So I was reading about simple random walks and came across this:
Suppose $(X_{n})_{n=0}^{N}$ is a sequence of i.i.ds adapted to a filtration $(\mathcal{F}_{n})_{n=0}^{N}$ with $E(X_{n})=0$ for all $n$, then $E(X_{n+1}\mathbb{I}_{A_{n}})=0$, where $\mathbb{I}_{A_{n}}$ is the characteristic function on $A_{n} \in \mathcal{F}_{n}$.
Can anyone help on how the result is obtained?
As stated this is false. Let $\{X_n\}$ be i.i.d. with mean $0$ on $(\Omega, \mathcal F,P)$ and $\mathcal F_n=\mathcal F$ for all $n$. Then the hyppothesis holds. But if the conclusion is true it would follow that $X_{n+1}=0$ almost surely!.
The conclusion is true if you assume that $\mathcal F_n=\sigma (X_1,X_2,\cdots,X_n)$ for each $n$. In that case $EX_{n+1} I_{A_n} =EX_{n+1} EI_{A_n} (=0) $ because $X_{n+1}$ and $I_{A_n}$ are independent.