$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ Consider a [[Brownian Motion with Filtration]]. >[!claim] Simple Markov Property >If $X_t$ is standard BM, for all $s\geq 0$, $(B_{s+t} - B_s)_{t\geq 0}$ is a standard BM independent of $\FFF_s$. This is obvious. >[!claim] Less simple Markov Property >Actually, $(B_{s+t} - B_s)_{t\geq 0}$ is a standard BM independent of $\FFF_{s+}$. >[!proof]- By continuity, plus details. > By continuity, $B_{s+t} - B_s = \lim_{\eps\to 0} B_{s+t+\eps} - B_{s+\eps}$ pointwise. The claim is thus that limits preserve independence. This is basically true, but you have to be careful to talk about the measure-$0$ events. Observe that$ > \{B_{s+t} - B_s \in (a,b)\}\subset \lim_{\eps\to 0} \{B_{s+t+\eps} - B_{s+\eps} \in (a,b)\}\subset \lim_{\eps\to 0} \{B_{s+t+\eps} - B_{s+\eps} \in [a,b]\}\subset \{B_{s+t} - B_s\in [a,b]\}. > $But the LHS and the RHS have the same measure. Thus, for any $A\in \FFF_{s+}$,$\PP(B_{s+t} - B_s\in (a,b), A) = \lim_{\eps\to 0} \PP\left(B_{s+t+\eps} - B_{s+\eps}\in (a,b), A\right) = \lim_{\eps\to 0} \PP\left(B_{s+t+\eps} - B_{s+\eps}\right)\PP(A) = \PP\left(B_{s+t} - B_s\in (a,b)\right)\PP(A).$This concludes. > [!theorem] Blumenthal > Suppose $A\in \FFF^X_{0+}$. Then $\PP(A)\in \{0,1\}$. We just showed $\FFF_{0+}\pperp \sigma(B_t: t>0)$. By continuity, this $\sigma$-algebra is the same as $\sigma(B_t: t\geq 0)$. > [!proof]- Really? > The functions $f_n = B_{\frac1n}$ are $\FFF_{0+}$-measurable. Thus their limit is as well. But by path-continuity, this limit is just $B_0$. Thus $\{B_0\in A\}$ is $\FFF_{0+}$-measurable for all $A$. But $\FFF_{0+} = \bigcap_\eps \FFF_{\eps}\subset \sigma(B_t: t \geq 0)$, so $\FFF_{0+} \pperp \FFF_{0+}$, thus just like [[Kolmogorov's Zero-One Law]] you win. >[!claim] Strong Markov Property >For any [[Stopping Time]] $T$ with $\PP(T < \infty) > 0$, set $B_t^{(T)} = \id_{T < \infty}(B_{T+t} - B_T)_{t\geq 0}$. Under the probability measure $\PP(\bullet | T < \infty)$, $B_t^{(T)}$ is a standard Brownian motion $\pperp \FFF_T$. > [!proof]- Dyadic on $T$ to make it countable. (Sketch) > First take $T < \infty$ everywhere. Take $T^k = 2^{-k}\ceil{2^k T}\geq T$ and $B^{(T,k)}_t = B_{T^k+t} - B_{T^k}$. It is immediate that $B^{(T,k)}_t$ are BM $\pperp \FFF_T$. They converge path-wise pointwise to $B^{(T)}_t$, and arguments similar to the less simple MP above conclude. > > The $0 < \PP(T < \infty) < 1$ modification is harmless. These are generalized to arbitrary Markov processes [[Strong Markov Property|here]]. # Example Applications >[!example] Brownian motion crosses $0$ i.o. near $0$. >Almost surely: > - For all $\eps > 0$, $\sup(B_s: s\in [0,\eps) > 0$ and $\inf(B_s: s\in [0,\eps) < 0$. > - $\forall a\in \RR$, $T_a = \inf_{t\in \RR} B_t = a$ is finite (i.e. $\lim\sup B_t = +\infty$ and $\lim\inf B_t = -\infty$). The first one is trivial. The second one observes that$1 = \lim_{\delta\to 0} \PP\left(\sup\left\{B_s:s\in [0,1]\right\}\right)\geq \delta) = \lim_{\delta\to 0} \PP\left(\sup\left\{B_s:s\leq \left(\frac{a}{\delta}\right)^2\right\}\right)\geq a = \PP(T_a < \infty).$Very cool! An application of the strong markov property: >[!example] Reflection Principle >Let $S_t = \sup\{B_s: s\leq t\}$. Then, for all $0\leq a$ and $b\leq a$, $\PP(S_t \geq a, B_t\leq b) = \PP(B_t \geq 2a - b)$. This is trivial, cool though.