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Let $X,Y$ be two measurable spaces, $\mu,\nu$ two probability measures on $X$, and $\kappa$ a transition kernel from $X$ to $Y$. Define $\tilde\mu(dy)=\int_X\kappa(dy|x)\mu(dx)$ and $\tilde\nu(dy)=\int_X\kappa(dy|x)\nu(dx)$.

Denote by $\Gamma(\mu,\nu)$ the set of couplings of $\mu$ and $\nu$ and by $\Gamma(\tilde\mu,\tilde\nu)$ the corresponding set for $\tilde\mu$ and $\tilde\nu$.

Furthermore, let $S_{\kappa}:\Gamma(\mu,\nu)\to\Gamma(\tilde\mu,\tilde\nu)$ be the mapping defined by

$$S_{\kappa}(\pi)(dy_1,dy_2)=\int_{X\times X}\kappa(dy_1|x_1)\kappa(dy_2|x_2)\pi(dx_1,dx_2),\quad \pi\in\Gamma(\mu,\nu).$$

Question: Is $S_{\kappa}$ onto $\Gamma(\tilde\mu,\tilde\nu)$?

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  • $\begingroup$ Looks like you have two transition kernels. E transition from kernel from $X$ to $Y$ is a map $\newcommand{\eE}{\mathscr{M}}$ $k: \eE_Y\times X\to[0,\infty]$, where $\eE_Y$ is the $\sigma$-algebra of events in $Y$. The formul{\ae} that you wrote in the 2nd line cannot simultaneously make sense if $X\neq Y$. $\endgroup$
    – Liviu Nicolaescu
    Commented Nov 23, 2018 at 13:15
  • $\begingroup$ Do you mean the definitions of $\tilde\mu$ and $\tilde\nu$? $\mu$ and $\nu$ are measures on $X$, and the kernel is used to produce two measures on $Y$. Am I missing something? $\endgroup$
    – S.Surace
    Commented Nov 23, 2018 at 13:31

1 Answer 1

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It's not true. If $\mu \mapsto \tilde\mu$ is injective and $X$ has at least two points then the transition kernels have to be deterministic, i.e. there is a measurable map $T:X\to Y$ such that $\kappa(dy|x)=\delta_{T(x)}$.

To illustrate this take the trivial example $X = \{1,2\}$, $Y=\{(1,1),(1,2),(2,1),(2,2)\}$ and $\kappa(dy|i) = \frac{1}{2}\left(\delta_{(i,1)} + \delta_{(i,2)}\right).$ Now choose $\mu = \delta_1$ and $\nu = \delta_2$. Observe that $\tilde\mu = \frac{1}{2}\left(\delta_{(1,1)} + \delta_{(1,2)}\right)$ and $\tilde\nu = \frac{1}{2}\left(\delta_{(2,1)} + \delta_{(2,2)}\right)$ However, there is only one coupling between $\mu$ and $\nu$ but there are uncountably many couplings between $\tilde\mu$ and $\tilde\nu$.

This construction generalizes to arbitrary $X$ and $Y$ by observing that there are uncountably many couplings between two transition kernels $\kappa(dy|x_1)$ and $\kappa(dy|x_2)$ but the map $S_\kappa$ only picks the product coupling (choose $\mu=\delta_{x_1}$ and $\nu=\delta_{x_2}$ for $x_1\ne x_2$). If $\mu$ and $\nu$ are non-atomic then the construction of the counterexample is a bit messier but still works.

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  • $\begingroup$ Is it possible to give a counterexample for the non-injective case, or does one need the injectivity? $\endgroup$
    – S.Surace
    Commented Nov 27, 2018 at 14:40
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    $\begingroup$ The counterexample works when there are at least two measures with different image and where $\kappa$ is a delta measure on a set of full measure of those. Observe that two delta measure have the same image if and only if $\kappa(dy|x_1)=\kappa(dy|x_2)$. If the kernel is not a delta measure then the construction works and the image measures of the two deltas have uncountably many couplings (there are many self-couplings for non-delta measures). $\endgroup$
    – Martin Kell
    Commented Nov 28, 2018 at 8:52
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    $\begingroup$ CONTINUED: Note that this also means the counterexample is true if $\kappa$ is not a delta measure and $X=\{x_0\}$. Hence the only case left is all $\kappa$s are mapped to the same delta measure $\delta_{y_0}$. In this case $S_\kappa$ is obviously onto as the image is just the trivial and unique coupling of $\delta_{y_0}$ with itself. $\endgroup$
    – Martin Kell
    Commented Nov 28, 2018 at 8:52

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