Consider $X= \left( X_t \right)_{t\geq 0}$ is a Lévy process whose characteristic triplet is $\left( \gamma, \sigma ^2, \nu \right)$ and where its Lévy measure is

$$ \nu \left( dx\right) = A \sum_{n=1} ^{\infty} p^n \delta_{-n} \left( dx \right) + Bx^{\beta-1}\left( 1+x \right)^{-\alpha -\beta}e^{-\lambda x } \mathbf{1}_{\left ]0,+\infty \right[}\left( x\right)dx.$$

I'd like to know how to show that $Z_t = Z_0 \exp\left( \mu t + X_t \right)$ is well defined and admits first and second order moments.

I'm kind of lost here. I don't see what is the problem with this definition. Could someone please enlighten me ?

Must I show that $Z_t < \infty \ a.s.$ ?

Or maybe aplly Itô-Lévy lemma for derive the SDE $Z_t$ satisfies and so conclude that it's well defined as the unique strong solution of this SDE?

Or maybe another thing I've not even think about?


1 Answer 1


I don't think there should be any problem with the definition of $Z_t$, but of course showing that moments are finite can be tricky. You are trying to take exponential moments of $X_t$ and so you need fast enough exponential decay for the (right) tails of $X_t$. I'd need to review some stuff on Levy triples to be sure, but I suspect that the $e^{-\lambda x}$ term in the Levy measure $\nu$ implies that $Z_t$ has finite first moment if $\lambda>1$ and finite second moment if $\lambda > 2$.

  • $\begingroup$ @Jon Peterson : The question seams to be tricky so because it asks first of all to justifie why is it well defined. And after asks about the moments. Anyway could you explain please how did you get this conditions about $\lambda$ ? Thank you! $\endgroup$
    – Paul
    Commented Jan 31, 2013 at 19:09

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