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!


1 Answer 1


$$\{\tau_x\leq t\}=\{\max_{k\leq t}B_k\geq x\}.$$

$$\mathcal{F}_t=\sigma(B_s: 0\leq s\leq t).$$

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.

  • 2
    $\begingroup$ This argument is for $x\ge 0$. Turn it upside down for $x\le 0$. The measurability of $(\omega,t)\mapsto B_t(\omega)$ is not enough to guarantee that the supremum $\sup_{k\le t}B_k$ over an uncountable set is $\mathcal F_t$-measurable. But continuity of $k\mapsto B_k(\omega)$ saves the day: $\sup_{k\le t}B_k(\omega)=\sup_{k\le t, k\in\Bbb Q}B_k(\omega)$. $\endgroup$ Oct 13, 2016 at 19:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .