Calculating the Stationary Distribution of a Markov Chain I am working on 2.27 from Resnick's "Adventures in Stochastic Processes". 
The problem states,
If $\{X_n\}$ is Markov with stationary distribution $\pi$, transition matrix $P$ and state space $S$, show that $\{(X_n,X_{n+1}), n \geq 0)\}$ is Markov. Give its stationary distribution.
Showing that $\{Y_n = (X_n,X_{n+1}), n \geq 0)\}$ was straightforward but I wasn't able to come up with an approach to calculate $\pi_{Y}$. I think I managed to guess a stationary distribution and verified whether it satisfied,
$$\pi_Y(i_1,i_2) = \sum_{(x,y) \in S \times S} \pi_Y(x,y)\mathbb P \{(x,y) \to (i_1,i_2)\}$$
I guessed that $\pi_Y(i_1,i_2) = \pi(i_1)P(i_1,i_2)$ and it seems to satisfy the above criteria but I don't like this approach at all. Could some one please illustrate a nice method for this question.
 A: Using Did's hint,
Let us assume that ${X_n}$ starts off with it's stationary distribution $\pi$.
$$
\begin{split}
\implies \mathbb{P}\{X_0 = i\} &= \pi(i) \\
\implies \mathbb{P}\{Y_0 = (i,j)\} &= \mathbb{P}\{X_0 = i, X_1 = j\} \\
                   &= \mathbb{P}\{X_0 = i\}\mathbb{P}\{X_1 = j|_{X_0 = i}\}\\ 
                   &= \mathbb{P}\{X_0 = i\}\mathbb{P}_i\{X_1 = j\}\\
                   &= \pi(i)P(i,j)\\  
\end{split}
$$
Now we should examine $\mathbb{P}\{Y_1\}$. 
$$
\begin{split}
 \mathbb{P}\{Y_1\} &= \mathbb{P}\{Y_1|_{Y_0}\}\mathbb{P}\{Y_0\} \\
\implies \mathbb{P}\{Y_i = (x,y)\} &= \sum_i\mathbb{P}\{Y_1 = (x,y)|_{Y_0 =(i,x)}\}\mathbb{P}\{Y_0=(i,x)\} \\
                   &= \sum_i P(x,y)\pi(i)P(i,x) \\
                   &= P(x,y)\sum_i \pi(i)P(i,x) \\
                   &= P(x,y)\pi(x) \\
\mathbb{P}\{Y_i = (x,y)\} &= \pi(x)P(x,y)=\pi_Y(x,y)\\
\end{split}
$$
Therefore we can see that ${Y_n}$ is also stationary if ${X_n}$ is stationary and $\pi_Y$ is the stationary distribution.
