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Question:

Is there a rule of thumb for setting a non-diagonal covariance matrix for your Metropolis-Hastings proposal distribution? References are appreciated.

Background:

Say I have some posterior distribution I am interested in obtaining samples from $p(\theta|y)$. I choose an initial proposal distribution for a Metropolis-Hastings algorithm. It is $$ q(\theta^*|\theta) = \text{N}(\theta^*; \theta, \text{diag}[\sigma_1^2, \ldots, \sigma_p^2]). $$

Notice that the covariance matrix is diagonal. I choose the elements of this diagonal covariance matrix according to some other rule of thumb (e.g. $\sigma_i \overset{set}{=}2.38/\sqrt{d}).$

It goes for a while, but then it gets stuck. I assume that this is because of the fact that elements of $\theta$ are correlated. Is there any justification for setting $q$'s correlation matrix to the sample correlation matrix? Here's a pairwise scatterplot of all the iterations:

enter image description here

Edit:

I just opened my copy of a Bayesian textbook, and it describes something similar to what I suggested. It's a little light on references, though. So I am primarily interested in references, here.

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  • $\begingroup$ If you set the proposal distribution correlation matrix as the sample correlation matrix up until that time, your proposal distribution will depend on all the steps before the current step, and not just the previous step. Thus, this will not lead to a time homogenous Markov chain, and the usual theoretical conclusions cannot be made. However, these proposal distributions lead to Adaptive MCMC methods, which have proven to be useful in certain situations this. $\endgroup$ – Greenparker Jun 2 '17 at 9:27
  • $\begingroup$ If you don't want to use the adaptive MCMC theory, you could run your native proposal for a long time, calculate the correlation matrix, and then independently start another sampler with this correlation matrix in the proposal distribution. $\endgroup$ – Greenparker Jun 2 '17 at 9:29
  • $\begingroup$ @Greenparker thanks for the guidance. I'll do a little digging :) $\endgroup$ – Taylor Jun 2 '17 at 20:21

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