<p>Assuming the auction has a pure strategy equilibrium, we can set up the problem as follows:</p>

<p>Each bidder has a value $v_i\sim F$ on support $[\underline{v},\overline{v}]$. The symmetric equilibrium, we posit, is an increasing continuous function $b(v)$.</p>

<p>Now, imagine bidder $i$ bids <em>as if</em> his value were $\widetilde{v}$. Then he wins iff $v_j&lt;\widetilde{v}$ and $i$'s expected payoff is therefore</p>

<p>$$U_i=-[1-F(\widetilde{v})]b(\widetilde{v})+\int_0^{\widetilde{v}}\! \left(v_i-b(v_j)\right)F'(v_j)\,dv_j$$.</p>

<p>The first term is the payoff if $i$ loses (he then pays his own bid, which is $b(\widetilde{v})$ sinc $i$ is using $\widetilde{v}$'s bidding strategy). The second term is the payoff if he wins (he pays his rival's bid, provided his rival's value is less than $\widetilde{v}$).</p>

<p>Now we compute the best $\widetilde{v}$ for $i$ to choose by computing a first-order condition:
$$\frac{\partial U_i}{\partial \widetilde{v}}=-[1-F(\widetilde{v})] b'(\widetilde{v})+v F'(\widetilde{v})=0.$$</p>

<p>We know that, in equilibrium, it must be optimal for $i$ to choose $\widetilde{v}=v_i$ (i.e., it must be a best response for $i$ to use his intended strategy rather than deviate to someone else's strategy):</p>

<p>$$-[1-F(v)] b'(v)+v F'(v)=0.$$</p>

<p>Thus,
$$b(v)=b(\underline{v})+\int_{\underline{v}}^v\!\frac{xF'(x)}{1-F(x)}\,dx.$$</p>

<p>We can check that $$b'(v)=\frac{vF'(v)}{1-F(v)}&gt;0$$ as required. All we need is a boundary condition to pin-down $b(\underline{v})$. For example, we might think it natural that $b(\underline{v})=0$ because a $\underline{v}$ type knows he can never win the auction.</p>

CC BY-SA 3.0
2018-01-27T13:00:44.313Z
2018-01-27T13:06:38.673Z
2018-01-27T13:06:38.673Z
Ubiquitous
<p>I am trying to find Nash Equilibrium of an auction with two bidders in which the highest bidder wins the object but both bidders pay the losing bid. Here every bidder follows the same bidding strategy β which is a strictly increasing and differentiable map from values to bids.</p>

<p>I am unable to find the Nash equilibrium of the game. Can someone help me with this? Is this some kind of standard Auction?</p>

CC BY-SA 3.0
2018-01-27T12:26:23.137Z
Is it a sealed-bid auction or an open one?
Ubiquitous
CC BY-SA 3.0
2018-01-27T12:56:54.933Z
@Ubiquitous It's sealed Bid auction.
guest4321
<game-theory><nash-equilibrium><auctions>
Nash Equilibrium in 2 bidder auction
guest4321
# Nash Equilibrium in 2 bidder auction

(Asked by: username_0 on 2018-01-27)

I am trying to find Nash Equilibrium of an auction with two bidders in which the highest bidder wins the object but both bidders pay the losing bid. Here every bidder follows the same bidding strategy β which is a strictly increasing and differentiable map from values to bids.

I am unable to find the Nash equilibrium of the game. Can someone help me with this? Is this some kind of standard Auction?

**Tags:** game-theory, nash-equilibrium, auctions
> **Comments:**
> **username_1**: Is it a sealed-bid auction or an open one?  
> **username_0**: @Ubiquitous It's sealed Bid auction.  

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## 1 answers

### Answer by username_1 on 2018-01-27. Score: 3

Assuming the auction has a pure strategy equilibrium, we can set up the problem as follows:

Each bidder has a value $v\_i\sim F$ on support $[\underline{v},\overline{v}]$. The symmetric equilibrium, we posit, is an increasing continuous function $b(v)$.

Now, imagine bidder $i$ bids *as if* his value were $\widetilde{v}$. Then he wins iff $v\_j<\widetilde{v}$ and $i$'s expected payoff is therefore

$$U\_i=-[1-F(\widetilde{v})]b(\widetilde{v})+\int\_0^{\widetilde{v}}\! \left(v\_i-b(v\_j)\right)F'(v\_j)\,dv\_j$$.

The first term is the payoff if $i$ loses (he then pays his own bid, which is $b(\widetilde{v})$ sinc $i$ is using $\widetilde{v}$'s bidding strategy). The second term is the payoff if he wins (he pays his rival's bid, provided his rival's value is less than $\widetilde{v}$).

Now we compute the best $\widetilde{v}$ for $i$ to choose by computing a first-order condition:
$$\frac{\partial U\_i}{\partial \widetilde{v}}=-[1-F(\widetilde{v})] b'(\widetilde{v})+v F'(\widetilde{v})=0.$$

We know that, in equilibrium, it must be optimal for $i$ to choose $\widetilde{v}=v\_i$ (i.e., it must be a best response for $i$ to use his intended strategy rather than deviate to someone else's strategy):

$$-[1-F(v)] b'(v)+v F'(v)=0.$$

Thus,
$$b(v)=b(\underline{v})+\int\_{\underline{v}}^v\!\frac{xF'(x)}{1-F(x)}\,dx.$$

We can check that $$b'(v)=\frac{vF'(v)}{1-F(v)}>0$$ as required. All we need is a boundary condition to pin-down $b(\underline{v})$. For example, we might think it natural that $b(\underline{v})=0$ because a $\underline{v}$ type knows he can never win the auction.


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