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Let $\pi$, $\tilde \pi$ be two probability measures on the same measurable space $(\Xset,\Xsigma)$.
We draw jointly the couple of random variables $(\tilde Y,Y)$ according to the following procedure:
Proposition. $(\tilde Y,Y)$ is a coupling of $(\tilde \pi,\pi)$.
Obviously, $Y|_{\tilde Y} \sim K(\tilde Y,\cdot)$ where $$ K(\tilde y,\rmd y)=(1-\alpha(y,\tilde y)) \pi(\rmd y) + \lrb{\int \pi(\rmd x) \alpha(x,\tilde y)} \delta_{\tilde y} (\rmd y). $$
We now show that $(\tilde Y,Y)$ is a coupling of $(\tilde \pi,\pi)$. To do so, it is sufficient to check that for any bounded or non-negative function $f$, $\int \tilde \pi(\rmd \tilde y) K(\tilde y, \rmd y) f(y)=\pi(f)$.
Indeed, write, using the detailed balance condition in the second line : \begin{align*} \int \tilde \pi(\rmd \tilde y) K(\tilde y,\rmd y) f(y)&= \int \pi(\rmd y) f(y) \lrcb{\int \lrb{1-\alpha(y,\tilde y)}\tilde \pi(\rmd \tilde y)} + \int \tilde \pi(y) f(y)\rmd y \int \pi(x) \alpha(x,y) \rmd x \\ & = \int \pi(\rmd y) f(y) \lrcb{\int \lrb{1-\alpha(y,\tilde y)}\tilde \pi(\rmd \tilde y)} + \int f(y) \rmd y \lrcb{\int \underbrace{\tilde \pi(y) \pi(x) \alpha(x,y)}_{\tilde \pi(x) \pi(y) \alpha(y,x)} \rmd x} \\ & = \int \pi(\rmd y) f(y) \lrcb{\int \lrb{1-\alpha(y,\tilde y)}\tilde \pi(\rmd \tilde y)} + \int f(y) \rmd y \lrcb{\int \tilde \pi(x) \pi(y) \alpha(y,x) \rmd x}\\ & = \int \pi(\rmd y) f(y) \lrcb{1-\int \alpha(y,\tilde y)\tilde \pi(\rmd \tilde y)} + \int f(y) \pi(\rmd y) \lrcb{\int \alpha(y,x) \tilde \pi(\rmd x)} \\ & = \pi(f) \end{align*} which completes the proof.
The coupling probability is given by: $$ \PP(\tilde Y=Y)=\int \tilde \pi(\rmd \tilde y) \pi(\rmd x) \alpha (x,\tilde y)=\int \lrb{\pi(x) \tilde \pi(\tilde y) \wedge \pi(\tilde y) \tilde \pi(x)} \rmd x \rmd \tilde y $$
Question: we know that $\PP(\tilde Y=Y) \leq \int \pi(x) \wedge \tilde \pi(x) \rmd x$. But I can't see how to prove $$ \int \lrb{\pi(x) \tilde \pi(\tilde y) \wedge \pi(\tilde y) \tilde \pi(x)} \rmd x \rmd \tilde y \leq \int \pi(x) \wedge \tilde \pi(\tilde x) \rmd x $$