Knowing that if you have two independent $X$ and $Y$, and $ f $ and $ g $ measurable functions, how to show that then $ U = f (X) $ and $ V = g (Y) $ are still independent.
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2$\begingroup$ Jonas: This is a typical homework question, so please see these guidelines that apply to such questions, and edit your question accordingly. Otherwise, your question may be closed. $\endgroup$– Nate EldredgeCommented Jul 14, 2013 at 21:33
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2$\begingroup$ A hand-waving proof starts with the observation that independence of $X$ and $Y$ means that knowing the value of $X$ tells you nothing that you did not already know about the value of $Y$. For example, the conditional distribution $F_{Y\mid X}(y\mid X=x)$ is the same as the unconditional distribution $F_Y(y)$. If $X$ and $Y$ are independent but $g(X)$ and $h(Y)$ were dependent, then knowing the value of $X$ (and therefore $g(X)$) would mean that we could make some inference about $h(Y)$ and hence possibly about $Y$ which contradicts the independence of $X$ and $Y$. $\endgroup$– Dilip SarwateCommented Jul 14, 2013 at 21:45
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$\begingroup$ math.stackexchange.com/q/8742/321264 $\endgroup$– StubbornAtomCommented May 7, 2020 at 20:53
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You said measurable so I am going to assume you want a measure-theoretic answer, and that your definition of independence is $X$ and $Y$ are independent iff $\sigma(X)$ is independent of $\sigma(Y)$, i.e. that $P[X \in B_1, Y \in B_2] = P[X \in B_1]P[Y \in B_2]$ for all borel sets $B_1,B_2$. You must assume $f,g$ are borel functions (this is so that $f(X),g(Y)$ are measurable, so asking the question of independence makes sense). The $\sigma$-algebra generated by $f(X)$ is a sub-$\sigma$-algebra of the $\sigma$-algebra generated by $X$, and similarly for $g(Y)$ and $Y$. To see this note that for any borel set $B$ we have $(f\circ X)^{-1}(B) = X^{-1}(f^{-1}(B)) = X^{-1}(\text{some borel set}) \in \sigma(X)$. Since $\sigma(X) \perp \sigma(Y)$ it follows that $\sigma(f(X)) \perp \sigma(g(Y))$.
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2$\begingroup$ My sincere thanks for your help. $\endgroup$– ZbigniewCommented Jul 14, 2013 at 23:41
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$\begingroup$ @Zbigniew and nullUser, do you know the elementary probability answer for more than 2 random variables? math.stackexchange.com/questions/3944284/… $\endgroup$– BCLCCommented Dec 18, 2020 at 0:27