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.

  • 2
    $\begingroup$ $\lambda $ is not but $\tau$ is a stopping time. $\endgroup$
    – Surb
    Oct 6, 2021 at 10:52
  • 1
    $\begingroup$ $\lambda$ 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? $\endgroup$ Oct 6, 2021 at 11:26
  • $\begingroup$ 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. $\endgroup$
    – finquant75
    Oct 6, 2021 at 11:37
  • $\begingroup$ 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 $\mathcal F_T$. $\endgroup$ Oct 6, 2021 at 12:07

1 Answer 1


$\lambda$ is a stopping time: $\{\lambda\le t\}$ is empty if $t\le 2$; if $t>2$ then $\{\lambda\le t\} =\{B_1^2\ge -\log(t-1)\}\in\mathcal F_1\subset\mathcal F_t$.

  • $\begingroup$ How do you know that {B21≥−log(t−1)}∈F1 ? I mean if i'm right we can say it is a stopping time if B21+log(t+1) is a continuous adapted process ? $\endgroup$
    – finquant75
    Oct 12, 2021 at 15:32
  • $\begingroup$ Because $B_1$ is $\mathcal F_1$-measurable. $\endgroup$ Oct 12, 2021 at 21:59

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