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.

  • $\begingroup$ He gets it exactly as you say: $\mathbb P^0(\exists t>1/n:B_t=0)=\mathbb P^0(\exists t>0:B_{t+1/n}=0)=\mathbb E^0\mathbb P^{B_{1/n}}(\exists t>0:B_{t}=0).$ $\endgroup$
    – Kurt G.
    Apr 6, 2022 at 17:12
  • $\begingroup$ Thanks for the reply. The second equality that you have is what confuses me and I don't know how $\mathbb{P}^0(B_{1/n}=0)=0$ is used. If $A$ is the event $(\exists t>0: B_{t+1/n}=0)$ then I know the middle term is $\mathbb{E}^0[\mathbb{1}_A]$ but I don't understand how to get the second equality from here. $\endgroup$
    – raba_123
    Apr 6, 2022 at 22:31
  • $\begingroup$ The second equality is precisely the Markov property which says when we restart $B$ at $1/n$ we can subract that term from $B_{t+1/n}$ so that $B_{t+1/n}=0$ becomes $B_t=0$. Sloppily spoken: Markov property is restart plus a time shift. $\endgroup$
    – Kurt G.
    Apr 7, 2022 at 5:56


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