# In Metropolis Algorithm, if draws $\theta^{t-1}$ and $\theta^t$ have the same marginals, why is the target is the same as the stationary distribution?

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.

• $\theta^{t-1}$ and $\theta^t$ having the same marginal distribution, $p(\theta \mid y)$, by definition means that this marginal distribution, $p(\theta \mid y)$, is the stationary distribution of the Markov chain, so it is unclear what needs to be explained. Feb 16, 2017 at 18:58
• Does my comment answer the question? Feb 27, 2017 at 19:04
• So you are saying that it is something we already assumed before doing the algorithm that it is the stationary distribution? Feb 28, 2017 at 2:50
• What do you understand "stationary distribution" to mean? Feb 28, 2017 at 5:57