Suppose we have two random variables $X$ and $Y$ that are independent. Does that mean $X^2$ and Y are also independent?
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Yes, of course . Independence still exists – tommik Feb 07 '21 at 18:04
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@tommik it doesn't feel right to me, is there any proof you could provide? would be much appreciated. – Hijaw Feb 07 '21 at 18:07
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@Semiclassical any two constant random variables, even two of the same constant, are independent. (And more generally any constant is independent of any other random variable.) – spaceisdarkgreen Feb 07 '21 at 18:31
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@spaceisdarkgreen so you confirm what tommik says? – Hijaw Feb 07 '21 at 18:32
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1@Hijaw Yes, more generally, if $X$ and $Y$ are independent then $f(X)$ and $g(Y)$ are independent for any (measurable) functions $f$ and $g$. – spaceisdarkgreen Feb 07 '21 at 18:35
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https://stats.stackexchange.com/questions/94872/functions-of-independent-random-variables – Semiclassical Feb 07 '21 at 18:35
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In short: $P(f(X)\in A,g(Y)\in B)=P(X\in f^{-1}(A),Y\in g^{-1}(B))=P(X\in f^{-1}(A))P(Y\in g^{-1}(B))=P(f(X)\in A)P(g(Y)\in B)$. The second equality is a consequence of independence of $X$ and $Y$. This shows that also $f(X)$ and $g(Y)$ are independent. – drhab Feb 07 '21 at 18:54
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https://math.stackexchange.com/q/8742/321264 – StubbornAtom Feb 08 '21 at 12:08
