MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there an equivalent of martingale representation theorem for Levy processes in some form? I believe there is no such theorem in generality, but maybe there are some specific cases?

share|cite|improve this question
Have you taken a look at ? There seems to be some kind of result for jump processes. – Paul Tupper Jul 24 '11 at 3:29
up vote 3 down vote accepted


Here is a theorem that might answer your question (it is coming from Chesnay, Jeanblanc-Piqué and Yor's book "Mathematical Methods for Financial Markets").

It is theorem ( page 621) here it is :

(edit note : be carefull as mentioned by G. Lowther there's a typo in the book regarding the domain of integration in the conditions over $\psi$ (defined hereafter) )

Let $X$ be an $R^d$ valued Lévy Process and $F^X$ its natural filtration. Let $M$ be an $F^X$-local Martingale. Then there exist an $R^d$-valued predictable process $\phi$ and an predictable function $\psi : R^+ \times \Omega \times R^d\to R$ such that :

-$\int_0^t \phi^i(s)^2ds <\infty$ almost surely

-$\int_0^t \int_{|x|> 1} |\psi(s,x)|ds\nu(dx) <\infty$ almost surely

-$\int_0^t \int_{|x|\le 1} \psi(s,x)^2ds\nu(dx) <\infty$ almost surely


$M_t=M_0+ \sum_{i=0}^d \int_0^t \phi^i(s)dW^i_s + \int_0^t \int_{R^d} \psi(s,x)\tilde{N}(ds,dx)$

Where $\tilde{N}(ds,dx)$ is the compensated measure of the Lévy process $X$ and $\nu$ the associated Lévy measure.

Moreover if $(M_t)$ is square integrable martingale then we have :

$E[(\int_0^t \phi^i(s)dW^i_s)^2]=E[\int_0^t \phi^i(s)^2ds]<\infty$


$E[(\int_0^t \int_{R^d} \psi(s,x)\tilde{N}(ds,dx))^2]=E[ \int_0^t ds \int_{R^d} \psi(s,x)^2\nu(dx)]<\infty$

and $\phi$ and $\psi$ are essentially unique.

The theorem is not proved in the book but there is a reference to the following parpers :

1/H. Kunita and S. Watanabe. On square integrable martingales. Nagoya J. Math., 30:209–245, 1967

2/H. Kunita. Representation of martingales with jumps and applications to mathematical finance. In H. Kunita, S. Watanabe, and Y. Takahashi, editors, Stochastic Analysis and Related Topics in Kyoto. In honour of Kiyosi Itô, Advanced studies in Pure mathematics, pages 209–233. Oxford University Press, 2004.


share|cite|improve this answer
That sounds perfectly reasonable, except I think you might the inequalities $\vert x\vert\le1$ and $\vert x\vert > 1$ the wrong way round in the conditions for $\psi$. – George Lowther Jul 25 '11 at 17:37
Also, this answer does not seem to directly answer the question, although you can use the result you state to give a necessary and sufficient condition for a Levy process to satisfy the martingale representation property (satisfied by Brownian motion and by compensated Poisson processes, but not by general Levy processes). – George Lowther Jul 25 '11 at 17:40
@George Lowther : Hello George unless there's a typo in the book I quote I think I got it right. – The Bridge Jul 26 '11 at 10:05
@The Bridge: If that's what it says, then I think there must be a typo. The sum of squares of the jumps of a martingale is finite, $\sum\_{s\le t}(\Delta M_s)^2 < \infty$, because it has finite quadratic variation. That should correspond to $\int\_0^t\int\_{\vert x\vert\le1}\psi(s,x)^2ds\nu(dx) < \infty$. Also, the integrability of the martingale should correspond to $\int\_0^t\int\_{\vert x\vert > 1}\vert\psi(s,x)\vert ds\nu(dx) < \infty$. This is also required so that $\psi$ is $\tilde N$ integrable (almost-surely). – George Lowther Jul 26 '11 at 22:08
Actually, I think it does answer the question fully (once the possible typo is sorted), as it is the correct form of the martingale representation theorem for Levy processes. I was just thinking he meant representation as a stochastic integral wrt the original process, which is clearly not possible except for special cases. But he does say "in some form", and this seems to be the best form. – George Lowther Jul 26 '11 at 23:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.