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Theorem (Dambis, Dubins-Schwarz). If $M$ is a $\left(\mathscr{F}_t, P\right)$-continuous martingale vanishing at 0 and such that $\langle M, M\rangle_{\infty}=\infty$ and if we set $$ T_t=\inf \left\{s:\langle M, M\rangle_s>t\right\}, $$ then, $B_t=M_{T_t}$ is a $\left(\mathscr{T}_{T_t}\right)$-Brownian motion and $M_t=B_{\langle M, M\rangle_t}$.

Here is the Dambis Dubins Schwarz theorem. I am wondering if it holds when M becomes discrete martingale?

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  • $\begingroup$ What do you mean by "a discrete version of this theorem"? What do you mean by "it" in your "does it hold for discrete martingale?"? $\endgroup$
    – Iosif Pinelis
    Commented Apr 18, 2023 at 14:23
  • $\begingroup$ @IosifPinelis Does it also hold when M becomes a discrete martingale? $\endgroup$
    – neveryield
    Commented Apr 18, 2023 at 14:30
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    $\begingroup$ The answer is yes, with the cost that your filtration is large enough. By mathoverflow.net/questions/159496/… every discrete martingale can be embedded into a continuous one $\endgroup$
    – Fawen90
    Commented Apr 18, 2023 at 14:51
  • $\begingroup$ @Fawen90 That answers my question, thanks! $\endgroup$
    – neveryield
    Commented Apr 18, 2023 at 14:55

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