# 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.

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 $$\mathop{A}\left(x\rightarrow y\right)=\min\left(1, \dfrac{\mathop{P}\left(y\right)\mathop{q}\left(y\rightarrow x\right)}{\mathop{P}\left(x\right)\mathop{q}\left(x\rightarrow y\right)}\right)$$.

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, Tjelmeland 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.

• Thanks for the answer. There could be many acceptance methods but Metropolis happened to use $min( 1, \frac {P(y)}{P(x)} )$ by setting A(y->x) to 1 because it is OK to do so?
– mon
Dec 12, 2022 at 21:44