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I understand that:

$X\perp Y$, for any $x$ and any $y$, $x\perp y$.

Then $P(X=x) = P(X=x|Y=y)$.

But I am struggling on giving a reason how $x^2\perp y^2$.

cdc9c9c9
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    I don't understand your $Val(X)$ set. Moreover, if $X$ is continuous, $\mathbb P(X=x)=0$ for all $x$, so, it won't take you anywhere. What is you definition of independence ? – joshua Mar 24 '22 at 08:06
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    $X$ and $Y$ are independent iff for every $A\in\sigma(X)$ and every $B\in\sigma(Y)$ we have $P(A\cap B)=P(A)P(B)$. For Borel-measurable functions $g$ and $h$ we have $\sigma(g(X))\subseteq\sigma(X)$ and $\sigma(h(Y))\subseteq\sigma(Y)$. So independence of $X$ and $Y$ implies independence of $g(X)$ and $h(Y)$. – drhab Mar 24 '22 at 08:46
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    What is the meaning of $x\perp y$ for real numbers $x$ and $y$? – drhab Mar 24 '22 at 09:33
  • https://math.stackexchange.com/questions/8742/are-functions-of-independent-variables-also-independent –  Mar 30 '22 at 06:42

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You don't really want to work with something like $P(X=x|Y=y)$ in general because it is possible that $P(X=x) = 0$ and $P(Y=y)=0$ for all $x,y \in \mathbb{R}$, e.g. if $X$ and $Y$ are both normally distributed. It's usually easier to work with the CDFs of $X$ and $Y$.

One property equivalent to $X \perp Y$ is $P(X \le x, Y \le y) = P(X \le x)P(Y \le y)$ for all $x,y \in \mathbb{R}$. So to show $X^2 \perp Y^2$, we can show $P(X^2 \le x, Y^2 \le y) = P(X^2 \le x)P(Y^2 \le y)$ for all $x,y \in \mathbb{R}$. If either $x$ or $y$ is negative, both sides will be $0$, so we just need to handle the case $x \ge 0$ and $y \ge 0$. Then we have \begin{align*} P(X^2 \le x, Y^2 \le y) &= P(|X| \le \sqrt{x}, |Y| \le \sqrt{y}) \\ &= P(X \le \sqrt{x}, Y \le \sqrt{y}) \\ &= P(X \le \sqrt x)P(Y \le \sqrt y) \\ &= P(X^2 \le x)P(Y^2 \le y), \end{align*} where we used that $X$ and $Y$ are non-negative to conclude $|X| \le \sqrt x$ iff $X \le \sqrt x$.

user6247850
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