Problem about independence of RVs and tail sigma algebras Let $Y_0,Y_1,...$ be independent random variables with $P(Y_n=1)=P(Y_n=-1)=\frac{1}{2}$ and let $X_n=\prod_{i=0}^nY_i$. 
Let $\mathcal{Y}=\sigma(Y_1,Y_2,...)$
, $\mathcal{T}_n=\sigma(X_r, r>n)$, $\mathcal{L}=\bigcap_n\sigma(\mathcal{Y}, \mathcal{T}_n)$ and $\mathcal{R}=\sigma(\mathcal{Y}, \bigcap_n\mathcal{T}_n)$.
1) Prove $X_n$ are independent.
2) Prove $\mathcal{L}\neq \mathcal{R}$.
This is my attempt for the first part. 
$(X_n=1)$ happens when there is an even number of $Y$s equal to $-1$, so
$$P(X_n=1)=\left({n+1 \choose 0}+{n+1 \choose 2}+...+{n+1 \choose 2[\frac{n+1}{2}]}\right)\frac{1}{2^{n+1}}.$$ 
 I computed that the sum is equal to $2^n$, so
$P(X_n=1)=P(X_n=-1)=\frac{1}{2}$. Now the event $(X_n=1)\cap(X_{n+k}=1)$ consists of the number of cases when in the first $n+1$ terms we have an even 
number of $-1$ (which is $2^n$ by the previuos computation) times the number of cases where $\prod_{i=n+1}^kY_{i}=1$, which is $2^{k-1}$
because it's the same number of cases that would give $X_{k-1}=1$. Therefore $$P((X_n=1)\cap(X_{n+k}=1))=\frac{2^n2^k}{2^{1+n+k}}=2^{-2}$$ which 
is $P(X_n=1)P(X_{n+k}=1)$. Similarly I can prove that the $X_n$ are independent because when I intersect an $n_1$ number of such events I get probability $2^{-n_1}$.
Is this correct? Are there simpler ways to do it?
As for the second part I don't know how to proceed.
Any help would be appreciated, thank you.
 A: 1) $X_n$'s are independent: here

2)


*

*$\sigma(Y_0) \subseteq \mathscr{L}$

*$\sigma(Y_0)$ and $\mathscr{R}$ are independent
Now assume on the contrary that $\mathscr{L} = \mathscr{R}$.
If $\sigma(Y_0) \subseteq \mathscr{L}$, then $\sigma(Y_0) \subseteq \mathscr{R}$.
Since $\sigma(Y_0)$ and $\mathscr{R}$ are independent, $\sigma(Y_0)$ is independent of itself.
This means $\forall F \in \sigma(Y_0), P(F \cap F) = P(F)P(F) \ \to \ P(F) \in \{0,1\}$.
Choose $F = (Y_0 = 1)$ or $F = (Y_0 = -1)$. We have $P(F) = 1/2 \notin \{0,1\}$
A: You can use induction.  
For any cardinal $k$ then $P(X_k=1) = 1/2$.  
By reason of $P(X_1=1) = 1/2 \quad\wedge\quad P(X_k)=1/2 \to P(X_{k+1})=1/2$
Likewise, $P(X_k=-1)=1/2$ for any cardinal $k$

Then show that for any $k , h$ then $X_k$ is independent of $X_{k+h}$, because : for all $(\alpha, \beta) \in \{-1, 1\}^2$
$$\begin{align}
P(X_{k+h} = \beta\mid X_k= \alpha) & = P(\prod_{i=1}^k Y_k=\alpha, \prod_{j=1}^h Y_{k+j}=\beta/\alpha\mid \prod_{i=1}^k Y_k=\alpha)
\\ & = P(\prod_{j=1}^h Y_{k+j}=\beta/\alpha)
\\ & =1/2
\\[2ex]\therefore P(X_{k+h}=\beta \mid X_k=\alpha) & = P(X_{k+h}=\beta) & =1/2 
\end{align}$$
