> [!definition] > > Let $(\Omega, \cf, \bracs{\cf_n}_0^\infty, \bp)$ be a [[Filtration|filtered]] [[Probability|probability space]], and $T: \Omega \to \nat \cup \bracs{\infty}$ be a [[Random Variable|random variable]]. $T$ is a **$\bracs{\cf_n}_0^\infty$-stopping time** if for all $n \in \nat$, $\bracs{T \le n} \in \cf_n$. > > Let $\cf_{\infty} = \sigma\paren{\bigcup_{n \ge 0}\cf_n}$, then > $ > \cf_T = \bracs{A \in \cf_\infty: A \cap \bracs{T \le n} \in \cf_n \forall n \ge 0} > $ > is the **$T$-stopped $\sigma$-field**. > [!definition] > > Let $(\Omega, \cf, \bracs{\cf_t: t \ge 0}, \bp)$ be a filtered probability space and $T: \Omega \to [0, \infty]$ be a random variable. $T$ is a $\bracs{\cf_t: t \ge 0}$**-stopping time** (ST) if the event $\bracs{T \le t} \in \cf_t$ for all $t \ge 0$. $T$ is a $\bracs{\cf_t: t \ge 0}$-**optional time** (OT) if the event $\bracs{T < t} \in \cf_t$ for all $t \ge 0$. > [!theorem] > > Let $T$ be a $\bracs{\cf_t}$-stopping time. Define > $ > \cf_T = \bracs{A \in \cf: A \cap \bracs{T \le t} \in \cf_t} > $ > is the $T$**-stopped** $\sigma$-field. $T$ is measurable with respect to $\cf_T$. > [!theorem] > > Let $T$ be a $\bracs{\cf_t: t \ge 0}$-stopping time, then $T$ is also a $\bracs{\cf_t: t \ge 0}$-optional time. > > *Proof*. $\bracs{T < t} = \bigcup_{s_n \upto t}\bracs{T \le s_n} \in \cf_t$. > [!theorem] > > $T$ be a $\bracs{\cf_t}$-optional time if and only if it is a $\bracs{\cf_{t^+}}$-stopping time. > > *Proof*. Suppose that $T$ is a $\bracs{\cf_{t^+}}$-stopping time, then $\bracs{T < t} = \bigcup_{s_n \upto t}\bracs{T \le s_n} \in \cf_{s_n}$. > [!theorem] > > Let $T_1, T_2$ be $\bracs{\cf_t}$-stopping times, then $T_1 \wedge T_2 = \min(T_1, T_2)$, $T_1 \vee T_2 = \max(T_1, T_2)$, $T_1 + T_2$ are all $\bracs{\cf_t}$-stopping times. > > Let $\seq{T_n}$ be a family of $\bracs{\cf_t}$-stopping times, then $\sup_{n \in \nat}T_n$ is a $\bracs{\cf_t}$-stopping time. > [!theorem] > > Let $\bracs{X_t}$ be a $\bracs{\cf_t}$-progressively measurable stochastic process, and $E \in \cb(\real^d)$ be a Borel set. Define > $ > H_B: \Omega \to [0, \infty] \quad \omega \mapsto \inf\bracs{t \ge 0: X_t(\omega) \in E} > $ > as the **hitting time** of $X$ at $B$. Note that $H_B(\omega) = \infty$ if $X_t(\omega) \not\in E$ for all $t \ge 0$. > 1. If $X$ takes RCLL sample paths and $B$ is open, then $H_B$ is a $\bracs{\cf_t}$-optional time. > 2. If $X$ takes continuous sample paths and $B$ is closed, then $H_B$ is a $\bracs{\cf_t}$-stopping time. > [!theorem] > > Let $S, T$ be $\bracs{\cf_t}$-stopping times. > 1. If $S \le T$, then $\cf_S \subset \cf_T$. > 2. If $\cf_{S \wedge T} = \cf_S \cap \cf_T$. > > *Proof*. Let $A \in \cf_S$, then since $\bracs{S \le t} \subset \bracs{T \le t}$, intersecting $A$ with $\bracs{T \le t}$ does not change its value. Thus $A \in \cf_T$. > > Since $S \wedge T \le S, T$, $\cf_{S \wedge T} \subset \cf_S \cap \cf_T$. Let $A \in \cf_S \cap \cf_T$, then $A \cap (\bracs{S \le t} \cup \bracs{T \le t})$, so $A \in \cf_{S \wedge T}$. > [!theorem] > > Let $X$ be a $\bracs{\cf_t}$-progressively measurable process and $T$ be a $\bracs{\cf_t}$-stopping time. Define > $ > X_T(\omega) = \begin{cases} > X_{T(\omega)}(\omega) &T(\omega) < \infty \\ > \limv{t}X_{t}(\omega) &T(\omega) = \infty, \exists\limv{t}X_t(\omega) > \end{cases} > $ > Note that $X_T$ is not defined on places where $T$ and $X$ misbehave. > 1. $X_{T \wedge t}$ is $\bracs{\cf_t}$-progressively measurable. > 2. $X_T$ is measurable with respect to $\cf_T$. > > *Proof*. Let $t > 0$ and $c \ge 0$, and consider > $ > E = \bracs{(s, w) \in [0, t] \times \Omega: s \wedge T(\omega) \le c} > $ > If $c \ge t$, then $E = [0, t] \times \Omega$. If $c < t$, then we can express > $ > E = \paren{[0, c] \times \Omega} \cup \paren{(c, t] \times T^{-1}([0, c])} > $ > so $E$ is in $\cb([0, t]) \otimes \cf_t$. Thus the map $(s, \omega) \mapsto s \wedge T(\omega)$ is progressively measurable. This allows viewing the stopped process as a composition of two progressively measurable processes. > > Let $E \in \cb(\real^d)$ be a Borel set. For any $t \ge 0$, > $ > \bracs{X_t \in E} \cap \bracs{T \le t} = \bracs{X_{t \wedge T} \in E} \cap \bracs{T \le t} \in \cf_T > $ > [!theorem] > > Let $(\Omega, \cf, \bp)$ be a probability space, $B$ be the standard one-dimensional [[Brownian Motion]], and $\bracs{\cf_t}$ be the natural filter. Let $\alpha > 0$, let > $ > T_a = \inf\bracs{t \ge 0: B_t \ge a} > $ > be the hitting time for $B$ to rise above $a$. Since $B$ has continuous sample paths, $T_a$ is a stopping time, where $T_a \sim \text{abs}\gamma_{0, 1}$. > > *Proof*. Let $t \ge 0$, then $\bracs{T_a \le t} = \bracs{\max_{s \in [0, t]}B_t \ge a}$. Let $M_t = \max_{s \in [0, t]}B_t$ be the running maximum, then $M_t$ and $T_a$ have the same distribution. From here, we claim that > $ > \bp\bracs{X_a \le t} = \bp\bracs{M_t \ge a} = 2\bp\bracs{B_t \ge a} > $ > > Write > $ > M_t = \limv{n}M_t^{(n)} \quad M_t^{(n)} = \max_{q \in D_n}B_{q} > $ > where $D_n$ are the dyadic rational numbers with denominator $2^{-n}$. So > $ > \limv{n}\bp\underbrace{\bracs{M_t^{(n)} \ge a}}_{E_a^{(n)}} \le \bp\bracs{M_t \ge a}\le \limv{n}\underbrace{\bp\bracs{M_t^{(n)} \ge a - \eps}}_{E_{a - \eps}^{(n)}} > $ > Firstly, > $ > \begin{align*} > \bp\bracs{B_t \ge a - \eps} &= \bp\bracs{B_t \ge a - \eps, E_{a - \eps}^{(n)}} \\ > &= \sum_{q \in D_n}\bp\bracs{B_t \ge a - \eps, B_q \text{ is the first to exceed}\ a - \eps} \\ > &\ge \sum_{q \in D_n}\bp\bracs{B_t - B_{q} \ge 0, B_q \text{ is the first to exceed}\ a - \eps} \\ > &= \frac{1}{2}\bp\bracs{E_{a - \eps}^{(n)}} > \end{align*} > $ > On the other hand, let > $ > A_\eps^{(n)} = \bigcap_{q \in D_n}\bracs{\abs{B_{q} - B_{q^-}} \le \eps} > $ > then by i.i.d, $\bp\bracs{A_\eps^{(n)}} = \gamma_{0, 2^{-n}t}([-\eps, \eps])^{2^n} \to 1$ as $n \to \infty$, and > $ > \bp\bracs{B_t \ge 0, A_t^{(n)}} = \bp\bracs{B_t \ge a + \eps, A_\eps^{(n)}, E_{a}} > $ > From here, using the same techniques gives back the same probability.