math.illinoisstate.edumath.illinoisstate.edu/krzysio/1-17-9-ko-exercise.pdf · an insurance company...

Krzys’ Ostaszewski: http://www.krzysio.net Author of the “Been There Done That!” manual for Course P/1 http://smartURL.it/krzysioP (paper) or http://smartURL.it/krzysioPe (electronic) Instructor of online P/1 seminar: http://smartURL.it/onlineactuary If you find these exercises valuable, please consider buying the manual or attending the seminar, and if you can’t, please consider making a donation to the Actuarial Program at Illinois State University: https://www.math.ilstu.edu/actuary/giving/ Donations will be used for scholarships for actuarial students. Donations are tax- deductible to the extent allowed by law. If you have questions about these exercises, please send them by e-mail to: [email protected] Exercise for January 17, 2009 November 2000 Course 1 Examination, Problem No. 9, also Study Note P-09-08, Problem No. 70 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 mean 300. What is the 95-th percentile of actual losses that exceed the deductible? A. 600 B. 700 C. 800 D. 900 E. 1000 Solution. Let X stand for actual losses incurred. We are given that X follows an exponential distribution with mean 300, and we are asked to find the 95-th percentile of all claims that exceed 100. In other words, we want to find p 95 such that 0.95 = Pr 100 < x < p 95 ( ) PX > 100 ( ) = F X p 95 ( ) − F X 100 ( ) 1 − F X 100 ( ) , where F X is the cumulative distribution function of X. We know that F X x () = 1 − e − x 300 . Therefore, we end up with this equation 0.95 = 1 − e − p 95 300 − 1 − e − 100 300 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 − 1 − e − 100 300 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = e − 1 3 − e − p 95 300 e − 1 3 = 1 − e 1 3 e − p 95 300 . The solution is given by e − p 95 300 = 0.05e − 1 3 , or p 95 = −300 ln 0.05e − 1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≈ 999. Answer E. © Copyright 2004-2009 by Krzysztof Ostaszewski.

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Page 1: math.illinoisstate.edumath.illinoisstate.edu/krzysio/1-17-9-KO-Exercise.pdf · An insurance company sells an auto insurance policy that covers losses incurred by a ... Losses incurred

If you have questions about these exercises, please send them by e-mail to: [email protected]

Exercise for January 17, 2009

November 2000 Course 1 Examination, Problem No. 9, also Study Note P-09-08, Problem No. 70 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 mean 300. What is the 95-th percentile of actual losses that exceed the deductible?

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

Solution. Let X stand for actual losses incurred. We are given that X follows an exponential distribution with mean 300, and we are asked to find the 95-th percentile of all claims that exceed 100. In other words, we want to find p95 such that

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. We know that FX x( ) = 1− e−x300 .

Therefore, we end up with this equation

0.95 =1− e

−p95300 − 1− e

−100300

⎛⎝⎜

⎞⎠⎟

1− 1− e−100300

⎛⎝⎜

⎞⎠⎟

=e−13 − e

−p95300

e−13

= 1− e13e

−p95300 .

The solution is given by e−p95300 = 0.05e

−13 , or

p95 = −300 ln 0.05e−13

⎛⎝⎜

⎞⎠⎟≈ 999.

Answer E.

Page 2: math.illinoisstate.edumath.illinoisstate.edu/krzysio/1-17-9-KO-Exercise.pdf · An insurance company sells an auto insurance policy that covers losses incurred by a ... Losses incurred

All rights reserved. Reproduction in whole or in part without express written permission from the author is strictly prohibited. Exercises from the past actuarial examinations are copyrighted by the Society of Actuaries and/or Casualty Actuarial Society and are used here with permission.

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