Assume $X_{t}$ is a Levy process with triplet $(\sigma^{2}, \lambda, \nu)$, here $\nu$ is the Levy measure of $X_{t}$. Define $\tau_{1},\tau_{2},\dots$ be the time gap between the successive jumps happen.

There are two questions for me. First, is $\tau_{i}$ well defined?. Second, the answer of first one is yes, we know $\tau_{i}$ are i.i.d. Then how to find the distribution of $\tau_{i}$? Can we prove the expectation of $\tau_{i}$ is finite or even its variance is finite according to the information of Levy measure $\nu$?

I can do this when the Levy process $X_{t}$ is Poisson process or negative binomial process. I have difficulty when the probability density function of $\nu$ is continuous. For example, the case that $X_{t}$ is a Gamma process.

Any references would be very appreciated.


This is a good shorter reference: http://www.impan.pl/CZM/tankov.pdf. Cont and Tankov have also written a longer book about modelling with Levy processes that I think is really good.

There's going to be a strong connection between the sequence of jump times and the Levy measure $\nu$. In a single unit of time, $ \nu(dx)$ is a measure (not necessarily a probability) that records the expected number of jumps of size dx.

Now, if $\nu$ is not a finite measure, then I don't think it makes sense to talk about the jump times. They're basically happening all the time. So, for example, if you tried to define $\tau_1 = \inf \{ t \geq 0 : \Delta t \neq 0 \}$, $\tau_1$ would be equal to zero. This is why I say the notion isn't meaningful.

On the other hand, suppose that $\nu$ is a finite measure. This is closer to the setting of a Poisson process, where $\nu$ is $\lambda$ times a point mass at 1, or for a compound Poisson process, $\lambda$ times the pdf of the jump size distribution. So analogously, let $\lambda' = \nu(\mathbb{R})$. This is the expected number of jumps per unit of time. Then the jump times (ignoring size) will come like a Poisson process with intensity $\lambda'$, so the individual gaps have the corresponding exponential distribution.

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  • $\begingroup$ Very good answer. $\endgroup$ – Ric Feb 17 '14 at 8:44

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