Let $\pi$ be a target probability distribution on a measurable space $(E, \mathcal{E})$. MCMC obtains dependent samples from $\pi$ by using a Markov Chain with transition kernel $\mathrm{K}:E\times \mathcal{E}\to[0,1]$ that leaves $\pi$ invariant $$ \pi = \pi \mathrm{K} $$ In Metropolis-Hastings we check a weaker condition called detailed balance (which implies $\pi$-invariance). Detailed balance is a notion of reversibility and it is often defined by saying that the measures $\pi(d x) \mathrm{K}(x, dx') = \pi(dx')\mathrm{K}(x', dx)$ are equivalent, which people also write as $$ \int_A \pi(dx)\mathrm{K}(x, B) = \int_B \pi(dx)K(x, A) $$

In measure-theoretic terms, how can one define detailed balance?

My guess is that we define a product measure $\mu$ on $(E\times E, \mathcal{E}\otimes \mathcal{E})$ that is defined as $$ \mu(A\times B)= \pi(A) \mathrm{K}(x, B) \qquad A, B\in\mathcal{E} \qquad x\in A $$ and then the statement above essentially means this? $$ \mu(A\times B) = \mu(B\times A) $$ I guess this would make sense because we could write $$ \int_A \pi(dx) \mathrm{K}(x, B) = \int_{A} \pi(dx) \int_B \mathrm{K}(x, dy) = \int_{A\times B} \pi(dx) \mathrm{K}(x, dy) = \int_{A\times B} \mu(dx, dy) = \mu(A\times B) $$

  • $\begingroup$ I guess my key question is: is the measure they are talking about a product measure? $\endgroup$ Jun 29 at 13:34
  • 1
    $\begingroup$ Yes the measure as defined is symmetric. $\endgroup$
    – Xi'an
    Jun 29 at 14:04
  • $\begingroup$ @Xi'an Thank you! Is $\mu$ defined correctly here for Metropolis-Hastings then? $\endgroup$ Jun 29 at 14:09
  • 1
    $\begingroup$ Yes, I think so! $\endgroup$
    – Xi'an
    Jun 29 at 16:05
  • $\begingroup$ @Xi'an thank you! $\endgroup$ Jun 29 at 16:49

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