How to get the pdf of this transformation $Y=|X|$?

Question

This question is from George Casella textbook question 8.30(b). $$f(x\mid\theta)=\frac{\theta}{\pi}\frac{1}{\theta^2+x^2},\quad -\infty<x<\infty,\, \theta>0$$

Let $ Y=|X|$, how to get the pdf of $Y$? I know theorem 2.1.15, but it requires that $Y=g(X)$, where $g$ is a monotone function. Here I have $ Y=|X|$, obviously it is not monotone. So I am stuck here.

Here is the solution:

Answer 1

This case is discussed in Chapter 2 of Casella. You would simply write $$\Pr[Y \le y] = \Pr[|X| \le y] = \Pr[-y \le X \le y] = \int_{x=-y}^y \frac{\theta}{\pi} \frac{1}{\theta^2 + x^2} \, dx.$$ Note since $f(x \mid \theta) = f(-x \mid \theta)$ for all $x$, we can also write this as $$\Pr[Y \le y] = 2 \Pr[0 \le X \le y],$$ hence $$f_Y(y \mid \theta) = 2 f_X(y \mid \theta) \mathbb 1 (y \ge 0).$$