In the Metropolis algorithm, suppose I start my algorithm at time $t-1$ with a draw $\theta^{t-1}$ from my target distribution $p(\theta|y)$.
It can be shown that $\theta^t$ and $\theta^{t-1}$ are symmetric in that
$$ P(\theta^t = \theta_a, \theta^{t-1} = \theta_b) = P(\theta^{t-1} = \theta_a, \theta^{t} = \theta_b) $$
Based on this, it is written in Bayesian Data Analysis 3 by Gelman that $\theta^{t}$ and $\theta^{t-1}$ have the same marginal distributions and so $p(\theta|y)$ is the stationary distribution of the Markov Chain of $\theta.$
Could someone explain why $\theta^{t}$ and $\theta^{t-1}$ having the same marginal distributions imply $p(\theta|y)$ becomes the stationary distribution of my chain? Thanks.