# Finiteness of the number of big jumps of a Lévy process on a finite interval

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

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 $$>r$$.

• 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? – Ivan Nov 25 '14 at 19:26
• @Ivan Usually, a Lévy process has by definition (almost surely) càdlàg sample paths, but obviously this depends on the definition you are using. – saz Nov 25 '14 at 19:27
• 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? – 0xbadf00d Dec 1 '18 at 17:20
• @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 $r \in [s-\delta_2,s)$. – saz Dec 1 '18 at 17:29