Krzys’ Ostaszewski: http://www.math.ilstu.edu/krzysio/ Author of the “Been There Done That!” manual for Course P/1 http://smartURL.it/krzysioP (paper) or http://smartURL.it/krzysioPe (electronic) Instructor for Course P/1 online seminar: http://smartURL.it/onlineactuary If you find these exercises valuable, please consider buying the manual or attending our 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 taxdeductible to the extent allowed by law. Questions about these exercises? E-mail: [email protected] Exercise for May 12, 2007 An insurance policy has a deductible of 10. Losses follow a probability distribution with density f X ( x ) = xe! x for x > 0 and f X ( x ) = 0 otherwise. Find the expected payment. A. e!10 B. 2e!10 C. 10e!10 D. 12e!10 E. 100e!10 Solution. Let X be the random variable describing losses, and let Y be the amount of payment. Then X ! 10, # 0, Y =$ % X " 10, X > 10. We can calculate the expected value of Y directly +" +" +" v = !e! x = E (Y ) = # ( x ! 10 ) xe dx = # x e dx ! # 10xe dx = du = 2xdx dv = e! x dx 10 10 10 !####"#### $ !x 2 !x u = x2 !x Integration by parts for the first integral ( 2 !x = !x e = u = 8x ) x$" x =10 +" + # 2xe !x 10 v = !e! x !x du = 8dx dv = e dx !### #"#### $ Integration by parts for the first integral +" dx ! # 10xe !x dx = 100e !10 +" ! 10 # 8xe !x dx = 10 ( = 100e!10 + 8xe! x ) x$" x =10 +" !8 #e !x dx 10 !" # # $ = Survival function of exponential distribution with hazard rate evaluated at x =10 = 100e!10 ! 80e!10 ! 8e!10 = 12e!10 . Answer D. We could also do this problem by applying the Darth Vader Rule. We have +( if y = 0, % " Pr ( X > 10 ) , sY ( y ) = Pr (Y > y ) = # = Pr X > y + 10 = xe! x dx, = ( ) & ) Pr X ! 10 > y , if y > 0, ) $ ( ' 10 + y = u = !x v = e! x = !xe! x !x dx = !dx dv = !e !###"###$ +( x*( + x =10 + y Integration by parts ) e! x dx = 10 + y ! #"# $ Survival function of exponential with hazard rate 1 evaluated at 10 + y = (10 + y ) e!10 ! y + e!10 ! y = (11 + y ) e!10 ! y for any nonnegative y. As the payment random variable is non-negative almost surely, the expected payment is +" # (11 + y ) e 0 !10 ! y dy = 11e !10 +" $ #e !y dy 0 !" # # $ This equals one because e! y for y>0 is a density of an exponential random variable with hazard rate 1 +e !10 +" $ # ye !y dy 0 ! #"# $ = 12e!10 . This equals one because the integrand is the density in this problem. Answer D. © Copyright 2007 by Krzysztof Ostaszewski. 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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