I'm trying to follow the following computation:
For a fixed $h >0$ and $x\in D \subset \mathbb{R}^d$ we have (let $\tau$ be an exit time from $D$ and $B_t$ be a brownian motion started from $x \in D$):
$$\textbf{P}^x(\tau > h) = \textbf{P}^x(\forall t \in (o,h]: B(t) \in D) \\ = \inf_n \textbf{P}^x(\forall t \in (1/n,h]: B(t) \in D) \\ = \inf_n \textbf{E}^x \big[ \textbf{P}^x(\forall t \in (1/n,h]: B(t) \in D) \big] \\ = \inf_n \textbf{E}^x \big[ \textbf{P}^{B_{1/n}}(\forall t \in (0,h-1/n]: B(t) \in D) \big] $$
I get lost at the last two lines:
- Where does the expectation suddenly come from in Line 3?
- I think Strong Markov Property was used in the 4th line, but I've only seen this formulated in terms of expectations. So I'm having trouble deciphering what's going on here?
Thoughts?
I don't see the need for line three in the original computation. To get line four, I would write: \begin{align*} \mathbb{P}^{x}\left\{B(t) \in D \, \, \text{for all} \, \, t \in (\frac{1}{n},h]\right\} &= \mathbb{E}^{x}(1_{\{B(t) \in D \, \, \text{for all} \, \, t \in (\frac{1}{n},h]\}}) \\ &= \mathbb{E}^{x} \left(\mathbb{P}^{B_{1_{n}}} \left\{B(t) \in D \, \, \text{for all} \, \, t \in \left(0,h - \frac{1}{n}\right]\right\} \right) \end{align*} by the Markov property.