# Counter Example for a couple for random variables $(U,V)$ and $(X,Y).$

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.

• In what sense do you mean $U=X$? That $U$ and $X$ have the same distribution, or that $U=X$ with probability one? Commented Oct 21, 2019 at 18:34
• @Math1000 $p_U=p_X.$ They have the same probability density/mass function. Commented Oct 21, 2019 at 18:46
• 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. Commented Oct 21, 2019 at 18:51

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 $$1/2$$.