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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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Not necessarily the case. Suppose $f \equiv g \equiv 0$. Then $U=V=0$, which means that they certainly aren't independent. –  Omnomnomnom Jul 14 '13 at 21:01
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@Omnomnomnom I don't think independent means what you think it means. –  Thomas Andrews Jul 14 '13 at 21:09
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@Omnomnomnom: That works, and I think you'll find $U=0$ and $V=0$ are independent by this definition. It may sound strange, but a constant random variable is independent of every random variable, even itself. (Moreover, if a random variable is independent of itself, it must be constant.) –  Nate Eldredge Jul 14 '13 at 21:32
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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. –  Nate Eldredge Jul 14 '13 at 21:33
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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 $Y$. –  Dilip Sarwate Jul 14 '13 at 21:45
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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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My sincere thanks for your help. –  Zbigniew Jul 14 '13 at 23:41
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