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Now, if there is a PBNE where player one chooses either to play B or C, or randomizes between the two, his expected payoff has to be greater than ? from such a move. Player 1’s expected payoff on the other hand depends upon player 2’s strategies and whether he randomizes and with what probabilities. Now, player 2’s expected payoff from playing D is PB-(1- PB) and that from playing E is -PB+(1- PB). He will pick D iff PB-(1- PB)> -PB+(1- PB)4PB>2 PB>1/2.He will be indifferent between playing D and E if PB=1/2 (and hence randomize between the two) and will pick E if PB1/2 is invalidated.
Thus, we cannot have a PBNE with Player 1 choosing to play C and player 2 choosing D since this invalidates the required belief criterion. Similarly it can be shown that there cannot be a PBNE where 1 chooses B and 2 plays E that is consistent with the required beliefs for player 2. Finally, if player 2 randomizes, the maximal expected payoff for player 1 is 0 as well. Thus, in all PBNE, player 1 picks A. Since in the equilibrium path 2’s turn never comes, he can pick either D or E or choose to randomize in this equilibrium.
So we can have the following PBNE: 1. (A, D) with PB>1/2, 2.(A,E) with PB v2. This is true since by bidding b2= v2 he has already won the auction and the final payment he has to make depends only on player 1’s bid. On the other hand, bidding less than v2 doesn’t change anything as long as the bid is greater than player 1’s bid = player 1’s valuation. However if b2
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