4. Bayesian Games

4.2 Bayesian Games: Bayesian Nash Equilibrium (Chapter 26)

### Static Bayesian Games

Nash eqbm is readily applied to Bayesian games and well-suited to static games

The Nash eqbm of Bayesian games is called *Bayesian Nash equilibrium* (BNE)

We study two methods of computing BNE of static games

(1) Compute Nash eqbm of the **Bayesian normal form**

1 \ 2 | C | D |

\(A^{12}A^{0}\) | 8, 9 | 3, 6 |

\(A^{12}B^{0}\) | 10, 6 | 4, 7 |

\(B^{12}A^{0}\) | 4, 3 | 5, 8 |

\(B^{12}B^{0}\) | 6, 0 | 6, 9 |

The normal-form game is then solvable by ISD: there is a unique strategy profile that survives ISD,

\((B^{12}B^{0}, D)\).

It is the unique BNE of Example 4.4

EXAMPLE 4.5 (EXERCISE 7 ON PAGE 357).

Nature selects the type (c) of player 1, where c = 2 with probability 2/3 and c = 0 with probability 1/3

Player 1 observes c, whereas player 2 does not observe c

The players make a simultaneous and independent choice and receive a payoff as follows:

1 \ 2 | X | Y |

A | 0, 1 | 1, 0 |

B | 1, 0 | c, 1 |

(a) Draw the Bayesian normal form of this game

\(S_{1} = {A^{2}A^{0}, A^{2}B^{0}, B^{2}A^{0}, B^{2}B^{0}}\)

\(S_{2} = {X, Y}\)

X | Y | |

\(A^{2}A^{0}\) | 0, 1 | 1, 0 |

\(A^{2}B^{0}\) | 1/3, 2/3 | 2/3, 1/3 |

\(B^{2}A^{0}\) | 2/3, 1/3 | 5/3, 2/3 |

\(B^{2}B^{0}\) | 1, 0 | 4/3, 1 |

(b) Compute the Bayesian Nash equilibrium

strategy \(A^{2}A^{0}\), \(A^{2}B^{0}\) are dominated by \(B^{2}B^{0}\) (relative to player 1)

strategy X is dominated by Y (relative to player 2

strategy \(B^{2}B^{0}\) is dominated by \(B^{2}A^{0}\) (relative to player 1)

Therefore, the prediction by NE is (\(B^{2}A^{0}\), Y).

(2) The second method is to treat the types of each player as separate players; more suitable for *infinite* games

EXAMPLE 4.6 (COURNOT DUOPOLY WITH UNKNOWN COSTS).

Firm 1’s production cost \(c_{1}\) = 0 is commonly known, while firm 2’s

\(c_{2}\) = \(\begin{cases} & \text{ 0 with prob 1/2 (type L) } \\ & \text{ 4 with prob 1/2 (type H) } \end{cases}\)

Market demand function is P = 10 – \(q_{1} – q_{2}\)

Denote by \(q_{2}^{H}\) and \(q_{2}^{L}\) the quantities produced by firm 2

Writing each type’s payoff function, we can find firm 2’s type-dependent best response from the FOC

2’s profit function where \(c_{2}\) = 0

\((10-q_{1}-q_{2}^{L}) \cdot q_{2}^{L}\)

∴ \(q_{2}^{L} = \frac{10-q_{1}}{2}\)

2’s profit function where \(c_{2}\) = 4

\((10-q_{1}-q_{2}^{H}) \cdot q_{2}^{H} – 4q_{2}^{H}\) ⇔ (differentiate over \(q_{2}^{H}\)) \(6 – q_{1} – 2q_{2}^{H}\) = 0

∴ \(q_{2}^{H} = \frac{6-q_{1}}{2}\)

\(c_{2}\) = 0 → \(BR_{2}^{L}(q_{1}) = \frac{10-q_{1}}{2}\) and \(c_{2}\) = 4 → \(BR_{2}^{H}(q_{1}) = \frac{6-q_{1}}{2}\)

The expected profit accuring to firm 1 is

\(\frac{1}{2}(10-q_{1}-q_{2}^{2})q_{1} + \frac{1}{2}(10-q_{1}-q_{2}^{H})q_{1}\)

(differentiate over \(q_{1}\)) \(\frac{1}{2}(10-2q_{1}-q_{2}^{L}) + \frac{1}{2}(10-2q_{1}-q_{2}^{H}) = 0\)

The FOC with respect to \(q_{1}\) gives us

\(BR_{1}(q_{2}) = 5 – \frac{1}{4}(q_{2}^{L} + q_{2}^{H})\)

\(q_{1} = 5 – \frac{1}{2}\bar{q_{2}}\) (∵ \(\bar{q_{2}} = \frac{1}{2}q_{2}^{L} + \frac{1}{2}q_{2}^{H}\))

\(\bar{q_{2}} = 10 – 2q_{1}\)

```
f(q) = (10-q)/2, f(q) = (6-q)/2, f(q) = 10-2q, q=(0,10)
x-axis: q1, y-axis: q2
```

10-2q : Firm 1’s BR

5-q/2: Firm 2’s BR when c2 = 0

3-q/2: Firm 2’s BR when c2 = 4

→ 4-q/2: Firm 2’s BR perceived by firm 1

\(q_{1} = 5 – \frac{1}{4}(\frac{10-q_{1}}{2} + \frac{6-q_{1}}{2})\)

Bayesian Nash eqbm

\(q_{1}^{*} = 4, q_{2}^{L*} = 3, q_{2}^{H*} = 1\)

Thus, (4, 1), (4, 3)

- Reference: Chang-Koo Chi, (29/50) Game Theory and Applications 8 – Bayesian Nash equilibrium, Jul 10, 2020, https://youtu.be/HNF01aurERg