How to determine the distribution iof this transformed variable $X_c = X ({1_{\{|X|>c\}}}-{1_{\{|X|\le c\}}})$ I have one question from econometric class which asks to find the probability distribution of the following: 
$$X_c = X ({1_{\{|X|>c\}}}-{1_{\{|X|\le c\}}}) $$
where $X \sim N(0,1)$ and $c > 0$. 
I am not even sure what $X_c$ is referring to. Does anyone have any idea? Thanks!
 A: As you suggest ${1_{\{|X|>c\}}}-{1_{\{|X|\le c\}}}$ will be $1$ if $|X|>c$ and $-1$ otherwise. 
I suggest you start by drawing a picture. Draw yourself a normal density. 

Colour it in (say) blue where the expression ${1_{\{|X|>c\}}}-{1_{\{|X|\le c\}}}$ is $1$ and red where it's $-1$. Mark few values along the x-axis in both the blue and red regions. What will happen to the blue parts? What will happen to the red parts - where do they map to? If you can't see it yet, join each point up to its image on the axis (with a curved arrow). Can you see what's going on? (it should be quite obvious once you draw the diagram) Now go back and see if you can do it algebraically.

Ahem. I was looking for something more or less like this:

A: Hints: Consider some generic $x\in\mathbb{R}$ and let $Z=2\cdot 1_{|X|\geq c}-1$. $Z=1$ means $X>c$ or $X<-c$ and $Z=-1$ otherwise. Let $\Phi$ denote the CDF of $N(0,1)$. Then, $\Pr(X_c\leq x)=\Pr(XZ\leq x)$ and
\begin{aligned}
\Pr(XZ\leq x)=\Pr(X Z\leq x\;\cap\; Z=1)+\Pr(X Z\leq x\;\cap\; Z=-1)\equiv \color{red}{A}+\color{blue}{B}.
\end{aligned} 
There are 3 cases:


*

*$x<-c$: show $A=\Phi(x)$ and $B=\ldots$ (To see why $A=\Phi(x)$ here, note $Z=1$ and $XZ\leq x$ for $x<-c$ is equivalent to $X<x$.)

*$-c\leq x\leq c$: show $A=\Phi(-c)$ and $B=\ldots$

*$x>c$: show $A=\Phi(x)-\Phi(c)+\Phi(-c)$ and $B=\ldots$


Show then that in all cases, $A+B$ is always $\ldots$ and conclude.
Edit: this kind of construction provides a way to demonstrate uncorrelation does not imply independence: you will find $E(X_c)=0$. And, therefore, whereas $X$ and $Z$ are clearly dependent, we have
$$
\text{Cov}(X,Z)=E(XZ)-E(X)E(Z)=E(X_c)-E(X)E(Z)=0-0E(Z)=0.
$$
There are easier constructions of course (e.g., take $Z=X^2$ instead.)
