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?
Relation between $\text{d}M_t$ and $\text{d}B_t$ when $M_t=\max_{0\leq s\leq t}B_s$
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stochastic-processes
brownian-motion
stochastic-calculus
stochastic-integrals
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0Do 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 $\lim_{t\rightarrow t_0} \frac{\mathbb{E}[f(M_t)]-f(M_{t_0})}{t-t_0}$ – 2012-12-21
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0No, since $M_t=\max_{0\leq s \leq t}B_s$, I am not sure how to find the distribution of $M$. What is $f$? – 2012-12-22
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0The 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. – 2012-12-28
2 Answers
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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$
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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
$$ \int_0^{\infty} \int_0^m f(M_t = m, \; B_t = b) \; db \; dm = 1 \; .$$
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0"I argue that $dM_t$ and $dB_t$ are independent" Argue all you want, this will still be false. – 2014-08-13