Intuition behind a set of a assumptions about Lévy processes.

Question

I'm reading this paper in mathematical finance where they start with three assumptions, which I cannot really relate to anything I know so I hope you can help to give me some intuition about what they mean so that I have the possibility to show them fulfilled in specific circumstances. I'm not unfamiliar (albeit a bit rusty) with Lévy processes, I know up to about the level of the representation theorems.

Let $X$ be the Lévy process with no determistic drift and denote $\kappa(\theta) = \log (E[e^{\theta X_1}]) $ the cumulant transform then the assumptions are:

1. $\kappa$ is on the form $\kappa(\theta)=\int_0^\infty (e^{\theta x } - 1 ) w(x) dx$ for $w$ a density.
2. Assume $\hat{\theta}>0$ with $ \hat{\theta} = \sup \{ \theta \in \mathbb{R} : \kappa(\theta) < \infty \}$
3. $\lim_{\theta \to \hat{\theta}} \kappa(\theta) = \infty$

Thank you in advance

Answer 1

1. This assumptions means in particular that the jump measure $\nu$ of the Lévy process $(X_t)_{t \geq 0}$ has a density $w$ with respect to the Lebesgue measure. Moreover, the form of the integral shows that $(X_t)_{t \geq 0}$ has only positive jumps and that the sample paths are of bounded variation (see e.g. Sato's book on Lévy processes), i.e. it is a so-called subordinator.
2. It is not clear that $\kappa(\theta)$ is well-defined for $\theta \neq 0$. So, this assumption basically means that the Lévy process has exponential moments, i.e. $\mathbb{E}e^{\theta X_1}$ for some $\theta>0$. Since $(X_t)_{t \geq 0}$ is a Lévy process, this is equivalent to $$\int_1^{\infty} e^{\theta x} w(x) \, dx < \infty,$$ see Theorem 25.3 in Sato's book.
3. Sounds like a rather technical assumption. It implies in particular that $\kappa(\hat{\theta}) = \infty$, i.e. $$\int_1^{\infty} e^{\hat{\theta} x} w(x) \, dx = \infty.$$ In some sense, this means that the decay of the density is dominated by $$e^{-\hat{\theta} x} \frac{1}{x}.$$ For example $$w(x) := e^{-\hat{\theta} x} \frac{1}{x^2}$$ does not satisfy this assumption, whereas $$w(x) := e^{-\hat{\theta} x} \frac{1}{\log x}$$ does.