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!
On
This question is old, but based on the previous answer by Lukas Geyer I think I came up with a satisfying proof:
Start with $sup_{t \in [c,d]}B_t = sup_{t \in [0,d-c]}B_{t+c} - B_c + B_c = (sup_{t \in [0,d-c]}B_{t+c} - B_c ) + B_c $ a.s. . The expression in brackets is independent of $\sigma(B_s ; s \leq c)$ because of Brownian Motion properties.
Now,
$ \{sup_{t \in [a,b]}B_t = sup_{t \in [0,d-c]}B_{t+c} - B_c + B_c \} = $ $ \{ sup_{t \in [a,b]}B_t - B_c = sup_{t \in [0,d-c]}B_{t+c} - B_c \}$.
The L.H.S. is measurable w.r.t. $\sigma(B_s ; s \leq c)$, so independent of the R.H.S. . Both sides have gaussian densities (the sup turns into a max due to the continuity of brownian paths), and so the result follows by the reasoning from the previous answer. Namely, the sum of these two r.v. has a continuous density (as the convolution of two continuous densities) and therefor no point masses. Therefor, $P(sup_{t \in [a,b]}B_t - B_c -sup_{t \in [0,d-c]}B_{t+c} - B_c = 0) = 0 $
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 $c=d$.)