# Why is this probability equal to 1?

Consider for one car owner the insurance policy with the following clauses:

• Deductible: If the loss $X>d$, then the insurer pays only for loss above $d>0$.

• Coverage Limit: If the loss $X>l$, then the insurer pays only for loss below $l>d$.

Now, we are to assume that the potential loss $X$ to the owner is of Pareto distribution with density $\displaystyle f(x)=\frac{ab^a}{x^{a+1}}$, for $x \geq b$, $a>0$, $d>b>0$.

If we let $Y$ be the potential loss to the insurer, then $$Y = \begin{cases}0, & \text{if}\,b<x\leq d \\ (X-d)_{+}, & \text{if}\, d<x \leq l \\ l-d, & \text{if}\, x>l \end{cases}$$

Now, I have been told that $P(Y=l-d)=1$; however that doesn't make any sense to me, since because of how we defined $Y$, $P(Y=l-d)=P(X>l)$, right? Which according to our density function should come out to be $\displaystyle \left(\frac{b}{l}\right)^{a}$, so what am I missing here?

Unless I just misunderstood them. They said $F(l-d)=1$, which I took to mean $P(Y=l-d)=1$...

Thanks.

• Maybe it was supposed to be $P(Y\leq l-d)=1$" instead? – jdods Nov 17 '17 at 0:48
• Usually capital F means cumulative distribution function. $F(x)=P(X \leq x)$. – jdods Nov 17 '17 at 0:56

$F(x)$ usually stands for cumulative distribution function; the definition is
$$F(x) = P(Y \le x)$$
So $F(l-d) = P(Y \le l-d) = 1$