Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
@Colin: thanks, fixed –  Ilya Jan 19 '13 at 11:47
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 $X_1'$. –  Anthony Quas Jan 20 '13 at 19:00
@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). –  Ilya Jan 21 '13 at 13:12
add comment

1 Answer

up vote 2 down vote accepted
  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: $$P(B)=\int_{x\in E_1} K_{x}(\{y\in E_2\colon (x,y)\in B\}) d\mu(x)\qquad (B\in\mathcal B(E_1\times E_2)).$$

share|improve this answer
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? –  Ilya Jan 19 '13 at 11:49
I've corrected the $P$'s. I'm afraid I don't know an instructive reference, sorry. –  Colin McQuillan Jan 19 '13 at 16:40
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/… –  Colin McQuillan Jan 19 '13 at 16:42
Thanks a lot for this reference! –  Ilya Jan 20 '13 at 10:20
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 $\Bbb P(X\in A,\hat X\in \hat A)$? –  Ilya Jan 20 '13 at 10:45
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.