# Are $1+\exp(-B_{1}^2)$ and $\inf\{t \geq 0 : B_t \geq W_t + \exp(-t)\}$ stopping times?

I have got difficulties with an exercise on stochastic processes.

Let $$B$$ and $$W$$ be two independent Brownian motions on filtration $$(\mathcal{F}_t)_{t\geq 0}$$

Are $$\lambda$$ = $$1+\exp(-B_{1}^2)$$ and $$\tau$$ = $$\inf\{t \geq 0 : B_t \geq W_t + \exp(-t)\}$$ stopping times ?

For the first one it seems to me that this is a constant so a constant is a stopping time right ? The second implies two brownian motion so I have no idea where to start.

• $\lambda$ is not but $\tau$ is a stopping time.
– Surb
Oct 6, 2021 at 10:52
• $\lambda$ is not a constant, it depends upon $B_1$ which is not a constant. But do you think that $\lambda$ is a stopping time? How can $\lambda$ be determined if you only know about $B_t$ and $W_t$ till time $0.5$, for example? Oct 6, 2021 at 11:26
• Thank you for your answers. I don't understand very well your answer Teresa. $\lambda$ depends on $B_1$ but my intuition is to think that $B_1$ is bounded at time $t=1$ because it follows a normal law so$\lambda$ is finite ? For $\tau$ I don't see why it is a stopping time. Oct 6, 2021 at 11:37
• It is not sufficient for $\lambda$ to be finite. For example, if $\lambda$ is a stopping time, then you should know whether or not $\{\lambda \leq 0.5\} \in \mathcal F_{0.5}$. Now, this can't be true, because $1+exp(-B_1^2)$ depends on $B_1$, which obviously can't be predicted based on the observations up till time $0.5$, right? About $\tau$ : for any $T$ try writing $\tau > T$ as (first) an uncountable union of events by using the definition of $\inf$, then use the fact that $B_t,W_t$ are continuous, so the uncountable union can be made countable. Each subevent is in $\mathcal F_T$. Oct 6, 2021 at 12:07

$$\lambda$$ is a stopping time: $$\{\lambda\le t\}$$ is empty if $$t\le 2$$; if $$t>2$$ then $$\{\lambda\le t\} =\{B_1^2\ge -\log(t-1)\}\in\mathcal F_1\subset\mathcal F_t$$.
• Because $B_1$ is $\mathcal F_1$-measurable. Oct 12, 2021 at 21:59