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Let $X_1,\,X_2,\,Y_1,\,Y_2$ be random variables (not necessarily defined on the same probability space) such that $X_1\overset{d}{=}Y_1$ and $X_2\overset{d}{=}Y_2$, i.e. $X_1,\,Y_1$ are identically distributed (i.d.), that is $F_{X_1}=F_{Y_1}$ (cdf's) and the same for $X_2,\,Y_2$. Is true that $(X_1,X_2)\overset{d}{=}(Y_1,Y_2)$?

Attempt. In general I believe the answer is no. The special case where $X_1,\,X_2$ are independent and $Y_1,\,Y_2$ are independent is pretty straightforward, since: $$\mathbb{P}_{(X_1,X_2)}\overset{\textrm{indep.}}{=}\mathbb{P}_{X_1}\otimes\mathbb{P}_{X_2}\overset{\textrm{i.d.}}{=}\mathbb{P}_{Y_1}\otimes\mathbb{P}_{Y_2} \overset{\textrm{indep.}}{=}\mathbb{P}_{(Y_1,Y_2)}.$$ Regarding the general case I haven't been able to give a counterexample.

Thanks in advance.

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Take $X_1\sim \mathcal N(0,1)$ and $Y_1=-X_1$ (then $Y_1\sim \mathcal N(0,1)$ as well). Now, you can build a space $(\Omega ',\mathcal F',\mathbb P')$ and $X_2$, $Y_2$ s.t. $X_2\sim X_1$, $Y_2\sim Y_1$, but $X_2$ and $Y_2$ are independents (this is a classical exercise. So if you don't know it, try to do it).

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