As in the title, I have to prove that for a Brownian motion B the time
$$
\tau_x = \inf \{t\geq 0 : B(t)=x \}
$$
is a stopping time. It seems obvious intuitively but I struggle with the formal proof. I know that the random variable $\tau$ is a stopping time for a given filtration $(\mathcal{F}_t)$, $t\in T$ if
$$
\{\tau \leq t \}\in \mathcal{F_t} \hspace{0.2cm} \forall t\in T.
$$
However I can't seem to conduct a formal proof (still new to the probability theory).
I also have to proof that collection $\mathcal{F_t}$ associated with a stopping time $\tau$ is a $\sigma$-algebra. I have the same problem here: I know what a $\sigma$-algebra is but I don't know how to show the required property.
Thanks in advance for any help!
Best Answer
$$\{\tau_x\leq t\}=\{\max_{k\leq t}B_k\geq x\}.$$
$$\mathcal{F}_t=\sigma(B_s: 0\leq s\leq t).$$
By definition of filtration, $\{B_k\geq x\}\in \mathcal{F}_t$ for all $k\in [0,t]$. So it's a stopping time.
More details:
The function $B(t,\omega):=B_t(\omega)$ is measurable with respect to the product sigma algebra $\mathcal{B}\times \mathcal{F}_t$, where $\mathcal{B}$ is the borel sigma algebra (generated by open intervals). Hopefully this is something you've proven in your class (or perhaps taken as a given). Otherwise, see Lemma 1.1 of this for details. Then just like in classical real analysis, if $f_n$ is measurable, then so is $\sup_n f_n, \max_n f_n$, etc.