A Markov chain indexed by the integers My definition of a markov chain (in discrete time with countable state space), is the following:
Let $S$ be a countable set, $P$ be a stochastic matrix indexed by $S$ and , $\lambda$ a distribution probability on $S$.
A sequence of random variables $\{X_n\}_{n \geq 0}$ is a markov chain Markov$(\lambda, P)$ if $X_0\sim \lambda$ and given any $i_0,i_1,..., i_{n+1} \in S$ it is true that
$$P(X_{n+1} = i_{n+1}|X_0=i_0,...,X_n = i_n) = P(X_{n+1}=i_{n+1}|X_n = i_n) = p_{i_n,i_{n+1}}$$
where $p_{i_n,i_{n+1}}$ is an entry of the matrix $P$.
Now, my markov chain is indexed by the natural numbers ($n\geq 0$). Is there a way to formally define a markov chain indexed by the integers $\mathbb{Z}$? There's no longer an initial value of the markov chain, so how could it work?

${\color{red}{\mbox{Thought:}}}$
Suppose we take a stochastic matrix $P$ that is irreducible and positive recurrent, so that there exists an invariant distribution $\pi$.
Given any $k \in \mathbb{N}$, if I take $\{X_n\}_{n \geq 0}\sim Markov(\pi,P)$ and $\{Y_n\}_{n \geq -k} \sim Markov(\pi,P)$ (in this case we begin the $Y$ chain at $n=-k$, so $Y_{-k}\sim \pi$)
Then we have $X_n \overset{d}{=}Y_n \forall n\geq 0$.
Does it make sense then to think about a "doubly infinite Markov chain" $\{X_n\}_{n \in \mathbb{Z}}\sim Markov(P)$  as a sequence of random variables such that
$$P(X_{n+1}=i_{n+1}|X_n = i_n,... X_0=i_0,...X_{-n} = i_{-n},...) = P(X_{n+1} = i_{n+1}|X_n=i_n) = p_{i_n,i_{n+1}}$$
and such that $P(X_n = j) = \pi_j \forall n$ where $\pi_j$ is the invariant distribution of the matrix $P$?
Is there a correct/standard way to define this object (if it exists)?

Any information or references about the subject would be appreciated. Thanks in advance.
 A: Yes.  Suppose $\{X_n\}_{n=0}^{\infty}$ is DTMC on a finite or countably infinite state space $S$ with transition probabilities $(P_{ij})$. Suppose it is irreducible and it has a stationary mass function $\pi = (\pi(s))_{s \in S}$, so that
$$ \pi(j) = \sum_{i \in S} \pi(i) P_{ij} \quad \forall j \in S $$
Since the DTMC is irreducible it can be shown that $\pi(j)>0$ for all $j \in S$.
Suppose that $X_0$ starts in this stationary distribution:
$$ P[X_0=s] = \pi(s) \quad \forall s \in S$$
Now define a "backwards" DTMC $\{Z_n\}_{n=0}^{\infty}$ with some transition probabilities $\tilde{P}_{ji}$ (defined below) such that $Z_0=X_0$, and where all transitions in $Z_n$ are conditionally independent of $\{X_n\}_{n=0}^{\infty}$ (given $X_0$):
$$ \tilde{P}_{ji} = P[Z_{n+1} = i | Z_n=j] = \frac{\pi(i)P_{ij}}{\pi(j)} \quad \forall i, j \in S$$
You can verify these are valid transition probabilities (nonnegative, and sum to 1).
Define $\{Y_n\}_{n=-\infty}^{\infty}$ by
$$ Y_n = \left\{\begin{array}{cc}
X_n & \mbox{ if $n \in \{0, 1, 2, ...\}$} \\
Z_{-n} & \mbox{ if $n \in \{-1, -2, -3, ...\}$}
\end{array}\right.$$
You can verify that
$$P[Y_n=s] = \pi(s) \quad \forall n \in \mathbb{Z}, \forall s \in S$$
and for all $n \in \mathbb{Z}$:
$$P[Y_{n+1}=j|Y_n=i, Y_{n-1}=i_{n-1}, Y_{n-2} = i_{n-2}, ...] = P_{ij} \quad \forall i, j \in S$$
