# Uniqueness of marginal measures

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 $$\eta\in\Pi(\nu_1,\nu_2)$$?

• $\mu_1(A):=\eta(A\times Y)$, $A\in\mathscr{B}(X)$ 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 $\mu_2$. Sep 23, 2020 at 15:51
• What @OliverDiaz said is that the measure determines the marginals (although the marginals do not determine the measure).
– Ruy
Sep 24, 2020 at 0:28
• @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 $\mu_2$. Sep 24, 2020 at 0:41
• @OliverDiaz, sure but not uniquely.
– Ruy
Sep 24, 2020 at 0:47
• I think you meant "previous", didn't you ;-) but we're surely on the same page!
– Ruy
Sep 24, 2020 at 1:17

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 $$η$$),

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