ps1sol - 14.12 Game Theory Muhamet Yildiz Fall 2005...

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14.12 Game Theory Muhamet Yildiz Fall 2005 Solution to Homework 1 Solution to Problem 1 a) (ten points) Let’s fi rst fi nd appropriate payo ff s for Alice. The information stated in the question shows us that the payo ff s will be of the form Bob Alice A M H P a b b G b a c We can arbitrarily assign b to be 0 and (using the information that b > c) c to be -1. All that remains is to solve for a. To do this, we’ll use the information that Alice prefers Penn Station to Grand Central Station i ff p > q-r/2. In other words, we have ap + 0 q + 0(1 p q ) > 0 p + aq + ( 1)(1 p q ) i ff p > q (1 p q ) / 2 . We can rearrange these inequalities to see that we want a ( p q ) > ( 1)(1 p q ) i ff p q > (1 p q ) / 2 . In other words, we want a ( p q ) > ( 1)(1 p q ) i ff 2( p q ) > (1)(1 p q ) . So we see that we need a = 2 . Now let’s look at Bob’s payo ff s. The information stated in the question shows us that they must be of the form Bob Alice A M H P x z w G z y w Moreover, when the probability of Alice waiting at Penn Station is 1/2, Bob prefers Amtrak to Metroliner, which tells us that x > y. Using this information, we can arbitrarily assign x to be 1 and y to be 0. Now we need to solve for z and w. Using s to denote the probability of Alice waiting at Penn Station, the fact that Bob prefers Amtrak to Metroliner i ff s > 1/3 means that 1 s + z (1 s ) > zs + 0(1 s ) i ff s > 1/3. This will hold if z=-1. Finally, the fact that Bob prefers Amtrak to Home i ff s > 2/3 means that 1 s + ( 1)(1 s ) > w i ff s > 2/3. This means that w=1/3. Thus, in the end, our normal form game is: Bob Alice A M H P 2,1 0,-1 0,1/3 G 0,-1 2,0 -1,1/3 1
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b) ( fi ve points) We can perform any positive a ne transformation to the payo ff function of either player and preserve the structure of the game. So let’s add 1 to Alice’s payo ff s and multiply Bob’s payo ff s by 3. We end up with: Bob Alice A M H P 3,3 1,-3 1,1 G 1,-3 3,0 0,1 c) (ten points) The question is asking for the rationalizable strategy pro fi les. Note that H strictly dominates M. Removing M from the game gives us: Bob Alice A H P 3,3 1,1 G 1,-3 0,1 In this new game above, P strictly dominates G. Removing G gives us: Bob Alice A H P 3,3 1,1 And in the game where Alice goes to Penn Station and Bob chooses between A and H, A strictly dominates H. So the only possible outcome given that it’s common knowledge that both players are expected utility maximizers with the stated preferences is for Bob to take Amtrak and for Alice to meet him at Penn Station. Solution to Problem 2 a) (six points) 2 1 Ll λ Ll ρ Lr λ Lr ρ Rl λ Rl ρ Rr λ Rr ρ Axa 3,0 3,0 3,0 3,0 0,2 0,2 0,2 0,2 Axb 3,0 3,0 3,0 3,0 0,2 0,2 0,2 0,2 Aya 1,5 1,5 1,5 1,5 0,2 0,2 0,2 0,2 Ayb 1,5 1,5 1,5 1,5 0,2 0,2 0,2 0,2 Bxa 1,2 1,2 2,1 2,1 1,2 1,2 2,1 2,1 Bxb 1,2 1,2 2,1 2,1 1,2 1,2 2,1 2,1 Bya 1,2 1,2 2,1 2,1 1,2 1,2 2,1 2,1 Byb 1,2 1,2 2,1 2,1 1,2 1,2 2,1 2,1 Cxa 1,0 2,1 1,0 2,1 1,0 2,1 1,0 2,1 Cxb 0,1 0,0 0,1 0,0 0,1 0,0 0,1 0,0 Cya 1,0 2,1 1,0 2,1 1,0 2,1 1,0 2,1 Cyb 0,1 0,0 0,1 0,0 0,1 0,0 0,1 0,0 2
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b) (seven points) For player 1, Cxa (among other things) strictly dominates Cxb and Cyb. Moreover, the mixed strategy that involves playing Axa and Bxa
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