PDF for $\frac1a Uniform(−a,a)$

Question

Problem: Let $X$ be $Uniform(−a,a)$ distrubuted. Calculate the PDF for $Z = \frac1a abs(X)$.

Attempt: I think graphically here. $X$ is $U(-a,a)$ with $PDF = \frac1{2a}$ so $abs(X)$ is $U(0,a)$ with $PDF = \frac1a$ and thus PDF for $Z = \frac1a \cdot \frac1a = \frac1{a^2}$ with support on $x \in [0,a]$.

I do know that I could use the inverse tranform method and do like this question Does the square of uniform distribution have density function?. My first thought was to do this and put calculate PDF for $Z^2$ since I do not like working with abs() but I canno tdo that since $\sqrt{Z^2} \neq Z$ due to the fact that Z har support on $[-a,a]$.

In your attempt, the PDF you derived does not integrate to one. — Jano Kakara, Oct 10 '15 at 11:08
@JanoKakara No it does not. I cannot figure out the gap in my reasoning. — jacob, Oct 10 '15 at 11:13
Hint: If $U$ is uniform on $(0,a)$, what is the range of $U/a$? — Jano Kakara, Oct 10 '15 at 11:16

score 1 · Accepted Answer · answered Oct 10 '15 at 11:33

In your attempt, the first step is correct, that is really $\left\vert X \right\vert$ is distributed as $U \sim Uniform(0,a)$ with distribution function $$ F_U(u) = \begin{cases} u/a & \mbox{ for }0 \leq u \leq a, \\ 0 & \mbox{ for }u < 0, \\ 1 & \mbox{ for }u > a. \end{cases}$$ However, then you must obtain the distribution of $Z = U/a$, and the easiest way to do this is through its distribution function $$ F_Z(z) = P(Z \leq z) = P(U/a \leq Z) = P(U \leq a z) = F_U(a z) = \begin{cases} a z /a & \mbox{ for }0 \leq a z \leq a, \\ 0 & \mbox{ for }z<0/a, \\ 1 & \mbox{ for }z>a/a. \end{cases}$$ Thus, $U/a$ is uniform on $(0,1)$.

PDF for $\frac1a Uniform(−a,a)$

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