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Let $\{B_t:t\geq0\}$ be a $d$-dimensional Brownian motion. $D\in\mathbb{R}^d$ is a bouded open set, $\sigma:=\inf\{t\geq0:B_t\in\partial D\}$.

I want to prove that $\mathrm{P}_x(\sigma<\infty)=1$. My proof is correct?

To prove this, it is enough to show $\mathrm{P}_x(\tau<\infty)=1$. $\tau:=\inf\{t\geq0:B_t\notin\mathcal{B}(x,R)\}$, $\mathcal{B}(x,R)$ is a open ball which contain $D$. \begin{align} \mathrm{P}(\tau=\infty)&=\mathrm{P}({}^\forall t\geq0, B_t\in\mathcal{B}(x,R))\\ &\leq \mathrm{P}(B_t\in\mathcal{B}(x,R))\\ &=\int_{|y|\leq R}\frac{1}{\sqrt[d]{2\pi t}}e^{-|y-x|^2/2t}dy. \end{align} By changing of variable $z_i=\frac{y_i-x_i}{\sqrt{t}}$, \begin{align} \mathrm{P}(\tau=\infty) &\leq\int_{|\sqrt{t}z+x|\leq R}\frac{1}{\sqrt[d]{2\pi}}e^{-|z|^2/2}dz\\ &=\int_{\mathbb{R}^d}{\bf 1}_{|\sqrt{t}z+x|\leq R}\frac{1}{\sqrt[d]{2\pi}}e^{-|z|^2/2}dz. \end{align} Because of $\int_{\mathbb{R}^d}\frac{1}{\sqrt[d]{2\pi}}e^{-|z|^2/2}dz<\infty$ and $t$ is arbitrary, we can use dominated convergence theorem as $t\to\infty$. Therefore, $\mathrm{P}(\tau=\infty)=0$, this completes proof.

Thank you for your cooperation.

sate
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    When $z$ is close to the origin and $t$ is large you don't have control over $|\sqrt tz+x|$ so your DC T argument fails. – Kavi Rama Murthy Dec 14 '19 at 11:20
  • @KaboMurphy Thank you for reply. Well then, how is monotone convergence theorem instead of DC? – sate Dec 14 '19 at 11:29
  • Monotone Convernce Theorem also is not applicable. I am not sure if your approach works at all. – Kavi Rama Murthy Dec 14 '19 at 11:31
  • @KaboMurphy Well, why MC is not applicable? I think ${\bf 1}_{|\sqrt{t}z+x|\leq R}\to 0$ in a monotone. And, if you know other proof, please tell me web page or pdf about this proposition. – sate Dec 14 '19 at 11:37
  • @KaboMurphy Oh, I see why MC is not applicable for this problem. In fact, ${\bf 1}_{|z+\frac{x}{\sqrt{t}|\leq \frac{R}{\sqrt{t}}}$ is a bit strange. Thank you for pointing out. – sate Dec 14 '19 at 13:35

1 Answers1

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Here is a proof using martingale techniques: Set $$\tau := \inf\{t \geq 0; |B_t-x| \geq R\}.$$ By the optional stopping theorem, $$M_t := \exp\left(\xi B_{t \wedge \tau}-\frac{1}{2} \xi^2 t \wedge \tau \right),\qquad \xi \in \mathbb{R}^d,$$ is a martingale and so $$\mathbb{E}^x(M_t)=\mathbb{E}^x(M_0)=e^{\xi x}.$$ As $|B_{t \wedge \tau}| \leq R+|x|$ and

$$\lim_{t \to \infty} \exp\left(\xi B_{t \wedge \tau}-\frac{1}{2} \xi^2 t \wedge \tau \right) = \begin{cases} \exp(\xi B_{\tau}- \frac{1}{2} \tau \xi^2), &\tau<\infty,\\ 0, & \tau=\infty \end{cases}$$

it follows from the dominated convergence theorem that

$$e^{\xi x} = \mathbb{E}^x\exp\left(\xi B_{t \wedge \tau}-\frac{1}{2} \xi^2 t \wedge \tau \right) \xrightarrow[]{t \to \infty} \mathbb{E}^x\left[ 1_{\{\tau<\infty\}} \exp \left( \xi B_{\tau} - \frac{\tau}{2} \xi^2 \right) \right].$$

Letting $\xi \to 0$ yields

$$1 = \mathbb{P}^x(\tau<\infty).$$

saz
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