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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!

MJD
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Higgs88
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2 Answers2

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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 $c=d$.)

Lukas Geyer
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  • 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 –  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
  • Why $\mathbb{P}[A-C = B_c - B_a]=0$? Could you formalize this argument? – lele Nov 08 '15 at 11:09
  • @leticia: I guess the easiest way to formally argue is that the distribution of the difference $(A-C) - (B_c - B_a)$ is the convolution of the distributions of $A-C$ and $B_a - B_c$ (by independence), and that the convolution of two continuous random variables is also continuous. This shows that it has no atoms and so it is $0$ with probability $0$. – Lukas Geyer Nov 09 '15 at 21:23
  • Why is $(X_t)_t$ independent of $B_a, B_c$? – unwissen Oct 31 '21 at 16:10
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    @unwissen: Looking at it again, independence of $(X_t)$ from $B_a$ is just the basic property that increments of Brownian motion are independent of the past. However, independence from $B_c$ might actually not be true, so the argument might have to be modified. – Lukas Geyer Nov 01 '21 at 19:51
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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 $

focusonimprove333
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  • I don't agree that both sides have gaussian densities. For instance if $a=0$, the first supremum is non negative. – Kieran McShane Oct 26 '23 at 15:59
  • However, $\sup_{t\in [a,b]} B_t= \sup_{t\in [0,b-a]} (B_{t+a}-B_a)+B_a$ which has the same distribution as $|B_{b-a}| + B_a$ (due to simple Markov property and reflection principle). So it has a density with no atoms. – Kieran McShane Oct 26 '23 at 16:12