0
$\begingroup$

I have the following definition of a stopping time:

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and $(\mathcal{F}_n)_{n \geq 0}$ be a filtration. A random variable $\tau$ taking values in $\mathbb{N} \cup \{\infty\}$ is called a stopping time with respect to $(\mathcal{F}_n)_{n \geq 0}$ if $\{\tau = n\} \in \mathcal{F}_n$ for all $n$.

Intuitively this means that the time we choose to stop playing the game only depends on information known up to the current time, and does not depend on the future information.

Now I have seen that it is equivalent to use $\{\tau \leq n\}$ instead.

Intuitively this also means that the time we choose to stop playing the game does not depend on future information. However, I struggle to see why these two are equivalent mathematically.

If I use the example of a hitting time: $\tau_B = \inf\{n\geq 0: X_n \in B\}$, then why does $\{\tau \leq n\} \in \mathcal{F}_n$ imply that $\{\tau = n\} \in \mathcal{F}$ and vice versa.

I have that $\{\tau \leq B\} = \bigcup_{k=0}^n \{X_n \in B\}$ and $\{\tau = B\} = (\bigcap_{k=0}^{n-1} \{X \in B^c \}) \cup \{X_n \in B\}$. Both are in $\mathcal{F_n}$ but why does one imply the other? Maybe I am not understanding the definition intuitively either?

Thanks

$\endgroup$

1 Answer 1

1
$\begingroup$

Recall $\mathscr{F}_{n-1}\subseteq \mathscr{F}_n,\forall n$. Suppose $\{\tau\leq n\}\in \mathscr{F}_n,\forall n$. Then, $\{\tau=n\}=\{\tau>n-1\}\cap\{\tau\leq n\}=\{\tau\leq n-1\}^c\cap\{\tau\leq n\}\in \mathscr{F}_n,\forall n$. Suppose $\{\tau=n\}\in \mathscr{F}_n,\forall n$. Then $\{\tau\leq n\}=\cup_{k\leq n}\{\tau=k\}\in \mathscr{F}_n,\forall n$.

$\endgroup$

You must log in to answer this question.

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