# All-pay auction question problem

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?

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 $$2\displaystyle\int_0^1\frac{v^2}{2}dv=\frac{1}{3}$$

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.