If $X$ and $Y$ are independent and identically distributed (iid) random variables, does it imply that $|X|$ and $|Y|$ are iid? How would you go about proving this?
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$X,Y$ are independent if and only if $$P\left\{ X\in A\wedge Y\in B\right\} =P\left\{ X\in A\right\} P\left\{ Y\in B\right\} $$ is true for measurable sets $A,B$.
If this is the case and $f,g:\mathbb{R}\rightarrow\mathbb{R}$ are measurable functions then:
$$P\left\{ f\left(X\right)\in A\wedge g\left(Y\right)\in B\right\} =P\left\{ X\in f^{-1}\left(A\right)\wedge Y\in g^{-1}\left(B\right)\right\} =P\left\{ X\in f^{-1}\left(A\right)\right\} P\left\{ Y\in g^{-1}\left(B\right)\right\} =P\left\{ f\left(X\right)\in A\right\} P\left\{ g\left(Y\right)\in B\right\} $$ showing that $f\left(X\right)$ and $g\left(Y\right)$ are independent.
drhab
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Yes. In fact one can show that $X$ and $Y$ are independent if and only if $f(X)$ and $g(Y)$ are independent for every measurable function $f, g$
I don't know how you introduced independence, but X and Y independent means that the $\sigma$-algebrae generated by them are independent; it is easy to see that the $\sigma$-algebra generated by $f(X)$ is smaller that the one generated by $X$, hence independence is preserved
Ant
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Silly: $\sigma$-algebrae or $\sigma$-algebras? – snar Aug 19 '14 at 20:43
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@snarski well I'm not sure, since algebra is a Latin word I used algebrae, but it is possible that the form that is actually used is algebras.. I am not a native English speaker so I improvised :-D – Ant Aug 20 '14 at 00:31
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The first paragraph is dangerously ambiguous (actually, if one sticks to the rules of English grammar, this statement is false because of a misplaced comma). The equivalence holds between the properties that [X and Y are independent] and [for every measurable f and g, f(X) and g(Y) are independent]. – Did Aug 22 '14 at 14:18
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@Did that is what I meant. I removed the comma, is it okay now? – Ant Aug 22 '14 at 14:20
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Perfect. (To tell you the truth, not being a native speaker, I was rather amused by the importance of this poor comma in your sentence when I discovered it... Hope there is no offense.) – Did Aug 22 '14 at 14:28
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@Did Of course not! I should have been more careful ;) – Ant Aug 22 '14 at 15:00
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Evidently so. See this link
Are functions of independent variables also independent?
which addresses independence. It's clear they are identically distributed, isn't it?
