本章导读 本章证明布朗运动具有马尔可夫性。§4.1 搭建 \(\sigma\)-代数 框架（\(\mathcal F_t\) 由 \(\{B_s:s\le t\}\) 生成、\(\mathcal D_t\) 由未来增量 \(\{B_{s+t}-B_t\}\) 生成、由增量独立性 \(\mathcal F_t\perp\mathcal D_t\)、整条 BM 的 \(\sigma\)-代数 \(\mathcal F=\mathcal F_\infty\)、右极限 \(\mathcal F_{t+}\)）。§4.2 马尔可夫性（命题 4.1：平移并去基点后的 \(\{Y_t\}=\{B_{t+t_0}-B_{t_0}\}\) 仍是布朗运动且独立于 \(\mathcal F_{t_0}\)；Def 4.1：更一般的马尔可夫过程定义——给定 \(X_t\) 后，未来只依赖当前）。无图。

σ-代数框架 / The σ-algebra framework 设 \((\Omega,\tilde{\mathcal F},\mathbb P)\) 是一个足够大的概率空间，使我们能在其上定义标准布朗运动 \(\{B_t\}\)。记号约定如下。Let \((\Omega,\tilde{\mathcal F},\mathbb P)\) be a probability space big enough to define a standard Brownian motion \(\{B_t\}\) on it. The notation is as follows.

• \(\mathcal F_t\)：由 \(\{B_s:s\le t\}\) 生成的 \(\sigma\)-代数（到 \(t\) 为止路径的信息）。\(\mathcal F_t\): the \(\sigma\)-algebra generated by \(\{B_s:s\le t\}\) (the information of the path up to \(t\)).
• \(\mathcal D_t\)：由 \(\{B_{s+t}-B_t:s\ge0\}\) 生成的 \(\sigma\)-代数（\(t\) 之后的未来增量信息）。\(\mathcal D_t\): the \(\sigma\)-algebra generated by \(\{B_{s+t}-B_t:s\ge0\}\) (the future increments after \(t\)).
• 由增量独立性，给定 \(t\)，有 \(\mathcal F_t\perp\mathcal D_t\)。By increment independence, given \(t\), \(\mathcal F_t\perp\mathcal D_t\).

\(\mathcal F\) 是由整条布朗运动生成的 \(\sigma\)-代数，即\(\mathcal F\) is the \(\sigma\)-algebra generated by the whole Brownian motion, i.e.

$$\mathcal F=\mathcal F_\infty=\bigcup_{t\ge0}\mathcal F_t,\qquad\text{and also}\qquad\mathcal F=\bigcup_{t\ge0}\mathcal D_t.$$

最后，记右极限 \(\sigma\)-代数 \(\mathcal F_{t+}\equiv\bigcap_{s>t}\mathcal F_s\)。Finally, denote the right-limit \(\sigma\)-algebra \(\mathcal F_{t+}\equiv\bigcap_{s>t}\mathcal F_s\).

命题 4.1（布朗运动的马尔可夫性）/ Proposition 4.1 (Markov property for Brownian motion) 设 \(\{B_t\}\) 是布朗运动。对某 \(t_0>0\)，令 \(Y_t=B_{t+t_0}-B_{t_0}\)（\(\forall t\ge0\)）。则 \(\{Y_t\}\) 是一个独立于 \(\mathcal F_{t_0}\) 的布朗运动。我们称布朗运动 \(\{B_t\}\) 具有马尔可夫性。Let \(\{B_t\}\) be a Brownian motion. For some \(t_0>0\), let \(Y_t=B_{t+t_0}-B_{t_0}\) (for all \(t\ge0\)). Then \(\{Y_t\}\) is a Brownian motion independent of \(\mathcal F_{t_0}\). We say Brownian motion \(\{B_t\}\) has the Markov property.

命题 4.1 证明 / Proof of Proposition 4.1 要证 \(\{Y_t\}\) 是布朗运动，只需注意到 \(\{B_{t+t_0}-B_{t_0}\}\) 与 \(\{B_t\}\) 同分布，而 \(\{B_t\}\) 是布朗运动。要证 \(\{Y_t\}\) 与 \(\mathcal F_{t_0}\) 独立，注意支撑 \(Y_t\) 的最小 \(\sigma\)-代数是 \(\sigma\{Y_t:t\ge0\}=\mathcal D_{t_0}\perp\mathcal F_{t_0}\)，证毕。\(\blacksquare\)To show \(\{Y_t\}\) is a Brownian motion, we only need to note that \(\{B_{t+t_0}-B_{t_0}\}\) has the same distribution as \(\{B_t\}\), and \(\{B_t\}\) is a Brownian motion. To show independence of \(\{Y_t\}\) with \(\mathcal F_{t_0}\), note that the smallest \(\sigma\)-algebra supporting \(Y_t\) is \(\sigma\{Y_t:t\ge0\}=\mathcal D_{t_0}\perp\mathcal F_{t_0}\), and we are done. \(\blacksquare\)

定义 4.1（更一般的马尔可夫性）/ Definition 4.1 (More general Markov property) 称 \(\{X_t\}\) 是马尔可夫过程，若对 \(s\ge t\)，给定 \(\{X_\tau:\tau\le t\}\) 时 \(X_s\) 的条件分布只依赖于 \(X_t\)。直觉：给定"现在"\(X_t\)，"未来"与"过去"条件独立。We say \(\{X_t\}\) is a Markov process if the conditional distribution of \(X_s\) for \(s\ge t\) given \(X_\tau\) for \(\tau\le t\) depends only on \(X_t\). Intuition: given the "present" \(X_t\), the "future" is conditionally independent of the "past".