 Solution

Let $D_n=M_n-M_{n-1}$. Given ${\mathcal{F}}_{n-1}$, the new information at time $n$ is only the independent symmetric sign ${\xi }_n$. Thusfor some ${\mathcal{F}}_{n-1}$-measurable $C_n,B_n$. Taking conditional expectation and using the martingale property gives $C_n=0$. More explicitly, one can takewhich is predictable. ThereforeThis is the predictable representation in a Rademacher filtration.