# Coupling of non-probability/sub-probability measures

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.

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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
@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

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).

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