Stopped Sigma-Algebra is Sigma-Algebra
Theorem
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let $\FF_T$ be the stopped $\sigma$-algebra associated with $T$.
Then $\FF_T$ is a $\sigma$-algebra.
Proof
We show that $\Omega \in \FF_T$, that $\FF_T$ is closed under countable intersection, and relative complement.
For each $t \in \Z_{\ge 0}$ we have:
- $\Omega \cap \set {\omega \in \Omega : \map T \omega \le t} = \set {\omega \in \Omega : \map T \omega \le t}$
Since $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$ we have:
- $\Omega \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
So $\Omega \in \FF_T$.
Now let $\sequence {A_n}_{n \mathop \in \N}$ be a sequence of sets in $\FF_T$.
Then for each $t \in \Z_{\ge 0}$, we have:
- $\ds \paren {\bigcap_{n \in \N} A_n} \cap \set {\omega \in \Omega : \map T \omega \le t} = \bigcap_{n \in \N} \paren {A_n \cap \set {\omega \in \Omega : \map T \omega \le t} }$
from Intersection is Associative.
Since $A_n \in \FF_T$ for each $n \in \N$, we have:
- $A_n \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for each $n \in \N$.
Since $\FF_t$ is closed under countable intersection, we have:
- $\ds \paren {\bigcap_{n \in \N} A_n} \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for each $t \in \Z_{\ge 0}$.
So:
- $\ds \paren {\bigcap_{n \in \N} A_n} \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_T$
Finally we look to prove that if $A \in \FF_T$ then $A^c \in \FF_T$.
Let $t \in \Z_{\ge 0}$.
Note that:
\(\ds \set {\omega \in \Omega : \map T \omega \le t} \cap \paren {A^c \cup \set {\omega \in \Omega : \map T \omega > t} }\) | \(=\) | \(\ds \paren {A^c \cap \set {\omega \in \Omega : \map T \omega \le t} } \cup \paren {\set {\omega \in \Omega : \map T \omega \le t} \cap \set {\omega \in \Omega : \map T \omega > t} }\) | Intersection Distributes over Union | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {A^c \cap \set {\omega \in \Omega : \map T \omega \le t} } \cup \O\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds A^c \cap \set {\omega \in \Omega : \map T \omega \le t}\) | Union with Empty Set |
From De Morgan's Laws (Set Theory): Set Difference: Family of Sets: Difference with Intersection:
- $A^c \cup \set {\omega \in \Omega : \map T \omega > t} = \paren {A \cap \set {\omega \in \Omega : \map T \omega \le t} }^c$
Since $A \in \FF_T$ we have:
- $A \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
Since $\FF_t$ is closed under relative complement we have:
- $\paren {A \cap \set {\omega \in \Omega : \map T \omega \le t} }^c \in \FF_t$
Since $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$, we have:
- $\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
Since $\FF_t$ is closed under finite intersection, we have:
- $\set {\omega \in \Omega : \map T \omega \le t} \cap \paren {A^c \cup \set {\omega \in \Omega : \map T \omega > t} } \in \FF_t$
so that:
- $A^c \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
Since this holds for all $t \in \Z_{\ge 0}$, we have $A^c \in \FF_T$.
$\blacksquare$