Jeanblanc M., Yor M., Chesney M. Mathematical Methods for Financial Markets
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600 11 L´evy Processes
Definition 11.2.1.3 AL´evy process that takes values in [0, ∞[ (equivalently,
which has increasing paths) is called a subordinator.
In this case, the parameters in the L´evy-Khintchine formula are m ≥ 0,σ =0
and the L´evy measure ν is a measure on ]0, ∞[ with
]0,∞[
(1 ∧ x)ν(dx) < ∞.
This last property is a consequence of Proposition 11.2.3.11.
Proposition 11.2.1.4 (Strong Markov Property.) Let X be an F-L´evy
process and τ an F-stopping time. Then, on the set {τ<∞} the process Y
t
=
X
τ+t
− X
τ
is an (F
τ+t
,t ≥ 0)-L´evy process; in particular, Y is independent
of F
τ
and has the same law as X.
Proof: Let us set ϕ(t; u)=E(e
iuX
t
). Let us assume that the stopping time τ
is bounded and let A ∈F
τ
.Letu
j
,j =1,...,n be a sequence of real numbers
and 0 ≤ t
0
< ··· <t
n
an increasing sequence of positive numbers. Then,
applying the optional sampling theorem several times, for the martingale
Z
t
(u)=e
iuY
t
/E(e
iuY
t
),
E
1
A
e
i
P
n
j=1
u
j
(Y
t
j
−Y
t
j−1
= E
⎛
⎝
1
A
j
Z
τ+t
j
(u
j
)
Z
τ+t
j−1
(u
j
)
ϕ(t
j
− t
j−1
,u
j
)
⎞
⎠
= P(A)
j
ϕ(t
j
− t
j−1
,u
j
) .
The general case is obtained by passing to the limit.
Proposition 11.2.1.5 Let X be a one-dimensional L´evy process. Then, for
any fixed t,
(X
u
,u≤ t)
law
=(X
t
− X
(t−u)−
,u≤ t) . (11.2.1)
Consequently, (X
t
, inf
u≤t
X
u
)
law
=(X
t
,X
t
− sup
u≤t
X
u
). Moreover, for any
α ∈ R, the process (e
αX
t
t
0
du e
−αX
u
,t≥ 0) is a Markov process and for any
fixed t,
e
X
t
,e
X
t
t
0
e
−X
s−
ds
law
=
e
X
t
,
t
0
e
X
s−
ds
.
Proof: It is straightforward to prove (11.2.1), since the right-hand side has
independent increments which are distributed as those of the left-hand side.
The proof of the remaining part can be found in Carmona et al. [141]and
Donati-Martin et al. [258].
Proposition 11.2.1.6 If X is a L´evy process, for any t>0, the r.v. X
t
is
infinitely divisible.
Proof: Using the decomposition X
t
=
n
k=1
(X
kt/n
−X
(k−1)t/n
)weobserve
the infinitely divisible character of any variable X
t
.
11.2 L´evy Processes 601
We shall prove later (see Corollary 11.2.3.8)thataL´evy process is a
semi-martingale. Hence, the integral of a locally bounded predictable process
with respect to a L´evy process is well defined.
Proposition 11.2.1.7 Let (X, Y ) be a two-dimensional L´evy process. Then
U
t
= e
−X
t
u +
t
0
e
X
s−
dY
s
,t≥ 0
is a Markov process, whose semigroup is given by
Q
t
(u, f )=E
f(ue
−X
t
+
t
0
e
−X
s−
dY
s
)
.
Comment 11.2.1.8 The last property in Proposition 11.2.1.5 is a particular
case of Proposition 11.2.1.7.
Exercise 11.2.1.9 Prove that (e
αX
t
t
0
du e
−αX
u
,t≥ 0) is a Markov process
whereas
t
0
du e
αX
u
is not. Give an explanation of this difference.
Hint: For s<t,write
Y
t
= e
αX
t
t
0
du e
−αX
u
= e
α(X
t
−X
s
)
e
αX
s
Y
s
+ e
α(X
t
−X
s
)
t
s
e
−α(X
u
−X
s
)
du
and use the independence property of the increments of X.
