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:

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 2Let 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

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 1Consider a 1-d Levy process of the formwhere 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