Paley Wiener stochastic integral

Question

Sorry for the stupid question, no answers necessary anymore!

let $(B_t)_{t\in [0,1]}$ be a standard Brownian motion and $F\in C[0,1]$ differentiable. Then the sequence (which is an easy version of the Paley Wiener stochastic integral)

$\xi_n=2^n\sum_{j=1}^{2^n}[F(\frac{j}{2^n}-F(\frac{j-1}{2^n})][B(\frac{j}{2^n})-B(\frac{j-1}{2^n})]$

Peres & Moerters (Brownian Motion) proof that $\xi_n$ is a martingale. They show that

$\xi_n-\xi_{n-1}=2^n\sigma_n\sum_{j=1}^{2^{n-1}}[2F(\frac{2j-1}{2^n})-F(\frac{2j-2}{2^n})-F(\frac{2j}{2^n})]Z(\frac{2j-1}{2^n})$

where $\sigma_n=2^{-(n+1)/2}$ and $Z(t)$ for $t$ binary rational are iid standard normal random variables. Then they only mention that this result shows that $\xi_n$ is a martingale (i think with respect to the Brownian filtration). Why is that so obvious? Thanks, nbt

EDIT: If we had $\xi_0=0$, then $\xi_n=\sum_{k=1}^n\xi_{k}-\xi_{k-1}$ and since $E_P[\xi_n-\xi_{n-1}|\mathcal{F}_{n-1}]=0$ we would have a martingale. But for $\xi_0=0$ I would have to demand $F(1)=0$ which was not mentioned by Peres and Moerters...

Answer 1

The argument is that the random variables $Z((2j-1)/2^n)$ are not just any random variables and the sigma-algebras $\mathcal F_n$ are not just any sigma-algebras... Each $Z((2j-1)/2^n)$ is proportional to $$Y=B(u+2v)+B(u)-2B(u+v),\qquad u=(j-1)/2^n,\qquad v=1/2^{n+1},$$ and $\mathcal F_n\subseteq\mathcal G_n$ where $$\mathcal G_n=\sigma(B(k/2^n);1\leqslant k\leqslant 2^n).$$ The crucial fact is the following.

Let $r\lt s\lt t$, then $E(B(s)-B(r)\mid B(r),B(t))=\dfrac{s-r}{t-r}(B(t)-B(r))$.

In particular, $B(u)$ and $B(u+2v)$ are $\mathcal G_n$-measurable (but not $B(u+v)$) hence $$E(B(u+2v)-B(u+v)\mid\mathcal G_n)=E(B(u+v)-B(u)\mid\mathcal G_n)=\tfrac12(B(u+2v)-B(u)).$$ This implies $E(Y\mid\mathcal G_n)=0$, and finally, $$E(Y\mid\mathcal F_n)=E(E(Y\mid\mathcal G_n)\mid\mathcal F_n)=0.$$