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====== A new coupling technique ======

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:

* Draw $\tilde Y \sim \tilde \pi$. 
  * Draw $X \sim \pi$ and set $Y=X$ with probability $1 -\alpha(X,\tilde Y)$ where $\alpha(x,y)=\frac{\pi(y) \tilde \pi(x)}{\pi(x) \tilde \pi(y)} \wedge 1$. Otherwise set $Y=\tilde Y$.

<WRAP center round tip 80%>  **Proposition. ** $(\tilde Y,Y)$ is a coupling of $(\tilde \pi,\pi)$.   
</WRAP>

<WRAP center round important 80%>
  * What is nice is that we are able to couple these random variables whereas their densities are known only up to a multiplicative constant. I  wonder if it is better to couple in that way: $\pi(y) \propto Q(x,\rmd y) \alpha_{MH}(x,y)$ and $\tilde \pi(y) \propto Q(x',\rmd y) \alpha_{MH}(x',y)$. Up to some tricks, can we deduce a way of coupling two MH starting from different initial distributions, maybe with delayed coupling? Can we compare it to the coupling of Pierre Jacob et al.?
  * Moreover, if we look at the problem today, if $(\tilde X_0,X_0)=\tilde \pi \otimes \pi$ then, $(\tilde X_1,X_1)$ is a coupling of $(\tilde \pi,\pi)$, no?   
</WRAP>

==== Proof ==== 
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. 
    
==== Some comments ====

The coupling probability is given by: 
$$
\PP(\tilde Y=Y)=\int \tilde \pi(\rmd \tilde y) \pi(\rmd x) \alpha (x,\tilde y)=\int \lrb{\pi(x) \tilde \pi(\tilde y) \wedge \pi(\tilde y) \tilde \pi(x)} \rmd x \rmd \tilde y
$$

<WRAP center round tip 80%>
**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 
$$
\int \lrb{\pi(x) \tilde \pi(\tilde y) \wedge \pi(\tilde y) \tilde \pi(x)} \rmd x \rmd \tilde y \leq \int \pi(x) \wedge \tilde \pi(\tilde x) \rmd x
$$
</WRAP>

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