Let $X(t)$ be a martingale w.r.t. filtration generated by Brownian motion $B(t)$. There is a well-known theorem that states that there is a unique adapted process $H(t)$ such that $$ X(t) = \int_0^t H(s)dB(s).$$ Let's define step process $$M(t)=2 \sum_{j=1}^{\lfloor t \rfloor} 1_{\{ B(j)-B(j-1) > 0 \}} - \lfloor t \rfloor.$$ It's just a simple martingale, generated by fair coin tosses, just written in another way. One can see that $B(t)$ is the only source of uncertainty. What is $H(t)$ for $M(t)$ then? Or am I missing something and M.R.T. is not applicable here?
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$\begingroup$ $M(t)$ is not continuous. But for distrete time, we can use the stoping time $T=\inf{t:B_t =\pm 1}$ and $H(s)=1_{s\leq T}$ $\endgroup$ – RaphaelB4 Dec 4 '18 at 10:06
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I've caught myself into a trap here: the reason why I couldn't find a martingal representation is that $M(t)$ is not a martingale. For example: $$ \mathbb{E}[M(1) | M(1/2)] = 2 \mathbb{P}(B(1) > 0 |B(1/2)) - 1. $$ And if $B(1/2) \neq 0$, which is true a.s., $\mathbb{E}[M(1) | M(1/2)] \neq M(1/2).$
