Consider a measurable function $g$ mapping a probability space $\left(\Omega,\mathcal{F},\mu\right)$ to a product measurable space $\left(T,\mathcal{T}\right)$ with cartesian product $T = X \times Y$ and product sigma-algebra $\mathcal{T} = \mathcal{X} \otimes \mathcal{Y}$; $g$ thus induces a push-forward joint probability measure $\mu_T$ on $\mathcal{T}$ with marginals $\mu_X$, $\mu_Y$ respectively on $\mathcal{X}$, $\mathcal{Y}$.

Let $g, \mu_X, \mu_Y$ be fixed and define $\Pi\left(\mu_X,\mu_Y\right)$ as the class of couplings of $\mu_X$, $\mu_Y$ on $\mathcal{T}$, i.e. all joint probability measures with marginals $\mu_X$, $\mu_Y$.

Is it true that any of such coupling is obtainable as the push-forward via $g$ of an appropriate probability measure $\nu$ on $\left(\Omega,\mathcal{F}\right)$, i.e. if $\Gamma \in \Pi\left(\mu_X,\mu_Y\right)$ then $\Gamma = \nu g^{-1}$ for some $\nu$?

No invertibility hypothesis are made on $g$. Applications are on $T = \mathbb{R}^{k}$ but I am mostly interested in the general case.

  • 1
    $\begingroup$ Although I am not sure, this does’t look promising. I think you are asking too much. $\endgroup$
    – b00n heT
    Aug 14, 2018 at 19:18
  • $\begingroup$ @b00nheT you were right, see my own answer below, don't know why it took me so much to see it. I am still unsure about a weaker version (closer to problem I'm looking at), will open another question $\endgroup$
    – KRao
    Aug 15, 2018 at 6:55

1 Answer 1


The statement as presented is false. For example if $g$ is defined by $x = \omega$, $y = \omega$, then the only attainable coupling via $\nu g^{-1}$ is the maximal one no matter the $\nu$.


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