I am trying to understand a proof in my book where the setup is:

Let $(Y_t)_{t \geq 0}$ be a $(\mathcal{F}_t)_{t \geq 0}$-Lévy-proces and $\tau$ an extended stopping time wrt. $(\mathcal{F}_t)_{t \geq 0}$. Assume that $\tau$ is finite almost surely and put $D_\tau Y_t = (Y_{\tau + t} - Y_\tau) \textbf{1}_{\{\tau < \infty\}}$. Then I have to argue that $(D_\tau Y_t)_{t \geq 0}$ is a Lévy-proces wrt. $(\mathcal{F}_{(\tau + t)})_{t \geq 0}$

Therefore, for one of the conditions, I have to prove that $t \mapsto D_\tau Y_t(\omega)$ is cadlag for every $\omega \in \Omega$.

Since $(Y_t)_{t \geq 0}$ is a Lévy-process, I know that $t \mapsto Y_t(\omega)$ is cadlag for every $\omega \in \Omega$. However, what is the argument then that $Y_{\tau + t}$ is cadlag (I know that cadlag is simply right-continuity with left limits)? Is there some basic properties that I need to know of? For continuity I know that if both $f,g$ is continuous then their sum is continuous. Is this also what is used here? Or do we use that $\tau$ is an extended stopping time which is finite almost surely?


1 Answer 1


This does not involve Probability Theory at all. Fix $\omega$ with $\tau (\omega)<\infty$ and note that $ t \to Y_{\tau (\omega)+t}(\omega)-Y_{\tau (\omega)}(\omega)$ is right continuous with left limits. [If $f$ is cadlag so is $f(x+y)-c$ for any fixed $y,c$].


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