any stationary sequence can be embedded in a two-sided stationary sequence

Question

In Durrett's probability theory and examples, it asks us to prove the following:

Any stationary sequence $\{ X_n, n \ge 0\}$ can be embedded in a two-sided stationary sequence $\{ Y_n : n \in \mathbf{Z} \}$.

I know that $X_1, X_2, ... $ is a stationary sequence if $(X_1, X_2, ..., X_n)$ has the same distribution as $(X_k, X_{k+1}, ..., X_{k+n})$ for all $k > 0$. From google search, it seems like I need to use Kolmogorov's extension theorem to solve this, but what does it mean that a sequence can be "embedded" in another sequence? Additionally, what does "two sided sequence" mean?

Theorem 2.1.21 (Kolmogorov's extension theorem) Suppose we are given probability measures $\mu_n$ on $(\mathbf{R}^n, \mathcal{R}^n)$ that are consistent, that is, $$\mu_{n+1}((a_1, b_1] \times \cdots \times (a_n, b_n] \times \mathbf{R}) = \mu_n ((a_1, b_1] \times \cdots \times (a_n, b_n]).$$ Then there is a unique probability measure $P$ on $(\mathbf{R}^N, \mathcal{R}^N)$ with $$P(\omega: \omega_i \in (a_i, b_i], 1 \le i \le n ) = \mu_n((a_1, b_1] \times \cdots \times (a_n, b_n])$$

Answer 1

The precise statement of this result is as follows:

For any stationary process $\{X_n\}_{n= 1}^{\infty}$, there exists a doubly-infinite stationary sequence $\{Y_n\}_{n=-\infty}^{\infty}$ s.t. $X_1,\ldots$ and $Y_1,\ldots$ have the same distribution.

Proof. By Kolmogorov's extension theorem, it suffices to define the finite dimensional distributions of $\{Y_n\}$. Specifically, for $m,n\ge 0$ set \begin{align*} &\mathsf{P}(Y_{-m}\le y_{-m}, Y_{-m+1}\le y_{-m+1},\ldots, Y_{n-1}\le y_{n-1},Y_{n}\le y_n) \\ &\qquad=\mathsf{P}(X_{1}\le y_{-m}, X_2\le y_{-m+1},\ldots, X_{n+m}\le y_{n-1},X_{n+m+1}\le y_n). \end{align*}