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 ) =&gt; .4+.1U($50)+.4U($20)=.56 U(P C )= U(P D ) =&gt; .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. Jiongs preference: alpha=.668 U($400)=1, U($0)=0 A~$300, B~$200 and C~$100 =&gt; U(P A )=U($300)=.75+.25U($200), U(P B )=U($200)=.6, U(P C )= U($100)=.55U($200) 2 U($100)=.33, U($200)=.6 and U($300)=.9 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)&gt;U(A)&gt;U(B), the ranking of preference is C&gt;A&gt;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 ) =&amp;amp;gt;...

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