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!
 A: 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 $t_1<t_2<\ldots<t_n$ 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 $t_1<t_2$ 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 $t_1<t_2$, so certainly as processes
$$M-B \stackrel{d}{\ne} M\,.$$
A: 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.)
