博弈論(Game Theory)------經典問題詳解(1)

Nash Equilibria(納什均衡)

Definition

A strategy profile $x^*∈ S$ is a Nash equilibrium (NE) if no unilateral deviation in strategy by any single player is profitable for that player, that is

${\displaystyle \forall i,x_{i}\in S_{i}:f_{i}(x_{i}^{*},x_{-i}^{*})\geq f_{i}(x_{i},x_{-i}^{*}).}$
When the inequality above holds strictly (with > instead of ≥) for all players and all feasible alternative strategies, then the equilibrium is classified as a strict Nash equilibrium. If instead, for some player, there is exact equality between ${\displaystyle x_{i}^{*}}$ and some other strategy in the set ${\displaystyle S}$, then the equilibrium is classified as a weak Nash equilibrium.

納什均衡在經濟學上的定義：所謂納什均衡，指的是參與人的這樣一種策略組合，在該策略組合上，任何參與人單獨改變策略都不會得到好處。換句話說，如果在一個策略組合上，當所有其他人都不改變策略時，沒有人會改變自己的策略，則該策略組合就是一個納什均衡。

Practice

1. Find all Nash equilibria and their payoffs for the strategic game shown in the table. Hint: do not forget to check for dominated strategies.

	L	M	R
U	6,4	3,3	0,5
M	3,1	4,4	1,2
D	1,3	6,1	0,4

Solution:
It’s easy to see that L is strictly dominated by R. With L removed, U is strictly dominated by M. Removing U we are left with the following game:

	M	R
M	4,4	1,2
D	6,1	0,4

We see that there is no pure strategy NE. Using the indifference criterion we can find
the unique mixed strategy NE of this reduces game: $σ_1(M) = \frac{3}{5}$ and $σ_2(M) = \frac{1}{3}$. Thus we
obtain the following NE of the original game:
$σ^∗ = ((0, \frac{3}{5}, \frac{2}{5}), (0,\frac{1}{3}, \frac{2}{3}))$ with payoff of $(2, \frac{14}{5} )$

2. Consider the stage games G1 and G2 whose strategic form looks like

	A	B
A	1,1	2,0
B	0,2	3,3

and

	A	B
A	1,1	2,2
B	2,2	3,3

a) Find all Nash equilibria in pure strategies of these two stage games.

b) Suppose a dynamic game is played in which the stage game G1 is repeated twice. Find all SPE and their payoffs of the repeated game assuming that no discount is used and the players use only pure strategies.

c) Suppose a dynamic game is played in which the stage game G2 is repeated twice. Find all SPE and their payoffs of the repeated game assuming that no discount is used and the players use only pure strategies.

Solution:
a) It’s easy to see that G1 has two pure strategy NE: (A,A) and (B,B) and G2 has only one: (B,B).

b) If G1 is repeated twice then, first there will be four SPE in which a NE is palyed in every period: (A,A) followed by (A,A) with payoff (2,2); (A,A) followed by (B,B) with payoff (4,4); (B,B) followed by (A,A) with payoff (4,4); (B,B) followed by (B,B) with payoff (6,6).

In addition, there will be some SPE in which a non-NE is played in the first period and a punishment scheme for a deviator is used in the second period. (Note that in this case both the default and the punishment have to be a NE of the stage in the second period.) There will be two such SPE: (A,B) followed by (B,B) ((A,A) is used as a punishment if player 2 deviates) with payoff (5,3); (B,A) followed by (B,B) ((A,A) is used as a punishment if player 1 deviates) with payoff (3,5).

c) Since the stage NE is unique the only SPE in this case is (B,B) followed by (B,B) with payoff (6,6).

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