Size of the jumps in Itô-Lévy processes

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I am trying to make sense of the Lévy Itô decomposition, in particular, of a note I have found regarding the size of the jumps.

From the Lévy decomposition we know that any Lévy process is a semimartingale, such that $$ X_t = b_t + \sigma B_t + \int_0^t\int_{\mathbb R} x \,\hat{N}(dt,dx) $$ where $B$ is a Brownian Motion started at $0$, and $$ \hat{N}(dt,dx) = \left\{ \begin{array}{ll} N(dt,dx) - \nu(dx)dt & \mbox{if } |x| < R \\ N(dt,dx) & \mbox{if } |x| \geq R \end{array} \right. $$ for some $R\in[0,\infty]$

What I am trying to make sense of is the following statement:

"If $R = 0$ then $\hat{N} = N$ everywhere, and if $R = \infty$ then $\hat{N} = \tilde{N}$ everywhere" (where $\tilde{N}$ is the compensated Poisson random measure).

Obviously, this make sense if one simply realizes how $\hat{N}$ is defined, but I would like to get a more intuitive idea. In other words, why when $R = \infty$ we have a martingale, and we have 'something else' if $R = 0$?

Thanks to anyone who bothers reading!

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Basically, it's the fact that stochastic integrals with respect to a compensated random measure are martingales whereas stochastic integrals with respect to a non-compensated random measure are not martingales.

Just the same as for processes: The Poisson process is not a martingale, but the compensated (or centered) Poisson process is a martingale.