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Let

  • $(E_i,\mathcal E_i)$ be a measurable space
  • $\kappa$ be a Markov kernel with source $(E_1,\mathcal E_1)$ and target $(E_2,\mathcal E_2)$

Assume $\kappa$ has a positive density with respect to a measure $\mu$ on $(E_2,\mathcal E_2)$, i.e. there is a $\mathcal E_1\otimes\mathcal E_2$-measurable $f:E_1\times E_2\to(0,\infty)$ with $$\kappa(x,\;\cdot\;)=f(x,\;\cdot\;)\mu\;\;\;\text{for all }x\in E_1.$$ Now, let $\nu$ be a probability measure on $(E_1,\mathcal E_1)$ and $$\overleftarrow\kappa_\nu(y,\;\cdot\;):=\frac1{c(y)}f(\;\cdot\;,y)\nu\;\;\;\text{for }y\in E_2,$$ where $c(y):=\int\nu({\rm d}x)f(x,y)$ (and we assume that $c(y)<\infty$) for $y\in E_2$.

How can we show that $\overleftarrow\kappa_\nu$ is the reverse kernel of $\kappa$ with respect to $\nu$ (see Definition 2.1.2), i.e.$^1$ $$\int\nu({\rm d}x)\int\kappa(x,{\rm d}y)g(x,y)=\int\nu\kappa({\rm d}y)\int\overleftarrow\kappa_\nu(y,{\rm d}x)g(x,y)\tag1$$ for all bounded and $\mathcal E_1\otimes\mathcal E_2$-measurable $g:E_1\times E_2\to\mathbb R$?

Let $$\pi:E_1\times E_2\to E_2\times E_1\;,\;\;\;(x,y)\mapsto(y,x).$$ It's easy to observe that the left-hand side of $(1)$ is equal to$^2$ $$\int g\:{\rm d}(\nu\otimes\kappa)\tag2$$ and the right-hand side is equal to $$\int g\circ\pi^{-1}\:{\rm d}(\nu\kappa\otimes\overleftarrow\kappa_\nu).\tag3$$

Now, it's easy to see that$^3$ $$\pi_\ast(\nu\otimes\kappa)=\mu\otimes\overleftarrow\kappa_\nu\tag4$$ and $$\nu\otimes\kappa=\left(\pi^{-1}\right)_\ast(\mu\otimes\overleftarrow\kappa_\nu)\tag5.$$

$(1)$ is claimed in the linked document below the Definition. Could it be the case that their definition of "reverse kernel" is broken? From a terminological point of view it would make more sense to me if $\nu\kappa$ on the right-hand side of $(1)$ would be replaced by $\nu$.


$^1$ $\nu\kappa$ denotes the composition of $\nu$ and $\kappa$.

$^2$ $\nu\otimes\kappa$ denotes the product of $\nu$ and $\kappa$.

$^3$ $\pi_\ast(\nu\otimes\kappa)$ denotes the pushforward measure of $\pi$ with respect to $\nu\otimes\kappa$.

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