1

Spin-off from here.

The solution given is that

$$E[X_{n+1}|X_n] = 1/2\times 2X_n + 1/2\times 0 = X_n$$

How about using indicator functions?

I was thinking that $X_n = 2^n 1_{A_1}$, but I guess that's wrong.

What I tried:

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

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

since

$X_0 = 1$

$X_1 = 2 X_0 \times 1_{A_0} + 0 \times 1_{A_0^c} = 2^1 \times 1_{A_0}$

$X_2 = 2 X_1 \times 1_{A_1} + 0 \times 1_{A_1^c} = 2^2 \times 1_{A_0} \times 1_{A_1}$

and so on.

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

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

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

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

Is that correct?

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

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

$\sigma(X_0) = \{ \emptyset, \Omega \}$

$\sigma(X_1) = \sigma(A_0) \because X_1 = 2 \times 1_{A_0}$

$\sigma(X_2)$...$\subseteq \sigma(\sigma(A_0) \cup \sigma(A_1)) = \sigma(A_0, A_1)$

Community
  • 1
BCLC
  • 13,459

1 Answers1

0

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 $\mathscr F_m$

BCLC
  • 13,459