MCMC - How to derive the acceptance ratio from Markov chain detailed balance? Please explain how the acceptance ratio:
$$\rm A(x\to y) =\min\left(1,\frac{P(y)q(y\to x) }{P(x) q(x\to y) }\right)$$
is derived from the detailed balance:
$$\rm \frac{A(x\to y) }{A(y\to x)}=\frac{P(y)q(y\to x) }{P(x) q(x\to y) }  .$$
Reading the articles but not sure how the ratio is logically derived from the detailed balance.

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*Markov Chain Monte Carlo (MCMC)


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*Wikipedia Metropolis–Hastings algorithm



*MCMC - Metropolis Hasting: formal derivation of detailed balance
 A: If $\mathop{P}\left(x\right)\mathop{q}\left(x\rightarrow y\right) > \mathop{P}\left(y\right)\mathop{q}\left(y\rightarrow x\right)$, then the process moves from $x$ to $y$ too often.
Setting $\mathop{A}\left(x\rightarrow y\right)=\dfrac{\mathop{P}\left(y\right)\mathop{q}\left(y\rightarrow x\right)}{\mathop{P}\left(x\right)\mathop{q}\left(x\rightarrow y\right)}$ and $\mathop{A}\left(y\rightarrow x\right)=1$ achieves detailed balance by reducing the probability that a move from $x$ to $y$ is made such that $\mathop{P}\left(x\right)\mathop{q}\left(x\rightarrow y\right)\mathop{A}\left(x\rightarrow y\right)=\mathop{P}\left(y\right)\mathop{q}\left(y\rightarrow x\right)$ holds.
Analogously, if $\mathop{P}\left(x\right)\mathop{q}\left(x\rightarrow y\right) \leq \mathop{P}\left(y\right)\mathop{q}\left(y\rightarrow x\right)$, take $\mathop{A}\left(y\rightarrow x\right)=\dfrac{\mathop{P}\left(x\right)\mathop{q}\left(x\rightarrow y\right)}{\mathop{P}\left(y\right)\mathop{q}\left(y\rightarrow x\right)}$ and $\mathop{A}\left(x\rightarrow y\right)=1$.
Written more compactly, we thus have $\mathop{A}\left(x\rightarrow y\right)=\min\left(1, \dfrac{\mathop{P}\left(y\right)\mathop{q}\left(y\rightarrow x\right)}{\mathop{P}\left(x\right)\mathop{q}\left(x\rightarrow y\right)}\right)$.
A: The acceptance probability $A(x\to y)$ is not derived from detailed balance. It is the opposite: given this choice of acceptance probability the Markov kernel satisfies detailed balance
$$
p(x)\int q(x\to y)A(x\to y)\,\text dy=\int p(y) q(y\to x) A(y\to x)\,\text dy
$$
But so do (infinitely) other choices of transition kernels.
For instance, Tjemeland defines a multiple choice proposal with acceptance probability
$$A_\ell(x\to y_\ell)=p_\ell(y_\ell)\Big/ p(x)+\sum_{j=1}^m p_j(y_j)$$
that generalises Barker’s (1965) acceptance probability,
$$A(x\to y)=p(y)\Big/ p(x)+p(y)$$
Delayed acceptance is another version of a valid transition.
