NTU_paper

國立台灣大學數學系碩士班碩士論文
指導教授: 姜祖恕博士
On uniqueness and existence of stochastic
differential equations with non-Lipschitz
coefficients and Lévy noise
在非 Lipschitz 係數條件及 Lévy noise 下隨機
微分方程解存在性及唯一性
研究生: 黃勝郁撰
學號:R94221039
中華民國九十六年六月

On uniqueness and existence of stochastic
differential equations with non-Lipschitz
coefficients and Lévy noise
在非 Lipschitz 係數條件及 Lévy noise 下隨機
微分方程解存在性及唯一性
研究生: 黃勝郁 Student:Sheng-Yu Huang
指導教授: 姜祖恕 Advisor:Tzuu-Shuh Chiang
國立台灣大學
數學系碩士班
碩士論文
A Thesis
Submitted to Department of Mathematics
National Taiwan University
in Partial Fulfillment of the Requirements
for the Degree of
Master
in
June,2007
Taipei,Taiwan
中華民國九十六年六月

Contents
謝辭 ii
中文摘要 iii
Abstract iv
1 Introduction 1
2 Lévy Processes and its Properties 4
2.1 Definition and Characteristic of Lévy process . . . . . . . . . . . . . . . . . 4
2.2 Analytic view of Lévy processes . . . . . . . . . . . . . . . . . . . . . . . . 7
3 Stochastic integration and Itô’s formula 12
3.1 Stochastic integrals with respect to compensated Poisson processes . . . . 12
3.2 Iˆto’s formula for Lévy diffusion . . . . . . . . . . . . . . . . . . . . . . . . 14
4 SDE with Lévy noise 16
4.1 Definition of the SDE with Lévy noise . . . . . . . . . . . . . . . . . . . . 16
4.2 Existence and uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5 The Coefficients of the SDE with Lévy Noise 24
5.1 Life time of SDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
5.2 Non Lipschitz Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
5.2.1 Some studies on pathwise uniqueness . . . . . . . . . . . . . . . . . 33
i

謝辭
命運的手總是在最意想不到的時候推你一把。
對於非數學系出生的我能在這裡發表碩士論文, 是我想也沒有想過的事。若沒有大三時在數學系
所修的線性代數, 讓我一頭栽進數學這個深不見底的黑洞, 也不會有這篇論文的產生了吧! 感謝黃漢
水老師讓我由線性代數進入數學的殿堂。
有別於一般的應用科學系, 數學系強調的是邏輯上的完美, 不容許一絲差錯。但從機率的角度來
看世界, 似乎佔著越來越重要的地位。感謝 Kolmogorov 能將兩者合諧地安排在一起。
許許多多的金融問題能輕鬆的轉化成數學問題, 用數學的角度思考。感謝 Palmer 老師讓我了解
到許多的金融數學。
接下來感謝跟這篇論文比較有關的人。
感謝姜祖恕教導我做數學方式、帶領我進入 L´evy 過程的世界以及這篇論文裡面大大小小的細
節處理。
感謝口試委員許順吉、吳慶堂老師對於論文的指教, 且讓我通過論文口試。
感謝林敬堯、鄒忠男、黃士晏、劉音宏同學及王俊傑、邱奕泰、李俊璋學長們為繁瑣孤獨碩士生
涯增添了許多的色彩, 也幫助我在論文不少的問題。
最後要謝我的父母親養育我。
來不及感謝及忘記感謝的人們, 我會感謝上天, 請祂轉達給你 (妳)。
ii

中文摘要
我們在這篇論文主要探討的是 Lévy 擾動型隨機微分方程解的存在與唯一性的關係。我們更專注在
非 Lipshcitz 條件下其解路徑唯一性的條件。其後介紹及比較近來有關路徑惟一在隨機微分方程相
於對稱穩定過程的研究。
關鍵字: Lévy 過程, Lévy 型隨機微分方程, pathwise uniqueness.
iii

Abstract
In this paper, we devote our attention to the relation of existence and uniqueness of
stochastic differential equations with Lévy noise. Especially, we shall be concerned with
the pathwise uniqueness of SDE with Lévy noises under non-Lipschitzian coefficients.
We also describe, do and compare some of the resent work on pathwise uniqueness on
stochastic differential equations with symmetric α-stable process, 1 < α < 2.
Keywords: Lévy process, SDE driven by Lévy process, pathwise uniqueness.
iv

Chapter 1
Introduction
In the past research, there are fruitful results on the stochastic differential equation
of the diffusion form
(1.1) dXt = b(Xt−)dt + σ(Xt−)dBt
The famous result of Yamada and Watanabe have shown that if (1.1) has the pathwise
uniqueness, then it admits a unique strong solution[7]. We also know two things about
weak solution. The first one is the existence of weak solution is equivalents to the solution
of the martingale problem. The second is that if the coefficients of martingale problem
is bounded and continuous then the problem has solution. So it is crucial to study when
the pathwise uniqueness holds.
To have infinite life time, the typical conditions of the coefficients is linear growth:
(LG)
|b(x)| ≤ C(|x| + 1)
σ(x) 2
≤ C(|x|2
+ 1)
where we denote by σ the Hilbert-Schmidt norm: σ 2
= ij σ2
ij.
The classical conditions for which pathwise uniqueness holds is Lipschitz conditions:
the coefficients function satisfy
(L)
|b(x) − b(y)| ≤ C|x − y|
σ(x) − σ(y) ≤ C|x − y|2
Shizan Fang and Tusheng Zhung [13] generalize both the linear growth and Lipschitz
conditions:
(NLG Z)
|b(x)| ≤ C(|x|γ (|x|2
) + 1)
σ(x) 2
≤ C(|x|2
γ (|x|2
) + 1)
where γ is a strictly positive function and γ ∈ C1
([K, ∞)) 1
for some K > 0.The examples
of γ that satisfy the conditions are γ (s) = log s, γ (s) = log log s,. . .
(i) lim
s→∞
γ(s) = ∞ (ii) lim
s→∞
sγ′
(s)
γ (s)
ds = 0 (iii)
∞
k
ds
sγ(s) + 1
= ∞
1
C1
([K, ∞)) the set of all C1
function from [K, ∞) to R
1

CHAPTER 1. INTRODUCTION 2
(NL Z)
|b(x) − b(y)| ≤ C|x − y|ρ (|x − y|2
)
σ(x) − σ(y) ≤ C|x − y|2
ρ (|x − y|2
)
where ρ is a strictly positive function and ρ ∈ C1
((0, ε]), satisfying: for each ε > 0,
(i) lim
s→0
ρ(s) = ∞ (ii) lim
s→0
sρ′
(s)
ρ (s)
ds = 0 (iii)
ε
0
ds
sρ(s)
= ∞
If (1.1) is one dimension, T. Yamada and S. Watanabe [14] showed more general condition
for pathwise uniqueness:
(NL Y)
|b (x) − b (y)| ≤ κ (|x − y|)
|σ (x) − σ (y)| ≤ ρ (|x − y|)
for |x − y| ≤ ε, ε > 0
where ρ be a strictly increasing function on [0, ∞) and κ be a increasing and concave
function on [0, ∞) such that
(i) ρ (0) = 0 (ii)
ε
0
1
ρ2 (s)
ds = ∞ and (i) κ (0) = 0 (ii)
ε
0
1
κ (s)
ds = ∞
The main purpose of this paper is to extend the results to the more general form : the
stochastic differential equation driven by Lévy noise
(SDE 1)
dXt = b(Xt−)dt + σ(Xt−)dBt +
|z|<1
F(Xt−, z) Ñ(dt, dz) +
|z|≥1
G(Xt−, z)N(dt, dz)
Here are the main results:
• We prove that weak existence and pathwise uniqueness imply unique strong solution
in (SDE 1).
• The following conditions:
(NLG)



|b(x)| ≤ C(|x|γ (|x|2
) + 1)
σ(x) 2
≤ C(|x|2
γ (|x|2
) + 1)
|z|<1
|F (x, z) |2
ν (dz) ≤ C(|x|2
γ (|x|2
) + 1)
where γ is same as (NLG Z), implies (SDE 1) has infinite life time.
• The following conditions:
(NL 1)



|b(x) − b(y)| ≤ C|x − y|ρ (|x − y|2
)
σ(x) − σ(y) ≤ C|x − y|2
ρ (|x − y|2
)
|z|<1
|F (x, z) − F (y, z) |2
ν (dz) ≤ C|x − y|2
ρ (|x − y|2
)
where ρ is same as (NL Z ), implies (SDE 1) has pathwise uniqueness.
• The following conditions: Let all set up of the SDE be one dimensional and
(NL 2)



|b (x) − b (y)| ≤ κ (|x − y|)
|σ (x) − σ (y)| ≤ ρ (|x − y|)
|z|<1
|F (x, z) − F (y, z) |ν (dz) ≤ κ (|x − y|)
for |x − y| ≤ ε, ε > 0
where κ and ρ are same as (NL Y), implies (SDE 1) pathwise uniqueness.

CHAPTER 1. INTRODUCTION 3
There are five chapter in this paper.
In chapter 2, the first section, we introduce Lévy processes and give an intuition of
constructing Lévy-Itô decomposition. The second section, we use the theory of semigroups
to analyze Lévy processes.
In chapter 3, we define the integral with the integrator compensated Poisson pro-
cesses. Itô’s formula for Lévy type diffusion will be stated. We end this chapter with an
application of Itô’s formula.
In chapter 4, the SDE with Lévy noise will be introduced. We describe two king of
solutions: strong and weak, and then we give the proof of weak existence and pathwise
uniqueness implying a unique strong one.
In chapter 5, we discuss life time and the pathwise uniqueness property of SDE with
Lévy noise. We also make some note of papers Bass[2] and Komastu[8].

