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M20.L05 — Sequential Games: Backward Induction and Stackelberg

Module: Module 20 — Monopoly and Game Theory (Advanced) Lesson: L05 of 06 Duration: ~30 minutes Level: Intermediate Provenance: Intermediate Microeconomics with Excel (Barreto) | MIT OCW 14.01 Principles of Microeconomics

Learning Objective

Solve sequential games using backward induction to find subgame perfect equilibria.

Stackelberg Model

Leader (L) moves first, then follower (F). Solve backward:

  1. F's problem: max π_F(y_F; y_L) → reaction function y_F(y_L)
  2. L anticipates F's reaction: max π_L(y_L, y_F(y_L))

Australian example: BHP (leader) and Fortescue (follower) in iron ore market:

P = 100 - Q, MC_L = MC_F = 20

F's reaction: y_F = (100 - y_L - 20)/2 = 40 - 0.5y_L L's profit: (100 - y_L - (40 - 0.5y_L) - 20)y_L → y_L* = 40 → y_F* = 20

Worked Example

[Australian context]

An Australian pharmaceutical firm (leader) and generic competitor (follower) face P = 200 - Q with MC=50. Find Stackelberg equilibrium.

Solution: 1. Follower: max (200 - y_L - y_F - 50)y_F → y_F = (150 - y_L)/2 2. Leader: max (200 - y_L - (150 - y_L)/2 - 50)y_L → y_L* = 75 → y_F* = 37.5 P* = 200 - 112.5 = $87.5

Common Misconception

"First movers always have advantage"

In some games (e.g., centipede game), first movers may be worse off due to credible threats.

Key Takeaways

  • SPNE requires optimal play in all subgames
  • Backward induction eliminates non-credible threats
  • Stackelberg leaders produce more than Cournot
  • Commitment is key to first-mover advantage

Practice

  1. Solve Stackelberg model with P=120-Q, MC_L=30, MC_F=20
  2. Draw game tree for sequential entry game
  3. Explain why Cournot differs from Stackelberg

Further Resources

Provenance: Barreto | MIT OCW