How to compute the unique disintegration w.r.t. the first coordinate?

Question

Set $\pi=\frac{1}{4}(\delta_{(1,0)}++\delta_{(1,3)}+\delta_{(1,1)}+\delta_{(2,2)})$. Suppose that $\pi\in\Pi(\mu,\nu)$.

How to get the disintegration of $\pi$ with respect to $\mu$?

Answer 1

By disintegration is meant that $\pi$ can by broken down as $$ \pi(dx,dy) =\mu(dx)\pi_{x}(dy), $$ with $\pi_x$ a Borel measurable family of sub-probability measures on $X$.

In your example: $\pi_{x_1}={1\over 2}(\delta_{0}+\delta_{1})$ if $x_1=1$; $=\delta_{2}$ if $x_1=2$; and $=0$ otherwise.

Then you have $W_1(\pi_{x_1},\nu) = 1/2$ if $x_1=1$, and $W_1(\pi_{x_1},\nu)=1$ if $x_1=2$.

Answer 2

1. Suppose $\pi$ is the joint distribution of random variables $Y$ and $Z$. (The marginal distribution of $Y$ is therefore $\mu$.) Using the definition of conditional expectation you have $$ \eqalign{ E[E[f(Z)\mid Y] g(Y)]]= &=E[f(Z)g(Y)] =\int_{X\times X} g(y)f(z)\pi(dy,dz)\cr &=\int_Xg(y)\mu(dy)\left[\int_X f(z) \pi_y(dz)\right]\cr &=\int_X g(y)h(y)\mu(dy)\cr &=E[g(Y)h(Y)], } $$ where $$ h(y):=\int_Xf(z)\pi_y(dz). $$ Thus, the random variable $h(Y)$ has the properties required of $E[f(Z)\mid Y]$, so we must have $E[f(Z)\mid Y]=\int_X f(z)\,\pi_Y(dz)$, a.s., for each bounded measurable $f$. This amounts to saying that $(\omega, B)\mapsto\pi_{Y(\omega)}(B)$ is a regular conditional distribution for $Z$ given $Y$.

2. If $\pi$ were absolutely continuous with respect to the product measure $\mu\otimes\nu$, say $$ \pi(dy,dz)=\rho(y,z)\mu(dy)\nu(dz), $$ then clearly $$ \pi_y(dz) = \rho(y,z)\,\nu(dz). $$ Notice that because $\mu$ is the distribution of $Y$, you have $\int_X\rho(y,z)\nu(dz)=1$ for $\mu$-a.e. $y$.

3. To check that $W_1(\pi_{x_1},\nu)=1/2$ one can guess at a most efficient way to transport the mass of $\pi_{x_1}$ to that of $\nu$ (namely, take $1/6$ from $\pi_{x_1}$ at $0$ and at $1$ and move the combined weight to $2$ to obtain $\nu$. This gives a transport cost of $(1/6)\cdot 1+(1/6)\cdot 2 = 1/2$.