# Erroneous expression for Metropolis-Hastings acceptance ratio in a paper

Let

• $$(E,\mathcal E)$$ be a measure space;
• $$\rho:E\to[0,\infty)$$ be $$\mathcal E$$-measurable, $$p:E^2\to[0,\infty)$$ be $$\mathcal E^{\otimes2}$$-measurable, $$r(x,y):=\left.\begin{cases}\displaystyle\frac{\rho(y)p(y,x)}{\rho(x)p(x,y)}&\text{, if }\rho(x)p(x,y)>0\\1&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }x,y\in E$$ and $$\overline\rho(x,y):=\left.\begin{cases}\displaystyle\frac{\rho(y)}{p(x,y)}&\text{, if }\rho(x)p(x,y)>0\\0&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }x,y\in E.$$

Assuming $$\forall y\in E:\left(p(y)>0\Rightarrow\forall x\in G:q(x,y)>0\right),\tag1$$ are we able to show that $$\tilde r(x,y):=\left.\begin{cases}\displaystyle\frac{\overline\rho(x,y)}{\overline\rho(y,x)}&\text{, if }\overline\rho(y,x)>0\\1&\text{, otherwise}\end{cases}\right\}=r(x,y)\tag2$$ for all $$x,y\in E$$?

This claim is made in this paper on page 8.$$^1$$ However, it should hold $$\tilde r(x,y)=\left.\begin{cases}\displaystyle\frac{\rho(y)p(y,x)}{\rho(x)p(x,y)}&\text{, if }\rho(x)\rho(y)>0\\1&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for all }x,y\in E\tag3$$ and hence, for example, if $$x,y\in E$$ with $$\rho(x)p(x,y)>0$$ and $$\rho(y)=0$$, then $$r(x,y)=0$$, but $$\tilde r(x,y)=1$$.

Am I missing something? If not, can we fix this?

$$^1$$ They actually claim that $$\left.\begin{cases}\displaystyle\frac{\overline\rho(y,x)}{\overline\rho(x,y)}&\text{, if }\overline\rho(x,y)>0\\1&\text{, otherwise}\end{cases}\right\}=r(x,y)\;\;\;\text{for all }x,y\in E,\tag4$$ but since this is obviously wrong, I suspected that they mean $$\tilde r$$ instead.

• I do not think this is of importance: while the Markov chain remains in the exterior of the support of $\rho$ it is free to do whatever it wants. The sooner it leaves this transient region the better. Jan 25 at 7:01

There is an error in the paper, indeed.

I think you state that the paper is wrong with the following "claim":

There's really no proof or derivation of the expression. All they did was to plug the definition of $$\bar \rho$$ on the same page into the definition of $$r(x,y)$$ on p.5. Unfortunately, while doing so they messed up. Here's why.

Both definitions are in your question, first two equations. I can re-write $$r(x,y)$$ as follows: $$\frac{\rho(y)}{p(x,y)}\frac 1 {\left(\frac{\rho(x)}{p(y,x)}\right)}=\bar \rho(x,y)\frac{1}{\bar \rho(y,x)}$$

I don't think this error impacts the rest of the paper though, because it's not used anywhere further in the text explicitly.

• Thank you for your answer. Your last displayed equation is my equation $(2)$. My problem is that I don't understand why this equation holds, since we not have the equivalence $\overline\rho(y,x)>0\Leftrightarrow\rho(x)p(x,y)>0$. Feb 11 '20 at 7:41
• @0xbadf00d, $\rho(x)>0\implies p(y,x)>0\implies\bar r(y,x)>0$, see the statement on Assumption 1 on p.8 of the paper Feb 12 '20 at 15:09
• @Aksakal Yes, but this yields only one implication. What about the other direction? Feb 13 '20 at 5:06
• @0xbadf00d, why do you need it in other direction? as I wrote, this result is not important for the paper anyways Feb 13 '20 at 15:32
• @Aksakal We need the other direction since otherwise the claimed equality only holds on a subset. (I know that this result is not important for the paper, but it's important in my application.) Feb 13 '20 at 17:04