Let $I\subseteq\overline{\mathbb R}$, $(\mathcal F_t)_{t\in I}$ be a filtration on a measurable space $(\Omega,\mathcal A)$ and $\sigma,\tau:\Omega\to I\cup\{\sup I\}$ be $(\mathcal F_t)_{t\in I}$-stopping times. Moreover, let $\mathcal F_\tau$ denote the $\sigma$-algebra of the $\tau$-past, i.e. $$\mathcal F_\tau=\{A\in\mathcal A:A\cap\{\tau\le t\}\in\mathcal F_t\},$$ and $\left.\mathcal F\right|_A:=\{A\cap B:B\in\mathcal F\}$ denote the trace of a $\sigma$-algebra $\mathcal F$ on a set $A$.
Are we able to show that $$\left.\mathcal F_{\sigma}\right|_{\{\:\sigma\:\le\:\tau\:\}}=\left.\mathcal F_{\sigma\:\wedge\:\tau}\right|_{\{\:\sigma\:\le\:\tau\:\}}\tag1$$, where $a\wedge b:=\min(a,b)$?
I only was able to show that $$\left.\mathcal F_{\sigma}\right|_{\{\:\sigma\:\le\:\tau\:\}}\subseteq\mathcal F_{\sigma\:\wedge\:\tau}=\mathcal F_\sigma\cap\mathcal F_\tau\tag2,$$ but I struggle to obtain the desired claim.
Best Answer
Drop the equality in (2). If $$ E \in\left.\mathcal F_{\sigma}\right|_{\{\:\sigma\:\le\:\tau\:\}}\subseteq\mathcal F_{\sigma\:\wedge\:\tau},$$ then $E=E \cap \{\sigma \leq \tau\}$ so $E \in \mathcal F_{\sigma \wedge \tau}|_{\sigma \leq \tau}$. The reverse inclusion is immediate.