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?


1 Answer 1


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

  • $\begingroup$ Thanks for your answer! But how to understand $ \mathbb E_{B_T}[f(B_t)]$? Which are averaged over there? $\endgroup$
    – Hermi
    Jun 15, 2020 at 23:34
  • $\begingroup$ 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 $B_{T(\omega)}(\omega)$ $\endgroup$ Jun 15, 2020 at 23:44
  • $\begingroup$ How to apply $2$ getting: $\mathbb P_x ( (B_{T+t}) \in A_0 | \mathcal F(T)) = \mathbb P_{B_T}( (B_t) \in A_0)$? $\endgroup$
    – Hermi
    Jun 16, 2020 at 0:43
  • $\begingroup$ apply $2$ with function $f= 1_{A_0}$ (indicator function of set $A_0$). $\endgroup$ Jun 16, 2020 at 0:50
  • 2
    $\begingroup$ 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 $f(B_t) = 1_{\{B_t - h \ge 0\}}$ $\endgroup$ Jun 16, 2020 at 19:54

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