Consider a first-price auction. Suppose that we have $N$ bidders, and they believe that their opponents' values is drawn from a uniform distribution on interval $[0,1]$.

Let us eliminate weakly dominated strategies. The 1st round will clearly eliminate all bids higher than the private value $x$. But what range of prices will be eliminated in a 2nd round?

My conjecture: after elimination of bids higher than private values, bidder $i$'s objective function in a 2-bidder situation will be $(v_i-b_i)\Pr(b_{-i}\leq b_i)$. The probability $\Pr(b_{-i}\leq b_i)$ is maximized when $b_{-i}$ is approaching $v_{-i}$. So maximized form of objective functions of bidder $i$ is $(v_i-b_i)\Pr(v_{-i}\leq b_i)$, which is $(v_i-b_i)b_{i}$. (Since we assume uniform distribution on values) So after 1st round of rationalizability, a bidder's maximized payoff will be $\frac{v^2_i}{4}$. This means that in 2nd round of rationalizability, any bidder will not be bidding higher than $v_i-\frac{v^2_i}{4}$.

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    $\begingroup$ Take any strategy of a player that always bids less than $1$. This strategy is a best response to the strategy of the other player of always bidding $1$. It will, therefore, not be eliminated in the first round, even if it bids higher than the actual value. $\endgroup$ Sep 9, 2020 at 16:52
  • $\begingroup$ But this is a 1st price auction. If one bidder bids higher than private value, there will be chances that he has to pay that price. So the highest he would bid is his private value. $\endgroup$
    – ask
    Sep 9, 2020 at 17:23
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    $\begingroup$ No. If the other bidder bids $1$ and you bid less, you will never have to pay anything- even if your bid is higher than your valuation. $\endgroup$ Sep 9, 2020 at 17:47
  • $\begingroup$ I see in literature that some authors impose an additional condition like "bidders expect positive bids to win with positive probability". Will this condition solve the issue you mentioned? Probably I need to add this condition to my original post? $\endgroup$
    – ask
    Sep 9, 2020 at 18:06
  • $\begingroup$ Yes. You can also avoid the problem if you eliminate weakly dominated strategies. $\endgroup$ Sep 9, 2020 at 18:08

1 Answer 1


Elimination of (weakly) dominated actions will not get you far. In fact, all bids strictly between 0 and the value are undominated. Hence, a buyer's optimal bid in a first-price auction can't be determined without knowing the bidder's belief (unless the buyer's value is 0).

Claim: For any type $x>0$, all bids $c \in (0,x)$ are undominated.

  • $c$ is not dominated by any bid $b>c$: $b$ yields a strictly smaller than $c$ if all other bidders bid $b_j<c$.
  • $c$ is not dominated by any bid $b<c$: $b$ yields a strictly smaller utility than $c$ if all other bidders bid $b_j=c$.

You can show that, for all types $x>0$, bidding zero is dominated by any $b\in (0,\frac{N-1}{N} x)$, though.

Suppose $p$ is the probability that all other $(N-1)$ bidders bid zero. Then, bidding zero yields utility $\frac{p}{N}x$: winning by a randomization because everyone bids the same and then winning the lottery and paying nothing. However, when bidding $b$ from the interval above, your exepected payoff is at least $$\geq p(x-b) > p (x - \frac{N-1}{N} x) > \frac{p}{N}x.$$

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    $\begingroup$ But this is about iterated elimination of weakly dominated strategies. If you know the other bidder never bids above their valuation and that this valuation is independently uniformly distributed, you get additional restrictions. $\endgroup$ Sep 10, 2020 at 18:58
  • $\begingroup$ I see, I made a general statement ignoring the uniform distribution belief. $\endgroup$
    – Bayesian
    Sep 10, 2020 at 19:50

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