Relation between $\text{d}M_t$ and $\text{d}B_t$ when $M_t=\max_{0\leq s\leq t}B_s$

Question

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?

Answer 1

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$

Answer 2

You can look up the joint density of $B_t$ and $M_t$ in Shreve's book, Vol. II: $$ f(M_t = m, \; B_t = b) = \frac{2(2m-b)}{t\sqrt{2 \pi t}} e^{-\frac{(2m-b)^2}{2t}}, \text{where } b \leq m, m > 0 $$

I argue that $dM_t$ and $dB_t$ are independent as

$$ \int_0^{\infty} \int_0^m f(M_t = m, \; B_t = b) \; db \; dm = 1 \; .$$