Independence of the absolute value of a standard normal distribution to its positive indicator

Question

Suppose that $X$ is identically distributed to $N(0,1)$. Show that $Y=|X|$ and $Z=1(X>0)$ are independent. (Note: $Z=1$ if $X>0$ and $Z=0$ otherwise.)

Answer 1

Because Z is a discrete random variable it's enough to show that for $c\in\mathbb{R}$

$$P(|X| \le c, Z=0) = P(|X| \le c)P(Z=0)$$ and $$P(|X| \le c, Z=1) = P(|X| \le c)P(Z=1)$$

But $$P(|X| \le c, Z=0) = P(-c \le X \le 0) = \frac{1}{2}P(|X| \le c) = P(Z=0)P(|X| \le c)$$

due to the symmetry of the normal distribution w.r.t to the origin and the identity

$$\begin{align*}P(|X| \le c) &= P(-c \le X \le c) \\&= P(-c \le X \le 0) + P(0 \le X \le c) \\&= 2 P(-c \le X \le 0)\\&= 2 P(0 \le X \le c)\end{align*}$$