Introduction
What is a Lévy process?
In a nutshell, Lévy processes are continuous-time random walks that are
“mathematically viable.” We are about to describe these processes in greater
depth, but it might help to keep in mind Brownian motion as a central example of a “mathematically viable” continuous-time random walk.
A stochastic process X := {X� }�≥0 [with values in R� ] is a continuoustime random walk if X0 = 0 [the process starts at the origin at time 0] and
X has i.i.d. increments. The latter property means that for all �� � ≥ 0:
(1) X�+� − X� and X� have the same distribution; and
(2) X�+� − X� is independent of {X� }�∈[0��] .
While the preceding definition makes perfect sense in discrete time, it
does not lend itself to a rich mathematical theory as the following example
might suggest:
Consider the following deterministic [i.e., nonrandom] equation:
�(� + �) = �(�) + �(�)
for all �� � ≥ 0�
(1)
It is a fundamental fact that all Borel-measurable solutions to equation (1)
have the form �(�) = �� for some � ≥ 0; see Theorem 9 below. But
it is also known (Hamel, 1905) that, under the axiom of choice, (1) has
nonmeasurable nonlinear solutions [which can be shown are nowhere
continuous also]; see Theorem 10. Choose and fix one such badly-behaved
solution, call it �, and observe that X� := �(�) is a [nonrandom] continuoustime random walk! The nonlinear solutions to the nonrandom equation
(1) have very bad measurability properties, and therefore the class of all
1
2
1. Introduction
continuous-time random walks contains such badly-behaved objects that it
is hopeless to study them seriously. Fortunately, there is a fix that is simple
to describe:
Definition 1. A Lévy process X := {X� }�≥0 is a continuous-time random
walk such that the trajectories, or paths, of X [i.e., the function � �� X� as
a function of ω ∈ Ω] are right-continuous with left limits everywhere.1 �
This is a course about Lévy processes. Some standard references are
the following books: Bertoin (1996); Kyprianou (2006); Sato (1999); see also
the survey monograph by Fristedt (1974). The notation of this course is
on the whole borrowed from Math. 6040 (Khoshnevisan, 2007).
Infinite divisibility
Suppose X := {X� }�≥0 is a Lévy process on R� , and let µ� denote the
distribution [or “law”] of X� for every � ≥ 0. For all � ≥ 1� � > 0,
X� =
�
�
�
�=1
X��/� − X(�−1)�/�
�
(2)
is a sum of � i.i.d. random variables. For example, we set � = 1 to find that
µ1 = µ1/� ∗ · · · ∗ µ1/�
(� times)�
where “∗” denotes convolution. Equivalently, we can write this using the
�
Fourier transform as µ̂1 = µ̂1/�
; in particular, µ̂11/� is the Fourier transform
of a probability measure.
We can also apply (2) with � := � to find that
µ� = µ1 ∗ · · · ∗ µ1
(� times)�
or equivalently, (µ̂1 )� is also the Fourier transform of a probability measure. Thus, to summarize, (µ̂1 )� is the Fourier transform of a probability
measure for all rationals � ≥ 0. In fact, a little thought shows us that (µ̂1 )�
is the Fourier transform of µ� for all rational � ≥ 0. Thus, we can write
probabilistically,
�
��
µ̂1 (ξ) = Ee�ξ·X� = µ̂� (ξ)
for all ξ ∈ R� �
for all rational � ≥ 0. And because X has right-continuous trajectories, for
all � ≥ 0 we can take rational � ↓ � to deduce that the preceding holds
for all � ≥ 0. This shows (µ̂1 )� is the Fourier transform of a probability
measure for every � ≥ 0.
1In other words, X = lim X and X := lim X exists for all � > 0.
�
�↓� �
�−
�↑� �
The Lévy–Khintchine formula
3
Definition 2. A Borel probability measure ρ on R� is said to be infinitely
divisible if (ρ̂)� is the Fourier transform of a probability measure for every
� ≥ 0.
�
Thus, if X is a Lévy process then the distribution of X1 is infinitely
divisible. In fact, the very same reasoning shows that the distribution of
X� is infinitely divisible for all � ≥ 0. A remarkable fact, due to Lévy and
then Itô, is that the converse is also true: Every infinitely-divisible measure
ρ on R� corresponds in a unique way to a Lévy process X in the sense that
the law [i.e., the distribution] of X1 is ρ. Thus, we can see immediately that
the standard-normal law on R corresponds to one-dimensional Brownian
motion, and Poiss(λ) to a rate-λ Poisson process on the line. In other words,
the study of Lévy processes on R� is in principle completely equivalent to
the analysis of all infinitely-divisible laws on R� .
