SMU H3 Map

Content map: SMU H3 Game Theory Map

Choice Games with Continuous Strategies

Comments

Strategies are subsets of real numbers
Make use of calculus whenever possible

Price Competition with Imperfect Substitutes (Duopoly)

Coke vs. Pepsi: Price competition or quantity competition
Electoral competition: Campaign spending to win votes
Mediacorp Subaru Impreza Challenge: Last person to take hand off car wins it

Price Competition Game

Consider the duopoly pricing game between Coke and Pepsi

Notes

Price rigid market structure: Demand does not change dramatically with a change in price of substitute

Setup

\text{let } x \text{ be the quantity of Coke that buyers are willing to buy given } p \text{ and } q \\ x = 44 - 2 p + q

\text{let } y \text{ be the quantity of Pepsi that buyers are willing to buy given } p \text{ and } q \\ y = 44 - 2 q + p

\text{MC (marginal cost) is } 8 \text{ for each firm}

Payoffs

\text{payoff} \equiv \text{profits}

\pi = \text{total revenue (TR)} - \text{total cost (TC)}

\text{TR}_\text{coke} = p \cdot x, \quad \text{TC}_\text{coke} = 8x \\ \pi_\text{coke} = p x - 8x = (p - 8) (44 - 2p + q)

\text{TR}_\text{pepsi} = q \cdot y, \quad \text{TC}_\text{pepsi} = 8y \\ \pi_\text{coke} = q x - 8x = (p - 8) (44 - 2q + p)

Best Response Analysis

Derive each firm’s best response
Find $(p, q)$ that satisfy both best responses

Suppose $q = 20$ , what is the value of $p$ that maximises $\pi_\text{coke}$

\pi_\text{coke} = (p - 8) (44 - 2p + 20) = (p - 8)(64 - 2p) \\ \pi_\text{coke} = -2(p - 8)(p - 32)

\frac{d \pi_\text{coke}}{dp} = -4p + 80 \\ 4p = 80 \\ p = 20 \\

\therefore p_\text{max} = 20 \\ \therefore \pi_\text{coke} = 288

Suppose $q = 40$ , what is the value of $p$ that maximises $\pi_\text{coke}$

\pi_\text{coke} = (p - 8) (44 - 2p + 40) = (p - 8)(84 - 2p) \\ \pi_\text{coke} = -2(p - 8)(p - 42)

\frac{d \pi_\text{coke}}{dp} = -4p + 100 \\ 4p = 100 \\ p = 25 \\

\therefore p_\text{max} = 25 \\ \therefore \pi_\text{coke} = 34

Conclusion

You can be repeated over and over, OR
You can simply compute the best $p$ for an unspecified $q$ and then express the best $p$ as a “function” of $q$

First-order condition for Coke

\text{FOC:} \quad \pi_\text{coke}' = [(44 - 2p + q) - 2(p - 8)] \\

p = \frac{1}{4}(q + 60)

First-order condition for Pepsi

\text{FOC:} \quad \pi_\text{pepsi}' = [(44 - 2q + p) - 2(q - 8)] \\

q = \frac{1}{4}(p + 60)

Diagram representation

Nash equilibrium is obtained when when $p = q$ and it meets both FOCs

p = \frac{1}{4}(q + 60), \quad q = \frac{1}{4}(p + 60)

p = 20, \quad q = 20

Nash equilibrium is (20, 20)
Associated profits are (288, 288)
Could the firms do better? (Collusion)

Collusive Game

Comments

Fixing the prices to make bigger profits
One model of collusion is: both Coke and Pepsi act as if they are under the same ownership
Monopoly chooses both $p$ and $q$ to maximise $\pi$ (profits)

Optimisation Goal

p, q = \argmax_{p, q} {[(p-8)(44-2p+q) + (q-8)(44-2q+p)]}

First-order condition w.r.t. $p$	First-order condition w.r.t. $q$
$\frac{\partial \pi}{\partial p} = 0$	$\frac{\partial \pi}{\partial q} = 0$
$60 - 4p + q + (q - 8) = 0$	$(p - 8) + 60 - 4 q + p = 0$

Outcome

p_\text{max} = 26 > 20, \quad q_\text{max} = 26 > 20

\pi = (26 - 8)(44 - 26 \cdot 2 + 26) = 18^2 = 324 > 288

Clearly, firms prefer to collude to maximise profits
However, this is not a Nash equilibrium

\text{if } q = 26, p = 21.5

Even if firms agree to collude, they can gain from deviating if the other firm remains at $q = 26$

Positional Externalities

Definition

People feel worse off when others obtain better or more of the same scarce goods
This creates an “arms race” where everyone spends more just to keep up in relative standing
The result is inefficient overconsumption and lower social welfare (e.g., bigger houses or expensive strollers)

Setup

2 candidates are running for office in a presidential election
Suppose P1 spends $x$ in campaigning, P2 spends $y$ in campaigning
This is a simultaneous choice game
P1’s and P2’s payoffs are:

