Thanks again. We are currently revising the paper towards submitting to a new journal and we will include references to your solution. Can you please let me know how you would like to be quoted on this?

]]>Thanks again for revisiting this!

Philip ]]>

Thank you Philip. Without checking anything yet (!), I am surprised at managing to keep the cheap versus expensive separation and yet achieving detailed balance.

]]>In the decomposition, let the rho_i (i=1:c) be the cheap factors to compute, phi_i (i=1:d) be the expensive factors. I.e., [pi*q(theta->eta)]/pi*q(eta->theta)]= [Prod{rho}]*[Prod{phi}].

Take as acceptance probability:

min_(j=1:c){1,Prod_(i=1:j){rho_i}} * min_(k=1:d){1,Prod_(i=1:k){phi_i}}. You get to compute the cheap terms first, expensive terms last, but there is still a symmetry requirement on ordering of the set of cheap terms, and a symmetry requirement on the set of expensive terms. I have algebra that (appears to) confirm that this satisfies detailed balance, but it is too cumbersome to type here.

Cheers,

Philip ]]>