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?



Martingale representation theorem for symmetric random walk
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., $\mathbb{E}[M(1) | M(1/2)] \neq M(1/2).$


