Independence of continuous random variables squared

Question

If I know that two continuous random variables $X$ and $Y$ are independent, are $X^2$ and $Y$ necessarily independent? Are $X^2$ and $Y^2$ also independent?

Answer 1

As we will see, the result has nothing to do with the fact that $X$ and $Y$ are continuous or not.

We will use the following definition of independence: $X$ and $Y$ are independent if $$\forall s,t\in\mathbb R\quad P(X\leq s,Y\leq t)=P(X\leq s)P(Y\leq t).$$ (in fact it's equivalent to $P(X\in A,Y\in B)=P(X\in A)P(Y\in B)$ for all $A,B$ Borel measurable subsets of $\mathbb R$). With the first definition, we have for $s\geq 0, t\in\mathbb R$: \begin{align*} P(X^2\leq s,Y\leq t)&=P(|X|\leq \sqrt s,Y\leq t)\\ &=P(X\leq\sqrt s,Y\leq t)-P(X< -\sqrt s,Y\leq t)\\ &=P(X\leq \sqrt s)P(Y\leq t)-P(X<-\sqrt s)P(Y\leq t)\\ &=P(|X|\leq \sqrt s)P(Y\leq t)\\ &=P(X^2\leq s)P(Y\leq t), \end{align*} and the last inequality is true if $s<0$, since all these probabilities are $0$. It shows that $X^2$ and $Y$ are independent.

For the independence of $X^2$ and $Y^2$, we have shown that if $X_1$ and $X_2$ are independent then so are $X_1^2$ and $X_2$. So apply it to $X_1=Y$ and $X_2=X^2$.

If we work with the second definition, and take $f,g$ two Borel measurable functions and $A,B$ two Borel-measurable sets, since $f^{-1}(A)$ and $g^{-1}(B)$ are still Borel measurable, we have \begin{align*}P(f(X)\in A,g(Y)\in B)&=P(X\in f^{-1}(A),Y\in g^{-1}(B))\\ &=P(X\in f^{-1}(A))P(Y\in g^{-1}(A))\\ &=P(f(X)\in A) P(g(Y)\in B). \end{align*} So $f(X)$ and $g(Y)$ are still independent.