Let $I\subseteq\overline{\mathbb R}$ and $(\mathcal F_t)_{t\in I}$ be a filtration on a measurable space $(\Omega,\mathcal A)$.

Remember that $\tau:\Omega\to I\cup\{\sup I\}$ is called $(\mathcal F_t)_{t\in I}$-stopping time if $$\forall t\in I:\{\tau\le t\}\in\mathcal F_t\tag1.$$

Now it's natural (think about $I=\mathbb N_0$ and $J=[0,\infty)$, for example) ask whether we can show the following: Let $J\subseteq\overline{\mathbb R}$ with $I\cup\{\sup I\}\subseteq J\cup\{\sup J\}$ and $(\mathcal G_t)_{t\in I}$ be a filtration on $(\Omega,\mathcal A)$ with $$\forall t\in I:\mathcal F_t=\mathcal G_t\tag2.$$ Assuming $\tau$ is an $(\mathcal F_t)_{t\in I}$-stopping time, is it an $(\mathcal G_t)_{t\in I}$-stopping time as well?

Let $t\in J$. We weed to show $\{\tau\le t\}\in\mathcal G_t$.

  1. If $t\in I$, then $\{\tau\le t\}\in\mathcal F_t=\mathcal G_t$.
  2. If $t\ge\sup I$, then $\{\tau\le t\}=\Omega\in\mathcal G_t$.
  3. If $t<\inf I$, then $\{\tau\le t\}=\emptyset\in\mathcal G_t$.

Now, I struggle to show $\{\tau\le t\}\in\mathcal G_t$ in the remaining cases. Clearly, it will boil down to argue that there must be a $s\in I$ with $s<t$ and $$\{\tau\le t\}=\{\tau\le s\}\in\mathcal F_s=\mathcal G_s\subseteq\mathcal G_t\tag3.$$ But can we really find such an $s$? (If we can but only under some additional assumption, please feel free to add this assumption.)


1 Answer 1


(I'm going to assume that $\sup I\in I$ and $\sup J\in J$; this entails no real loss of generality.)

Let $I_0$ be a countable dense subset of $I$. If $t\in J\setminus I$ then $$ \{\tau\le t\}=\cup_{\{s\in I_0: s<t\}}\{\tau\le s\}\in\vee_{\{s\in I_0: s<t\}}\mathcal F_s=\vee_{\{s\in I_0: s<t\}}\mathcal G_s\subset\mathcal G_t. $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .