EC487 Advanced Microeconomics, Part I: Lecture 8 Leonardo Felli - - PowerPoint PPT Presentation

EC487 Advanced Microeconomics, Part I: Lecture 8

Leonardo Felli

32L.LG.04

17 November 2017

Extensive Form Game

Consider the following extensive form game:

❝❝❝❝❝❝❝❝❝❝❝ ❜ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ r ✁ ✁ ✁ ✁ ✁ ✔ ✔ ✔ ✔ ✔ ❚ ❚ ❚ ❚ ❚ r ❙ ❙ ❙ ❙ ❙

1 2 2 U D L1 L2 R1 R2 (4, 1) (0, 0) (−1, 1) (3, 2)

r r r r

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 2 / 61

Formal description of an extensive form game

◮ The set of players:

N = {1, 2}

◮ The set of histories of the game:

H =

• ∅, U, D, [U, L1], [U, R1], [D, L2], [D, R2]
• the (sub)set of terminal histories:

Z =

• [U, L1], [U, R1], [D, L2], [D, R2]
• ◮ The player function that determines which player chooses an

action after each non-terminal history: P(∅) = 1, P(U) = P(D) = 2.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 3 / 61

Formal description of an extensive form game (cont’d)

◮ The players’ payoffs associated to each terminal history:

u1([U, L1]) = 4 u2([U, L1]) = 1 u1([U, R1]) = u2([U, R1]) = 0 u1([D, L2]) = −1 u2([D, L2]) = 1 u1([D, R2]) = 3 u2([D, R2]) = 2

◮ The strategies for each player:

s1 = a1(∅) ∈ {U, D} s2 = (a2(U), a2(D)) ∈ {(L1, L2), (L1, R2), (R1, L2), (R1, R2)}

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 4 / 61

Nash Equilibrium

Definition

A Nash Equilibrium is a strategy profile s∗ such that for every player i ∈ N: ui(o(s∗

i , s∗ −i)) ≥ ui(o(si, s∗ −i))

∀si The strategy s∗

i is the best reply to s∗ −i for every player i ∈ N.

Where o(s) denotes the outcome of the game, that is the terminal history that results when each player follows strategy si: o(s) ∈ Z.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 5 / 61

Associated Normal Form

◮ Notice that the set of Nash equilibria of the extensive form

game can be identified by looking at the Nash equilibria of the normal form game associated with the extensive form game: Γ = {N, Si, ui(o(·))}

◮ In the previous example this associated normal form game is:

1/2 L1, L2 L1, R2 R1, L2 R1, R2 U 4, 1 4, 1 0, 0 0, 0 D −1, 1 3, 2 −1, 1 3, 2

◮ The Nash equilibria of this game are:

{(U, (L1, L2)); (U, (L1, R2)); (D, (R1, R2))}

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 6 / 61

Credible Threats

◮ Clearly there exist several extensive form games with the same

normal form representation.

◮ A serious problem with Nash equilibria of extensive form

games is that there exist outcomes o(s) that are not affected by the actions that the strategy si specifies after contingencies that are inconsistent with the history that leads to the terminal node.

◮ In other words, there exists Nash Equilibria that are built on

non-credible threats.

◮ Example: the outcome o(D, (·, R2)) is independent of the

action chosen following player 1’s choice U.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 7 / 61

Non-Credible Threat

❝❝❝❝❝❝❝❝❝❝❝ ❜ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ r ✁ ✁ ✁ ✁ ✁ ✔ ✔ ✔ ✔ ✔ ❚ ❚ ❚ ❚ ❚ r ❙ ❙ ❙ ❙ ❙

1 2 2 U D L1 L2 R1 R2 (4, 1) (0, 0) (−1, 1) (3, 2)

r r r r ⑦ ✇ ✇

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 8 / 61

Subgame Perfect Equilibrium (Selten 1965):

We need a stronger equilibrium concept to make predictions that are credible. Define a proper subgame of a given extensive form game with perfect information as the subgame that follows any non-terminal history.

