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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<t, |X_t-X_{\tau}| \ge r/2) &= \int_{0}^{t}E_{x}[P_{X_s}(|X_{t-s}-X_0| \ge r/2);\tau \in ds]. \tag{1} \end{align*} However, I do not know the proof.

By the strong Markov property, \begin{align} P_{x}(\tau<t, |X_t-X_{\tau}| \ge r/2) &=P_{x}(\tau<t, |X_{(t-\tau)+\tau}-X_{\tau}| \ge r/2) \\ &=E_{x}[P_{X_\tau}(|X_{t-\tau}-X_0| \ge r/2);\tau<t] \tag{2}. \end{align}

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

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  • $\begingroup$ What exactly do you mean by the notation $E_x(f(X_s); \tau \in ds)$? $\endgroup$ – saz Apr 21 at 6:58
  • $\begingroup$ @saz I think that $E_{x}[f(X_s); \tau \in ds]$ denotes the measure induced by $s \mapsto E_{x}[f(X_s); \tau \le s]$ $\endgroup$ – sharpe Apr 21 at 7:05
  • $\begingroup$ I do not understand why is (1) not a precise equality. It should be one, by the strong Markov property. $\endgroup$ – zhoraster Apr 21 at 8:26
  • $\begingroup$ @zhoroaster Sorry. (1) should be an equality. $\endgroup$ – sharpe Apr 21 at 10:35
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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<t, |X_{t}-X_{\tau}| \geq r/2) &= \mathbb{E}^x \big[ \mathbb{E}^x( 1_{\{\tau <t\}} 1_{\{|X_{t}-X_{\tau}| \geq r/2\}} \mid \mathcal{F}_{\tau+}) \big] \\ &= \mathbb{E}^x \big[ 1_{\{\tau <t\}} \mathbb{E}^x( 1_{\{|X_{t}-X_{\tau}| \geq r/2\}} \mid \mathcal{F}_{\tau+}) \big] \end{align*}

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<t, |X_{t}-X_{\tau}| \geq r/2) &= \int_{\Omega} 1_{\{\tau <t\}}(\omega) \mathbb{P}^{X_{\tau}(\omega)}(|X_{t-\tau(\omega)}-X_0| \geq r/2) \, d\mathbb{P}^x(\omega) \\ &= \int_{(0,t)} \mathbb{E}^x(\mathbb{P}^{X_s}(|X_{t-s}-X_0| \geq r/2); \tau \in ds) \end{align*}

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

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