Independence of two normally distributed random variables 
Let $X \sim \mathcal{N}(0, 1)$ and $Y$ be a random variable independent of $X$ such that
  \begin{align*}
P(Y=y) = \begin{cases}
\frac{1}{2} & y = -1\\
\frac{1}{2} & y = 1\\
0 & otherwise
\end{cases}
\end{align*}
  If $Z = XY$, are $Z$ and $X$ independent?

I've found that the correlation between $X$ and $Z$ is 0; however, I know that this does not say anything about their independence.
From the problem, it is quite obvious that the two variables are not independent; however, I am having trouble showing this formally. I want to find some way to show $f_{X,Z}(x,z)\neq f_X(x) \cdot f_Z(z)$ for some $(x, z)$, but I'm not sure what the joint density function actually is.
Since both $X$ and $Z$ follow $\mathcal{N}(0,1)$, I assume that the density function for both variables is the normal density. What, then, would be the joint density?
 A: Based on: $$P\left(X\leq x\wedge Z\leq x\right)=$$$$P\left(X\leq x\wedge Z\leq x\mid Y=-1\right)P\left(Y=-1\right)+P\left(X\leq x\wedge Z\leq x\mid Y=1\right)P\left(Y=1\right)$$
we find for $x<0$:
$$P\left(X\leq x\wedge Z\leq x\right)=\frac{1}{2}P\left(X\leq x\wedge-X\leq x\right)+\frac{1}{2}P\left(X\leq x\wedge X\leq x\right)=\frac{1}{2}P\left(X\leq x\right)$$
So independence demands that $P\left(Z\leq x\right)=\frac{1}{2}$
for each $x<0$, which cannot be true.
A: It's easy to specific the value of $x$ and $z$ to disprove the independence.
Let $x=1$ and $z=3$. Firstly we have $f_{XZ}(1,3)=0$ because that $z$ can only be equal to $x$ or $-x$.
While $f_X(1)*f_Z(3) = f_X(1)*(0.5f_X(3)+0.5f_X(-3))=f_X(1)*f_X(3) \ne0 $
So $X$ and $Z$ are not independent.
A: A faster way to notice lack of independence is as follows. If $X$ and $Z$ are independent, then knowing the value of $X$ you still need $Z$ to have the same distribution. It is easy to notice that $Z$ has continuous distribution, but given $X = x$, you know for sure that $Z\in\{-x,x\}$ so that conditional distribution of $Z$ much be different from the unconditional one. If you feel uncomfortable with such statements, you may still want to verify it formally as in drhab's answer, but I think even before that it's a good mental check to know which result to expect from formal computations.