Definition

A Nash equilibrium s∗

i of a game Γ is a subgame perfect equilibrium

if and only if it is a Nash equilibrium for every proper subgame. Notice that this way to proceed eliminates any Nash equilibrium sustained by a non-credible threat.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 9 / 61

Unique Subgame Perfect Equilibrium

The unique subgame perfect equilibrium in the example above is : [U, (L1, R2)]

❝❝❝❝❝❝❝❝❝❝❝ ❜ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ r ✁ ✁ ✁ ✁ ✁ ✔ ✔ ✔ ✔ ✔ ❚ ❚ ❚ ❚ ❚ r ❙ ❙ ❙ ❙ ❙

1 2 2 U D L1 L2 R1 R2 (4, 1) (0, 0) (−1, 1) (3, 2)

r r r r ✴ ✇ ✢

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 10 / 61

Properties of SPE:

◮ How to construct Subgame Perfect equilibria: ◮ Do subgame perfect equilibria exist? ◮ The answer to both questions for finite horizon games is

provided by the following result.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 11 / 61

Kuhn Theorem:

Theorem (Kuhn Theorem)

Every finite extensive form game with perfect information has a subgame perfect equilibrium. We will go through the construction and the use of the algorithm that proves this theorem — known as backward induction — in an example.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 12 / 61

Example of the use of Backward Induction

❜ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ r

• r

r r r ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗

1 1 2 A B C1 D1

❅ ❅ ❅ ❅ ❅

E1 (1, −1)

★ ★ ★ ★ ★ ★ ★ r r r r r

(1, 3) (2, 0) (0, 2) (4, 1) (3, 7) C2 D2 F1 E2 F2 2 1

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 13 / 61

Example of the use of Backward Induction (cont’d)

◮ Notice first that the strategies for the two players are:

s1 = (a1(∅), a1(A, C1), a1(B, D2)), s2 = (a2(A), a2(B))

◮ Start from the last two subgames in which player 1 chooses

between E1 and F1 at one node and E2 and F2 at the other node.

◮ Compute the Nash equilibrium of these two extremely simple

subgames.

◮ This Nash equilibrium is: E1 and E2 respectively. ◮ Hence, all Subgame Perfect strategy of player 1 have the two

components: s∗

1 = (·, E1, E2)

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 14 / 61

Example of the use of Backward Induction (cont’d)

We can now transform the game tree from:

❜ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ r

• r

r r r ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗

1 1 2 A B C1 D1

❅ ❅ ❅ ❅ ❅

E1 (1, −1)

★ ★ ★ ★ ★ ★ ★ r r r r r

(1, 3) (2, 0) (0, 2) (4, 1) (3, 7) C2 D2 F1 E2 F2 2 1

❄ ✴

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 15 / 61

Example of the use of Backward Induction (cont’d)

To:

❜ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅

D1 (2, 0)

r r r r

(4, 1) 1 2 A B C1 C2 (1, 3) (1, −1) D2

r r

2

❏❏❏❏❏ ✡ ✡ ✡ ✡ ✡

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 16 / 61

Example of the use of Backward Induction (cont’d)

◮ Consider the last two subgames of this new game tree in

which player 2 chooses between C1 and D1 at one node and C2 and D2 at the other node.

◮ The Nash equilibrium of these two subgames is: C1 and C2

respectively.

◮ The Subgame Perfect strategy of player 2 is then:

s∗

2 = (C1, C2)

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 17 / 61

Example of the use of Backward Induction (cont’d)

We can once again transform the game tree:

❜ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅

D1 (2, 0)

r r r r

(4, 1) 1 2 A B C1 C2 (1, 3) (1, −1) D2

r r

2

❏❏❏❏❏ ✡ ✡ ✡ ✡ ✡ ✴ ✴

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 18 / 61

Example of the use of Backward Induction (cont’d)

In the following one:

❜ ✱ ✱ ✱ ✱ ✱ ✱

(2, 0) (1, 3)

❅ ❅ ❅ ❅ ❅ r r

1 A B

✴

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 19 / 61

Example of the use of Backward Induction (cont’d)

◮ The Nash equilibrium of this subgame is for player 1 to choose

A.

◮ This completes the description of the Subgame Perfect

equilibrium strategy for player 1: s∗

1 = (A, E1, E2). ◮ The unique Subgame Perfect equilibrium of this game is then:

s∗ = (s∗

1, s∗ 2) =

• (A, E1, E2), (C1, C2)
• .