Chapter 2
Lévy Processes and its Properties
In section 2.1, we first give the definition of Lévy processes, and then describe some
their properties which mainly include Lévy-Itô decomposition (Theorem 4). The next
section, we take analytic view to it such as infinitesimal generator, resolvent. In the end
of this chapter, we see a example of Lévy process: symmetry α-stable process.
2.1 Definition and Characteristic of Lévy process
Definition 1 (Lévy process)
Let Xt be Rn
value cádlág stochastic process starting at X0 = 0 on a probability space
(Ω, F, P) satisfied the following condition, then {Xt}t≥0 is a Lévy process
1. Independent increments:
For each 0 ≤ t1 ≤ t2 ≤ · · · ≤ tn < ∞ , Xti
− Xti−1 i∈N
are independent
2. Stationary increments:
Xti
− Xti−1
D
= Xti−ti−1
∀i ∈ N, where
D
= means same in law.
3. Stochastically continuous:
∀ε > 0, s ≥ 0, lim
t→s
P (|Xt − Xs| > ε) = 0.
Briefly speaking, Lévy processes are hybrid of Brownian motion and compound Pois-
son processes. Without taking out the stationary increments assumption, it merely a
Brownian motion1
plus small jump and large jump. Small jump part may be considered
as a limit of compensated centered compound Poisson processes; Large jump part is a
compound Poisson process2
.
In order to isolate the compound Poisson process from Lévy process, we need to
introduce the following notations:
We say that A is bounded below if A ∈ B (Rn
{0}) and 0 /∈ A, the closure of A.
Assume a counting process on a probability space (Ω, F, P)
N (t, A) = # {0 ≤ s ≤ t : ∆X (s) ∈ A, A is bounded below}
where # means the number of the set. Now, fix t > 0, ω ∈ Ω and A bounded below.
With the notation, we have:
1
It can be considered as the representation of continuous stationary increments process.
2
The typical representation of discontinuous stationary increments process.
4

CHAPTER 2. LÉVY PROCESSES AND ITS PROPERTIES 5
Proposition 2 Let A be a bounded below set, then
1. N (t, A) < ∞ a.s. for all t ≥ 0.
2. (N (t, A) , t ≥ 0) is a Poisson process with intensity ν (A) = E [N (1, A)].
3. If A1, . . ., Am ∈ B (Rn
{0}) are bounded below and disjoint, then the random vari-
ables N(t, A1), . . . , N(t, Am) are independent
Proof. See [1] Lemma 2.3.4,Theorem 2.3.5
Be aware that if A fails to be bounded below then the Theorem may longer hold, be-
cause of the accumulation of infinite numbers of small jumps. There is another observation
that ν (·) is a measure associated with X. We call it intensity measure.
Now, we define the Poisson integral of a Borel measurable function f from Rn
to Rn
as a random finite sum:
A
f (z) N (t, dz) (ω) =
z∈A
f (z) N (t, {z}) (ω)
If we vary t, it gives rise to a càdlàg stochastic process and can be written as
A
f (z) N (t, dz) (ω) =
0≤u≤t
f (∆Xu) 1A (∆Xu) =
n∈N
f ∆XTA
n
1[0,T] ∆XTA
n
where TA
n , n ∈ N be the arrival times for the Poisson process (N (t, A) , t ≥ 0). We
investigate the property of the Poisson integral.
Proposition 3 Let A be a bounded below set, then for each t ≥ 0, denote Nt,f :=
A
f (z) N (t, dz) and has a compound Poisson distribution with the characteristic function
E ei(u·Nt)
= exp t
Rn
ei(u·z)
− 1 µf,A (dz) , µf,A (B) = µ A ∩ f−1
(B) , ∀B ∈ B (Rn
) ,
µ (A) is the distribution of N (t, A), (·) means usual innerproduct.
Proof. See [1] Theorem 2.3.8
Now we can say the big jump (here we consider the jump size large than one) part of
a Lévy process Xt can be written as the form Nt = |z|≥1
zN (t, dz), and it is a compound
Poisson process. Let’s keep going for the decomposition. There is a difficulty to isolate the
small jump that (jump size small than one), because |z|<1
zN (t, dz) may be infinite. We
need to subtract the expectation from the small jump part. We define the compensated
Poisson integral by
Ñt,f =
A
f (z) Ñ (t, dz) (ω) =
z∈A
f (z) N (t, {z}) (ω) − t
A
f (z) µ (dz)
A straightforward argument shows that Ñt,f is a martingale. Let Ñε
t = ε≤|z|<1
z Ñ (t, dz),
so it is also a martingale. We can turn to our main decomposition of X.

Theorem 4 (Lévy-Itô decomposition)
Every Rn
-value Lévy process Xt can be represent in the form
Xt = bt + BA (t) + lim
ε→0
Ñε
t + Nt
= bt + BA (t) +
|z|<1
z Ñ(t, dz) +
|z|≥1
zN(t, dz)
where b ∈ Rn
, BA is a Brownian motion with covariance matrix A and intensity measure
ν with the property3
Rn−{0}
|z|2
∧ 1 ν (dz) < ∞
Moreover, all the components are independent.
Proof. The remaining thing is to show
1. The existence of limε→0
Ñε
t
2. Xt − limε→0
Ñε
t − Nt is a diffusion
3. The property of the intensity measure ν
4. The independence
We refer to [1] section 2.4
Above theorem also shows the if part of Lévy-Khintchine formula:
Corollary 5 (Lévy-Khintchine)
A random variable X is infinite divisible4
if and only if its characteristic have the form
E ei(u·X)
= exp i (b · u) −
1
2
(u · Au) +
Rn−{0}
ei(u·y)
− 1 − i (u · y) 1|y|<1 (y) ν (dy)
:= eη(u)
where b ∈ Rn
, A is a positive definite symmetric n × n matrix and ν is a Lévy measure.
Proof. It is easy to see that if Xt is a Rn
-value Lévy process then it is infinite divisible.
For every t ∈ [0, ∞), n ∈ N, write
X (t) = X
t
n
− X (0) + X
3t
n
− X
2t
n
+· · ·+ X
nt
n
− X
(n − 1) t
n
by independent and stationary increments assumptions we get the statement. By Lévy-Itô
decomposition , the characteristic function of X (t) has the form
E ei(u·Xt)
= exp t i (b · u) −
1
2
(u · Au) +
Rn−{0}
ei(u·y)
− 1 − i (u · y) 1|y|<1 (y) ν (dy)
:= etη(u)
We have proved the if part of corollary. To check the characteristic function is infinitely
divisible we refer to [1]
3
We also call ν Lévy measure.
4
If X
D
= Y1 + Y2 + · · · + Yn for n ∈ N, Y1, Y2, · · · , Yn are i.i.d. random variables, then X is called
infinite divisible.

Remark 6 We denote some things.
1. We call η (u) Lévy symbol.
2. Above theorem ensure us to use the triplet (b, A, ν) to represent a Lévy process.
2.2 Analytic view of Lévy processes
In this section, we take an analytic diversion into semigroup theory and state the
important concepts of generator and resolvent, then we obtain key representations for
the generator: as a pseudo-differential operator and as sum of a second-order elliptic
differential operator and a (compensated) integral of difference operators.
To start analytic view of Lévy process, we point out a fact: every Lévy process
is a homogeneous Markov process. Now, let Xt be a Lévy process and ph,t+h (x, A) =
P (Xt+h ∈ A|Xh = x) be the transition probability. We have
ph,t+h (x, A) = p0,t (x, A) := pt (x, A)
and
E (f (Xt+h) |Xh = x) = E (f (Xt) |X0 = x) = E (f (Xt + x)) := (Ttf) (x)
where f is continuous function from Rn
to R and f (±∞) = 0.5
By Chapman-Kolmogorov
equation and doing some calculus, we know Tt is a semigroup associated with Lévy process
X. i.e., Ts+t = TsTt.
Remark 7 A useful observation is that
(2.1) (Ttf) (x) = f (y) pt (x, dy)
Definition 8 (infinitesimal generator)
Let Tt be an arbitrary semigroup in a Banach space B. let
DL = ψ ∈ B : ∃φψ ∈ B such that lim
t→0
Ttψ − ψ
t
− φψ = 0
By the prescription, let
Lψ := φψ = lim
t→0
Ttψ − ψ
t
we call L the infinitesimal generator.
The domain of T may be different from L. In order know the relation between T and
L, we introduce the following concept.
Definition 9 (resolvent)
Let T be a linear operator in B with domain DT . Define resolvent set ρ (T) =
{λ ∈ C : (λI − T) is invertible}. If λ ∈ ρ (T), we call the linear operator
Rλ (T) = (λI − T)−1
the resolvent of T.
5
With appropriate norm of the function space, it may be considered as Banach space. i.e., f =
sup {|f (x) , x ∈ R|}

There is no a prior reason why ρ (T) should be non-empty. Fortunately, we have the
following:
Theorem 10 (Hille-Yosida)
If L is the infinitesimal generator of a semigroup (Tt, t ≥ 0) associate a Lévy process,
then we have:
1. The interval (0, ∞) ⊆ ρ (L)
2. for each λ > 0,
(2.2) Rλ (L) =
∞
0
e−λt
Ttdt
Proof. See David [1] theorem 3.2.9.
We now turn our attention to the infinitesimal generators of Lévy processes. We
may get the infinitesimal generators with analytic way. We first introduce the concept of
pseudo-differential operator.
Let α = (α1, · · · , αn) be a multi-index. and define |α| = α1 + · · · + αn,
Dα
=
1
i|α|
∂α1
∂xα1
1
· · ·
∂αn
∂xαn
n
and xα
= xα1
· · · xαn
Define Schwartz space S (Rn
, C) to be linear space of all f ∈ C∞
(Rn
, C) for which for all
multi-index α, β
sup
x∈Rn
xβ
Dα
f (x) < ∞
Let f ∈ S (Rn
, C). Denote its Fourier transform Ff is ˆf ∈ S (Rn
, C), where
(Ff) (u) = ˆf (u) = (2π)− d
2
Rn
e−i(u·x)
f (x) dx
for all u ∈ Rn
, and the Fourier inversion F−1
f yields
f (x) = F−1
Ff (x) = (2π)− d
2
Rn
ˆf (u) ei(u·x)
du
for each x ∈ Rn
.
For each cα ∈ C∞
(Rn
) define p (x, u) = |α|≤k cα (x) uα
and P (x, D) = |α|≤k cα (x) Dα
.
Using Fourier inversion and dominated convergence, we find that
(P (x, D) f) (x) =
1
(2π)
d
2 Rn
p (x, u) ˆf (u) ei(u·x)
du
Now, we replace p by a more general function σ : Rn
× Rn
→ C
Definition 11 (pseudo-differential operator)
If σ satisfies
(σ (x, D) f) (x) =
1
(2π)
d
2 Rn
σ (x, u) ˆf (u) ei(u·x)
du
then we call the operator σ (x, D) pseudo-differential operator and σ (x, u) be its symbol.

