+x) ( t + 1 + x )(H− α
1 ) −λx ( t + x )(H− α
−λt(1− α
α e
−
e
e
e
1
1
tα
tα
1)
( t + x )(H− 1 ) α
α −λαx −λt(α−1) −λ ( t + 1 + x )(H− α
α
= θ1 e
e
e
−
1
1
tα
tα
α −λαx −λ
α
1
1
≤ θ1 e
e (1 + 1 + x)H− α − (1 + x)H− α

= θ1

]
α −λαx [ −λα
Hα−1
Hα−1
≤ θ1 e
e
(2 + x)
+ (1 + x)
α

≤ 2 θ1 e−λαx (2 + x)Hα−1 .
(4.30)
On the other hand,

α at bt

α−1

= α θ1 θ2

α−1

( t + x )(H− 1 )
( t + 1 + x )(H− 1 )
α
α
−λx
−λ(1+x)
−e
× K(x)
e
t

t

:=S(t)

where
1

1

K(x) = e−λ(1+x) (1 + x)(H− α ) − e−λx (x)(H− α )

68

α−1

.

(4.31)

Note that S(t) is a decreasing function and hence

sup α at bt

α−1

t≥1

= α θ1 θ2

α−1

(4.32)
1
1
e−λ(1+x) (2 + x)(H− α ) − e−λx (1 + x)(H− α ) × K(x)

where K(x) is given by (4.31). From (4.29), (4.30) and (4.32)
α
α−1
1
sup eλt t−(H− α ) gt,t+1,0,1 ≤ 2 θ1 e−λαx (2 + x)Hα−1 + α θ1 θ2
t≥1

(4.33)
1
1
e−λ(1+x) (2 + x)(H− α ) − e−λx (1 + x)(H− α ) × K(x)

which belongs to L1 (0, ∞), since Hα > 1. Then, the Dominated Convergence Theorem
implies that
∫ ∞
0

α−1

1

gt,t+1,0,1 (x) dx → αθ1 θ2
e−λt t(H− α )
∫ ∞
1
1 α−1
e−λ(1+x) − e−λx e−λ(1+x) (1 + x)H− α − e−λx xH− α
dx
0

1

= C2 (α, λ, θ1 , θ2 )e−λt t(H− α ) ,
(4.34)
as t → ∞, where
α−1

C2 (α, λ, θ1 , θ2 ) = αθ1 θ2
(4.35)
∫ ∞
1
1 α−1
−λ(1+x)
H−
−λ(1+x)
−λx
−λx
H−
α
α
e
(1 + x)
e
−e
dx
−e
x
0

is a constant that is independent of t. Therefore from (4.28) and (4.34)

1

I1 (t) ∼ C2 (α, λ, θ1 , θ2 )e−λt t(H− α ) ,
69

(4.36)

as t → ∞. Finally, recall that
∫ 1

α

θ1 [gt+1 (x) − gt (x)] + θ2 g1 (x) dx
0
∫ 1
∫ 1
α
α
−
θ1 [gt+1 (x) − gt (x)] dx −
θ2 g1 (x) dx,

I2 (t) =

0

0

and that ut (x) and v(x) are given by (4.22) and (4.23) respectively. Then
∫ 1
I2 (t) =

0

ξ(ut (x) + v(x)) − ξ(ut (x)) − ξ(v(x)) dx

where ξ(g(x)) is given by (4.24). Lemma 3.1 in [2] implies that, using argument similar to
(4.26), we have
∫ 1

ξ(ut (x) + v(x)) − ξ(ut (x)) − ξ(v(x)) dx
∫ 1
∫ 1
α−1
α
≤
Rα ut (x) v(x)
dx + Sα
ut (x) dx

I2 (t) =

0

0

≤ Rα θ1

∫ 1[

0
1
α−1
H−
1
α −λ(t−1)
e
v(x)
dx
H − +λ t−1

]

