stat155_Summer_8_solutions.pdf - Game Theory Worksheet#8 player I 1(7.2 from KP Consider the following symmetric game as played by two drivers both

# stat155_Summer_8_solutions.pdf - Game Theory Worksheet#8...

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This preview shows page 1 out of 3 pages. Unformatted text preview: Game Theory Worksheet #8, 07/20/2018 player I 1. (7.2 from KP) Consider the following symmetric game as played by two drivers, both trying to get from Here to There (or two computers routing messages along cables of dierent bandwidths). There are two routes from Here to There; one is wider and therefore faster, but congestion will slow them down if both take the same route. Denote the wide route W and the narrower route N. The payo matrix is player II W N W (3, 3) (5,4) N (4,5) (2,2) Find all Nash equilibria and determine which ones are evolutionarily stable. Solution. First, let's nd the pure equilibria. Leave only the maximal payo in a column for the rst player and in a row for the second:   (∗, ∗) (5, 4) (4, 5) . (∗, v) We see that there are two pure NE. Next, we search for the fully mixed strategies. The pair of fully mixed strategies is a Nash equilibrium i the strategies are both equalizing. Let's nd such equilibria: The second game: 3x1 + 5(1 − x1 ) = 4x1 + 2(1 − x1 ) =⇒ x1 = 3/4. Since the game is symmetric, the equalizing strategy of the second player is the same. Note that there is no equalizing pure strategy of any of the players. Thus, there are no Nash equilibria where the strategy of one of the players is pure and of the other is fully mixed.   We've just found all the Nash equilibria: (0.75, 0.25)T , (0.75, 0.25)T , (0, 1)T , (1, 0)T , (1, 0)T , (0, 1)T . Now we need to check, which of the equilibria are evolutionary stable. Recall the denition - the symmetric equilibrium (x, x) is evolutionary stable i it is a Nash equlibrium and for any pure strategy z 6= x s.t. zT Ax = xT Ax holds zT Az < xT Az, where A is the payo matrix of the rst player. First, note that all the pure equilibria in this game are not symmetric. So, they don't correspond to ESS. Thus, we only need to check the equilibrium (0.75, 0.25)T , (0.75, 0.25)T . Since x = (0.75, 0.25)T is an equalizing strategy, the condition zT Ax = xT Ax holds for any pure z, so we need to check if there is such pure z that zT Az ≥ xT Az. • z = (0, 1)T : z T Az = 2, xT Az = 0.75 ∗ 5 + 0.25 ∗ 2 = 4.25. • z = (1, 0)T : z T Az = 3, xT Az = 0.75 ∗ 3 + 0.25 ∗ 4 = 3.25. In both cases zT Az < xT Az, which means that the strategy (0.75, 0.25)T is evolutionary stable. 2. In the following general-sum game nd a correlated equilibrium, which is not a NE.  (0, 0) (1, 0)  (0, 1) (−10, −10) Compare the payos in your correlated equilibrium with the payos in the (unique) symmetric Nash equilibrium. Solution We will use the following facts: • Each NE is a correlated equilibrium. • The set of correlated equilibria is convex. First we see that (e1 , e2 ) and (e2 , e1 ) are pure NE in this game, which means (by the rst fact) that the following distributions are correlated equilibria:     0 0 1 , 0 0 1 0 . 0 By the second fact, any convex combination of those distributions is still a correlated equilibrium. For example,  0 0.5   0.5 0 = 0.5 ∗ 0 0   1 0 + 0.5 ∗ 0 1  0 . 0 player I This probability distribution does not correspond to a NE: indeed, both players pick the rst move with probability 0.5, so if they acted independently they would both play the rst move with probability 0.25. However, they never both play the rst move at the same time. The only thing left is to compute the payos: • In our correlated equilibriun the payos are 0.5 ∗ (1.0) + 0.5 ∗ (0, 1) = (0.5, 0.5).  • In the unique symmetric NE (10/11, 1/11)T , (10/11, 1/11)T the payo to both players is 0. 