Using the inversion formula to the semigroup associated with the Lévy process, we
can get the following results.
Theorem 12 Let X be a Lévy process with Lévy symbol η and characteristics (b, A, ν),
(Tt, t ≥ 0) be the associated semigroup and L be its infinitesimal generator. then we have
the following identity
1. ∀t ≥ 0, f ∈ S (Rn
), x ∈ Rn
,
(Ttf) (x) = (2π)− d
2
Rn
ei(u·x)
etη(u) ˆf (u) du
so that Tt is a pseudo-differential operator with symbol etη(u)
.
2. ∀ f ∈ S (Rn
), x ∈ Rn
,
(2.3) (Lf) (x) = (2π)− d
2
Rn
ei(u·x)
η (u) ˆf (u) du
so that L is a pseudo-differential operator with symbol η (u).
Proof. See David[1] theorem 3.3.3
From above theorem and Lévy-Khintchine formula, we can get
Corollary 13 ∀ f ∈ S (Rn
, C) , x ∈ Rn
Lf =
n
j=1
bj∂jf (x) +
1
2
n
j=1
n
k=1
ajk∂j∂kf (x)
+
Rn−{0}
f (x + y) − f (x) −
n
j=1
yj∂jf (x) 1(|y|<1) (y) ν (dy)
where (ajk) = A.
Proof. By Lévy-Khintchine formula, we have
η (u) = i
n
k=1
bkuk −
1
2
n
j=1
n
k=1
ajkujuk +
Rn−{0}
ei(u·y)
− 1 − i
n
j=1
yjuj1(|y|<1) (y) ν (dy)
= η1 (u) − η2 (u) + η3 (u)
It is easy to see that η1 (u) − η2 (u) is the symbol of
n
j=1
bj∂j +
1
2
n
j=1
n
k=1
ajk∂j∂k
Using Fourier inversion to f (y) , we have
f (y) ν (dy) =
Rn
ei(u·y) ˆf (u) duν (dy) =
Rn
ei(u·y) ˆf (u) ν (dy) du

So
[f (x + y) − f (x)] ν (dy) =
Rn
ei(u·x+y)
− ei(u·x) ˆf (u) ν (dy) du
=
Rn
ei(u·y)
− 1 ν (dy) ˆf (u) ei(u·x)
du
and
iyj∂jf (x) ν (dy) = yj
Rn
uj
ˆf (u) ei(u·x)
duν (dy) =
Rn
yjujν (dy) ˆf (u) ei(u·x)
du
Combining above equalities, we get the desired identity.
Example 14 (symmetric stable process)
Let Xt be a one dimensional Lévy process with the triple (0, 0, ν (dz)), where
ν (dz) = 1(0,∞) (z)
dz
|z|1+α + 1(−∞,0) (z)
dz
|z|1+α ,
1 < α < 2. We call Xt is a symmetric stable process. By Lévy-Itô decomposition , we
have
dXt =
|z|<1
z Ñ (dt, dz) +
|z|≥1
zN (dt, dz)
the Lévy symbol
η (u) =
∞
0
eiuy
− 1 − iuy1|y|<1 (y)
1
|y|1+α dy +
0
−∞
eiuy
− 1 − iuy1|y|<1 (y)
1
|y|1+α dy.
By [12] Lemma 14.11, we have
C =
∞
0
eiy
− 1 − iy
1
|y|1+α dy = Γ (−α) e− iπα
2
Let z = −y
0
−∞
eiy
− 1 − iy
1
|y|1+α dy =
∞
0
e−iz
− 1 + iz
1
|z|1+α dz = C, C is conjugate of C
and from the property of gamma function,
Γ (α) Γ (1 − α) =
π
sin πα
thus Γ (−α) = −
π
Γ (α + 1) sin πα
So
η (u) = uα
C + uα ¯C = Re2uα
Γ (−α) e− iπα
2
= Re 2Γ (−α) e− iπα
2 uα
= −
2π
cos
πα
2
uα
= −
π
2
uα
= −c1uα
, c1 =
π
Γ (α + 1) sin απ
2

By (2.3) and above equalities
(Lf) (x)
= (2π)− 1
2
R
eiux
η (u) ˆf (u) du
= −c1F−1
[uα
(Ff) (u)] (x)(2.4)

Chapter 3
Stochastic integration and Itô’s
formula
In the first section, we give the definition of every integral form
|z|<1
F(s, z) Ñ(ds, dz) and
|z|≥1
G(s, z)N(ds, dz)
so it is meaningful to say a Lévy diffusion:
(3.1)
Xt = X0+
t
0
btds+
t
0
σtdBs+dXt+
t
0 |z|<1
F(s, z) Ñ(ds, dz)+
t
0 |z|≥1
G(s, z)N(ds, dz)
The next section we give Itô’s formula to compute f (Xt), Xt satisfying above identity.
In the end of the chapter, we give an application of Itô’s formula.
3.1 Stochastic integrals with respect to compensated
Poisson processes
Like the way to define stochastic integral with Brownian motion, we start off with pre-
dictable simple process. Consider a simple predictable process1
F : Ω × [0, T] × E → R :
F (t, z) =
m
j=1
n
i=1
Fi (tj) 1(tj ,tj+1] (t) 1Ai
(z)
where, E ∈ B (Rn
), 0 ≤ t1 < t2 < · · · < tm+1 = T, Fi (tj) is a bounded Ftj
-measurable
random variable and disjoint Borel subsets A1, A2, . . . , An of E. Define the stochastic
integral with respect to Ñ (t, E)
¯IT (F) :=
T
0 E
F (t, z) Ñ (dt, dz) =
m,n
i,j=1
Fi (tj) Ñ ((tj, tj+1], Ai)
we list properties of IT (F).
1
We denote S (T, E) be the space of the simple predictable process.
12

CHAPTER 3. STOCHASTIC INTEGRATION AND IT Ô’S FORMULA 13
1. ¯It (F) is Ft-adapted and square-integrable local martingale.
2. Linear combination: if F, G ∈ S (T, E) and α, β ∈ R then ¯IT (αF + βG) = α¯IT (F)+
β ¯IT (G)
3. E ¯IT (F) = 0
4. E ¯IT (F)2
=
T
0 E
E |F (t, z)|2
ν (dz) dt
Of course, we want to extend the domain of S (T, E). We use L2
theory to enlarge the
domain to P2 (T, E). P2 (T, E) is the space of functions F : [0, T] × E × Ω → R which
• F is predictable
• P
T
0 E
E F (t, z)2
ν (dz) dt < ∞ = 1
For each F ∈ P2 (T, E), define
IT (F) :=
T
0 E
F (t, z) Ñ (dt, dz) = lim
n→∞
¯IT (Fn)
where Fn ∈ S (T, E) and Fn
L2
→ F.
There may be questions to the definition that
1. Does sequence Fn exist?
2. Can it preserve properties of ¯IT ?
The answer is yes! We refer to [1] section 4.2.
Similarly, we can define
T
0 E
G (t, z) N (dt, dz). Take E = {z ∈ B (Rn
) : |z| ≥ 1}, we
have the relation
T
0 |z|≥1
G (t, z) Ñ (dt, dz) =
T
0 |z|≥1
G (t, z) N (dt, dz) −
T
0 |z|≥1
G (t, z) ν (dz) dt
Remark 15 The set E of
T
0 E
G (t, z) N (dt, dz) must be bounded below in B (Rn
{0})
or it will have no sense.
Now let
Xi
(t) = Xi
(0)+
t
0
bi
(s) ds+
t
0
σi
j (s) dBs+
t
0 |z|<1
Fi
(s, z) Ñ(ds, dz)+
t
0 |z|≥1
Gi
(s, z)N(ds, dz)
where, for each 1 ≤ i ≤ n, 1 ≤ j ≤ r, t > 0, and all the functions satisfy the condition
which make the integral meaningful. We rewrite it to multi-dimensional form as (3.1). So
the Lévy type diffusion (3.1) make sense.

3.2 Iˆto’s formula for Lévy diffusion
There also is Iˆto’s formula for Lévy diffusion.
Theorem 16 (Itô’s formula for Lévy noise)
If Xt is a Lévy-type diffusion of (3.1), then for each f is C2
continuous function from
Rn
to R, t ≥ 0, we have
f (Xt) − f (X0) =
n
i=1
t
0
∂f
∂xi
(Xs) ds +
1
2
n
i,j=1
t
0
∂2
f
∂xi∂xj
(Xs) d Xi
c, Xj
c (s)
+
t
0 |z|<1
[f (Xs− + F (s, z)) − f (Xs−)] Ñ (ds, dz)
+
t
0 |z|<1
f (Xs− + F (s, z)) − f (Xs−) −
n
i=1
Fi
(s, z)
∂f
∂xi
(Xs−) ν (dz) ds
+
t
0 |z|≥1
[f (Xs− + G (s, z)) − f (Xs−)] N (ds, dz)
where d [Xi
c, Xj
c ] (s) is the quadratic covariation of the continuous part of Xt
Proof. see [1] Theorem 4.4.7
We can use Itô’s formula to compute the infinitesimal generator. Let us see a example.
Example 17 (symmetric stable process again)
Let Xt be a one dimension Lévy process with the triple (0, 0, ν (dz)), where ν (dz) =
|z|−1−α
dz, 1 < α < 2. We call Xt is a symmetric stable process. It has the decomposed
form
dXt =
|z|<1
z Ñ (dt, dz) +
|z|≥1
zN (dt, dz)
where Ñ (dt, dz) = N (dt, dz) − ν (dz) dt. By Itô’s formula, we have
f (Xt + x) = f (x) +
t
0 |z|<1
f (Xs + x + z) − f (Xs + x) Ñ (ds, dz)
+
t
0 |z|<1
[f (Xs + x + z) − f (Xs + x) − zf′
(Xs + x)]
1
|z|1+α dzds
+
t
0 |z|≥1
[f (Xs + x + z) − f (Xs + x)] N (ds, dz) .
Denote the last term is Nt, we have Nt =
t
0 |z|≥1
f (Xs + x + z)−f (Xs + x) Ñ (ds, dz)+
t
0 |z|≥1
[f (Xs + x + z) − f (Xs + x) − zf′
(Xs + x)] 1
|z|1+α dzds
+
t
0
|z|≥1
f′
(Xs + x)
1
|z|α dz
Cs
ds. Observe that Cs = 0. So replacing to f (Xt + x), get
f (Xt + x) = f (x) +
t
0 R{0}
f (Xs + x + z) − f (Xs + x) Ñ (ds, dz)
+
t
0 R{0}
[f (Xs + x + z) − f (Xs + x) − zf′
(Xs + x)]
1
|z|1+α dzds

