If $(\mathcal{F}_n)_{n \in \mathbb{N}}$ is the canonical filtration of a stochastic process $(X_n)_{n \in \mathbb{N}}$, then $\mathcal{F}_n$ contains all the information about the process up to time $n$. After observing realizations $X_1(\omega),\ldots,X_n(\omega)$ of the stochastic process, we can decide whether an event $A_n \in \mathcal{F}_n$ has happened, i.e. whether
$$\omega \in A_n \qquad \text{or} \qquad \omega \notin A_n.$$
Since $\{\tau \leq n\} \in \mathcal{F}_n$ this means, in particular, that we can decide whether the stopping has occurred up to time $n$ given the observations $X_1(\omega),\ldots,X_n(\omega)$.
This intuition can be made precise:
Let $(\mathcal{F}_n)_{n \in \mathbb{N}}$ be the canonical filtration of a stochastic process $(X_n)_{n \in \mathbb{N}}$, and let $\tau: \Omega \to \mathbb{N} \cup \{\infty\}$. Then the following statements are equivalent:
- $\tau$ is a stopping time
- If $\omega,\omega' \in \Omega$ are such that $\tau(\omega) \leq k$ and $X_j(\omega) =X_j(\omega')$ for all $j=1,\ldots,k$, then $\tau(\omega') \leq k$.
Summary: $\tau$ is a stopping time if the decision to stop before or at time $n$ (i.e. $\tau(\omega) \leq n$) depends only on $X_1(\omega),\ldots,X_n(\omega)$.
Let's turn to $\mathcal{F}_{\tau}$. Fix observations $X_1(\omega),\ldots,X_n(\omega)$. As we have seen in the first part, we then know whether the stopping has occured up to time $n$, i.e. whether
$$\tau(\omega) \leq n.$$
Suppose for the moment being that the stopping has indeed occured before or at time $n$. Then a set $A \in \mathcal{F}$ is in $\mathcal{F}_{\tau}$ if, and only if, we can decide whether $A$ has occurred (given our observations $X_1(\omega),\ldots,X_n(\omega)$).
Example 1: Let $X_n = \sum_{j=1}^n \xi_j$ for random variables $\xi_j$ which are Gaussian with mean $0$ and variance $1$. Define $$\tau := \inf\{n \in \mathbb{N}; X_n < 0\}.$$ Then the set $$\{X_{\tau} \in B\}$$ is in $\mathcal{F}_{\tau}$ for any Borel set $B$. Indeed: Given that we know that the stopping has occured up to time $n$, we can say which values $X_{\tau}(\omega)$ takes, given the observations $X_1(\omega),\ldots,X_n(\omega)$. In contrast, if the stopping has not occured up to time $n$, the observations $X_1(\omega),\ldots,X_n(\omega)$ don't tell us anything about $X_{\tau}(\omega)$.
Example 2: Let $X_n = \sum_{j=1}^n \xi_j$ for random variables $\xi_j$ such that $\mathbb{P}(\xi_j = 1)= 1/4$ and $\mathbb{P}(\xi_j = -1) = 3/4$. If we define
$$\tau := \inf\{n \in \mathbb{N}; X_n = 100\}$$
then
$$A := \{ \exists k \in \mathbb{N}; X_k =95\} \in \mathcal{F}_{\tau};$$
however, for instance,
$$B := \left\{ \max_{k \geq 0} X_k \leq 100 \right\} \notin \mathcal{F}_{\tau}.$$
Note that $$\min(x,y) \leq n \iff x \leq n \quad \text{or} \quad y \leq n.\tag{1}$$ Applying this for $x=\sigma(\omega)$ and $y = \tau(\omega)$ proves \begin{align*} \{\min(\sigma,\tau) \leq n\} &= \{\omega \in \Omega; \min(\sigma(\omega),\tau(\omega)) \leq n\} \\ &\stackrel{(1)}{=} \{\omega \in \Omega; \sigma(\omega) \leq n \, \, \text{or} \, \, \tau(\omega) \leq n\} \\ &= \{\sigma \leq n\} \cup \{\tau \leq n\}.\end{align*} The reasoning for $\max$ is similar; use that $$\max(x,y) \leq n \iff x \leq n \quad \text{and} \quad y \leq n.$$
Best Answer
I have read the book solution incorrectly. Indeed the maximum of two stopping times is a stopping time, as I showed in the question.