Take the 2-minute tour ×  Let $B_t$ be a standard Wiener motion.
What can we say about $\text{d}M_t$ and $\text{d}B_t$ when
$M_t=\max_{0\leq s\leq t}B_s$? Is there a relation?
share|improve this question Do you know how to relate $P(M_t\leq
x)$ to probabilities involving $B_t$? If you have this relation,
you can deduce the generator of $M_t$ by looking at
$\lim_{t\rightarrow t_0}
\frac{\mathbb{E}[f(M_t)]-f(M_{t_0})}{t-t_0}$ –  Alex R. Dec 21
'12 at 22:05 No, since $M_t=\max_{0\leq s \leq t}B_s$, I am not
sure how to find the distribution of $M$. What is $f$? –  Nick
Papadopoulos Dec 22 '12 at 17:34 The question is quite vague
(and, as such, should probably be closed) but the OP might be
referring to theorems relating (|B|,M) to a Brownian motion and
its local time at zero. –  Did Dec 28 '12 at 12:07 add comment
2 Answers  Actually, $\mathbb{P} ( M_t \geq k) = \mathbb{P}
(\left| B_t \right| \geq k)$ that means $ M_t
\overset{\mathcal{L}}{=} \left|B_t \right| \ \ , \forall t \geq
0$  share|improve this answer