Let $\mathcal{S}_x$ and $\mathcal{S}_y$ be a finite discrete sets, such that

$$ 0 < |\mathcal{S}_x| < \infty, \qquad 0 < |\mathcal{S}_y| < \infty, \qquad \mathcal{S}_x \cap \mathcal{S}_y = \emptyset $$

Let $\mathcal{T}_x$ and $\mathcal{T}_y$ be non-empty sets such that

$$ \mathcal{T}_x \subseteq \mathcal{S}_x, \qquad \mathcal{T}_x \neq \emptyset; \qquad \qquad \qquad \mathcal{T}_x \subseteq \mathcal{S}_y, \qquad \mathcal{T}_y \neq \emptyset $$

Let $\mathcal{X} = \left\{ x_i \right\}_{i=1}^m$, and $\mathcal{Y} = \left\{ y_i \right\}_{j=1}^n$ be partitions of $\mathcal{S}_x$ and $\mathcal{S}_y$ respectively, i.e.

$$ \sqcup_{i=1}^m x_i = \mathcal{S}_x; \qquad \qquad \qquad \sqcup_{j=1}^n y_j = \mathcal{S}_y $$

where $\sqcup \cdot$ is a disjoint union.

Define measures $\; \mu_x: \mathcal{X} \to \left[0,\, |\mathcal{T}_x|\right]\;$ and $\;\mu_y: \mathcal{Y} \to \left[0,\, |\mathcal{T}_y|\right]\; $ as

$$ \mu_x(A) = |A \cap \mathcal{T}_x|, \qquad \forall\, A \subseteq \mathcal{X} $$

$$ \mu_y(B) = |B \cap \mathcal{T}_y|, \qquad \forall\, B \subseteq \mathcal{Y} $$

Now I want to design a measure $\mu : \mathcal{X} \times \mathcal{Y} \to \left[0,\, |\mathcal{T}_x||\mathcal{T}_y| \right]$, which cannot be factorized as a product of measures $\mu_x$ and $\mu_y$, i.e.

$$ \mu(A, B) \neq \mu_x(A) \mu(B) $$ except when on the boundaries, or more precisely such that

  1. $$ \forall\, A \subseteq \mathcal{X}: \qquad \mu(A, \emptyset) = 0 \qquad \text{and} \qquad \mu(A, \mathcal{Y}) = \mu_x(A) |\mathcal{T}_y| $$

  2. $$ \forall\, B \subseteq \mathcal{Y}: \qquad \mu(\emptyset, B) = 0 \qquad \text{and} \qquad \mu(\mathcal{X}, B) = |\mathcal{T}_x| \mu_y(B) $$

  • $\begingroup$ I think what you're asking for is a essentially a non-trivial coupling of the counting measure on $\mathcal X\times\mathcal Y$. This is usually done with probability measures, although it's not essential. If you write the entries as a matrix, you're asking that $\sum_j \mu(x_i,y_j)=n$ and $\sum_i \mu(x_i,y_j)=m$. There are lots of ways to do this: you're satisfying $n+m$ equations in $nm$ variables. $\endgroup$ Mar 7 '17 at 14:13
  • $\begingroup$ One interesting way to do this is to set $\mu(x_i,y_i)=m$ for $i\le m$, $\mu(x_i,y_j)=0$ for $j\le m$ with $i\ne j$ , and $\mu(x_i,y_j)=1$ for $j>m$. $\endgroup$ Mar 7 '17 at 14:19
  • $\begingroup$ By the way: a measure is a product measure if and only if the matrix has rank 1. This one has rank $m$. $\endgroup$ Mar 7 '17 at 17:28
  • 1
    $\begingroup$ I just noticed that you posted the exact same question at essentially the same time on math.stackexchange.com. You did the same thing with a previous question. Please don't do this. It leads to wasted effort. If in doubt, post to MSE first. After a few days, if there is no answer, you can post to MO, along with a note that its cross-posted. For reference, this question is probably closer to the MSE level. $\endgroup$ Mar 7 '17 at 17:33

I suppose you can just randomly perturb the independent distribution. That is, let $$\mu(x_i,y_j)=\mu_x(x_i)\mu_y(y_j)+\epsilon_{i,j}$$ where the $\epsilon_{i,j}$ form a random matrix (sufficiently random), with the conditions that $$\sum_i\epsilon_{i,j}=0=\sum_{j}\epsilon_{i,j}\qquad \forall i,j$$ (all columns and rows add to 0), $$0\le \mu(x_i,y_j)\le 1\qquad \forall i,j$$ (entries $\epsilon_{i,j}$ are sufficiently small).

In theory you could pick these randomly with respect to Lebesgue measure on the set of all such matrices.


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