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Let $W$ denote a Wiener process, $\displaystyle M_t = \max_{0 \le s \le t} W_s$ its running maximum. The martingale representation of $M$ is known explicitly:

$$M_T = \sqrt{\frac{2T} \pi} + \int_0^T 2\left(1 - \phi\left(\frac{M_t-W_t} {\sqrt{T-t}} \right)\right) \, \mathrm dW_t, $$

where $\phi$ denotes the pdf of a standard normal variable.

Question: do we have an explicit martingale representation for the running maximum of the Brownian bridge (conditioned to be zero at both ends of $[0,1]$)?

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  • $\begingroup$ Seems like this ought to be obtainable from the Clark-Ocone formula. $\endgroup$
    – Nate Eldredge
    Mar 29, 2017 at 19:49
  • $\begingroup$ Thanks for your answer. How do you compute the Malliavin derivative of the running maximum? Is there a way to link it to the distribution of the first hitting time, like for Brownian motion? $\endgroup$
    – Tartrate
    Mar 29, 2017 at 21:26
  • $\begingroup$ I don't know what it is off the top of my head, but one possible approach might be to compute the Malliavin derivative of $\max(X_{t_1}, \dots, X_{t_n})$ or some smoothed version of it, which would be a cylinder function, and then pass to the limit as mesh size tends to zero. I don't know what the relationship with first hitting time would be. $\endgroup$
    – Nate Eldredge
    Mar 30, 2017 at 14:45
  • $\begingroup$ If we recognize that a Brownian bridge can be written as $B(t)=W(t)-\frac{t}{T}W(T)$ where $W(t)$ is a Wiener process, and the maximum process is equivalent to the minimum of reflected process, is it just a simple linear combination or am I missing something? $\endgroup$
    – Henry.L
    May 4, 2017 at 15:30
  • $\begingroup$ What I mean is that $max_{[0,T]}B(t)=max_{[0,T]}W(t)-min_{[0,T]}\frac{t}{T}W(T)$, and the $-min_{[0,T]}\frac{t}{T}W(T)=max_{[0,T]}\frac{t}{T}W(T)$ $\endgroup$
    – Henry.L
    May 4, 2017 at 15:32

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