Now we compute the infinitesimal generator of Xt
Lf (x) = lim
t→0
E (f (Xt + x)) − f (x)
t
=
R{0}
Elim
t→0
t
0
f (Xs + x + z) − f (Xs + x) − zf′
(Xs + x) 1
|z|1+α ds
t
dz − f (x)
=
R{0}
[f (x + z) − f (x) − zf′
(x)]
dz
|z|1+α .(3.2)
Now let
Zt =
t
0
HsdXs
then using Itô’s formula again, we have
(3.3) f (Zt) = f (Z0) + Mt +
t
0 |z|<1
[f (Zs− + Hsz) − f (Zs−) − Hszf′
(Zs−)]
dz
|z|1+α ds
Let y = Hsz, we have
Rn−{0}
[f (Zs− + Hsz) − f (Zs−) − Hszf′
(Zs−)]
dz
|z|1+α
=
Rn−{0}
[f (Zs− + y) − f (Zs−) − yf′
(Zs−)] |Hs|α dy
|y|1+α
= |Hs|α
Lf (Zs−)(3.4)
fix λ > 0, let
gλ (x) =
∞
0
e−λt
pt (0, x) dt and f =
∞
0
h (y) gλ (x − y) dy
by (2.1) and Hille-Yosida theorem we have
(λI − L) f = (λI − L)
∞
0
h (y) gλ (x − y) dy
= (λI − L)
∞
0
∞
0
h (y) e−λt
pt (x, dy) dt
= (λI − L)
∞
0
e−λt
Tth (y) dt
= (λI − L) Rλ (L) h = h
That is
(3.5) Lf = λf − h
We will use the above equality in the chapter 5.

Chapter 4
SDE with Lévy noise
At the beginning, we introduce stochastic differential equations with Lévy noise. In
the next section, we concern on the existence and uniqueness, so we give the formal
definition of strong and weak solutions, and give the pathwise uniqueness and unique in
law. We point out that pathwise uniqueness imply unique in law (Theorem 29). Finally
,we provide a method to identify unique strong solution in Theorem 30
4.1 Definition of the SDE with Lévy noise
Let (Ω, F, P) be a probability space equipped with a filtration {Ft, t ≥ 0} which is right
continuous and complete. We say a Stochastic differential equation with Lévy noise, which
means a stochastic process Xt on Ω such that
(SDE 1)
dXt = b(Xt−)dt + σ(Xt−)dBt +
|z|<c
F(Xt−, z) Ñ(dt, dz) +
|z|≥c
G(Xt−, z)N(dt, dz)
where
dXt = (dXi (t))n×1 ,
b (x) = (bi (x))n×1 , bi : Rn
→ R B (Rn
) /B (R) measurable
dBt = (dBi (t))n×1 , Bt be Ft-Brownian motion in Rn
σ(x) = (σij (x))n×n , σij : Rn
→ R B (Rn
) /B (R) measurable
F(Xt−, z) = (Fi (x, z))n×1 , G(x, z) = (Gi (x, z))n×1 ,
Fi and Gi : Rn
×Rn
→ R B (Rn
× Rn
) /B (R) measurable. N (t, dz) be the Poisson process
on [0, ∞) × (Rn
− {0}) with intensity measure ν. All the mappings are assumed to be
measurable for 1 ≤ i ≤ n, 1 ≤ j ≤ n. The rigorous interpretation of (SDE 1) is the
integral form,
Xt = X0+
t
0
b(Xs−)ds+
t
0
σ(Xs−)dBs+
t
0 |z|<c
F(Xs−, z) Ñ(ds, dz)+
t
0 |z|≥c
G(Xs−, z)N(ds, dz)
Remark 18 We denote some things.
1. For convenience, we take c = 1.
16

CHAPTER 4. SDE WITH LÉVY NOISE 17
2. We can also define one dimensional SDE with symmetric stable process which will
be discussed at next chapter:
dYt = H (Yt) dXt, H : R → R measurable
=
|z|<1
H (Yt) z Ñ(dt, dz) +
|z|≥1
H (Yt) zN(dt, dz)(SDE α)
where Ñ(dt, dz) = N(dt, dz) − |z|−1−α
dzdt
There are two kinds of the solutions of (SDE 1), strong solution and weak solution.
We first introduce strong solution, and leave the weak to the next section.
Definition 19 (strong solution)
We call Xt a strong solution of (SDE 1) on the given probability space (Ω, F, P) and
with respected to the fixed Ft- Brownian motion Bt and Poisson random measure such
that be Ft-compound Poisson process and initial condition X0 = ξ if and only if Xt with
càdlàg path and satisfies the following properties:
1. X is adapted to the filtration {Ft},
2. P (X0 = ξ) = 1,
3.
Xt = X0 +
t
0
b(Xs−)ds +
t
0
σ(Xs−)dBs+
t
0 |z|<c
F(Xs−, z) Ñ(ds, dz) +
t
0 |z|≥c
G(Xs−, z)N(ds, dz) a.s.
Remark 20 We may define another general type SDE with Lévy noise
¯Xi (t) = ¯Xi (0) +
t
0
¯bi( ¯Xs−)ds +
t
0
¯σij( ¯Xs−)dBj (s)(SDE 2)
+
t
0 |z|<c
¯Fi( ¯Xs−, z) Ñ(ds, dz) +
t
0 |z|≥c
¯Gi( ¯Xs−, z)N(ds, dz)
for 1 ≤ i ≤ n, 1 ≤ j ≤ r1, where Bt is r1-value Brownian motion and |z|<c
z Ñ(t, dz) is
r2-value compensated compound Poisson process. The two perturbations are in different
dimension and also different from Xt. But in another way, we can modify the coefficient
of (SDE 1) or enlarge the dimension of Xt to get the different conditions . i.e. If n >
r1 > r2, let σ(Xt−) = ¯σ(Xt−)n×r1 0n×(n−r1) and same method to construct F in ¯F.
If r2 = n < r1, let X (t) =
¯X (t)n×1
0(r1−n)×1 n×n
, F (t) =
¯F (t)n×1
0(r1−n)×1 n×n
.
4.2 Existence and uniqueness
To discuss solvability for the SDE (SDE 1), we may loose the definition of strong and
introduce the following kind of solution.

Definition 21 (weak solution)
A weak solution Xt of (SDE 1) means that
1. There exists a probability space (Ω, F, P) and filtration {Ft} of F ,
2. Xt is Ft-adapted càdlàg process, Lt is an Ft-adapted Lèvy process.
3. Xt = b(Xt−)dt+σ(Xt−)dBt+ |z|<1
F(Xt−, z) Ñ(dt, dz)+
t
0 |z|≥1
G(Xs−, z)N(ds, dz),
X0 = x
The main difference of strong and weak strong solution is that the measurability of
the solution. If Xt is strong solution, then Xt must the adapted to the given filtration
which make Lt a Lévy process in the given probability space. Weak one dose not request
Xt to be adapted to the given filtration. The classical example that weak solution don’t
imply strong one is Takana type equation [7].
The existence of weak solution is equivalent to the martingale problem. i.e. Let L be
the infinitesimal associate with a Lévy process, we write it again as in Theorem 12
(Lf) (x) =
n
i=1
bi∂if (x) +
1
2
n
i=1
n
j=1
aij∂i∂jf (x)
+
Rn−{0}
f (x + y) − f (x) −
n
i=1
yi∂if (x) 1(|x|<1) (y) ν (dy)
where aij(x) = n
k=1 σik (x) σjk (x). The martingale problem for L is that of finding for
each (s, x) a probability measure P on D[0, ∞)n
(the space of càdlàg function from [0, ∞)
to Rn
) such that P (Xs = x) = 1 and
Mt = f (Xt) − f (X0) −
t
0
(Lf) (Xs) ds
is a martingale for all f ∈ C∞
0 (Rn
) (the C∞
functions having compact support). For
more discussion we refer to [5]
The following theorem provides the existence of martingale problem. It also give
conditions to the corresponding SDE with Lévy noise that exists of weak solution. We
just give the description without proving.([5] theorem2.2)
Theorem 22 (existence of martingale problem)
If the coefficient of martingale problem satisfies the following condition
1. a are bounded and continuous,
2. b admit the decomposition:b = σσ∗
c1 + c2,
where c1 is bounded and measurable and c2 is bounded and continuous,
3. F are bounded and continuous,
then the SDE exists weak solution.

