Proving Y is a martingale using indicator functions

Question

Prove the stochastic process $Y = (Y_n)_{n \geq 0}$ is a martingale w/rt the filtration $\mathscr{F} = (\mathscr{F_n})_{n \geq 0}$, where $\mathscr{F_n} = \mathscr{F_n}^{Y} \doteq \sigma(Y_0, Y_1, ..., Y_n)$ where $Y$ is given by:

$Y_0 = 1$

and for $n \geq 1$

$Y_{n+1} = 2Y_n$ w/ prob 1/2

$Y_{n+1} = 0$ w/ prob 1/2

1. $\sigma(Y_n) \subseteq \mathscr{F_n}$, obviously

2. $Y_n$ is integrable because it is bounded: $Y_n \leq 2^n$

3. $E[Y_{n+1} | \mathscr{F_n}] = Y_n$

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What I tried doing:

Let $A_n = \{ \omega \in \Omega | Y_{n+1}(\omega) = 2 Y_n(\omega) \} \in \mathscr{F}$. Note that $P(A_n) = 1/2$.

$Y_n$ is a product of indicator functions namely: $Y_n = 2^n \prod_{0 \leq i \leq n-1} 1_{A_i}$

since

$Y_0 = 1$

$Y_1 = 2 Y_0 \times 1_{A_0} + 0 \times 1_{A_0^c} = 2^1 \times 1_{A_0}$

$Y_2 = 2 Y_1 \times 1_{A_1} + 0 \times 1_{A_1^c} = 2^2 \times 1_{A_0} \times 1_{A_1}$

and so on.

Now $E[Y_{n+1} | \mathscr{F_n}] = E[2^{n+1} \prod_{0 \leq i \leq n} 1_{A_i} | \mathscr{F_n}]$

$= 2^{n+1} \prod_{0 \leq i \leq n-1} 1_{A_i} E[1_{A_{n}} | \mathscr{F_n}]$

$= 2^{n+1} \prod_{0 \leq i \leq n-1} 1_{A_i} P[{A_{n+1}} | \mathscr{F_n}]$

$= 2^n \prod_{0 \leq i \leq n-1} 1_{A_i} = Y_n$.

Is that correct?

I seemed to assume $\sigma(1_{A_{n}}) = \sigma(A_n)$ and $\mathscr{F_n}$ are independent. Are they?

What is $\sigma(Y_n)$ in terms of $A_n$ anyway? I guess that:

$\sigma(Y_0) = \{ \emptyset, \Omega \}$

$\sigma(Y_1) = \sigma(A_0) \because Y_1 = 2 \times 1_{A_0}$

$\sigma(Y_2)$...$\subseteq \sigma(\sigma(A_0) \cup \sigma(A_1)) = \sigma(A_0, A_1)$

Answer 1

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, $X_n = \prod_{i=0}^{n} V_i$.

1. $X_n$'s are bounded and hence integrable.
2. $X_n$'s are adapted to their natural filtration.
3. $E[X_n \mid \mathscr{F_m}] = X_m$

\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 $\mathscr F_m$