α
0
[
]α
α
Hα−1 −λα(t−1)
1
+ Sα θ1 H − + λ t − 1
e
α
1
]
[
H− α
1
= Rα θ1 H − + λ t − 1
e−λ(t−1)
α
∫ 1
1 α−1
·
θ2 e−λ(1−x) (1 − x)H− α
dx
0

+ Sα θ1

α[

]α
Hα−1 −λα(t−1)
1
H − +λ t−1
e
α

1
H− α
:= C3 (α, λ, θ1 ) t − 1
e−λ(t−1)

+ Sα θ1

α[

H−

]α
Hα−1 −λα(t−1)
1
e
,
+λ t−1
α

70

(4.37)

where
[
]
1
C3 (α, λ, θ1 ) = Rα θ1 |H − | + λ
α
∫ 1
1 α−1
·
θ2 e−λ(1−x) (1 − x)H− α
dx
0

)
(
πα
is a constant. Recall that Rα = α 1 + tan( πα
2 ) , and Sα = (α + 1) + (α + 3) tan( 2 ) (see
[2], Page 11) are also constants. Note that the upper bound which is obtained by (4.37) is
of the same order as the upper bound for I1 (t), given by (4.34). Hence

1

I(t) ≍ e−λt t(H− α )

and this completes the proof.
Definition 4.1.8. A symmetric α-stable stationary process {Yt } has long memory if r(θ1 , θ2 , t)
defined in (4.10) satisfies
∞
∑

r(θ1 , θ2 , n) = ∞

(4.38)

n=0

Lemma 4.1.9. The LTFSM process does not have long memory property in the sense of
(4.38).
Proof. From Theorems 4.1.6 and 4.1.7 we have
∞
∑

r(θ1 , θ2 , n) < ∞

n=0

which proves the statement.
Remark 4.1.10. Theorem 4.1.7 gives the fact that when the tempering parameter λ is
sufficiently small, TFSN exhibits semi-long range dependence, with the asymptotic rate I(t)
71

1
that falls off like tH− α for moderate values of t > 1, and eventually that rate falls off like
1

e−λt tH− α for t sufficiently large.

4.2

Tempered fractional harmonizable stable motion

Let X = X1 + iX2 be a complex-valued random variable. We say X is isotropic SαS if the
vector (X1 , X2 ) is SαS and for any θ = θ1 + iθ2

E [exp {i(θ1 X1 + θ2 X2 )}] = exp {−c|θ|α } ,

for some constant c > 0. A complex-valued stochastic process {X(t)} is called isotropic SαS
if all complex linear combinations

∑n

j=1 θj X(tj ) are complex-valued isotropic SαS random

variables. Let (R, B(R), dx), where B(R) is the σ-algebra of Borel subsets of R. We say
the process {Zα (B), B ∈ B(R)} is a complex-valued isotropic SαS random measure with
Lebesgue control measure dx if
[

{
}]
E exp iRe(θZα (B)) = exp {−|B||θ|α } ,

for any θ ∈ C where |B| denotes the Lebesgue measure of B (B ∈ B(R)).
Definition 4.2.1. Let {Zα (B), B ∈ B(R)} be a complex-valued isotropic SαS random measure with Lebesgue control measure dx. Then the stochastic integral
∫ +∞
I(f ) := Re

−∞

72

f (k)Zα (dk),

where f ∈ Lα (R) is the complex-valued SαS random variable such that
[

{

E exp iRe

∫ +∞
−∞

}]
{ ∫ +∞
}
α
f (k)Zα (dk)
= exp −
f (k) (dk)
−∞

and

I(f )

α
α

∫ +∞
= Re
(
:=

−∞

f (k)Zα (dk)

α

[
])
− log E exp{iRe I(f )}

∫ +∞
=

α

−∞

f (k)

α

(4.39)

dk,

for any 0 < α < 2.
Definition 4.2.2. The real harmonizable tempered fractional stable motion (HTFSM) is the
process

∫ +∞
XH,α,λ (t) = Re

e−ikt − 1

1
−∞ (λ − ik)H+ α

Zα (dk)

