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.


1 Answer 1


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.

  • $\begingroup$ 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 $M_t=\int_0^t a_s dL_s$? $\endgroup$
    – Fred Li
    Jan 16, 2022 at 12:03
  • 1
    $\begingroup$ 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. $\endgroup$
    – Kurt G.
    Jan 16, 2022 at 16:35

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