Can a martingale always be written as the integral with regard to Brownian motion?

Question

Let $(M_t)_{t \in \mathbb R^+}$ be a continuous martingale with regard to a filtration $(\mathcal F_t)$ generated by a continuous stochastic process $(Y_t)$.

Is it true that there exists a Brownian motion $(B_t)$ adapted to $(\mathcal F_t)$ and a stochastic process $(X_t)$ such that $M_t=M_0+\int_0^t X_s d B_s$ ?

This is nearly the martingale representation theorem: Let $(B_t)$ be a standard Brownian motion defined on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and $\left\{\mathcal{F}_{t}\right\}_{t \geq 0}$ be its natural filtration. Then, every $\left\{\mathcal{F}_{t}\right\}$-local martingale $M$ can be written as \begin{equation*} M_t=M_{0}+\int_0^t \xi_s d B_s \end{equation*} for a predictable, $B$-integrable, process $\xi$ (statement taken from: https://almostsuremath.com/2010/05/25/the-martingale-representation-theorem/).

In my statement, $(\mathcal F_t)$ is not necessarily generated by Brownian motion. Does it still work?

Answer 1

If a continuous martingale $(M_t)_{t\in\mathbb{R}_+}$ could be written as a stochastic integral with respect to Brownian motion $(B_t)$: \begin{equation*} M_t=M_0+\int^t_0 X_s\,\text{d}B_s, \end{equation*} where $X$ is predictable. Then \begin{equation*} \langle M\rangle_t=\int_0^tX_s^2\,\text{d}s. \end{equation*} Furthermore, from above equality and Radon-Nikodym theorem, for Lebesgue measure $\lambda$, \begin{equation*} \text{d}\langle M\rangle \ll \text{d}\lambda. \quad(\text{absolute continuous of measures})\tag{1} \end{equation*} Now suppose $F$ is a continuous increasing function with $F(0)=0$ and $\text{d}F\bot\text{d}\lambda$(mutual singular), then there exist $\{Z(t), t\ge 0\}$ a continuous Gaussian process with \begin{equation*} \mathsf{E}[Z_t]=0,\qquad \mathsf{E}[Z_sZ_t]=F(s\wedge t), \quad s,t\ge0. \end{equation*} Since $Z$ is a continuous process with independent increments and $\mathsf{E}[Z_t]=0 $, hence $Z$ is also a continuous martingale with $\langle Z\rangle =F $, it doesn't satisfy (1), so continuous martingale $Z$ doesn't be written as a stochastic integral w.r.t. BM.