33 – Brownian motion: Markov property, right–continuous filtration, time inversion, tail events
We work with standard Brownian motion starting at 0.
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Sample space
\(\Omega = C[0,\infty) = \{\omega : [0,\infty) \to \mathbb R \text{ continuous}\}.\) -
Canonical process
\(B_t(\omega) = \omega(t), \qquad t \ge 0.\) -
Natural filtration
\(\mathcal F_t^0 = \sigma\{B_s : 0 \le s \le t\}, \qquad t \ge 0.\) -
Laws $P_x$ on $(\Omega,\mathcal F)$ with \(P_x(B_0 = x) = 1, \qquad P_x(A) = P_0(A - x)\) (spatial shift of Wiener measure).
Markov property (recall)
For bounded measurable $Y : \Omega \to \mathbb R$ and $t \ge 0$, \(E^{x}_{\mathcal F_t^0}(Y \circ \theta_t) = E^{B_t^x}(Y) =: \psi(B_t^x),\) where the shift operator $\theta_t : \Omega \to \Omega$ is \((\theta_t \omega)(s) = \omega(t+s), \qquad s \ge 0.\)
Equivalently, \(E^x(Y \circ \theta_t) = E^x\big[\,E^{B_t^x}(Y)\,\big].\)
Right–continuous augmentation of the filtration
Define \(\mathcal F_t^+ := \bigcap_{n=1}^\infty \mathcal F_{t+1/n}^0, \qquad t \ge 0.\)
Then:
- $\mathcal F_t^0 \subset \mathcal F_{t+1/n}^0$ for each $n$, hence $\mathcal F_t^0 \subset \mathcal F_t^+$.
- The family ${\mathcal F_t^+}_{t \ge 0}$ is right–continuous: \(\mathcal F_t^+ = \bigcap_{s>t} \mathcal F_s^+.\)
We want to prove \(\mathcal F_t^0 = \mathcal F_t^+ \quad \text{up to null sets under } P_x,\ \forall x \in \mathbb R.\)
In particular for $t=0$ we get Blumenthal 0–1 law: \(\mathcal F_0^0 = \mathcal F_0^+ = \sigma\{B_0\}\) up to null sets and \(P^x(A) \in \{0,1\} \quad \text{for all } A \in \mathcal F_0^+,\, x \in \mathbb R.\)
Hitting times of the positive and negative half–lines
Let ${B_t}_{t\ge0}$ be SBM under $P_0$.
Define the first hitting times \(T = \inf\{t>0 : B_t > 0\}, \qquad S = \inf\{t>0 : B_t < 0\}.\)
Claim. \(P_0(T = 0) = 1, \qquad P_0(S = 0) = 1.\)
Sketch of idea:
- On the event ${T>0}$ we would have a sequence $t_n \downarrow 0$ with $B_{t_n} \le 0$ for all $n$, contradicting oscillation of Brownian paths.
- Similarly for $S$.
From the two claims we deduce that almost surely there exist sequences $t_n \downarrow 0$ with $B_{t_n} > 0$ and also $s_n \downarrow 0$ with $B_{s_n} < 0$; hence Brownian motion visits both sides of 0 arbitrarily close to time 0.
This oscillatory behavior is key in the proof that the augmented and natural filtrations coincide up to null sets.
Equality of $\mathcal F_t^0$ and $\mathcal F_t^+$
Theorem.
For SBM ${B_t}_{t\ge0}$,
\(\mathcal F_t^0 = \mathcal F_t^+ \quad \text{up to null sets under } P_x,\ \forall x \in \mathbb R.\)
Equivalently, for any bounded r.v. $Y$ on $\Omega$, \(E_{\mathcal F_t^+}^x(Y) = E_{\mathcal F_t^0}^x(Y) \quad \text{a.s.}\)
Proof strategy:
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First show this for $Y$ of the special form \(Y = X \cdot (Z \circ \theta_t), \qquad X \in \mathcal F_t^0,\) where $Z$ is bounded measurable on $\Omega$.
