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When looking back at the Metropolis-Hastings algorithms, I think I have been missing the most important point:

How do Metropolis-Hastings algorithms ensure the constructed MC has a limiting distribution same for all initial distributions?

From what I have read, the algorithms try to construct a MC that has the target distribution being a stationary distribution and with respect to which the MC is reversible. However, a stationary distribution may not be the limiting distribution, and it is the limiting distribution only when the limiting distribution exists. For example, for a finite state MC, the limiting distribution exists if and only if the MC is irreducible and aperiodic. However, I haven't seen a description of the Metropolis-Hastings algorithms that states explicitly how existence of the limiting distribution can be ensured by properly choosing the proposal/auxiliary MC and accept/reject probabilities at each iteration.

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@Procrastinator: My post is asking this: how do you know the constructed MC by MH has the limiting distribution? In other words, what does MH do to ensure the distribution of $X_t$ will converge as $t \to \infty$? Without knowing that, no matter how long you run the algorithms, the distribution may even not converge! –  Tim Dec 12 '12 at 22:39
    
Yes, but this is the first thing you have to check when you propose a new sampler. And this is, of course, well-known. Then, Hastings (1970) is the best reference is you want the technical details. –  user10525 Dec 12 '12 at 22:44
    
How do you "check" this first thing? Thanks for the discussion and the reference. –  Tim Dec 12 '12 at 22:47
    
The basic idea is to prove that the algorithm produces an ergodic Markov chain that satisfies the detailed balance property with respect to the desired distribution and therefore it produces the desired samples. –  user10525 Dec 12 '12 at 22:58
    
What do the MH algorithms do to ensure the constructed MC is ergodic (in finite state case, ergodic iff irreducible and aperiodic)? Without having taken any step for that purpose, it is not meaningful to prove the constructed MC is ergodic because it may not be, right? –  Tim Dec 12 '12 at 23:01

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