Given two probability measures $\mu \in \mathcal P(X)$ and $\nu \in \mathcal P(Y)$. The set of admissible couplings is given by
$$\Pi(\mu,\nu) =\{\pi \in \mathcal P(X\times Y) : \mathrm{proj}_{\#}^X\pi=\mu, \mathrm{proj}_{\#}^Y\pi=\nu \}$$ where $\mathrm{proj}_{\#}^X:X\times Y \to X$ is the projection onto the $X$-component , i.e. $\mathrm{proj}_{\#}^X(x,y)=x$ and similarly for $\mathrm{proj}_{\#}^Y$. It is often stated (as a probably very obvious fact) that this set is always non-empty, since the product measure $\mu(X)\otimes\nu(Y)$ is always contained. However, I am not very versed in anything that goes with measures – can anyone explain to me, maybe by example, that the product measure fulfills the conditions of an admissible coupling?
Best Answer
As a reminder, the product measure $\mu\otimes\nu$ is by definition the only measure on $X\times Y$ (endowed with the product $\sigma$-algebra) which satisfies: $$ \forall A \textrm{ measurable }\subset X,\quad\forall B\textrm{ measurable }\subset Y,\quad(\mu\otimes\nu)(A\times B)=\mu(A)\nu(B). $$
Let $\pi=\mu\otimes\nu$. Then for all measurable set $A\subset X$, $$ \mathrm{proj}_{\#}^X\pi(A)=\pi(A\times Y)=\mu(A)\nu(Y)=\mu(A), $$ hence $\mathrm{proj}_{\#}^X\pi=\mu$. Similarly we have $\mathrm{proj}_{\#}^Y\pi=\nu$, hence $\pi\in\Pi(\mu,\nu)$.
You can also see it with random variables. Let $U:\Omega\to X$ and $V:\Omega\to Y$ be two random variables, and $\pi$ be the distribution of $(U,V)$. Then the distribution of $U$ is $\mathrm{proj}_{\#}^X\pi$, and that of $V$ is $\mathrm{proj}_{\#}^Y\pi$. Therefore, $\pi\in\Pi(\mu,\nu)$ iff $U$ and $V$ are respectively distributed according to $\mu$ and $\nu$. So $\Pi(\mu,\nu)$ can be interpreted as the set of all distributions of any bivariate random variable whose first component is distributed according to $\mu$, and the second one according to $\nu$. Take two independent random variables $U\sim\mu$ and $V\sim\nu$. Then the distribution of $(U,V)$ belongs to $\Pi(\mu,\nu)$. Actually, the distribution of $(U,V)$, if $U$ and $V$ are independent, is precisely $\mu\otimes\nu$.