Are products of independent random variables independent? Let $Z_0, Z_1, Z_2,...$ be independent and identically distributed such that
$P(Z_n = 1) = P(Z_n = -1) = 1/2$ for $n = 0, 1, 2, ...$
Let $X_0 = Z_0$, $X_1 = X_0 Z_1$, $X_2 = X_1 Z_2$, ...
Are $X_0, X_1, X_2, ...$ independent?

What I tried:
We must prove that for Borel sets $B_1, ..., B_n$,
$$P(X_0 \in B_0, X_1 \in B_1, ..., X_n \in B_n) = \prod_{i=0}^{n} P(X_i \in B_i) \ (*) $$
since $X_0, X_1, X_2, ..., X_n$ are independent $\forall n \in \mathbb{N}$ iff $X_0, X_1, X_2, ...$ are independent.


*

*$\{ X_n \}_{n=0}^{\infty}$ is Markov i.e.


$$P[X_n \in B| X_m] = P[X_n \in B| \mathscr{F}_m]$$
$\forall m \in [0,n], B \in \mathscr B$
[see proof in answer below]
This implies that LHS of (*) is equivalent to:
$$P(X_0 \in B_0) P(X_1 \in B_1 | X_0 \in B_0) ... P(X_n \in B_n | X_{n-1} \in B_{n-1})$$


*$P(X_{n+1} = 1) = P(X_{n+1} = -1) = 1/2$ can be proven by induction by noting the recurrence relations:


$$P(X_{n+1} = 1) = P(X_n = 1)P(Z_{n+1} = 1) + P(X_n = -1)P(Z_{n+1} = -1)$$
$$P(X_{n+1} = -1) = P(X_n = 1)P(Z_{n+1} = -1) + P(X_n = -1)P(Z_{n+1} = 1)$$.
I made use of the fact that $X_n$ and $Z_{n+1}$ are independent, which follows from $X_n = Z_0Z_1 \dots Z_n$ and $Z_1, ..., Z_n$ and $Z_{n+1}$ are independent.
This makes the RHS of (*) to be $(1/2)^{n+1}$


*$P(X_i \in B_i | X_{i-1} \in B_{i-1}) = 1/2$ because for $a_{n+1} \in \{-1, +1\}$


$P(X_{n+1} = a_{n+1} | X_n) = E[1_{X_{n+1} = a_{n+1}} | X_n] = E[1_{X_{n+1} = a_{n+1}} | X_n = 1]P(X_n = 1) + E[1_{X_{n+1} = a_{n+1}} | X_n = -1]P(X_n = - 1)$
This makes the LHS of (*) to be $(1/2)^{n+1}$ as well. QED
Any mistakes or missing steps?
 A: You are making this problem a lot harder than it needs to be because the
random variables in question are two-valued, and the problem can be
treated as one of independence of events rather than independence of
random variables.  In what follows, I will treat the independence of
events even though the events will be stated in terms of random variables.

Let $Z_0,Z_1,Z_2,\cdots$ be independent random variables $\ldots$

I will take this as the assertion that the countably infinite collection of events $A_i = \{Z_i = +1\}$ is a collection of independent events.  Now, a countable collection of events is said to be a collection of
independent events if each finite subset (of cardinality $2$ or
more) is a collection of independent events.  Recall that
$n\geq 2$ events $B_0, B_1, \cdots, B_{n-1}$ are said to be independent events
if
$$P(B_0\cap B_1\cap \cdots \cap B_{n-1}) 
= P(B_0)P(B_1) \cdots P(B_{n-1})$$
and every finite subset of two or more of these events is a
collection of independent events.  Alternatively, 
$B_0, B_1, \cdots, B_{n-1}$ are said to be independent events
if the following $2^n$ equations hold:
$$P(B_0^*\cap B_1^*\cap \cdots \cap B_{n-1}^*)
= P(B_0^*)P(B_1^*)\cdots  P(B_{n-1}^*)\tag{1}$$
Note that in $(1)$, $B_i^*$ stands for $B_i$ or $B_i^c$
(same on both sides of $(1)$) and the $2^n$ choices 
($B_i$ or $B_i^c$) give us the $2^n$ equations.
For our application, $A_i = \{Z_i = +1\}$ and $A_i^c = \{Z_i=-1\}$,
and so checking whether the $2^n$ equations
$$P(A_0^*\cap A_1^*\cap \cdots \cap A_{n-1}^*)
= P(A_0^*)P(A_1^*)\cdots  P(A_{n-1}^*)\tag{2}$$
hold or not, is equivalent to checking that the
joint probability mass function (pmf) of $Z_0, Z_1, \cdots, Z_{n-1}$
factors into the product of the $n$ marginal pmfs at each and
every one of the points $(\pm 1, \pm 1, \cdots, \pm 1)$ which is
what you would be doing if you had never heard of independent
events, just about independent random variables.
Thus, the statement

