# Mathematical reference for the convergence in distribution of the Gibbs sampler

This question is in some sense the intersection of this question and this question. I have read up on the Gibbs sampler, and am now asking for an introduction to the Gibbs sampler for mathematicians. I like plausibility arguments and simulations as much as the next guy, but would like to know more about the math behind it.

I have read Casella/George and some other review articles, which all give nice heuristics and plausibility arguments for why the Gibbs sample should work, but I lack a mathematical argument. I tried to patch one up yesterday, but it appears to demand some knowledge I don't yet possess. I have the intuition, but I lack the mathematics with regards to the Gibbs sampler.

What is a good reference for a proof of the convergence of the Gibbs sampler for a nice class of distributions? I don't care much for rate of convergence (yet), although if that comes for essentially free, I'll be interested in that as well. :) Casella/George refer to Gelfand/Smith, who in turn refer to Geman/Geman. Geman/Geman write for image processing people and seem to defer some of the proofs to other sources, so it is not an ideal source. Hopefully someone has written a review article, book or lecture notes containing a nice proof since.

I know a "fair amount" of mathematics, so Lebesgue integrals, topology and other stuff don't scare me per se, although of course all else being equal I prefer simple proofs.

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If I recall correctly you only need to prove that you have a Markov chain with nice properties, then the convergence proof comes from Markov chain theory. Maybe that is the reason why everybody keeps referring instead of giving exact proof. –  mpiktas Jul 29 '11 at 8:01
That is my also suspicion, which is why I am hoping that somewhere someone has written a Markov chain theory text which culminates with proving some properties of a Gibbs sampler. But perhaps using full-blown Markov chain theory is using a sledgehammer to kill a mosquito, and there is a simpler proof (in the sense that it doesn't depend on too much Markov chain machinery). –  Har Jul 29 '11 at 8:11
In my personal experience, mathematicians are quite happy to use sledgehammers for mosquito killing, as long as it kills mosquito for sure. Afterwards some investigate whether there are alternative easier ways, but this does not always happen. I think that convergence result for Markov chain is not considered complicated, especially in simple cases. For finite-state Markov chain if I recall the proof is not that complicated. For more complicated Markov chains the proofs should go along similar lines. Asking this on mathoverflow.net or math.SE might give more answers. –  mpiktas Jul 29 '11 at 8:23
In my personal experience, mathematicians also take a delight in avoiding sledge hammers (non-complex proofs of the prime number theorem comes to mind)! :) I would imagine that a finite-state Markov chain could be proven along the lines of the Bernoulli case outlined by Casella-George. So I guess Casella-George suitably adapted is a proof of convergence for distributions on finite (and perhaps discrete) sets (which is a nice, but a bit small, class of distributions admittedly). –  Har Jul 29 '11 at 8:33