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?
1 Answer
Hi,
Here is a theorem that might answer your question (it is coming from Chesnay, JeanblancPiqué and Yor's book "Mathematical Methods for Financial Markets").
It is theorem (11.2.8.1 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
and
$M_t=M_0+ \sum_{i=0}^d \int_0^t \phi^i(s)dW^i_s + \int_0^t \int_{R^d} \psi(s,x)\tilde{N}(ds,dx)$
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 :
$E[(\int_0^t \phi^i(s)dW^i_s)^2]=E[\int_0^t \phi^i(s)^2ds]<\infty$
and
$E[(\int_0^t \int_{R^d} \psi(s,x)\tilde{N}(ds,dx))^2]=E[ \int_0^t ds \int_{R^d} \psi(s,x)^2\nu(dx)]<\infty$
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.
Regards

$\begingroup$ 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 $\psi$. $\endgroup$ Commented Jul 25, 2011 at 17:37

$\begingroup$ 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). $\endgroup$ Commented Jul 25, 2011 at 17:40

$\begingroup$ @George Lowther : Hello George unless there's a typo in the book I quote I think I got it right. $\endgroup$ Commented Jul 26, 2011 at 10:05

$\begingroup$ @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 (almostsurely). $\endgroup$ Commented Jul 26, 2011 at 22:08

1$\begingroup$ The answer given by The Bridge, was useful for me. Thank you. But I have a question, can this theorem be given for X being not just R^n valued but having values in any Hilbert space. i.e., can this theorem be extended to infinite dimensional setting. If there are any reference can you please provide me with the same. Thank you. $\endgroup$ Commented Jul 18, 2016 at 7:39