Independence of continuous random variables squared 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?
 A: 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.
