solution4

# solution4 - outcomes U(P A)= U(P B...

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1 Eco 354K Game Theory Solutions to Problem Set No. 4 1. Ronald’s preference (a) The ranking is B > C > A. U(\$50)=.25U(\$500), U(\$25)=.15U(\$500). So let U(\$500)=1, U(\$0)=0, Then U(\$50)=.25 and U(\$25)=.15. U(P A )=.05U(\$500)+.3U(\$50)+.3U(\$25)+.35U(\$0)=.17 U(P B )=.75U(\$50)+.1U(\$25)+.15U(\$0)=.2025 U(P C )=.02U(\$500)+.16U(\$50)+.82U(\$25)=.183 Since U(P B )> U(P C )> U(P A ), lottery B is the most preferred. (b) A fourth lottery’s utility value U(P) Note that U(\$0)=0<=U(P)<=1=U(\$500) for any lotteries P and the sum of probability equals to 1. That is, P=(a, b, c, d) with a+b+c+d=1 and a, b, c, d in [0,1]. U(P)=a+.25b+.15c 2. Amy’s preference (a) Amy would prefer A to D, because (i) B is better than C by the monotonicity axiom, and (ii) Amy is indifferent between A and B and between C and D which implies A is preferred over D by transitivity of preferences (axiom 1) (b) p=.8 (c) q=.2 U(\$100)=1, U(\$5)=0, since \$100 and \$5 are the “best” and “worst”

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Unformatted text preview: outcomes. U(P A )= U(P B ) => .4+.1U(\$50)+.4U(\$20)=.56 U(P C )= U(P D ) => .1+.4U(\$50)+.1U(\$20)=.44 So that, U(\$50)=.8 and U(\$20)=.2 Therefore, U((p,0,0,1-p))= p =U(\$50)=.8; U((q,0,0,1-q))= q =U(\$20)=.2 (d) P=(a, b, c, d) with a+b+c+d=1 and a, b, c, d in [0,1]. U(P)=a+.8b+.2c 3. Jiong’s preference: alpha=.668 U(\$400)=1, U(\$0)=0 A~\$300, B~\$200 and C~\$100 => U(P A )=U(\$300)=.75+.25U(\$200), U(P B )=U(\$200)=.6, U(P C )= U(\$100)=.55U(\$200) 2 Ö U(\$100)=.33, U(\$200)=.6 and U(\$300)=.9 Ö U(P D )=.35U(\$400)+.18U(\$300)+.15U(\$200)+.2U(\$100)+.12U(\$0)=.668= alpha 4. Dutta, Ch. 27: 19 U(x)=log(10+x) U(A)=.5log(10+1)+.5log(10-1)=.9978 U(B)=.5log(10+5)+.5log(10-5)=.9375 5. Dutta, Ch. 27: 20 U(C)=.25log(10+13)+.25log(10+16)+.5log(10-4)=1.083 Since U(C)>U(A)>U(B), the ranking of preference is C>A>B. Lottery C is the most preferred. 6. E.g. \$100 \$0 \$50...
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solution4 - outcomes U(P A)= U(P B...

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