Then \(\begin{aligned} E_{\mathcal F_t^+}^x\big(X \cdot (Z \circ \theta_t)\big) &= X\, E_{\mathcal F_t^+}^x(Z \circ \theta_t) = X\,E_{\mathcal F_t^0}^x(Z \circ \theta_t) \\ &= E_{\mathcal F_t^0}^x\big(X \cdot (Z \circ \theta_t)\big), \end{aligned}\) using the Markov property and right–continuity.
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Then extend to all bounded $Y$ in $\mathcal F_t^+$ via the monotone class theorem, starting from indicator functions and using linearity and monotone convergence.
Thus conditional expectations with respect to $\mathcal F_t^0$ and $\mathcal F_t^+$ agree almost surely for all bounded $Y$, hence the σ–algebras coincide up to null sets.
Time inversion of Brownian motion
Theorem.
Let ${B_t}_{t\ge 0}$ be standard Brownian motion. Define
\(\tilde B(t) =
\begin{cases}
t\, B\!\left(\dfrac{1}{t}\right), & t>0, \\
0, & t=0.
\end{cases}\)
Then ${\tilde B(t)}_{t \ge 0}$ is also standard Brownian motion.
Checks:
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Mean: \(E[\tilde B(t)] = t\,E\big[B(1/t)\big] = 0.\)
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Covariance, for $s,t>0$: \(\begin{aligned} \operatorname{Cov}\big(\tilde B(t), \tilde B(s)\big) &= ts\,\operatorname{Cov}\Big(B\!\left(\tfrac{1}{t}\right), B\!\left(\tfrac{1}{s}\right)\Big) \\ &= ts \Big(\tfrac{1}{t} \wedge \tfrac{1}{s}\Big) = t \wedge s. \end{aligned}\)
So the finite–dimensional distributions are centered Gaussian with covariance $t\wedge s$, the same as SBM.
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Continuity of sample paths: since $B(\cdot)$ is continuous and the map $t \mapsto 1/t$ is continuous on $(0,\infty)$, and we set $\tilde B(0)=0$, one checks that $\tilde B(t)$ is continuous on $[0,\infty)$ almost surely.
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Limit at infinity: \(\lim_{t\to\infty} \tilde B(t) = \lim_{t\to\infty} t B(1/t) = 0 \quad \text{a.s.}\) (from continuity of $B$ at 0).
Hence $\tilde B$ is a version of SBM. This “time inversion” will be used to study tail events.
Tail events of Brownian motion
Let ${B_t}_{t\ge 0}$ be standard Brownian motion and let \(\mathcal T = \bigcap_{T>0} \sigma\{B_t : t \ge T\}\) be its tail σ–field.
Theorem (tail 0–1 law for Brownian motion).
If $A \in \mathcal T$ is a tail event of Brownian motion, then
\(P(A) \in \{0,1\}.\)
Moreover, for the Brownian family ${P_x}_{x\in\mathbb R}$,
\(P_x(A) = P_0(A), \qquad \forall x \in \mathbb R.\)
Idea of proof:
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Work first under $P_0$ (SBM starting at 0). Using time inversion, tail events in the future correspond to events near time 0, hence are in $\mathcal F_0^+$.
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By Blumenthal 0–1 law, any event in $\mathcal F_0^+$ has probability 0 or 1 under $P_0$.
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Use Markov property and the family ${P_x}$ to show $P_x(A) = P_0(A)$ for all $x$.
Example of the mechanism: if $A \in \mathcal F_1^+$ and $B = A \circ \theta_{-1} \in \mathcal F_0^+$, then \(P_0(A) = E^0[1_B \circ \theta_1] = E^0\big[E^{B_1}(1_B)\big] = \int_{\mathbb R} P_y(B)\,P_0(B_1 \in dy),\) so if $P_y(B) = 0$ for all $y$ then $P_0(A)=0$, etc.
This completes the discussion of the Markov property, right–continuous filtrations, time inversion, and tail events for Brownian motion.
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