11.2.2 Poisson Point Processes, L´evy Measures
Let X be a L´evy process. For every bounded Borel set Λ ∈ R
d
, such that
0 /∈
¯
Λ,where
¯
Λ is the closure of Λ, we define
N
Λ
t
=
0<s≤t
1
Λ
(ΔX
s
)
to be the number of jumps up to time t which take values in Λ.Themap
Λ → N
Λ
t
(ω) defines a σ-finite measure on R
d
\ 0 denoted by N
t
(ω, dx).
Definition 11.2.2.1 The σ-additive measure ν defined on R
d
\ 0 by
ν(Λ)=E(N
Λ
1
)
is called the L´evy measure of the process X.
We shall soon show (see Remark 11.2.3.3) that the L´evy measure is indeed
the same measure which occurs in the L´evy-Khintchine representation of the
r.v. X
1
.
602 11 L´evy Processes
Proposition 11.2.2.2 Let X be a L´evy process with L´evy measure ν.
(i) Assume ν(Λ) < ∞. Then, the process
N
Λ
t
=
0<s≤t
1
Λ
(ΔX
s
),t≥ 0
is a Poisson process with intensity ν(Λ).
(ii) Let Λ
i
be a finite collection of sets and assume that, ∀i, ν(Λ
i
) < ∞.
The processes N
Λ
i
are independent if and only if ν(Λ
i
∩ Λ
j
)=0,i = j,
in particular if the sets Λ
i
are disjoint. If Λ = ∪
i
Λ
i
, with disjoint Λ
i
, then
N
Λ
=
i
N
Λ
i
is a Poisson process with intensity
i
ν(Λ
i
).
Proof: (i) The process N
Λ
has independent and stationary increments,
its paths are increasing and right continuous and increase only by jumps of
size 1, hence it is a Poisson process (Watanabe’s characterization, Proposi-
tion 8.3.3.1).
(ii) We give the proof for two sets Λ and Γ .
(a) We first assume that Λ and Γ are disjoint. The processes N
Λ
and N
Γ
are Poisson processes in the same filtration and they never jump at the same
time. Hence, they are independent (see Proposition 8.3.6.2).
(b) Conversely, assume that N
Λ
and N
Γ
are independent. From the
independence property, for any pair (λ, μ) of positive real numbers
E
exp
−(λN
Γ
t
+ μN
Λ
t
)
= E
exp
−λN
Γ
t
E
exp
−μN
Λ
t
,
hence
exp
−t
ν(dx)
1 − e
−(λ1
Γ
(x)+μ1
Λ
(x))
=exp
−t
ν(dx)
1 − e
−λ1
Γ
(x)
exp
−t
ν(dx)
1 − e
−μ1
Λ
(x)
.
It follows that
Λ∩Γ
ν(dx)(1 − e
−(λ+μ)
)+
Λ\Γ
ν(dx)(1 − e
−λ
)+
Γ \Λ
ν(dx)(1 − e
−μ
)
=
Γ
ν(dx)(1 − e
−λ
)+
Λ
ν(dx)(1 − e
−μ
) ,
therefore
Λ∩Γ
ν(dx)(1 − e
−(λ+μ)
)=
Λ∩Γ
ν(dx)(1 − e
−λ
)+
Λ∩Γ
ν(dx)(1 − e
−μ
)
which implies that ν(Γ ∩Λ)=0.
It follows that, if ν(Γ ) < ∞, the process (N
Γ
t
− ν(Γ )t, t ≥ 0) is a
martingale. Note that if Γ and Λ are disjoint, N
Γ
and N
Λ
are independent,
11.2 L´evy Processes 603
and the processes N
Γ
t
− ν(Γ )t and N
Λ
t
− ν(Λ)t are orthogonal martingales.
The jump process of a L´evy process is a Poisson point process (see Section 8.9).
Let Λ be a Borel set of R
d
with 0 /∈
¯
Λ,andf a positive Borel function
defined on Λ. We have, by definition of N
t
,
Λ
f(x)N
t
(ω, dx)=
0<s≤t
f(ΔX
s
(ω))1
Λ
(ΔX
s
(ω)) .
Proposition 11.2.2.3 (Compensation Formula.) Let f be a positive
function such that f(0) = 0.Then,
E
0<s≤t
f(ΔX
s
)
= t
R
d
f(x)ν(dx) .
