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Given a filtered probability space $(\Omega, \mathscr{F}, \{\mathscr{F_n}\}, \mathbb{P})$ where $\mathscr{F_n} = \mathscr{F_n}^{Z} \doteq \sigma(Z_0, Z_1, \ldots, Z_n)$, show that $Z = (Z_n)_{n \geq 0}$ is a $(\mathscr{F}_n^Z, \mathbb{P})$-martingale where $Z$ is given by:

$Z_{n+1} = 2Z_n$ w/ prob 1/2

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

and $Z_0 = 1$.


What I tried:

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

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

\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[Z_m \prod_{i=m+1}^{n} V_i\mid \mathscr{F_m}\right] \\ & = Z_m E\left[\prod_{i=m+1}^{n} V_i\mid \mathscr{F_m}\right] \\ & = Z_m E\left[\prod_{i=m+1}^{n} V_i\right] \tag{*} \\ & = Z_m \prod_{i=m+1}^{n} E\left[V_i\right] \text{ by the independence of the $V_i$'s} \\ & = Z_m \prod_{i=m+1}^{n} E\left[V_i\right] \\ & = Z_m \prod_{i=m+1}^{n} (1) \\ & = \text{RHS} \quad \text{QED} \end{align}

(*) I guess this is true if $\sigma(Z_0, Z_1, \ldots, Z_n) = \sigma(V_0, V_1, \ldots, V_n)$? Is it that $\sigma(Z_0, Z_1, \ldots, Z_n) = \sigma(V_0, V_1, \ldots, V_n)$?

Any mistakes or missing steps?

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$\mathscr G_m = \sigma(V_1,\ldots,V_m)$ is strictly larger than $\mathscr F_m = \sigma(Z_1,\ldots,Z_m)$. ($\mathscr G_m$ has $2^{2^m}$ measurable sets while $\mathscr F_m$ only has $2^{m+1}$)

But $\mathscr F_m$ is a subset of $\mathscr G_m$ and so $E[X \mid \mathscr F_m] = E[E[X \mid \mathscr G_m] \mid \mathscr F_m]$.

Now, with $X$ being independant of $\mathscr G_m$, $E[X \mid \mathscr G_m] = E[X]$ which is a real number, it is more a constant than a random variable.

Then $E[E[X] \mid \mathscr F_m] = E[X]$.

Or you could also argue that being independant of $\mathscr G_m$ implies being independant of $\mathscr F_m$ and conclude that $E[X \mid \mathscr F_m] = E[X]$.

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