Game Theory

Question 1

What is a game?

Answer 1

A game in strategic form is given by
- the set of players
- for each player, a set of possible strategies
- for each player, at every outcome, a payoff

Each player picks a strategy simultaneously; the strategy profile (=one strategy from each player) determines the outcome; payoffs represent the agents’ preference orderings over the outcomes

Each agent is fully aware of the game (players, strategies, payoffs), acts so as to maximize own payoff knowing others do the same

Question 2

What is a strictly dominant strategy?

Answer 2

A strictly dominant strategy is one that yields a greater payoff to the player than any other strategy no matter what the other player(s) do.

In games where each agent has a dominant strategy our prediction is that they will play accordingly, each using his/her dominant strategy.

Question 3

What is Iterated Elimination of Strictly Dominated Strategies?

Answer 3

• eliminate strictly dominated strategies (if any) for each player
• this may “unlock” further rounds of elimination
• repeat until no strategy is strictly dominated in reduced game

Question 4

When is a strategy profile Nash Equilibrium?

Answer 4

Strategy profile (a strategy from each player) is Nash equilibrium if each player best-responds to the other players’ equilibrium strategies

Self-fulfilling prophecy: if all players anticipate that a specific Nash equilibrium will be played then no-one has an incentive to deviate. Also known as rational expectations equilibrium.

Question 5

What is Nash’s theorem (1950)

Answer 5

Nash’s Theorem (1950): Every game with finitely many strategies has at least one Nash equilibrium in pure or mixed strategies

Question 6

What about weak dominance?

Answer 6

• playing weakly dominant strategies is a good prediction
• iterated elimination of weakly dominated strategies is less commonly used because it may delete Nash equilibria…
• …moreover, the order of deletion may affect prediction
  (whereas order does not matter in iterated strict dominance)

Question 7

Recipe for finding all Nash Equilibria in 2x2 games?

Answer 7

• if solvable by IESDS/ISD then unique Nash, done
• otherwise find each player’s best response as a function of the other’s mixing probability, and see where best replies intersect

Question 8

How to find mixed Nash equilibria?

Answer 8

• compute and graph row’s utility from T and B as a function of r (where r is the probability of column playing R)
• Then find Row’s best reply t*(r), the optimal probability weight on T given Column’s mixing probability r
• DO the same for Column based on t, Row’s probability of playing T
• Flip the axes in the graph showing Column’s best reply so that t is measured vertically and r ∗(t) horizontally, and then superimpose the graphs of the best-reply functions in the (r , t) space
• Nash Equilibria will be where lines cross

Question 9

What is extensive form given by?

Answer 9

The extensive form is given by
- a directed tree graph (nodes, directed edges, no cycles)
- a player assigned to each node, actions corresponding to edges
- payoffs written at terminal nodes
- “who knows what” indicated via information sets

Question 10

What is backward induction and when can it be used?

Answer 10

• replace each penultimate node with the decision that is optimal for the player making the decision at that node
• repeat this in the reduced game as long as possible

Backward induction can be used in every sequential-moves game with perfect information (=players observing all earlier moves)

Question 11

What is a sub-game perfect equilibrium?

Answer 11

The resulting game plan from backwards induction, which specifies a move at every node in the game tree, is called a subgame-perfect equilibrium or SPE.

Every SPE is Nash equilibrium (hence also called “SPNE”), moreover, SPE induces Nash equilibrium in every continuation (subgame) of the perfect-information, sequential-moves game

Question 12

What type of Nash equilibria does SPE rule out

Answer 12

SPE rules out Nash equilibria sustained by non-credible threats

Question 13

What does it mean for an action to be sequentially rational?

Answer 13

It is credible and time consistent

Question 14

What happens in Cournot competition as opposed to Stackelberg competition?

Answer 14

In Cournot competition two firms simultaneously decide what quantity to produce. Price is a function of total supply.

In Stackelberg one firm picks q before the other so it is a sequential game. Firm 2 sets q2 given firm 1’s quantity, and firm 1 anticipates this and substitutes q2*(q1) into its own objective. The first mover is strictly better off in quantity-setting duopoly

Question 15

What happens in a finitely repeated prisoner’s dilemma game?

Answer 15

The unique SPE is to play D in every period.

In the final round each player has a strictly dominant strategy, D - no point playing C as it has no effect on future play (game ends)
Hence in the final period each player plays D without conditioning on the history of play up to that point
Replace the final period with (D,D) and payoffs (0,0); step back one period: now this is the final period where D is again strictly dominant

Question 16

What happens when the Prisoner’s Dilemma is played infinitely many times by
players who maximize their discounted present-value payoffs?

Answer 16

Theorem: In the infinitely-repeated Prisoner’s Dilemma the outcome “(C,C) forever” can be sustained in SPE for δ sufficiently close to 1 (where δ is the discount rate)

proof:
- Consider the strategy “play C in the first period then C as long as the outcome has been (C,C), otherwise switch to D forever”
- If anyone deviates from C then mutual “D forever” is of course Nash equilibrium in the continuation (subgame) for any δ
- Check deviation from C: playing D pays 2 once then 0 forever, whereas staying on-path pays 1 forever; the latter is better iff
2 < 1 + δ + δ2 + . . . = 1/(1−δ) ⇐⇒ δ > 1/2 , i.e., true for δ close to 1

Playing “D forever” is a credible threat under infinite repetition, and it helps sustain (C,C) forever if the long-run gain from cooperation exceeds the short-run gain from defection, here 1 + δ + δ2 + . . . > 2

The strategy “play C as long as C is played, otherwise switch to D forever” is informally called “grim trigger” or “trigger” strategy or “Nash reversion” (as play reverts to one-shot Nash upon deviation)

The result on the previous slide is referred to as a “folk theorem” because various versions have been around without attribution

Question 17

What happens in a one-shot Bertrand game?

Answer 17

-Bertrand game: N firms, a homogeneous product, price competition
-whichever firm charges the lowest price takes the entire market, if all set the same p they split the market & profit (if any) equally
- Assume a constant unit cost, c (could even normalize c = 0 )
- Denote the monopoly price be pM , the monopoly profit by ΠM > 0

Equilibrium in the one-shot Bertrand game
- charging price p > c cannot be Nash, each firm has a strict incentive to undercut its rivals and “steal” the whole market
- hence unique Nash in one-shot game is p∗ = c

Question 18

What happens in an infinitely repeated Bertrand model?

Answer 18

Bertrand competition is a Prisoner’s Dilemma for the firms:
- setting p = pM is like action C (cooperative but dominated)
- undercutting pM is like action D: rational but leads to a Pareto-inferior outcome for the firms (ultimately p = c, zero profits)

Firms can sustain cooperation (setting pM and sharing ΠM equally) in SPE in the long run by threatening to revert to marginal-cost pricing (and here: no profits) forever in case anyone deviates

Formally, in infinitely-repeated Bertrand, p = pM can be sustained for δ close to 1 via “grim trigger” (Nash reversion) strategies

The condition to satisfy is (ΠM /N) 1/(1−δ) ≥ ΠM ⇐⇒ δ ≥ (N−1)/N

More firms, or more impatient firms, find it harder to collude