Bayesian Games
\[ \def\BB#1{{\mathbb{#1}}} \def\BF#1{{\mathbf{#1}}} \]
yedlu, Winter 2024
Bayesian Nash Equilibrium
A Bayesian normal form game can be described as:
\[ G = (N, (S_i), (T_i), (u_i), \BB{P}). \]
- \(N = \{1, 2, ..., n\}\) be the set of players.
- \(S_i\) be the set of actions available to player \(i \in N\).
- \(T_i\) be the finite set of types to player \(i \in N\). This is private information to player \(i\).
- \(u_i: S \times T_i \to \BB{R}\) be \(i\)’s utility function representing the preference relation.
- \(\BB{P}\) be the belief, which is a probability measure over types \(\BF{t} = (t_1, t_2, ..., t_n)^T\).
Bayesian Belief System
Question: If I only learn my type (\(t_i\)), what is the probability that other players’ types are \(\BB{t}_{-i}\)?
Answer (Click to Expand)
By Bayes rule,
\[\begin{aligned} \BB{P}(t_{-i} \mid t_i) &= \frac{\BB{P}(t_i, t_{-i})}{\BB{P}(t_i)} \\ &= \frac{\BB{P}(t_i, t_{-i})}{\sum_{t^{'}_{-i} \in T_{-i}} \BB{P}(t_i \mid t^{'}_{-i})} \end{aligned}\]
Bayesian NE
Dfn (1) A strategy profile \(\sigma = (\sigma_1, ..., \sigma_n)\) is a Bayesian NE if
- for each player \(i\)
- and their each type \(t_i \in T_i\)
the strategy maximizes the expected utility:
\[ \sum_{t_{-i} \in T_{-i}} u_i(\sigma_i (t_i), \sigma_{-i} (t_{-i}) \times \BB{P}(t_{-i} \mid t_i)) \]
Dfn (2) A pure strategy for a player \(i\) is a function:
\[ \sigma_i: T_i \to S_i \]
Dfn (3) A pooling strategy means that for each and every player, her strategy is fixed regardless of types. In other words, \(\sigma_i(t_i) = \sigma_i\) for all \(t_i \in T_i\).
Example: Cutoff Strategy
Consider the following normal-form game.
| b | s | |
|---|---|---|
| B | \(2 + t_1, 1\) | \(0, 0\) |
| S | \(0, 0\) | \(1, 2 + t_2\) |
Suppose \(t_1\) and \(t_2\) are i.i.d. with \(t_i \sim U[0,x]\) for some arbitrarily small \(x>0\).
Answer (Click to Expand)
Consider a cutoff strategy.
\[ \sigma_1(t_1) = \begin{cases} B &, t_1 \geq C_1 \\ S &, t_1 < C_1 \end{cases} \]
\[ \sigma_2(t_2) = \begin{cases} s &, t_2 \geq C_2 \\ b &, t_2 < C_2 \end{cases} \]
Player 2 best-responds to player 1’s expected strategy: \[\begin{aligned} U_2 (b, \sigma_1, t_2) &= 1 \times \BB{P} \{\sigma_1(t_1) = B\} + 0 \times \BB{P} \{\sigma_1(t_1) = S\} \\ &= 1 \times \BB{P} \{t_1 \geq C_1\} + 0 \times \BB{P} \{t_1 < C_1\} \\ &= \frac{x - cC_1}{x} \\ \\ U_2 (s, \sigma_1, t_2) &= 0 \times \BB{P} \{\sigma_1(t_1) = B\} + (2 + t_2) \times \BB{P} \{\sigma_1(t_1) = S\} \\ &= 0 \times \BB{P} \{t_1 \geq C_1\} + (2 + t_2) \times \BB{P} \{t_1 < C_1\} \\ &= \frac{C_1}{x} \times (2 + t_2) \end{aligned}\]
Suppose \(\sigma_2 (t_2) = s\), then: \[\begin{aligned} U_2 (b, \sigma_1, t_2) & \leq U_2 (s, \sigma_1, t_2) \\ \\ \frac{x - C_1}{x} & \leq \frac{c_1}{x} \times (2 + t_2) \\ t_2 & \geq \frac{x}{C_1} - 3 \\ \\ \implies C_2 &= \frac{x}{C_1} - 3 \end{aligned}\]
Similarly for player 1, her cutoff point is: \[\begin{aligned} C_1 &= \frac{x}{C_2} - 3 \end{aligned}\]
Solving the system \[\begin{cases} C_1 &= \frac{x}{C_2} - 3 \\ C_2 &= \frac{x}{C_1} - 3 \end{cases}\]
yields \[\begin{aligned} C_1 = C_2 = \frac{\sqrt{9 + 4x} - 3}{2} \end{aligned}\]
If the maximum payoff shock shrinks to 0, \[\begin{aligned} \lim_{x \to 0^+} \BB{P} \{\sigma_1 (t_1) = B\} &= \lim_{x \to 0^+} \frac{x - C_1}{x} \\ &= \lim_{x \to 0} 1 + \frac{3 - \sqrt{9 + 4x}}{2x} \\ &= \frac{2}{3} \quad \text{(L'Hôpital Rule)} \end{aligned}\]