I am having some trouble understanding how to interpret $\sigma$-algebras related to a sequence $\{X_n : n \geq 1\}$ of random variables, especially tail events. I have found some great explanations online about why certain random variables (such as $\{\underset{{n \to \infty}}{\limsup}X_n \ \text{exists} \}$ are tail events, but I am having trouble expanding my understanding beyond these basic examples.
In particular, I found this great example which I am unable to answer completely, but I think if I can understand how it works, then I will be better able to understand these concepts more generally.
Suppose that $\{Y_0,Y_1,Y_2,\ldots\}$ are i.i.d. random variables on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ with the following distribution:
$$ \mathbb{P}(Y_k = -1) = \mathbb{P}(Y_k = 1) = \frac{1}{2}, \ \ k \geq 0.$$
Furthermore, for any $n \in \mathbb{N}$, $n \geq 1$, we define:
$$X_n := Y_0 Y_1 \cdots Y_n. $$
Now, suppose I have the following $\sigma$-algebras:
$$\mathcal{Y} := \sigma(\{Y_1, Y_2, Y_3 \ldots \}) \ \text{and} \ \mathcal{T_n} := \sigma(\{X_n, X_{n+1}, X_{n+2} \ldots \}), \ n \geq 1. $$
Furthermore, let:
$$\mathcal{L} := \bigcap_{n \geq 1} \sigma (\mathcal{Y} \cup \mathcal{T_n}) \ \text{and} \ \mathcal{R} := \sigma \bigg(\mathcal{Y} \cup \bigg ( \bigcap_{n \geq 1} \mathcal{T_n} \bigg) \bigg). $$
The goal is to show that $\mathcal{L} \neq \mathcal{R}$. One suggested approach is to show that $Y_0 \in m \mathcal{L}$ (i.e. $Y_0$ is measurable w.r.t. $\mathcal{L}$) and $Y_0$ is independent of $\mathcal{R}$.
One problem I am having is understanding why $\mathcal{Y} \neq \mathcal{T_n}$, as $Y_n$ and $X_n$ have the same probability distribution and $\{X_n : n \geq 1 \}$ are independent.
I truly appreciate any help with solving this problem (hints, solutions, references etc.).
Best Answer
For why $\mathcal Y \ne \mathcal T_n$, you can think about the easiest case where $n = 1$. Then $\mathcal T_1 = \sigma(X_1) = \sigma(Y_0Y_1)$, so knowing $Y_1$ and $\mathcal T_1$ tells you $Y_0$. This is the reason $Y_0$ is $\mathcal L$ measurable - $Y_0$ is measurable with respect to $\sigma(\mathcal Y \cup \mathcal T_n)$ for all $n$ because knowing $Y_i$ for $i \ge 1$ and any $X_n$ is enough to find $Y_0$.