Past exam of the martingales and financial mathematics course of the University of Oxford / 2008 / A / 1 / b

The whole space belongs to ${\mathcal{F}}_T$, since $\Omega \cap \{T\le n\}=\{T\le n\}\in {\mathcal{F}}_n$. If $A\in {\mathcal{F}}_T$, thenIf $A_j\in {\mathcal{F}}_T$, then $({\bigcup }_j{A}_j)\cap \{T\le n\}={\bigcup }_j(A_j\cap \{T\le n\})$ is also in ${\mathcal{F}}_n$. These are precisely the sigma-algebra closure properties, so ${\mathcal{F}}_T$ is a sigma-algebra.

For every integer $m\ge 0$,Hence $\{T\le m\}\in {\mathcal{F}}_T$. Such threshold sets generate the Borel sigma-algebra on ${\mathbb{N}}_0\cup \{\infty \}$, proving $T$ is ${\mathcal{F}}_T$-measurable. This is the stopping-time sigma-algebra.

For $A\in {\mathcal{F}}_S$, writeThe first two factors together belong to ${\mathcal{F}}_k$, since $A\cap \{S=k\}$ is the difference of its intersections with $\{S\le k\}$ and $\{S\le k-1\}$. Also $\{k\le T\le n\}\in {\mathcal{F}}_n$ by the stopping time property. The filtration inclusion puts every term, and therefore the union, in ${\mathcal{F}}_n$. By definition,

If $S\le T$, then $\{S\le T\}=\Omega$. The previous result says that every $A\in {\mathcal{F}}_S$ equals $A\cap \{S\le T\}\in {\mathcal{F}}_T$. ThusInformation known at the earlier stopping time remains known at the later stopping time.