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  1. In a discrete-time Markov chain, having constructed the transition distributions, the initial distribution should not affect the limiting distribution (when it exists) in theory. So in MCMC, how to choose the initial state should not affect that the distribution of $X_n$ will converges to the target distribution.

    But I wonder if the initial state can really be arbitrarily chosen? Is there some consideration for picking the initial state?

  2. If we want to get multiple sample points from the same discrete-time Markov chain, which one is better:

    • Starting from a state, after sufficiently long time, all sample points will be kept.

    • Starting from a state, after sufficiently long time, the first sample point will be kept. Then start over again from a state, after sufficiently long time, the second sample point will be kept. So on for the third, fourth, ..., sample points.

      I saw in a note uses the second way, and the initial state is fixed over all runs that start over. I am not sure what benefits can using the same or different initial state bring?

Thanks!

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1 Answer

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  1. If you choose a very bad starting state then it will take longer (perhaps a very long time) to converge to the limiting distribution, but as long as your chain is ergodic it will converge eventually. In MCMC we typically throw away some portion at the beginning of the chain as "burn-in," so having a very bad starting point might mean you need a longer burn-in.

  2. I'm not sure what you want exactly one sample point from each run for--if you're going to go the second way you might as well keep all the draws from each run and use them to estimate the density. Draws from a single run generally aren't independent of each other (which is one reason you might want multiple runs), but there's no reason to just throw away all that information.

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Thanks! (1) Does "a very bad starting state" mean a state with very small probability under the limiting distribution? (2) So in a single sample path (i.e. a single run of MCMC), after burn in, the points are not independently but approximately identically distributed according to the limiting distribution? In different sample paths (i.e. different runs of MCMC), points from different sample paths will be independently and approximately identically distributed according to the limiting distribution? – Tim Dec 9 '12 at 10:14
(1) A state with very small probability, or separated from most of the mass of the distribution (for example, it might be a small secondary mode a long ways off or something). (2) With most MCMC methods there is a positive autocorrelation, so two consecutive draws are often obviously not independent, but two distant draws (how distant depends on how strong the autocorrelation is) can be considered basically independent. Two separate runs starting at the same starting point are also not completely independent even after burnin, but assuming the (P)RNGs are independent they can be considered so. – Jonathan Christensen Dec 9 '12 at 20:34
Thanks! Is "the same starting point" "the same initial state"? "assuming the (P)RNGs are independent", do you mean the (P)RNGs used for both sampling the transition distributions are independent? So even if the initial states are the same, if the (P)RNGs used in sampling the transition distributions are independent, then the two runs can be considered independent? Not necessarily require the initial states to be different? – Tim Dec 9 '12 at 20:40
"starting point" = "initial state," yes. Your interpretation of my RNG comment is also correct. Two chains with the same initial state might technically still not be independent after any finite number of draws, but assuming the chains have mixed reasonably well by the time burn-in is over they can be considered independent for all practical purposes. – Jonathan Christensen Dec 9 '12 at 20:45
Thanks! What does "mixed" and "mixed well" mean? Does "mixed" apply to a single chain, instead of two chains "mixed" together? Does "mixed well" mean that the distribution of $X_t$ is close enough to the limiting distribution? – Tim Dec 9 '12 at 21:01

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