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?


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

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


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

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

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.