Briefly speaking, pathwise uniqueness means the existence of two weak solutions of
stochastic differential equation with same initial data are almost surely same for all time;
uniqueness in law means two weak solutions are only same in law for all time. Here are
formal definition:
Definition 23 (pathwise uniqueness)
We say that the pathwise uniqueness of (SDE 1) holds if whenever X and Y are any
two weak solutions defined on same probability space (Ω, F, P) and filtration {Ft} of
F. such that X0 = Y0 a.s., then P (Xt = Yt, ∀t) = 1
Remark 24 Using regular conditional probability (definition28), we need only consider
non-random initial values; i.e., X0 = Y0 = x a.s., for some fixed x ∈ Rn
. This is because
that if X is a solution of (SDE 1) on the space (Ω, F, P) and filtration {Ft} with initial
random variable then setting P0
= P (·|F0) then we get constant initial data and the unique
solution is same in (Ω, F, P0
).
Proposition 25 Using interlacing, it makes sense to begin by omitting large jumps. In
other words, if
(SDE 3) dXt = b(Xt−)dt + σ(Xt−)dBt +
|z|<1
F(Xt−, z) Ñ(dt, dz)
has pathwise unique strong solution Xt, then (SDE 1) also pathwise unique strong solution
Yt.
Proof. Let τn be the sequence of arrival times for the jumps of the compound Poisson
process P (t) = |z|≥c
zN(t, dz). Construct Yt by following steps
Y (t) =



X (t)
X (τ1−) + G (X (τ1−) , ∆P (τ1))
X1 (t − τ1)
X1 (τ2−) + G (X1 (τ2−) , ∆P (τ2))
...
for 0 ≤ t < τ1
for t = τ1
for τ1 < t < τ2
for t = τ2
...
recursively. Where X1 (t) on [0, ¯σ1] is the pathwise unique solution of
dX1 (t) = b(X1 (t))dt + σ(X1 (t−))d ¯Bt +
|z|<1
F(X1 (t−) , z) ¯N(dt, dz)
with initial X1 (0) = Y (τ1), where ¯τ1 = τ2 − τ1, ¯B (t) = B (t + τ1) − B (τ1), ¯N (t, dz) =
Ñ (t + τ1, dz) − Ñ (τ1, dz) , ∆P (τ1) = P (τ1) − P (τ1−), Y is clearly adapted, càdlàg,
pathwise uniqueness and solves (SDE 1).
By above remark, from now on, we can simplify and concentrate on the SDE in the
form (SDE 3) to get existence and uniqueness property.
Definition 26 (uniqueness in law)
We say that the solution of (SDE 3) is unique in law if whenever Xt and Yt are
solution of (SDE 3), {Xt} and {Yt} have the same distribution.

Remark 27 For the same reason of pathwise uniqueness, we can only consider constant
initial data.
It seems like that uniqueness in law imply pathwise uniqueness. But in a rigorous
proof, we need following concepts to complete the statement.
Definition 28 (regular conditional probability)
Let (Ω, F, P) be a probability space and G a sub -σ-algebra of F. A function Q (ω, A) :
Ω × F → [0, 1] is called a regular conditional probability for F given G if
1. for each fixed ω ∈ Ω, Q (ω, ·) is a probability measure on (Ω, F),
2. for each fixed A ∈ F, Q (·, A) is G-measurable
3. for each A ∈ F, Q (ω, A) = P (A|G) (ω), P-a.e. ω ∈ Ω
Under the usual set up of P (A|G) (ω) := E (1A|G), P (A|G) (ω) may not be a proba-
bility measure. So we must define the regular conditional probability to ensure if it is a
probability measure. Regular conditional probabilities do not always exist. Fortunately,
If Ω is a completely separate space1
, then it uniquely exists [7].
Theorem 29 Pathwise uniqueness in (SDE 3) implies uniqueness in law
Proof. Let X(1)
, L(1)
, X(2)
, L(2)
be two weak solutions of (SDE 3). For simplicity,
we may assume the SDE with zero constant initial. If the two solutions on the same
probability space, it is trivial that pathwise uniqueness imply unique in law. Now set
X(i)
, L(i)
is on probability space (Ωi
, Fi
, µi
), i = 1, 2, and induce probability distribution
P(i)
on the space
(S, B (S)) = (D ([0, ∞), Rn
) × Dn
0 , B(D[0, ∞)n
) ⊗ B(Dn
0 ))
where D ([0, ∞), Rn
) is càdlàg function from [0, ∞) to Rn
, Dn
0 is {f ∈ D ([0, ∞), Rn
) : f (0) = 0}.2
Such that
(4.1) P(i)
(A) = µi
X(i)
, L(i)
∈ A
Our first task is to bring them together on the same canonical space, preserving their
joint distributions. To do this, set
Qw
i (A) = Qi (w, A) : Dn
0 × B(D ([0, ∞), Rn
)) → [0, 1]
as the regular conditional probability for B(D ([0, ∞), Rn
)) given w. That is the regular
conditional probability enjoys the following properties:
for each w ∈ Dn
0 , Qi (w, ·) is a probability measure on (D ([0, ∞), Rn
) , B(D ([0, ∞), Rn
)) ,
(4.2)
for each A ∈ B(D ([0, ∞), Rn
)), Qi (·, A) is B(Dn
0 )-measurable, and
(4.3)
P(i)
(A × C) =
C
Qw
i (A) PL
(dw) ; A ∈ B(D[0, ∞)n
), C ∈ B(Dn
0 )
(4.4)
1
Alternatively, we call it Polish space.
2
For more discuss for topology induced from D[0, ∞) and Dn
0 we refer to [4] Chapter3

where PL
is the measure on Dn
0 induce by the given L´evy process3
. Now consider
the measurable space (Ω, F), where Ω = D ([0, ∞), Rn
) × S, F is the completion of the
σ-algebra B(D ([0, ∞), Rn
)) ⊗ B (S) by the N of null sets under the probability measure
(4.5) P (dω) = Qw
1 (dx1) Qw
2 (dx2) PL
(dw)
We have set ω = (x1, x2, w) ∈ Ω. Take
Gt = σ {x1 (s) , x2 (s) , w (s) ; 0 ≤ s ≤ t} , Gt = σ (Gt ∨ N ) , Ft = Gt+,
for 0 ≤ t < ∞. By(4.1),
(4.6) P [ω ∈ Ω: (xi, w) ∈ A] = µi X(i)
, L(i)
∈ A , A ∈ B (S) , i = 1, 2,
By Lemma 32, we have the distribution of (xi, w) under P is the same as the distribution
of X(i)
, L(i)
under µi. Applying pathwise uniqueness, we get
(4.7) P [ω = (x1, x2, w) ∈ Ω; x1 = x2] = 1
It develops from (4.7), (4.6) that
µ1 X(1)
, L(1)
∈ A = P [ω =∈ Ω; (x1, w) ∈ A]
= P [ω =∈ Ω; (x2, w) ∈ A]
= µ2 X(2)
, L(2)
∈ A
and this is the desired statement
Within the above experience of constructing the probability space, we can prove the
main theorem:
Theorem 30 Pathwise uniqueness and existence of weak solution in (SDE 3) is equiva-
lent to uniqueness of strong solution.
Proof. We begin the proof under the same set-up as above theorem. Now we conclude
from Lemma 32 that (x1, w) and (x2, w) are solutions on the same space (Ω, F, P) with
same reference family Ft. Hence the pathwise uniqueness implies that x1 = x2 P-a.e..
Now deﬁne a measure
Qw
(dx1, dx2) := Q (w, dx1, dx2) = Qw
1 (dx1) Qw
2 (dx2)
on
(S′
, B (S′
)) := (D ([0, ∞), Rn
) × D ([0, ∞), Rn
) , B(D ([0, ∞), Rn
)) ⊗ B(D ([0, ∞), Rn
)))
From (4.1), we have
P (A × B) =
B
Qw
(A) PL
(dw) A ∈ B (S′
) , B ∈ B(Dn
0 )
Take A = {(x1, x2) ∈ S : x1 = x2} and B = Dn
0 , comparing (4.7), we have that there
is a PL
-null set N ∈ B(Dn
0 ), such that Qw
(A) = 1 for w /∈ N. That is given any
3
The existence of PL
, we also refer to [4] Chapter3 example 13.1 in weak convergence method

w then x1 = x2 = some constant depended on = h (w)4
and by Lemma 31 h (w) is
Bt (Dn
0 )/ Bt(D ([0, ∞), Rn
)) measurable, since the entry of Qw
i must satisfy measurability.
Moreover, X = h (L) solves the given SDE. Now if there are any SDE with given space
˜Ω, ˜F, ˜P and ˜Ft- Lévy process ˜L. then using same construction, we have X = h ˜L is
a strong solution.
Lemma 31 For every fixed t ≥ 0, and A ∈ Bt(D[0, ∞)n
),5
then Qi (·, A) is Bt (Dn
0 )-
measurable, where Bt (Dn
0 ) is the augmentation of the filtration Bt(Dn
0 ) by the null sets
of PL
.
Proof. Let (ϕtw) (s) = w (t ∧ s).Consider the regular conditional probabilities up to time
t
Qw,t
i (A) = Qt
i (w, A) : Dn
0 × Bt(D ([0, ∞), Rn
)) → [0, 1]
for Bt(D[0, ∞)n
), given ϕtw. These enjoy properties analogous to Qi (w, A) such that
for each w ∈ Dn
0 , Qt
i (ϕtw, ·) is a probability measure on (Dt ([0, ∞), Rn
) , Bt(D ([0, ∞), Rn
)) ,
for each A ∈ Bt(D[0, ∞)n
), Qt
i (·, A) is Bt(Dn
0 )-measurable, and
P(i)
(A × C) =
C
Qw,t
i (A) PL
(dw) ; ∀ A ∈ Bt(D ([0, ∞), Rn
)), C ∈ Bt(Dn
0 )
(4.8)
If we can show (4.8) holds for all C ∈ B(Dn
0 ), then this implies that Qt
i (w, A) = Qi (w, A)
for PL
-a.s.(w). and the conclusion follows. Note that
C1 = {C ∈ B(Dn
0 ) : C satisfies (4.8)}
is a λ-system and
C2 = C ∈ B(Dn
0 ) : C = ϕ−1
t C1 ∩ θ−1
t C2
is a π-system, where (θtw) (s) = w (t + s) − w (s), C1, C2 ∈ B(Dn
0 ). Obviously, we have
C2 ⊆ C1 and σ (C2) = B(Dn
0 ). By the Dynkin π-λ system theorem, we can just prove (4.8)
holds for C ∈ C2. For such a C, we have
C
Qw,t
i (A) PL
(dw)
=
{ϕ−1
t C1}
Qw,t
i (A) PL
θ−1
t C2|B(Dn
0 )
=
{ϕ−1
t C1}
Qw,t
i (A) PL
(dw) PL
θ−1
t C2
= P(i)
A × ϕ−1
t C1 PL
θ−1
t C2
Observe that θ−1
t C2 is independent of B(Dn
0 ) under PL
. From (4.1), we have
P(i)
A × ϕ−1
t C1 = µi
X(i)
∈ A, ϕtL(i)
∈ C1
PL
θ−1
t C2 = P(i)
[(x, w) ∈ S : θtw ∈ C2] = µi
θtL(i)
∈ C2
4
we used the fact:
X = Y a.s. X ⊥ Y =⇒ X = Y =constant
5
In this prove we always use subscript to denote the space or σ-field generated by the function up to
time t.