(i) If
R
d
f(x)ν(dx) < ∞, then
0<s≤t
f(ΔX
s
) is a L´evy process, with
L´evy measure ν ◦ f
−1
, the image of ν by f. The process
M
f
t
:=
0<s≤t
f(ΔX
s
) − t
R
d
f(x)ν(dx)
is a martingale. More generally, if H is a positive predictable function (i.e.,
H : Ω × R
+
× R
d
→ R
+
is P×Bmeasurable) such that H
s
(ω, 0) = 0
E
s≤t
H
s
(ω, ΔX
s
)
= E
t
0
ds
dν(x)H
s
(ω, x)
.
(ii) Let Λ beaBorelsetinR
d
such that 0 /∈
¯
Λ and g := f1
Λ
.Thenif
g ∈ L
1
(dν) ∩L
2
(dν), the process (M
g
t
)
2
−t
Λ
f
2
(x)ν(dx) is a martingale. In
particular,
E
Λ
f(x)N
t
(·,dx) − t
Λ
f(x)ν(dx)
2
= t
Λ
f
2
(x)ν(dx) .
Proof: In a first step, these results are established for functions f of the
form f (x)=
α
i
1
A
i
(x) where the sets A
i
are disjoint, using properties
of compound Poisson processes. Then, it suffices to use the dominated
convergence theorem.
Example 11.2.2.4 TheGammaprocessisaL´evy process with L´evy measure
dx x
−1
e
−x
,andforα>0, the process
0<s≤t
(Δγ
s
)
α
is a subordinator with
L´evy measure ν
α
(dy)=α
−1
dy y
−1
e
−y
1/α
.
In the particular cases of Poisson processes and compound Poisson processes,
the compensation formula in Proposition 11.2.2.3 was already obtained
respectively as a direct consequence of (8.2.4) for Poisson processes and as
a consequence of Proposition 8.6.3.6 for compound Poisson processes.
604 11 L´evy Processes
One can extend the exponential formula (8.6.2)toL´evy processes:
Proposition 11.2.2.5 (Exponential Formula.) Let X be a L´evy process
and ν its L´evy measure.
(i) For any t ∈ R
+
, u
i
≥ 0,andf satisfying
Λ
(1 − e
−u f(x)
)ν(dx) < ∞,
one has:
E
exp
−u
Λ
f(x)N
t
(·,dx)
=exp
−t
Λ
(1 − e
−u f(x)
)ν(dx)
.
(ii) For every t and every Borel function f defined on R
+
× R
d
such that
t
0
ds
|1 − e
f(s,x)
|ν(dx) < ∞, one has
E
⎡
⎣
exp
⎛
⎝
s≤t
f(s, ΔX
s
)1
{ΔX
s
=0}
⎞
⎠
⎤
⎦
=exp
−
t
0
ds
R
(1 − e
f(s,x)
)ν(dx)
.
Exercise 11.2.2.6 Let X be a L´evy process with finite variance. Check that
E(X
t
)=tE(X
1
)andVar(X
t
)=t Var(X
1
).
Exercise 11.2.2.7 Prove that, if f ∈ L
2
(ν1
Λ
),
Var
Λ
f(x)N (t, dx)=t
Λ
f
2
(x)ν(dx) .
Prove that E((N
Λ
t
)
2
)=t
2
(ν(Λ))
2
+ tν(Λ).
Exercise 11.2.2.8 This exercise will provide another proof of Proposition
11.2.2.2. Assuming that N
Γ
and N
Λ
are independent, prove that ν(Γ ∩Λ)=0.
Hint: The integration by parts formula leads to
N
Γ
t
N
Λ
t
=
t
0
N
Γ
s
−
dN
Λ
s
+
t
0
N
Λ
s
−
dN
Γ
s
+
s≤t
ΔN
Γ
s
ΔN
Λ
s
=
t
0
N
Γ
s
−
dN
Λ
s
+
t
0
N
Λ
s
−
dN
Γ
s
+
s≤t
ΔN
Γ ∩Λ
s
.
Hence
E(N
Γ
t
N
Λ
t
)=ν(Λ)E
t
0
N
Γ
s
ds
+ ν(Γ )E
t
0
N
Λ
s
ds
+ tν(Γ ∩ Λ)
= t
2
ν(Γ )ν(Λ)+tν(Γ ∩ Λ) .