The Lévy–Khintchine formula
We need to introduce some terminology before we can characterize infinitelydivisible laws on R� .
Definition 3. A Lévy triple is a trio (� � σ� �) where � ∈ R� , σ is a (� × �)
matrix, and � is a Borel measure on R� such that
� �
�
�({0}) = 0 and
1 ∧ ���2 �(d�) < ∞�
(3)
σ �σ
R�
The matrix
is called the diffusion [or Gaussian] covariance matrix,
and � is called the Lévy measure.
�
Although �(R� ) might be infinite, (3) ensures that �(A) < ∞ for every
open set A that does not contain the origin. Indeed, the fact that � ��
�/(1 + �) is decreasing on R+ implies that
�
� 1 + �2 �
���2
�
� � ∈ R : ��� > � ≤
·
�(d�)
2
�2
R� 1 + ���
(4)
� �
�
1 + �2
2
≤
·
1 ∧ ��� �(d�) < ∞�
�2
R�
�
A similar argument shows that R� (� ∧ ���2 ) �(d�) < ∞ for all � > 0.
Definition 4. A Lévy exponent is the function Ψ : R� → C, where
� �
�
1
2
1 − e�ξ·� + �(ξ · �)1l(0�1) (���) �(d�)�
Ψ(ξ) := �(� · ξ) + �σξ� +
2
R�
where (� � σ� �) is a Lévy triple.
(5)
�
Lemma 5. The integral in (5) is absolutely convergent, and Ψ is continuous with Ψ(0) = 0.
4
1. Introduction
Proof. Because 1 − cos θ ≤ θ 2 /2 and θ − sin θ ≤ θ 2 /6 for all θ ∈ R [Taylor’s
theorem with remainder],
�
�
�
� 1 − e−�ξ·� − �(ξ · �)1l (���) � �
1
(0�1)
�
�
2
�
�
� ≤ ��� ∧
�
�
�ξ�2
�ξ�2
The definition of a Lévy measure then tells us that the integral in Ψ is
absolutely convergent. And the remaining assertions follow easily from
this.
�
The following result is called the Lévy–Khintchine formula; it provides
the reason for introducing all this terminology.
Theorem 6 (Khintchine, 1938; Kolmogorov, 1932; Lévy, 1934). A Borel
probability measure ρ on R� is infinitely divisible if and only if ρ̂(ξ) =
exp(−Ψ(ξ)) for all ξ ∈ R� , where Ψ is a Lévy exponent. The corresponding
triple (� � σ� �) determines ρ uniquely.
See also Lévy (1937, pp. 212–220).
The uniqueness portion will follow immediately from Fourier analysis;
see also the proof of Theorem 3 below [page 29]. That proof also implies
the more important half of the theorem; namely, that if ρ̂ = e−Ψ for a
Lévy exponent Ψ, then ρ is infinitely divisible. The proof of the remaining
half is a difficult central-limit-type argument, and does not concern our
immediate needs; you can find it in Sato (1999, pages 42–45).
There are many other ways of writing the
� Lévy–Khintchine formula.
Here is one that is used frequently: Suppose R� (1 ∧ ���) �(d�) < ∞. Then
�
�
�ξ·� ) �(d�) and
R� (1 − e
R� �(ξ · �) �(d�) both converge absolutely; this can
be seen from a Taylor expansion
� similar to the one in the proof of Lemma
5. Therefore, in the case that R� (1 ∧ ���) �(d�) < ∞, we can also write
� �
�
1
1 − e�ξ·� �(d�)�
Ψ(ξ) = �(� · ξ) + �σξ�2 +
2
R�
�
(6)
where � := � −
� �(d�)�
���<1
Let us conclude with a few basic properties of Lévy exponents.
Lemma 7. If Ψ is a Lévy exponent, then Ψ and ReΨ are also Lévy exponents. Moreover, Ψ(ξ) = Ψ(−ξ) and ReΨ(ξ) ≥ 0 for all ξ ∈ R� .