Therefore,
P(i)
(A × C) = µi
X(i)
∈ A, L(i)
∈ C
= µi
X(i)
∈ A, ϕtL(i)
∈ C1, θtL(i)
∈ C2
= µi
X(i)
∈ A, ϕtL(i)
∈ C1 µi
θtL(i)
∈ C2
=
C
Qw,t
i (A) PL
(dw)
because X(i)
∈ A, ϕtL(i)
∈ C1 ∈ F
(i)
t ⊥ θtL(i)
∈ C2 , where ⊥ means independent of.
We get desired statement.
Lemma 32 w = w (t) is an n-dimensional Ft-Lévy process on (Ω, F, P).
Proof. Since w is certainly a Lévy process on Dn
0 under PL
. We need only to show inde-
pendent and stationary increment property of w. Using Lemma 31 and Lévy-Khintchine
formula we have
EP
ei(u·w(t)−w(s))
1A×B×C
=
C
ei(u·w(t)−w(s))
Qt
1 (w, A) Qt
2 (w, B) PL
(dw)
= e(t−s)η(u)
C
Qw,t
1 (A) Qw,t
2 (B) PL
(dw)
= e(t−s)η(u)
P (A × B × C)
for A, B ∈ Bs (D[0, ∞)n
) and C ∈ Bs (Dn
0 ). It shows w have independent and stationary
increment. So w is a Lévy process on (Ω, F, P).

Chapter 5
The Coefficients of the SDE with
Lévy Noise
The mainly purpose if in this chapter is to establish under some coefficients with non-
liner growth and non-Lipschitz condition that the life time of SDE is infinite and pathwise
uniqueness holds. We also collect some other results under non-Lipschitz condition at the
end of the chapter.
5.1 Life time of SDE
Theorem 33 Let γ be a strictly positive function and γ ∈ C1
([ε, ∞)) , satisfying
lim
s→∞
γ(s) = ∞(5.1)
lim
s→∞
sγ′
(s)
γ (s)
ds = 0(5.2)
∞
ε
ds
sγ(s) + 1
= ∞(5.3)
and the coefficient of stochastic differential equation with Lévy noise SDE 3 satisfy
(NLG)



|b(x)| ≤ C(|x|γ (|x|2
) + 1)
σ(x) 2
≤ C(|x|2
γ (|x|2
) + 1)
|z|<1
|F (x, z) |2
ν (dz) ≤ C(|x|2
γ (|x|2
) + 1)
then (SDE 3) has no explosion. i.e. P (ζ = ∞) = 1, where ζ = inf {t > 0 : |Xt|2
= ∞} .
Proof. Let (Xt, Lt) be a solution of (SDE 3) that the coefficients satisfy (NLG) .Although
γ is defined on [ε, ∞), we may extend it to [0, ∞) to simplify proof. Let ξt = |Xt|2
,by
Itô’s formula we have
ξt = ξo +
t
0
2 (σ(Xs−)∗
Xs− · dBs) +
t
0
2 (Xs− · b(Xs−)) + σ(Xs−) 2
ds(5.4)
+
t
0 |z|<1
|Xs− + F(Xs−, z)|2
− |Xs−|2 Ñ (ds, dz)
+
t
0 |z|<1
|Xs− + F (Xs−, z)|2
− |Xs−|2
− 2 (F (Xs−, z) · Xs−) ν (dz) ds
24

CHAPTER 5. THE COEFFICIENTS OF THE SDE WITH LÉVY NOISE 25
where ( · ) means scalar product and * means transpose of the matrix. Thus
d [ξ, ξ]t = 4 |σ (Xt)∗
Xt|
2
dt
where [·, ·]t means quadratic variation.
Let ψ (ξ) =
ξ
0
ds
sρ(s)+1
and ϕ (ξ) = eψ(ξ)
, ξ ≥ 0.Using Itô’s formula again, we have
ϕ (ξt) = ϕ (ξ0)
+ 2
t
0
ϕ′
(ξs−) (σ (Xs−)∗
Xs− · dBs)
+
t
0 |z|<1
ϕ ξs− + |Xs− + F(Xs−, z)|2
− |Xs−|2
− ϕ (ξs−) Ñ (ds, dz)
+ 2
t
0
ϕ′
(ξs−) (Xs · b (Xs)) ds +
t
0
ϕ′
(ξs−) σ(Xs−) 2
ds +
t
0
ϕ′′
(ξs−) |σ (Xs−)∗
Xs|
2
ds
+
t
0 |z|<1
[ϕ ξs− + |Xs− + F(Xs−, z)|2
− |Xs−|2
− ϕ (ξs−)
− |Xs− + F(Xs−, z)|2
− |Xs−|2
ϕ′
(ξs−) ]ν (dz) ds
+
t
0 |z|<1
ϕ′
(ξs−) |Xs− + F (Xs−, z)|2
− |Xs−|2
− 2 (F (Xs−, z) · Xs−) ν (dz) ds
= ϕ (ξ0) + Mt + I1 (t) + I2 (t) + I3 (t) + I4 (t) + I5 (t)
where
Mt = 2
t
0
ϕ (ξs−) (σ (Xs−)∗
Xs− · dBs)
+
t
0 |z|<1
ϕ ξs− + |Xs− + F(Xs−, z)|2
− |Xs−|2
− ϕ (ξs−) Ñ (ds, dz)
I1 (t) = 2
t
0
ϕ′
(ξs−) (Xs− · b (Xs−)) ds, I2 (t) =
t
0
ϕ′
(ξs−) σ(Xs−) 2
ds,
I3 (t) =
t
0
ϕ′′
(ξs−) |σ (Xs−)∗
Xs−|
2
ds
I4 (t) =
t
0 |z|<1
[ϕ ξs− + |Xs− + F(Xs−, z)|2
− |Xs−|2
− ϕ (ξs−)
− |Xs− + F(Xs−, z)|2
− |Xs−|2
]ϕ′
(ξs−) ν (dz) ds
I5 (t) =
t
0 |z|<1
ϕ′
(ξs−) |Xs− + F (Xs−, z)|2
− |Xs−|2
− 2 (F (Xs−, z) · Xs−) ν (dz) ds
Using (NLG), It is easy to see Mt is a martingale; therefore E (Mt) = 0.Observe
(5.5) ϕ′
(ξ) =
ϕ (ξ)
ξγ (ξ) + 1
and ϕ′′
(ξ) =
ϕ (ξ) (1 − γ (ξ) − ξγ′
(ξ))
(ξγ (ξ) + 1)2
By (5.1) and (5.2) , we can choose large constant C1 such that |1 − γ (ξ) − ξγ (ξ)| ≤
C1γ (ξ).So that for all ξ ≥ 0, for some large C1
(5.6) ϕ′′
(ξ) ≤ C1
ϕ (ξ) γ (ξ)
(ξγ (ξ) + 1)2 ≤ C1
ϕ (ξ) γ (ξ)
(ξγ (ξ) + 1) ξγ (ξ)
= C1
ϕ (ξ)
(ξγ (ξ) + 1) ξ

Using Cauchy-Schwartz inequality, (NLG) and (5.5), we get
|ϕ′
(ξs−) (Xs− · b (Xs−))| ≤ |ϕ′
(ξs−)| |(Xs−)| |b (Xs−)|
≤ CC1
C2
ϕ (ξs)
ξs−γ (ξs−)
ξγ (ξs−) + 1
|Xs−| (|Xs−| |γ (ξs−) + 1|)
= C2ϕ (ξs−)
ξs−γ (ξs−) + ξ
1
2
s−
ξγ (ξs−) + 1
Since γ ≥ 1, for ξ ≥ 0
ξγ (ξ) + ξ
1
2
ξγ (ξ) + 1
= 1 +
ξ
1
2 − 1
ξγ (ξ) + 1
≤ 1 +
ξ
1
2 − 1
ξ + 1
≤ 2
we get
(5.7) E (|I1 (t)|) ≤ 4C2
t
0
E (ϕ (ξs−)) ds
Use same method,
E |I2 (t)| ≤
t
0
E |ϕ′
(ξs)| σ (Xs) 2
≤
t
0
E C1
ϕ (ξs)
ξsγ (ξs) + 1
C (ξsγ (ξs) + 1)
= C2
t
0
Eϕ (ξs) ds(5.8)
Now, by (5.6) and (NLG)
ϕ′′
(ξs) |σ (Xs)∗
Xs|
2
≤ |ϕ′′
(ξs)| σ (Xs) 2
|Xs|2
≤ C2
ϕ (ξs)
(ξsγ (ξs) + 1) ξs
ξs (ξsγ (ξs) + 1)
= C2ϕ (ξs)
so that
(5.9) E (I3 (t)) ≤ 2C2
t
0
Eϕ (ξs) ds
I4 (t) + I5 (t) =: I6 (t)
=
t
0 |z|<1
[ϕ ξs− + |Xs− + F(Xs−, z)|2
− |Xs−|2
− ϕ (ξs−)(5.10)
− 2ϕ′
(ξs−) · F (ξs−, z) Xs−ν (dz) ]ds
Consider I7 (t) = ϕ ξs− + |Xs− + F(Xs−, z)|2
− |Xs−|2
, by Taylor’s expansion and
ϕ′′
≤ 0
I7 (s) = ϕ ξs− + 2 (Xs− · F (Xs−, z)) + |F(Xs−, z)|2
= ϕ (ξs−) + ϕ′
(ξs−) 2Xs− · F (Xs, z) + |F (Xs−, z)|2
+
1
2
ϕ′′
(θ) 2Xs− · F (Xs, z) + |F (Xs−, z)|2 2
≤ ϕ (ξs−) + ϕ′
(ξs−) 2Xs− · F (Xs, z) + |F (Xs−, z)|2
(5.11)

