A Course In Game Theory: solutions to exercises

Player i has valuation vi and vi > vi+1 > 0. The set of actions for each player is a bid bi ∈ [0,∞]. The price paid for the item is p = maxi{bi} and the player of minimum index bidding this price wins. The payoff for player i is vi − p if i wins, 0 else. A Nash equilibrium is denoted b∗ = {bi }. If b∗ is a Nash equilibrium, then player 1 wins. Suppose I have a strategy, that is, a vector of bids, subject to the constraint that player one did not win. I infer that this strategy is not a N.E. Let j 6= 1 be the winner. Since j can get 0 by bidding 0, the strategy is not a N.E. if vj − p < 0. I am therefore left to consider only strategies for which v1 > vj ≥ p > b1. These are not N.E.’s since player 1 can raise his bid to slightly above vj . A N.E. would be b1 = v2, some other player bidding v2 and every other player bidding bi ∈ [0, v2]. By this strategy player 1 gains v1 − v2 and would not change, other bids being fixed. Any other player has a payoff of 0, and could only change that by increasing his bid above v2, in which case the payoff would be negative, all other bids being fixed.