# Stopping time $\tau_k$ measurable w.r.t. $\mathcal{F}_k$

We have $(\Omega,\mathcal{F},\mathbb{P})$ stochastic basis. Let $\tau: \Omega \to \mathbb{N}\cup \{\infty\}$ is a $(\mathcal{F}_k，k \in \mathbb{N} )$ be stopping time and define another stopping time $\tau_k = \tau \wedge k$ How would I prove that $\tau_k$ is a $(\mathcal{F}_k，k \in \mathbb{N} )$ stopping time?

$\tau_k$ is taking on values of either $\tau$ or $k$. We know that $[\tau \le k]$ is in $\mathcal{F}_k$ by the very definition of a stopping time. And $k$ is obviously $\mathcal{F}_k$-measurable. Does a proof like this make sense?

Also apologies to Did, for my formerly incorrectly posed question. Writing Latex on an iPad is a royal pain.

• The question you ask makes no sense since $\tau_k$ is a random variable, not an event (wasn't this precise point already explained to you à propos another question?). Idem for item 1. The question is probably whether $\eta_k$ is an $(\mathcal F_n)$-stopping time. You might want to transcribe correctly what this would mean. – Did May 31 '13 at 17:10
• sorry I meant if its F_k measurable (of course) – shimee May 31 '13 at 17:44
• Then please correct your question. – Did May 31 '13 at 18:07
• The assertion that $\tau_k$ is a $\mathcal{F}_k$ stopping time is meaningless since $\mathcal{F}_k$ is not a filtration. (See, these are exactly these improper uses of the definitions which prevent you to reach a solution.) – Did May 31 '13 at 18:34
• I'm sorry I know that filtration is an increasing sequence of sigma algebras...I revised the definitions, originally merely abbreviating – shimee May 31 '13 at 19:02

Let us avoid the confusions remaining in the question, due to some conflicting uses of the index $k$, and let us recall what is to prove. Fix some index $k$ and let $\mathfrak F$ denote the filtration $\mathfrak F=(\mathcal F_n)_n$.
Show that the random time $\tau_k$ is a stopping time for the filtration $\mathfrak F$. This means that, for every $n$, the event $[\tau_k\leqslant n]$ is in $\mathcal F_n$.
To show this, note that, for every $n\lt k$, $[\tau_k\leqslant n]=$ $____$ and that $\tau$ is a stopping time for the filtration $\mathfrak F$ hence $[\tau\leqslant n]$ is in $\mathcal F_n$, hence $____$, while, for every $n\geqslant k$, $[\tau_k\leqslant n]=$ $____$ hence $[\tau_k\leqslant n]$ is in $\mathcal F_n$. QED.
• I'm not sure if my analysis is correct. For every $n\lt k$, $[\tau_k\leqslant n]= [\tau \le n]$ and $\tau$ is a stopping time for the filtration $\mathfrak F$ hence $[\tau\leqslant n]$ is in $\mathcal F_n$, hence an $\mathcal F_n$ stopping time, while, for every $n\geqslant k$, $[\tau_k\leqslant n]=$ $[\tau \le k]$ hence $[\tau_k\leqslant n]$ is in $\mathcal F_n$. Works? – shimee Jun 1 '13 at 8:15
• If $n\geqslant k$, $[\tau_k\leqslant n]$ is not what you write. Otherwise, this is fine. – Did Jun 1 '13 at 9:50