# 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!

• 1. You should add the self-study tag (as a signal to potential answerers) and see the discussion on homework-style questions in the help center. 2. Do you understand the meaning of the indicator notation $\mathbf{1}_A$? – Glen_b Oct 7 '17 at 3:18
• I understand notation $1_A$ as the function that gives $1$ if $|X|>c$ and $-1$ otherwise. I thought it would be $P(|X|>c) * X + P(|X|\le c)*-X$ but was not 100% sure. Could you let me know if I am going to the right direction? – David Kim Oct 7 '17 at 4:11
• Your response seems a bit muddled. $\mathbf{1}_A$ is a function defined on a set $X$ that indicates membership of an element in a subset $A$ of $X$, having the value $1$ for all elements of $A$ and the value $0$ for all elements of $X$ not in $A$ (wikipedia). I think you understood that already but you didn't convey it. – Glen_b Oct 7 '17 at 4:49

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.)

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: • Thank you very much for your response. Unfortunately, I am not quite sure whether I followed your advice correctly. For $1_{\{|X|>c\}}$, assuming a specific number for $c$ (i.e. $c = 1$), I highlighted the region of the pdf where it is greater than 1 and less than -1 (the outer regions of the straight line of pdf you posted). I put the red color on the inside, where it is the region for when $|X|<c$. It seems to me that it is just the addition of two separate regions of pdf that will give the whole pdf, which I believe I am misunderstanding something.. – David Kim Oct 7 '17 at 15:12
• Please disregard the previous comment(for some reason I can't edit my comment.) How I understood is that for $1_{\{|X|>c\}}$, I highlighted the region of the pdf where it is less than 1. I put the red color where it is the region is less than -1. With this, the blue region and red region overlaps where it is less than -1. But I am not sure if this is what you intended me to follow.. – David Kim Oct 7 '17 at 15:23
• If $X=-2$, what is $X^c$? If $X=0.5$ what is $X^c$? Note that the center (between the two vertical lines above) should be red, and the outer sections blue. – Glen_b Oct 7 '17 at 23:05
• Dear Glen, thank you for your comment! If I understand correctly, for $X = -2$, then $X_c = -2 * 1 = -2$ assuming that $c = 1$ so it should be on the outer region blue. For $X=0.5$, since $|X|\le c = 1$, it should be $0.5 * -1 = -0.5$ which is in the inner section. To determine the probability determine function, since $X ~ N(0,1)$, is it correct for me to assume that I have to work on with the below? $$\frac1{\sqrt 2\pi}exp(\frac{-x^2}2)$$ – David Kim Oct 8 '17 at 1:16
• This is just to give you the intuition for what's occurring. You should be able to do the derivation of the correct answer by working directly with the indicator functions. Presumably you have some of the results for manipulating indicator functions you need already (though they're not so hard to derive anyway). There are a few basic results here but I expect you've been shown (or have been asked to show) some others – Glen_b Oct 8 '17 at 1:41