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Prove that if $X$ and $Y$ are independent discrete variables, for $f: \mathbb{R} \rightarrow \mathbb{R}$, then $f(X)$ and $f(Y)$ are independent.

Here is the exact same question. I define independence identically to this question. My main question is regarding one of the answers.

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

Why is that true? I understand the rest of the proof. Just not that part.

Bob the Builds
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  • It's worth mentioning that $f$ needs to be measurable, otherwise $f(X)$ and $f(Y)$ aren't even random variables. – Math1000 May 11 '15 at 23:06
  • "I define independence identically to this question." Already on the other page they explain why this definition works only for discrete distributions. One gets the feeling that you did not read the other page before posting this duplicate. – Did May 30 '15 at 08:07

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Your definition works only for discrete variables. A nearly identical one that works for every variable is

$$P(X \in A, Y \in B) = P(X \in A)P(Y \in B)$$

So that statement is just a restatement of the above definition; $X \in A$ is a shorthand for the event $X^{-1}(A)$. Both mean the same thing, that more explicitly is $\{\omega \in \Omega: X(\omega) \in A\}$

So saying that for all $A, B$ the events $X^{-1}(A)$ and $Y^{-1}(B)$ are independent amounts to saying that

$$P(X^{-1}(A), Y^{-1}(B)) =P(X^{-1}(A))P(Y^{-1}(B)) \iff P(X \in A, Y \in B) = P(X\in A)P(Y \in B)$$

Ant
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