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Let $Y(t)$, $0 \leq t \leq 1$, be a Lévy process. Denote by $\{ \Delta Y_i \}$ the jumps of the process. I would like to show that the set $\{i: |\Delta Y_i| > r\}$ is finite almost surely. However, I do not entirely understand what should be shown and how to approach this problem. I would be grateful for any help.

Thank you.

Regards, Ivan

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Let $(Y_t)_{t \in [0,1]}$ be a Lévy process and denote by $\Delta Y_t := Y_t-Y_{t-}$ the jump height at time $t$. The aim is to show that $(Y_t)_{t \in [0,1]}$ has only finitely many jumps with jump size larger than $r$, i.e. that

$$\{t \in [0,1]; |\Delta Y_t(\omega)|>r\}$$

is a finite set for (almost) all $\omega \in \Omega$. This follows from the following lemma (applied to the sample paths $[0,1] \ni t \mapsto Y_t(\omega)$ for fixed $\omega \in \Omega$):

Lemma Let $f:[0,1] \to \mathbb{R}$ be a càdlàg function (i.e. right-continuous with left-hand limits). Then the set $$\{t \in [0,1]; |\Delta f(t)|>r\}$$ is finite for each $r>0$.

Proof:

  • Suppose that the set is not finite. Then we can choose a sequence $(t_n)_{n \in \mathbb{N}} \subseteq [0,1]$ such that $|\Delta f(t_n)|>r$ for each $n \in \mathbb{N}$. Since $[0,1]$ is (sequentially) compact, we can pick a subsequence $(t_{n(k)})_{k \in \mathbb{N}}$ such that $t_{n(k)} \to s$ for some $s \in [0,1]$.

  • Since $f$ is right-continuous at $s$, there exists $\delta_1>0$ such that $$|f(s)-f(t)| < \frac{r}{4}$$ for any $t \in [s,s+\delta_1]$. This means in particular that there cannot exist $t \in [s,s+\delta_1]$ such that $|\Delta f(t)|>r$. Similarly, one can show that the existence of the left-limit implies the existence of $\delta_2>0$ such that $|\Delta f(t)| \leq r$ for all $t \in [s-\delta_2,s)$. Obviously, this contradicts the first step of our proof where we constructed a sequence converging to $s$ with jumps heights $>r$.

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  • $\begingroup$ Thank you for the answer. What is left to do is to mention that every Lévy process has a version whose paths are càdlàg, right? $\endgroup$ – Ivan Nov 25 '14 at 19:26
  • $\begingroup$ @Ivan Usually, a Lévy process has by definition (almost surely) càdlàg sample paths, but obviously this depends on the definition you are using. $\endgroup$ – saz Nov 25 '14 at 19:27
  • $\begingroup$ Sorry for necroing this old answer, but in a first step, we only obtain the existence of $\delta_2>0$ such that $|f(t)-f(s-)|\le r$ for all $t\in[s-\delta_2,s]$ by the existence of the left-limit at $t$. How do you proceed? $\endgroup$ – 0xbadf00d Dec 1 '18 at 17:20
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    $\begingroup$ @0xbadf00d If $|f(t)-f(s-)| \leq r/4$ for $t \in [s-\delta_2,s)$, then $|f(t)-f(u)| \leq r/2$ for any $u,v \in [s-\delta_2,s)$, and this clearly implies $|\Delta f(t)| \leq r$ for any $r \in [s-\delta_2,s)$. $\endgroup$ – saz Dec 1 '18 at 17:29

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