Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
LaTeX fixed${}{}{}$ –  Gerald Edgar Jan 31 '13 at 18:24

1 Answer 1

up vote 5 down vote accepted

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].

share|improve this answer
    
@Yuri Bakhtin: Thank you very much for your answer! Nice solution. –  Paul Feb 18 '13 at 0:50
    
@Paul: You are welcome. –  Yuri Bakhtin Feb 18 '13 at 4:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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