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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 Questions? E-mail: [email protected] Exercise for February 10, 2007 May 1983 Course 110 Examination, Problem No. 19 Let X1 , X2 , and X 3 be a random sample from a normal distribution with mean µ ! 0 1 and variance ! 2 = . What are the values of a and b, respectively, in order for 24 L = aX1 + 4 X2 + bX 3 to have a standard normal distribution? A. a = !2, b = !2 B. a = !2, b = 2 C. a = !1, b = !3 D. a = 2, b = 2 E. Cannot be determined from the given information Solution. The random variable L, as a linear combination of independent normal variables, is normal itself. We have E ( L ) = aE ( X1 ) + 4E ( X2 ) + bE ( X 3 ) = ( a + 4 + b ) µ. In order for L to be standard normal, we must have E ( L ) = 0, so that a + b = !4. Also, a 2 + 16 + b 2 . 24 In order for L to be standard normal, we must have Var ( L ) = 1, so that a 2 + b 2 = 8. This gives 2 a 2 + ( !a ! 4 ) = 8, so that a 2 + a 2 + 8a + 16 = 8, and 2a 2 + 8a + 8 = 0, resulting in 2 a 2 + 4a + 4 = ( a + 2 ) = 0, i.e., a = !2. We also get b = !a ! 4 = 2 ! 4 = !2. Answer A. Var ( L ) = a 2Var ( X1 ) + 16Var ( X 2 ) + b 2Var ( X 3 ) = © 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.