as372f08a1soln

# as372f08a1soln - Solution to Actsc 372 F08 Assignment...

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Unformatted text preview: Solution to Actsc 372 F08: Assignment 1 (total: 40 marks) 1. (5 marks) Expected utility of gamble: ∞ X k =1 2- k (2 k ) 1 / 2 = ∞ X k =1 2- k/ 2 = 1 / (1- . 7071)- 1 = 2 . 414 max amount = 2 . 424 2 = 5 . 828 2. (8 marks) (a) E [1 A ] = 1 million E [1 B ] = 0 . 89(1 million ) + 0 . 1(5 million ) + 0 . 01(0) = 1 . 39 million E [2 A ] = 0 . 11(1 million ) + 0 . 89(0) = 0 . 11 million E [2 B ] = 0 . 10(5 million ) + 0 . 90(0) = 0 . 5 million An expected value maximizer would choose Gamble 1B and Gamble 2B. (b) Solution 1: Choosing 1A (over 1B) and 2B (over 2A), respectively, implies u (1) > . 89 u (1) + 0 . 1 u (5) + 0 . 01 u (0) (1) . 1 u (5) + 0 . 9 u (0) > . 11 u (1) + 0 . 89 u (0) (2) (1) leads to . 11 u (1) > . 1 u (5) + 0 . 01 u (0) while (2) yields . 11 u (1) < . 1 u (5) + 0 . 01 u (0) . Hence, a contradiction. Solution 2: • Let C ∼ L (5 , , 10 11 ) and D ∼ L (1 , , 1). Then 1 A ∼ L ( D,C, 1) and 1 B ∼ L ( D,C, . 89). By monotonicity axiom, 1 A 1 B implies D C and hence...
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## This note was uploaded on 09/18/2011 for the course ACTSC 372 taught by Professor Maryhardy during the Winter '09 term at Waterloo.

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as372f08a1soln - Solution to Actsc 372 F08 Assignment...

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