Let $(X\_t)\_{0\le t\le 1}$ be a continuous Markov martingale (with respect to its natural filtration) s.t. $X\_0=0$ and $X\_1\in\{-1,1\}$. Can we prove the existence of some measurable function $\sigma: [0,1]\times \mathbb R\to\mathbb R\_+$ s.t.


$$X\_t=\int\_0^t\sigma(s,X\_s)dW\_s,\quad \forall 0\le t\le 1?$$


Here $(W\_t)\_{0\le t\le 1}$ denotes some Brownian motion.



Question on the martingale representation theorem
No. There are at least two reasons for that.


First, the underlying probability space $(\Omega, \mathcal F)$ and the natural filtration can be too small for the Brownian motion to exist. Indeed: consider the case when $X\_t$ is constant for $t \geqslant \tfrac12$, and $\Omega$ is just the space of paths of $X\_t$. If $W\_t$ and $\sigma$ existed, then $\sigma\{W\_t : t < \tfrac12\}$ would contain $\sigma\{X\_t : t < \tfrac12\} = \mathcal F \supseteq \sigma\{W\_t : t < 1\}$, and so $\sigma\{W\_t : t < \tfrac12\} = \sigma\{W\_t : t < 1\}$, a contradiction.


Another, more important, reason is that $X\_t$ can be too rough to be an Itô integral with respect to a Brownian motion. To see this, consider an arbitrary martingale $X\_t$, and modify its time using the Cantor's (devil's staircase) function $\phi(t)$: the process $X\_{\phi(t)}$ is again a continuous martingale, but it is piecewise constant on a set of full Lebesgue measure, so the corresponding function $\sigma$ would have to be zero almost everywhere, a contradiction.




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That said, I believe the following is likely to be true: if $X\_t$ is a continuous martingale *with respect to some Brownian filtration*, and additionally $X\_t$ is a Markov process, then it is an Itô integral with respect to the underlying Brownian motion, with the integrand of the form $\sigma(s, X\_s)$.


