Consider a game in which an auctioneer sells one dollar to the highest bidder. The high bidder wins the dollar, but every bidder pays their bid. Concretely, assume that there are two bidders $\{1,2\}$; a strategy for bidder $i$ is a nonnegative number $b_i \ge 0$. The payoff to bidder $i$ is given by $\pi_i(b_i,b_j)-b_i$, where $$\pi_i(b_i,b_j)= \begin{cases} 1 & \text{if } b_i>b_j\\ 0.5 & \text{if } b_i=b_j \\ 0 & \text{if } b_i<b_j \end{cases} .$$ That is, when there is a tie the two bidders split the dollar equally.
I am able to find a symmetric (Nash) equilibrium of randomized strategies given by each player playing the uniform distribution supported by the interval $[0,1]$ independently. I am also able to show that there does not exist an equilibrium of pure strategies (either symmetric or otherwise) in this two-bidder setting.
My current question is whether there exists an asymmetric equilibrium (of randomized strategies) in the two-bidder setting. If not, are there any asymmetric equilibria when the number of bidders is more than 2? Thank you!
Best Answer
I do not think that there exists any mixed strategy asymmetric equilibria with two players.
The trick is always the same when looking for mixed strategy equilibria : if a player plays different pure strategies with strictly positive probability at an equilibrium, these pure strategies must yield the same expected payoff given how the other player plays.
Now, I am not 100% positive how this applies to cases with a continuum of possible strategies, but I am pretty confident one can prove that "playing different pure strategies with positive probability" simply extends to "playing different pure strategies with positive probability mass". Then you can see that the only way for player $1$ to put strictly positive probability mass on two different pure strategies while being at an equilibrium is for player $2$ to play a uniform distribution over $[0;1]$.
Regarding the existence of asymmetric equilibria with more bidders, I do not think there exists any either, at least when players play continuous strategies. This time, for any player $i$, the relevant information regarding other players action is only what is the maximum of others bids (as we assumed away discrete strategies). So for $i$ to play different pure strategies with strictly positive probability, it needs to be the case that
$$ \max \{ b_1, \dots , b_{i-1}, b_{i+1} ,\dots b_n\} \sim U[0,1] $$
which is equivalent to
$$ P(b_1 \leq p, \dots , b_{i-1} \leq p, b_{i+1}\leq p ,\dots b_n \leq p) = p $$ $$ F(b_1 \leq p) * \dots * F(b_{i-1} \leq p) * F(b_{i+1}\leq p) *\dots* F(b_n \leq p) = p$$
This work for every $i$ if $F(b_j \leq p) = p^{1/n-1}$ for all $j \in \{1,\dots,n\}$, but I do not see how it could work with asymmetric bidding function. In effect, assume that $F_j \neq F_k$. We must have
$$ F_j \prod_{i \neq j,k} F_i = p$$.
and
$$ F_k \prod_{i \neq j,k} F_i = p $$
but this contradicts $F_j \neq F_k$.
You might still want to consider the possibility of there being asymmetric equilibria with discrete mixed strategies. The former argument does not apply to these cases, but my guess would be that there are none.