as372w10a1soln

# as372w10a1soln - αw e αG(3 1945 The maximum premium is...

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Solution to Actsc 372 W01: Assignment 1 (total: 28 marks) 1. (6 marks) Expected utility of gamble: X k =1 2 - k (2 k ) γ = X k =1 2 - k (1 - γ ) = 1 / (1 - 2 - (1 - γ ) ) - 1 Therefore CE = [1 / (1 - 2 - (1 - γ ) ) - 1] 1 Example, γ = 0 . 2 ,CE = 4 . 47; γ = 0 . 6 ,CE = 6 . 70. More risk averse with smaller γ 2. (8 marks) (a) E [ A ] = 99 E [ B ] = 0 . 5 * 150 + 0 . 3 * 100 + 0 . 2 * 20 = 109 Hence choose Project B. (b) ﬁrst need to ﬁnd the parameter α , which is the solution to u (100) = 0 . 5 u (200) + 0 . 5 u (50) α = 1 / 450 = 0 . 002222 E [ u ( A )] = u (99) = 77 . 22 E [ u ( B )] = 0 . 5 * u (150) + 0 . 3 * u (100) + 0 . 2 * u (20) = 77 . 155 Hence choose Project A. Note a change in the choice of project due to the risk aversion. 3. (8 marks) Expected Utility without insurance: E [ u ( w 0 - X )] = Z 5000 0 - e - 0 . 002(5000 - x ) 1 5000 dx = - 0 . 099995 Expected Utility with insurance with insurance premium G : E [ u ( w 0 - G - X 5 )] = - e - αw 0 e αG · E [ e 1 5 αX ] , where α = 0 . 002 = - e - αw 0 e αG Z 5000 0 e 0 . 002 x/ 5 1 5000 dx = - e -

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Unformatted text preview: αw e αG (3 . 1945) The maximum premium is found by equating E [ u ( w-X )] = E [ u ( w-G-X 5 )] which leads to-. 099995 =-e-αw e αG (3 . 1945) ⇒ G = 3267 . 97 2 4. (6 marks) (a) Note that u ( x ) = x-γ and u 00 ( x ) =-γx-γ-1 . u ( x ) > 0 and u 00 ( x ) < 0 for x > imply that γ > 0 and γ 6 = 1. (b) If Y ∼ N ( μ,σ 2 ), then m.g.f. M Y ( t ) = E[ e tY ] = e μt + 1 2 σ 2 t 2 . u ( CE ) = E[ u ( X )] ⇒ CE 1-γ 1-γ = E " X 1-γ 1-γ # ⇒ CE 1-γ = E[ X 1-γ ] = E[ e (1-γ ) ln X ] = exp ± μ (1-γ ) + 1 2 σ 2 (1-γ ) 2 ² ⇒ CE = exp ± μ + 1 2 σ 2 (1-γ ) ² 3...
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as372w10a1soln - αw e αG(3 1945 The maximum premium is...

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