# $X_1,…X_4$ independent, then $(X_1,X_2)$ and $(X_3,X_4)$ independent

If $$X_1,…,X_4$$ are independent random variables how can I show that $$(X_1,X_2)$$ and $$(X_3,X_4)$$ are independent? Intuitively, I would say this is trivial but is there a formal proof?

This classical property is a sort of associativity property of independence. It follows from a corollary of the monotone convergence theorem saying that two probability measures coinciding on some collection $$\mathcal{S}$$ of sets which is closed under finite intersections coincide on $$\sigma(\mathcal{S})$$.
For example, the collection $$\mathcal{S}_1 := \{A_1 \cap A_2 : A_1 \in \sigma(X_1), A_2 \in \sigma(X_2)\}$$ is stable under intersection and generates $$\sigma(X_1,X_2)$$. Proving that $$\sigma(X_1,X_2)$$ is independent of $$\sigma(X_3,X_4)$$ is equivalent to prove that for every event $$B \in \sigma(X_3,X_4)$$ with positive probability, the probability measures $$P$$ and $$P[\cdot|B]$$ coincide on $$\sigma(X_1,X_2)$$. Hence, we only need to check those probability measures coincide on $$\mathcal{S}_1$$.
Given $$A_1 \in \sigma(X_1)$$ and $$A_2 \in \sigma(X_2)$$, we have to prove that $$\forall B \in \sigma(X_3,X_4), \quad P[A_1 \cap A_2 \cap B] = P[A_1 \cap A_2] P[B].$$ If $$P[A_1 \cap A_2]=0$$, it is obvious. Otherwise, we are led to check that the probability measures $$P$$ and $$P[\cdot|A_1 \cap A_2]$$ coincide on $$\sigma(X_3,X_4)$$. In the same way as above, we only need to check the probability measures $$P$$ and $$P[\cdot|A_1 \cap A_2]$$ coincide on $$\mathcal{S}_2 := \{A_3 \cap A_4 : A_3 \in \sigma(X_3), A_4 \in \sigma(X_4)\}$$, which follows directly by the independence of $$X_1,X_2,X_3,X_4$$.