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.

  • 3
    $\begingroup$ $\theta^{t-1}$ 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. $\endgroup$ – Juho Kokkala Feb 16 '17 at 18:58
  • $\begingroup$ Does my comment answer the question? $\endgroup$ – Juho Kokkala Feb 27 '17 at 19:04
  • $\begingroup$ So you are saying that it is something we already assumed before doing the algorithm that it is the stationary distribution? $\endgroup$ – user321627 Feb 28 '17 at 2:50
  • $\begingroup$ What do you understand "stationary distribution" to mean? $\endgroup$ – Juho Kokkala Feb 28 '17 at 5:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.