Yes no pure strategy eqbm, this is what i had in mind.. so here can we have nash equilibrium in mixed strategy?? If yes how to assign appropriate probabilities..?
vandita
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, voter can at least avoid voting costs by not voting. 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.
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