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 20 / 61

Mutiple SPE

The Kuhn algorithm gives you the whole set of SPE (uniqueness not guaranteed). Recall the simple bargaining game:

❝❝❝❝❝❝❝❝❝❝❝ ❜ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ r r ✁ ✁ ✁ ✁ ✁ ✔ ✔ ✔ ✔ ✔ ❚ ❚ ❚ ❚ ❚ ✔ ✔ ✔ ✔ ✔ ❚ ❚ ❚ ❚ ❚ r ❙ ❙ ❙ ❙ ❙

1 2 2 2 (2, 0) (1, 1) (0, 2) y y y n n n (2, 0) (0, 0) (1, 1) (0, 0) (0, 2) (0, 0)

r r r r r r

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 21 / 61

Mutiple SPE

The Kuhn algorithm gives you the whole set of SPE (uniqueness not guaranteed). Recall the simple bargaining game:

❝❝❝❝❝❝❝❝❝❝❝ ❜ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ r r ✁ ✁ ✁ ✁ ✁ ✔ ✔ ✔ ✔ ✔ ❚ ❚ ❚ ❚ ❚ ✔ ✔ ✔ ✔ ✔ ❚ ❚ ❚ ❚ ❚ r ❙ ❙ ❙ ❙ ❙

1 2 2 2 (2, 0) (1, 1) (0, 2) y y y n n n (2, 0) (0, 0) (1, 1) (0, 0) (0, 2) (0, 0)

r r r r r r ❂ ✴ ✴ ✴

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 22 / 61

Mutiple SPE

The Kuhn algorithm gives you the whole set of SPE (uniqueness not guaranteed). Recall the simple bargaining game:

❝❝❝❝❝❝❝❝❝❝❝ ❜ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ r r ✁ ✁ ✁ ✁ ✁ ✔ ✔ ✔ ✔ ✔ ❚ ❚ ❚ ❚ ❚ ✔ ✔ ✔ ✔ ✔ ❚ ❚ ❚ ❚ ❚ r ❙ ❙ ❙ ❙ ❙

1 2 2 2 (2, 0) (1, 1) (0, 2) y y y n n n (2, 0) (0, 0) (1, 1) (0, 0) (0, 2) (0, 0)

r r r r r r ❄ ✴ ✇ ✴

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 23 / 61

Mutiple SPE: Example

◮ Using backward induction we obtain the following: ◮ the SPE with payoffs (1,1) such that:

◮ player 1 chooses (1, 1); ◮ player 2 chooses: n if (2, 0), y if (1, 1), y if (0, 2);

◮ the SPE with payoffs (2,0) such that:

◮ player 1 chooses (2, 0); ◮ player 2 chooses: y if (2, 0), y if (1, 1), y if (0, 2). Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 24 / 61

The Centipede Game

✱✱✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ❅ ❅ ❅ ❧❧❧ ❧❧❧ q q q q q q

2 2 1 C S C S C S (2, 4) (5, 3) (4, 6) (6, 5)

❅ ❅ ❅ ✱ ✱ ✱❧ ❧ ❧ ✱✱✱❅ ❅ ❅ q q q q q q

1 2 1 C C C S S S (1, 0) (0, 2) (3, 1)

❛

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 25 / 61

The Centipede Game: SPE

✱✱✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ❅ ❅ ❅ ❧❧❧ ❧❧❧ q q q q q q

2 2 1 C S C S C S (2, 4) (5, 3) (4, 6) (6, 5)

❅ ❅ ❅ ✱ ✱ ✱❧ ❧ ❧ ✱✱✱❅ ❅ ❅ q q q q q q

1 2 1 C C C S S S (1, 0) (0, 2) (3, 1)

❛ ✇ ✇ ✇ ✇ ✇ ✇

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 26 / 61

SPE of the Centipede Game: a Concern

◮ This game has a unique SPE in which each player chooses S

at every history.

◮ A concern with this game is whether the SPE prediction is

intuitively appealing.

◮ In particular, the issue is which prediction should a player

make after he has observed repeated and clear violations of rationality.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 27 / 61

Infinite Horizon Extensive Form Games

◮ The main problem with backward induction is that it only

applies to finite extensive form games.

◮ What about infinite horizon extensive form games? ◮ How do we verify that a strategy profile s∗ is a SPE ? ◮ According to the definition of SPE we need to check that for

every player i ∈ N and for every subgame Γ(h) there does not exist any strategy (deviation) that leads to an outcome that player i strictly prefers.

◮ Since strategies are fully contingent plans the set of possible

strategies is, in general, quite a big set.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 28 / 61

Infinite Horizon Extensive Form Games (cont’d)

◮ Therefore there are quite a lot of possible deviations to be

checked.

◮ We will provide a result that in a game with finite (at first)

horizon restricts the set of deviations that need to be considered.

◮ In particular, the following one deviation principle will allow us

to restrict attention to deviations that entail a different action choice in the next step of the considered strategy and from that point one proceeds according to the given strategy s∗ we are testing.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 29 / 61

One Deviation Principle.

