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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?

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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 $X_1,X_2,X_3,X_4$.

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