Let $X\neq \{0\}$ and $Y\neq \{0\}$ be complete separable metric spaces, $M(A)$ the set of Borel measures with finite total variation in $A$ and $\Pi(\rho_1,\rho_2)$ the set of all Borel measures in $X\times Y$ with marginals $\rho_1$ in $X$ and $\rho_2$ in $Y$.

Given a measure $\eta\in M(X\times Y)$, is it possible in any case to find two different pair of measures $\mu_1,\nu_1\in M(X)$, $\mu_2,\nu_2\in M(Y)$ such that $\eta\in\Pi(\mu_1,\mu_2)$ and $\eta\in\Pi(\nu_1,\nu_2)$?

  • $\begingroup$ $\mu_1(A):=\eta(A\times Y)$, $A\in\mathscr{B}(X)$ and $\mu_2(B)=\eta(X\times B)$, $B\in\mathscr{B}(Y)$, are probability measures and $\eta$ is in the coupling of $\mu_1$ and $\mu_2$. $\endgroup$ Sep 23, 2020 at 15:51
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    $\begingroup$ What @OliverDiaz said is that the measure determines the marginals (although the marginals do not determine the measure). $\endgroup$
    – Ruy
    Sep 24, 2020 at 0:28
  • $\begingroup$ @Ruy: One can construct a measure on $(X\times Y,\mathscr{B}(X)\otimes\mathscr{B}(Y))$ with described marginals $\mu_1$ on $X$ and $\mu_2$ on $Y$ too: The product measure $\eta:=\mu_1\otimes\mu_2$ for example. There may be there couplings with such marginals $\mu_1$ and $\mu_2$. $\endgroup$ Sep 24, 2020 at 0:41
  • $\begingroup$ @OliverDiaz, sure but not uniquely. $\endgroup$
    – Ruy
    Sep 24, 2020 at 0:47
  • $\begingroup$ I think you meant "previous", didn't you ;-) but we're surely on the same page! $\endgroup$
    – Ruy
    Sep 24, 2020 at 1:17

1 Answer 1


I guess the question has already been answered in the comments but it might be a good idea to summarize it in a formal answer.

The fact that the OP refers to metric spaces is perhaps irrelavant, so let us simply assume that $(X, \mathcal A)$ and $(Y, \mathcal B)$ are measurable spaces, that is, $X$ and $Y$ are sets, while $\mathcal A$ and $\mathcal B$ are $\sigma$-algebras of subsets of $X$ and $Y$, respectively.

One then defines $\mathcal A\times \mathcal B$ to be the smallest $\sigma$-algebra of subsets of $X\times Y$ containing all sets of the form $A\times B$, where $A\in \mathcal A$, and $B\in \mathcal B$.

Given any measure $\eta $ on $(X\times Y, \mathcal A\times \mathcal B)$, be it finite or infinite, probability or not, positive or signed, the marginal measures $\mu $ and $\nu $ are defined in terms of $\eta $ by $$ \mu (A) = \eta (A\times Y), \text { for all } A\in \mathcal A, $$ and $$ \nu (B) = \eta (X\times B), \text { for all } B\in \mathcal B. $$

It is easy to see that $\mu $ and $\nu $ are measures on $X$ and $Y$, respectively. They are clearly unequivocally determined by $\eta $, in the same way that the derivative of a smooth function $f$ is determined by $f$.

Thus, the question of whether there is a measure $\eta $ lying simultaneously in $\Pi(\mu_1,\mu_2)$ and in $\Pi(\nu_1,\nu_2)$, for two different pairs of measures $\mu_1,\nu_1\in M(X)$, $\mu_2,\nu_2\in M(Y)$ therefore has an immediate negative answer because both $\mu _1$ and $\nu _1$ must be the marginal of $\eta $ relative to $X$, so necessarily $\mu _1=\nu _1$, and similarly $\mu _2=\nu _2$.

