Question

Second Price sealed bid-auction: Assume n players are bidding in an auction in order to obtain an indivisible object. Denote by vi the value player i attaches to the object; if she obtains the object at the price p her payoff is vi −p. Assume that the players’ valuations of the object are all different and all positive; number the players 1 through n in such a way that v1 > v2 > · · · > vn > 0. Each player i submits a (sealed) bid bi . If player i’s bid is higher than every other bid, she obtains the object at a price equal to the second-highest bid, say bj , and hence receives the payoff vibj . If some other bid is higher than player i’s bid, player i does not obtain the object, and receives the payoff of zero. If player i is in a tie for the highest bid, her payoff depends on the way in which ties are broken. A simple (though arbitrary) assumption is that the winner is the player among those submitting the highest bid whose number is smallest (i.e. whose valuation of the object is highest). (If the highest bid is submitted by players 2, 5, and 7, for example, the winner is player 2.) Under this assumption, player i’s payoff when she bids bi and is in a tie for the highest bid is vi − bi if her number is lower than that of any other player submitting the bid bi , and 0 otherwise. Date: 1st April, 2019. 1 (a) Formulate this situation as a strategic (normal-from) game. (b) Find a pure strategy Nash equilibrium in which player 1 obtains the object. (c) Find a pure strategy Nash equilibrium in which player n obtains the object. (d) Show that bidding own valuation is a weakly dominant strategy for each player. (e) Find ALL pure strategy Nash equilibria of this game when n = 2.

Answer 1

In a second price sealed bid auction an object is assigned to the bidders and the one with the highest bid wins and has to pay an amount equal to the second highest bid.

Each bidder has a valuation v_i of his/her own bid and we further assume that

v₁> v₂> v₃ >........>v_n >0

We know assume that everybody knows all the valuations and this is a complete information game.

Now, the players simultaneously bid b₁, b₂,.....b_n.

The object is given to the highest bidder and he/she pays the amount equal to the second highest bid.

a) The payoff of this person is v_i-b_j and all others get a payoff of 0.

b) If everyone plays according to the strategy of the game then the player 1 receives the object and pays b₂. Thus the payoff is v₁-b₂.

For everyone person to win with assurance it is better to bid equal to the valuation. Thus the payoff now becomes v₁-v₂  >0 and all other payoffs will be 0.

Player 1 now has no incentive as the utility of the player can only decrease. Also, for all other players, having valuation v_i not equal to v₁, in order to change the payoff from 0 the bidder need to bid more than v₁ and thus the payoff would become v_i-v₁<0 and thus no incentive to deviate from the strategy of the game.

Thus the Nah Equilibrium in this game is to bid equal to the valuation i.e. b_i= v_i. This is the optimal bid because no other choice makes the players better off.

c) Similarly, for the player n to obtain the object, the bid by player n shall be the highest. Thus the player n wins and has to pay the amount equal to the second highest bid. Payoff is v_n-v_j  >0 and all other get 0 payoff. Where, v_j is the second highest bid.

Thus the Nash equilibrium in this case would be b_n= v_n and b_-n<v_n.

In this case the player n will obtain the object.

d) In the case when the highest bidder wins and valuation is v₁ and the other players bid 0. None of the other players can increase their payoff as they value the item below the valuation of player 1. This makes it a weakly dominant strategy. Thus the Nash equilibrium b_i= v_i is the weakly dominant Nash equilibrium.

e) When n=2, we have v₁>v₂. And the player with the highest bid wins the object. The payoff of the player 1 is then v₁-v₂>0 and the payoff of other player is 0 as he does not win.

Thus the Nash equilibrium is ( v₁-v₂, 0).