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We can show the following: If $E$ is a normed $\mathbb R$-vector space, $x:[0,\infty)\to E$ is càdlàg, $B\subseteq E\setminus\{0\}$, $\tau_0:=0$ and $$\tau_n:=\inf\underbrace{\{t>\tau_{n-1}:\Delta x(t)\in B\}}_{=:\:I_n}$$ for $n\in\mathbb N$, then

  1. $\tau_1\in(0,\infty]$;
  2. If $n\in\mathbb N$, then either $I_n=\emptyset$ and hence $\tau_n=\infty$ or $\tau_n\in I_n$.

Now if $X$ is any $E$-valued càdlàg Lévy process on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$, we can similarly define $I_n$ and $\tau_n$ with $x$ replaced by $X$.

$(\Delta X_t)_{t\ge0}$ is clearly $\mathcal F$-adapted and $(X_{t_1+s}-X_{t_1})_{s\ge0}$ is a Lévy process with respect to the filtration $(\mathcal F_{t_1+s})_{s\ge0}$ with the same distribution as $X$ for all $t_1>0$.

Question: Are we able to show that $\tau_1$ is $\mathcal A$-measurable? Or are we even able to show that $\tau_1$ is measurable with respect to the right-continuous filtration $\mathcal F_{t+}:=\bigcap_{\varepsilon>0}\mathcal F_{t+\varepsilon}$? The latter is equivalent to showing that $\{\tau_1<t\}\in\mathcal F_t$ for all $t>0$. Can we show this?

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  • $\begingroup$ What is $\Delta X(t)$? $\endgroup$ Commented Oct 24, 2021 at 11:34
  • $\begingroup$ @KaviRamaMurthy $\Delta x(t):=x(t)-x(t-)$ and $x(t-):=\lim_{s\to t-}x(s)$. $\endgroup$
    – 0xbadf00d
    Commented Oct 24, 2021 at 13:56

1 Answer 1

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The process of jumps $\Delta X$ is progressively measurable since it is the difference of two progressively measurable processes. Then, for a measurable subset $B\subseteq E$ the hitting time $\tau_1=\inf\{t>0:\Delta X(t)\in B\}$ is a stopping time according to the Début theorem. Note: The theorem assumes that the probability space is complete.

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  • $\begingroup$ Thank you for your answer. I was hoping that there would be a more elementary answer which doesn't rely on the Début theorem and hence doesn't require to assume completness (and right-continuity, which I guess you've forgotten to mention) of the filtration. $\endgroup$
    – 0xbadf00d
    Commented Oct 31, 2021 at 11:14
  • $\begingroup$ What can we say in the canonical setting? Please take a look at the corresponding question I've asked for this: math.stackexchange.com/q/4292571/47771. $\endgroup$
    – 0xbadf00d
    Commented Oct 31, 2021 at 11:15
  • $\begingroup$ Oh, and might there be a more elementary proof available, if we assume $0\not\in\overline B$? $\endgroup$
    – 0xbadf00d
    Commented Oct 31, 2021 at 13:40

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