◮ It allows us when testing whether a strategy profile s∗ is a

SPE to restrict attention to a deviation si for player i that differs from s∗

i in the immediate action it prescribes and

coincides with s∗

i in the remaining actions. ◮ This principle is very useful when constructing SPE of finite

and (more importantly) infinite horizon games.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 30 / 61

Example of One Deviation Principle

❜ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ❅ ❅ ❅ ❅ ❧❧❧❧❧ ❧❧❧❧❧ r r r r r r

1 1 2 A B C D E F (1, 3) (1, −1) (0, 2) (2, 0)

✴ ✴ ✴ ✴ ✴ ✇

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 31 / 61

Example of One Deviation Principle (cont’d)

◮ We test the strategy profile: [(A, E), C]. ◮ In general, player 1’s possible deviations are: (B, E), (B, F),

(A, F).

◮ According to one deviation principle we can restrict attention,

when looking at player 1’s decision at the initial node to the

• nly deviation (B, E).

◮ We can avoid considering the other deviations: (B, F) and

(A, F).

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 32 / 61

Example of One Deviation Principle (cont’d)

◮ Can we use the one deviation principle to solve extensive form

games with perfect information and infinite horizon?

◮ The answer is yes provided an extra condition is satisfied:

continuity at infinity: lim

t→∞

sup

{(h,˜ h)|ht=˜ ht}

• ui(σ|h) − ui(σ′|˜

h)

• = 0

◮ The issue is an infinite sequence of deviations, the condition

makes sure that these do not have too big of an impact on payoffs.

◮ Continuity at infinity is satisfied if each player’s payoff is the

discounted sum of per-period payoffs.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 33 / 61

Back to the Oddness Theorem:

◮ Recall the oddness theorem. ◮ Since every extensive form game is associated with a normal

form game representation the oddness theorem holds.

◮ However, the theorem could be meaningless once one refers

back to the original extensive form game.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 34 / 61

Simple Extensive Form Game

❜ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱

(0, 0)

❅ ❅ ❅ ❅ ❅ ❧❧❧❧❧❧ r r r r

1 2 D U R L (2, 2) (1, 1)

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 35 / 61

Simple Extensive Form Game (cont’d)

◮ The associated normal form is:

L R U 2, 2 2, 2 D 1, 1 0, 0

◮ There exist two Nash equilibria of this game that correspond

to the strategy profiles (U, L) and (U, R) both with payoffs (2, 2).

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 36 / 61

Simple Extensive Form Game (cont’d)

◮ Consider now a perturbation of the payoffs of the is game

such as: L R U 2 − ε, 2 − ε 2, 2 D 1, 1 0, 0

◮ For any arbitrary small ε > 0 the Nash equilibrium (U, R) is

unique.

◮ However the perturbation is clearly meaningless on the

extensive form game.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 37 / 61

Stackelberg Duopoly (Stackelberg 1934):

◮ Consider an environment in which two firms compete in the

same market producing quantities q1 ≥ 0 and q2 ≥ 0.

◮ The firms’ technologies are identical:

c(qi) = c qi ∀i ∈ {1, 2}.

◮ The consumer behavior is represented by the inverse demand:

P(q1 + q2) = a − (q1 + q2) if q1 + q2 ≤ a if q1 + q2 ≥ a where c < a.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 38 / 61

Stackelberg Duopoly (cont’d):

◮ Assume that there exists a dominant firm (the leader) that

competes in its choice of the quantity produced with a subordinate firm (the follower).

◮ The key difference with the Cournot model of duopoly is that

the leader moves first and chooses its quantity q1 ≥ 0.

◮ Then the follower observes the leader’s quantity choice and

choose its quantity q2 ≥ 0.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 39 / 61

Stackelberg Duopoly (cont’d)

◮ We solve the model using backward induction. ◮ Start from the last period in which for any given quantity q1

chosen by firm 1 firm 2 decides how much to produce.

◮ This choice of q2 is clearly the solution to the following

problem: max

q2∈R+ q2 [a − (q1 + q2) − c]

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 40 / 61

Stackelberg Duopoly (cont’d)

◮ The solution to this problem is represented by the first order

conditions which are both necessary and sufficient condition for a maximum. a − 2 q2 − q1 − c = 0

◮ Solving for q2 we obtain:

q2 = 1 2 (a − q1 − c) .