Let $Z_0,Z_1,Z_2,\cdots$ be independent random variables $\ldots$

does mean, among other things, that $Z_0,Z_1,Z_2,\cdots, Z_{n-1}$
is a finite collection of independent random variables. But, 
does the assertion

For all $n \geq 2$, $\{Z_0,Z_1,Z_2,\cdots, Z_{n-1}\}$ is a set 
  of $n$ independent random variables 

imply that the
countably infinite set $\{Z_0,Z_1,Z_2,\cdots \}$ is a
collection of independent random variables? 
The answer is Yes, because we know by hypothesis
that some specific finite
subsets of $\{Z_0,Z_1,Z_2,\cdots \}$ are independent random
variables, while any other finite subset, say $\{Z_2, Z_5, Z_{313}\}$,
is a subset of $\{Z_0, Z_1, \cdots, Z_{313}\}$ which are independent
per the hypothesis and so the subset is also a set of independent
random variables.
In your question, with each $a_i \in \{+1, -1\}$ and
defining $b_i = \prod_{j=0}^i a_j$ which is also in $\{+1,-1\}$,
\begin{align}
P(X_0 = a_0, X_1 = a_1, \cdots, X_n = a_n)
&= P(Z_0 = a_0, Z_1 = a_0a_1, Z_2 = a_0a_1a_2, \cdots, Z_n = a_0a_1...a_n)\\
&= P(Z_0=b_0, Z_1 = b_1, \cdots, Z_n = b_n)\\
&= \prod_{i=0}^n P(Z_i = b_i)\\
&= 2^{-(n+1)}\\
&= \prod_{i=0}^n P(X_i = a_i),
\end{align}
that is, all $2^{n+1}$ equations of the form $(2)$ hold.
Thus, for each $n \geq 1$, $X_0, X_1, \cdots, X_n$ are
independent random variables, and therefore the
countably infinite collection $\{X_0, X_1, \cdots\}$
of random variables is a collection of independent
random variables.

After reading over my revised answer, perhaps it is I who is
making the problem much harder than necessary. My apologies.
A: 

*

*... How do I state this precisely, if it is right? $\forall i \leq n, \sigma(X_i) \subseteq \sigma(X_n)$ ?


Your have the right idea, but I would recommend using the definition of the Markov property to state this, namely that we have $P(X_n\mid X_0,\dots,X_{n-1})=P(X_n \mid X_{n-1})$. There is nothing imprecise about this as long as you have a precise definition of conditional probabilities. The $\sigma-$algebra condition you wrote is not correct.



*...It seems like I assumed $X_n$ and $Z_{n+1}$ are independent. Are they?


Hint: measurable functions of independent random variables are independent (you decide if you need to prove this).



*...I'm stuck.

Is what I've done right so far? Which parts are wrong? Where do I go from here?

Try structure your answer some more. Specify the events $B_i$ under consideration, e.g. notice that since each variable only takes 2 values there are not that many different types of events to consider.
First solve for the right hand side using, e.g., the argument that
$$P(X_i = 1)=\mathbb E P(X_i = 1 \mid X_{i-1})=\mathbb E 1/2=1/2;$$ you have the right value.
Then solve for the left hand side using the Markov property as you have attempted.
