Question

2. (2.5 Points) An insurance company sells an autoinsurance policy that covers losses incurred by a policyholder, subject to a deductible of $100. Losses (in $) incurred have cumulative distribution function (cdf) F(t) where F(t) 0, if t <0; and F(t) 1 - exp(-t/300), if t>O (a) What is the 95th percentile of losses incurred? (b) What is the 95th percentile of the actual losses that exceed the deductible?

Answer 1

Question 2

Here cumulative distribution of losses incurred is given

F(t) = 1 - exp(-t/300), if t > 0

so here we have to find the 95th percentile of losses incurred

so here that means we have to find t for which

F(t) = 0.95 = 1 - exp(-t/300)

0.05 = e^-t/300

ln (0.05) = -t/300

t= $898.72

(b) so here we have to find the actual losses that exceed the deductibles.

that means we have to take only those losses that are greater than $100.

so here first we will identify the percentage of losses greater than $ 100

F(t > 100) = 1- (1 - exp(-100/300)) = 0.7165

now we have to identify 95% of this which is = 0.7165 * 0.95 = 0.6807

so we have to find the value of t where

F(t) = (1 - 0.7165) + 0.6807 = 0.9642

0.9642 = 1 - e^-t/300

t = $ 998.72