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Let $(X_{n}), n= 1,...4$ independent real valued random variables. Are $(X_1,X_2)$ and $(X_3,X_4)$ independent? If so please prove why.

Sebastian
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  • You cannot prove this with the information given. Are you also assuming the $(\mathbf{X}_n)$ are independent where $\mathbf{X}_n = (X_{n,1}, \ldots, X_{n,r})$? If that's the case, your question is asked and answered at https://stats.stackexchange.com/questions/94872. If it's not the case, then what are you assuming? – whuber Feb 06 '18 at 19:20
  • no i do not assume this. – Sebastian Feb 06 '18 at 19:24
  • are you sure that this cannot be proven with the information given? also not if $X_i$ are Unif(0,1) distributed? – Sebastian Feb 06 '18 at 19:33
  • How could you possibly conclude that a set of variables is independent if you know absolutely nothing about them? – whuber Feb 06 '18 at 20:40
  • Aaah One knows what that $X_{n,j} $ are Independent – Sebastian Feb 06 '18 at 20:42
  • OK: in that case your question is answered already. – whuber Feb 06 '18 at 20:49
  • The question i have is the following: Why are $X_n=(X_{n,1}, ... X_{n,r})$ independent? How does this follow from the answer on the other post? – Sebastian Feb 06 '18 at 20:56
  • It may be that the multiple subscripting is creating ambiguities. As I read your question, it appears to ask whether measurable functions of independent random variables are independent. That's what is asked and answered in the apparent duplicate. It that's not what you're trying to ask, then consider editing your question to clarify how it differs from that one. – whuber Feb 06 '18 at 21:01
  • i simplified the question – Sebastian Feb 06 '18 at 21:07
  • Thank you: it's much clearer. The other thread answers it. – whuber Feb 06 '18 at 21:10
  • i really dont see how the other thread answers the question or can $(X_1,X_2) $ be interpreted as the composition of functions $X_1, X_2$? – Sebastian Feb 06 '18 at 21:21
  • It seems to me that I am making an incredible stupid mistake, can you please just complete: $P((X_1,X_2) \in A_1 \cap (X_3,X_4) \in A_2) = .... = P((X_1,X_2) \in A_1) \times P((X_3,X_4) \in A_2)$ – Sebastian Feb 06 '18 at 21:22
  • I dont see why $(X_1,X_2)^{-1}(A_1)$ and $(X_3,X_4)^{-1}(A_2)$ should be contained in the sigma algebras generated bx $X_i, i = 1...4$ – Sebastian Feb 06 '18 at 21:33

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