On equation (1)
This section is not covered in the lectures. Nonetheless it would be a
shame to say nothing indepth about the functional equation (1). Therefore,
we close this chapter with a discussion on (1).
On equation (1)
5
Suppose � solves (1). Then �(0) = 2�(0), whence we have �(0) = 0.
Therefore, we can extend � to a function F on all of R as follows: F(�) =
�(�) if � ≥ 0; and F(�) = −�(−�) for all � < 0. Note that F solves “Cauchy’s
functional equation,”
F(� + �) = F(�) + F(�)
for every �� � ∈ R�
(7)
The preceding reduces the analysis of (1) to an analysis of (7). Therefore,
we investigate the latter equation from now on. The following is immediate,
but important.
Proposition 8. If F solves (7), then F(��) = �F(�) for all � ∈ R and all
integers � ≥ 2. In particular, F(�) = �F(1) for all rationals �.
Suppose F : R → R and G : R → R are right-continuous functions that
have left limits everywhere and agree on the rationals. Then it is easy to
see that F = G on all of R. As a consequence of Proposition 8, we find that
if F solves (7) and F is right continuous with left limits everywhere, then
F(�) = �F(1) for all �; i.e., F is a linear function. As it turns out, “right
continuous with left limits” can be reduced to the seemingly-stronger condition “Borel measurable,” without changing the content of the preceding
discussion. More precisely, we have the following.
Theorem 9. Every Lebesgue-measurable solution to (7) is linear.
This is classical; we follow a more recent proof—due to Fitzsimmons
(1995)—that is simpler than the classical one.
Proof. Let F be a Lebesgue-measurable solution to (1), consider the Cvalued function G(�) := exp(�F(�)), defined for every � ∈ R. It is clear
that
G(� + �) = G(�)G(�)
for all �� � ∈ R;
(8)
compare with (1). Because
|G(�)| = 1, G is locally integrable and never
��
vanishes. Therefore, 0 G(�) d� �= 0 for almost every � ≥ 0. We can now
integrate (8) over all � ∈ [0 � �] [for almost all � ≥ 0] to obtain the following:
For almost all � ∈ R, we have
� �+�
� �
G(�) d�
for every � ∈ R�
(9)
G(�) d� = G(�) ·
�
0
The dominated convergence theorem implies that the left-hand side is a
continuous function of �, and hence so is the right-hand side; i.e., G is
continuous. Thanks to Proposition 8 every continuous solution to (8) is
a complex exponential. Therefore, there exists θ ∈ R such that G(�) =
e�θ� for all � ∈ R. From this it follows that F(�) = θ� + 2πN(�), where
N : R → Z is measurable. It suffices to prove that N(�) = 0 for all �.
But this is not too difficult to establish. Indeed, we note that N solves (7).
6
1. Introduction
Therefore, N(�)/� = N(�/�) for all � ∈ R and integers � ≥ 2 (Proposition
8). Consequently, N(�)/� is an integer for all � ∈ R and � ≥ 2; and this
implies readily that N(�) = 0 [for otherwise, we could set � = 2|N(�)|]. �
Theorem 10 (Hamel, 1905). Assume the axiom of choice. Then, there are
uncountably-many nonmeasurable solutions to (7).
In fact, the proof will show that there are “many more” nonmeasurable
solutions to (7) than there are measurable ones.
Proof (sketch). Let H—a socalled “Hamel basis”—denote the maximal
linearly-independent subset of R, where R is viewed as a vector space
over the field Q of rationals. The existence of H follows from the axiom
of choice (Hewitt and Stromberg, 1965, (3.12), p. 14). And it follows fairly
easily from the axiom of choice (Hewitt and Stromberg, 1965, (3.20), p. 18)
that for every � ∈ R there exists a unique function ξ� : H →�
Q such that:
(i) ξ� (�) = 0 for all but a finite number of � ∈ H; and (ii) � = �∈H �ξ� (�).
Because a countable union of countable sets is itself countable [this follows,
for instance, from the axiom of choice], we can deduce that H has the cardinality c of the continuum. [For if the cardinality of H were < c, then H
would be countable.]
Now let � denote the collection of all functions φ : H → R; the cardinality of � is 2c > c.
Define
Fφ (�) :=
�
�∈H
φ(�)ξ� (�)
for all � ∈ R and φ ∈ �.
