Let $M(t)$ be the maximum of Brownian motion $B(t)$ up to time $t$, show that ${M(t)-B(t)}_{t\ge 0}$ and ${|B(t)|}_{t\ge 0}$ have the same distributions.

I have been working on this for a long time and can't get anywhere. Any help would be appreciated. Thank you.

  • $\begingroup$ Is your question how to prove that $M_t-B_t$ has the same distribution as $\left| B_t\right|$ for each $t\geq 0$? $\endgroup$ – peer Apr 10 '17 at 11:22
  • $\begingroup$ @peer The formulation of the question clearly points at the identity in distribution of the processes, not only of the random variables as in your answer. $\endgroup$ – Did Apr 10 '17 at 13:26

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.