Show that if $X\sim$ uniform$(-\theta, \theta)$ then $|X|\sim$ uniform$(0, \theta)$.

Question

I am trying to show that if $X\sim$ Uniform$(-\theta, \theta)$ then $|X|\sim$ Uniform$(0, \theta)$. I start by using the Jacobian but I end up with $$ f(y)=\left|\frac{\partial x}{\partial y} \right|f(x) = \frac{y}{2\theta|y|}$$ but I do not think this is correct. Can some one help me?

Answer 1

Hint:

$Y= |X| $ is not one to one transformation. So

$$f_Y(y)=\left| \frac{d}{dy} (-y) \right| f_X(-y)+\left| \frac{d}{dy} (y) \right| f_X(y) $$

Answer 2

The distribution function of Unif$(-a,a)$ is the following: $$F_X(x)= \frac{1}{2a} x+ \frac{1}{2}$$ for $x \in [-a, a]$ which we can get by integrating the density function (which is constant with value $\frac{1}{2a}$ on $[-a,a]$). Now we know by definition of the distribution function: $$F_{|X|}(x)= P(|X| \leq x) = P(-x \leq X \leq x) = F_X(x)-F_X(-x) = \frac{x} {a}$$ for $x \in [0,a]$ because negative values don't make sense here. The absolute value can never be negative. The distribution with the density function $F_{|X|}(x)= \frac{x} {a}$ on $[0,a]$ is precisely Unif$(0,a)$. (which we could again check by differentiating and getting the constant function $\frac{1}{a}$ on $[0, a]$ which is the density function of Unif$(0,a)$).

Answer 3

Why make it so complicated? This can be done without thinking about Jacobians.

For $0<a<b<\theta$ we have \begin{align} & \Pr(a<|X|<b) = \Pr(a<X<b) + \Pr(a<-X<b) \\[8pt] = {} & \frac{b-a}{2\theta} + \frac{b-a}{2\theta} = \frac{b-a}\theta. \end{align}

Also, note that $\dfrac y{|y|} = 1$ when $y>0,$ so what you did is correct.