# Social choice problem set solutions

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Social Choice Class, Prof. Juan Andrade 2011
Final Problem Set Solutions, Due Dec. 8

1. Prove that if a 2 × 2 matrix game has a unique strict Nash equilibrium, then the game can be solved via iterated strict dominance. In particular, consider the game below and suppose that (T,L) is the unique Nash equilibrium and that unilateral deviations from it would leave the deviator strictly worseoff. Show that (T,L) is the only strategy profile surviving iterated elimination of strictly dominated strategies.

Table 1: Problem 1
Player 2
| L | R |
T | u1(T,L), u2(T,L) | u1(T,R), u2(T,R) |
B | u1(B,L), u2(B,L) | u1(B,R), u2(B,R) |

Player 1

Solution: Let (T,L) be a unique strict NE.
u1(T,L) > u1(B,L) and u2(T,L) > u2(T,R)

Now, assume the game cannot be solved viaiterated elimination of strictly dominated strategies.
u1(T,R) ≤ u1(B,R) and u2(B,L) ≤ u2(B,R)

This implies (B,R) is a NE which contradicts the uniqueness of (T,L). Therefore, the assumption that the game cannot be solved via iterated elimination of strictly dominated strategies is false.

2. Consider the game depicted in Table 2. Specify the set of pure-strategy profiles satisfying eachsolution concept. (Use mixed strategies to eliminate pure ones when necessary.)

Table 2: Problem 2
Player 2
| X | Y | Z |
X | (2, 2) | (0, 2) | (3, 1) |
Y | (2, 0) | (1, 1) | (0, 0) |
Z | (1, 3) | (0, 0) | (1, 1) |

Player 1

A. Iterated elimination of strictly dominated strategies.

Solution: For both players, mixing over X and Y with probability 0.5 strictly dominates Z. Thesurviving profiles are {(X,X), (X,Y), (Y,X), (Y,Y)}.

B. Iterated elimination of weakly dominated strategies.

Solution: First eliminate Z because it is strictly dominated. Then Y weakly dominates X. This leaves us with (Y,Y).

C. Rationalizable.

Solution: Since Z is strictly dominated it cannot be a best response. X and Y, however, can both be best responses so they cannot be eliminated.The surviving profiles are {(X,X), (X,Y), (Y,X), (Y,Y)}.

D. Nash equilibrium.

Solution: (X,X) and (Y,Y)

E. Pareto-efficient Nash equilibrium.

Solution: (X,X)

3. Consider a game between a tax collector (player 1) and a tax payer (player 2). Player 2 has income of 200 and may either report his income truthfully or lie. If he reports truthfully, he pays 100 to player 1 and keeps therest. If player 2 lies and player 1 does not audit, then player 2 keeps all his income. If player 2 lies and player 1 audits, then player 2 gives all his income to player 1. The cost to player 1 of conducting an audit is 20. Suppose that both parties are risk neutral and that they move simultaneously (i.e., player 1 must decide whether to audit before he knows player 2’s reported income).

A.Depict this situation in a matrix game.

Solution:

Player 2
| Truth (T) | Lie (L) |
Audit (A) | (80,100) | (180,0) |
No Audit (N) | (100,100) | (0,200) |

Player 1

B. Calculate the probability, σ1*, that player 1 performs an audit in the Nash equilibrium of this game. Calculate the probability σ2* that player 2 lies about his income.

Solution: Player 1 mixes to make Player 2indifferent between playing T or L. This implies
100σ1+1001-σ1=2001-σ1
σ1=12

Player 2 mixes to make Player 1 indifferent between playing A or N. This implies

801-σ2+180σ2=1001-σ2
σ2=110

C. What are the expected equilibrium payoffs to each player.

Solution: E(U1) = 90, E(U2) = 100.

D. Is theNash equilibrium pareto efficient? (Explain.) Does the Nash equilibrium maximize social surplus? (Explain.)

Solution: The NE is not pareto efficient. (N,T) would give player 1 a higher payoff without making player 2 worse off. For this same reason, the NE does not maximize social either.

4. Consider the game depicted in table 3, where ai> 0 and bi∈R for i = 1, 2.
Table 3: Generalized...