(10)
Since H is linearly independent, it follows that if φ and ψ are two different
elements of �, then Fφ �= Fψ . Consequently, the cardinality of {Fφ }φ∈� is
2c [one for every φ ∈ �].
The definition of the ξ� ’s implies that Fφ solves (7) for every φ ∈ �.
It follows from Proposition 8 that if Fφ were continuous, then Fφ would
be linear; in fact, Fφ (�) = �Fφ (1) for all � ∈ R. Note that the collection of
all numbers Fφ (1) such that Fφ is continuous is at most c. Therefore, the
cardinality of all linear/continuous solutions to (7) that have the form Fφ
for some φ ∈ � is at most c. Because {Fφ }φ∈� has cardinality 2c > c, it
follows that there are at least 2c − c = 2c different solutions to (7) none of
which are continuous. Theorem 9 then implies that every discontinuous
solution Fφ to (7) is nonmeasurable, and this completes the proof.
�
Interestingly enough, Lévy (1961)—after whom the stochastic processes
of this course are named—has used the Hamel basis H of the preceding
Problems for Lecture 1
7
proof in order to construct an “explicitly constructed” set in R that is not
measurable.
Problems for Lecture 1
1. Verify Lemma 7.
2. Prove that every Lévy process X on R� is continuous in probability; i.e., if
� → � then X� → X� in probability. [Our later examples show that convergence
in probability cannot in general be improved to almost-sure convergence.]
3. Verify that if X is a Lévy process on R� , then {X�� }∞
�=0 is a �-dimensional
random walk for every fixed � > 0.
4. Let µ be an infinitely-divisible distribution on R� , so that for every integer
� ≥ 1 there exists a Borel probability measure µ� on R� such that µ̂1/� = µ̂� .
Observe that lim�→∞ |µ̂� |2 is the indicator of the set {ξ ∈ R� : µ̂(ξ) �= 0}. Use
this to prove that µ̂ is never zero. (Hint: You may use the following theorem of
P. Lévy without proof: If {µ� }�≥1 is a sequence of probability measures on R�
such that � := lim�→∞ µ̂� exists and � is continuous in an open neighborhood
of the origin, then there exists a probability measure µ such that � = µ̂, and
µ� � µ, in particular, µ̂� → µ̂ everywhere.)
5. Is Unif (� � �) infinitely divisible? (Hint: Exercise 4!)
6. Verify the following:
(1) Y is infinitely divisible iff its law [or distribution] is;
(2) Constants are infinitely-divisible random variables;
(3) N(µ � σ 2 ) is infinitely divisible for every fixed µ ∈ R and σ 2 > 0;
(4) Poiss(λ) is infinitely divisible for every λ > 0 fixed;
(5) Gamma (α � λ) is infinitely divisible for all α� λ > 0 fixed. Recall that the
density function of Gamma (α � λ) is �(�) := λ α � α−1 e−λ� / Γ(α) · 1l(0�∞) (�).
Gamma laws include Exp (λ) = Gamma (1 � λ) and ξ�2 = Gamma (� � 1/2)
for � ≥ 1.
7. One can combine Lévy processes to form new ones:
(1) Prove that if X 1 � � � � � X � are independent Lévy processes with values in
R� , then Y� := X�1 +· · ·+X�� defines a Lévy process on R� as well. Identify
the Lévy triple and exponent of Y in terms of those of X � ’s.
(2) Prove that if X 1 � X 2 � · · · � X � are independent Lévy processes with respective values in R�1 � � � � � R�� , then Z� := (X�1 � � � � � X�� ) defines a Lévy
process with values in R� , where � = �1 + · · · + �� . Identify the Lévy
triple and exponent of Z in terms of those of X � ’s. In particular, prove
that if X is a Lévy process on R� , then Y� := (� � X� ) defines a Lévy process on R+ × R� . Identify the Lévy triple and exponent of Y in terms of
those of X � ’s.
8
1. Introduction
8. Prove that lim sup�ξ�→∞ |Ψ(ξ)|/�ξ�2 < ∞ for all Lévy exponents Ψ.
9 (Time reversal). Verify that if
defines a Lévy process. Identify
those of X. Furthermore, prove
{X(�−�)− − X� }�∈[0��] and {X̃� }�∈[0��]
X is a Lévy process on R� , then X̃� := −X�
its Lévy triple and exponent of X̃ in terms of
that for every fixed time � ≥ 0, the processes
have the same finite-dimensional distributions.