(4.40)

where 0 < α < 2, H > 0, λ > 0 and Z˜α is a complex isotropic SαS random measure.
Remark 4.2.3. The stochastic integral in (4.40) is well defined, since
∫ +∞
−∞

α

e−ikt − 1
1

(λ − ik)H+ α

dk < ∞,

(4.41)

for any H > 0 and 0 < α < 2 (as |k| → ∞, the integrand behaves like |k|−Hα−1 , which is
integrable for any 0 < α < 2 and H > 0; as |k| → 0, the integrand tends to zero).
Definition 4.2.4. Given a HTFSM (4.40), we define tempered fractional harmonizable sta-

73

ble noise (TFHSN)

YH,α,λ (t) := XH,α,λ (t + 1) − XH,α,λ (t) for integers −∞ < t < ∞.

(4.42)

It follows easily from (4.40) that TFHSN has the harmonizable representation
∫ +∞
YH,α,λ (t) = Re

−∞

eikt Ψ(dk),

(4.43)

Zα (dk)

(4.44)

where
Ψ(dk) =

eik − 1
1

(λ + ik)H+ α

is a complex symmetric α-stable (SαS) random measure with the control measure

m(dk) =

|eik − 1|α
dk,
|λ + ik|Hα+1

(4.45)

for any 0 < α < 2, λ > 0 and H > 0.
Theorem 4.2.5. The tempered fractional stable motion (LTFSM) defined in (4.5) and tempered fractional harmonizable stable motion (HTFSM) defined in (4.40) are different processes.
Proof. To prove that the processes {XH,α,λ (t)} and {XH,α,λ (t)} are different it is enough
to show that
lim rY

t→∞

(θ1 , θ2 , t) = 0

(4.46)

(θ1 , θ2 , t) > 0,

(4.47)

H,α,λ

and
lim r

t→∞ YH,α,λ

74

where {YH,α,λ (t)} and {YH,α,λ (t)} are the increments of {XH,α,λ (t)} and {XH,α,λ (t)} respectively. Lemma 6.1 in [33] shows that if
∫ +∞
Y (t) =

−∞

f (t − x)Zα (dx),

for f ∈ Lα (R), {Zα } is the SαS random measure on R, then

lim rY (θ1 , θ2 , t) = 0.

t→∞

1

1

H−
H−
Along the same lines, we now define f (x) := (x+1)+ α e−λ(x+1)+ −(x)+ α e−λ(x)+ . Then

YH,α,λ (t) = XH,α,λ (t + 1) − XH,α,λ (t)
∫ +∞ [
]
H− 1
H− 1
=
(t − x + 1)+ α e−λ(t−x+1)+ − (t − x)+ α e−λ(t−x)+ Zα (dx)
−∞

∫ +∞
=

−∞

f (t − x) Zα (dx)

and hence by applying Lemma 6.1 [33], as described above,

lim rY

t→∞

H,α,λ

(θ1 , θ2 , t) = 0

which is Property (4.46). Theorem 3.1 in [37] states that if
∫ +∞
Y (t) = Re

−∞

eikt Ψ(dk) for −∞ < t < ∞

75

is a stationary real harmonizable SαS process, then
(
)
∫ T
1
1
lim inf
r (θ1 , θ2 , t) dt ≥ K(θ1 , θ2 )c0 m{0}F0 +
m(R − {0})F1 > 0,
2π
T →∞ 2T −T Y
where m is the control measure of the isotropic complex-valued random measure Ψ, F0 ∈ R
and F1 > 0 are some constants depends on α, m, θ1 and θ1 . Along the same lines, we now
define
YH,α,λ (t) = XH,α,λ (t + 1) − XH,α,λ (t)
∫ +∞ −ik(t+1)
e
− e−ikt
= Re
Zα (dk)
1
−∞
(λ − ik)H+ α
∫ +∞
eikt Ψ(dk),
= Re
−∞

where Ψ(dk) and m(dk) are given by (4.44) and (4.45) respectively. Hence by applying
Theorem 3.1 in [37], as explained above, we get
(
)
∫ T
1
1
r
(θ1 , θ2 , t) dt ≥ K(θ1 , θ2 )c0 m{0}F0 +
lim inf
m(R − {0})F1 > 0
2π
T →∞ 2T −T YH,α,λ
and consequently
lim r

t→∞ YH,α,λ

(θ1 , θ2 , t) > 0,

which is Property (4.2) and this completes the proof of the theorem.

