Let $E$ be a normed $\mathbb R$-vector space and $(X_t)_{t\ge0}$ be an $E$-valued Lévy process on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$, i.e.
- $X_0=0$;
- $X$ is $\mathcal F$-adapted and continuous in probability;
- $X_{s+t}-X_s$ is independent of $\mathcal F_s$ for all $s,t\ge0$;
- $X_{s+t}-X_s\sim X_t$ for all $s,t\ge0$.
Assume $X$ is right-continuous. We can show that if $\tau$ is a finite $\mathcal F$-stopping time, $$Y_t:=X_{\tau+t}-X_\tau$$ and $$\mathcal G_t:=\mathcal F_{\tau+t}$$ for $t\ge0$, then $Y$ is $\mathcal G$-Lévy and $X\sim Y$.
It was shown in this answer that this result can be extended to arbitrary (i.e. not necessarily finite) $\mathcal F$-stopping times. However, why do we argue in a rather complicated way to obtain this result?
What I mean is the following: Let $\tau$ be an arbitrary $\mathcal F$-stopping time and $A:=\{\tau<\infty\}$. If $\operatorname P[A]=0$, I don't see any reasonable extension of the former result. So, assume $\operatorname P[A]>0$.
Now, if I'm not missing something crucial, the process $(\left.X_t\right|_A)_{t\ge0}$ should be a $\left(\left.\mathcal F\right|_A\right)_{t\ge0}$-Lévy process on $\left(A,\left.\mathcal A\right|_A,\frac{\left.\operatorname P\right|_A}{\operatorname P[A]}\right)$, where $\left.\mathcal A\right|_A:=\{B\in\mathcal A:B\subseteq A\}$ and $\left.\operatorname P\right|_A[B]=\operatorname P\left[B\right]$ for $B\in\left.\mathcal A\right|_A$ denote the trace of $\mathcal A$ and $\operatorname P$, respectively.
By definition, $\left.\tau\right|_A$ is a finite $\left(\left.\mathcal F\right|_A\right)_{t\ge0}$-stopping time and hence the former result immediately applies.
Is the result obtained in this way somehow weaker than what is shown in the answer linked above?
