<p>I think you're overcomplicating things. Say bidder values are iid on the unit interval <span class="math-container">$[0,1]$</span>. Then wlog each bidder <span class="math-container">$i$</span>'s strategy space is <span class="math-container">$S_i=[0,1]$</span>, and <span class="math-container">$i$</span>'s rival bid profile is simply <span class="math-container">$s_{-i}\in[0,1]^{N-1}$</span>. Moreover, in a second price auction, the only rival bid that matters in <span class="math-container">$s_{-i}$</span> is its highest value. Let's denote that as <span class="math-container">$\overline b=\max s_{-i}$</span>.</p>
<p>To show the weak dominance of truthful bidding <span class="math-container">$s_i=v_i$</span>, we just need to consider two kinds of &quot;untruthful&quot; bidding: over-bidding <span class="math-container">$s_i'&gt;v_i$</span> and under-bidding <span class="math-container">$s_i''&lt;v_i$</span>.</p>
<p>The table below shows that <span class="math-container">$s_i'&gt;v_i$</span> is weakly dominated by <span class="math-container">$s_i=v_i$</span>:</p>
<p><span class="math-container">\begin{matrix}
&amp; u_i(s_i,s_{-i}) &amp; u_i(s_i',s_{-i}) \\\hline
\overline b\ge s_i'&gt;v_i &amp; 0 &amp; 0 \\
s_i'&gt;\overline b&gt;v_i &amp; 0 &amp; v_i-\overline b &lt; 0 \\
s_i'&gt;v_i\ge\overline b &amp; v_i-\overline b &amp; v_i-\overline b \\
\end{matrix}</span></p>
<p>In particular, when the highest rival bid---which can be delivered by any rival bidding profile that features <span class="math-container">$\overline b\in(v_i,s_i')$</span> as the highest bid---is between <span class="math-container">$i$</span>'s true value and his actual bid, over-bidding does strictly worse.</p>
<p>The case for under-bidding is symmetric: when the highest rival bid is between <span class="math-container">$i$</span>'s true value and his actual bid, <span class="math-container">$\overline b\in(s_i'',v_i)$</span>, under-bidding does strictly worse.</p>

CC BY-SA 4.0
2024-08-12T16:49:39.747Z
2024-08-12T16:49:39.747Z
Herr K.
<p>In most books on auction theory, it is left as exercise to prove that the truthful strategy <span class="math-container">$s_{i} = v_i$</span> is a weakly dominant strategy and not a very weakly dominant strategy. To eliminate confusion, I am writing the definition of both as under:</p>
<p>(Def 1a): A strategy <span class="math-container">$s_i \in S_i$</span> is a weakly dominant strategy if it weakly dominates all strategies <span class="math-container">$t_i \in S_i \setminus \{s_i\} $</span></p>
<p>(Def 1b): A strategy <span class="math-container">$s_i \in S_i$</span> weakly dominates strategy <span class="math-container">$t_i$</span> if  we have,
<span class="math-container">$u_i(s_i, s_{-i}) \ge u_i(t_i,s_{-i}) \space \forall s_{-i} \in S_{-i}$</span> and for at least one <span class="math-container">$s_{-i}$</span> the inequality is strict.</p>
<p>(Def 1c): A strategy <span class="math-container">$s_i \in S_i$</span> very weakly dominates strategy <span class="math-container">$t_i$</span> if  we have,
<span class="math-container">$u_i(s_i, s_{-i}) \ge u_i(t_i,s_{-i}) \space \forall s_{-i} \in S_{-i}$</span></p>
<p>Now to form a mathematical definition definition (involving utility numbers) of a weakly dominant strategy I proceed as follows:</p>
<p>(OP's own def ): A strategy <span class="math-container">$s_i \in S_i$</span> is a weakly dominant strategy if</p>
<p>(a) <span class="math-container">$u_i(s_i, s_{-i}) \ge u_i(t_i,s_{-i}) \space \forall s_{-i} \in S_{-i}$</span></p>
<p>(b) <span class="math-container">$\forall t_i \in S_i \setminus \{s_i\} \space \exists \space  t_{-i} \in S_{-i} : u_i (s_i, t_{-i})&gt;u_i(t_i, t_{-i}) $</span></p>
<p>We can denote <span class="math-container">$t_{-i}= t_{-i}(t_{i})$</span> to remind ourselves we need to find an opponent strategy profile (not necessarily unique) for each non-truthful strategy.</p>
<p>The framework is that of IID private values and in case of tie, each auction winner receives the object with a positive and equally likely probability and he pays his own bid value. (which is also the second highest bid value)</p>
<p>My question is how to do this? I do not need help in proving that the truthful strategy is very weakly dominant strategy. But I am stuck it is in proving a weakly dominant strategy.</p>
<p>For a lot of time I was trying to find <strong>the</strong> rival bidding profile where the above holds for all non-truthful strategy (all at once!) but was unsuccessful. And then realised I need an approach to find <strong>a</strong> rival bidding profile for each  non-truthful strategy.</p>
<p>Dear experts,</p>
<p>A) Could you tell exactly what is the (N-1) dimensional rival bid vector is (if there exists a unique one?) so that the strict inequality holds for all bidding strategies of the ith player not equal to his value?</p>
<p>B) Could you check and fill in the blanks for my proof of existence?</p>
<p>I am outlining  proof of existence of a unique rival bidding profile where the inequality is strict for all non-truthful strategies:</p>
<p>(Proof by contradiction) Suppose there is no rival strategy (function) vector <span class="math-container">$s_{-i}\in S_{-i}$</span> such that <span class="math-container">$u_i (v_i,s_{-i})&gt;u_i(s^{'}_{i},s_{-i})  \space \forall s^{'}_{i}\in S_i$</span></p>
<p>that is ,for all non-truthful bids of the ith player, <span class="math-container">$b_i \ne v_i$</span>, it is always the case that <span class="math-container">$u_i (v_i,s_{-i})\le u_i(b_{i},s_{-i})$</span> irregardles of what the opponent bid vector is.</p>
<p>Case A): <span class="math-container">$u_i(b_{i},s_{-i})=0$</span></p>
<p>Then the ith bidder loses :<span class="math-container">$ \space b_i &lt; max_{j \ne i} b_j $</span></p>
<p>If for some insane reason, all rival bidders bid, <span class="math-container">$b_j =0$</span>, then it is in the ith bidder's interest to bid <span class="math-container">$v_i$</span>. That is, <span class="math-container">$u_i(v_{i},s_{-i}) &gt;0$</span></p>
<p>Case B): <span class="math-container">$0&lt; u_i(b_{i},s_{-i}) &lt; v_i -max_{j \ne i}b_j$</span></p>
<p>I can't come up with a rival strategy profile here.</p>
<p>Case C) <span class="math-container">$ u_i = v_i -max_{j \ne i}b_j$</span></p>
<p>Here also, I can't come up with a rival strategy profile.</p>

