TITLE: Martingale representation theorem discrete
QUESTION [4 upvotes]: Let $X_1,X_2,...$ be an iid sequence of random variables such that $P(X_i=1)=p$, $P(X_i=0)=1-p$, defined on a probability space $(\Omega, \mathcal{F}, P)$. Take $\mathcal{F}_0$ to be be the trivial $\sigma$- algebra and let $\mathcal{F_n}=\sigma(X_1,...,X_n)$. Suppose that $M$ is a martingale adapted to this filtration. Show that there exists a constant $m$ and a predictable process $Y$ such that $M_n=m+\sum_{k=1}^nY_k(X_k-p)$.
The things I have thought of so far is that since $\mathcal{F}_0$ is the trivial $\sigma$-algebra, $M_0$ is a constant. Therefore we should take $m=M_0$. For any predictable process $Y$, $m+\sum_{k=1}^nY_k(X_k-p)$ is also a martingale because the $X_i$ form an independent sequence. It seems then that $M_1$ can take on only two distinct values since it is $\mathcal{F_1}$-measurable.. 
I do not know how to proceed and what to take for the predictable process $Y$. Could anyone give me a hint?

REPLY [3 votes]: Hints:

Since $M_n$ is $\mathcal{F}_n = \sigma(X_1,\ldots,X_n)$-measurable, there exists a Borel-measurale function $f_n$ such that $$ M_n = f_n (X_1,\ldots,X_n).\tag{1}$$
This implies $$\begin{align*}M_n &= f_n(X_1,\ldots,X_{n-1},0) 1_{\{X_n=0\}} + f_n(X_1,\ldots,X_{n-1},1) 1_{\{X_n=1\}} \\ &= f_n(X_1,\ldots,X_{n-1},0) + (f_n(X_1,\ldots,X_{n-1},1)-f_n(X_1,\ldots,X_{n-1},0)) \underbrace{1_{\{X_n=1\}}}_{=X_n}. \end{align*} \tag{2}$$
Use the martingale property of $(M_n)_{n \in \mathbb{N}}$ to show that $$M_{n-1} = f_n(X_1,\ldots,X_{n-1},0) (1-p) + f_n(X_1,\ldots,X_{n-1},1) p. \tag{3}$$
By $(2)$ and $(3)$, $$M_n-M_{n-1}= \underbrace{(f_n(X_1,\ldots,X_{n-1},1)-f_n(X_1,\ldots,X_{n-1},0)}_{=:Y_n} (X_n-p).$$
Conclude that $$M_k-M_0 = \sum_{n=1}^k Y_n (X_n-p)$$ for any $k \geq 1$.
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