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 41 / 61

Stackelberg Duopoly (cont’d)

◮ A strategy for firm 2 is then a choice of quantity q2 for any

quantity q1 that firm 1 might choose.

◮ In other words a strategy for firm 2 is the following function

s∗

2(q1):

s∗

2(q1) =

1 2 (a − q1 − c) if q1 ≤ a − c if q1 ≥ a − c

◮ We associate this choice of quantity for firm 2 to every choice

• f quantity for firm 1 and move backward.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 42 / 61

Stackelberg Duopoly (cont’d)

◮ Firm 1’s choice of the quantity q1 is obtained by solving the

following problem: max

q1∈R+ q1 [a − (q1 + s∗ 2(q1)) − c] ◮ In other words:

max

q1∈R+ q1

• a −
• q1 + 1

2 (a − q1 − c)

• − c
• ◮ The necessary and sufficient condition for the solution to this

problem are: a − 2 q1 − c = 0.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 43 / 61

Stackelberg Duopoly (cont’d)

◮ Solving for q1 we obtain the optimal strategy s∗ 1 for player 1:

s∗

1 = (a − c)

2 .

◮ Therefore the unique SPE of the Stackelberg duopoly model is

represented by the strategies: s∗

1 = (a − c)

2 s∗

2(q1) =

(1/2) (a − q1 − c) if q1 ≤ a − c if q1 ≥ a − c

◮ The outcome associated with this SPE is:

• q∗

1 = (a − c)

2 , q∗

2 = (a − c)

4

• Leonardo Felli (LSE)

EC487 Advanced Microeconomics, Part II 17 November 2017 44 / 61

Cournot vs. Stackelberg

◮ Clearly it is important to compare the outcome above with

the outcome associated with Cournot game we solved before:

• qc

1 = (a − c)

3 , qc

2 = (a − c)

3

• ◮ In terms of profits:

Π1(q∗

1, q∗ 2) = (a − c)2

8 > Π1(qc

1, qc 2) = (a − c)2

9 . Π2(q∗

1, q∗ 2) = (a − c)2

16 < Π2(qc

1, qc 2) = (a − c)2

9 .

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 45 / 61

Cournot vs. Stackelberg (cont’d)

✻ ✲

q2 q1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

q ❍❍❍❍❍❍❍❍❍❍❍❍❍❍

s∗

2(q1)

q

a − c

(a−c) 2

a − c s∗

1 = (a−c) 2

q

(0, 0)

✻

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

✻

Cournot equilibrium

• (a−c)

3

, (a−c)

3

• Stackelberg equilibrium
• (a−c)

2

, (a−c)

4

• Leonardo Felli (LSE)

EC487 Advanced Microeconomics, Part II 17 November 2017 46 / 61

Leadership

Question: is it always a good idea to be the first to move? Consider the following simple prize competition game: the consumer demands only one unit of the homogeneous good produced by either firm. Let v be the utility that the consumer derives from this unit of a good. A consumer will buy the good from the firm that charges the lowest price pi if v ≥ pi. When the price is the same he will buy the good from firm 2.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 47 / 61

Leadership (cont’d)

Assume that firm i produces this unit of the good at a cost ci and c1 > c2. where v > c1. The payoffs are: Π1(p1, p2) = p1 − c1 if p1 < p2 if p1 ≥ p2 Π2(p1, p2) = p2 − c2 if p2 ≤ p1 if p2 > p1

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 48 / 61

Leadership (cont’d)

Assume the extensive form of the game is:

◮ Firm 1 choose p1 first, ◮ Firm 2 observes p1 and then chooses p2.

We solve the Subgame Perfect equilibria of the game using backward induction.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 49 / 61

Leadership (cont’d)

Start from the last move of the game. For any given p1 firm 2’s

• ptimal strategy is:

s∗

2(p1) =

   p2 > p1 if p1 < c2 p2 ≥ p1 if p1 = c2 p2 = p1 if p1 > c2 Clearly it is now optimal for firm 1 to choose any strategy s∗

1 = p1 ≥ c2

In such a case firm 1’s payoff is zero while it is negative for any

• ther choice of p1.

The outcome of the SPE is that the consumer will always buy from firm 2 at the price p∗ ≥ c2.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 50 / 61

Leadership (cont’d)

Notice first that there exist a whole continuum of subgame perfect equilibria in such a case. Moreover, being the first to move is certainly not an advantage in this case. The reason being that in this game prices are strategic complements, while in the Stakelberg game quantities are strategic substitutes.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 51 / 61

Bargaining

◮ Consider the following simple extensive form bargaining game. ◮ Two players, i ∈ {A, B} are trying to share a surplus. ◮ The size of the surplus is normalized to 1. ◮ Payoffs to the players in case of disagreement are normalized

to 0.

