Question

An insurance company sells an auto insurance policy that covers losses incurred by a policyholder, subject to a deductible of 100. Losses incurred follow an exponential distribution with a mean of 300.

Calculate the 95th percentile of actual losses that exceed the deductible. Give your answer rounded to the nearest whole number. Hint: You can either do this one ‘directly’ or you can use the ‘memoryless’ property of the exponential distribution.

A) 600 B) 700 C) 800 D) 900 E) 1000

Answer 1

Option E - 1000

Explanation:

Let X stand for actual losses incurred.

Given that X follows an exponential distribution with mean 300,

To find the 95-th percentile of all claims that exceed 100.

In other words,

0.95 = Pr (100 < x < p95 ) / P(X > 100)

= Fx( P95) − Fx(100 ) / 1− Fx (100)

, where Fx is the cumulative distribution function of X

since, Fx(x) = 1 - e^ (-x/300)

0.95 = 1 - e^ (-P95/300) - [ 1 - e^ ( -100/300) ] / 1 - [ 1 - e^ ( -100/300) ]

= e^ ( -1/3 ) - e^ ( - P95//300) / e^(-1/3)

= 1 - e^1/3 e^ (-P95/300)

The solution is given by , e^ ( - P95/300) = 0.05e^(-1/3)

P95 = -300 ln ( 0.05e^(-1/3) )

= 999

= 1000