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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?

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    $\begingroup$ math.stackexchange.com/questions/4422881/… $\endgroup$
    – user408858
    Mar 21 at 8:10
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    $\begingroup$ 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. $\endgroup$
    – user408858
    Mar 21 at 8:19
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    $\begingroup$ @user408858 Thank you. Feel free to write your comment as an answer for me to accept. $\endgroup$ Mar 21 at 8:26

2 Answers 2

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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 $.

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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.

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