This is a follow-up to my previous question on MathOverflow.

Is there a way to combine the Dambis-Dubins-Schwarz theorem and the Martingale Representation Theorem to get the following result?

Let $(M_t, t\ge 0)$ be a continuous $(\mathcal{F}_t)_{t \ge0}$ local martingale, then there exists some Brownian motion $B$ and some progressively measurable process $\xi$ on some probability space such that $$M_t = \int_0^t \xi_s dB_s = \large{ B_{\int_0^t \xi_s^2 ds}} \quad \forall t \in [0,\infty)$$

In particular it would follow that $\langle M \rangle_t=\int_0^t \xi_s^2 ds$ (by Ito's Isometry?).

This is a follow-up to my previous question on MathOverflow.

Is there a way to combine the Dambis-Dubins-Schwarz theorem and the Martingale Representation Theorem to get the following result?

Let $(M_t, t\ge 0)$ be a continuous $(\mathcal{F}_t)_{t \ge0}$ local martingale, then there exists some Brownian motion $B$ and some progressively measurable process $\xi$ on some probability space such that $$M_t = \int_0^t \xi_s dB_s = \large{ B_{\int_0^t \xi_s^2 ds}} \quad \forall t \in [0,\infty)$$

In particular it would follow that $\langle M \rangle_t=\int_0^t \xi_s^2 ds$ (by Ito's Isometry?).

This is a follow-up to my previous question on MathOverflow.

Is there a way to combine the Dambis-Dubins-Schwarz theorem and the Martingale Representation Theorem to get the following result?

Let $X$ be a continuous local martingale (under the natural filtration), then there exists some Brownian motion $B$ and some progressively measurable process $\xi$ on some probability space such that $$X_t = \int_0^t \xi_s dB_s = \large{ B_{\int_0^t \xi_s^2 ds}} \quad \forall t \in [0,\infty)$$

In particular it would follow that $\langle X \rangle_t=\int_0^t \xi_s^2 ds$.

This is a follow-up to my previous question on MathOverflow.

Is there a way to combine the Dambis-Dubins-Schwarz theorem and the Martingale Representation Theorem to get the following result?

Let $(M_t, t\ge 0)$ be a continuous $(\mathcal{F}_t)_{t \ge0}$ local martingale, then there exists some Brownian motion $B$ and some progressively measurable process $\xi$ on some probability space such that $$M_t = \int_0^t \xi_s dB_s = \large{ B_{\int_0^t \xi_s^2 ds}} \quad \forall t \in [0,\infty)$$

In particular it would follow that $\langle M \rangle_t=\int_0^t \xi_s^2 ds$ (by Ito's Isometry?).