where θ belongs to line segment of ϕ (ξs−) (ω) and 2Xs− · F (Xs, z) + |F (Xs−, z)|2
(ω).Placing
(5.11) into (5.10), we get
I6 (t) ≤ ϕ′
(ξs−) |F (Xs−, z)|2
So that
E |I6 (t)| ≤ 2
t
0
Eϕ′
(ξs−) |F (ξs−, z)|2
ν (dz) ds
≤ 2C2
t
0
Eϕ (ξs) ds(5.12)
Combining (5.7) (5.8) (5.11) (5.12), and set
τR = inf t > 0 : ξ2
t > R, R > 0
we can get
(5.13) E (ϕ (ξt∧τR
)) ≤ ϕ (ξ0) + K
t
0
E (ϕ (ξs∧τR
)) ds for some K > 0
Applying Gronwall inequality to (5.13) we get that for all t ≥ 0 and R ≥ 0,
(5.14) E (ϕ (ξt∧τR
)) ≤ ϕ (ξ0) eKt
Let R → ∞ in (5.14), by monotone convergence theorem, we get
(5.15) E (ϕ (ξt∧ζ)) ≤ ϕ (ξ0) eKt
If P (ζ < ∞) > 0, then for a large T > 0, P (ζ ≤ T) > 0. Taking t = T in (5.15), we get
(5.16) E ϕ 1(ζ<T)ξζ ≤ ϕ (ξ0) eKt
Since ϕ (ξζ) = ϕ (∞), it contradicts to ϕ (ξ0) eKt
< ∞. Therefore P (ζ = ∞) = 1
Lemma 34 (Gronwall’s inequality)
Let [a, b] ⊂ R, and ϕ (t) : [a, b] → [0, ∞) and satisfy ∃C > 0, such that ∀t ∈ [a, b]
ϕ (t) ≤ C + K
t
a
ϕ (s) ds
then we have
ϕ (t) ≤ CeKt
Proof. Let h : [a, b] → [0, ∞) be deﬁne by
h (t) = C + K
t
a
ϕ (s) ds ∀t ∈ [a, b]
Then we have
h′
(t) = Kϕ (t) ≤ Kh (t)

Since h > 0, we get
h′
(t)
h (t)
≤ K, h (0) = C
Integrating both side we get
h (t) ≤ CeKt
So
ϕ (t) ≤ C + K
t
a
ϕ (s) ds = h (t) ≤ CeKt
5.2 Non Lipschitz Coefficients
Theorem 35 Let ρ be a strictly positive function and ρ ∈ C1
((0, ε]) , satisfying
lim
s→0
ρ(s) = ∞(5.17)
lim
s→0
sρ′
(s)
ρ (s)
ds = 0(5.18)
ε
0
ds
sρ(s)
= ∞(5.19)
and the coefficient of stochastic differential equation with Lévy noise (SDE 3) satisfy
(NL 1)



|b(x) − b(y)| ≤ C|x − y|ρ (|x − y|2
)
σ(x) − σ(y) ≤ C|x − y|2
ρ (|x − y|2
)
|z|<1
|F (x, z) − F (y, z) |2
ν (dz) ≤ C|x − y|2
ρ (|x − y|2
)
for some constant C > 0 then (SDE 3) have pathwise uniqueness property.
Proof. Let Xt and Yt be solutions of (SDE 3) with Xt = x = Yt. Set Zt = Xt − Yt and
ξt = |Zt|2
.Then we have:
Zt =
t
0
(b(Xs) − b(Ys))ds +
t
0
(σ (Xs) − σ (Ys)) dBs +
t
0 |z|<1
(F(Xs−, z) − F (Ys−, z)) Ñ(ds, dz)
=:
t
0
˜bsds +
t
0
˜σsdBs +
t
0 |z|<1
˜Fs−
Ñ(ds, dz)
(5.20)
where ˜bt = b(Xt) − b(Yt), ˜σt = σ (Xt) − σ (Yt) , and ˜Ft = F(Xt, z) − F (Yt, z). Same as
(5.4), we get
ξt =
t
0
2 (˜σ∗
s Xs− · dBs) +
t
0
2Xs− · ˜bs + ˜σs
2
ds
+
t
0 |z|<1
Xs− + ˜Fs
2
− |Xs−|2 Ñ (ds, dz)
+
t
0 |z|<1
Xs− + ˜Fs
2
− |Xs−|2
− 2 ˜Fs · Xs− ν (dz) ds(5.21)

Let τ = inf {t > 0, ξt ≥ ε2
}. Define for δ > 0,
ψδ (ξ) =
ξ
0
ds
sρ (s) + δ
and ϕδ (ξ) = eψδ(ξ)
So as in Theorem 33 we can get
ϕ′
δ (ξ) =
ϕδ (ξ)
ξρ (ξ) + δ
and ϕ′′
δ (ξ) =
ϕδ (ξ) (1 − ρ (ξ) − ξρ′
(ξ))
(ξρ (ξ) + δ)2
And we may assume ρ ≥ 1, or the condition will worse than Lipschitzian coefficients. By
(5.17), (5.18) on ρ, there exists a large constant C1 > 0 such that
|1 − ρ (ξ) − ξρ (ξ)| ≤ C1ρ (ξ)
Again applying Itô’s formula, we get
ϕδ (ξt) = ϕδ (ξ0) + 2
t
0
ϕδ (ξs−) (˜σ∗
s Xs · dBs)
+
t
0 |z|<1
ϕδ ξs− + Xs− + ˜Fs−
2
− |Xs−|2
− ϕδ (ξs−) Ñ (ds, dz)
+ 2
t
0
ϕ′
δ (ξs−) Xs · ˜bs ds +
t
0
ϕ′
δ (ξs−) ˜σs
2
ds +
t
0
ϕ′′
δ (ξs−) |˜σ∗
s Xs|2
ds
+
t
0 |z|<1
[ϕδ ξs− + Xs− + ˜Fs−
2
− |Xs−|2
− ϕδ (ξs−)
− Xs− + ˜Fs−)
2
− |Xs−|2
ϕ′
δ (ξs−) ]ν (dz) ds
+
t
0 |z|<1
ϕ′
δ (ξs−) Xs− + ˜Fs−
2
− |Xs−|2
− 2 ˜Fs− · Xs− ν (dz) ds
= ϕδ (ξ0) + Mt + I1 (t) + I2 (t) + I3 (t) + I4 (t) + I5 (t)
Since all the conditions are similar to the theorem 33,we can get the same estimating on
ϕδ (ξt)
E (ϕδ (ξt∧τ )) ≤ eC2t
Letting δ → 0 in the above inequality and use Fatou’s lemma we get
E eψ0(ξt∧τ )
≤ eC2t
which implies that for t given,
(5.22) ξt∧τ = 0 almost surely
If P (τ < ∞) > 0 , then we get ξτ = 0 almost surely for all t ∈ Q ∩ [0, T], for some large
T > 0. It contradict to the assumption of the stopping time τ. So P (τ = ∞) = 1, it
means Xt = Yt almost surely.
We can make the proof of theorem easy but not so general.

Theorem 36 Let κ be a increasing and concave function on (0, ∞) such that
(i) κ (0) = 0 (ii)
ε
0
1
κ (s)
ds = ∞
and the coefficient of stochastic differential equation with Lévy noise (SDE 3) satisfy
(NL 2)
2 (x − y)·(b (x) − b (y))+|σ (x) − σ (y)|2
+
|z|<1
|F (x, z)−F (y, z) |2
ν (dz) ≤ Cκ |x − y|2
for some constant C > 0 . Then (SDE 3) have pathwise uniqueness property.
Proof. Let Xt and Yt be solutions of (SDE 3) with Xt = x = Yt. Set Zt = Xt − Yt and
ξt = |Zt|2
. By (5.21), we have
ξt =
t
0
2 (˜σ∗
s Xs− · dBs) +
t
0
2Xs− · ˜bs + ˜σs
2
ds
+
t
0 |z|<1
Xs− + ˜Fs
2
− |Xs−|2 Ñ (ds, dz)
+
t
0 |z|<1
˜Fs
2
ν (dz) ds
Let τN = inf {t ≥ 0 : ξt > N} and ζt∧τN
= E (ξt∧τN
), by (NL 2) and Jensen’s inequality
we have
ζt∧τN
= E
t∧τN
0
2 (˜σ∗
s Xs− · dBs) + 2Xs− · ˜bs + ˜σs
2
+
|z|<1
˜Fs
2
ν (dz) ds
≤ E
t∧τN −
0
Cκ (ξs) ds ≤
t
0
Cκ (ζs∧τN
) ds.
Hence by the following Lemma for any T < ∞, ζt∧τN
= 0 a.s., ∀t ∈ [0, ∞). Letting
N → ∞ we get
Zt = Xt − Yt = 0, ∀t ∈ [0, ∞) a.s.
(We have used the assumption that the solutions of the SDE have infinite life time.)
Lemma 37 If
0 ≤ f (t) ≤ C
t
0
κ(f (s) ds
where ∞ > C > 0, f is continuous function, κ (s) > 0 is increasing for s > 0, and has
ε
0
1
κ(s)
ds = ∞ ∀ε > 0 then f ≡ 0
Proof. Let
gε (t) = ε + C
t
0
κ (gε
(s)) ds
then gε (t) ≥ f (t) ≥ 0 by definition. Since g′
ε (t) = κ (gε
(t)) > 0, there exists hε (t) =
g−1
ε (t)
h′
ε (t) =
1
g′
ε (hε (t))
=
1
Cκ (gε (hε (t)))
=
1
Cκ (t)
, for t > ε

and gε (t) is decreasing with respected to ε. We also have
lim
ε→0
gε (hε (t)) = t
Then we claim lim
ε→0
gε (t) = 0 which implies f (t) = 0. Suppose lim
ε→0
gε (t) = a > 0,
lim
ε→0
hε (a) = lim
ε→0
a
ε
1
Cκ (t)
dt =
a
0
1
Cκ (t)
dt = ∞
It contradicts to lim
ε→0
gε (hε (a)) = a. So we get f (t) ≡ 0.
Let’s focus on some one dimensional cases.
Theorem 38 Let d = r = 1 ,ρ be a strictly increasing function on (0, ∞) such that
ρ (0) = 0(5.23)
ε
0
1
ρ2 (s)
ds = ∞(5.24)
and κ be a increasing and concave function on (0, ∞) such that
κ (0) = 0(5.25)
ε
0
1
κ (s)
ds = ∞(5.26)
Assume the coeﬃcient of (SDE 3) σ (x) , b (x) and F (x, z) ν (dx) are bounded and satisfy
(NL 3)



|b (x) − b (y)| ≤ κ (|x − y|)
|σ (x) − σ (y)| ≤ ρ (|x − y|)
|z|<1
|F (x, z) − F (y, z) |ν (dz) ≤ κ (|x − y|)
for |x − y| ≤ ε, ε > 0
Then the pathwise uniqueness of solutions holds for (SDE 3)
Proof. Let Xt and Yt be solutions of (SDE 3) with Xt = x = Yt and Zt = Xt − Yt. By
(5.24), we can set 1 > a1 > a2 > · · · > an > · · · > 0 such that
1
a1
1
ρ2 (s)
ds = 1,
a2
a1
1
ρ2 (s)
ds = 2, · · · ,
an−1
an
1
ρ2 (s)
ds = n, · · ·
and ψn (s), n = 1, 2, . . .be a continuous function such that its support is contained in
(an, an−1), such that
(5.27) 0 ≤ ψn (s) ≤
2
ρ2 (s) n
and
an−1
an
ψn (s) ds = 1
Set
ϕn (x) =
|x|
0
y
0
ψn (s) dsdy
Doing some calculus, we know that ϕn ∈ C2
(R),
|ϕ′
n (x)| =
|x|
0
ψn (s) ds ≤ 1(5.28)
|ϕ′′
n (x)| = ψn (x) ≤
2
ρ2 (x) n
(5.29)
ϕn (x) ր |x| as n → ∞(5.30)

