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Let $X(.)$ be a (strictly) $\alpha$-stable process (with $\alpha \in (1,2)$). Assume also that $X(.)$ is spectrally positive (its Lévy measure is concentrated in $[0,+\infty)$).

I am looking for a result that qualitatively says that the set of jumps heights of $X(.)$ is unbounded. More formally, define $J_t(x(.)) := \sup_{0\leq s \leq t} \{\vert x(s) - x(s^-)\vert\}$. Is it true that

\begin{equation} J_t(X(.)) \stackrel{t\rightarrow\infty}{\longrightarrow} + \infty\qquad \mathrm {a.s.} \end{equation} or any other suitable convergence?

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Denote by $N$ the jump measure of the Lévy process, i.e. $$N_t(B) := N([0,t] \times B) := \sharp \{s \in [0,t]; \Delta X_s := X_s-X_{s-} \in B\},$$ and by $\nu$ its Lévy measure. It is widely known that $(N_t(B))_{t \geq 0}$ is a Poisson process with intensity $\nu(B)$. In particular, we have

$$\mathbb{P}(N_t(B) >0) = 1- \mathbb{P}(N_t(B)=0)= 1-e^{-\nu(B) t}.$$

For any set $B$ such that $0<\nu(B)<\infty$ this implies

$$\mathbb{P}(\exists s \in [0,t]: \Delta X_s \in B) = \mathbb{P}(N_t(B) >0) \stackrel{t \to \infty}{\to} 1.$$

Applying this for $B = [n,n+1)$, we get

$$\mathbb{P}(\exists t \geq 0: \Delta X_t \in [n,n+1)) = 1.$$

Hence,

$$\mathbb{P}(\forall N \geq 1 \exists t \geq 0: \Delta X_t \geq N) = 1.$$

This shows that the jump heights are (almost surely) unbounded.

Remark: The proof applies to any Lévy process with unbounded Lévy measure.

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  • $\begingroup$ This is extremely elegant, thanks. I suspect it would be easy to obtain a modified version of the very last statement like $\mathbb P (\forall N \geq 1 \forall T\geq0\exists t\geq T:\Delta X_t \geq N) = 1$, for example by taking a slightly different definition for the jump measure, counting the jumps in $[T,t]$ instead of $[0,t]$. Am I correct? $\endgroup$ – Indigo Aug 3 '15 at 19:52
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    $\begingroup$ @Indigo No need to do so. Just note that the restarted process $\tilde{X}_t := X_{t+T}-X_T$ is again a Lévy process with Levy measure $\nu$ for any fixed $T>0$. $\endgroup$ – saz Aug 3 '15 at 19:57
  • $\begingroup$ Yes, yes of course. That was very insightful and helpful, thanks! $\endgroup$ – Indigo Aug 4 '15 at 7:57
  • $\begingroup$ @Indigo You are welcome. $\endgroup$ – saz Aug 4 '15 at 8:11

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