Let $P$ be a transition matrix on a discrete state space with $N$ elements. $P_{i,j}$ is the probability of going from state $i$ to state $j$. Let $\pi$ be the stationary distribution.
Let $\{X_n\}$ be a Markov chain with this transition matrix. Given these $X_n$, which have values in $1, \cdots, N$ for some $N$, can form the random variables $Y_n = P_{X_n}$. Each $Y_n$ is the probability vector that gives the transition probabilities out of state $X_n$.
I am wondering how does the long-time behavior of the $\{Y_n\}$ relate to the stationary distribution of the $\{X_n\}$. My hypothesis is that if $\gamma$ is the stationary distribution (assuming it exists) of the $\{Y_n\}$ then $$\mathbb{E}_\gamma[Y] =\pi$$ (We're dealing with discrete space so $\gamma$ can be regarded as a probability measure over $N$ dimensional vectors with non-negative entries that sum to one)
I reasoned thusly, and I am wondering if I'm making any mistakes. The joint process $\{ (X_n,Y_n)\}$ can be described as follows: $$X_{n+1} \sim Y_n$$ $$Y_{n+1} = P_{X_{n+1}}$$ Let $q$ be the stationary distribution of this process; the projection of $q$ onto $X$ should yield $\pi$ while the projection onto $Y$ should yield $\gamma$. The transition probabilities (Let's use big $Q$ for that) should satisfy $$Q(X_1,Y_1|X_0,Y_0) = P(X_1|X_0)1_{Y_1=P_{X_1}}$$ Then I use the standard identity for stationary expectations. For any $f:X\times Y \to \mathbb{R}$ we have $$\int\limits_{(x,y)}q(x,y)f(x,y) = \int\limits_{(x,y)}\int\limits_{(x',y')}q(x',y')Q( x,y | x',y')f(x,y)$$ I was able to obtain the conclusion, by setting $f(x,y) =y$ and using various properties of probability, including the definition of $Q$ and the projection properties mentioned above.
Is the basic hypothesis sound?