3
$\begingroup$

How to show that

$$ \lim_{\alpha \rightarrow \infty} \sup_{t \in \left [0,T \right]} \left | e^{-\alpha t} \int _ 0 ^t e^{\alpha s} ~ dB_s \right | =0, \ \ \text{a.e.}$$

where $\left (B_s \right)_{s\geq 0}$ is a real standard brownian motion starting from zero ?

I'd like to have some ideas to deal with this problem. After all, I'll show some solutions that I propose and somme colegues also but that i believe be all wrong. (I just don't show know to don't interffer in your ideas.

Thank you all.

$\endgroup$
1
  • $\begingroup$ LaTeX fixed${}{}{}$ $\endgroup$ Jan 31, 2013 at 18:24

1 Answer 1

9
$\begingroup$

Here's one way of dealing with it. Integrate by parts to see that the expression under the sup is $$ \Bigl|B(t)-\alpha\int_0^t e^{\alpha(s-t)}B(s)ds\Bigr|\le\alpha\int_0^t e^{\alpha(s-t)}|B(t)-B(s)|ds +e^{-\alpha t}|B(t)|. $$

Now the result follows since $B$ is a.s.-bounded and a.s.-Holder on [0,T].

$\endgroup$
2
  • $\begingroup$ @Yuri Bakhtin: Thank you very much for your answer! Nice solution. $\endgroup$
    – Paul
    Feb 18, 2013 at 0:50
  • $\begingroup$ As the fact that B is Holder concludes? $\endgroup$ Nov 2, 2016 at 13:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.