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.
Best Answer
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 $\mathfrak F=(\mathcal F_n)_n$.
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.