# Martingale representation theorem for symmetric random walk

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?

• $M(t)$ is not continuous. But for distrete time, we can use the stoping time $T=\inf{t:B_t =\pm 1}$ and $H(s)=1_{s\leq T}$ – RaphaelB4 Dec 4 '18 at 10:06

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., $$\mathbb{E}[M(1) | M(1/2)] \neq M(1/2).$$