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Let $(B_t)_{t\ge 0}$ be a one-dimensional Brownian motion and set $M_t = \sup_{s\le t} B_s$. Denote the $\xi_t$ the largest zero of $B_s$ before time $t$ and by $\eta_t$ the largest zero of $Y_s = M_s - B_s$ before time $t$. Show that $\xi_t \sim \eta_t$.

I know that $M-B$ is a Markov process with the same transition as $|B|$. Hence they have the same finite dimensional distributions.

But, how do we conclude from this that $\xi_t$ has the same distribution as $\eta_t$?

I am not fully convinced with the above argument because, e.g. $\xi_t = \sup\{s \le t: |B_s| = 0\}$, so to find out $\xi_t$ we need to know $|B_s|$ at all $s \in [0,t]$, which is uncountable. And similarly for $M_s - B_s$, $s\in [0,t]$. Hence, how does the equivalence of the distributions of $|B|$ and $M-B$ at all finite steps in $[0,t]$ conclude that $\xi_t$ and $\eta_t$, which require knowledge of continuous times at $[0,t]$, have the same distribution?

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By definition, $$\xi_t(\omega) < s \iff \forall r \in [s,t]: |B_r(\omega)| \neq 0.$$ Since Brownian motion has (a.s.) continuous sample paths, it attains its minimum on compact sets. Thus, $$\xi_t(\omega)<s \iff \inf_{r \in [s,t]} |B_r(\omega)|>0 \iff \inf_{r \in [s,t] \cap \mathbb{Q}} |B_r(\omega)|>0.$$ Consequently, $$\mathbb{P}(\xi_t < s) = \mathbb{P} \left( \inf_{r \in [s,t] \cap \mathbb{Q}} |B_r| > 0 \right),$$ i.e. the distribution of $\xi_t$ is uniquely determined by the finite-dimensional distributions of $|B|$. An analogous reasoning works for $\eta_t$ and $M-B$, which gives, in conclusion, that $\xi_t \sim \eta_t$.

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  • $\begingroup$ Why is $P(\inf_{r\in [s,t] \cap \mathbb{Q}} |B_r| > 0)$ determined by the finite-dimensional distributions of $|B|$ when it involves countably many $|B_r|$'s? $\endgroup$ Jul 6, 2020 at 9:47
  • $\begingroup$ And does it mean that such arguments only apply to processes with continuous paths? That is, if say, $X$ and $Y$ are processes with the same fdd's, but without continuous paths, then we may not have $\xi_t \sim \eta_t$? $\endgroup$ Jul 6, 2020 at 9:48
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    $\begingroup$ @nomadicmathematician Take an enumeration $(q_k)_k$ of all rationals in $[s,t]$, then by the monotone convergence theorem $$\mathbb{P} \left( \inf_{r \in [s,t] \cap \mathbb{Q}} |B_r| \geq m \right) = \lim_{N \to \infty} \mathbb{P} \left( \inf_{k \leq N} |B_{q_k}| \geq m \right).$$ This shows that the distribution of $\inf_{r \in [s,t] \cap \mathbb{Q}} |B_r|$ is uniquely detrmined by the finite dimensional distributions of $|B|$. $\endgroup$
    – saz
    Jul 6, 2020 at 10:06
  • $\begingroup$ I understand your argument now, infimum over countable is just the limit of minimum over finite and we can take the limit outside of the probability, so it is determined by fdd's. However, I just found from this link: math.stackexchange.com/questions/3194284/… for $\mathbb{R}^d$-valued processes the fdd's determine the law of the whole process either by Kolmogorov's extension theorem, or Yuval's answer, so for the second question I had, I think we don't need the restriction of continuous paths. $\endgroup$ Jul 6, 2020 at 10:07
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    $\begingroup$ And yes, typically you will need some regularity assumptions on the sample paths to make such a reasoning work. (Otherwise you might run into trouble concerning measurability, e.g. already showing that $\xi_t$ is measurable is getting messy without any assumptions on the regularity of the sample pathjs.) $\endgroup$
    – saz
    Jul 6, 2020 at 10:08

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