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AMS410_EXAMSOL2

# AMS410_EXAMSOL2 - AMS 410 Actuarial Mathematics Fall 2009...

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Unformatted text preview: AMS 410 Actuarial Mathematics Fall 2009 Exam 2 solution: Univariate distributions 10/27/2009 1. Let X be a random variable with moment generating function M X ( t ) = ( 2 + e t 3 ) 9 ,-∞ < t < ∞ Calculate the variance of X . (A) 11 (B) 9 (C)8 (D) 3 (E)2 Solution: M ( t ) = 9 · ( 2+ e t 3 ) 8 · e t 3 , and M (0) = EX = 9 · 1 3 = 3 . M 00 ( t ) = 72 · ( 2+ e t 3 ) 7 · ( e t 3 ) 2 + 9 · ( 2+ e t 3 ) 8 · e t 3 , and M 00 (0) = EX 2 = 72 · 1 9 +9 · 1 3 = 11 . V AR ( X ) = EX 2- ( EX ) 2 = 11- 3 2 = 2 2. An insurance policy on an electrical device pays a bene t of 4000 if the device fails during the rst year. The amount of the bene t decreases by 1000 each successive year until it reaches 0. If the device has not failed by the beginning of any given year, the probability of failure during that year is 0.4. What is the expected bene t under this policy? (A) 2234 (B)2400 (C) 2500 (D) 2667 (E) 2694 Solution: De ne X : year in which the device fails, Y : bene t under the insurance policy. Then Y = 4000 if X = 1 , 3000 if X = 2 , 2000 if X = 3 , 1000 if X = 4 , if X ≥ 5 . Since if the device has not failed by the beginning of any given year, the probability of failure during that year is 0.4 , X follows a Geom(0.4) distribution. So P ( X = 1) = 0 . 4 , P ( X = 2) = 0 . 6 · . 4 , P ( X = 3) = 0 . 6 2 · . 4 , P ( X = 4) = 0 . 6 3 · . 4 , and P ( X ≥ 5) = 1- P (1)- P (2)- P (3)- P (4) . Thus the expected bene t is....
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AMS410_EXAMSOL2 - AMS 410 Actuarial Mathematics Fall 2009...

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