I am working with a sequence $Z_i=G(\mathcal{F}_i)$ where $\mathcal{F}_k=(\dots,\epsilon_{-1},\epsilon_0,\epsilon_1,\dots,\epsilon_{k-2},\epsilon_{k-1},\epsilon_k)$ where each $\epsilon_i$ are iid random variables, and $G$ is a smooth function. For example you could have $Z_i=\epsilon_i+\epsilon_{i-1}$.
I am interested in $$\lim_{k\to-\infty} E[Z_i|\mathcal{F}_k]$$
from my reading I believe that what I really need is
$$E[Z_i|\mathcal{T}]$$ where $\mathcal{T}=\cap_{\ell\geq 0} \mathcal{F}_{-\ell}$, the tail $\sigma$-field.
Based on Kolmogorov's 0-1 law we have that since the $\epsilon_i$'s are independent, then for any $A\in\mathcal{T}$ $P(A)=0$ or $P(A)=1$. Also, by the definition of conditional expectation, $E[Z_i|\mathcal{T}]$ is a random variable such that
- $Y\in\mathcal{T}$ (that is, is $\mathcal{T}$ measurable).
- For all $A\in\mathcal{F}$, $\int_A Z_i dP=\int_A Y dP$
I believe it should be true then that $E[Z_i|\mathcal{T}]=E[Z_i]$. This is because $E[Z_i]$ is a constant and hence $\mathcal{T}$ measurable, and for all $A\in\mathcal{T}$, for any $A$ where $P(A)=1$ then $$\int_A Z_i dP=\int_\Omega Z_i dP=E[Z_i]=E[Z_i] \int_A 1dP=\int_A E[Z_i]dP$$ and for any $A$ where $P(A)=0$ then $$\int_A Z_i dP=0=E[Z_i] \int_A 1 dP= \int_A E[Z_i] dP$$
Anything look off or is my reasoning sound?