# Integral of $M^\text{*} - M$ with respect to $M^\text{*}$ is zero for $M^\text{*}$ the running maximum of $M$ a continuous local martingale

Given $$M$$ a continuous local martingale, and $$M^\text{*} = \sup_{0 \leq s \leq t} M_s$$ its running maximum, we consider the finite variation integral $$I_T:= \int_0^T (M^\text{*}_s - M_s) \, \text{d}M^\text{*}_s$$ I'm seeking to show this integral is 0 a.s.. Intuitively, this is since $$(M^\text{*}_s - M_s) \, \text{d}M^\text{*}_s$$ is zero: if we're currently at our maximum then the bracket term zero, and otherwise $$M^\text{*}_s$$ is constant near $$s$$.

A little bit more rigorously, we have that is $$I_T = \lim_{n \to \infty} \sum_{[s,t]\in \pi_n} (M^\text{*}_{t,s} - M_{t,s})M^\text{*}_{t,s}$$ (where the limit is in probability, and for a sequence of partitions $$\pi_n$$ with mesh tending to 0, and where we write $$M_{t,s} = M_t - M_s$$ and similarly for $$M^\text{*}$$) and that eventually we'll have "sufficiently small" (this is the part which I think needs more details) $$t-s$$. Then either we're reaching a new maximum, so that $$M^\text{*}_t = M_t$$ and $$M^\text{*}_s = M_s$$ and the difference in brackets is $$0$$. Or we're not reaching a new maximum, and so $$M^\text{*}_{t,s} = 0$$.

I think this isn't sufficient at the moment, since even if we could suppose $$M$$ is monotone on sufficiently small intervals we still need to choose our partitions $$\pi_n$$ uniformly in $$\omega \in \Omega$$ (for the convergence in probability).

I've also tried applying Itô's on $$MM^\text{*}$$, but it reduces this to a similar problem of reasoning for sufficiently small $$t-s$$ for a partition dependent on the outcome for the $$\int M^\text{*} \, \text{d}M$$ term.

(I've asked this question on MSE here, but gotten little traction. I thought Overflow might be more appropriate since I'm seeking to formalize and fill the gaps of the intuitive ideas I have - let me know if this isn't appropriate, for instance if the question particularly belongs on either site)

• This problem for Brownian motion is I think prop vi.1.3 in revuz and yor, although it is stated for local time
– mike
Commented Jun 14, 2023 at 15:45
• @mike Thank you for your comment! I'm not seeing how to apply it in this case - I'm not very familiar with local times. It feels like I would want to apply this with (in their notation) $a=0$ and $X_t = M^\text{*}_t - M_t$ but I'm not sure how this corresponds to $dL_t^a$... Do you have any quick words about intuition in interpreting this local time? Commented Jun 14, 2023 at 16:11
• I thought , without saying, that if you use the reflected process $W^*_t - W_t$, for which the running max, $W^*_t$ is the local time, then it was exactly the same problem for brownian motion.
– mike
Commented Jun 15, 2023 at 5:14

The integral with regard do $$\mathrm{d}M^*$$ is a pathwise Stieltjès integral, so the question is an analysis problem.
Let $$f : \mathbb{R}_+ \to \mathbb{R}$$ be any continuous function, $$F$$ its current maximum, and $$\mu$$ the Stieltjès measure associated to $$F$$
One checks that $$F$$ is also continuous, so $$O := \{s \in \mathbb{R}_+ : F(s)-f(s)>0\}$$ is open subset in $$\mathbb{R_+}$$ and contained in $$\mathbb{R_+}^*$$ since $$F(0)=f(0)$$. Hence it is an at most countable union of disjoint open intervals. On each one of these open intervals, $$F$$ remains constant, so the $$\mu$$-measure of this interval is $$0$$. As a result $$\mu(O)=0$$, so $$F-f$$ is null $$\mu$$-almost everywhere and the integral $$\int(F-f) \mathrm{d}\mu$$ is $$0$$.