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Let $M_t$ be a continuous time martingale, and assume its quadratic variation is identically zero with some positive probability less than $1$.

To be more precise, assume there exists some event $E$ with $0 < \mathbb P(E) < 1$ such that for almost every $\omega \in E$, $\langle M, M\rangle_t (\omega) = 0$ for all $t \geq 0$.

Question: Does it follow that $M$ is almost surely constant in time on $E$? That is, $M_t = M_0$ for all $t > 0$ a.s. on $E$.

Remark: For continuous martingales, this follows directly from the Dambis-Dubins-Schwartz theorem.

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    $\begingroup$ how about we use the Ito isometry for martingales $$\operatorname{E}\left(\left(\int_0^t H_{s}\,dM_{s}\right)^2\right) = \operatorname{E}\left(\int_0^tH_{s}^2\,d[M]_{s}\right).$$ but for $H$ constantly equal to the indicator of $E$ at least in the case of E being measurable with respect to the filtration of M? $\endgroup$
    – Thomas Kojar
    Commented Oct 20, 2022 at 22:11
  • $\begingroup$ Oh damn, that works indeed.. $\endgroup$
    – Nate River
    Commented Oct 20, 2022 at 22:58
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    $\begingroup$ @Thomas Kojar: Such an $H$ is not likely to be adapted. $\endgroup$
    – John Dawkins
    Commented Oct 22, 2022 at 16:48
  • $\begingroup$ @JohnDawkins I agree. Perhaps some conditional version might do the trick i.e. $H(t):=E[\chi_E|Ft]$. $\endgroup$
    – Thomas Kojar
    Commented Oct 23, 2022 at 22:35

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To answer your question, the following Lengart's inequality is useful: (Please refer to S. W. He et al., Semimartingale Theory and Stochastic Calculus, Sci. Press and CRC(1992), p.239, Theorem 9.23.)

Theorem Let $ X $ be an adapted cadlag process, dominated by an predictable process $ A $. Then for arbitrary constants $C>0, d>0$, stopping time $ T $ and measurable set $ H $ we have \begin{equation*} \mathsf{P}(H\cap [X^\ast_T\ge C])\le \frac{1}{C}\mathsf{E}[A_T\wedge d]+ \mathsf{P}(H\cap [A_T\ge d]). \tag{1} \end{equation*} Hence for the continuous time martingale $M$(if $ M $ is also a locally square integrable martingale), $M^2$ is dominated by its pridictable quadratic variation $ \langle M \rangle $. Using (1) for $ H=E $, it follows \begin{align*} \mathsf{P}(E\cap [M^{\ast2}_T\ge C])&\le \frac{1}{C}\mathsf{E}[\langle M \rangle_T\wedge d]+ \mathsf{P}(H\cap [\langle M \rangle_T\ge d])\\ & = \frac{1}{C}\mathsf{E}[\langle M \rangle_T\wedge d] \le \frac{d}{C}, \quad \forall C>0, d>0. \end{align*} Now let $d\downarrow0$ to get \begin{equation*} \mathsf{P}(E\cap [M^{\ast2}_T\ge C])=0, \qquad \forall C>0. \end{equation*} Further more letting $ C\downarrow 0 $ to get that $ M=0 $ almost surely on $ E $.

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  • $\begingroup$ Hi, how do you obtain that $M^2$ is dominated by its predictable QV? $\endgroup$
    – Nate River
    Commented Nov 2, 2022 at 8:36
  • $\begingroup$ @Nate River Thank you for your question. If $ M $ is a square integrable martingale, then $ M^2-\langle M\rangle $ is a local martingale and $\mathsf{E}[M^2_T]=\mathsf{E}[\langle M\rangle_T] $ for every bounded stopping time $ T $. $\endgroup$
    – JGWang
    Commented Nov 3, 2022 at 1:56
  • $\begingroup$ Hmm, but I don’t see necessarily that this means that $M^2$ is a.s. dominated by $\langle M \rangle$. Sorry, I may be missing something obvious. $\endgroup$
    – Nate River
    Commented Nov 3, 2022 at 10:24
  • $\begingroup$ @NateRiver Pls refer to the book, mentioned above, p.238, Example 9.21. $\endgroup$
    – JGWang
    Commented Nov 4, 2022 at 1:58

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