CC BY-SA 4.0
2024-08-09T18:10:17.433Z
The conventional definition of a weakly dominant strategy is slightly different from the one you put here. In particular, the quantifier $\forall s_i'\in S_i$ in your definition puts a more stringent requirement than the conventional one, which only requires that all $s_i'$ be weakly dominated by the truthful strategy $s_i=v_i$. Thus, the conventional definition implies that $\forall s_i'$ $\exists s_{-i}$ such that $u_i(v_i,s_{-i})>u_i(s_i',s_{-i})$, whereas your definition reverses the two quantifiers, namely, $\exists s_{-i}$ $\forall s_i'$ such that the previous strict inequality holds.
Herr K.
CC BY-SA 4.0
2024-08-10T04:57:45.257Z
@HerrK. Thank you for pointing that out. I have made the necessary edits.
Karan Kumar
<game-theory><bayesian-game><auctions>
Proving Truthful bidding is weakly dominant strategy and not just very weakly dominant strategy in second price sealed bid auction
Karan Kumar
# Proving Truthful bidding is weakly dominant strategy and not just very weakly dominant strategy in second price sealed bid auction

(Asked by: username_0 on 2024-08-09)

In most books on auction theory, it is left as exercise to prove that the truthful strategy $s\_{i} = v\_i$ is a weakly dominant strategy and not a very weakly dominant strategy. To eliminate confusion, I am writing the definition of both as under:

(Def 1a): A strategy $s\_i \in S\_i$ is a weakly dominant strategy if it weakly dominates all strategies $t\_i \in S\_i \setminus \{s\_i\} $

(Def 1b): A strategy $s\_i \in S\_i$ weakly dominates strategy $t\_i$ if we have,
$u\_i(s\_i, s\_{-i}) \ge u\_i(t\_i,s\_{-i}) \space \forall s\_{-i} \in S\_{-i}$ and for at least one $s\_{-i}$ the inequality is strict.

(Def 1c): A strategy $s\_i \in S\_i$ very weakly dominates strategy $t\_i$ if we have,
$u\_i(s\_i, s\_{-i}) \ge u\_i(t\_i,s\_{-i}) \space \forall s\_{-i} \in S\_{-i}$

Now to form a mathematical definition definition (involving utility numbers) of a weakly dominant strategy I proceed as follows:

(OP's own def ): A strategy $s\_i \in S\_i$ is a weakly dominant strategy if

(a) $u\_i(s\_i, s\_{-i}) \ge u\_i(t\_i,s\_{-i}) \space \forall s\_{-i} \in S\_{-i}$

(b) $\forall t\_i \in S\_i \setminus \{s\_i\} \space \exists \space t\_{-i} \in S\_{-i} : u\_i (s\_i, t\_{-i})>u\_i(t\_i, t\_{-i}) $

We can denote $t\_{-i}= t\_{-i}(t\_{i})$ to remind ourselves we need to find an opponent strategy profile (not necessarily unique) for each non-truthful strategy.