◮ Denote (δA, δB) the parties’ discount factors:

0 ≤ δi ≤ 1, ∀i ∈ {A, B};

◮ Let:

◮ x the share of the pie to party A; ◮ (1 − x) the share of the pie to party B. Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 52 / 61

Take-it-or-leave-it offer by A

Extensive form:

◮ A makes an offer x ∈ [0, 1] to B; ◮ B observes the offer x and decides whether to accept it (y) or

reject it (n).

◮ If the offer is accepted the game ends and the players payoffs

are: ΠA(x, y) = x ΠB(x, y) = (1 − x)

◮ If the offer is rejected the game ends and the players’ payoffs

are: ΠA(x, n) = 0 ΠB(x, n) = 0.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 53 / 61

Take-it-or-leave-it offer by A (cont’d)

◮ Nash equilibria:

any share of the surplus x ∈ [0, 1]

◮ Strategies:

◮ A offers share x ∈ [0, 1]; ◮ B accepts any share x′ such that

1 − x′ ≥ 1 − x and rejects any share such that 1 − x′ < 1 − x.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 54 / 61

Take-it-or-leave-it offer by A (cont’d)

◮ Subgame Perfect Equilibria Outcome:

Shares: x = 1 (1 − x) = 0

◮ SPE Strategies:

◮ A offers share x = 1; ◮ B accepts any share x′ ≤ 1. Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 55 / 61

Take-it-or-leave-it offer by A (cont’d)

Proof: backward induction:

◮ In the last stage of the game B gets 0 if he rejects A’s offer,

hence B accepts any offer x′ such that (1 − x′) ≥ 0;

◮ In the first stage of the game A makes the offer that gives A

the highest payoff. A’s payoff is x hence the unique equilibrium offer is x = 1.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 56 / 61

Two periods alternating offers bargaining

Extensive form:

◮ Period 1:

◮ A makes an offer xA ∈ [0, 1] to B; ◮ B observes the offer xA and decides whether to accept it (y) or

reject it (n).

◮ If the offer is accepted the game ends and the players payoffs

are: ΠA = xA ΠB = (1 − xA)

◮ If the offer is rejected the game moves to the next period Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 57 / 61

Two periods alternating offers bargaining (cont’d)

◮ Period 2:

◮ B makes an offer xB ∈ [0, 1] to A; ◮ A observes the offer xB and decides whether to accept it (y) or

reject it (n).

◮ If the offer is accepted the game ends and the players payoffs

are: ΠA(xB, y) = δA xB ΠB(xB, y) = δB (1 − xB)

◮ If the offer is rejected the game ends and the players’ payoffs

are: ΠA(xB, n) = 0 ΠB(xB, n) = 0.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 58 / 61

Two periods alternating offers bargaining (cont’d)

◮ Nash equilibria:

any share of the surplus x ∈ [0, 1]

◮ Strategies:

◮ A offers share xA = x ∈ [0, 1] in the first period; ◮ B accepts any share x′ ≤ x and rejects any share x′ > x in the

first period;

◮ B offer share xB = x in the second period; ◮ A accepts any offer xB ≥ x in the second period. ◮ A rejects any offer xB < x in the second period. Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 59 / 61

Two periods alternating offers bargaining (cont’d)

◮ Subgame Perfect Equilibrium Outcome:

Agreement is reached in the first period with payoffs: (1 − δB, δB)

◮ SPE Strategies:

◮ A offers share xA = 1 − δB in the first period; ◮ B accepts any share x′ ≤ 1 − δB in the first period; ◮ B rejects any share x′ > 1 − δB in the first period; ◮ B offers share xB = 0 in the second period; ◮ A accepts any share x′ ≥ 0 in the second period. Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 60 / 61

Two periods alternating offers bargaining (cont’d)

Proof: backward induction:

◮ In the last stage of the second period of the game B makes a

take-it-or-leave-it offer to A gets all the surplus that is worth to B exactly δB.

◮ In the first period of the game B will therefore accept any

• ffer 1 − xA ≥ δB or xA ≤ 1 − δB.

◮ Therefore A will make the offer that maximizes her payoff and

is accepted: xA = 1 − δB.

Leonardo Felli (LSE) EC487 Advanced Microeconomics, Part II 17 November 2017 61 / 61