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Let

  • $I$ be a finite nonempty set
  • $\zeta$ denote the counting measure on $(I,2^I)$
  • $(E,\mathcal E,\lambda)$ be measure space
  • $p_i:E\to[0,\infty)$ be $\mathcal E$-measurable with $$\int p_i\:{\rm d}\lambda=1\tag1$$ for $i\in I$
  • $w_i:E\to\mathbb R$ be $\mathcal E$-measurable for $i\in I$ with $$\sum_{i\in I}w_i=1\tag2$$

In my application, I'd like to run the Metropolis-Hastings algorithm to sample from the measure $\mu$ on $(I\times E,2^I\otimes\mathcal E)$ defined by $$\mu(\{i\}\times B):=\int_Bw_ip_i\:{\rm d}\lambda\;\;\;\text{for }i\in I\text{ and }B\in\mathcal E.\tag3$$

If we're assuming that each $w_i$ maps into $[0,1]$, then it's easy to see from $(2)$ that $\mu$ is an ordinary probability measure. However, I'd like to allow negative "weights" $w_i$ and hence don't want to impose this assumption. Does the Metropolis-Hastings algorithm still work in this setting, i.e. is $\mu$ still invariant with respect to the transition kernel of the generated Markov chain and does this chain still converge to equilibrium?

0xbadf00d
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    To be clear - are you assuming that the weights can be negative, but the total density will still be nonnegative? – πr8 Aug 11 '19 at 19:18
  • @πr8 Thank you for your comment. I'm only assuming that the weights can be negative (map to $\mathbb R$) and that $(2)$ holds. I'm not sure what you mean by "total density". The density of $\mu$ wrt $\lambda$ is $(i,x)\mapsto w_ip_i(x)$. – 0xbadf00d Aug 12 '19 at 04:17
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    Okay, I misread - I assumed you wanted samples from the $x$-marginal, i.e. from $ p (x) = \sum_i w_i (x) p_i (x) $ (which might be possible). What would it mean to sample from $ p (i, x) = w_i ( x ) p_i ( x )$? If the measure is not necessarily nonnegative, what would the interpretation of the samples be? – πr8 Aug 12 '19 at 08:04
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    Separately - I would generally recommend simplifying the statement of your questions somewhat. The machinery of measure theory seems (in my opinion) to be overkill for the sort of tasks which you describe, and make the questions appear more complicated than is necessary. – πr8 Aug 12 '19 at 08:06
  • @πr8 Say I want to sample from the $x$-marginal. Wouldn't I still need to run the algorithm for $\mu$ given on $(I\times E,2^I\otimes\mathcal E)$? The idea is that I want the algorithm to move to sampling techniques $p_i$ with a large weight $w_i$. That's why I incorporated $i$ into the state space. This is somehow related to multiple importance sampling, but the problem with MIS is that we always generate samples from all techniques $p_i$ even when they have a very small weight. Hope you get what I mean. – 0xbadf00d Aug 12 '19 at 10:13
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    If you are able to write down the marginal for $p(x)$ directly, then you can run a Metropolis-Hastings algorithm directly on $p$. What sort of an application do you have in mind? – πr8 Aug 12 '19 at 13:19
  • @πr8 The application is described here in section 4.2: https://cgg.mff.cuni.cz/~jaroslav/papers/2018-mcmc-survey/2018-sik-mcmc-survey-paper.pdf. (It's the same as in [my other question](https://stats.stackexchange.com/q/420368/222528) you've respondet to.) – 0xbadf00d Aug 12 '19 at 13:21
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    This question - https://stats.stackexchange.com/questions/326351/exact-sampling-from-improper-mixtures - may be relevant. – πr8 Aug 12 '19 at 13:21
  • @πr8 Thank you for the link to the question, but I think that the scenario described there is almost ordinary mixture importance sampling as described, for example, [here](https://statweb.stanford.edu/~owen/mc/Ch-var-is.pdf) on page 27. – 0xbadf00d Aug 12 '19 at 13:26
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/97343/discussion-between-r8-and-0xbadf00d). – πr8 Aug 12 '19 at 13:28
  • @πr8 (Took me a while to respond to your last question properly. Please take a look in the chat; don't know if you get notified.) – 0xbadf00d Aug 12 '19 at 14:52
  • @πr8 After our discussion last week, I've rephrased the problem in this question: https://stats.stackexchange.com/q/422838/222528. Maybe you can take a look. – 0xbadf00d Aug 19 '19 at 19:17

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