Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question:

Given two disjoint intervals $[a,b]$ and $[c,d]$, how to prove almost surely we have

$$\sup_{t\in[a,b]}B_t\neq\sup_{t\in[c,d]}B_s$$

where $B$ is a standard brownian motion. I have no idea about this problem. Does someone have an idea? Thansk a lot!

share|improve this question
add comment

1 Answer

If you write $X_t = B_{a+t}-B_a$ for $0 \le t \le b-a$ and $Y_s =B_{c+s}-B_c$ for $0 \le d-c$, then $(X_t)_{t\in[0,b-a]}$ and $(Y_s)_{s\in[0,d-c]}$ are independent Brownian motions, also independent of $B_a$ and $B_c$. So their suprema $$A = \sup\limits_{t\in [0,b-a]}X_t \text{ and } C = \sup\limits_{s\in [0,d-c]}Y_s$$ are independent random variables, which are also independent of $B_a$ and $B_c$. Now $$\mathbb{P}\left[\sup\limits_{t\in[a,b]} B_t = \sup\limits_{t\in[c,d]} B_t\right] =\mathbb{P}[B_a + A = B_c + C] = \mathbb{P}[A-C = B_c - B_a].$$ This is always $0$ since $B_c-B_a$ is continuous (actually, $N(0,c-a)$-distributed) and independent of $A-C$ (which is also continuous unless both $a=b$ and $c=d$.)

share|improve this answer
    
Just a query about the last line of reasoning, isn't A-C a constant and hence we have that a continuous r.v taking a constant value is a 0 probability event? Thanks –  user77485 May 12 '13 at 13:01
    
Why would $A-C$ be constant? Since they are independent, that would only be the case if both $A$ and $C$ are constant, i.e., iff $a=b$ and $c=d$. –  Lukas Geyer May 14 '13 at 16:24
add comment

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.