Show that a dynamical system is not ergodic (finding the invariant sets)

Question

Given $(\Omega, \mathcal F, \mathbb P)$ a probability space. Consider the iid sequence of random variables $(X_n)_{n\in \mathbb N}$ defined on $(\Omega, \mathcal F, \mathbb P)$ : $$X_n \overset{iid}{\sim} N(0,1).$$ Let $B:(\Omega, \mathcal F, \mathbb P)\to \{\hbox{head},\hbox{tail}\}$ be a Bernoulli random variable: $B \sim \hbox{Bernoulli}(p)$. Define $(Y_n)_{n\in \mathbb N}$ as follows. Flip a coin according to $B$ and:

1. If it is head, set $Y_n=X_n$, for all $n$;
2. If it is tail, set $Y_n=0$, for all $n$.

Notice that $(Y_n)_{n\in \mathbb N}$ is defined on $(\Omega, \mathcal F, \mathbb P)$ with trajectories in $(\mathbb R^\mathbb N, \mathcal B^\mathbb N, P_Y)$, where $P_Y$ is the law of the stochastic process $(Y_n)_{n\in \mathbb N}$. Moreover, it is straightforwar to show that $(Y_n)_{n\in \mathbb N}$ is stricty stationary.

According to Chapter V, Probability, Albert Shiryaev, $(Y_n)_{n\in \mathbb N}$ generates a dynamical system $(T, P_Y)$ as follows. First, define $T: \mathbb R^\mathbb N \to \mathbb R^\mathbb N $ as: $$T(x_1,x_2,...)= (x_2,x_3,...).$$ Since $(Y_n)_{n\in \mathbb N}$ is stricty stationary, we have that $P_Y$ is $T$-invariant in the following sense: $$P_Y(A)=P_Y(T^{-1}(A)), \quad \forall \, A \in \mathcal{B}^\mathbb N$$

We say that $(T, P_Y)$ is ergodic if: $$P_Y(A)\in \{0,1\}, \quad \forall A \in \mathcal{I}_T$$ where $\mathcal{I}_T$ is the class of $T-$invariant sets (indeed, a $\sigma-$algebra), defined as : $$\mathcal{I}_T:=\{A \in \mathcal B^\mathbb N : A = T^{-1}(A)\}$$

I want to show that $(Y_n)_{n\in \mathbb N}$ is not ergodic. For this, I want to:

1. Find $\mathcal{I}_T$ for the dynamical system generated by $(Y_n)_{n\in \mathbb N}$;
2. Show that $P_Y(A)\notin \{0,1\}$ for some $A \in \mathcal{I}_T$.

Your help in these two points can help me to show non-ergodicity in other more complex cases.

Update question

I am indeed interested in these two points above, but I would like to focus on showing whether it is true that $\mathcal{I}_T \subset \mathcal B^\mathbb N$ is related (in some sense) to $\sigma(B) \subset \mathcal F$. Here, $\sigma(B)$ is the sigma algebra generated by the Bernoulli $B$.

Answer 1

Here are what my considerations about it. To show that the dynamical system generated by $(Y_n)_{n\in \mathbb N}$ is not ergodic, you can try to find a set $A \in \mathcal{I}_T$ such that $P_Y(A)\notin \{0,1\}$. One way to do this is to consider the set $A = \{(x_n)_{n\in \mathbb N} : x_1 = 0\}$. Since $T(x_1,x_2,...)= (x_2,x_3,...)$, we have that $T^{-1}(A) = A$, so $A$ is a T-invariant set. Now we need to compute the probability of this set.

Since the coin flip determines whether all the $Y_n$ are equal to 0 or equal to the corresponding $X_n$, we have that $$P_Y(A) = P(B=\text{tail}) = 1-p.$$ Thus, if $0 < p < 1$, then we have found a T-invariant set with probability strictly between 0 and 1, which shows that the dynamical system generated by $(Y_n)_{n\in \mathbb N}$ is not ergodic.