It remains to note that E(N
Γ
t
N
Λ
t
)=t
2
ν(Λ)ν(Γ ), because of our hypothesis of
independence, thus ν(Λ ∩ Γ ) = 0. We might also have obtained the result by
noting that
s≤t
ΔN
Γ
s
ΔN
Λ
s
= 0, since the independent processes N
Γ
and
N
Λ
do not have common jumps.
11.2 L´evy Processes 605
Exercise 11.2.2.9 Let X be a real-valued L´evy process. Prove the following
assertions:
1. If E(|X
1
|) < ∞,then(X
t
− tE(X
1
),t≥ 0) is a martingale.
2. If E(exp(λX
t
)) < ∞,then(exp(λX
t
)[E(exp(λX
t
))]
−1
,t ≥ 0) is a
martingale. (See also Subsection 11.3.1.)
3. If μ
t
(u)=E(e
iuX
t
), the process (e
iuX
t
/μ
t
(u),t≥ 0) is a martingale.
4. For t, s ≥ 0, μ
t+s
= μ
t
μ
s
and μ
t
does not vanish.
Exercise 11.2.2.10 Let Λ be such that 0 /∈
¯
Λ and f ∈ L
1
(ν1
Λ
). Prove that
the process
Λ
f(x)N
t
(·,dx) is a compound Poisson process.
Exercise 11.2.2.11 The Ornstein-Uhlenbeck Process Driven by a
L´evy Process. The OU process driven by the L´evy process (X
t
,t≥ 0) with
initial state U
0
and parameter c is the solution of
U
t
= U
0
+ X
t
− c
t
0
U
s
ds .
Check that
U
t
= e
−ct
U
0
+
t
0
e
cs
dX
s
.
Hint: This is a particular case of Proposition 11.2.1.7. See Novikov [678]and
Hadjiev [416] for a study of these OU processes.
Exercise 11.2.2.12 The Structure of Compound Poisson Processes.
Let (Y
n
) be a sequence of random variables, and (T
n
) an increasing sequence
of random times such that the series
n
|Y
n
|1
{T
n
≤t}
converges and that
the process X
t
=
n
Y
n
1
{T
n
≤t}
is a L´evy process. Prove that the random
variables (S
n
= T
n
−T
n−1
,n≥ 1) are i.i.d. with exponential law and that the
(Y
n
,n≥ 0) are i.i.d. and independent of (S
n
,n≥ 1).
Exercise 11.2.2.13 Let X be a L´evy process and Q
t
=
s≤t
(ΔX
s
)
2
.Prove
that
E(e
−λQ
t
) = exp
−t
∞
−∞
(1 − e
−λx
2
)ν(dx)
.
See Carr et al. [151] for some applications to finance.
Exercise 11.2.2.14 (a) Prove that, if (X
t
,t≥ 0) is a L´evy process, then for
every n ∈ N, one has
(X
nt
,t≥ 0)
law
=(X
(1)
t
+ X
(2)
t
+ ···+ X
(n)
t
,t≥ 0) (11.2.2)
where the X
(i)
t
are n independent copies of X.
606 11 L´evy Processes
(b) Let (X
t
,t ≥ 0) be a L´evy process starting from 0, and let a, b > 0.
Define
Y
t
=
bt
at
du
u
X
u
,t≥ 0.
Prove that Y also satisfies (11.2.2), although the process (Y
t
,t ≥ 0) is not a
L´evy process, unless X
u
= cu, u ≥ 0.
(c) Prove that, if the process (X
t
,t≥ 0) satisfies (11.2.2), then the process
X considered as a r.v. with values in D([0, ∞[) is infinitely divisible.
Let us call processes that satisfy (11.2.2) IDT processes (infinitely divisible
in time). This exercise shows that there are many IDT processes which are
not L´evy processes. See Mansuy [620] and Es-Sebaiy and Ouknine [334].
