I am currently reading through Brownian Motion by Schilling and am confused by something in the section on transience and recurrence. He says,
"By the Markov Property and the fact that $\mathbb{P}^0(B_{1/n}=0)=0$,
we have
$\mathbb{P}^0(\exists t> 1/n: B_t=0)=\mathbb{E}^0\mathbb{P}^{B_{1/n}}(\exists t>0: B_t=0)$"
Intuitively, I understand that by the Markov Property, moving $t+1/n$ in time has the same distribution as starting at time $1/n$ and then moving $t$ units. But I am not quite sure how he gets the right-hand side of the equality from the left-hand side.
Any help is appreciated!
$\textbf{Edit}$: After referencing this post (Brownian motion: Strong Markov versus translation invariance) I came up with something that I think works. I still don't see where $\mathbb{P}^0(B_{1/n}=0)=0$ is used; I think probably so there are strict inequalities but I don't use it below so there may be some mistakes in my argument below but its an idea:
Let $f_t(x)=\begin{cases}1 & t>0, x=0\\0& \text{else} \end{cases}$.$\quad$ Then $f_t(B_{t+1/n})=\begin{cases}1 & t>0, B_{t+1/n}=0\\0& \text{else} \end{cases}$.
So by the Markov property, $\mathbb{E}^0[f_t(B_{t+1/n})|\mathcal{F}_{1/n}]=\mathbb{E}^{B_{1/n}}[f_t(B_t)] = \mathbb{P}^{B_{1/n}}(t>0,B_t=0)$.
Taking expectations of both sides we have $\mathbb{E}^0[\mathbb{E}^0(f_t(B_{t+{1/n}})|\mathcal{F}_{1/n})]=\mathbb{E}^0[\mathbb{P}^{B_{1/n}}(t>0,B_t=0)]$.
Note by the tower property, the left hand side equals $\mathbb{E}^0[f_t(B_{t+1/n})]$ which is $\mathbb{E}^0[\mathbb{1}_{\{t>0:B_{t+1/n}=0\}}]=\mathbb{P}^0(t>1/n, B_t=0)$ which is what we needed to show.
If you see any flaws in the argument, please point them out.