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32 — Markov Property of Brownian Motion

Let

  • $\Omega = C[0,\infty)$
  • $B_t(\omega) = \omega(t)$
  • $\mathcal{F}_t^0 = \sigma{B_s: 0 \le s \le t}$ the natural filtration
  • ${P_x}_{x\in\mathbb{R}}$ a family of probability measures with
    • $P_x(B_0 = x)=1$
    • $P_x(A)=P_0(A - x)$
      so $P_x$ is the distribution of Brownian motion starting at $x$.

The laws $P_x$ differ only in initial conditions.

Markov Property (Main Statement)

For any bounded measurable functional
\(Y : C([0,\infty)) \to \mathbb{R},\) and for any $t \ge 0$,
\(E^{x}\big( Y\circ \theta_t \,\big|\, \mathcal{F}_t^0 \big) \;=\; E^{B_t^x}(Y) \qquad\text{a.s.}\)

Shift operator

Define the shift map \(\theta_t(\omega)(s)=\omega(t+s),\quad s\ge 0.\)

Then

  • $Y\circ\theta_t$ depends only on the increments after time $t$.
  • By independent increments of Brownian motion, these increments are independent of $\mathcal{F}_t^0$.

Examples of $Y$

  1. Integral functional \(Y=\int_0^1 B_s\,ds.\)

  2. Cylinder functional \(Y=\prod_{k=1}^d f_k(B_{h_k}), \qquad 0<h_1<\cdots<h_d, \quad f_k:\mathbb{R}\to\mathbb{R} \text{ bounded}.\)

Then \(Y\circ\theta_t = \prod_{k=1}^d f_k\big(B_{t+h_k}\big).\)

Lemma (Lehenthal’s Lemma)

If

  • $X$ is $\mathcal{F}$-measurable,
  • $\vec{Y}=(Y_1,\dots,Y_d)$ is independent of $\mathcal{F}$,
  • $f:\mathbb{R}^{d+1}\to\mathbb{R}$ is bounded measurable,

then \(E\big[ f(X,\vec{Y}) \mid \mathcal{F} \big] = g(X), \quad g(x)=E\big[f(x,\vec{Y})\big].\)

This is a direct application of Fubini and independence.

Proof of Markov Property for Cylinder Sets

Let

  • $X = B_t^x$,
  • $Y_k = B_{t+h_k}^x - B_t^x$ (increments),
  • $\vec{Y} = (Y_1,\dots,Y_d)$.

Because Brownian increments are independent of $\mathcal{F}_t^0$, \(Y_k \perp\!\!\!\perp \mathcal{F}_t^0.\)

Take
\(f(x,\vec{y}) = \prod_{k=1}^d f_k(x+y_k).\)

Then by the lemma, \(E^x\big( Y\circ\theta_t \mid \mathcal{F}_t^0 \big) = g(B_t^x),\) where \(g(x)=E^x\Big[\prod_{k=1}^d f_k(B_{h_k})\Big].\)

Thus \(E^x( Y\circ\theta_t \mid \mathcal{F}_t^0 ) = E^{B_t^x}(Y),\) establishing the Markov property.

Example: Hitting Time After a Shift

Let
\(T_0=\inf\{t\ge 0 : B_t=0\}, \qquad R=\inf\{t>1 : B_t=0\}.\)

Then \(Y = \mathbf{1}_{\{T_0>1\}}\circ\theta_1 = \mathbf{1}_{\{R>1+t\}}.\)

Using Markov property, \(P_x(R>1+t) = \int_{-\infty}^\infty P_1(x, y)\, P_y(T_0>t)\, dy,\) where \(P_1(x,y) = p_t(x,y)=\frac{1}{\sqrt{2\pi t}} \exp\!\left(-\frac{(y-x)^2}{2t}\right)\) is the Brownian transition density.

Example: Last Zero Before Time 1

Let \(L=\sup\{t\le 1: B_t = 0\}.\)

Then for $0\le t\le 1$, \(P_0(L\le t) = \int_{-\infty}^{\infty} p_t(0,y)\, P_y(T_0>1-t)\, dy.\)

Note: \(\mathbf{1}_{\{T_0>1-t\}}\circ\theta_t = \mathbf{1}_{\{L\le t\}}.\)

Summary

  • Brownian motion has the strong Markov property.
  • Conditioning future functionals after time $t$ depends only on $B_t$.
  • The shift operator $\theta_t$ encodes this.
  • Lévy’s lemma makes the proof work for cylinder sets.
  • Examples: hitting times and last zero before 1.
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