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Let $X$ and $Y$ be independent random variables. Prove that $f(X)$ and $g(Y)$ are independent for any choice of measurable functions $f$ and $g$.

This sounds very obvious, but I have no idea how to approach it.

EDIT: Two random variables are independent if $\Pr\{X = x \text{ and } Y = y\} = \Pr\{X = x\} \cdot \Pr\{Y = y\}$

BCLC
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Stefan Kanev
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  • Write up the definition of independentness.. – Berci Oct 23 '13 at 21:15
  • I've added the definition of independence that I have. – Stefan Kanev Oct 23 '13 at 21:19
  • This definition must only be for discrete $X$ and $Y$. For continuous $X$ and $Y$, all of these probabilities are generally $0$. In that situation, this definition makes $X$ and $Y$ independent no matter what. – 2'5 9'2 Oct 23 '13 at 22:06
  • I might suggest this definition: $X$ and $Y$ are independent iff the distribution of $X$ is identical to the distribution of $X$ under any condition put on $Y$ (like $Y=y$). – 2'5 9'2 Oct 23 '13 at 22:09
  • f and g have to be measurable functions and not just 'any' thing right? – BCLC Nov 14 '15 at 08:13

1 Answers1

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$X,Y$ are independent iff for all measurable $A,B$, the events $X^{-1}(A)$ and $Y^{-1}(B)$ are independent.

Suppose $C,D$ are measurable, and consider $(f \circ X)^{-1} ( C)$ and $(g \circ Y)^{-1} (D)$. Since $(f \circ X)^{-1} (C) = X^{-1} (f^{-1}(C))$ and $(g \circ Y)^{-1} (D)= Y^{-1} (g^{-1}(D))$. Since $X,Y$ are independent, we see that $X^{-1} (f^{-1}(C))$ and $Y^{-1} (g^{-1}(D))$ are independent and since $C,D$ were arbitrary, we see that $f \circ X$ and $g \circ Y$ are independent.

copper.hat
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