Given a random variable X with uniform distribution on [-b,b], I want to compute the probability density function of Y = g(X) with
$$g(x) = \begin{cases} 0, ~~~x\in[-c,c]\\ x, ~~~\text{else} \end{cases}$$,
and $b>c$.
There is a discontinuity at $x = c$ and $x=-c$. Therefore, the distribution of Y has both an absolutely continuous component
$f_Y(y)=\frac{1}{2b}$ on $y\in (-b,-c)$ and $y\in (c,b)$
and a discrete component.
This is the point, where I am not sure...my idea was the following:
The discrete component consists of two mass points at $-c$ and $c$. This results in
$$P_Y = \frac{b-c}{b} P_Y^{AC} + \frac{c}{b}P_Y^{D}$$
with
$AC$...absolutely continuous
$D$...discrete
Could this be right?
The density function of Y, as you already observed, is a mixed density with a discrete component. This component has probability $P(X \in[-c;c])=[c-(-c)]\cdot\frac{1}{2b}=\frac{c}{b}$
This because when $X \in [-c;c]$ Y concentrate all this mass of probability in one single value: $Y=0$
Thus this mixed density (not absolute continuous) is the following
$$ f_Y(y) = \begin{cases} \frac{c}{b}, & \text{if $y=0$} \\ \frac{1}{2b}, & \text{if $y \in[-b;-c] \cup [c;b]$} \\ 0, & \text{elsewhere} \end{cases}$$