35 — Strong Markov Property and Reflection Principle

1. Formal Setup

We work on a family of probability spaces $(\Omega,\mathcal F, P_x)_{x\in\mathbb R}.$

Let

$Y:\ (\mathbb R^t \times \Omega)\to\mathbb R$ be measurable,
$S:\Omega\to \mathbb R_+ \cup \{\infty\}$ be a stopping time.

Strong Markov Property (SMP) for Brownian Motion

For any measurable $Y_s$, $E_x\!\left(Y_s\circ\theta_s \mid \mathcal F_s\right) = E_{B^x(s)}\!\left(Y_s\right)\mathbf 1_{\{s<\infty\}}.$

Equivalently, $E_x\!\left[(Y_s\circ\theta_s)\mathbf 1_{\{s<\infty\}}\right] = E_x\!\left(E_{B(s)}(Y_s)\mathbf 1_{\{s<\infty\}}\right).$

Here $\theta_s$ is the usual shift operator.

2. More Intuitive Formulation of SMP

Let $S$ be a stopping time with $P(S<\infty)=1.$

Then

Future after a stopping time is a Brownian motion started at the hitting position $\{B(S+t)\}_{t\ge 0}\ \overset{D}{=}\ \{B_t^{\,B(S)}\}_{t\ge 0}.$
Brownian increments after a stopping time are independent of the past $\{B(S+t) - B(S)\}_{t\ge 0} \perp\!\!\!\perp \mathcal F_S,$ and form a BM starting at zero.

3. Example: First Hitting Time

Let $S = T_a := \inf\{t>0: B_t = a\},\qquad a>0.$ Then
$P_x(T_a<\infty)=1$ for every starting point $x\in\mathbb R$.

Recall from last class: $\overline{\lim}_{t\to\infty}\frac{B_t}{\sqrt t} = +\infty\quad \text{a.s.}$ which implies the BM almost surely crosses every level eventually.

Thus:

\[\{B(T_a + t)\}_{t\ge 0} \overset{D}{=} \{B_t^{\,a}\}_{t\ge 0}.\]
\[\{B(T_a+t)\}_{t\ge 0} \perp\!\!\!\perp \mathcal F_{T_a}.\]

Note: One should subtract $a$ to get increments, but this is just a constant shift.

4. Sketch of Proof of SMP

Step 1: Discrete-valued stopping times

Assume $S(\omega)\in\{t_n\}_{n\ge1},\quad \sum_n P(S=t_n)=1.$ On the event ${S=t_n}$ apply the Markov property at deterministic times, then combine using disjointness.

This handles the case where the stopping time takes countably many values.

Step 2: Approximation of a general stopping time

Let $S_n \downarrow S$ with dyadic approximations $\delta_n \in \left\{\frac{k}{2^n}\right\}_{k\ge0},\qquad \delta_n = \frac{k+1}{2^n} \text{ if } \frac{k}{2^n}\le S < \frac{k+1}{2^n}.$ Then $\delta_n \to S$ and $\mathcal F_{\delta_n}\downarrow \mathcal F_S.$

Choose simple $Y$ of the form $Y_S(\omega) = \prod_{k=0}^m f_k\!\big(B_{t_k+S}\big),$ with $f_k$ bounded and continuous.

Apply the conditional DCT to pass to the limit $S_n\to S$.

5. Reflection Principle

Let $T_a = \inf{t>0: B_t=a}$.
Then for every $t>0$, $P_0(T_a \le t) = 2 P_0(B_t \ge a).$

Intuition (diagram on page 3 of PDF) :contentReference[oaicite:1]{index=1}

Conditional on the event ${T_a < t}$, Brownian motion has a 50% chance of being above $a$ and 50% chance of being below $a$ at time $t$.
The path is “reflected’’ after hitting $a$.

Formally, if a path hits $a$ before $t$, the post–$T_a$ segment is symmetric around the line $a$.

Thus: $P_0(B_t>a, T_a<t) = P_0(B_t<a, T_a<t)$ and so $2P_0(B_t>a, T_a<t)=P_0(T_a<t).$ But $P_0(B_t>a, T_a<t)=P_0(B_t>a)$, yielding the formula.

6. Formal Proof via SMP

Step 1: Term A in SMP

Let $Y_s(\omega)=\mathbf 1_{\{B_{t-s}(\omega)>a\}}\mathbf 1_{\{s<t\}}.$ Then $E_x(Y_s)=P_x(B_{t-s}>a)\mathbf 1_{\{s<t\}}.$

At $s=T_a$, $B(T_a)=a,$ and therefore $E_{B(T_a)}(Y_{T_a}) = \frac12\mathbf 1_{\{T_a<t\}}.$

Taking expectations: $E_0\!\left[E_{B(T_a)}(Y_{T_a})\right] = \frac12 P_0(T_a<t).$

Step 2: Term B in SMP

Evaluate $Y_s\circ\theta_s = \mathbf 1_{\{B_t>a\}}\mathbf 1_{\{s<t\}}.$

Thus $E_0\!\left(Y_{T_a}\circ\theta_{T_a}\right) = P_0(B_t>a,\, T_a<t) = P_0(B_t>a).$

Combine Steps 1 and 2 to obtain $2P_0(B_t>a) = P_0(T_a<t).$

7. Distribution of the Hitting Time

From $P_0(T_a<t)=2 P_0(B_t>a)= 2P(Z>\tfrac{a}{\sqrt{t}}),$ differentiate to obtain the density $f_{T_a}(t) = \frac{a}{\sqrt{2\pi t^3}} \exp\!\left(-\frac{a^2}{2t}\right),\qquad t>0.$