11.2.3 L´evy-Khintchine Formula for a L´evy Process
Proposition 11.2.3.1 Let X be a L´evy process taking values in R
d
.Then,
for each t, the r.v. X
t
is infinitely divisible and its characteristic function is
given by the L´evy-Khintchine formula: for u ∈ R
d
,
E(exp(iu X
t
))
=exp
t
iu m −
u Au
2
+
R
d
(e
iu x
− 1 − iu x1
|x|≤1
)ν(dx)
,
where m ∈ R
d
, A is a positive semi-definite matrix, and ν is a L´evy measure
on R
d
\{0}.Ther.v.X
t
admits the characteristic triple (tm, tA, tν). We shall
say in short that X is a (m, A, ν) L´evy process.
Proof: Let X be a L´evy process. We have seen that X
t
is an infinitely
divisible random variable. In particular the distribution of X
1
is infinitely
divisible and admits a L´evy-Khintchine representation (see Proposition
11.1.1.4). The L´evy-Khintchine representation for X
t
is then obtained from
that of X
1
,firstfort ∈ N,thenfort ∈ Q
+
and for any t, using the continuity
in law w.r.t. time for L´evy processes.
Exercise 11.2.3.2 Express the L´evy-Khintchine representation of the r.v.
n
i=1
λ
i
(X
t
i+1
− X
t
i
) in terms of the L´evy-Khintchine representation of the
process X,orofther.v.X
1
.
Remark 11.2.3.3 Let us check in a particular case that the L´evy measure
which appears in the L´evy-Khintchine formula is the same as the one which
appears in Definition 11.2.2.1 when X is a real-valued L´evy process with
bounded variation (i.e., the difference of two subordinators) and without drift
term (see Bertoin [78] for the general case). From
e
iuX
t
=1+
s≤t
e
iuX
s
− e
iuX
s−
=1+
s≤t
e
iuX
s−
(e
iuΔX
s
− 1) ,
11.2 L´evy Processes 607
we obtain
E(e
iuX
t
)=1+E
⎛
⎝
s≤t
e
iuX
s−
(e
iuΔX
s
− 1)
⎞
⎠
=1+E
t
0
ds e
iuX
s−
ν(dx)(e
iux
− 1)
,
where, in the second equality, we have used the compensation formula in
Proposition 11.2.2.3. Therefore, setting μ
t
(u)=E(e
iuX
t
), we get the linear
integral equation
μ
t
(u)=1+
t
0
ds μ
s
(u)
ν(dx)(e
iux
− 1) .
(The integral
ν(dx)(1 ∧|x|) is finite.) Solving this linear equation leads to
the L´evy-Khintchine formula.
Definition 11.2.3.4 The continuous function Φ : R
d
→ C such that
E[exp(iu X
1
)] = exp(−Φ(u))
is called the characteristic exponent (sometimes the L´evy exponent) of the
L´evy process X.
If E
"
e
λ X
1
#
< ∞ for any λ with positive components, the function Ψ
defined on [0, ∞[
d
, such that
E[exp(λ X
1
)] = exp(Ψ (λ))
is called the Laplace exponent of the L´evy process X.
It follows that
E[exp(iu X
t
)] = exp(−tΦ(u))
and, if Ψ(λ) exists,
E[exp(λ X
t
)] = exp(tΨ (λ))
and
Ψ(λ)=−Φ(−iλ) .
From Proposition 11.2.3.1,
Φ(u)=−iu m +
1
2
u Au −
(e
iu x
− 1 − iu x1
|x|≤1
)ν(dx)
Ψ(λ)=λ m +
1
2
λ Aλ +
(e
λ x
− 1 − λ x1
|x|≤1
)ν(dx)
(11.2.3)
If ν(R
d
\0) < ∞, the process X has a finite number of jumps in any finite
time interval. In finance, when ν(R
d
) < ∞, one refers to finite activity.
608 11 L´evy Processes
If moreover A = 0, the process X is a compound Poisson process with
“drift.”
If ν(R
d
\ 0) = ∞, the process corresponds to infinite activity. Assume
moreover that A =0.Then
• If
|x|≤1
|x|ν(dx) < ∞, the paths of X are of bounded variation on any
finite time interval.
• If
|x|≤1
|x|ν(dx)=∞, the paths of X are no longer of bounded variation
on any finite time interval.
Example 11.2.3.5 We present examples of L´evy processes and their char-
acteristics.
