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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.

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1 Answer 1

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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.


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 $\sigma(s, X_s)$.

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