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