Levy-Ito decomposion

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 ${X:\mathbb{R}^+\rightarrow \mathbb{R}}$ is cadlag. Let ${\Delta X(t) = X(t) - X(t-)}$ (it is well defined due to existence of ${X(t-)}$ ), and define Poisson Random Measure

$\displaystyle N(t,A)= \#\{ \Delta X(s) \in A: s\in [0,t] \}.$

Then, 1) ${N(1,B_\varepsilon^c(0)) <\infty}$ , 2) ${N(1,\mathbb{R}\setminus \{0\})}$ is countable.

Let ${A}$ be bounded below, i.e. ${0 \notin \bar A}$ . Then ${(N(t,A), t\ge 0)}$ is a Poisson process with intensity ${\mu(A)}$ , where ${\mu(\cdot) = \mathbb{E} [N(1,\cdot)]}$ . It follows immediately that ${\mu(A) < \infty}$ whenever ${A}$ is bounded below, hence the measure ${\mu}$ is ${\sigma}$ -finite.

Proposition 1 (Properties of Poisson random measure) Following properties are useful:

For each ${t>0}$ , ${\omega \in \Omega}$ , ${N(t, \cdot)(\omega)}$ is a counting measure on ${\mathcal{B}(\mathbb{R}^d\setminus \{0\}).}$
For each ${A}$ bounded below, ${(N(t,A), t\ge 0)}$ is a Poisson process with intensity
$\displaystyle \mu(A) = \mathbb{E}[N(1,A)]. \ \ \ \ \ (1)$

The compensator ${(\tilde N(t,A), t\ge 0)}$ is a martingale-valued measure where
$\displaystyle \tilde N(t,A) = N(t,A) - t \mu(A)$

for ${A}$ bounded below, i.e. For fixed ${A}$ bounded below, ${\tilde N(t,A)}$ is a martingale.

Let ${f}$ be a Borel measurable function ${f: \mathbb{R}^d \rightarrow \mathbb{R}^d}$ and let ${A}$ be bounded below, then for each ${t>0, \omega \in \Omega}$ , we may define the Poisson integral of ${f}$ as a random finite sum by

$\displaystyle \int_A f(x) N(t, dx) (\omega) = \sum_{x\in A} f(x) N(t, \{x\})(\omega). \ \ \ \ \ (2)$

Note that each ${\int_A f(x) N(t, dx)}$ is an ${\mathbb{R}^d}$ -valued r.v. and gives rise to a cadlag stochastic process as we vary t. In the sequel, we will sometimes use ${\mu_A}$ to denote the restriction to ${A}$ of the measure ${\mu}$ .

Theorem 2 Let ${A}$ be bounded below, then

${(\int_A f(x) N(t,dx), t\ge 0)}$ is a compound Poisson process, with characteristic function
$\displaystyle \mathbb{E}\Big[\exp \Big\{i u \cdot \int_Af(x)N(t,dx) \Big\} \Big) = \exp \Big\{ t \int_{A} (e^{iu\cdot x} - 1) \mu_{f} (dx) \Big\}$

for each ${u \in \mathbb{R}^d}$ , where ${\mu_{f} = \mu \circ f^{-1}}$ .

If ${f\in L^1(A, \mu_A),}$ then
$\displaystyle \mathbb{E}\Big[\int_A f(x) N(t, dx)\Big] = t \int_A f(x) \mu(dx). \ \ \ \ \ (3)$

If ${f\in L^2(A, \mu_A), }$ then
$\displaystyle Var \Big( \Big|\int_A f(x) N(t, dx)\Big| \Big) = t \int_A |f(x)|^2 \mu(dx).$

From Theorem 2, a Poisson integral may fail to have a finite mean if ${f\notin L^1(A, \mu)}$ . Thus, for each ${f\in L^1(A, \mu_A)}$ , we define the compensated Poisson integral by

$\displaystyle \int_A f(x) \tilde N(t,dx) = \int_A f(x) N(t,dx) - t \int_A f(x) \mu(dx). \ \ \ \ \ (4)$

A straightforward argument about ${\tilde N(t,dx)}$ shows that,

${(\int_A f(x) \tilde N(t,dx), t\ge 0)}$ is a martingale
Characteristic function is
$\displaystyle \mathbb{E} \Big[\exp \Big\{ i u \cdot \int_A f(x) \tilde N(t, dx) \Big\} \Big] = \exp \Big\{t \int_A (e^{i u\cdot x} - 1 - i u \cdot x ) \mu_f(dx)\Big\}. \ \ \ \ \ (5)$
For ${f\in L^2(A, \mu_A)}$ ,
$\displaystyle \mathbb{E} \Big[ \Big| \int_A f(x) \tilde N (t, dx) \Big|^2 \Big] = t \int_A |f(x)|^2 \mu (dx).$

Let’s turn to sketch the Levy-Ito decomposition.

