# Minimum bids in first-price auction

Suppose there are two bidders $i=1,2$ who may either have a low valuation $V_L$ or a high valuation $V_H.$ Bidders do not know others' valuations. In a first-price auction, where ties are broken by coin-flip, what is the the minimum bids (i.e. the lowest bid in the support $\underline{b_i}$ in $[\underline{b_i},\overline{b_i}]$ for $i=1,2$) for the $V_L$ type bidder?

In these auctions, strategies are some probability over the domain $[\underline{b_i},\overline{b_i}]$ (i.e. a mixed strategy). My guess is that the support for both the $V_L$ type player is $0$, i.e. that $$[\underline{b_i},\overline{b_i}]=(0,V_L].$$ Furthermore, the probability of bidding close to $0$ is quite trivial, since the bidder's chance of winning is also close to $0$. However, my lecturer argues that the support $\underline{b_i}=V_L$ for both players, but this would not make bidding worthwhile for the $V_L$ type, since she would receive $0$ surplus.

• What is the distribution of types? Jan 6, 2018 at 22:13
• @MichaelGreinecker, they're both possible with some nonzero probability. I believe no particular distribution is necessary for this. Jan 6, 2018 at 22:15
• In that case, bidding every type $V_L$ is certainly not a Bayes-Nash equilibrium Jan 6, 2018 at 22:17
• @MichaelGreinecker I don't claim that. I'm only interested in the minimum bid in both players' supports. Jan 6, 2018 at 22:17
• "The support...are...0." makes sense neither grammatically nor mathematically. Jan 6, 2018 at 22:44

Did you learn game theory? with a mixed strategy, all strategy must yield the same expected unity. If the strategy of lower type is over $[0,v_l]$, no one will bid $v_l$ since bidding $v_l$ will give 0, but bidding lower than $v_l$ gives more than 0 in expectation. In this way, you can show that any interval $[0,x < v_l]$cannot be the support. Only a single point can be a support. now any singel point less than $v_l$ cannot be a support. so $v_l$ is the only support.