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