Suppose $M_t=\sup_{0\leq s\leq t}\{B_s\}$, where $\{B_t\}_0^{\infty}$ is a standard Brownian Motion. I would like to know if it is true that $M_t e^{-t}$ converges to 0 almost surely?


  • $\begingroup$ taking expectations shows $\sum_n M_n e^{-n} < \infty $, and it should be straightforward although maybe not easy to deal with the difference between integers and reals. $\endgroup$ – mike Apr 8 '13 at 11:21
  • $\begingroup$ Yes, I think I am able to show that $M_t e^{-t}$ converges to 0 in $L^1$ because each $M_t$ has the same distribution as $|B_t|$, but pointwise convergence seems a more difficult question. $\endgroup$ – user69818 Apr 8 '13 at 19:36
  • 1
    $\begingroup$ my argument implies pt wise convergence of $M_n e^{-n}$ to $0$ because it is implied by $\sum M_n e^{-n} < \infty, (n^{th}$ term goes to $0$), but on second thought, you ought to be able to get it from $\frac {B_t}t \rightarrow 0 $. $\endgroup$ – mike Apr 9 '13 at 11:10
  • $\begingroup$ Yea, thanks. I've already got it. $\endgroup$ – user69818 Apr 9 '13 at 22:50

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.