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.

  • $\begingroup$ I think you should re-define maximal coupling to be any coupling satisfying 1. It's maximal in the sense that it puts maximum weight on the diagonal. This is very much non-unique. But now there is a maximal coupling of $(X_1,X_2)$ and $(X_1',X_2')$ which is also a maximal coupling of $X_1$ and $X_1'$. $\endgroup$ Jan 20, 2013 at 19:00
  • $\begingroup$ @Anthony: I agree - and I am pretty sure that one come up with a maximal coupling of $P$ and $\hat P$ doing it sequentially - I just wondered whether it's possible for the maximal coupling which satisfies one more additional assumption (aka $\gamma$-coupling according to Lindvall). $\endgroup$
    – SBF
    Jan 21, 2013 at 13:12

1 Answer 1

  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: $$P(B)=\int_{x\in E_1} K_{x}(\{y\in E_2\colon (x,y)\in B\}) d\mu(x)\qquad (B\in\mathcal B(E_1\times E_2)).$$

  • $\begingroup$ Thank you. Do you have a reference for the proof of the last formula? It's like a Fubini theorem, but I've never seen a proof of its version for kernels. Also, in your first example, do you mean that $P$ and $\hat P$ in place of $\mu$ and $\hat \mu$ - just to keep it consistent with the notation in OP? $\endgroup$
    – SBF
    Jan 19, 2013 at 11:49
  • $\begingroup$ I've corrected the $P$'s. I'm afraid I don't know an instructive reference, sorry. $\endgroup$ Jan 19, 2013 at 16:40
  • $\begingroup$ You could try Propositions 3.5.4 and 3.5.6 of An Introduction to Measure and Probability by John Christopher Taylor: books.google.co.uk/… $\endgroup$ Jan 19, 2013 at 16:42
  • $\begingroup$ As far as I've checked your example, it is correct - thanks again. I guess, I shall accept you answer - but could you suggest how to compute $\Bbb P(X_1 = \hat X_1)$ or maybe you know how to express $\Bbb P(X\in A,\hat X\in \hat A)$? $\endgroup$
    – SBF
    Jan 20, 2013 at 10:45

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