# If $X_1\overset{d}{=}Y_1$ and $X_2\overset{d}{=}Y_2$, then $(X_1,X_2)\overset{d}{=}(Y_1,Y_2)$?

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.

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).