Independence of two transformed variables from a symmetric distribution Let $X$ be a continuous random variable with a symmetric probability density function centered at zero. Let, $Y=|X|$ and $Z=\mathrm{sgn}(X)$, where $\mathrm{sgn}(X)=1$ if $X&gt;0$ and $\mathrm{sgn}(X)=-1$ if $X\leq 0$. Are $Y$ and $Z$ independent?
Yes, $Y$ and $Z$ are independent. Since $X$ is symmetric, if $y\geq 0$ then
$$ \mathbb{P}(Y\leq y,Z=1)=\mathbb{P}(-y\leq X\leq y,X&gt;0)=\mathbb{P}(0&lt;X\leq y)=\frac{1}{2}\mathbb{P}(-y\leq X\leq y)=\mathbb{P}(Y\leq y)\mathbb{P}(Z=1)$$

because $\mathbb{P}(Z=1)=\mathbb{P}(X&gt;0)=\frac{1}{2}$. Similarly, $\mathbb{P}(Y\leq y,Z=-1)=\mathbb{P}(Y\leq y)\mathbb{P}(Z=-1)$ for all $y\geq 0$. 

Since $Y$ only takes non-negative values and $Z$ only takes the values $\pm1$, this is enough to show that $Y$ and $Z$ are independent.
They are almost everywhere independent.

Due to symmetry of the distribution of $X$, knowing its sign does not tell you anything about its magnitude, nor vice versa, except at one fixed point: $0$.

When given that $~Y=0~$, then surely $~Z=0~$ and vice versa. 

$$\operatorname {sgn}(X)=0 ~\iff~ \lvert X\rvert = 0$$

Depending on the distribution of $X$, this may be important, or it may be negligible. &nbsp; (Ie: What is the probability measure of the event that $~X=0~$?)