El hombre moderno frente a los Andes

Kuhn’s Theorem constitutes a generalization of Theorem4.43, which states that every two-player zero-sum game with perfect information has a value. The proof of the theorem is similar to the proof of Theorem1.4(page3), and involves induction on the number of vertices in the game tree. Every child of the root of a game tree defines a subgame containing fewer vertices than the original game (a fact that follows from the assumption that the game has perfect recall) and the induction hypothesis then implies that the subgame has an equilibrium. Choose one equilibrium for each such subgame. If the root of the original game involves a chance move, then the union of the equilibria of all the subgames defines an equilibrium for the entire game. If the root involves a decision taken

119 4.13 Games with perfect information

v0

x1 _x2 _xL

Γ(x1_{) Γ(x}2_{) Γ(x}L₎

Figure 4.38 The game tree and subgames starting at the children of the root

by player i, then that player will survey the subgames that will be played (one for each child that he may choose), calculate the payoff he will receive under the chosen equilibrium in each of those subgames, and choose the vertex leading to the subgame that grants him the maximal payoff. These intuitive ideas will now be turned into a formal proof.

Proof of Theorem4.49: It is convenient to assume that if a player in any particular game has no action available in any vertex in the game tree, then his strategy set consists of a single strategy denoted by∅.

The proof of the theorem is by induction on the number of vertices in the game tree. If the game tree is comprised of a single vertex, then the unique strategy vector is (∅, . . . , ∅) (so a fortiori there are no available deviations), and it is therefore the unique Nash equilibrium. Assume by induction that the claim is true for each game in extensive form containing fewer than K vertices, and consider a game  with K vertices. Denote by x1, . . . , xLthe children of the root v0_{, and by (x}l_{) the subgame whose root is x}l _{and whose vertices}

are those following xl_{in the tree (see Figure4.38). Because the game is one with perfect}

information, (xl) is indeed a subgame. If we had not assumed this then (xl) would not necessarily be a subgame, because there could be an information set containing vertices that are descendants of both xl1_{and x}l2 _{(where l}

1= l2) and we would be unable to make

use of the induction hypothesis.

The payoff functions of the game  are, as usual, ui :

×

1, 2, . . . , L, the payoff functions in the subgame (xl) are ul_i :

×

is player i’s set of strategies in the subgame (xl_).

For any l∈ {1, . . . , L}, the root v0 of the original game  is not a vertex of (xl), and therefore the number of vertices in (xl_{) is less than K. By the induction hypothesis,}

for each l∈ {1, 2, . . . , L} the game (xl) has an equilibrium s∗l= (s_i∗l)i∈N (if there are

Case 1: The root v0_{is a chance move.}

For each l∈ {1, 2, . . . , L} denote by pl the probability that child xl is chosen. For each player i consider the strategy s_i∗in the game  defined as follows. If vertex xl _{is chosen}

in the first move of the play of the game, implement strategy s_i∗lin the subgame (xl). By definition it follows that ui(s∗)=

L

l=1pluli(s∗l).

We will show that the strategy vector s∗= (s_i∗)i∈N is a Nash equilibrium. Suppose

that player j deviates to a different strategy sj. Let sjl be the restriction of sj to the

subgame (xl_{). The expected payoff to player j under the strategy vector (s}l

j, s−j∗l) is L l=1plu l j(s l j, s−j∗l).

Since s∗l is an equilibrium of (xl), ul_j(s_jl, s_−j∗l )≤ ul_j(s∗l) for all l= 1, . . . , L, and therefore uj(sj, s_−j∗ )= L l=1 plul_js_jl, s_−j∗l ≤ L l=1 plul_j(s∗l)= uj(s∗). (4.67)

In other words, player j does not profit by deviating from s_j∗to sj. Since this holds true

for every player j ∈ N, the strategy vector s∗is indeed a Nash equilibrium. Case 2: The root is a decision vertex for player i0.

We first define a strategy vector s∗= (s_i∗)i∈N and then show that it is a Nash equilibrium.

For each player i, i= i0, consider the strategy si∗defined as follows. If vertex xlis chosen

in the first move of the play of the game, in the subgame (xl_{) implement strategy s}∗l i .

For player i0define the following strategy si∗0: at the root choose the child x

l0_{at which the} maximum max1≤l≤Lul_i(s∗l) is attained. For each l ∈ {1, 2, . . . , L}, in the subgame (xl) implement7the strategy s_i∗l. The payoff under the strategy vector s∗= (s_i∗)i∈Nis ul0(s∗l0).

The proof that each player i, except for player i0, cannot profit from a deviation from

s_i∗is similar to the proof in Case 1 above. We will show that player i0 also cannot profit

by deviating from s_i∗

0, thus completing the proof that the strategy vector s

∗ _{is a Nash}

equilibrium.

Suppose that player i0 deviates by selecting strategy si0. Let x

l_{be the child of the root}

selected by this strategy, and for each child xlof the root let s_il₀be the strategy si0restricted to the subgame (xl_).

r If l = l0, since s∗l0is an equilibrium of the subgame (xl0), the payoff to player i0is

ui0(si0, s ∗ −i0)= u l0 i0  sl0 i0, s ∗l0 −i0  ≤ ul0 i0(s ∗l0₎= u i0(s ∗_{). (4.68)}

In other words, the deviation is not a profitable one.

r If l = l0, since s∗lis an equilibrium of the subgame (xl) and using the definition of l0

we obtain ui0(si0, s ∗ −i0)= u l i0  sl_i 0, s ∗l −i0  ≤ uli0(s ∗l_{)≤ u}l0 i(s∗l0)= ui0(s ∗_{). (4.69)}

This too is not a profitable deviation, which completes the proof.  • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

7 Since defining a strategy requires defining how a player plays at each node at which he chooses an action, we also need to define s_i∗

0in the subgames (x

l_{) which the first move of the play of the game does not lead to (l= l}

121 4.14 Games on the unit square

Remark 4.50 In the course of the last proof, we proceeded by induction from the root