# running maximum of brownian motion and reflected brownian motion

Hi I am learning the theory of Brownian Motion using Morters and Peres' book (http://www.stat.berkeley.edu/~peres/bmbook.pdf).

Let $B$ be 1-dim standard Brownian motion and $M(t):=\max_{0\le s\le t} B(s)$.

In the book Theorem 2.18 says $\mathbb{P}\{M(t)>a\}=\mathbb{P}\{|B(t)|>a\}$ for any $a>0$.

To me this tells that $M\overset{d}{=}|B|$.

On the other hand, Theorem 2.31 says that the process $M-B$ is a reflected Brownian motion, in particular $M-B\overset{d}{=}|B|$.

Combining these two results together, does it mean that $M\overset{d}{=}M-B$. To me this seems weird but I am pretty sure I must mess up with something fundamental. Can anyone please point that out? Thanks!

• Although it seems weird to you, your argumentation is correct. :) (And, actually, the result is not that suprising. It follows directly from the invariance of Brownian motion under time reversal.)
– saz
Commented Mar 5, 2015 at 14:32
• Could you please elaborate a little bit more on the direct argument of using time reversal?
– Tim
Commented Mar 5, 2015 at 14:49

First of all: Yes, your argumentation is correct; the statement

$$M-B \stackrel{d}{=} M$$

holds true.

A direct proof goes like that: Let $(B_t)_{t \geq 0}$ be a Brownian motion. For fixed $T>0$, the process $(W_t)_{t \leq T}$ defined by

$$W_t := B_{T-t}-B_T, \qquad t \leq T,$$

is also a Brownian motion. Consequently,

\begin{align*} M_T -B_T &= \sup_{t \leq T} B_t - B_T = \sup_{t \leq T} (B_t-B_T) = \sup_{t \leq T} (B_{T-t} - B_T) \\ &= \sup_{t \leq T} W_t \stackrel{d}{=} \sup_{t \leq T} B_t = M_T. \end{align*}

(In the last step we have used that both $(B_t)_{t \leq T}$ and $(W_t)_{t \leq T}$ are Brownian motions and therefore the supremum is equal in distribution.)

• Note that this establishes a correct identity at a fixed time $T$, but the two processes do not have the same joint laws. Commented Jun 25, 2022 at 7:39

What makes the Levy identity $$M-B \stackrel{d}{=} |B| \tag{*}$$ so remarkable is that it holds for the two processes, i.e. the LHS of (*) has the same joint distribution at any sequence $$t_1 as the RHS, and both have continuous paths a.s.

But $$M \stackrel{d}{\ne} |B|$$ as processes, since the LHS is constant on intervals and the RHS is not. What does hold is for each fixed $$t$$, we have $$M(t) \stackrel{d}{=} |B|(t)$$ but the joint distributions on two points $$t_1 differ: $$(M(t_1),M(t_2)) \stackrel{d}{\ne} (|B (t_1)|,|B(t_2)|) \,.$$

Similarly, $$M(T)-B(T) \stackrel{d}{=} M(T)$$ holds for each fixed $$T$$ as shown in the answer given by @saz, but $$(M(t_1)-|B (t_1)|,M(t_2)-|B(t_2)|) \stackrel{d}{\ne} (M(t_1),M(t_2))$$ for $$t_1, so certainly as processes $$M-B \stackrel{d}{\ne} M\,.$$