# Are there multiple equilibria in the second price auction?

Suppose that $$n\geq 2$$ bidders compete in a second price auction. Each bidder $$i$$ knows their own valuation $$v_i$$, but only knows the distribution generating the valuations of the other players. Valuations are independently, continuously and symmetrically distributed. Define a (pure) strategy of a player as a function mapping from each valuation that they might have to a bid $$b$$; and define a (pure strategy) equilibrium as a set of (pure) strategies, one for each player, such that each player's strategy maximises their expected payoff given the strategies of the other players.

In this context, it is well known that it is weakly dominant for each player to bid their valuation, i.e. set $$b(v_i) = v_i$$. However, are there are other (pure strategy) equilibria of this game?

One point that perhaps should be clarified: in some discussion of second price auctions that I have seen, people define each player's strategy as their bid (a number) not their bidding functions. Here I am interested in the game obtained by defining strategies as functions.

Edit: in response to some helpful examples from @Giskard, we may wish to restrict attention to equilibria that (i) involve deviations from truthful bidding that occur with positive measure (ii) survive natural refinements that we might want to apply.

• Please explicitly define your natural refinements, as 'natural' seems to be a subjective word. Mar 26, 2019 at 5:10
• I don't have any particular requirements in mind, I am simply wondering whether it is possible to get uniqueness via a suitable refinement of one kind or another
– user17900
Mar 26, 2019 at 11:25
• You can look into affiliated signals. Mar 26, 2019 at 11:31

Sure. An example: if both valuations are drawn from the $$[0,1]$$ interval then the strategies $$b_1(v_1) = v_1$$ and $$b_2(v_2) = \left\{\begin{array}{cc} v_2 & \text{ if } v_2 < 1 \\ 5 & \text{ if } v_2 = 1. \end{array}\right.$$ Another, slightly more annoying equilibrium for $$v_1,v_2 \in [0,1]$$: \begin{align*} b_1(v_1) & = 0 \\ b_2(v_2) & = 2. \end{align*}

EDIT: The truth-telling bidding strategies are weakly dominant strategies. If we only allow trembling hand perfect equilibria, players assign positive probabilities (in this case densities) to all truncated strategy profiles. In such a setting, the weakly dominant strategies become strictly dominant. (Because of sequential rationality even zero measure deviations such as my first example are eliminated.) In a game where all players have strictly dominant strategies, there is only one equilibrium.

• Some nice ideas here... I wonder can we rule out the second with a refinement (e.g. a trembling hand)?
– user17900
Mar 25, 2019 at 22:47
• @afreelunch Seems so. At $v_2 = 0$ player two would not bid 2 if there is a trembling hand. Mar 26, 2019 at 5:09
• Yes this seems correct. However, could you explain how this eliminates your first example? I am a bit confused by the mention of sequential rationality since this is a one-shot game
– user17900
Mar 26, 2019 at 11:48
• @afreelunch Sequential rationality is one of the properties of Bayesian equilibrium. It means that in every information set each player plays a best response given their beliefs. In my example having a valuation of 1 ($v_2 = 1$) is such an information set. The bidding strategy has to specify a best response here as well, given trembling hand beliefs. Mar 26, 2019 at 15:00
• Ah I see, I was under the impression we were discussing BNE (not PBE) but now I understand what you mean
– user17900
Mar 26, 2019 at 15:09