asgn1_sol - CSC 5350 Assignment 1 1. (a). (b). (a1, a2, a3)...

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Unformatted text preview: CSC 5350 Assignment 1 1. (a). (b). (a1, a2, a3) (c) i. (a1, a2, a3) ii. Assume that an outcome ω which maximizes the total utilities of all players is not Pareto optimal. Then, there must be another outcome ω ′ , such that at least one player i gets a higher payoff in ω ′ than in ω , while no player gets a lower payoff. Then we have ∑i∈N ui (ω ) < ∑i∈N ui (ω′) which contradict the fact that ω maximizes the total utilities. Consequently, ω is Pareto optimal. 2. (a) N = {Tom, Mary}, A1 = {2, 1, 0}, where 2 means that Tom put 2 dollars in the box. A2 = {accept, reject} The payoff matrix accept reject Tom Mary 2 0,2 0,0 1 1,1 0,0 0 2,0 0,0 (b) The pure strategy Nash equilibriums are (0, reject) and (0, accept). Justification: UT is the utility of Tom, UM is the utility of Mary Since UT(0, accept)> UT(1, accept)> UT(2, accept) UM(0, accept)= UM(0, reject) So (0, accept) is the Nash equilibrium. Since UT(0, reject)= UT(1, reject)= UT(2, reject) UM(0, accept)= UM(0, reject) So (0, reject) is the Nash equilibrium. (c) The completely mixed strategy Nash equilibria is (e(0), (p, 1‐p)) where 0<p<1. Justification: Let (a*, b*) be the mixed strategy Nash equilibrium UT(e(0),b*)=UT(e(1),b*)=UT(e(2),b*) 2b*(accept) +0b*(reject) =1b*(accept) +0b*(reject) =0b*(accept) +0b*(reject) b*(accept) +b*(reject) =1 Then b*(accept)=0 b*(reject)=1 Since UM(a*, e(accept))=UM(a*,e(reject)) 0a*(0) +1a*(1) +2a*(2) = 0a*(0) + 0a*(1) + 0a*(2) a*(0)+ a*(1)+ a*(2)=1 Then a*(0)=1 a*(1)=0 a*(2)=0 Though the result is ((1, 0, 0), (0, 1)), it conflict with the concept of completely mixed strategy, which the probability for each action should be between 0 and 1, and not equal to 0 or 1. However, if Tom chooses the mixed strategy (1, 0, 0), no matter what Mary plays, Mary’s utility is 0. Consider Mary’s strategy as (p, 1‐p) where 0<p<1, since Tom can always get the maximum utility by using the strategy (1, 0, 0), so the completely mixed strategy Nash equilibria is (e(0), (p, 1‐p)) where 0<p<1. (d) Tom should put $0 in the box. As discussed in (c), no matter what Mary plays, Tom can always gets the maximum utility if he put $0 in the box. 3. (a). N= {1, 2} A1 = { n1 | x ≤ n1 ≤ y and n1 is an integer}, A2 = { n2 | x ≤ n2 ≤ y and n2 is an integer} ⎧n1 + x, ⎪ u1 = ⎨n1 , ⎪n − x , ⎩2 ⎧ n2 + x , ⎪ u 2 = ⎨ n2 , ⎪n − x, ⎩1 n1 < n2 n1 = n2 n1 > n2 n1 > n2 n1 = n2 n1 < n2 (b) The pure strategy Nash equilibrium is (x, x) Justification: If traveler 1 chooses x, traveler 2 will not change his option from x otherwise his payoff will be 0. If traveler 2 chooses x, traveler 1 will also not change his option otherwise his payoff will be 0. So (x, x) is a Nash equilibrium. In addition, if x<=1, the Nash equilibrium is (n, n), x<=n<=y, and n is an integer because any traveler cannot get better payoff through decreasing or increasing the value he write down since u1(n, n)=n >= u1(n‐1, n)= n‐1+x >= u1(n+1, n)=n‐x u2(n, n)=n >= u2(n, n‐1)= n‐1+x >= u2(n, n+1)=n‐x 4. Model the game as follows: a b If w > y then (a, a) is a strict equilibrium, so that it is an ESS. If z > x then (b, b) is a strict equilibrium, so that it is an ESS. If w < y and z < x then this game has a mixed strategy Nash equilibrium (m*, m*), m*(a)=p and m*(b) = 1‐p, where 0<p<1. Since U(e(a), m*)=U(e(b), m*) pw + (1‐p)x = py + (1‐p)z then p= (x‐z)/(x ‐ z + y ‐ w) 1‐p = (y‐w)/(x – z + y – w) We consider another mixed strategy m(a)=q, m(b)=1‐q. U(m*, m) ‐ U(m, m) a w, w y, x b x, y z, z = pqw + p(1‐q)x + (1‐ p)qy + (1‐p)(1‐q)z ‐qqw ‐ q(1‐q)x ‐ q(1‐q)y ‐ (1‐q)(1‐q)z = (p‐q)(qw + x – qx – qy – z + qz) = (p – q)2(x – z + y ‐ w) >0 (since w < y and z < x) Then (m*, m*) is an ESS In conclusion, there is a mixed strategy that is an evolutionarily stable strategy. ...
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This note was uploaded on 04/23/2010 for the course CSC CSC5350 taught by Professor Leunghofung during the Winter '09 term at CUHK.

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