Let $\sigma_i$ be the strategy profile for bidder $i$ that indicates how they should bid based on their value. As we know, if there are 2 bidders both with their values $v_1,v_2$ on $U[0,1]$ in a First-Price Auction, they will bid $\frac{v_1}{2}$ and $\frac{v_2}{2}$ respectively, so $\sigma_1(v_1)=\frac{v_1}{2},\sigma_2(v_2)=\frac{v_2}{2}$.

However, if bidder 1's value is still $U[0,1]$ but bidder 2's value is $U[0,2]$, there does not exist a Bayes-Nash Equilibrium. Why is this? Specifically, why is the following not a BNE?

\begin{equation*} \begin{split} \sigma_1(v_1) &= \frac{v_1}{2} \\ \sigma_2(v_2)&= \begin{cases} \frac{v_2}{2} & \mbox{ if }v_2\in[0,1] \\ \frac{1}{2} & \mbox{ if }v_2>1 \end{cases} \end{split} \end{equation*}

Equilibrium says for all bidders $i$ and values $v_i$, $\sigma_i(v_i)$ is the optimal bid. Is there a bidder and value for which this is not the case?


1 Answer 1


It is easy to see why the strategy profile does not form a BNE. Since bidder 2 never bids above $\frac12$, for a sufficiently high $v_1$, an alternative strategy $\sigma_1'(v_1)=\frac12+\epsilon$ is a better response to $\sigma_2$ than the proposed $\sigma_1$.

For example, suppose bidder 1 observes $v_1=1$. Following $\sigma_1$, his probability of winning is $\Pr(\sigma_2\le \frac12)=\frac12$ and so his expected payoff is $\frac12(1-\frac12)=\frac14$. But if he bids slightly above $\frac12$, his probability of winning jumps to $1$, and his expected payoff is just slightly below $\frac12$, which is greater than $\frac14$.

In fact, there does exist a BNE in the asymmetric two-bidder first price auction where values are uniformly distributed on $[0,\omega_i]$, with $\omega_1\ne \omega_2$. The equilibrium bidding strategy is given by \begin{equation} \sigma_i(v_i)=\frac{1}{k_iv_i}\left(1-\sqrt{1-k_iv_i^2}\right),\qquad\text{where }k_i=\frac1{\omega_i^2}-\frac1{\omega_j^2}. \end{equation} For the details of derivation, I'd refer you to Section 4.3 of Krishna (2010).


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