# Two definitions of Strong Markov property for Brownian motion

Strong Markov property for Brownian motion:

(Def 1)For every almost surely finite stopping time $$T$$, the process $$\{B(T+t)-B(T): t\geq 0\}$$ is a standard Brownian motion independent of $$\mathcal{F}(T)$$.

(Def 2) $$\mathbb{E}_x[f(B(t))|\mathcal{F}(T)]=\mathbb{E}_{B(T)}[f(B(t-T))]$$ on $${T\leq t}$$.

Why are these two definitions of SMP equivalent?

From $$2$$ to $$1$$:

Let $$V=(V_t) = (B_{T+t} - B_t)$$ be that process, let $$A \in \mathcal B(\mathbb R^{[0,\infty)})$$ (in cylinder $$\sigma-$$field). Let $$A_0 = \{ x \in \mathbb R^{[0,\infty)} : x-x(0) \in A \}$$ ( we translate every function in $$A$$ by minus it's value at $$0$$). Note that:

$$\mathbb P_x( V \in A | \mathcal F(T)) = \mathbb P_x( (B_{T+t} - B_T) \in A | \mathcal F(T)) = \mathbb P_x ( (B_{T+t}) \in A_0 | \mathcal F(T))$$ (since $$B_T$$ is value at $$0$$ of process $$(B_{T+t})_{t \ge 0}$$. Now apply $$2$$, getting: $$\mathbb P_x ( (B_{T+t}) \in A_0 | \mathcal F(T)) = \mathbb P_{B_T}( (B_t) \in A_0)$$

Note that for any $$y \in \mathbb R$$ we have: $$\mathbb P_y( (B_t) \in A_0 ) = \mathbb P_y( (B_t - y) \in A) = \mathbb P_0 ( (B_t) \in A)$$

Taking $$y = B_T$$ it finally gives us:

$$\mathbb P_x( V \in A | \mathcal F(T)) = \mathbb P_0 ( (B_t) \in A)$$

in particular $$\mathbb P_x (V \in A) = \mathbb E_x[\mathbb P_x(V \in A |\mathcal F(T))]= \mathbb P_0( ( B_t) \in A)$$ so we showed that $$V$$ has the same distribution (under $$\mathbb P_x$$ measure) as standard brownian motion $$(B_t)$$ (cause it's under $$\mathbb P_0$$ measure))

Now to show independence, Take any $$B \in \mathcal F(T)$$ we get:

$$\mathbb P_x(B \cap \{ V \in A\}) = \mathbb E_x [ 1_B \mathbb P_0 ( (B_t) \in A)) = \mathbb P_x(B)\mathbb P_0( (B_t) \in A) = \mathbb P_x(B)\mathbb P_x( V \in A)$$

From 1 to 2:

$$\mathbb E_x [ f(B_{T+t}) | \mathcal F(T)] = \mathbb E_x [ f(B_{T+t} - B_{T} + B_{T}) | \mathcal F(T)]$$

I don't know what information you possess, but it can be shown that as adapted, right continuous process, Brownian Motion is progresivelly measurable, hence $$B_T$$ is $$\mathcal F(T)$$ measurable. By 1. we have that $$B_{T+t} - B_T$$ is independent of $$\mathcal F(T)$$ so by conditional expected value property, the last one is equal to: $$\mathbb E_x[ f(B_{T+t} - B_T + p)] |_{p = B_T}$$ Again using $$1$$, we know that $$B_{T+t} - B_T$$ under $$\mathbb P_x$$ is distributed as standard (so under $$\mathbb P_0$$) brownian motion, so: $$\mathbb E_x[ f(B_{T+t} - B_T + p)] |_{p = B_T} = \mathbb E_0 [ f(B_t + p)]|_{p = B_T} = \mathbb E_p[f(B_t)]|_{p =B_T} = \mathbb E_{B_T}[f(B_t)]$$

So we proved $$\mathbb E_x [ f(B_{T+t}) | \mathcal F(T)] = \mathbb E_{B_T}[f(B_t)]$$.

• Thanks for your answer! But how to understand $\mathbb E_{B_T}[f(B_t)]$? Which are averaged over there? Jun 15, 2020 at 23:34
• Exaclty as $Y=\mathbb E_{B_T}[f(B_t)]$ is a random variable given by $Y(\omega) = \mathbb E_{B_{T(\omega)}(\omega)}[f(B_t)]$. So you average $f(B_t)$ under $\mathbb P_x$ measure but when $x$ is somehow random, that is in point $\omega$ it is $B_{T(\omega)}(\omega)$ Jun 15, 2020 at 23:44
• How to apply $2$ getting: $\mathbb P_x ( (B_{T+t}) \in A_0 | \mathcal F(T)) = \mathbb P_{B_T}( (B_t) \in A_0)$? Jun 16, 2020 at 0:43
• apply $2$ with function $f= 1_{A_0}$ (indicator function of set $A_0$). Jun 16, 2020 at 0:50
• No, you cant define $f(B_t)$. You can define $f(x) = ...$ and then plug $B_t$ in that formula. So you can't have $h_t$ which depends on $t$. You can for fixed number $h$ have $f(x) = 1_{\{x-h \ge 0\}}$ and then $f(B_t) = 1_{\{B_t - h \ge 0\}}$ Jun 16, 2020 at 19:54