$Y = |X|$, where $X \sim \text{N}(\mu, \sigma^2)$: How does the PDF and CDF change? I just encountered the random variable $Y = |X|$, where $X \sim \text{N}(\mu, \sigma^2)$. Now, based on what we know about the absolute value function, this random variable is still continuous; however, the absolute value function means that there exists a cusp at $X = 0$, and so the derivative is undefined at this point. 
This makes me wonder: How does this affect the PDF and CDF? How would we go about calculating such things in this case? 
I would greatly appreciate it if people could please take the time to clarify this situation.
 A: $$P(\lvert X\rvert\le\alpha)=\begin{cases}P(-\alpha\le X\le\alpha)&\text{if }\alpha>0\\ 0&\text{if }\alpha<0\end{cases}$$
 Therefore the cdf is $$F_{\lvert X\rvert}(\alpha)=\begin{cases}0&\text{if }\alpha<0\\ F_X(\alpha)-\sup_{\beta<-\alpha} F_X(\beta)&\text{if }\alpha\ge 0\end{cases}$$
Since $F_X$ is continuous, $F_{\lvert X\rvert}(\alpha)=F_X(\alpha)-F_X(-\alpha)$ for all $\alpha\ge 0$.
$F_{\lvert X\rvert}$ will therefore be differentiable on $\Bbb R\setminus\{0\}$, because $F_X$ is. Since $F_{\lvert X\rvert}$ also turns out to be continuous on the whole $\Bbb R$, any function in this form $$\begin{cases}F'_{\lvert X\rvert}(\alpha)&\text{if }\alpha\ne0\\ c&\text{if }\alpha=0\end{cases}$$ will be a pdf of $\lvert X\rvert$.
A: New CDF:if $z\le 0$, clearly $F_Y(z)=0$.Otherwise,
$$
F_Y(z)=P\{|X|\le z\}=P\{-z\le X\le z\}=2\Phi(z)-1
$$
where $\Phi$ is the CDF of the normal variable. Your PDF is zero for $z<0$, and $2f_X(z)$ for $z> 0$ (just take the derivative). At $z=0$ it has a jump.
Well, the above is for $X\sim \mathcal{N}(0,1)$ but you can easily adapt it to the general case.
