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I have read the Gelman-Rubin method for check the convergence in MCMC on $m\geq 2$ chain, but when I work with only one chain, what can i do to check the convergence?

Is there any method that works fine with $m=1$ chain?

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  • $\begingroup$ Is there some reason you have to use just one chain? The standard procedure is to run multiple chains, observe the diagnostic, and then combine the chains at the end for one long chain. $\endgroup$ – user44764 May 22 '14 at 15:15
  • $\begingroup$ I want to test a simple implementation (for my problem) of Metropolis-Hasintgs with only one start point (one chain), only if I don't have good results then go to implement another version with multiple start point (more chain). But without a convergence method i can't test the one chain case. $\endgroup$ – Neptune May 22 '14 at 15:22
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    $\begingroup$ There several other options that I'm aware of for chain diagnostics besides the Gelman-Rubin test. See here for a brief overview of them. I would be cautious on using any diagnostic though, as they can be overly sensitive (especially the Geweke diagnostic). I recommend scrutinizing trace plots if you can (although this is quite risky itself if you're not experienced). $\endgroup$ – user44764 May 22 '14 at 15:34
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First, the Gelman-Rubin test does not check convergence of an MCMC Markov chain but simply an agreement between several parallel chains: if all chains miss a highly concentrated but equally highly important mode of the target distribution, the Gelman-Rubin criterion concludes to the convergence of the chains. Using multiple chains to check for convergence is quite reasonable if costly, but one can never "be sure to have reached stationarity". Simulated tempering can help, though.

Second, to check convergence or stationarity on a single Markov chain $(x_t)_{t=1,\ldots,T}$, one needs to know a lot about the target distribution $\pi(x)$ because, otherwise, all you can judge from the sequence of values $x_1,x_2,\ldots,x_T$ is their stability. Hence only the ability of the MCMC sampler to explore the current region of the support of $\pi$. To go beyond that requires an assessment of this support and of the "missing mass", i.e. the mass under $\pi$ of the remainder of the space. This is an extremely rare occurrence.

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