MCqMC 2016 [#1]

mcqmc1This week, I attend the MCqMC 2016 conference in Stanford, which is quite an exciting gathering of researchers involved in various aspects of Monte Carlo methods. As Art Owen put it in his welcoming talk, the whole Carlo family is there! (Not to mention how pleasant the Stanford Campus currently is, after the scorching heat we met the past week in Northern California inlands.) My talk is on folded Markov chains, which is a proposal Randal and I have been working on for quite a while, with Gareth joining us more recently. The basic idea was inspired from a discussion I had about a blog post, so long ago that I cannot even trace it! Namely, when defining an inside set A and an outside set, such that the outside set can be projected onto the inside set, one can fold both the target and the proposal, essentially looking at a collection of values for each step of the Markov chain. In other words, the problem can be reduced to A at essentially no cost and with the benefits of a compact support A and of a possibly uniformly ergodic Markov chain. We are still working on the paper, but the idea is both cool and straightforward, so we decided to talk about it at Nordstat 2016 and now MCqMC 2016.

3 Responses to “MCqMC 2016 [#1]”

  1. Dan Simpson Says:

    This is interesting! To my eyes, it looks like you’re choosing A_0 as the “bulk” of the distribution, and its complement as the tail. In that case, maybe choosing the boundary of A_0 as the point where the second derivative of the density vanishes would be useful.

    For the Cauchy that would be +/- 3^{-1/2}.

    • Thanks, Dan: it however does not seem to matter very much provided I catch a high enough density region as A⁰ and as I try to produce this set automatically from an early if imperfect MCMC run I am afraid I cannot afford the luxury of looking at optimal sets. Too bad you could not make it here, I am sure you would have enjoyed the conference!

      • Dan Simpson Says:

        It looks really fun! It’s just a shame it’s been such a busy summer :(

        I was thinking that you don’t need the normalised density for the zero contour of the second derivative, so it may be cheaper than your current method. (In 2 or more dimensions, you’d have to approximately solve det(H(x))=0, so that may be harder than a pilot run…)

        My other thought (which is not very deep), is that given the folding can be seen as the Lie group action, there’s probably some natural way to combine these moves with a HMC method. You should grab Betancourt…

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s