4.3

Sample path properties

In this section, we present some results about the sample path properties of LTFSM and
HTFSM. The path behavior of the linear fractional stable motion (LFSM) process XH,α
76

50

20

X(t)

0

0

−100

−50

−40
−80

X(t)

0

100

200

300

400

500

0

t

100

200

300

400

500

t

Figure 4.1: Left panel: Sample paths of LTFSM (thick black line) with λ = 0.03 and H = 0.3,
and LFSM (thin black line) with H = 0.3. Both graphs use the same noise realization Zα (t).
The right panel shows the same plots for λ = 0.001, H = 0.7 and α = 1.5.
1

H−
(see Definition 7.4.1 [61]) depends on the structure of the kernel gα,t (x) := (t − x)+ α −
1

H−
(−x)+ α , t, x ∈ R (see [62]). When H − α1 < 0, the function gα,t (x), x ∈ R, has singu-

larities at x = 0 and x = t. By the same argument, The paths behavior of the LTFSM
1

H−
process XH,α,λ depends on the structure of the kernel gα,λ,t (x) := (t − x)+ α e−λ(t−x)+ −
1

H−
(−x)+ α e−λ(t−x)+ . In fact, The function gα,λ,t (x) has singularities at x = 0 and x = t too.

These singularities magnify the stable noise processes Zα (dx) and cause large spikes in the
paths of the LFSM and LTFSM processes. The left panel in Figure 4.1 compares a typical
sample path of both processes, simulated using the same noise realization Zα (t), in the case
H − α1 < 0. In the case H − α1 > 0, (since 0 < H < 1 it follows that α > 1) the paths
of the fractional stable motion can be made continuous with probability one (see Chapter
10 in [61]), since its kernel is bounded and positive for all t > 0. Similarly, the kernel of
LTFSM, gα,λ,t (x), is bounded and positive for all t > 0 and hence the paths of the process
XH,α,λ can be made continuous with probability one when H − α1 > 0. The right panel in
Figure 4.1 shows the corresponding sample paths in the case α = 1.5. These simulations use
a discretized version of the moving average representation of LTFSM (4.5).
Theorem 4.3.1. Let 0 < α < 2, 0 < H < α1 and X = {XH,α,λ (t)}t∈R be the LTFSM
77

∗
process defined in (4.5). Then, for any separable version X ∗ = {XH,α,λ
(t), t ∈ (a, b)} of the

process X, we have that
(
P {ω : sup
t∈(a,b)

)
∗
(t, ω) = ∞} = 1,
XH,α,λ

That is, every version of the process X = {XH,α,λ (t)}t∈R has unbounded paths.
Proof. We apply Theorem 10.2.3 in [61]. Indeed, consider the countable set T ∗ := Q ∩ [a, b],
where Q denotes the set of rational numbers. Since T ∗ is dense in [a, b], there exists a
sequence {tn }n∈N ∈ T ∗ such that tn → x as n → ∞. Therefore
f ∗ (T ∗ ; x) := sup gα,λ,t (x)
t∈T ∗

≥ sup gα,λ,tn (x) =: fn∗ (T ∗ ; x) = ∞,
tn ∈T ∗

∫
as n → ∞ and hence ab f ∗ (T ∗ ; x) dx = ∞, and this contradicts Condition (10.2.14) of
Theorem 10.2.3 in [61]. Therefore, the stochastic process {XH,α,λ } does not have a version
with bounded paths on the interval (a, b), and this completes the proof.
Lemma 4.3.2. Let 1 < α < 2, α1 < H < 1 and λ > 0. Then there exist a positive constant
C1 such that the LTFSM (4.5) satisfies

XH,α,λ (t) − XH,α,λ (s)

locally uniformly in s, t ∈ [0, 1].

