Game Theory 8 – chap14. Bayesian Nash equilibrium example

EXAMPLE 4.7 (PUBLIC GOOD PROVISION, EXERCISE 2 ON PAGE 355).

• Two players simultaneously and independently decide how much to contribute to a public good
• If player 1 contributes \(x_{1}\) and player 2 contributes \(x_{2}\), then the value of the public good is

\(2(x_{1} + x_{2} + x_{1}x_{2})\)

• Player 1 must pay a cost \(x_{1}^{2}\) of contributing \(x_{1}\). Player 2 must pay a cost \(tx_{2}^{2}\) of contributing \(x_{2}\), where the realization of t ∈ {2, 3} is player 2’s private information. Player 1 only knows that each type is equally likely.

Compute the Bayesian Nash equilibrium of this game

	excludability	less excludable
rivalry	private goods	common resource
less rival	club goods	public goods

when t = 2,
\(2(x_{1} + x_{2} + x_{1}x_{2}) – 2x_{2}^{2}\) ⇒ (differentiate over \(x_{2}\)) \(2 + 2x_{1} – 4x_{2}\) = 0
∴ \(x_{2}^{L} = \frac{1+x_{1}}{2}\)

when t = 3,
\(x_{3}^{H} = \frac{1+x_{1}}{3}\)

player 2,
\(\frac{1}{2}(2(x_{1} + x_{2}^{L} + x_{1}x_{2}^{L})) + \frac{1}{L}(2(x_{1} + x_{2}^{H} + x_{1}x_{2}^{H})) – x_{1}^{2}\)
⇒ (differentiate over \(x_{1}\)) \(1 + x_{2}^{L} + 1 + x_{2}^{H} – 2x_{1}\) = 0
\(x_{1} = 1 + \frac{1}{2}(x_{2}^{L} + x_{2}^{H}\)

f(x) = (1+x)/2, f(x) = (1+x)/3
x-axis: x1, y-axis: x2

• Reference: Chang-Koo Chi, (30/50) Game Theory and Applications 8 – Bayesian Nash equilibrium example, Jul 10, 2020, https://youtu.be/l3HqFpbyifw