In Metropolis-Hastings sampling, if every draw of my proposal distribution (Q) is independent from the previous draw, is the convergence to the stationary distribution still guaranteed? To be more precise, every draw is a fresh draw from the proposal (Q) and the acceptance rate is calculated as follows:

 a = min{1,(P(X')Q(X))/(P(X)Q(X'))}

Where X' is the new drawn sample, and X is the current sample, and P is the actual distribution. Also, can we still call this an MCMC sampling technique?


1 Answer 1


Yes, this is a special case of the Metropolis-Hastings algorithm, so it works and can be called a MCMC algorithm.

Nothing forbids setting the proposal distribution $Q(x_{\mathrm{new}} \mid x_{\mathrm{old}})$ to be independent of $x_{\mathrm{old}}$ so the proof for detailed balance condition in Metropolis-Hastings directly holds for this proposal distribution.

Note that to guarantee convergence to the correct stationary distribution, we must further assume that the proposal $Q$ has positive probability for all regions where $P$ has positive probability.

Edit: as Tom Minka pointed out, this is known as independence chain or independence sampler and is discussed in this answer and this webpage.

  • 2
    $\begingroup$ To add to Juho's answer, this is known as an independence chain and is described nicely by this answer and this webpage. $\endgroup$
    – Tom Minka
    Commented Oct 27, 2014 at 14:55
  • $\begingroup$ Thanks for the answer. My confusion was based on the fact that the notion of chain in Markov chain can only be defined if states are dependent on the previous ones. But as you said, this is called Independent MH Sampling, and is still a Markov chain because of the way we calculate the acceptance rate. $\endgroup$ Commented Oct 28, 2014 at 1:02
  • $\begingroup$ I added the useful additional information by @TomMinka into the answer. $\endgroup$ Commented Oct 28, 2014 at 9:50
  • $\begingroup$ @user3639557 Note that even if all the variables were independent, it would still be a Markov chain (though not have the correct stationary distribution). The Markov property required conditional independence of the current state and the history before the previous state conditional on the previous state. The current state and the previous state can be independent, which is a special case of Markov chains. If you want to continue discussion about this, you may want to post a separate question about that, as this is rather unrelated to the original question here. $\endgroup$ Commented Oct 28, 2014 at 9:53

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