# Strong Markov property, Brownian motion

I have a question about the strong Markov property of Brownian motions.

Let $$(\{X_t\}_{t \ge 0}, P_x)$$ be a $$d$$-dimensional Brownian motion starting from $$x \in \mathbb{R}^d$$.

Let $$\tau=\inf\{t>0 \mid |X_t-x|>r\}$$, $$r>0$$. Here, $$|\cdot|$$ denotes the $$d$$-dimensional Euclidean norm.

In a paper, the authors claim that \begin{align*} P_{x}(\tau However, I do not know the proof.

By the strong Markov property, \begin{align} P_{x}(\tau

Can we obtain the equality (1) from the identity (2)?

• What exactly do you mean by the notation $E_x(f(X_s); \tau \in ds)$? – saz Apr 21 at 6:58
• @saz I think that $E_{x}[f(X_s); \tau \in ds]$ denotes the measure induced by $s \mapsto E_{x}[f(X_s); \tau \le s]$ – sharpe Apr 21 at 7:05
• I do not understand why is (1) not a precise equality. It should be one, by the strong Markov property. – zhoraster Apr 21 at 8:26
• @zhoroaster Sorry. (1) should be an equality. – sharpe Apr 21 at 10:35

You didn't apply the strong Markov property correctly. This is indicated by the term

$$\mathbb{P}_{X_{\tau}}(|X_{t-\tau}-X_0| \geq r/2)$$

which appears on the right-hand side of $$(2)$$, note that this term is not well-defined since $$t-\tau$$ is not a non-negative random variable (it is only non-negative on $$\{\tau \leq t\}$$), and therefore $$X_{t-\tau}$$ is not well-defined.

There is the following stronger version of the strong Markov property which is needed here, you can find it for instance in the book by Schilling & Partzsch on Brownian motion (Theorem 6.11)

Theorem Let $$(X_t)_{t \geq 0}$$ be a $$d$$-dimensional Brownian motion with canonical filtration $$(\mathcal{F}_t)_{t \geq 0}$$. Let $$\tau$$ be a stopping time, and let $$\eta \geq \tau$$ be an $$\mathcal{F}_{\tau+}$$-measurable random variable. Then it holds for any bounded Borel measurable function $$u$$ that $$\mathbb{E}^x (u(X_{\eta}) \mid \mathcal{F}_{\tau+})(\omega) = (T_{\eta(\omega)-\tau(\omega)}u)(X_{\tau}(\omega)) \quad \text{a.s.} \tag{3}$$ where $$(T_s u)(x):= \mathbb{E}^x(u(X_s))$$

The above theorem gives the following corollary:

Corollary Under the assumptions of the theorem it holds that $$\mathbb{E}^x(f(X_{\eta},X_{\tau}) \mid \mathcal{F}_{\tau+})(\omega) = (G_{\eta(\omega)-\tau(\omega)} f)(X_{\tau}(\omega)) \quad \text{a.s.} \tag{4}$$ for any bounded Borel measurable function $$f$$ where $$(G_s f)(x) := \mathbb{E^x}(f(X_s,x)).$$

The idea is to prove $$(4)$$ first for functions of the form $$f(x,y) = u(x) v(y)$$ (... for such functions the assertion is a direct consequence of the above theorem and the pull out property), and then to use a density (or monotone class) argument to extend the identity to any bounded Borel measurable function $$f$$.

Now for the stopping time $$\tau$$ from your question define

$$\eta := \max\{t,\tau\}= \begin{cases} \tau, & \{\tau \geq t\}, \\ t, & \{\tau \leq t\}.\end{cases}$$

Then clearly $$\eta \geq \tau$$ and moreover $$\eta$$ is $$\mathcal{F}_{\tau+}$$-measurable. Using the tower property and the pull out property we find that

\begin{align*} \mathbb{P}^x(\tau

Since $$1_{\{|X_{t}-X_{\tau}| \geq r/2\}} = 1_{\{|X_{\eta}-X_{\tau}| \geq r/2\}} \tag{5}$$ we have $$\mathbb{E}^x( 1_{\{|X_{t}-X_{\tau}| \geq r/2\}} \mid \mathcal{F}_{\tau+}) = \mathbb{E}^x( 1_{\{|X_{\eta}-X_{\tau}| \geq r/2\}} \mid \mathcal{F}_{\tau+}),$$ and therefore it follows from the above corollary that

\begin{align*} \mathbb{P}^x(\tau

• Thank you for your kind proof. I understood. – sharpe Apr 22 at 11:45
• I could find the stronger version of the usual strong Markov property. – sharpe Apr 22 at 11:46