# Prove random variable $Y_0$ and $\sigma$-algebra $\mathscr{R}$ are independent. [closed]

Let $Y_0, Y_1, ...$ be independent and identically distributed random variables with

$P(Y_n = 1) = P(Y_n = -1) = 1/2$ for n = 0, 1, 2 ...

Define random variables $X_n = Y_0Y_1Y_2...Y_n = \prod_{i=0}^{n} Y_i$ for n = 0, 1, 2 ...

It can be shown that $X_0, X_1, X_2, ...$ are independent.

Define the $\sigma$-algebras:

$\mathscr{Y} \doteq \sigma(Y_1, Y_2, ...)$

$\mathscr{T_n} \doteq \sigma(X_r | r > n) = \sigma(X_{n+1}, X_{n+2}, ...)$

$\mathscr{R} \doteq \sigma(\mathscr{Y}, \bigcap_n \mathscr{T_n})$

Prove $Y_0$ and $\mathscr{R}$ are independent.

## closed as off-topic by Did, Leucippus, M. Vinay, choco_addicted, Daniel W. FarlowJun 17 '16 at 4:22

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1. $$\bigcap_n \mathscr{T_n} = \bigcap_n \sigma(X_{n+1}, X_{n+2}, ...) = \bigcap_n \sigma(Y_{n+1}, Y_{n+2}, ...) \tag{*}$$

2. By Kolmogorov 0-1 law, all the events in $\bigcap_n \mathscr{T_n}$ have probability 0 or 1, meaning $\bigcap_n \mathscr{T_n}$ is independent of any other collection of events.

3. $\mathscr{Y}$ and $\bigcap_n \mathscr{T_n}$ are independent by 2.

4. $\sigma(Y_0)$ and $\bigcap_n \mathscr{T_n}$ are independent by 2.

5. $\mathscr{Y}$ and $\sigma(Y_0)$ are independent since $Y_0, Y_1, ...$ are independent.

6. $\mathscr{Y}$, $\sigma(Y_0)$ and $\bigcap_n \mathscr{T_n}$ are pairwise independent by 3, 4 and 5.

7. $\forall \ A \in \sigma(Y_0), B \in \mathscr{Y}$ and $C \in \bigcap_n \mathscr{T_n}$, we have

$$P(A, B, C) \stackrel{2}{=} P(A, B) P(C) \stackrel{5}{=} P(A)P(B)P(C)$$

1. $\mathscr{Y}$, $\sigma(Y_0)$ and $\bigcap_n \mathscr{T_n}$ are independent by 6 and 7.

2. From 8, we conclude by this fact that $Y_0$ and $\mathscr R$ are independent.

QED

*You can think of it this way:

Suppose

$$A \in \sigma(Y_0,Y_1,Y_2,...)$$

$$A \in \sigma(Y_1,Y_2,...)$$

$$A \in \sigma(Y_2,...)$$

$$\vdots$$

Prove that

$$A \in \sigma(X_1,X_2,X_3,...)$$

$$A \in \sigma(X_2,X_3,...)$$

$$A \in \sigma(X_3,...)$$

$$\vdots$$

Also prove the converse.

Note that

1. $$\sigma(X_n) \subseteq \sigma(Y_0, Y_1, Y_2, ..., Y_n)$$

for reasons similar to $\sigma(X) \subseteq \sigma(Y,Z)$ if $X=YZ$

1. $$\sigma(X_1, X_2, ...) \subseteq \sigma(Y_0, Y_1, Y_2, ...)$$

for reasons similar to my other question

• Why does 1 hold? I don't see how it is true from the reason similar to your other questions. – takecare Jun 15 '16 at 15:41
• @takecare Edited! – BCLC Jun 16 '16 at 17:43