<p>To supplement Amit's answer, we can get the same answer using Myerson's trick. This trick is &quot;halfway&quot; between Amit's answer and using the Revenue Equivalence Theorem.</p>
<p>First, we may consider the all-pay auction in a mechanism design framework. A general mechanism is a set of transfer and allocation rules <span class="math-container">$(x_i,t_i)_{i=1}^2$</span> such that the mapping <span class="math-container">$[0,1]^2\owns (v_i,v_{-i})\mapsto x_i(v_i,v_{-i})\in[0,1]$</span> gives the probability agent <span class="math-container">$i$</span> gets the good and the mapping <span class="math-container">$[0,1]^2\owns (v_i,v_{-i})\mapsto t_i(v_i,v_{-i})\in \mathbb{R}$</span> gives the transfer to agent <span class="math-container">$i$</span> if the two agents announce their types to be <span class="math-container">$(v_i,v_{-i})$</span>. Implicitly, this description of the mechanisms has applied the revelation principle to restrict attention to direct mechanisms.</p>
<p>Myerson (1981) showed that any direct incentive compatible and individually rational mechanism satisfies the following equality <span class="math-container">$$\Pi_i(v_i)=\Pi_i(0)+\int_0^{v_i}P_i(x)dx \qquad \qquad (\dagger)$$</span> where <span class="math-container">$\Pi_i:[0,1]\to\mathbb{R}$</span> is the interim payoff function for agent <span class="math-container">$i$</span> reporting a certain type, and <span class="math-container">$P_i:[0,1]\to [0,1]$</span> maps from agent <span class="math-container">$i$</span>'s type report to interim probability agent <span class="math-container">$i$</span> receives the good. Formally, <span class="math-container">$P_i(v_i)=\mathbb{E}_{v_{-i}}[x_i(v_i,v_{-i})]$</span> and <span class="math-container">$\Pi_i(v_i)=v_iP_i(v_i)-\mathbb{E}_{v_{-i}}[t_i(v_i,v_{-i})]$</span>.</p>
<p>Consider now the case of the all-pay auction mechanism <span class="math-container">$(\hat{x}_i,\hat{t}_i)_{i=1}^2$</span>. Assuming a symmetric, strictly increasing, equilibrium bidding strategy <span class="math-container">$b_i(v_i):[0,1]\to \mathbb{R}_+$</span>, the allocation and transfer mechanisms are <span class="math-container">$$\begin{align*} \hat{x}(v_i,v_{-i})&amp;=\begin{cases} 1 &amp; \text{ if } v_i&gt;v_{-i} \\
 0 &amp; \text{ if } v_i&lt;v_{-i} \\
\text{Any }p\in[0,1] &amp; \text{ if } v_i=v_{-i} \end{cases} \\
\hat{t}_i(v_i,v_{-i}) &amp;= b(v_i) \end{align*}$$</span> Using these, we may calculate <span class="math-container">$$\begin{align*} \Pi_i(v_i) &amp;= \int_0^{v_i}v_i-b(v_i)dv_{-i}-\int_{v_i}^1 b(v_i)dv_{-i}=v_i^2-b(v_i) \\
P_i(x)&amp;=\mathbb{P}(v_{-i}&lt;x)=x\end{align*}$$</span> Additionally, since we have conjectured a strictly increasing bidding strategy, the <span class="math-container">$v=0$</span> type does not get the good with probability <span class="math-container">$1$</span>. Therefore, the dominant strategy for the <span class="math-container">$v=0$</span> type is to bid <span class="math-container">$0$</span>, so that <span class="math-container">$\Pi_i(0)=0$</span>. Therefore, plugging these expressions into <span class="math-container">$(\dagger)$</span>, we find <span class="math-container">$$\begin{align*} \Pi_i(v_i)&amp;=\Pi_i(0)+\int_0^{v_i}P_i(x)dx \\
 v_i^2-b(v_i) &amp;= \int_0^{v_i}xdx \\
v_i^2-b(v_i) &amp;= \frac{1}{2}v_i^2 \\
b(v_i) &amp;= \frac{1}{2}v_i^2\end{align*}$$</span> which is the same result as Amit's answer.</p>
<p>The reason this approach is &quot;halfway&quot; between Amit's approach and the Revenue Equivalence Theorem is that <span class="math-container">$(\dagger)$</span> is the key equation for deriving the Revenue Equivalence Theorem. From <span class="math-container">$(\dagger)$</span>, we may conclude <span class="math-container">$\mathbb{E}_{v_{-i}}[t_i(v_i,v_{-i})]=\Pi_i(0)-v_iP_i(v_i)+\int_0^{v_i}P_i(x)dx$</span>, from which we infer that if <span class="math-container">$\Pi_i(0)$</span> and <span class="math-container">$P_i(\cdot)$</span> are the same in two mechanisms, the expected transfers (and hence also the expected revenue) are the same in the two mechanisms. So, in some sense the approach in this answer is only one step removed from directly using the Revenue Equivalence Theorem.</p>

