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:

  1. For each {t>0}, {\omega \in \Omega}, {N(t, \cdot)(\omega)} is a counting measure on {\mathcal{B}(\mathbb{R}^d\setminus \{0\}).}
  2. 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)

  3. 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

  1. {(\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}}.

  2. 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)

  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,

  1. {(\int_A f(x) \tilde N(t,dx), t\ge 0)} is a martingale
  2. 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)

  3. 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}.


  1. Pingback: Levy Process-3: Ito’s formula « 01law's Blog

  2. 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.

  3. 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!

    • 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?

      • 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.

      • 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 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 \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).

      • 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?

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