78

α
α

≥ C1 t − s

Hα

,

Proof. We write

XH,α,λ (t) − XH,α,λ (s)

α

∫ t

1

|t − u|α(H− α ) e−λα|t−u| du
α
s
∫ t
−λα|t−s|
≥e
|t − u|Hα−1 du
≥

s

=

e−λα|t−s|
Hα

|t − s|Hα

= C1 |t − s|Hα
and this gives the lower bound.
Lemma 4.3.3. Let 1 < α < 2, α1 < H < 1 and λ > 0. Then there exist positive constants
C1 and C2 such that the HTFSM (4.40) satisfies

C1 t − s

Hα

α

≤ XH,α,λ (t) − XH,α,λ (s) ≤ C2 t − s
α

Hα

,

(4.48)

locally uniformly in s, t ∈ [0, 1].
Proof. To get the upper bound, write
∫ +∞ −ikt
|e
− e−iks |α
dk
XH,α,λ (t) − XH,α,λ (s) =
α
|λ − ik|Hα+1
−∞
∫ +∞
≤C
(1 ∧ |t − s|α |k|α ) |λ − ik|−Hα−1 dk
α

−∞

∫
[
α
= C |t − s|

|k|< 1
|t−s|

∫
+

|k|> 1
|t−s|

|k|α ||λ − ik|−Hα−1 dk

|λ − ik|−Hα−1 dk

]
[
≤ C |t − s|α I1 + I2

79

]

(4.49)

where

∫
I1 :=

|k|< 1
|t−s|

and

α

k

λ − ik

∫
I2 :=

−Hα−1

λ − ik

|k|> 1
|t−s|

−Hα−1

dk

dk

and C is a constant. Observe that
∫
I1 =

|k|< 1
|t−s|

k

∫
≤

|k|< 1
|t−s|

≤ t−s

k

Hα−α

α

α

·

−Hα−1
2

λ2 + k 2

k

dk

∫

−Hα−1

dk =

|k|< 1
|t−s|

k

−Hα−1+α

dk

(4.50)

2
α(1 − H)

and
∫
I2 =

λ2 + k 2

|k|> 1
|t−s|

∫
≤

k2

|k|> 1
|t−s|

≤ t−s

Hα

·

−Hα−1
2

dk
∫

−Hα−1
2

dk =

|k|> 1
|t−s|

k

−Hα−1

dk

(4.51)

2
.
Hα

Finally, from (4.49), (4.50) and (4.51) we get

XH,α,λ (t) − XH,α,λ (s)

α
α

≤C

[
[

α

t − s I1 + I2

]

2
2
≤C
+
α(1 − H) Hα
= C2 t − s

]
t−s

Hα

Hα

which gives the upper bound in (4.48). In order to get the lower bound, we use the fact that
80

there exist positive constants c1 , c2 such that |e−iy − 1| > c1 |y| for |y| < c2 . Therefore

XH,α,λ (t) − XH,α,λ (s)

α
α

∫ +∞
=

−∞
∫ +∞

=

−∞

e−ikt − e−iks

α

e−ik(t−s) − 1

α

∫

≥ c1

|k|<

c2
|t−s|

k

α

λ − ik
α

t−s

∫

= c1 |t − s|α

λ − ik

c2
|t−s|

−(Hα+1)

λ − ik

α

|k|<

−(Hα+1)

dk
dk

−(Hα+1)

k (λ2 + k 2 )

dk

−(Hα+1)
2
dk.

