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ACE 501 Fall, 2008
ACE 501 Lecture Notes — 9/03/
Carl H. Nelson
Problem Set 2 (due 9/12) MWG: 8.B.1,8.D.9,9.B.3,9.D.9,G: 2.1, 2.4, 2.6, 2.11, 2.13, 2.
1 Imperfect information
The following extensive form representation of a game can be used to discuss the distinction between perfect and imperfect information, as well as the distinction between backwards induction and sub-game perfection.
L R
L R
A B
C D C D
The dashed line means that when player 2 moves at that point she does not know if player 1 played A or B. This is the meaning of imperfect information — an extensive form game exhibits imperfect information if any player has an information set that consists of more than one node when the decision needs to be made. (The last information set of player 2 is an example of an information set that consists of more than one node.)
Furthermore, when an extensive form game exhibits imperfect information you can not solve the game with backwards induction because there will be points where the optimal decision depends upon the choice of the other player. (Re- member backwards induction means you can move backward through the game tree solving for each player’s optimal decision contingent on the decision of the opponent. But, this sort of reasoning does not apply to the game above.)
2 Sub-game perfection
At player 1’s second move the game is a zero sum simultaneous move game — matching pennies. The Nash equilibrium is the mixed strategy { 0. 5 , 0. 5 } with expected payoff 0. If we place payoffs (0, 0) at the node that initiates the proper sub-game, then we can solve for equilibrium by applying backwards induction. 2 will choose L, and 1 will choose R. Definition: A sub-game is part of an extensive form. It is a collection of nodes and branches satisfying three properties:
- starts from a single node.
- contains every successor node.
- if it contains any part of an information set, it contains all the nodes in that information set.
This definition guarantees that the sub-game is a well-defined game. You solve the game by backward induction starting with the terminal sub-games. Replace the sub-games with their Nash payoffs and move backwards.
3 Repeated Games
A special class of dynamic games of complete information that are solved with sub-game perfection are repeated games. These games are constructed by taking a static game of complete information, the stage game, and repeating play. This significantly expands the strategy spaces. It allows current strategies to be condi- tioned on past strategy profiles {S 1 , S 2 ,... , St− 1 }. But they don’t allow past play to change the feasible actions or payoffs in the stage game.
any Nash equilibrium of the stage game is a sub-game perfect equilibrium of the repeated game as well. Further, if discount rates are small enough, representing extreme impatience, the only equilibrium of the repeated game is to play the static game Nash equilibrium strategies at every stage. On the other hand, there are “folk theorems” which assert that if players are sufficiently patient, then any feasible, individually rational payoffs can be en- forced as an equilibrium. This is because deviation from desired play is punished. The following “folk theorem” is an example.
4 Minmax folk theorem
Folk Theorem 1: For every feasible payoff vector v with vi > vi for all players, there exists δ < 1 such that, for all δ ∈ (δ, 1) there is a Nash equilibrium with payoffs v. In this theorem vi is the reservation utility or minmax utility, defined by:
vi = M in a−i
M ax ai gi(ai, a−i)
This is the lowest payment that the opponents of player i can hold player i to. For example, in the following game:
L R
U -1,-1 -9,
D 0,-9 -6,-
Column can hold row to a payoff of − 6 and row can hold column to a payoff of − 6. Then a minmax strategy for cooperation is to play (U, L) until a deviation occurs, then play (D, R) for the rest of the game. Or consider the following example:
L R
U -2,2 1,-
M 1,-2 -2,
D 0,1 0,
To calculate the row player’s minmax payoffs to pure strategy choices U, M, D let the column player play L with probability q. Then the payoffs as a function of q are:
VU (q) = − 2 q + 1(1 − q) = − 3 q + 1 VM (q) = q − 2(1 − q) = 3q − 2 VD(q) = 0
So choose q to minimize max{− 3 q + 1, 3 q − 2 , 0 }. This appears as:
Thus any q ∈ [1/ 3 , 2 /3] enforces a payoff of 0.
Player 2’s minmax payoff can be expressed as a function of pU , pM.
vL = 2 (pu − pM ) + (1 − pU − pM ) vR = −2 (pu − pM ) + (1 − pU − pM )
Both equations are 0 at (1/ 2 , 1 / 2 , 0). Otherwise one or the other equation is posi- tive. No minmax equilibrium payoffs can be less than the minmax payoffs.
Then, working with normalized discounted payoffs, player i will not deviate if:
(1 − δ)mi + δei ≤ vi ⇒ mi − vi ≤ δ(mi − ei) ⇒
mi − vi mi − ei
≤ δ
Where ei and vi, which are normalized discounted payoffs from an infinite sum. They appear as they do because:
(1 − δ)
∑^ ∞
t=
δtvi = (1 − δ)
vi 1 − δ
and (1 − δ)
∑^ ∞
t=
δtvi = (1 − δ)
δvi 1 − δ
Consider the following example:
L R U 1,1 5, D 0,5 4,
Consider the strategy: play (D, R), if not play (U, L). Then the bound on δ is δ ≥ (5 − 4)/(5 − 1) = 0. 25. In infinitely repeated games, a large set of equilibria can often be supported as sub-game perfect equilibria. In finitely repeated games it is more difficult to support cooperation with sub- game perfect equilibria. The ability to support cooperation requires multiple equi- libria in the stage game. Consider the following game to be played twice.
L M R U 1,1 5,0 0, M 0,5 4,4 0, D 0,0 0,0 3,
In order to support playing (M, M ) in the first period promise to play (D, R) in the second period if (M, M ) in the first period, and (U, L) if anything else. This would produce the following normal form for the two period play of the game:
L M R U 2,2 6,1 1, M 1,6 7,7 1, D 1,1 1,1 4,
And (M, M ) is the equilibrium of this game. But, it should be noted that this equilibrium strategy, although it is sub-game perfect, is not renegotiation-proof. That is, in the second period if the strategy required play of the (U, L) stage Nash equilibrium, then the players would have an incentive to renegotiate and play the (D, R) stage Nash equilibrium in the second period. And this would break the incentive to cooperate.
