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

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?

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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

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
$$\int_0^{\infty} \int_0^m f(M_t = m, \; B_t = b) \; db \; dm = 1 \; .$$
"I argue that $dM_t$ and $dB_t$ are independent" Argue all you want, this will still be false. –  Did Aug 13 at 9:19
Actually, $\mathbb{P} ( M_t \geq k) = \mathbb{P} (\left| B_t \right| \geq k)$ that means $M_t \overset{\mathcal{L}}{=} \left|B_t \right| \ \ , \forall t \geq 0$