Prove that $Y := X|Z|$ has the same distribution as $Z$ where $Z$ is standard normal Let $Z\sim\mathcal{N}(0,1)$ and $X$ be a discrete random variable with pmf
$\mathbb{P}(X = -1) = \mathbb{P}(X = 1) = \frac{1}{2}$ and is independent of $Z$. Define $Y = X|Z|$ and show that $Y$ has the same distribution as $Z$.    
How would I start this?  
Following TheoreticalEconomist's suggestion:  
$$\mathbb{P}(Y\leq y) = \mathbb{P}(X|Z| \leq y) \\ = \mathbb{P}(-y \leq XZ \leq y) \\ = \mathbb{P}(-y \leq Z \leq y | X= 1) \mathbb{P}(X=1) + \mathbb{P}(-y\leq -Z \leq y|X=-1)\mathbb{P}(X=-1) \\ = \frac{1}{2}(2\mathbb{P}(-y\leq Z\leq y)) \\ = \mathbb{P}(-y \leq Z \leq y) 
$$
which doesn't seem to be the CDF of $Z$
 A: Suppose WLOG $y\geq 0,$ the other case is similar. You have:
$$\mathbb{P}(Y\leq y) = \mathbb{P}(X|Z| \leq y)= \\ =\mathbb{P}(\{X=-1, -|Z|\leq y\} \cup \{X=1, |Z|\leq y\})=\\=\mathbb{P}(X=-1,-|Z|\leq y)+\mathbb{P}(X=1, |Z|\leq y)=\\=\mathbb{P}(X=-1)\mathbb{P}(-|Z|\leq y)+\mathbb{P}(X=1)\mathbb{P}(|Z|\leq y)=\\=\frac{1}{2}\mathbb{P}(-|Z|\leq y)+\frac{1}{2}\mathbb{P}(|Z| \leq y)= \frac{1}{2}\cdot 1+ \frac{1}{2}\mathbb{P}(-y \leq Z \leq y)=\\=\frac{1}{2}(\mathbb{P}(Z\leq y)+P(Z \geq y)+ \mathbb{P}(-y \leq Z \leq y))=\\=\frac{1}{2}(\mathbb{P}(Z\leq y)+P(Z \leq -y)+ \mathbb{P}(-y \leq Z \leq y))=\\=\frac{1}{2}(2\cdot \mathbb{P}(Z\leq y))=\mathbb{P}(Z\leq y)
$$
where I used the independence of the variables to split the summands into 2 products, and the (centered) gaussianity to turn $\mathbb{P}(Z\geq -y)$ into $\mathbb{P}(Z\leq y).$ 
Notice that I've used the fact that $y\geq 0$ when I said that $\{-|Z|\leq y\}$ is an event of probability one, and I've split $1$ into $\mathbb{P}(Z\leq y)+P(Z \geq y).$
One last comment: notice that gaussianity hasn't really been used for this solution. What you really need is the independence of X and Z and the symmetry of the distribution of Z.
