Show that $$T = \inf \left \{n \geq 0|X_n \in \left \{0, N \right \} \right \}$$ is a stopping time with respect to $\mathcal{F}_n=\sigma(X_0,…,X_n)$ for $n\geq 0$.
I am fairly new to stopping times. From my understanding, I need to show that every set $\left \{T\leq n \right \}$ is a element in the filtration. My problem is that I intuitively do not know how to do this and why I am doing it other than thats the definition of the stopping time.
Thank you!
Best Answer
Yes, you need to show that for every $n,$ $\{T\le n\}\in \mathcal F_n.$
Observe that $\{T\le n\}=\bigcup_{k\le n}\{X_k \in\{0,N\}\}.$
Can you show that, for $k\le n,$ $\{X_k\in \{0,N\}\}\in \mathcal F_k\subseteq \mathcal F_n?$
Edit
A few more details as per the comments.
If we want to break it down further, $\{X_k\in \{0,N\}\}= \{X_k=0\}\cup \{X_k=N\},$ so we're done if we can show that in general, $\{X_k=j\}\in \mathcal F_k$ for any $k.$ But this is more-or-less clear from the definition of $\mathcal F_k,$ since by the definition of $\sigma(X_k)$ (i.e. the smallest sigma algebra with respect to which $X_k$ is measurable), we have $\{X_k=j\}\in\sigma(X_k) \subseteq \mathcal F_k.$