A critical proof or counterexample regarding independence Does independence of  $X^2$ and $Y^2$  imply independence of $X$ and $Y$?
 A: It seems likely that you could fairly easily construct a counter-example by assigning the negative roots of $Y^2$ to $Y$ for $X\leq1$ and to the positive roots for $X > 1$.
Y <- rnorm(1000)^2
X <- rnorm(1000)^2
sy[X1] <- sqrt(Y[X1])
sy[X <= 1] <- -sqrt(Y[X <= 1])
plot(sx, sy)
cor(sx, sy)
[1] 0.6537367


png()
plot(X, Y)
dev.off()
cor(X,Y)
[1] -0.05216765


A: The construction of a very general set of counterexamples illuminates this issue.  The underlying idea is that although $X^2$ and $Y^2$ might be independent, a choice of two square roots is available for every nonzero value taken on by these variables. By making those choices dependent, we create a counterexample where $X$ and $Y$ are not independent.  The details follow.

Because $X^2$ and $Y^2$ are independent, $|X|=\sqrt{X^2}$ and $|Y|=\sqrt{Y^2}$ are independent, too.  Let $I$ be a discrete random variable, independent of $|X|$ and $|Y|$, taking on the values $\pm 1$, so that $I^2=1$.  Let $p = \Pr(I=1)$ and assume $0\lt p\lt 1$.  $I$ will determine the choice of sign of the square roots, taking the positive sign with probability $p$.
One possibility for the random variables $X$ and $Y$ is
$$X_I= I|X|, \quad Y_I=I|Y|$$
because obviously $X_I^2 = I^2|X|^2 = X^2$ and $Y_I^2 = Y^2$ and the only thing we know about $X$ and $Y$ is that they are square roots of the given variables $X^2$ and $Y^2$. 
Let's check whether independence holds by looking at the event where both $X_I$ and $Y_I$ are nonnegative. Provided both $X$ and $Y$ have zero chance of equaling $0$, this is the event consisting of all values where the positive square root is chosen.  The calculations are easy because the signs of $X_I$ and $Y_I$ are entirely determined by the sign of $I$:
$$\eqalign{
&\Pr(I|X| \ge 0)\Pr(I|Y| \ge 0) = \Pr(I\ge 0)\Pr(I\ge 0) = p^2; \\ 
& \Pr(I|X|\ge 0\text{ and }I|Y|\ge 0) = \Pr(I\ge 0) = p.
}$$
Since $p\ne p^2$, $\Pr(X_I\ge 0)\Pr(Y_I\ge 0)\ne \Pr(X_I\ge 0\text{ and }Y_I\ge 0)$.  Therefore, by definition, $X_I$ and $Y_I$ are not independent, making a counterexample to the conjecture.
A: Let $X$ be the random variable with $P(X=1)=P(X=-1)=\frac12$ and let $Y=-X$. Then $X^2$ and $Y^2$ are independent but $X$ and $Y$ are not.
