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

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?

• My intuition says no. Feb 14, 2022 at 19:05

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.
• What does $\mathrm{d} F \perp \mathrm{d} \lambda$ mean? Also, where can I find a proof of $(1)$ ? Feb 15, 2022 at 14:44
• Thank you, could you give an example of $F$ that satisfy your conditions ? Feb 16, 2022 at 1:35
• $F$ --Cantor-Lebesgue function, c.f. Royden, H. L., Real Analysis, 4th Ed, (2010), p.50. Feb 16, 2022 at 3:56