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

A coupling of two probability measures $P,\tilde P$ on a Borel space $X$ is any probability measure on $X^2$ whose one-dimensional marginals are $P$ and $\tilde P$. In particular, for any such coupling $\Bbb P$ we have $$ 2\cdot \Bbb P(X^2\setminus\Delta_X)\geq\|P - \tilde P\| \tag{1} $$ where $\Delta_X = \{(x,x):x\in X\}$ is the diagonal, and $\|\cdot\|$ denotes the total variation norm. At the same time, there always exists a maximal coupling $\Bbb P^*$ such that the equality holds in $(1)$.

Extension of these results to the case when $P$ and $\tilde P$ are non-probability measures, or at least sub-probabilities does not seem to be complicated, however I wonder whether it has been already addressed somewhere. So far a search on google did not help.

share|improve this question
1  
Probably I've missed something: What do you mean by an extension of this result to non-probability measures? If $P$ and $\tilde P$ have the same total mass (but are not probabilities) anything is similar, but if the total mass is different, what should a coupling be? –  Dirk Aug 15 '13 at 15:12
1  
@Dirk, it should, I think, be a measure $\mu$ on $X^2$ such that $\int_{y \in X} d\mu(x,y) = P(x)$ and $\int_{x \in X} d\mu(x,y) = \tilde{P}(y)$. –  usul Aug 15 '13 at 23:04
    
(hope my notation makes sense.) @Ilya, I don't know of any literature, but maybe the keyword "Wasserstein metric" helps (although the metric is defined for probability distributions). en.wikipedia.org/wiki/Wasserstein_metric –  usul Aug 15 '13 at 23:13
    
The point is that if $P$ and $\tilde P$ have different mass there will be no coupling $\mu$: $\int_{x\in X}\int_{y\in X}d\mu(x,y) = \int_{x\in X}dP(x) \neq \int_{y\in X}d \tilde P(y) = \int_{y\in Y}\int_{x\in X}d\mu(x,y)$. For measures with different mass you may check this answer mathoverflow.net/a/120364/9652 which gives some hints (and a pointer to Gromov's "Metric structures…" Chapter $3\tfrac12$). –  Dirk Aug 16 '13 at 6:44
    
@Dirk: thanks for the comment, but for the case when $P$ and $\tilde P$ have different total masses, can't we yet define $\mu:= P\otimes\tilde P$ to be at least one coupling? –  Ilya Aug 16 '13 at 12:04
show 5 more comments

1 Answer

I thought, I could turn the comments into an answer…

The approach by couplings does not work without modifications and the reason is that couplings do not exist if the measures have different total mass: If $P$ and $Q$ are two measures on $X$ with different total masses which were coupled by $\mu$, then $$ \int_x\int_y d\mu(x,y) = \int_x dP(x) \neq \int_y d\tilde P(y) = \int_y\int_xd\mu(x,y). $$ However, for a given metric $d$ for probability measures one can build a metric for measures with different masses as follows: If $P$ has total mass $P(X) = p$ and $Q$ has total mass $Q(X) = q$, define $$ D(P,Q) = |p-q| + d(\tfrac{P}{p},\tfrac{Q}{q}) $$ (see Gromov's "Metric structures on Riemanian and Non-Riemannian Spaces", Chapter $3\tfrac12$.B).

There are also other approaches to metrics on measure spaces such as the Kantorovich-Rubinstein norm $$ W(P,Q) = \sup\bigg\{\int f\, dP - \int f\, dQ\ :\ f\ \text{Lipschitz with constant}\ \leq 1\bigg\} $$ and others (cf. Villani's "Optimal Transport - Old and New", Chapter 6).

share|improve this answer
add comment

Your Answer

 
discard

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.