Doob-Meyer Decomposition of the range of a standard Brownian Motion

I am curious about whether there exists a Doob-Meyer decomposition for the range process $$R_t$$, or the squared range $$R_t^2$$ of a standard Brownian motion $$B_t$$, defined as: $$R_t = M_t-m_t,$$ where $$M_t := \sup_{0\leq s\leq t} B_s$$ and $$m_t := \inf_{0\leq s\leq t} B_s$$. Clearly, $$R_t$$ and hence $$R_t^2$$ is monotonically increasing and continuous in $$t$$, thus they are by design submartingales which should possess unique Doob-Meyer decompositions.

I know that by the Tanaka's equation, we have: $$|B_t| = \int_0^t \mathrm{sgn}(B_s)dB_s + L_t,$$ where $$L_t$$ is the Brownian local time at zero. Consequently, this provides the Doob-Meyer decomposition for $$|B_t|$$. Also, for $$B_t^2$$ this is even simpler: $$B_t^2 = 2\int_0^t B_s dB_s + t,$$ directly from Ito's lemma. Therefore, I was thinking whether these results can be easily extended to describe $$R_t$$ or $$R_t^2$$ in light of the well-known relation that $$M_t\overset{d}{=} |B_t| \overset{d}{=} M_t-B_t \overset{d}{=}-m_t \overset{d}{=} -B_t-m_t$$? I spent hours trying to find relevant results in the literature but could not find anything relevant. Any suggestions or hints are highly appreciated.

Since $$m_t=-\max_{0\le s\le t}(-B_s)$$ ($$-m_t$$ is a disguised max) it is enough to find the Dooob-Meyer decomposition of $$M_t=\max_{0\le s\le t}B_s\,.$$ But that's trivial. Since $$M_t$$ is inreasing its decomposition is $$M_t=0+A_t$$ where the increasing process $$A_t$$ is $$M_t$$ itself and the martingale part is zero.
• Ah I see, thank you! But then there should be some process say $a_t$ that satisfy $M_t=\int_0^t a_s ds$ or $M_t=\int_0^t a_s dL_s$? Jan 16, 2022 at 12:03
• Correct . The running max $M_t$ is increasing and so has bounded variation. It is therefore a perfect inegrator to do Stieltjes integration with. The $a_s$ that comes to my mind which you are looking for is unfortunately quite boring again: $a_s\equiv 1$ and $M_t=\int_0^t\,dM_s\,.$ This does however not rule out that there is another less boring $a$ and another $L\,.$ You seem to have good sense of finding interesting questions. Please let me know if you find something. Jan 16, 2022 at 16:35