# Prove independence of $Y_0$ and $\sigma$-algebra $\mathscr{R}$. [duplicate]

Let $Y_0, Y_1, ...$ be iid RVs s.t. $P(Y_n = 1) = P(Y_n = -1) = 1/2$ for $n \geq 0$.

Define random variables $X_n = Y_0Y_1Y_2\cdots Y_n = \prod_{i=0}^{n} Y_i$ for $n \geq 0$. The $X_n$'s are independent.

Let $\mathscr{Y} \doteq \sigma(Y_1, Y_2, \ldots)$, $\mathscr{T}_n \doteq \sigma(X_r \mid r > n) = \sigma(X_{n+1}, X_{n+2}, \ldots)$ and $\mathscr{R} \doteq \sigma(\mathscr{Y}, \bigcap_n \mathscr{T_n})$

Prove $Y_0$ and $\mathscr{R}$ are independent $\iff \sigma(Y_0)$ and $\mathscr{R}$ are independent.

What I tried:

I think $Y_0$ and $\mathscr{R}$ are independent $\iff$

1. $Y_0$ and $\mathscr{Y}$ are independent

2. $Y_0$ and $\bigcap_n \mathscr{T}_n$ are independent

Is that right? If not, why, and how else might I be able to approach this?

Assuming it is:

1. $Y_0$ and $\mathscr{Y}$ are independent seems to be true since $\sigma(Y_0)$ and $\sigma(Y_1, Y_2, \ldots)$ are independent for reasons similar to step 1 here.

2. $Y_0$ and $\bigcap_n \mathscr{T}_n$ are independent:

I guess this is equivalent to showing that

$\forall n \in \mathbb{N}, \sigma(Y_0)$ and $\mathscr{T}_n$ are independent.

It looks like $\sigma(Y_0)$ and $\mathscr{T}_1$ being independent is equivalent to saying that

$\sigma(Y_0)$ and $\sigma(Y_0Y_1)$ are independent, $\sigma(Y_0)$ and $\sigma(Y_0Y_1Y_2)$ are independent and so on.

Thus, $\sigma(Y_0)$ and $\mathscr{T}_1$ being independent implies that $\forall n \in \mathbb{N}, \sigma(Y_0)$ and $\mathscr{T}_n$ are independent.

Assuming all that is right:

$\sigma(Y_0)$ and $\sigma(Y_0Y_1)$ are independent since $\sigma(X_0)$ and $\sigma(X_1)$ are independent

$\sigma(Y_0)$ and $\sigma(Y_0Y_1Y_2)$ are independent since $\sigma(X_0)$ and $\sigma(X_2)$ are independent

and so on.

## merged by user642796Oct 24 '15 at 5:23

This question was merged with Prove random variable $Y_0$ and $\sigma$-algebra $\mathscr{R}$ are independent. [closed] because it is an exact duplicate of that question.