Initializing a Markov chain with more than 1 sample is totally erroneous, right? If someone talks of initializing Markov chains (plural) with m samples, then they are necessarily talking about m chains, yeah?
cc @colindcarroll @jim_savage_
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Replying to @willwolf_ @jim_savage_
Sounds like not what was being discussed, but have been reading up on ensemble samplers (a la `emcee`) this week, but unclear to me whether we would even call the walkers "chains": https://msp.org/camcos/2010/5-1/camcos-v5-n1-p04-s.pdf …
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Glad to be taking the contrarian position against
@aloctavodia and@jim_savage_ . I'm sure this will go well.pic.twitter.com/gECKsmuKQS1 reply 0 retweets 2 likes -
I change my mind. If the chains comunicate somehow, then they are a single big chain being explored from several starting states.
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Replying to @aloctavodia @colindcarroll and
Err... Is this a semantics problem, grammar problem, or vocabulary problem?
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Replying to @ericmjl @colindcarroll and
A mathematical problem with a twist of semantics :) are the markov chains independent? If not they are just a single chain. They are generally not, otherwise they will be sampling from different distributions, like separated modes of multimodal distributions or nonconverged sampl
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Replying to @aloctavodia @ericmjl and
If we were being careful, I guess L walkers in R^n generate 1 (honest MCMC) chain in R^nL, which was initialized with a single point in R^nL. They happen to (asymptotically) map to samples from a particular distribution in R^n.
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I don't think that's quite right, because you propose and then accept/reject for each walker individually, rather than accepting all L or rejecting all L.
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