We will review some basic properties of Levy process, in particular Levy-Ito decomposition. See more details in [App04b]. This part is continuation of Levy Process-1.
If the filtration satisfies the “usual conditions” of right continuity and completion, then every Levy process has a cadlag modification which is itself a Levy process.
Function is cadlag. Let
(it is well defined due to existence of
), and define Poisson Random Measure
Then, 1) , 2)
is countable.
Let be bounded below, i.e.
. Then
is a Poisson process with intensity
, where
. It follows immediately that
whenever
is bounded below, hence the measure
is
-finite.
Proposition 1 (Properties of Poisson random measure) Following properties are useful:
- For each
,
,
is a counting measure on
![]()
- For each
bounded below,
is a Poisson process with intensity
- The compensator
is a martingale-valued measure where
for
bounded below, i.e. For fixed
bounded below,
is a martingale.
Let be a Borel measurable function
and let
be bounded below, then for each
, we may define the Poisson integral of
as a random finite sum by
Note that each is an
-valued r.v. and gives rise to a cadlag stochastic process as we vary t. In the sequel, we will sometimes use
to denote the restriction to
of the measure
.
Theorem 2 Let
be bounded below, then
From Theorem 2, a Poisson integral may fail to have a finite mean if . Thus, for each
, we define the compensated Poisson integral by
A straightforward argument about shows that,
Let’s turn to sketch the Levy-Ito decomposition.
First, note that for bounded below, for each
is the sum of all the jumps taking values in the set
up to the time
. Since the paths of
are cadlag, this is clearly a finite random sum. In particular,
is the sum of all jumps (finite many) of size bigger than one, and hence finite variation. It is a compound Poisson process with finite variation but may have no finite moments. Conversely, it can be shown that
is a Levy process having finite moments to all orders. But, it may have unbounded variation. Hence, we can define
then is well defined.
Now, let’s turn to the small jumps. We study compensated integrals, which we know are martingales. Introduce for
and
bounded below. For each
, let
. It can be shown that
as
in
, hence
is a martingale. Taking the limit in (5), we get
Finally, consider
The process is a centered martingale with continuous sample paths. Using Levy’s characterization of Brownian motion we have that
is Brownian motion with covariance
. Hence we have
Theorem 3 (Levy-Ito Decomposition) If
is a Levy process, then there exists
, a Brownian motion
with covariance matrix
in
, and an independent Poisson random measure
on
such that for each
![]()
The process is the compensated sum of small jumps. The compensation takes care of the analytic complications in the Levy-Khintchine formula in a probabilistically pleasing way, since it is an
-martingale.
Levy-Ito decomposition of Theorem 3 is provided with small jumps of and big jumps of
. In fact, for any given constant
, we can define small jumps and big jumps by
and
, respectively. The corresponding Levy-Ito decomposition is written by
for some constant defined similar to (6), i.e.
It can be shown is finite and well defined. One can compute
in terms of
using (3),
- If
, then
;
- If
, then
;
One possible question is that, can we simply remove upper bound for small jumps, and write Levy-Ito decomposition by
The decomposition of (10) is possible if and only if . In this case,
is finite and well-defined. Following example gives a Levy process which does not have decomposition of the form (10).
Example 1 Consider a 1-d Levy process
of the form
where
is a poisson random measure with Levy measure
given by
Show that,
does not admit decomposition of (10).
A Levy process has finite variation iff its Levy-Ito takes the form
where . In fact, A necessary and sufficient condition for a Levy process to be of finite variation is that there is no Brownian part (i.e.
in the Levy-Khinchine formula), and
.
A stochastic process is a semimartingale if it is an adapted process s.t. for each
,
, where
is a local martingale and
is an adapted process of finite variation. In particular, every Levy process is a semimartingale. To see this, use the Levy-Ito decomposition to write martingale part as
and bounded variation part as
The first three terms on the rhs of (7) have finite moments to all orders, so if a Levy process fails to have a moment, this is due to the large jumps with finite activity part. In fact, for all
iff
.
An interesting by-product of the Levy-Ito decomposition is the Levy-Khintchine formula for Levy process, which follows easily by independence in the Levy-Ito decomposition and (5).
Corollary 4 (Levy-Khintchine formula for Levy process) If
is a Levy process, then for each
,
,
and the intensity measure
of (1) is the Levy measure for
.
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I have been studying the Levy-Ito decomposition and came across your blog. I have a question. Perhpas you could answer it for me.
When discussing the small jumps, we needed to keep A bounded below.
QUESTION: If we do not keep A bounded below, what exactly is the problem?
Your question may be equivalent to ask, if a Levy process always admits a decomposition of equation (10) above. To my understanding, the answer is NO in general, and this is related to the integrability issue. Please see Example 1 in the above post.
Could you please enlighten me why we cannot deal with small jumps and big jumps at the same time using the formula for big jumps? Thanks!
Reblogged this on FinancialMathBlog.
I have a question. Say I give you a process
, such that
. Then what is the mean of this process at t = 1? I see two contradictory facts: one is that it is a process with bounded jumps, which should have a finite mean, the other is that the mean should be
, which is infinite. Could you please enlighten me on that?
I agree with you that a Levy process with bounded jumps has finite mean, and thus
. But I do not quite understand your second argument. For instance, if
, then it satisfies
. But, one can see
is symmetric distribution and
is a martingale, so
.
Thanks for your reply, what if the support of Mu is only the positive part of the real line?
Then, I guess it is subordinator (has increasing path almost surely), which must be of finite variation, and not belong to the class of processes you mentioned in the original question.
I agree that it is a subordinator and also a Levy process, then my question would be what is the Levy Ito decomposition of this process?
If $X(t)$ has only positive jumps, does the decomposition look like
?
If so, then according to our assumption, the first integral is
Infinite. So it cannot be the representation since we know the process has finite mean.
A subordinator is a process of finite variation, hence has
. Otherwise, if
(for instance the case
), the jump is so frequent that the corresponding
blows up to infinity quickly in any small time interval
.
Thanks a lot, I am still unclear of one thing, if we have the levy measure as you gave in the blowing up example, what is the corresponding levy Ito decomposition?
MATLAB code of Levy Ito decomposition