Mathematics Asked by nico_so on January 19, 2021
I have the following problem:
Given $X_1, X_2..X_N$ independent r.v.
${X_1,X_4 X_1}$ are independent of ${X_3,X_2X_1}$, but $X_4$ is not independent from $X_4 X_1$
Is that statement true or false? Why?
I believe it has to be False since both sets contain $X_1$ so if I had information on the second one, then the probability of the first would be altered, therefore they are not independent, the same with the second part.
I am not entirely sure, and i do not know how to prove it in a formal way. I believe it has something to with the $sigma$ algebra generated by the r.v., but honestly i dont know.
Thanks a lot in advance!
$$ P({X_1,X_1 X_4, X_3, X_2 X_1} = (a,b,c,d))=\ P(X_1 = a) P(X_3 = c) P({X_4, X_2} = (b/a,d/a)|X_1=a,X_3=c)\ P(X_1 = a) P(X_3 = c) P(X_4 = b/a) P (X_2 = d/a)\ P({X_1, X_1 X_4}=(a,b)) P({X_3, X_1 X_2}=(c,d)) $$
Therefore, ${X_1, X_1 X_4},{X_3, X_1 X_2}$ are independent $forall a$ such that $P(X_2 = d/a) = P(X_1 X_2 = d)$.
Answered by nico_so on January 19, 2021
Why should ${X_1, X_4 X_1}$ be independent of ${X_2, X_2 X_1}$? In particular, there is no reason for $X_1$ and $X_1 X_2$ to be independent, and it's easy to construct examples where they are dependent. For example, suppose both $X_1$ and $X_2$ have possible values $1$ and $2$. Then $X_1 X_2 = 1$ implies $X_1 = 1$.
Answered by Robert Israel on January 19, 2021
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