Prove X is a martingale Prove $X = (X_n)_{n \geq 0}$ is a martingale w/rt $\mathscr{F}$ where X is given by:
$X_0 = 1$
and for $n \geq 1$
$X_{n+1} = 2X_n$ w/ prob 1/2
$X_{n+1} = 0$ w/ prob 1/2
and $\mathscr{F_n} = \mathscr{F_n}^{X} \doteq \sigma(X_0, X_1, ..., X_n)$.
I think that $X_{n} = 2^{n} \prod_{i=1}^{n} 1_{A_i} \forall n \geq 0$ where $A_1 = \{ \omega \in \Omega | X_{2}(\omega) = 2 X_1(\omega) \} \in \mathscr{F}$ as follows:
$X_{1} = 2X_0* 1_{A_1} + 0*1_{A_1^c}$
$X_{2} = 2X_1* 1_{A_2} + 0*1_{A_2^c}$
$=2(2X_0* 1_{A_1} + 0*1_{A_1^c})*1_{A_2} + 0*1_{A_2^c}$
Is that right? If not, why? If so, here is my attempt:
(leaving out adaptability and integrability stuff)
We must show that $E[X_{n+1}|\mathscr{F_n}] \equiv E[2^{n+1} \prod_{i=1}^{n+1} 1_{A_i}|\mathscr{F_n}] = 2^n \prod_{i=1}^{n} 1_{A_i}$? Is that right?
help please?
 A: There are 3 things to check:


*

*$X_n$ is measurable with respect to $F_n$: This is ok by definition of $F$.

*$X_n$ is integrable: as $X_n$ is bounded, so that is ok.

*(the big part) $E[X_{n+1}|X_n] = X_n$:


conditionally to $X_n$, there are two options:


*

*with probability 1/2, $X_{n+1} = 2X_n$

*with probability 1/2, $X_{n+1} = 0$


hence $$E[X_{n+1}|X_n] = 1/2\times  2X_n + 1/2\times 0 = X_n
$$
A: Given a filtered probability space $(\Omega, \mathscr{F}, \{\mathscr{F_n}\}, \mathbb{P})$ where $\mathscr{F_n} = \mathscr{F_n}^{X} \doteq \sigma(X_0, X_1, \ldots, X_n)$, show that  $X = (X_n)_{n \geq 0}$ is a $(\mathscr{F}_n^X, \mathbb{P})$-martingale where $X$ is given by:
$X_{n+1} = 2X_n$ w/ prob 1/2
$X_{n+1} = 0$ w/ prob 1/2
and $X_0 = 1$.
Define the iid random variables $V_0 = 1$,
$V_1, V_2, \ldots \sim P(V_i = 0) = P(V_i = 2) = 1/2$. Then, $X_n = \prod_{i=0}^{n} V_i$.


*

*$X_n$'s are bounded and hence integrable.

*$X_n$'s are adapted to their natural filtration.

*$E[X_n \mid \mathscr{F_m}] = X_m$


\begin{align}
\text{LHS} & = E\left[\prod_{i=0}^{n} V_i \mid \mathscr{F_m}\right] \\
& = E\left[\prod_{i=0}^{n} V_i \mid \mathscr{F_m}\right] \\
& = E\left[\prod_{i=0}^{m} V_i \prod_{i=m+1}^{n} V_i\mid \mathscr{F_m}\right] \\
& = E\left[X_m \prod_{i=m+1}^{n} V_i\mid \mathscr{F_m}\right] \\
& = X_m E\left[\prod_{i=m+1}^{n} V_i\mid  \mathscr{F_m}\right] \\
& = X_m E\left[\prod_{i=m+1}^{n} V_i\right] \tag{*} \\
& = X_m \prod_{i=m+1}^{n} E\left[V_i\right] \text{ by the independence of the $V_i$'s} \\
& = X_m \prod_{i=m+1}^{n} E\left[V_i\right] \\
& = X_m \prod_{i=m+1}^{n} (1) \\
& = \text{RHS} \quad \text{QED}
\end{align}
(*)
$\mathscr G_m = \sigma(V_1,\ldots,V_m) \supset \mathscr F_m$
Being independent of $\mathscr G_m$ implies being independent of $\mathscr F_m$
