We define $T_0=0$ and $T_n=\min \{t>T_{n-1} :|B(t)-B(T_{n-1})|=1\}$, where $B $ deonotes Brownian motion. I have the following two questions:
(1) Show that for every $n ,\ \ T_n $ is a stopping time and (2) the waiting times $T_n - T_{n-1 } $ are independent and identically distributed.
For (1) we of course proceed by induction. We have to show that given any $0<s$, $\{T_n \le s \} \in \mathcal F_s$ where $F_s $ is a filtration for the brownian motion. Then I'm a bit lost: I can see the following. By sample continuity we could reduce the minimum over $\mathbb R $ to a minimum over a countable set. Also from sample continuity we have that for any $t $, $\{t< T_{n-1 } \} \in \mathcal F_{n-1 } $. And $\omega \mapsto |B(t)(\omega)-B(T_{n-1 })(\omega) |$ is $\mathcal F_{T_{n-1 } } $-measurable.
For (2) I know that from the strong Markov property for Brownian motion $T_n $ must be distributed as $T_{n-1 } $. Does the independence of the increments also follow from the Strong markov property? And how do we find the distribution?
Thanks in advance!
For (1), we can use this proposition.
$T_1<\infty$ is a stopping time by this proposition. Then we use induction. Note that $T_n-T_{n-1}=\inf\{t\geq0: |B_{t+T_{n-1}}-B_{T_{n-1}}|=1\}$. Let $X_t=B_{t+T_{n-1}}-B_{T_{n-1}}$, then for any $\omega\in\Omega$, $X_t(\omega)$ is a continuous function in $t$. An application of this proposition implies that $T_n-T_{n-1}<\infty$ is a stopping time, so $T_n$ is a stopping time by induction. (If $S$ and $T$ are stopping tmes then $S+T$ is a stopping time.
For (2), we use the strong Markov property, which aseerts that $(X_t)_{t\geq 0}$ defined above is a Brownian motion independent of $\mathscr F_{T_{n-1}}$( Here we need to use that $T_{n-1}<\infty$ a.s.) Since $T_{n-1},T_{n-2}\in \mathscr F_{T_{n-1}}$, we have $T_n-T_{n-1}\perp T_{n-1}-T_{n-2}$. Since $X_t$ is a Brownian motion, we have $T_n-T_{n-1}\overset{d}{=}T_1$. This completes the proof.