SMU H3 Notes Game TheoryEconomicsSMU H3Session

H3 Game Theory Session 4

SMU H3 Game Theory Session 4 Notes

Contents

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

let x be the quantity of Coke that buyers are willing to buy given p and qx=442p+q\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 let y be the quantity of Pepsi that buyers are willing to buy given p and qy=442q+p\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 MC (marginal cost) is 8 for each firm\text{MC (marginal cost) is } 8 \text{ for each firm}

Payoffs

payoffprofits\text{payoff} \equiv \text{profits} π=total revenue (TR)total cost (TC)\pi = \text{total revenue (TR)} - \text{total cost (TC)} TRcoke=px,TCcoke=8xπcoke=px8x=(p8)(442p+q)\text{TR}_\text{coke} = p \cdot x, \quad \text{TC}_\text{coke} = 8x \\ \pi_\text{coke} = p x - 8x = (p - 8) (44 - 2p + q) TRpepsi=qy,TCpepsi=8yπcoke=qx8x=(p8)(442q+p)\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

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

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

πcoke=(p8)(442p+20)=(p8)(642p)πcoke=2(p8)(p32)\pi_\text{coke} = (p - 8) (44 - 2p + 20) = (p - 8)(64 - 2p) \\ \pi_\text{coke} = -2(p - 8)(p - 32) dπcokedp=4p+804p=80p=20\frac{d \pi_\text{coke}}{dp} = -4p + 80 \\ 4p = 80 \\ p = 20 \\ pmax=20πcoke=288\therefore p_\text{max} = 20 \\ \therefore \pi_\text{coke} = 288

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

πcoke=(p8)(442p+40)=(p8)(842p)πcoke=2(p8)(p42)\pi_\text{coke} = (p - 8) (44 - 2p + 40) = (p - 8)(84 - 2p) \\ \pi_\text{coke} = -2(p - 8)(p - 42) dπcokedp=4p+1004p=100p=25\frac{d \pi_\text{coke}}{dp} = -4p + 100 \\ 4p = 100 \\ p = 25 \\ pmax=25πcoke=34\therefore p_\text{max} = 25 \\ \therefore \pi_\text{coke} = 34

Conclusion

First-order condition for Coke

FOC:πcoke=[(442p+q)2(p8)]\text{FOC:} \quad \pi_\text{coke}' = [(44 - 2p + q) - 2(p - 8)] \\ p=14(q+60)p = \frac{1}{4}(q + 60)

First-order condition for Pepsi

FOC:πpepsi=[(442q+p)2(q8)]\text{FOC:} \quad \pi_\text{pepsi}' = [(44 - 2q + p) - 2(q - 8)] \\ q=14(p+60)q = \frac{1}{4}(p + 60)

Diagram representation

p=14(q+60),q=14(p+60)p = \frac{1}{4}(q + 60), \quad q = \frac{1}{4}(p + 60) p=20,q=20p = 20, \quad q = 20

Collusive Game

Comments

Optimisation Goal

p,q=arg maxp,q[(p8)(442p+q)+(q8)(442q+p)]p, q = \argmax_{p, q} {[(p-8)(44-2p+q) + (q-8)(44-2q+p)]}
First-order condition w.r.t. ppFirst-order condition w.r.t. qq
πp=0\frac{\partial \pi}{\partial p} = 0πq=0\frac{\partial \pi}{\partial q} = 0
604p+q+(q8)=060 - 4p + q + (q - 8) = 0(p8)+604q+p=0(p - 8) + 60 - 4 q + p = 0

Outcome

pmax=26>20,qmax=26>20p_\text{max} = 26 > 20, \quad q_\text{max} = 26 > 20 π=(268)(44262+26)=182=324>288\pi = (26 - 8)(44 - 26 \cdot 2 + 26) = 18^2 = 324 > 288 if q=26,p=21.5\text{if } q = 26, p = 21.5

Positional Externalities

Definition

Setup

π1={10xx+yxif x0 or y05if x=y=0\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} π2={10yx+yxif x0 or y05if x=y=0\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

π1=5if x=y=0\pi_1 = 5 \quad \text{if } x = y = 0 π1=10(0.10.1)0.1=9.9>5if x=0.1, 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

FOC for P1

π1=10y(x+y)21=010y=(x+y)2\pi_1' = \frac{10 y}{(x + y)^2} - 1 = 0 \\ 10 y = (x + y)^2

FOC for P2

π2=10x(x+y)21=010x=(x+y)2\pi_2' = \frac{10 x}{(x + y)^2} - 1 = 0 \\ 10 x = (x + y)^2

Equality of FOC

x=yx = y 10x=(2x)2=4x2x=y=2.510x = (2x)^2 = 4x^2 \\ x = y = 2.5

Collusion

πt=π1+π2\pi_t = \pi_1 + \pi_2 πt=10xx+yx+10yx+yy=10xy\pi_t = \frac{10x}{x+y} - x + \frac{10y}{x+y}-y = 10 - x - y πt{10xyif x0, y010if x=y=0\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 \ P1RedBlack
Red2, 20, 0
Black0, 05, 5

Stag Hunt Game

Payoff Matrix

P2 \ P1StagHare
Stag5, 50, 3
Hare3, 04, 4

Chicken Game

Payoff Matrix

P2 \ P1SwerveStraight
Swerve1, 10, 2
Straight2, 0-1, -1

Evolutionary Biology

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