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Suppose that $X$ and $Y$ are independent random variables. I wish to prove that $X$ and $Z = Y^2$ are also independent.

I know from the definition of independence that $P({X \in A}$ and $Y \in B$) = $P({X \in A})$$P({Y \in B})$, but I'm not sure how to apply this to the above problem.

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Claim If $X$ and $Y$ are independent, then also $f(X)$ and $g(Y)$ are independent where $f,g: \mathbb{R} \to\mathbb{R}$ are measurable maps.

Proof claim: Let $A,B$ be Borel sets. Then

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

and this ends the proof. $\quad \square$

Apply this with $f: x \mapsto x$ and $g: x \mapsto x^2$ to obtain that $X=f(X)$ and $g(Y) = Y^2$ are independent.

J. De Ro
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  • Could you provide or link to a definition of Borel transformations? I could fine none. I find two meanings for Borel transforms, neither of which makes it obvious why the non-objective $y\mapsto y^2$ is OK. – J.G. Dec 08 '19 at 19:03
  • A Borel transformation is just a measurable map. Clearly $y \mapsto y^2$ is measurable since it is continuous. I edited my answer because a quick google search indeed suggests that 'Borel transformation' is non-standard terminology. The terminology comes from "Borel-measurable" which means that the inverse image of a Borel subset is again a Borel subset. – J. De Ro Dec 08 '19 at 19:03
  • beautiful proof. short and precise and clear and very useful. – user2550228 Apr 07 '21 at 07:13
  • @user2550228 Thank you. – J. De Ro Apr 07 '21 at 15:40