• Drifted Brownian Motion. The process (mt + σB
t
,t≥ 0) where B is a
BM is a L´evy process with characteristic exponent −ium + u
2
σ
2
/2, hence,
its L´evy measure is zero. We shall see that the family of L´evy processes
with zero L´evy measure consists precisely of all the L´evy processes with
continuous paths, which are exactly (mt + σB
t
,t≥ 0) for any m, σ ∈ R.
• Poisson Process. The Poisson process with intensity λ is a L´evy process
with characteristic exponent (see (8.2.1))
λ(1 − e
iu
)=
(1 − e
iux
)λδ
1
(dx)
hence its L´evy measure is λδ
1
,whereδ is the Dirac measure.
• Compound Poisson Process. Let X
t
=
N
t
k=1
Y
k
be a ν-compound
Poisson process. Its characteristic exponent is (see Proposition 8.6.3.4)
Φ(u)=
(1 − e
iux
) ν(dx) .
Its L´evy measure is ν(dx)=λF (dx), where F is the common law of all
the Y
k
’s, and λ the intensity of the Poisson process N.
• Process of Brownian Hitting Times. Let W be a BM. The process
(T
r
,r ≥ 0) where T
r
=inf{t : W
t
≥ r} is an increasing L´evy process.
The process (−T
r
,r ≥ 0) admits as Laplace exponent −
√
2λ. Hence, the
L´evy measure of the L´evy process T
r
is ν(dx)=dx1
{x>0}
/
√
2πx
3
.Wehave
recalled above that if a L´evy process is continuous, then, it is a Brownian
motion with drift. Hence, the process (T
r
,r ≥ 0) is not continuous.
• Stable Processes. With any stable r.v. of index α, we may associate a
L´evy process which will be called a stable process of index α. The process
T
r
is a stable process (in fact a subordinator) of index 1/2. The linear
Brownian motion (resp. the Cauchy process) is symmetric stable of index
2(resp.1).
Comment 11.2.3.6 The Laplace exponent is also called the cumulant
function. When considering X as a semi-martingale, the triple (m, σ
2
,ν)
11.2 L´evy Processes 609
is the triple of predictable characteristics of X (see Jacod and Shiryaev
[471]). Obviously, the characteristics of the dual process
X = −X are
(−m, σ
2
, ν)whereν(dx)=ν(−dx). The Laplace exponent of the dual process
is
Ψ(λ)=Ψ(−λ).
Proposition 11.2.3.7 (L´evy-Itˆo’s Decomposition.) If X is an R
d
-valued
L´evy process, it can be decomposed into
X = Y
(0)
+ Y
(1)
+ Y
(2)
+ Y
(3)
where Y
(0)
is a constant drift, Y
(1)
is a linear transform of a Brownian
motion, Y
(2)
is a compound Poisson process with jump sizes greater than
or equal to 1 and Y
(3)
is a L´evy process with jump sizes smaller than 1. The
processes Y
(i)
are independent.
Sketch of the proof: The characteristic exponent of the L´evy process can
be written
Φ = Φ
(0)
+ Φ
(1)
+ Φ
(2)
+ Φ
(3)
with
Φ
(0)
(λ)=−im λ, Φ
(1)
(λ)=
1
2
λ Aλ ,
Φ
(2)
(λ)=
(1 − e
iλ x
)1
{|x|>1}
ν(dx) ,
Φ
(3)
(λ)=
(1 − e
iλ x
+ iλ x)1
{|x|≤1}
ν(dx) .
Each Φ
(i)
is a characteristic exponent. The function Φ
(2)
can be viewed as the
characteristic exponent of a compound Poisson process with L´evy measure
1
{|x|≥1}
ν(dx) (see Proposition 8.6.3.4). In the case
{|x|≤1}
|x|ν(dx) < ∞,and
in particular if ν is finite, one can write
Φ
(3)
(λ)=−iλ ˜m
0
+
(1 − e
iλ x
)1
{|x|≤1}
)ν(dx)
and one can construct a L´evy process with characteristic exponent Φ
(3)
.
This approach fails if
{|x|<1}
|x|ν(dx)=∞. In that case, one constructs
aL´evy process with L´evy measure ν by approximating Φ
(3)
by
Φ
(3,)
(λ)=
(1 − e
iλ x
− iλ x) 1
{<|x|≤1}
ν(dx) ,
and letting → 0. See Chapter 4 in Sato [761] or Chapter 1 in Bertoin [79]
for details.