We now use the fact that
(

λ2 + k 2

) −(Hα+1)
2

≥

(

1 + c22

) −(Hα+1)
2

t−s

Hα+1

,

c

1 and |k| < 2 to continue the rest of the proof as follows:
for λ < |t−s|
|t−s|

c1 t − s

α

∫

)
α( 2
k
λ + k2

−(Hα+1)
2

dk
c2
|t−s|
∫ c2
) −(Hα+1)
(
α
Hα+1
|t−s| α
2
2
k dk
≥ 2c1 1 + c2
t−s t−s
0
Hα+α+1
−α−1
Hα
= C1 t − s
t−s
= C1 t − s
|k|<

and this gives the lower bound.

4.4

Local Times and Local nondeterminism

In this section, we prove the existence of the local time for LTFSM and HTFSM. We also show
that LTFSM and HTFSM are locally nondeterministic on every compact interval. We first
recall the definition of the local time (see [11] for more details). Suppose X = {X(t), t ≥ 0}
81

is a real-valued separable random process with Borel sample functions. For any Borel set
B ⊂ R+
µB (A) = η({s ∈ B, X(s) ∈ A})
is called the occupation measure of X on B, where η is the Lebesgue measure on R+ . If µB
is absolutely continuous with respect to the Lebesgue measure on R, then we say that X has
a local time on B and define its local time, L(B, .), to be the Radon-Nikodym derivative of
µB with respect to Lebesgue measure. We write L(t, x) instead of L([0, t], x).
Proposition 4.4.1. The LTFSM defined in (4.5) with 1 < α < 2 and α1 < H < 1 has a
square integrable local time L(t, x).
Proof. According to Theorem 3.1 in [11], a stochastic process {X(t), t ∈ [0, T ]} has a local
time L(t, x) which is continuous in t for a.e. x ∈ R and square integrable with respect to x
if {X(t), t ∈ [0, T ]} satisfies
• Condition (H): There exist positive numbers (ρ0 , H) ∈ (0, ∞) × (0, 1) and a positive
function ψ ∈ L1 (R) such that for all κ ∈ R, t, s ∈ [0, T ], 0 < |t − s| < ρ0 we have
( X(t) − X(s) )
E exp iκ
≤ ψ(κ).
|t − s|H

(4.52)

We prove that the LTFSM {XH,α,λ (t)} satisfies (H). Apply (4.7) and Lemma 4.3.2 to get
α)
)
(
( X
H,α,λ (t) − XH,α,λ (s)
α ∥XH,α,λ (t) − XH,α,λ (s)∥α
=
exp
−
|κ|
E exp iκ
|t − s|H
|t − s|αH
(
)
≤ exp − |κ|α C := ψ(κ)

where the function ψ(κ) belongs to L1 (R, dk) which means the HTFSM satisfies H and this
82

completes the proof.
Proposition 4.4.2. The HTFSM defined in (4.40) with 1 < α < 2 and α1 < H < 1 has a
square integrable local time L(t, x).
Proof. We prove that the HTFSM {XH,α,λ (t)} satisfies (H). Apply (4.39) and Lemma 4.3.3
to obtain
α)
( X
)
(
H,α,λ (t) − XH,α,λ (s)
α ∥XH,α,λ (t) − XH,α,λ (s)∥α
E exp iκ
=
exp
−
|κ|
|t − s|H
|t − s|αH
(
)
≤ exp − |κ|α C := ψ(κ),

where the function ψ(κ) belongs to L1 (R, dk) which means the HTFSM satisfies H and this
completes the proof.
We next show that HTFSM is locally nondeterministic on every compact interval [ϵ, T ],
for any 0 < ϵ < T < ∞. Recall that a stochastic process {X(t)}t∈T is locally nondeterministic (LND) if
1. ∥X(t)∥α > 0 for all t ∈ T
2. ∥X(t) − X(s)∥α > 0 for all t, s ∈ T sufficiently close; and
3. for any m ≥ 2,

lim inf
ϵ↓0

∥X(tm ) − span{X(t1 ), . . . , X(tm−1 )}∥α
> 0,
∥X(tm ) − X(tm−1 )∥α

where the lim inf is taken over distinct, ordered t1 < t2 < . . . < tm ∈ T with |t1 − tm | < ϵ,
[
]1/α
T ⊂ R and ∥X∥α = − log(E exp(iX))
for 0 < α ≤ 2 (see [51, 52, 67] for more details).
Next, we show that HTFSM also is LND.
83

