# A continuous local martingale $M$ is constant on an interval if $\langle M\rangle$ is

## Problem

Let $$M\in\mathbb{M}^{loc}_C$$, i.e. $$M$$ is an (a.s.) continuous local martingale with $$M_0=0$$. Show that $$M$$ is constant on an interval $$[a,b]$$ with $$0\leq a < b$$ if $$\langle M\rangle$$ is constant on $$[a,b]$$.

## My Attempt

Let $$\left\langle M\right\rangle$$ be constant on $$[a,b]$$, i.e. $$\forall t\in[a,b]: \left\langle M\right\rangle_t = \left\langle M\right\rangle_a \text{(a.s.)}$$. Since $$M$$ is an (a.s.) continuous (!) local martingale, there exists a reducing sequence $$(\tau_n)_{n\in\mathbb{N}}$$ of $$M$$ such that $$M^{\tau_n}$$ is a square integrable martingale. Let $$t\in(a,b]$$. Square integrable martingales like $$M^{\tau_n}$$ have the following useful property: \begin{align*} &\mathbb{E}\left[(M^{\tau_n}_t - M^{\tau_n}_a)^2\vert\mathcal{F}_a\right] = \mathbb{E}\left[\left(M^{\tau_n}_t\right)^2 - \left(M^{\tau_n}_a\right)^2 \vert \mathcal{F}_a\right]\\ \Rightarrow\ &\mathbb{E}\left[\mathbb{E}\left[(M^{\tau_n}_t - M^{\tau_n}_a)^2\vert\mathcal{F}_a\right]\right] = \mathbb{E}\left[\mathbb{E}\left[\left(M^{\tau_n}_t\right)^2 - \left(M^{\tau_n}_a\right)^2 \vert \mathcal{F}_a\right]\right]\\ \Leftrightarrow\ &\mathbb{E}\left[(M^{\tau_n}_t - M^{\tau_n}_a)^2\right] = \mathbb{E}\left[\left(M^{\tau_n}_t\right)^2 - \left(M^{\tau_n}_a\right)^2\right] \end{align*} $$M^{\tau_n}$$ being square integrable also implies that $$(\left(M^{\tau_n}_t\right)^2 - \left\langle M^{\tau_n}\right\rangle_t)_{t\geq 0}$$ is a martingale and therefore its martingale condition holds: \begin{align*} &\mathbb{E}\left[\left(M^{\tau_n}_t\right)^2 - \left\langle M^{\tau_n} \right\rangle_t \vert \mathcal{F}_a\right] = \left(M^{\tau_n}_a\right)^2 - \left\langle M^{\tau_n} \right\rangle_a\\ \Rightarrow\ &\mathbb{E}\left[\mathbb{E}\left[\left(M^{\tau_n}_t\right)^2 - \left\langle M^{\tau_n} \right\rangle_t \vert \mathcal{F}_a\right]\right] = \mathbb{E}\left[\left(M^{\tau_n}_a\right)^2 - \left\langle M^{\tau_n} \right\rangle_a\right]\\ \Leftrightarrow\ &\mathbb{E}\left[\left(M^{\tau_n}_t\right)^2 - \left\langle M^{\tau_n} \right\rangle_t\right] = \mathbb{E}\left[\left(M^{\tau_n}_a\right)^2 - \left\langle M^{\tau_n} \right\rangle_a\right]\\ \Leftrightarrow\ &\mathbb{E}\left[\left(M^{\tau_n}_t\right)^2\right] - \mathbb{E}\left[\left\langle M^{\tau_n} \right\rangle_t\right] = \mathbb{E}\left[\left(M^{\tau_n}_a\right)^2\right] - \mathbb{E}\left[\left\langle M^{\tau_n} \right\rangle_a\right]\\ \Leftrightarrow\ &\mathbb{E}\left[\left(M^{\tau_n}_t\right)^2\right] - \mathbb{E}\left[\left(M^{\tau_n}_a\right)^2\right] = \mathbb{E}\left[\left\langle M^{\tau_n} \right\rangle_t\right] - \mathbb{E}\left[\left\langle M^{\tau_n} \right\rangle_a\right]\\ \Leftrightarrow\ &\mathbb{E}\left[\left(M^{\tau_n}_t\right)^2-\left(M^{\tau_n}_a\right)^2\right] = \mathbb{E}\left[\left\langle M^{\tau_n} \right\rangle_t-\left\langle M^{\tau_n} \right\rangle_a\right]\\ \end{align*} Putting it both together yields $$\mathbb{E}\left[(M^{\tau_n}_t - M^{\tau_n}_a)^2\right] = \mathbb{E}\left[\left\langle M^{\tau_n} \right\rangle_t-\left\langle M^{\tau_n} \right\rangle_a\right]$$. It would be really nice if $$\left\langle M^{\tau_n} \right\rangle_t = \left\langle M \right\rangle_{t\wedge\tau_n}$$ were the case in order to make use of the fact that $$\left\langle M\right\rangle$$ is constant on $$[a,b]$$. We then would have $$\left\langle M^{\tau_n} \right\rangle_t-\left\langle M^{\tau_n} \right\rangle_a = \left\langle M \right\rangle_{t\wedge\tau_n} - \left\langle M \right\rangle_{a\wedge\tau_n}$$. For all $$c\geq 0$$ we have: \begin{align*} \left\langle M \right\rangle_{t\wedge c} - \left\langle M \right\rangle_{a\wedge c} = \begin{cases} \left\langle M \right\rangle_{c} - \left\langle M \right\rangle_{c} = 0 & (c < a < t)\\ \left\langle M \right\rangle_{c} - \left\langle M \right\rangle_{a} = \left\langle M \right\rangle_a - \left\langle M \right\rangle_{a} = 0 & (a \leq c < t)\\ \left\langle M \right\rangle_t - \left\langle M \right\rangle_a = \left\langle M \right\rangle_a - \left\langle M \right\rangle_{a} = 0 & (a < t \leq c) \end{cases} \end{align*} It follows that $$\left\langle M \right\rangle_{t\wedge\tau_n} - \left\langle M \right\rangle_{a\wedge\tau_n}=0$$. Putting everything together yields $$\mathbb{E}\left[(M^{\tau_n}_t - M^{\tau_n}_a)^2\right] = 0$$ meaning that $$M^{\tau_n}_t = M^{\tau_n}_a$$ (a.s.). Taking the limit on both sides would yield $$M_t = M_a$$ (a.s.) which is the desired result.