First, note that for ${A}$ bounded below, for each ${t\ge 0}$ ${ \int_A x N(t, dx) = \sum_{0\le u \le t} \Delta X(u) {\bf 1}_A (\Delta X(u)) }$ is the sum of all the jumps taking values in the set ${A}$ up to the time ${t}$ . Since the paths of ${X}$ are cadlag, this is clearly a finite random sum. In particular, ${\int_{|x|\ge 1} x N(t, dx)}$ 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 ${ X(t) - \int_{|x|\ge 1} x N(t,dx) }$ is a Levy process having finite moments to all orders. But, it may have unbounded variation. Hence, we can define

$\displaystyle b = \mathbb{E}\Big[X(1) - \int_{|x|\ge 1} x N(1, dx)\Big], \ \ \ \ \ (6)$

then ${b}$ is well defined.

Now, let’s turn to the small jumps. We study compensated integrals, which we know are martingales. Introduce ${M(t,A) = \int_A x \tilde N(t,dx)}$ for ${t\ge 0}$ and ${A}$ bounded below. For each ${m\in \mathbb{N}}$ , let ${A_m = \{x: \frac 1 {m+1} < |x| \le 1\}}$ . It can be shown that ${M(t, A_n) \rightarrow \int_{|x|<1} x \tilde N(t,dx)}$ as ${n \rightarrow \infty}$ in ${L^2}$ , hence ${\int_{|x|<1} x \tilde N(t,dx)}$ is a martingale. Taking the limit in (5), we get

$\displaystyle \mathbb{E} \Big [\exp \{ i u \cdot \int_A x \tilde N(t, dx)\} \Big] = \exp \Big\{t \int_A (e^{i u\cdot x} - 1 - i u \cdot x ) \mu(dx) \Big\}.$

Finally, consider

$\displaystyle W_A(t) = X(t) - bt - \int_{|x|<1} x \tilde N(t,dx) - \int_{|x|\ge 1} x N(t, dx),$

The process ${W_A}$ is a centered martingale with continuous sample paths. Using Levy’s characterization of Brownian motion we have that ${W_A}$ is Brownian motion with covariance ${A}$ . Hence we have

Theorem 3 (Levy-Ito Decomposition) If ${X}$ is a Levy process, then there exists ${b\in \mathbb{R}^d}$ , a Brownian motion ${W_A}$ with covariance matrix ${A}$ in ${\mathbb{R}^d}$ , and an independent Poisson random measure ${N}$ on ${\mathbb{R}^+\times (\mathbb{R}^d \setminus \{0\})}$ such that for each ${t\ge 0}$

$\displaystyle X(t) = bt + W_A(t) + \int_{|x|<1} x \tilde N (t, dx) + \int_{|x|\ge 1} x N(t,dx). \ \ \ \ \ (7)$

The process ${\int_{|x|<1} x \tilde N(t, dx)}$ 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 ${L^2}$ -martingale.

Levy-Ito decomposition of Theorem 3 is provided with small jumps of ${|x|<1}$ and big jumps of ${|x|\ge 1}$ . In fact, for any given constant ${R>0}$ , we can define small jumps and big jumps by ${|x|<R}$ and ${|x|\ge R}$ , respectively. The corresponding Levy-Ito decomposition is written by

$\displaystyle X(t) = b_R t + W_A(t) + \int_{|x|< R} x \tilde N (t, dx) + \int_{|x|\ge R} x N(t,dx), \ \ \ \ \ (8)$

for some constant ${b_R}$ defined similar to (6), i.e.

$\displaystyle b_R = \mathbb{E}\Big[X(1) - \int_{|x|\ge R} x N(1, dx)\Big]. \ \ \ \ \ (9)$

It can be shown ${b_R}$ is finite and well defined. One can compute ${b_R}$ in terms of ${b_1 = b}$ using (3),

If ${1<R<\infty}$ , then ${b_R = b + \int_{1\le |x| <R} x \mu (dx)}$ ;
If ${0<R<1}$ , then ${b_R = b - \int_{R \le |x| <1} x \mu (dx)}$ ;

One possible question is that, can we simply remove upper bound ${R}$ for small jumps, and write Levy-Ito decomposition by

$\displaystyle X(t) = b_\infty t + W_A(t) + \int_{|x|>0} x \tilde N (t, dx). \ \ \ \ \ (10)$

The decomposition of (10) is possible if and only if ${\mathbb{E} [X(1)] <\infty}$ . In this case, ${b_\infty = \mathbb{E}[X(1)]}$ 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 ${X(t)}$ of the form

$\displaystyle X(t) = \int_{|x|\ge 1} x N(t,dx),$

where ${N}$ is a poisson random measure with Levy measure ${\mu}$ given by

$\displaystyle \mu(x\le 1) = 0, \quad \mu(dx) = \frac{1}{x^2} dx, \ \hbox{ if } x>1.$

Show that, ${X(t)}$ does not admit decomposition of (10).