Applying Itô’s formula to ϕn (Zt) and (5.20),we have
ϕn (Zt) =
t
0
ϕ′
n (Zs) ˜σsdBs +
t
0 |z|<1
ϕn Zs− + ˜Fz (s) − ϕn (Zs−) Ñ (ds, dz)
+
t
o
ϕ′
n (Zs)˜bsds +
1
2
t
0
ϕ′′
n (Zs) ˜σ2
s ds
+
t
0 |z|<1
ϕn Zs− + ˜Fz (s) − ϕn (Zs−) − ϕ′
n (Zs−) ˜Fz (s) ν (dz) ds
= Mt + I1 (t) + I2 (t) + I3 (t)
where
Mt =
t
0
ϕ′
n (Zs) ˜σsdBs +
t
0 |z|<1
ϕn Zs− + ˜Fz (s) − ϕn (Zs−) Ñ (ds, dz)
I1 (t) =
t
o
ϕ′
n (Zs)˜bsds , I2 (t) =
1
2
t
0
ϕ′′
n (Zs) ˜σ2
s ds
I3 (t) =
t
0 |z|<1
ϕn Zs− + ˜Fz (s) − ϕn (Zs−) − ϕ′
n (Zs−) ˜Fz (s) ν (dz) ds
It is easy to see that Mt is a martingale. We have
(5.31) E (Mt) = 0
By (NL 3), (5.28) and Jensen’s inequality, we get
E (I1 (t)) ≤
t
0
E (|b (Xs) − b (Ys)|) ds
≤
t
0
E (κ (|Xs − Ys|)) ds(5.32)
By (NL 3), (5.29) and Jensen’s inequality we get
E (I2 (t)) ≤
1
2
t
0
E
2
nρ2 (Zt)
ρ2
(Zt) ds
≤
t
n
→ 0 as n → ∞(5.33)
Now consider I4 (s) = ϕn Zs− + ˜Fz (s) , by Taylor expansion, (NL 3) and (5.28) we have
I4 (s) = ϕn (Zs−) + ϕ′
n (θ) ˜Fz (t−)
≤ ϕn (Zs−) + κ (|Zs−|)(5.34)
Placing into I3 (t), we get
(5.35) E (|I3 (t)|) ≤ 2
t
0
E (κ (|Xs − Ys|)) ds

Combining (5.32), (5.33) and (5.35), we get
Eϕn (Xt − Yt) = 3
t
0
κ (E (|Xs − Ys|)) ds +
t
n
Let n → ∞, by (5.30)
E |(Xt − Yt)| = 3
t
0
κ (E (|Xs − Ys|)) ds
By above lemma we get E |(Xt − Yt)| = 0 and hence Xt = Yt almost surely. It proves
pathwise uniqueness for (SDE 3).
5.2.1 Some studies on pathwise uniqueness
Here we consider some special case of SDE to get more general non-Lipschitz condition.
We introduce some other peoples’ results of pathwise uniqueness.
Theorem 39 (Nakao)
Let b (x) and σ (x) be bounded Borel measurable. Suppose σ (x) is of bounded variation
on any compact interval. Further, suppose there exists a constant C > 0 such that σ (x) >
C for x ∈ R. Then, the pathwise uniqueness holds for one dimensional SDE with diffusion
(1.1).
Proof. For detail, see Nakao [9]. He use upcrossing estimate to get the fact.
Unfortunately, his method has difficulty to SDE with jump [3]. Now, consider the SDE
with symmetry α-stable process (SDE α). Komastu computes the symbol of infinitesimal
(2.4), and uses it to prove pathwise uniqueness.
Theorem 40 (Komastu)
Let 1 < α < 2 and ρ be an increasing function on [0, ∞) satisfying
(i) ρ (0) = 0 (ii)
ε
0
1
ρ (s)
ds = ∞, for ε > 0
If the coefficients of one dimensional SDE with symmetric α-stable process (SDE α) satisfy
(NL α) |H (x) − H (y)|α
≤ ρ (|x − y|) for |x − y| ≤ ε, ε > 0
then solution to (SDE α) is pathwise unique.
Proof. Komastu’s method
Let Yt and Y ′
t be solutions of (SDE 3) with Xt = x = Yt. From the assumption of ρ,
we can set 1 > a1 > a2 > · · · > an > · · · > 0 such that
1
a1
1
ρ (s)
ds = 1,
a2
a1
1
ρ (s)
ds = 2, · · · ,
an−1
an
1
ρ (s)
ds = n, · · ·
and ψn (s), n = 1, 2, . . .be a continuous function such that its support is contained in
(an, an−1), such that
0 ≤ ψn (s) ≤
1
ρ (s) n
and
an−1
an
ψn (s) ds = 1

Set
u (x) = |x|α−1
and un (x) = (u ∗ ψn) (x)
Since lim
n→∞
ψn = δ, the function un → u as n → ∞
uε
(x) = |x|α−1
eε|x|
and uε
n (x) = (uε
∗ ψn) (x)
Use the property of gamma function, we have
Fuε
n (y) = F |x|α−1
eε|x|
(y) Fψn (y)
=
∞
−∞
e−iyx
|x|α−1
eε|x|
dxFψn (y)
=
∞
0
e−iyx
xα−1
eεx
dx +
0
−∞
−e−iyx
xα−1
e−εx
dx Fψn (y)
=
∞
0
e−(iy−ε)x
xα−1
dx +
∞
0
e−(iy+ε)x
xα−1
dx Fψn (y)
= Γ (α) (ε − iy)−α
+ (ε + iy)−α
Fψn (y)
Note that
lim
εց0
|y|α
(ε − iy)−α
+ (ε + iy)−α
= |y|α
|y|−α
e
απ
2 + |y|−α
e
−απ
2
= 2 cos
απ
2
So we have
Lun = lim
εց0
Luε
n = − lim
εց0
c1Γ (α) F−1
[|yα
| (Fuε
n) (y)]
= − lim
εց0
c1Γ (α) F−1
|yα
| (ε + iy)−α
+ (ε − iy)−α
Fψn (y)
= −2
Γ (α) π
Γ (α + 1) sin απ
2
cos
απ
2
ψn (y)
= −2 cot
απ
2
ψn (y)
= cψn (y)
By (3.3) and (3.4),
un (Yt − Y ′
t ) − un (0)(5.36)
= Mt +
t
0 R{0}
|H (Ys) − H (Y ′
s )|
α
Lun (Ys − Y ′
s ) ds(5.37)
Set stopping time Tk = inf {t : |Yt − Y ′
t | > k}, by the assumption of coeﬃcients, we get
E un Yt∧Tk
− Y ′
t∧Tk
= E
t∧Tk
0 R{0}
|H (Ys) − H (Y ′
s )|
α
Lun (Ys − Y ′
s ) ds
≤ E
t∧Tk
0
ρ (Ys − Y ′
s ) cψn (Ys − Y ′
s ) ds
≤ E
t∧Tk
0
c
n
ds

Since un (x) → u (x) = |x|α−1
as n → ∞. Now let n → ∞, we have
E Yt∧Tk
− Y ′
t∧Tk
α−1
= 0
Let k → ∞, we conclude that Yt = Y ′
t a.s.
Unlike Komastu’s method, Bass[2] get the same result by using the resolvent of in-
ﬁnitesimal generator.
Proof. Bass’ method
Let Yt, Y ′
t and ψn be the same set-up of above proof. Let
fn =
∞
0
ψn (y) gλ (x − y) dy, gλ (x) =
∞
0
e−λt
pt (0, x) dt
and
At =
t
0
|H (Yt) − H (Y ′
t )|
α
ds
By (3.5), we have
Lfn = λfn − ψn
So by assumption and (3.5) , we get
E e−λAt
fn (Yt − Y ′
t ) − fn (0)
=
t
0
e−λAs
d [fn (Yt − Y ′
t )] −
t
0
e−λAs
λ |H (Yt) − H (Y ′
t )|
α
fn (Yt − Y ′
t )
=
t
0
e−λAs
|H (Yt) − H (Y ′
t )|
α
Lfn (Yt − Y ′
t ) ds −
t
0
e−λAs
λ |H (Yt) − H (Y ′
t )|
α
fn (Yt − Y ′
t )
= −
t
0
e−λAs
|H (Yt) − H (Y ′
t )|
α
ψn (Yt − Y ′
t )
≥ −
t
0
e−λAs
s
n
ds
Let n → ∞,
E e−λAt
gλ (Yt − Y ′
t ) − gλ (0) ≥ 0
we know gλ (0) > gλ (x) for x > 0 and e−λAt
< 1, so we must have
Yt − Y ′
t = 0 for each t > 0

Bibliography
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driven by stable processes for which pathwise uniqueness fails, Stoch. Proc. & their
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Inc, 2nd ed.
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North-Holland, Amsterdam.
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[9] Nakao, A. (1972) On the pathwise uniqueness of solutions of one-dimensional stochas-
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[10] Protter,P. (1990) Stochastic integration and differential equations.
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[11] Rong Situ. (2005) Theroy of stochastic differential equations with jumps and appli-
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36

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