\pi_1 = \begin{cases} \frac{10x}{x+y} - x & \text{if } x \ne 0 \text{ or } y \ne 0\\ 5 & \text{if } x = y = 0 \end{cases}

\pi_2 = \begin{cases} \frac{10y}{x+y} - x & \text{if } x \ne 0 \text{ or } y \ne 0\\ 5 \quad \text{if } x = y = 0 \end{cases}

Best Response Analysis

Verify that (0, 0) is not a Nash equilibrium

Does not require calculus
Just show that one player needs to deviate to gain

\pi_1 = 5 \quad \text{if } x = y = 0

\pi_1 = 10 \cdot (\frac{0.1}{0.1}) - 0.1 = 9.9 > 5 \quad \text{if } x = 0.1, \space y = 0

Perform best response analysis

Assuming $x \ne 0, \space y \ne 0$

FOC for P1

\pi_1' = \frac{10 y}{(x + y)^2} - 1 = 0 \\ 10 y = (x + y)^2

FOC for P2

\pi_2' = \frac{10 x}{(x + y)^2} - 1 = 0 \\ 10 x = (x + y)^2

Equality of FOC

By comparing $FOC_x$ and $FOC_y$ , we can conclude that

x = y

10x = (2x)^2 = 4x^2 \\ x = y = 2.5

Thus, (2.5, 2.5) is a Nash equilibrium

Collusion

\pi_t = \pi_1 + \pi_2

\pi_t = \frac{10x}{x+y} - x + \frac{10y}{x+y}-y = 10 - x - y

\pi_t \begin{cases} 10 - x - y \quad \text{if } x \ne 0, \space y \ne 0 \\ 10 \quad \quad \text{if } x = y = 0 \end{cases}

Assurance Game

Payoff Matrix

P2 \ P1	Red	Black
Red	2, 2	0, 0
Black	0, 0	5, 5

Nash Equilibria occur at $(Red, Red)$ and $(Black, Black)$
In this case, choosing $(Black, Black)$ might seem more logical since the payoff is higher than $(Red, Red)$
$(Black, Black)$ Pareto dominates $(Red, Red)$ , meaning that both players will be better off by doing so
Pareto Efficiency: We say that $(a, b)$ is Pareto efficient if it is to not possible to find another outcome in the game where at least 1 player is better off and no other player is worse off

Stag Hunt Game

Payoff Matrix

P2 \ P1	Stag	Hare
Stag	5, 5	0, 3
Hare	3, 0	4, 4

Nash equilibria at $(Stag, Stag)$ and $(Hare, Hare)$
$(Stag, Stag)$ gives the Pareto efficient outcome
$(Hare, Hare)$ is the risk-dominant Nash equilibrium
Assuming that the other player will choose $Stag$ $S t a g$ or $Hare$ $H a r e$ with the same frequency:
- Choosing $Stag$ : Expected payoff is $\frac{5 + 0}{2} = 2.5$
- Choosing $Hare$ : Expected payoff is $\frac{3 + 4}{2} = 3.5$

Chicken Game

Payoff Matrix

P2 \ P1	Swerve	Straight
Swerve	1, 1	0, 2
Straight	2, 0	-1, -1

Nash equilibria at $(Straight, Swerve)$ and $(Straight, Swerve)$
Both combinations of strategies are Pareto efficient

Evolutionary Biology

Payoffs indicate expected off-springs
Animals from same species (e.g. birds) meet at random
Birds are genetically programmed to choose one strategy
The strategy that gets more off-springs survives evolutionary pressures
This represents a form of a mixed strategy

H3 Game Theory Session 4

SMU H3 Map

Choice Games with Continuous Strategies

Comments

Price Competition with Imperfect Substitutes (Duopoly)

Price Competition Game

Consider the duopoly pricing game between Coke and Pepsi

Notes

Setup

Payoffs

Best Response Analysis

Suppose q=20q = 20q=20, what is the value of ppp that maximises πcoke\pi_\text{coke}πcoke​

Suppose q=40q = 40q=40, what is the value of ppp that maximises πcoke\pi_\text{coke}πcoke​

Conclusion

First-order condition for Coke

First-order condition for Pepsi

Diagram representation

Collusive Game

Comments

Optimisation Goal

Outcome

Positional Externalities

Definition

Setup

Best Response Analysis

Verify that (0, 0) is not a Nash equilibrium

Perform best response analysis

FOC for P1

FOC for P2

Equality of FOC

Collusion

Assurance Game

Payoff Matrix

Stag Hunt Game

Payoff Matrix

Chicken Game

Payoff Matrix

Evolutionary Biology

Suppose $q = 20$ , what is the value of $p$ that maximises $\pi_\text{coke}$

Suppose $q = 40$ , what is the value of $p$ that maximises $\pi_\text{coke}$