If $X$ and $Y$ are independent of Z, will any combination of $X$ and $Y$ be independent of Z?
$aX+bY \perp Z$?
Will that independence holds if $X \perp Y$?
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What happens when you try to prove it? – GEdgar Feb 25 '15 at 14:55
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1One approach: think of $\sigma$-algebras and note that the $\sigma$-algebra generated by $f(W)$ where $W=(X,Y)$ is a sub $\sigma$-algebra of the $\sigma$-algebra generated by $W$. This should allow you to show that $f(W)$ and $Z$ are independent whenever $W$ and $Z$ are. This is true for any measurable $f$, including $f(X,Y)=aX+bY$. – Kim Jong Un Feb 25 '15 at 14:58
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@KimJongUn Unfortunately, the approach you outline is wrong: one can have (X,Y) independent, (X,Z) independent, (Y,Z) independent, and yet, (X,Y,Z) not independent. Exercise: Provide an example. – Did Feb 25 '15 at 15:44
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@Vincent: No. Try (X,Y,Z) uniformly distributed on {(0,0,1),(0,1,0),(1,0,0),(1,1,1)}. – Did Feb 25 '15 at 15:50
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@Did I haven't thought about your example but I still think what I said is suitable to the problem of the OP. – Kim Jong Un Feb 25 '15 at 15:50
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@KimJongUn Where do you see in "the problem of the OP" the hypothesis that (X,Y) is independent of Z? – Did Feb 25 '15 at 15:51
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The very first phrase, that's how I read it. – Kim Jong Un Feb 25 '15 at 15:52
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@KimJongUn Please read better, it nowhere says that (X,Y) is independent of Z, only that X is independent of Z and that Y is independent of Z. You would not think that this implies that (X,Y) is independent of Z, by any chance, would you? (Unrelated: Please use @.) – Did Feb 25 '15 at 19:20