I try to solve a problem related with my previous post. I believe if I can show following inequality then I am done.
Let $D$ be $a$ bounded domain in $\mathbb{R}^{d}$, let $B_{t}$ be a standard Brownian motion starting at $x \in D,$ and let $$ T=\inf \left\{t: B_{t} \notin D\right\} $$ Let $q_{t}=\sup _{x \in D} \mathbb{P}^{x}\{T>t\}$, where superscript means that "starting at $x$".
By strong Markov property, it is easy to see that $q_{s+t} \leq q_{s} q_{t}$. I try to show there exists constant $c$ such that $q_{s+t} \geq c q_{s} q_{t}$, still by strong Markov property, if I can show: for any $x \in D$, there exist a constant $c_1$ such that
$$\mathbb{P}^{x}\{T>t\}\geq c_1 q_t$$ then I am done. But I have no idea how to show this inequality. Can anyone help? Really appreciate your help!
Unfortunately the inequality you want doesn't hold. The reason being that it would mean that for fixed $t$, $\mathbb{P}^{x}(T>t)$ is bounded below by some positive constant, independent of $x$. However, intuitively, it is clear that as $x$ tends towards the boundary of our domain $D$, $\mathbb{P}^{x}(T>t)$ tends to $0$.
More precisely, if $x^{*}\in{\partial{D}}$ is regular, then $\mathbb{P}^{x}(T>t)\rightarrow{0}$ as $x\rightarrow{x^{*}}$.