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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

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

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.

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

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

Define:

  1. 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

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

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

  1. probability measures $P_x : \mathcal{X} \to [0, 1]\;$ and $\;P_y : \mathcal{Y} \to [0, 1]$ as

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

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

Now I want to come up with joint probability mass function $P_{xy}$ of random variables $X$ and $Y$ such that

  1. Marginal distribution of $X$ is $P_x$ and marginal distribution of $Y$ is $P_y$, i.e.

$$ \sum_{y \in \mathcal{Y}} P_{xy}(x,y) = P_x(x), \qquad x \in \mathcal{X} $$

$$ \sum_{x \in \mathcal{X}} P_{xy}(x,y) = P_y(y), \qquad y \in \mathcal{Y} $$

  1. $X$ and $Y$ are not independent, i.e.

$$ P_{xy} \neq P_x P_y $$

So far I had 2 ideas:

  1. $$ P_{xy}(A,B) = \frac{\mu_x(A) + \mu_y(B)}{|\mathcal{T}_x| + |\mathcal{T}_y|}, \qquad \forall\, A \subseteq \mathcal{X},\, B \subseteq \mathcal{Y} $$ but I'm not sure whether condition 1 is not satisfied.

  2. $$ P_{xy}(A,B) = \frac{\mu_x(A) \mu_y(B)}{|\mathcal{T}_x| |\mathcal{T}_y|}, \qquad \forall\, A \subseteq \mathcal{X},\, B \subseteq \mathcal{Y} $$ but now looks like they are independent.

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1 Answer 1

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You need to specify a joint measure, $\mu_{x,y}(,)$, such that: $$ P_{x,y}(A,B) = \dfrac{\mu_{x,y}(A,B)}{\lvert{\mathcal T_x}\rvert\lvert{\mathcal T_y}\rvert}$$You've stated that $\mathcal {X, Y}$ are dependent but not specified what that dependency is.   There is not a single solution without knowing that.   Since there are many ways the variables could be dependent there are many joint measures they could have.

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  • $\begingroup$ Well, that's what I am actually asking for. To give an example of dependent random variables $X$ and $Y$, such that their marginal distributions are given as above $\endgroup$
    – aberdysh
    Commented Mar 6, 2017 at 23:18

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