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The goal is to find random variables $U, V, X$ and $Y$ such that the join distribution of $(U,V)$ and $(X,Y)$ is not the same, however, $U=X$ and $V=Y.$ From the statement, we see that all the random variables take values in the same in the set. I was thinking of working on the set $\Omega = \{0,1\}.$ Then perhaps we could take $U\sim B(p)$ and $X\sim B(p)$ Bernoulli and $V$ and $Y$ to be Bernoulli with parameter $q.$ Then we have the second condition satisfied. However, I want the joint distribution to be not equal and for that, I need to create some dependence between the variables $U$, $V$ and $X$, $Y$. This I am not sure of. Any hints will be much appreciated.

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  • $\begingroup$ In what sense do you mean $U=X$? That $U$ and $X$ have the same distribution, or that $U=X$ with probability one? $\endgroup$
    – Math1000
    Commented Oct 21, 2019 at 18:34
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    $\begingroup$ @Math1000 $p_U=p_X.$ They have the same probability density/mass function. $\endgroup$
    – Student
    Commented Oct 21, 2019 at 18:46
  • $\begingroup$ Let $X$ be uniformly distributed on $(0,1)$ and $U=1-X$, and let $Y$ be uniformly distributed on $(0,1)$ with $V=Y$. That should do it, I think - or if not, then something similar to this approach. $\endgroup$
    – Math1000
    Commented Oct 21, 2019 at 18:51

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Take $\varepsilon$, a random variable taking the values $-1$ and $1$ with probability $1/2$, $U=X=\varepsilon=V$ and $Y=-\varepsilon$. The marginals of the couples $(U,V)$ and $(X,Y)$ are the same, but the couple do not have the same distribution: the first one takes the values $(1,1)$ and $(-1,-1)$ with probability $1/2$; the second one the values $(1,-1)$ and $(-1,1)$ with probability $1/2$.

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