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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 $[d,b)$?

Avraham
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Castor
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    I think you are correct. That's a classic franchise deductible, so as long as $x \leq d$ the insurer pays nothing but once $x > d$ the insurer pays the full $x$ and not $x - d$. Maybe the PDF was thinking about the case where **payment** was 0 which occurs when $0 \leq x \leq d$? – Avraham Dec 22 '21 at 21:04
  • It's correct as is, as the left-hand side is $f_Y(x)$, the distribution of $Y$, and $Y$ can only take on values of $0$ and $> d$. The re-use of $x$ makes it less clear than it could be; it's not the same $x$ as the $X$ in $F_X$, writing imprecisely. The support of $Y$ has to include $0$, since there's a point mass there. – jbowman Dec 22 '21 at 22:48
  • Your asserted formula for the cost-per-loss does not match the definition in the linked source, which is $Y=(X-d)_+$. – Ben Dec 24 '21 at 02:35

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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.

jbowman
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  • I know that $x$ is not the same as $X$. My issue is that I don't understand why the pdf takes the value $F_X(d)$ at $0$, then skips values from $0$ to $d$, and then takes non-zero values from $d$ on – Castor Dec 22 '21 at 23:48
  • Because $Y$ can't take on any values between $0$ and $d$, so $F_Y$ is not defined over that range. Look at the definition of $Y$ in the first equation; is there any way, for example, if $d=2$, that $y$ can equal $1$? – jbowman Dec 23 '21 at 02:39