Density of cost-per-loss random variable Given a random variable $X$ with support $(a,b)$ and a value $d\in (a,b)$ (the "deductible"), the cost-per-loss random variable associated to $X$ is given by $$Y=\begin{cases}0&\text{if }X\le d\\ X&\text{if }X>d\end{cases}$$
According to this PDF, the pdf of $Y$ is $$f_Y(x)=\begin{cases}F_X(d)&\text{if }x=0,\\f_X(x)&\text{if }x>d\end{cases}$$ but I am having a hard time seeing why this is true. I have shown that the cdf of $Y$ is given by $$F_Y(x)=F_X(d)+F_X(t)-F_X(\min\{x,t\}).$$ So, I think the first condition in the pdf $f_Y$ above should be changed to $x\le d$ rather than $x=0$. Am I mistaken? Could someone please point out what the correct definition of $F_Y$ and $f_Y$ should be, in case I calculated it wrong? Also, is the support of $Y$ the interval $[d,b)$?
 A: The pdf as stated is correct; the issue is that the notation is not clear to those who aren't familiar with it.
In $f_Y(x)$, the random variable is $Y$ and the value it takes on is $x$.  $x$ is just a placeholder, as in integration, and has no relation to $X$, which refers to the original random variable, the one with support $(a,b)$.  It would have been clearer had the original writer used $y$ as a placeholder instead of $x$:
$$f_Y(y)=\begin{cases}F_X(d)&\text{if }y=0,\\f_X(y)&\text{if }y>d\end{cases}$$
where $Y$ has support $\{0\} \cup (d,b)$.
Due to the support of $Y$, $F_Y$ is not defined over the interval $(0,d]$ (nor is it defined over the intervals $(-\infty, 0)$ or $[b,\infty)$).  We could extend the range of $Y$ to include $(0, d]$ and define $F_y$ such that $f_Y(y) = 0$ for $y \in (0,d]$, but that's rather pointless in this case, just as extending the range of, say, a $U(0,1)$ variate to $(-\infty, \infty)$ and defining $f(x) = 0 \,\forall x \leq 0 \text{ or } x \geq 1$ doesn't add any value except perhaps in a few special situations.
