Re: Proble: Game theory- solve

Posted by Amit Goyal on Mar 13, 2014; 11:18am
URL: http://discussion-forum.276.s1.nabble.com/Proble-Game-theory-solve-tp7585164p7585225.html

Atika, This is incorrect. In this game there is no pure strategy equilibrium.
Here is the proof:
I will use the following terminology for explanation: I will call a voter "type i voter" if she supports candidate i.
And given a strategy profile s, n(i, s) denotes the number of voters voting for i in s.

Lemma 1: For type i voter, strategy of voting for candidate j ≠ i is strictly dominated by not voting.
Proof:
Clearly, such a voter can at least benefit by 1 (voting costs) by choosing not to vote. Thus, every type A voter will either choose to vote for A or will choose not to vote. And similarly a type B voter will either vote for B or none.

Lemma 2: Any strategy profile in which both candidates receive positive number of votes cannot be a Nash equilibrium i.e. s such that n(A, s) > 0 and n(B, s) > 0 is not a Nash equilibrium.
Proof:
Case 1: 0< n(A, s) < n(B, s): Consider a type A voter who choose to vote in s. Her payoff in strategy profile s is -11. If she does not vote, she gets -10. Hence, s is not a Nash equilibrium.
Case 2: 0< n(B, s) < n(A, s): Consider a type B voter who choose to vote in s. Her payoff in strategy profile s is -11. If she does not vote, she gets -10. Hence, s is not a Nash equilibrium.
Case 3: 0< n(B, s) = n(A, s): Consider a type A voter who choose not to vote in s. Her payoff in strategy profile s is 0. If she votes, she gets 9. Hence, s is not a Nash equilibrium.
Thus, s such that n(A, s)> 0 and n(B, s)>0 is not a Nash equilibrium.

Lemma 3: Any strategy profile in which any candidate receiving more than 1 vote cannot be a Nash equilibrium i.e. s such that n(A, s)> 1  or n(B, s) > 1 is not a Nash equilibrium.
Proof:
By Lemma 2, we just need to check these two cases:
Case 1: n(A, s) > 1 and n(B, s) = 0: Consider a type A voter who choose to vote in s. Her payoff in strategy profile s is 9. If she does not vote, she gets 10. Hence, s is not a Nash equilibrium.
Case 2: n(B, s) > 1 and n(A, s) = 0: Consider a type B voter who choose to vote in s. Her payoff in strategy profile s is 9. If she does not vote, she gets 10. Hence, s is not a Nash equilibrium.

Lemma 4: There is no pure strategy Nash equilibrium.
Proof:
By Lemma 2 and Lemma 3, we just need to analyse the following three cases:
Case 1: n(A, s) = 1 and n(B, s) = 0: Consider a type B voter who choose not to vote in s. Her payoff in strategy profile s is -10. If she votes, she gets -1. Hence, s is not a Nash equilibrium.
Case 2: n(A, s) = 0 and n(B, s) = 1: Consider a type A voter who choose not to vote in s. Her payoff in strategy profile s is -10. If she votes, she gets -1. Hence, s is not a Nash equilibrium.
Case 3: n(A, s) = 0 and n(B, s) = 0: Consider a type B voter who choose not to vote in s. Her payoff in strategy profile s is 0. If she votes, she gets 9. Hence, s is not a Nash equilibrium.

Hence, there is no Nash equilibrium in pure strategies.