<p>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).</p>
<p>Claim: For any type <span class="math-container">$x&gt;0$</span>, all bids <span class="math-container">$c \in (0,x)$</span> are undominated.</p>
<ul>
<li><span class="math-container">$c$</span> is not dominated by any bid <span class="math-container">$b&gt;c$</span>: <span class="math-container">$b$</span> yields a strictly smaller
than <span class="math-container">$c$</span> if all other bidders bid <span class="math-container">$b_j&lt;c$</span>.</li>
<li><span class="math-container">$c$</span> is not dominated by any bid <span class="math-container">$b&lt;c$</span>: <span class="math-container">$b$</span> yields a strictly smaller utility than <span class="math-container">$c$</span> if all other bidders bid <span class="math-container">$b_j=c$</span>.</li>
</ul>
<p>You can show that, for all types <span class="math-container">$x&gt;0$</span>, bidding zero is dominated by any <span class="math-container">$b\in (0,\frac{N-1}{N} x)$</span>, though.</p>
<p>Suppose <span class="math-container">$p$</span> is the probability that all other <span class="math-container">$(N-1)$</span> bidders bid zero. Then, bidding zero yields utility <span class="math-container">$\frac{p}{N}x$</span>: winning by a randomization because everyone bids the same and then winning the lottery and paying nothing.
However, when bidding <span class="math-container">$b$</span> from the interval above, your exepected payoff is at least
<span class="math-container">$$\geq p(x-b) &gt; p (x - \frac{N-1}{N} x) &gt; \frac{p}{N}x.$$</span></p>

CC BY-SA 4.0
2020-09-10T18:58:57.240Z
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.
Michael Greinecker
CC BY-SA 4.0
2020-09-10T19:50:40.730Z
I see, I made a general statement ignoring the uniform distribution belief.
Bayesian
CC BY-SA 4.0
2020-09-10T12:30:01.133Z
2020-09-10T12:30:01.133Z
Bayesian
<p>Consider a first-price auction.
Suppose that we have <span class="math-container">$N$</span> bidders, and they believe that their opponents' values is drawn from a uniform distribution on interval <span class="math-container">$[0,1]$</span>.</p>
<p>Let us eliminate weakly dominated strategies.
The 1st round will clearly eliminate all bids higher than the private value <span class="math-container">$x$</span>. But what range of prices will be eliminated in a 2nd round?</p>
<p>My conjecture: after elimination of bids higher than private values, bidder <span class="math-container">$i$</span>'s objective function in a 2-bidder situation will be <span class="math-container">$(v_i-b_i)\Pr(b_{-i}\leq b_i)$</span>. The probability <span class="math-container">$\Pr(b_{-i}\leq b_i)$</span> is maximized when <span class="math-container">$b_{-i}$</span> is approaching <span class="math-container">$v_{-i}$</span>. So maximized form of objective functions of bidder <span class="math-container">$i$</span> is <span class="math-container">$(v_i-b_i)\Pr(v_{-i}\leq b_i)$</span>, which is <span class="math-container">$(v_i-b_i)b_{i}$</span>. (Since we assume uniform distribution on values) So after 1st round of rationalizability, a bidder's maximized payoff will be <span class="math-container">$\frac{v^2_i}{4}$</span>. This means that in 2nd round of rationalizability, any bidder will not be bidding higher than <span class="math-container">$v_i-\frac{v^2_i}{4}$</span>.</p>

CC BY-SA 4.0
2020-09-09T16:52:25.103Z
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.
Michael Greinecker
CC BY-SA 4.0
2020-09-09T17:23:02.870Z
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.
ask
CC BY-SA 4.0
2020-09-09T17:47:02.270Z
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.
Michael Greinecker
CC BY-SA 4.0
2020-09-09T18:06:50.500Z
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?
ask
CC BY-SA 4.0
2020-09-09T18:08:17.520Z
Yes. You can also avoid the problem if you eliminate weakly dominated strategies.
Michael Greinecker
CC BY-SA 4.0
2020-09-11T19:35:49.453Z
Hi, I add some content, do you think that is enough for 2nd round of rationalizability?
ask
<microeconomics><game-theory><auctions>
What is 2nd round of rationalizability for a 1st price auction?
ask
# What is 2nd round of rationalizability for a 1st price auction?

(Asked by: username_0 on 2020-09-09)

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 $v\_i-\frac{v^2\_i}{4}$.

**Tags:** microeconomics, game-theory, auctions

> **Comments:**
> **username_1**: 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.  
> **username_0**: 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.  
> **username_1**: 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.  
> **username_0**: 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?  
> **username_1**: Yes. You can also avoid the problem if you eliminate weakly dominated strategies.  
> **username_0**: Hi, I add some content, do you think that is enough for 2nd round of rationalizability?  

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## 1 answers

### Answer by username_2 on 2020-09-10. Score: 2

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.
* $c$ is not dominated by any bid $b: $b$ 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
$$\geq p(x-b) > p (x - \frac{N-1}{N} x) > \frac{p}{N}x.$$

> **Comments:**
> **username_1**: 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.  
> **username_2**: I see, I made a general statement ignoring the uniform distribution belief.  

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