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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2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$
    – mon
    Dec 12, 2022 at 21:44

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