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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.

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  • $\begingroup$ 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. $\endgroup$
    – Xi'an
    Jan 25 at 7:01
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There is an error in the paper, indeed.

I think you state that the paper is wrong with the following "claim": enter image description here

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.

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  • $\begingroup$ 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 $\overline\rho(y,x)>0\Leftrightarrow\rho(x)p(x,y)>0$. $\endgroup$
    – 0xbadf00d
    Feb 11 '20 at 7:41
  • $\begingroup$ @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 $\endgroup$
    – Aksakal
    Feb 12 '20 at 15:09
  • $\begingroup$ @Aksakal Yes, but this yields only one implication. What about the other direction? $\endgroup$
    – 0xbadf00d
    Feb 13 '20 at 5:06
  • $\begingroup$ @0xbadf00d, why do you need it in other direction? as I wrote, this result is not important for the paper anyways $\endgroup$
    – Aksakal
    Feb 13 '20 at 15:32
  • $\begingroup$ @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.) $\endgroup$
    – 0xbadf00d
    Feb 13 '20 at 17:04

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