Determine the induced distribution for $|X|$ when $X$ is standard normal Suppose that $X \sim N(0,1)$ - i.e., it is of standard normal distribution. I need to determine the induced distribution for $|X|$.
According to my text, if an r.v. $X$ has distribution $\mu$ and $g: \mathbb{R} \mapsto \mathbb{R}$, then, for $Y = g(X)$, $$ P(Y \in A) \equiv P(g(X) \in A) = P (X \in g^{-1}(A)) = \mu(g^{-1}(A)) = \mu g^{-1} (A) $$
where $A$ is a Borel set.  Thus $Y = g(X)$ has the induced distribution $\mu g^{-1}$. 
Here, $g(X) = |X|$, but the absolute value function doesn't have an inverse in the usual way - you'd need to consider it piecewise, say for $x < 0$ and for $x \geq 0$, so how do I express this here? Also, I know that $\mu(x) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{x}e^{-t^2/2}dt$.
So, after I have figured out how to properly express $g^{-1}$ here (which I'm hoping you can help me with), how do I put it together with $\mu$? 
Thank you.
 A: You aren't taking the inverse in the usual sense; you are looking at the inverse of a set, so the inverse of (1,2) will be $(-2, -1) \cup (1,2)$. You can now find the inverse of any set of the form $(-\infty, x)$; think of what happens when $x < 0$, and $x > 0$. I am not sure how much you can simplify it, but if your final answer is allowed to be in terms of $\mu(x)$, then it looks like a very reasonable expression. 
A: Firstly, in order to directly address the point where you're stuck, note that in this context $g^{-1}$ refers to computing preimages rather than inverting the function: $|X|$ only takes nonnegative values, so it is sufficient to consider $A\in\mathcal{B}(\mathbb{R}^+)$. Then 
$$g^{-1}(A)=A\cup(-A)$$
and due to symmetry of $\mathcal{N}(0,1)$
$$\mathbb{P}(|X|\in A)=2\mathbb{P}(X\in A).$$
In order to explicitly describe the distribution, take $x\geq 0$ and observe
\begin{align}\mathbb{P}(|X|\leq x)&=2\mathbb{P}(0\leq X\leq x)\\
&=2\left(\mu(x)-\mu(0)\right)\\
&=2\left(\mu(x)-\frac{1}{2}\right)\\
&=2\mu(x)-1.\end{align}
This fully describes the distribution. You could also obtain the pdf through differentiation. 
