{"id":1194,"date":"2023-06-03T04:10:46","date_gmt":"2023-06-02T19:10:46","guid":{"rendered":"https:\/\/saraheee.com\/?p=1194"},"modified":"2023-07-02T22:53:40","modified_gmt":"2023-07-02T13:53:40","slug":"game-theory-8-chap13-bayesian-nash-equilibrium","status":"publish","type":"post","link":"https:\/\/saraheee.com\/ko\/2023\/06\/game-theory-8-chap13-bayesian-nash-equilibrium\/","title":{"rendered":"Game Theory 8 \u2013 chap13. Bayesian Nash equilibrium"},"content":{"rendered":"

4. Bayesian Games
4.2 Bayesian Games: Bayesian Nash Equilibrium (Chapter 26)<\/p>\n\n\n\n

Static Bayesian Games<\/h3>\n\n\n\n

Nash eqbm is readily applied to Bayesian games and well-suited to static games<\/p>\n\n\n\n

The Nash eqbm of Bayesian games is called Bayesian Nash equilibrium<\/em> (BNE)<\/p>\n\n\n\n

We study two methods of computing BNE of static games<\/p>\n\n\n\n

(1) Compute Nash eqbm of the Bayesian normal form<\/mark><\/strong><\/p>\n\n\n\n

1 \\ 2<\/td>C<\/td>D<\/td><\/tr>
\\(A^{12}A^{0}\\)<\/td>8, 9<\/td>3, 6<\/td><\/tr>
\\(A^{12}B^{0}\\)<\/td>10, 6<\/td>4, 7<\/td><\/tr>
\\(B^{12}A^{0}\\)<\/td>4, 3<\/td>5, 8<\/td><\/tr>
\\(B^{12}B^{0}\\)<\/td>6, 0<\/td>6, 9<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

The normal-form game is then solvable by ISD: there is a unique strategy profile that survives ISD,<\/p>\n\n\n\n

\\((B^{12}B^{0}, D)\\).<\/mark><\/p>\n\n\n\n

It is the unique BNE of Example 4.4<\/p>\n\n\n\n

EXAMPLE 4.5 (EXERCISE 7 ON PAGE 357)<\/mark>.<\/p>\n\n\n\n

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

Player 1 observes c, whereas player 2 does not observe c<\/p>\n\n\n\n

The players make a simultaneous and independent choice and receive a payoff as follows:<\/p>\n\n\n\n

1 \\ 2<\/td>X<\/td>Y<\/td><\/tr>
A<\/td>0, 1<\/td>1, 0<\/td><\/tr>
B<\/td>1, 0<\/td>c, 1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

(a) Draw the Bayesian normal form of this game<\/p>\n\n\n\n

\\(S_{1} = {A^{2}A^{0}, A^{2}B^{0}, B^{2}A^{0}, B^{2}B^{0}}\\)
\\(S_{2} = {X, Y}\\)<\/mark><\/p>\n\n\n\n

<\/td>X<\/td>Y<\/td><\/tr>
\\(A^{2}A^{0}\\)<\/td>0, 1<\/td>1, 0<\/td><\/tr>
\\(A^{2}B^{0}\\)<\/td>1\/3, 2\/3<\/td>2\/3, 1\/3<\/td><\/tr>
\\(B^{2}A^{0}\\)<\/td>2\/3, 1\/3<\/td>5\/3, 2\/3<\/td><\/tr>
\\(B^{2}B^{0}\\)<\/td>1, 0<\/td>4\/3, 1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

(b) Compute the Bayesian Nash equilibrium<\/p>\n\n\n\n

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)<\/mark><\/p>\n\n\n\n

Therefore, the prediction by NE is (\\(B^{2}A^{0}\\), Y).<\/mark><\/p>\n\n\n\n

(2) The second method is to treat the types of each player as separate players; more suitable for infinite<\/em> games<\/p>\n\n\n\n

EXAMPLE 4.6 (COURNOT DUOPOLY WITH UNKNOWN COSTS)<\/mark>. <\/p>\n\n\n\n

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

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

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<\/p>\n\n\n\n

2’s profit function where \\(c_{2}\\) = 0
\\((10-q_{1}-q_{2}^{L}) \\cdot q_{2}^{L}\\)
\u2234 \\(q_{2}^{L} = \\frac{10-q_{1}}{2}\\)<\/mark><\/p>\n\n\n\n

2’s profit function where \\(c_{2}\\) = 4
\\((10-q_{1}-q_{2}^{H}) \\cdot q_{2}^{H} – 4q_{2}^{H}\\) \u21d4 (differentiate over \\(q_{2}^{H}\\)) \\(6 – q_{1} – 2q_{2}^{H}\\) = 0
\u2234 \\(q_{2}^{H} = \\frac{6-q_{1}}{2}\\)<\/mark><\/p>\n\n\n\n

\\(c_{2}\\) = 0 \u2192 \\(BR_{2}^{L}(q_{1}) = \\frac{10-q_{1}}{2}\\) and \\(c_{2}\\) = 4 \u2192 \\(BR_{2}^{H}(q_{1}) = \\frac{6-q_{1}}{2}\\)<\/p>\n\n\n\n

The expected profit accuring to firm 1 is<\/p>\n\n\n\n

\\(\\frac{1}{2}(10-q_{1}-q_{2}^{2})q_{1} + \\frac{1}{2}(10-q_{1}-q_{2}^{H})q_{1}\\)<\/mark><\/p>\n\n\n\n

(differentiate over \\(q_{1}\\)) \\(\\frac{1}{2}(10-2q_{1}-q_{2}^{L}) + \\frac{1}{2}(10-2q_{1}-q_{2}^{H}) = 0\\)<\/mark><\/p>\n\n\n\n

The FOC with respect to \\(q_{1}\\) gives us<\/p>\n\n\n\n

\\(BR_{1}(q_{2}) = 5 – \\frac{1}{4}(q_{2}^{L} + q_{2}^{H})\\)<\/p>\n\n\n\n

\\(q_{1} = 5 – \\frac{1}{2}\\bar{q_{2}}\\) (\u2235 \\(\\bar{q_{2}} = \\frac{1}{2}q_{2}^{L} + \\frac{1}{2}q_{2}^{H}\\))
\\(\\bar{q_{2}} = 10 – 2q_{1}\\)<\/mark><\/p>\n\n\n\n

f(q) = (10-q)\/2, f(q) = (6-q)\/2, f(q) = 10-2q, q=(0,10)\nx-axis: q1, y-axis: q2<\/code><\/pre>\n\n\n
\n
\"\"<\/figure><\/div>\n\n\n
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
\u2192 4-q\/2: Firm 2’s BR perceived by firm 1<\/p>\n\n\n\n
\\(q_{1} = 5 – \\frac{1}{4}(\\frac{10-q_{1}}{2} + \\frac{6-q_{1}}{2})\\)<\/mark><\/p>\n\n\n\n
Bayesian Nash eqbm<\/p>\n\n\n\n
\\(q_{1}^{*} = 4, q_{2}^{L*} = 3, q_{2}^{H*} = 1\\)<\/mark><\/p>\n\n\n\n
Thus, (4, 1), (4, 3)<\/mark><\/p>\n\n\n\n
<\/p>\n\n\n\n