Let $X = (X\_1,X\_2)$ and $\hat X = (\hat X\_1,\hat X\_2)$ be two random variables where $X\_i,\hat X\_i$ are taking values over the Polish space $E\_i$ endowed with their Borel $\sigma$-algebras, where $i=1,2$. 


Let $X\_1$ has a distribution $\mu$ and $\hat X\_1$ has a distribution $\hat \mu$. Furthermore, suppose that $K$ is a conditional kernel on $E\_2$ given $E\_1$ which describe the distribution of $X\_2$ given $X\_1$, i.e.
$$
 P(X\_2\in B|X\_1) = K\_{X\_1}(B)
$$
where $B\in \mathfrak B(E\_2)$ is any Borel measurable set and $\mathsf P$ is a joint distribution of $X = (X\_1,X\_2)$. Let $\hat K$ be defined similarly for $\hat X\_2$ given $\hat X\_1$ and let $\hat P$ be the joint distribution of $\hat X = (\hat X\_1,\hat X\_2)$. 


Suppose that for the total variation it holds that $\|P - \hat P\|>0$. 
Let $E = E\_1\times E\_2$ be the product space. The measure $\Bbb P$ on $E^2$ is called a coupling of $\mathsf P$ and $\hat {\mathsf P}$ if
$$
 \Bbb P\circ\pi^{-1} = \mathsf P,\quad \Bbb P\circ\hat \pi^{-1} = \hat{\mathsf P}
$$
where $\pi,\hat \pi$ are the corresponding projections maps. The coupling $\Bbb P$ is called maximal if


1. it holds that $$\|P - \hat P\| = 2\Bbb P(X\neq \hat X)\tag{1} $$
2. $X$ and $X'$ are $\Bbb P$-independent conditional on $\{X\neq \hat X\}$, i.e.
$$
 \Bbb P(X\in A,\hat X\in \hat A|X\neq \hat X) = \Bbb P(X\in A|X\neq \hat X)\Bbb P(\hat X\in \hat A|X\neq \hat X)
$$
for any sets$A,\hat A\in \mathrm B(E)$.


The maximal coupling always exists and is unique.


I have two questions: 


* is that true that the maximal coupling of $X$ and $\hat X$ is also a maximal coupling of their coordinates $X\_1, \hat X\_1$ and $X\_2, \hat X\_2$? Here I mean the projection of $\Bbb P$ on the correspondent spaces. Or at least, does $(1)$ holds for the projected coupling measures?
* since the maximal coupling $\Bbb P$ is unique, can you suggest how to express $\Bbb P(X = \hat X)$ in terms of $\mu,\hat \mu$ and $K,\hat K$?


I am not experienced in conditioning, so any help is appreciated. I know that
$$
 \Bbb P(X = \hat X) = \Bbb P(X\_1 = \hat X\_1)\Bbb P(X\_2 = \hat X\_2|X\_1 = \hat X\_1)
$$
but I am not sure even how to compute the first term in the RHS.



Coupling of vectors
1. No, this is not true. For example, let $E\_1=E\_2=\{0,1\}$, let $P$ be the uniform distribution on $\{(0,0),(1,1)\}$ and let $\hat P$ be the uniform distribution on $\{(0,1),(1,0)\}$. By (2) the maximal coupling $\mathbb P$ is the product distribution $P\otimes\hat P$, so the pushforward measure on $E\_1\times E\_1$ (or $E\_2\times E\_2$) is the uniform distribution on $\{0,1\}^2$, which is never a maximal coupling.
2. The question already mentions how $\mathbb P(X=\hat X)$ is related to the total variation distance of $P$ and $\hat P$. And $P$ is given in terms of $K$. To make the dependence on $\mu$ more explicit, how about: $$P(B)=\int\_{x\in E\_1} K\_{x}(\{y\in E\_2\colon (x,y)\in B\}) d\mu(x)\qquad (B\in\mathcal B(E\_1\times E\_2)).$$


