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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 for Course P/1 online 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 taxdeductible to the extent allowed by law. If you have questions about these exercises, please send them by e-mail to: [email protected] Exercise for October 11, 2008 Study Note P-09-08, Problem No. 131 Let N1 and N 2 represent the numbers of claims submitted to a life insurance company in April and May, respectively. The joint probability function of N1 and N 2 is ⎧ 3 1 n1 −1 n2 −1 ⎪ ⎛ ⎞ e− n1 1 − e− n1 for n1 = 1, 2, 3…and n2 = 1, 2, 3,… fN1 , N2 ( n1 , n2 ) = ⎨ 4 ⎜⎝ 4 ⎟⎠ ⎪ otherwise. ⎩ 0 Calculate the expected number of claims that will be submitted to the company in May if exactly 2 claims were submitted in April. ( A. ( ) 3 2 e −1 16 B. 3 2 e 16 ) 3e 4−e C. D. e2 − 1 E. e2 Solution. We begin by finding the conditional probability function of N 2 given N1 = n1 , i.e., fN , N ( n1 , n2 ) fN2 ( n2 N1 = n1 ) = 1 2 . fN1 ( n1 ) We have ∞ 3⎛ 1⎞ fN1 ( n1 ) = ∑ fN1 , N2 ( n1 , n2 ) = ⎜ ⎟ 4 ⎝ 4⎠ n2 =1 n1 −1 ∞ ∑ e (1 − e ) ⋅ − n1 − n1 n2 −1 n2 =1 geometric series with ratio 1− e− n1 and first term e− n1 = 3⎛ 1⎞ ⎜ ⎟ 4 ⎝ 4⎠ n1 −1 Therefore fN2 ( n2 N1 = n1 ) = ⋅ e− n1 3⎛ 1⎞ = ⎜ ⎟ − n1 4 ⎝ 4⎠ 1− 1− e ( fN1 , N2 ( n1 , n2 ) fN1 ( n1 ) n1 −1 ) ( = e− n1 1 − e− n1 . ) n2 −1 . = This is actually the probability function of a geometric random variable with parameter 1 1 p = e− nt . The mean of this distribution is = − n1 = en1 , and that equals e2 when n1 = 2. p e Answer E. © Copyright 2004-2008 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.