CC BY-SA 4.0
2024-05-20T08:19:51.207Z
2024-05-20T08:19:51.207Z
Joseph Basford
<p>For the Bayesian all-pay-auction game above where valuations are private information, let <span class="math-container">$b:[0,1]\rightarrow \mathbb{R}_+$</span> be the symmetric equilibrium strategy which is strictly increasing and differentiable and is played by the two players. To find this strategy, consider an arbitrary realisation <span class="math-container">$v_1$</span> of player <span class="math-container">$1$</span>'s valuation. Given that player <span class="math-container">$2$</span> is playing according to the strategy <span class="math-container">$b:[0,1]\rightarrow \mathbb{R}_+$</span>, player <span class="math-container">$1$</span>'s expected payoff from bidding <span class="math-container">$b_1$</span> is as follows:</p>
<p><span class="math-container">\begin{eqnarray*} v_1\Pr(V_2\in\{v\in [0,1]|b(v)&lt;b_1\})-b_1 &amp; = &amp; v_1\Pr(b(V_2)&lt;b_1)-b_1 \\ &amp; = &amp; v_1\Pr(V_2&lt;b^{-1}(b_1))-b_1 \\ &amp; = &amp; v_1b^{-1}(b_1)-b_1\end{eqnarray*}</span></p>
<p>where <span class="math-container">$V_2$</span> is the valuation of player 2 and is a Unif(0,1) random variable.</p>
<p>To choose the <span class="math-container">$b_1$</span> that maximise the above expression, we differentiate it with respect to <span class="math-container">$b_1$</span> and get the FOC as:
<span class="math-container">$\frac{v_1}{b'(b^{-1}(b_1))}-1=0$</span></p>
<p>Using the fact that <span class="math-container">$b:[0,1]\rightarrow \mathbb{R}_+$</span> is the symmetric equilibrium strategy, we know that <span class="math-container">$b_1=b(v_1)$</span>. Substituting it in the above expression, the following is true about the equilibrium strategy <span class="math-container">$b:[0,1]\rightarrow \mathbb{R}_+$</span>:</p>
<p><span class="math-container">$\frac{v_1}{b'(b^{-1}(b(v_1)))}-1=0$</span></p>
<p>which gives</p>
<p><span class="math-container">$\frac{v_1}{b'(v_1)}-1=0$</span></p>
<p>i.e. <span class="math-container">$b'(v_1)=v_1$</span></p>
<p>Also note that <span class="math-container">$b(0)=0$</span> must hold.</p>
<p>This gives the equilibrium strategy <span class="math-container">$b(v_1)=\frac{v_1^2}{2}$</span>.</p>
<p>So, the two players use the equilibrium strategy: <span class="math-container">$b_i(v_i)=b(v_i)=\frac{v_i^2}{2}$</span> for <span class="math-container">$i\in\{1,2\}$</span>.</p>
<p>Expected Revenue from this auction is
<span class="math-container">$2\displaystyle\int_0^1\frac{v^2}{2}dv=\frac{1}{3}$</span></p>