The framework is that of IID private values and in case of tie, each auction winner receives the object with a positive and equally likely probability and he pays his own bid value. (which is also the second highest bid value)

My question is how to do this? I do not need help in proving that the truthful strategy is very weakly dominant strategy. But I am stuck it is in proving a weakly dominant strategy.

For a lot of time I was trying to find **the** rival bidding profile where the above holds for all non-truthful strategy (all at once!) but was unsuccessful. And then realised I need an approach to find **a** rival bidding profile for each non-truthful strategy.

Dear experts,

A) Could you tell exactly what is the (N-1) dimensional rival bid vector is (if there exists a unique one?) so that the strict inequality holds for all bidding strategies of the ith player not equal to his value?

B) Could you check and fill in the blanks for my proof of existence?

I am outlining proof of existence of a unique rival bidding profile where the inequality is strict for all non-truthful strategies:

(Proof by contradiction) Suppose there is no rival strategy (function) vector $s\_{-i}\in S\_{-i}$ such that $u\_i (v\_i,s\_{-i})>u\_i(s^{'}\_{i},s\_{-i}) \space \forall s^{'}\_{i}\in S\_i$

that is ,for all non-truthful bids of the ith player, $b\_i \ne v\_i$, it is always the case that $u\_i (v\_i,s\_{-i})\le u\_i(b\_{i},s\_{-i})$ irregardles of what the opponent bid vector is.

Case A): $u\_i(b\_{i},s\_{-i})=0$

Then the ith bidder loses :$ \space b\_i < max\_{j \ne i} b\_j $

If for some insane reason, all rival bidders bid, $b\_j =0$, then it is in the ith bidder's interest to bid $v\_i$. That is, $u\_i(v\_{i},s\_{-i}) >0$

Case B): $0< u\_i(b\_{i},s\_{-i}) < v\_i -max\_{j \ne i}b\_j$

I can't come up with a rival strategy profile here.

Case C) $ u\_i = v\_i -max\_{j \ne i}b\_j$

Here also, I can't come up with a rival strategy profile.

**Tags:** game-theory, bayesian-game, auctions
> **Comments:**
> **username_1**: The conventional definition of a weakly dominant strategy is slightly different from the one you put here. In particular, the quantifier $\forall s_i'\in S_i$ in your definition puts a more stringent requirement than the conventional one, which only requires that all $s_i'$ be weakly dominated by the truthful strategy $s_i=v_i$. Thus, the conventional definition implies that $\forall s_i'$ $\exists s_{-i}$ such that $u_i(v_i,s_{-i})>u_i(s_i',s_{-i})$, whereas your definition reverses the two quantifiers, namely, $\exists s_{-i}$ $\forall s_i'$ such that the previous strict inequality holds.  
> **username_0**: @HerrK. Thank you for pointing that out. I have made the necessary edits.  

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## 1 answers

### Answer by username_1 on 2024-08-12. Score: 3 | Accepted answer

I think you're overcomplicating things. Say bidder values are iid on the unit interval $[0,1]$. Then wlog each bidder $i$'s strategy space is $S\_i=[0,1]$, and $i$'s rival bid profile is simply $s\_{-i}\in[0,1]^{N-1}$. Moreover, in a second price auction, the only rival bid that matters in $s\_{-i}$ is its highest value. Let's denote that as $\overline b=\max s\_{-i}$.

To show the weak dominance of truthful bidding $s\_i=v\_i$, we just need to consider two kinds of "untruthful" bidding: over-bidding $s\_i'>v\_i$ and under-bidding $s\_i''.

The table below shows that $s\_i'>v\_i$ is weakly dominated by $s\_i=v\_i$:

\begin{matrix}
& u\_i(s\_i,s\_{-i}) & u\_i(s\_i',s\_{-i}) \\\hline
\overline b\ge s\_i'>v\_i & 0 & 0 \\
s\_i'>\overline b>v\_i & 0 & v\_i-\overline b < 0 \\
s\_i'>v\_i\ge\overline b & v\_i-\overline b & v\_i-\overline b \\
\end{matrix}

In particular, when the highest rival bid---which can be delivered by any rival bidding profile that features $\overline b\in(v\_i,s\_i')$ as the highest bid---is between $i$'s true value and his actual bid, over-bidding does strictly worse.

The case for under-bidding is symmetric: when the highest rival bid is between $i$'s true value and his actual bid, $\overline b\in(s\_i'',v\_i)$, under-bidding does strictly worse.


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