A Levy process has finite variation iff its Levy-Ito takes the form

$\displaystyle X(t) = \gamma t + \int_{x\neq 0} x N(t, dx) = \gamma t + \sum_{0\le s \le t} \Delta X(s),$

where ${\gamma = b - \int_{|x|<1} x \mu(dx)}$ . 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. ${A = 0}$ in the Levy-Khinchine formula), and ${\int_{|x|<1} |x| \mu (dx) < \infty}$ .

A stochastic process ${X}$ is a semimartingale if it is an adapted process s.t. for each ${t\ge 0}$ , ${X(t) = X(0) + M(t) + B(t)}$ , where ${M}$ is a local martingale and ${C}$ 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 ${ M(t) = W_A(t) + \int_{|x|<1} x \tilde N(t, dx) }$ and bounded variation part as ${ B(t) = b t + \int_{|x|\ge 1} x N(t, dx). }$

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, ${\mathbb{E} [|X(t)|^n]<\infty}$ for all ${t>0}$ iff ${\int_{|x|\ge 1} |x|^n \mu(dx)<\infty}$ .

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 ${X}$ is a Levy process, then for each ${u\in \mathbb{R}^d}$ , ${t\ge 0}$ ,

$\displaystyle \mathbb{E} [e^{iu\cdot X(t)}] = \exp \Big\{ t \Big[ ib\cdot u - \frac 1 2 u \cdot Au + \int_{\mathbb{R}^d\setminus \{0\}} (e^{iu\cdot y} - 1 - iu\cdot y {\bf 1}_{\{0< |y|<1\}}(y)) \mu (dy) \Big] \Big\}, \ \ \ \ \ (11)$

and the intensity measure ${\mu}$ of (1) is the Levy measure for ${X}$ .

16 comments

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Clark says:|

May 5, 2011 at 1:35 pm

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JB says:|

May 27, 2011 at 6:09 am

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?

- kenneth says:|
  
  May 30, 2011 at 12:12 pm
  
  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.
  
Jiantao says:|

July 13, 2013 at 5:37 pm

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!

cantiacatalin says:|

August 11, 2013 at 10:51 pm

Reblogged this on FinancialMathBlog.

- JT says:|
  
  September 28, 2013 at 2:58 am
  
  I have a question. Say I give you a process $X(t) = \int_{\{|x|\le1\}} x N(t, dx)$ , such that $\int_{\{|x| \le 1\}} |x| \mu(dx) = \infty$ . 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 $\int_{\{|x|\le 1\}} |x| \mu(dx)$ , which is infinite. Could you please enlighten me on that?
  
  - kenneth says:|
    
    September 28, 2013 at 10:33 pm
    
    I agree with you that a Levy process with bounded jumps has finite mean, and thus $E[X(1)]<\infty$ . But I do not quite understand your second argument. For instance, if $\mu(dx) = x^{-2} dx$ , then it satisfies $\int_{\{|x|\le 1\}} |x| \mu(dx) = \infty$ . But, one can see $\mu$ is symmetric distribution and $X(t)$ is a martingale, so $E[X(1)] = 0<\infty$ .
  - Jiantao says:|
    
    September 29, 2013 at 1:26 am
    
    Thanks for your reply, what if the support of Mu is only the positive part of the real line?
  - kenneth says:|
    
    September 29, 2013 at 1:28 pm
    
    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.
  - Jiantao says:|
    
    September 29, 2013 at 2:45 pm
    
    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?
  - kenneth says:|
    
    September 29, 2013 at 4:22 pm
    
    If $X(t)$ has only positive jumps, does the decomposition look like $X(t) = t \int_0^1 x \mu (dx) + \int_0^1 x \tilde N(t,dx)$ ?
  - Jiantao says:|
    
    September 30, 2013 at 1:16 am
    
    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.
  - kenneth says:|
    
    September 30, 2013 at 11:57 am
    
    A subordinator is a process of finite variation, hence has $\int_0^1 x \mu(dx) <\infty$ . Otherwise, if $\int_0^1 x \mu(dx) = \infty$ (for instance the case $\mu(dx) = x^{-2} 1_{(0,1)}(x) dx$ ), the jump is so frequent that the corresponding $X(t)$ blows up to infinity quickly in any small time interval $(0,\epsilon)$ .
  - Jiantao says:|
    
    September 30, 2013 at 3:00 pm
    
    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?
chencan says:|

September 18, 2015 at 4:26 pm

MATLAB code of Levy Ito decomposition

Omega

Deterministic Randomness

Levy-Ito decomposion

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