34 — Strong Markov Property, Hitting Times, Stopping Times
1. Example (From Book)
We know that for standard Brownian motion $B_t$,
\(\lim_{t\to\infty} \frac{B_t}{\sqrt{t}} = +\infty \quad \text{a.s.}\) and \(\lim_{t\to\infty} \frac{B_t}{\sqrt{t}} = -\infty \quad \text{a.s.}\)
By continuity of sample paths, this implies Brownian motion oscillates to $+\infty$ and $-\infty$, therefore:
\[T_a = \inf\{t>0 : B_t = a\} \quad \text{is finite a.s. for every } a\in\mathbb{R}.\]2. Proof of the first limit
Let \(B_n = \sum_{k=1}^{n} (B_k - B_{k-1}) = \sum_{k=1}^{n} Z_k,\) where $Z_k \overset{\text{iid}}{\sim} N(0,1)$.
Define \(S_n = \sum_{k=1}^{n} Z_k.\)
Then \(\frac{S_n}{\sqrt{n}} \sim N(0,1),\) so \(P\!\left(\frac{S_n}{\sqrt{n}} > 1\right) = P(Z>1) > 0.\)
By the Hewitt–Savage 0-1 law, for exchangeable events:
\[P\left( \limsup_{n\to\infty} \frac{S_n}{\sqrt{n}} > 1 \right) = P(Z>1) > 0 \quad\Rightarrow\quad P\left( \lim_{n\to\infty} \frac{S_n}{\sqrt{n}} = +\infty \right) = 1.\]Thus, \(\lim_{n\to\infty} \frac{B_n}{\sqrt{n}} = +\infty \quad \text{a.s.}\)
3. Stopping Times for Brownian Motion
We work on the canonical space: \(\Omega = C[0,\infty), \qquad B_t(\omega) = \omega(t).\)
Filtration: \(\mathcal{F}_t^0 = \sigma\{B_s : 0 \le s \le t\}.\)
Definition.
A random time $T:\Omega\to[0,\infty]$ is a stopping time (S.T.) if
\(\{T \le t\} \in \mathcal{F}_t^0 \quad \forall t\ge 0.\)
Equivalent condition: \(\{T < t\} \in \mathcal{F}_t^0 \quad \forall t>0.\)
Because Brownian motion filtration is right-continuous: \(\mathcal{F}_t^+ = \bigcap_{n=1}^\infty \mathcal{F}_{t + 1/n}^0, \qquad \mathcal{F}_t^0 = \mathcal{F}_t^+ \;\; \text{(up to null sets)}.\)
4. Properties of Stopping Times
(1) Increasing limits
If $T_n \uparrow T$ a.s. and each $T_n$ is a S.T., then $T$ is a S.T.
Proof: \(\{T \le t\} = \bigcap_{n=1}^\infty \{T_n \le t\} \in \mathcal{F}_t.\)
(2) Decreasing limits
If $T_n \downarrow T$ a.s. and each $T_n$ is a S.T., then $T$ is a S.T.
Proof: \(\{T < t\} = \bigcup_{n=1}^\infty \{T_n < t\} \in \mathcal{F}_t.\)
(3) Approximating a stopping time by rationals
If $T$ is a S.T., define \(T_n = \frac{k+1}{2^n} \quad\text{if}\quad \frac{k}{2^n} \le T < \frac{k+1}{2^n}.\)
Then $T_n \downarrow T$ and each $T_n$ is a S.T.
(4) Algebra of stopping times
If $S,T$ are S.T., then:
- $S\wedge T$
- $S\vee T$
- $S+T$
are all stopping times.
Example:
Using approximations $S_n\downarrow S,\; T_n\downarrow T$,
\(S_n + T_n \downarrow S+T,\)
and each $S_n+T_n$ is a S.T.
(5) Sigma-algebra at a stopping time
Define \(\mathcal{F}_T = \{A : A\cap\{T \le t\} \in \mathcal{F}_t,\;\forall t\ge 0\}.\)
Then:
- $T$ is measurable w.r.t. $\mathcal{F}_T$.
- The events ${S<T}, {S=T}, {S>T}$ all belong to $\mathcal{F}_S \cap \mathcal{F}_T$.
5. Application: Hitting Times of Brownian Motion
Let \(T_a = \inf\{t>0 : B_t = a\}.\)
Because Brownian motion is recurrent and oscillates to both $\pm\infty$,
Theorem:
\(T_a < \infty \quad \text{a.s. for every } a\in\mathbb{R}.\)
This follows from the earlier limit theorem: \(\limsup_{t\to\infty} \frac{B_t}{\sqrt t} = +\infty \quad\text{and}\quad \liminf_{t\to\infty} \frac{B_t}{\sqrt t} = -\infty.\)
6. Next Topic
We now have the tools to state and prove the Strong Markov Property formally.
This will use:
- Stopping times $T$
- Shift operator $\theta_T$
- Filtration right-continuity
- Brownian motion independent increments
This leads to: \(E\!\left[ Y \circ \theta_T \mid \mathcal{F}_T \right] = E^{B_T}(Y)\)
for all bounded measurable $Y$.
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