CC BY-SA 4.0
2024-05-20T05:45:04.250Z
2024-05-20T06:06:46.763Z
2024-05-20T06:06:46.763Z
Amit
<p>In a sealed bid all pay auction, the highest bidder receives the good but
every buyer pays the seller the amount of her bid regardless of whether she
wins. Suppose there are two bidders whose valuations are independently
drawn from a uniform distribution on [0, 1].
a) Find an equilibrium bidding strategy (without using the revenue equivalence principle).
b) Compute the expected revenue from this auction.</p>
<p>Initially, my first thought is the expected utility from this auction is:
<span class="math-container">$F(x)(v-b(x))+(1-F(x))(-b(x))$</span>,
where <span class="math-container">$b(x)$</span> is my bid, <span class="math-container">$v$</span> is the valuation, and <span class="math-container">$F(x)=x$</span>.</p>
<p>Then, I would take the first order conditions and solve for <span class="math-container">$b(x)$</span>. Is this correct?</p>

<microeconomics><game-theory><auctions>
All-pay auction question problem
GraceLynn87
# All-pay auction question problem

(Asked by: username_0 on 2024-05-19)

In a sealed bid all pay auction, the highest bidder receives the good but
every buyer pays the seller the amount of her bid regardless of whether she
wins. Suppose there are two bidders whose valuations are independently
drawn from a uniform distribution on [0, 1].
a) Find an equilibrium bidding strategy (without using the revenue equivalence principle).
b) Compute the expected revenue from this auction.

Initially, my first thought is the expected utility from this auction is:
$F(x)(v-b(x))+(1-F(x))(-b(x))$,
where $b(x)$ is my bid, $v$ is the valuation, and $F(x)=x$.

Then, I would take the first order conditions and solve for $b(x)$. Is this correct?

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

---

## 2 answers

### Answer by username_1 on 2024-05-20. Score: 1

To supplement Amit's answer, we can get the same answer using Myerson's trick. This trick is "halfway" between Amit's answer and using the Revenue Equivalence Theorem.

First, we may consider the all-pay auction in a mechanism design framework. A general mechanism is a set of transfer and allocation rules $(x\_i,t\_i)\_{i=1}^2$ such that the mapping $[0,1]^2\owns (v\_i,v\_{-i})\mapsto x\_i(v\_i,v\_{-i})\in[0,1]$ gives the probability agent $i$ gets the good and the mapping $[0,1]^2\owns (v\_i,v\_{-i})\mapsto t\_i(v\_i,v\_{-i})\in \mathbb{R}$ gives the transfer to agent $i$ if the two agents announce their types to be $(v\_i,v\_{-i})$. Implicitly, this description of the mechanisms has applied the revelation principle to restrict attention to direct mechanisms.

Myerson (1981) showed that any direct incentive compatible and individually rational mechanism satisfies the following equality $$\Pi\_i(v\_i)=\Pi\_i(0)+\int\_0^{v\_i}P\_i(x)dx \qquad \qquad (\dagger)$$ where $\Pi\_i:[0,1]\to\mathbb{R}$ is the interim payoff function for agent $i$ reporting a certain type, and $P\_i:[0,1]\to [0,1]$ maps from agent $i$'s type report to interim probability agent $i$ receives the good. Formally, $P\_i(v\_i)=\mathbb{E}\_{v\_{-i}}[x\_i(v\_i,v\_{-i})]$ and $\Pi\_i(v\_i)=v\_iP\_i(v\_i)-\mathbb{E}\_{v\_{-i}}[t\_i(v\_i,v\_{-i})]$.