3. (7.4 from Karlin and Peres) Occasionally, two parties resolve a dispute (pick a "winner") by playing a variant of Rock-Paper-Scissors. In this version, the parties are penalized if there is a delay before a winner is declared; a delay occurs when both players choose the same strategy. The resulting payo matrix is the following: player II Rock Paper Scissors Rock (-1, -1) (0,1) (1,0) Paper (1,0) (-1, -1) (0,1) Scissors (0,1) (1, 0) (-1,-1) player I Show that this game has a unique Nash equilibrium that is fully mixed, and results in expected payos of 0 to both players. Then show that the following probability distribution is a correlated equilibrium in which the players obtain expected payos of 1/2: player II Rock Paper Scissors Rock 0 1/6 1/6 Paper 1/6 0 1/6 Scissors 1/6 1/6 0 Solution. We need to show that the set of fully mixed equilibria consists of only one element. Fully mixed equilibria are equalizing pairs of strategies. Let's nd equalizing strategies in this game: x3 − x1 = x1 − x2 , x1 − x2 = x2 − x3 , x1 + x2 + x3 = 1. x3 + x2 + x1 = 3x1 , x1 + x2 + x3 = 3x2 , ⇐⇒ x1 + x2 + x3 = 1. ⇐⇒ x1 = x2 = x3 = 1/3. Thus there is unique equalizing strategy for the rst player. Since the game is symetric for both players (A = B T ), the same strategy is the unique equalizing strategy for the  rst player. So there is indeed only one fully mixed Nash equilibrium (x∗ , y∗ ) = (1/3, 1/3, 1/3)T , (1/3, 1/3, 1/3)T . Now we can compute x∗ Ay∗ = x∗ By∗ = 0, so the payos are indeed zero. Next, we check that the probability distribution given by the second matrix is a correlated equilibrium. Consider the pair of random variables (R, C) sampled from this distribution. Let's compute the conditional distribution of R given C . For example, let's compute P(R = ·|C = "Rock"): P(R = "Paper"&C = "Rock") P(R = "Paper"|C = "Rock") = , P(C = "Rock") P(C = "Rock") = P(R = "Rock"&C = "Rock") + P(R = "Paper"&C = "Rock") + P(R = "Scissors"&C = "Rock"), P(C = "Rock") = 0 + 1/6 + 1/6 = 1/3, 1/6 = 1/2. P(R = "Paper"|C = "Rock") = 1/3 Analogously we can compute P(R = "Scissors"|C = "Rock") = 1/2, P(R = "Rock"|C = "Rock") = 0. After we found the conditional distribution of R given C = "Rock", we need to check that if the second player picks moves according to that conditional distribution, then "Rock" is an optimal response of the rst player. So, player II plays the strategy y = (0, 1/2, 1/2)T . The vector of the average payos to the rst player is Ay = (1/2, −1/2, 0)T . As one can see, the rst coordinate (corresponding to the move "Rock") is indeed the largest. Thus, "Rock" is an optimal response. Note that neither the game matrix, nor the correlated equilibrium matrix do not change after cyclic relabelling "Rock"→"Paper"→"Scissors"→"Rock". Thus, the statement that we've just proved for the case C ="Rock" also holds for the other values of C . Furthermore, as the game is symmetric, all the analogous statements hold for the second player. Thus, we proved that the given probability distribution is a correlated equilibrium. The nal step is to compute the payos: denote the elements of the matrices by lower indices (Ai,j ), and denote the correlated equilibrium matrix as E . Then the payo for the player I is X Ai,j Ei,j = (−1) · 0 + 0 · 1/6 + 1 · 1/6 + 1 · 1/6 + (−1) · 0 + 0 · 1/6 + 0 · 1/6 + 1 · 1/6 + (−1) · 0+ = 1/2. i,j From the symmetry, the payo to the player II is the same. 4. (based on KP 4.8) Three rms (players I, II, and III) put three items on the market and advertise them either on morning or evening TV. A rm advertises exactly once per day. If more than one rm advertises at the same time, their prots are zero. If exactly one rm advertises in the morning, its prot is \$200K. If exactly one rm advertises in the evening, its prot is \$300K. Firms must make their advertising decisions simultaneously. Find a correlated equilibrium which is not a Nash equilibrium. Solution First, note that all the following triples of strategies are pure NE: (M, M, E) (M, E, M ), , , . Analogously to the problem 2, any convex combination of those gives a correlated equilibrium. (M, E, E) (E, M, E) (E, E, M ) (E, M, M ) , ...
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