0
$\begingroup$

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.

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Leucippus, M. Vinay, choco_addicted, Daniel W. Farlow
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$
  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

$\endgroup$
  • 2
    $\begingroup$ Why does 1 hold? I don't see how it is true from the reason similar to your other questions. $\endgroup$ – takecare Jun 15 '16 at 15:41
  • $\begingroup$ @takecare Edited! $\endgroup$ – BCLC Jun 16 '16 at 17:43

Not the answer you're looking for? Browse other questions tagged or ask your own question.