Question: Let $Z$ be the standard normal random variable. Find the value of the probability density function $f_{x}(x)$ of $X = |Z + 1|$ for x = 1.
My approach:
The standard normal RV is Z~N(0,1) and its pdf is $\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$
$$F_{x}(X) = P(X \le x) = P(|Z+1|\le x) = P(Z \le \pm x-1) = F_{z}(\pm x-1)$$
Then, $$f_{x}(x) = f_{z}(\pm x-1)\frac{d}{dx}(\pm x-1)$$ $$ f_{x}(x) = \frac{1}{\sqrt{2\pi}}e^{(\pm x-1)^2/2}\frac{d}{dx}(\pm x-1) $$
This is the part where I get stuck. When I substitute x=1 for both $-x-1$ and $x-1$ I get incorrect answers both times. Moreover, I am confused about finding the inverse of the absolute function in the beginning steps. I don't think there should be 2 cases (i.e. both $-x-1$ and $x-1$) but am unsure of how to proceed. Any help would be appreciated. Thanks!
Suppose $Z$ is an arbitrary continuous random variable with PDF $f_z$. Let's express PDF of $X_c = |Z - c|$ for any constant $c \in \mathbb{R}$ through it:
Suppose $F_{X_c}$ and $F_Z$ CDFs of $X_c$ and $Z$ respectively.
Then $\forall t > 0$
$$F_{X_c}(t) = P(X_c < t) = P(|Z - c| < t) = P(Z \in [c-t; c+t]) = F_Z(c+t) - F_Z(c-t)$$
If we differentiate both sides of the equality by $t$ we get:
$$f_{X_c}(t) = f_Z(c+t) + f_Z(c-t)$$