Proposition 4.4.3. The HTFSM (4.40) with 1 < α < 2 and α1 < H < 1 is LND on every
interval [ϵ, κ] for ϵ < κ < ∞.
Proof. Theorem 3.3 in [19] shows that a harmonizable multifractional stable motion is LND
[ϵ, κ] for ϵ < κ < ∞. Our proof is a modification of that Theorem. We need to verify
conditions (1), (2) and (3) as described above. The first and second condition follows from
Lemma 4.3.3
α

XH,α,λ (t) − XH,α,λ (s) α ≥ C1 t − s

Hα

,

where C1 is a positive constant. To prove the third condition, observe first that the inverse
Fourier transform of
fH,α,λ (t, k) :=

e−ikt − 1

(4.53)

1

(λ − ik)H+ α

on Lα (R) which is

F −1 fH,α,λ (t, k) =

α−1 ]
Γ(H + α1 ) [ −λ(t−x)
H− α−1
α − e−λ(−x)+ (−x)H− α
+ (t − x)
√
e
+
+
2π

(4.54)

by (2.11). In order to verify the third condition, we shall establish a lower bound for

XH,α,λ (tm ) −

m−1
∑
j=1

uj XH,α,λ (tj )

= fH,α,λ (tm , k) −
α

m−1
∑
j=1

uj fH,α,λ (tj , k)

Lα (R)

α . By applying the Hausdorff-Young
where fH,α,λ (t, k) is defined in (4.53). Let β = α−1

84

inequality (see Theorem 5.7 in [39]):

fH,α,λ (tm , k) −

m−1
∑

uj fH,α,λ (tj , k)

Lα (R)

j=1

≥ C F −1 fH,α,λ (tm , k) −

uj F −1 fH,α,λ (tj , k)

j=1

(∫
=C

m−1
∑

tm−1

−∞

∫ tm
+
tm−1

F −1 fH,α,λ (tm , k) −

m−1
∑

Lβ (R)

(4.55)
uj F −1 fH,α,λ (tj , k)

β

j=1

F −1 fH,α,λ (tm , k)

β

)1
β
dk

.

From (4.54) we have

F −1 fH,α,λ (tm , k) =

α−1 ]
Γ(H + α1 ) [ −λ(tm −x)
H− α−1
α − e−λ(−x)+ (−x)H− α
+ (tm − x)
√
e
+
+
2π
α−1

H−
and the second term, e−λ(−x)+ (−x)+ α , vanishes on the interval [tm−1 , tm ]. Hence we

can continue (4.55) as the following:
[∫
≥C

tm

tm−1

]1
1
β
β(H− ) −λβ(tm −x)
β e
dx
(tm − x)

(4.56)

H
≥ C e−λ(tm −tm−1 ) tm − tm−1 ≥ C e−λ(κ−ϵ) XH,α,λ (tm ) − XH,α,λ (tm−1 )
α

for tm and tm−1 close enough (and C is a constant). In the last line in (4.56), we used the
fact that |tm − tm−1 | < κ − ϵ and we also applied Lemma 4.3.3 to get the last inequality.

85

Therefore

XH,α,λ (tm ) − span{XH,α,λ , . . . , XH,α,λ (tm−1 )}

α

= XH,α,λ (tm ) −

m−1
∑

uj XH,α,λ (tj )

j=1

α

≥ C XH,α,λ (tm ) − XH,α,λ (tm−1 )
α
and consequently

lim inf
ϵ↓0

XH,α,λ (tm ) − span{XH,α,λ , . . . , XH,α,λ (tm−1 )}
XH,α,λ (tm ) − XH,α,λ (tm−1 )

α > C,

α

where C is a positive constant and this verifies Condition (3) of the LND property and this
completes the proof.

86

BIBLIOGRAPHY

87

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