Consider now the case of the all-pay auction mechanism $(\hat{x}\_i,\hat{t}\_i)\_{i=1}^2$. Assuming a symmetric, strictly increasing, equilibrium bidding strategy $b\_i(v\_i):[0,1]\to \mathbb{R}\_+$, the allocation and transfer mechanisms are $$\begin{align\*} \hat{x}(v\_i,v\_{-i})&=\begin{cases} 1 & \text{ if } v\_i>v\_{-i} \\
0 & \text{ if } v\_i Using these, we may calculate $$\begin{align\*} \Pi\_i(v\_i) &= \int\_0^{v\_i}v\_i-b(v\_i)dv\_{-i}-\int\_{v\_i}^1 b(v\_i)dv\_{-i}=v\_i^2-b(v\_i) \\
P\_i(x)&=\mathbb{P}(v\_{-i} Additionally, since we have conjectured a strictly increasing bidding strategy, the $v=0$ type does not get the good with probability $1$. Therefore, the dominant strategy for the $v=0$ type is to bid $0$, so that $\Pi\_i(0)=0$. Therefore, plugging these expressions into $(\dagger)$, we find $$\begin{align\*} \Pi\_i(v\_i)&=\Pi\_i(0)+\int\_0^{v\_i}P\_i(x)dx \\
v\_i^2-b(v\_i) &= \int\_0^{v\_i}xdx \\
v\_i^2-b(v\_i) &= \frac{1}{2}v\_i^2 \\
b(v\_i) &= \frac{1}{2}v\_i^2\end{align\*}$$ which is the same result as Amit's answer.

The reason this approach is "halfway" between Amit's approach and the Revenue Equivalence Theorem is that $(\dagger)$ is the key equation for deriving the Revenue Equivalence Theorem. From $(\dagger)$, we may conclude $\mathbb{E}\_{v\_{-i}}[t\_i(v\_i,v\_{-i})]=\Pi\_i(0)-v\_iP\_i(v\_i)+\int\_0^{v\_i}P\_i(x)dx$, from which we infer that if $\Pi\_i(0)$ and $P\_i(\cdot)$ are the same in two mechanisms, the expected transfers (and hence also the expected revenue) are the same in the two mechanisms. So, in some sense the approach in this answer is only one step removed from directly using the Revenue Equivalence Theorem.


---

### Answer by username_2 on 2024-05-20. Score: 1

For the Bayesian all-pay-auction game above where valuations are private information, let $b:[0,1]\rightarrow \mathbb{R}\_+$ be the symmetric equilibrium strategy which is strictly increasing and differentiable and is played by the two players. To find this strategy, consider an arbitrary realisation $v\_1$ of player $1$'s valuation. Given that player $2$ is playing according to the strategy $b:[0,1]\rightarrow \mathbb{R}\_+$, player $1$'s expected payoff from bidding $b\_1$ is as follows:

\begin{eqnarray\*} v\_1\Pr(V\_2\in\{v\in [0,1]|b(v)

where $V\_2$ is the valuation of player 2 and is a Unif(0,1) random variable.

To choose the $b\_1$ that maximise the above expression, we differentiate it with respect to $b\_1$ and get the FOC as:
$\frac{v\_1}{b'(b^{-1}(b\_1))}-1=0$

Using the fact that $b:[0,1]\rightarrow \mathbb{R}\_+$ is the symmetric equilibrium strategy, we know that $b\_1=b(v\_1)$. Substituting it in the above expression, the following is true about the equilibrium strategy $b:[0,1]\rightarrow \mathbb{R}\_+$:

$\frac{v\_1}{b'(b^{-1}(b(v\_1)))}-1=0$

which gives

$\frac{v\_1}{b'(v\_1)}-1=0$

i.e. $b'(v\_1)=v\_1$

Also note that $b(0)=0$ must hold.

This gives the equilibrium strategy $b(v\_1)=\frac{v\_1^2}{2}$.

So, the two players use the equilibrium strategy: $b\_i(v\_i)=b(v\_i)=\frac{v\_i^2}{2}$ for $i\in\{1,2\}$.

Expected Revenue from this auction is
$2\displaystyle\int\_0^1\frac{v^2}{2}dv=\frac{1}{3}$


---