This is a bit like asking whether there is a smooth function $f$ whose derivative is both equal to $g_1$ and to $g_2$, for different functions $g_1$ and $g_2$!

An entirely different (and highly relevant) question is whether or not there are two different measures $\eta _1$ and $\eta _2$ on $X\times Y$ whose marginals on $X$ and $Y$ coincide. In other words, whether $\Pi(\mu,\nu)$ contains more than one measure, once we are given measures $\mu $ and $\nu $ on $X$ and $Y$, respectively.

Restricting the discussion to probability measures from now on, and before discussing uniqueness, it is nice to know that, given $\mu $ and $\nu $, there always exists at least one measure in $\Pi(\mu,\nu)$, namely the product measure, variously denoted $\mu \times \nu $ or $\mu \otimes \nu $. This measure is characterized by the fact that $$ (\mu \times \nu )(A\times B) = \mu (A)\nu (B), $$ for all $A$ in $\mathcal A$, and $B$ in $\mathcal B$. Incidentally this property, together with the assumption that $\mu (X)=1=\nu (Y)$, immediately implies that the marginal measures for $\mu \times \nu $ are $\mu $ and $\nu $.

The concept of product measures sits squarely at the center on the notion of independent random variables: seeing the projections $$ x:X\times Y\to X $$ and $$ y:X\times Y\to Y $$ as random variables (this is specially relevant when $X=Y=\mathbb R$), then the probability $$ \mathbb P(x\in A\ \wedge\ y\in B) $$ is precisealy the measure of $A\times B$. So this always coincides with the product of probabilities $$ \mathbb P(x\in A) \ \mathbb P(y\in B) $$ a.k.a. $\mu (A)\nu (B)$, iff the random variables are independent, iff the probability measure on $X\times Y$ is the product measure.

The covariance of $x$ and $y$, namely, $$ \text{cov}(x,y) = \mathbb E(xy) -\mathbb E(x)\mathbb E(y), $$ can be computed by $$ \text{cov}(x,y) = \int_{X\times Y}xy\,d(\mu \times \nu ) - \left(\int_Xx\,d\mu\right)\left(\int_Yy\,d\nu \right), $$ and is easily seen to vanish due to the Fubini Theorem which allows for the iterated integration $$ \int_{X\times Y}xy\,d(\mu \times \nu ) = \int_Y\int_X xy\,d\mu \, d\nu . $$ In other words, if $x$ and $y$ are independent, then $\text{cov}(x,y) = 0$.

Back to the uniqueness question, suppose e.g. that $X=Y=[0,1]$, and that $\mu =\nu =\lambda $, where $\lambda $ is Lebesgue measure. As already seen, the product measure $\mu \times \nu $ (incidentally the 2-dimensional Lebesgue measure on the square) admits $\mu $ and $\nu $ as marginals.

So what would be another example of a measure in $\Pi(\mu,\nu)$? Well, here is one: given any Borel measurable subset $E\subseteq [0,1]^2$, set $$ \eta (E) = \lambda (\{x\in [0, 1]: (x, x)\in E\}). $$

It is very easy to see that the marginals of $\eta $ are still $\mu $ and $\nu $, but now the measure of a product set $A\times B$ can no longer be computed just in terms of $\mu (A)$ and $\nu (B)$. To see for yourself, try to prove that $$ \eta ([0,1/2]\times [1/2, 1]) = 0 $$ (the top left-hand quarter of the square has only one point in common with the diagonal!), while $$ \eta ([0,1/2]\times [0,1/2]) = 1/2. $$

Needless to say, $x$ and $y$ are not independent random variables. They are in fact so dependent to each other that $x=y$ almost surely, meaning that the set $$ \{(x,y)\in [0,1]\times [0,1]: x=y \} $$ has full measure (according to $η$),

  • $\begingroup$ @Ruy how does one interpret the space 𝐴×𝑌 defined in the marginal measures? $\endgroup$
    – xiA
    Apr 30, 2021 at 9:58

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