I am stuck with the following problem: let $B_t$ be a standard Brownian motion and let $S_{t}:=\sup_{0 \leq s \leq t} B_s$. Prove that for every $\lambda \geq 0$ and $\mu \leq \lambda$, $\mathbb{P}(B_t \leq \mu, S_t \geq \lambda)=\mathbb{P}(B_t \geq 2 \lambda-\mu)$.

I know that for any $a>0,\ \mathbb{P}(S_t \geq a)=2 \mathbb{P}(B_t \geq a)$ but I do not know how I can use this fact.

Thank you very much !!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.