## Questions

Does $$\left\langle M^{\tau_n} \right\rangle_t = \left\langle M \right\rangle_{t\wedge\tau_n}$$ hold? If yes, is my proof correct?

• math.stackexchange.com/questions/4422881/… Commented Mar 21 at 8:10
• Yes, the quadratic variation is unique by the Doob-Meyer-decomposition. Since $M^2-\langle M\rangle$ is a local martingale for a local martingale $M$, it necessarily holds $\langle M^{\tau_n } \rangle=\langle M \rangle^{\tau_n }$. Your proof looks correct to me. Commented Mar 21 at 8:19
• @user408858 Thank you. Feel free to write your comment as an answer for me to accept. Commented Mar 21 at 8:26

Yes, the quadratic variation is unique by the Doob-Meyer-decomposition. Since $$M^2-\langle M\rangle$$ is a local martingale for a local martingale $$M$$, it necessarily holds $$\langle M \rangle ^{\tau_n}=\langle M ^{\tau_n}\rangle$$.

As you indicate, the basic hypothesis is stable under stopping, so we can take $$M$$ to be a bounded martingale. Let $$S\le T$$ be stopping times such that $$\langle M\rangle_T = \langle M\rangle_S$$. then $$E\left(M^2_T|\mathcal F_S\right)=E\left(M^2_T-\langle M\rangle_T|\mathcal F_S\right)+\langle M\rangle_S=M^2_S-\langle M\rangle_S+\langle M\rangle_S=M^2_S.$$ Consequently, \eqalign{ E\left[(M_T-S_S)^2\right] &=E[M^2_T]+E[M^2_S]-2E[M_SM_T]\cr &=2E[M_S^2]-2E\left[ E[M_T|\mathcal F_S]M_S\right]\cr &=2E[M_S^2]-2E[M_S^2]=0,\cr} so $$M_T=M_S$$ a.s.