<p>Let's forget about the reserve price for a second.</p>

<p>Suppose there is a symmetric increasing equilibrium bidding strategy $b(v)$. We can think of a bidder with value $v$ as choosing which $\tilde{v}$ he will "pretend to be" by submitting bid $b(\tilde{v})$. The bidder wins iff $\tilde{v}&gt;v_j$ (where $v_j$ is his rival's value). Thus, the payoff is</p>

<p>$$(v-b(\tilde{v}))\Pr(\tilde{v}&gt;v_j)=(v-b(\tilde{v}))\tilde{v}.$$</p>

<p>Now compute the FOC for choice of $\tilde{v}$:</p>

<p>$$v-b(\tilde{v})-\tilde{v}b'(\tilde{v})=0.$$</p>

<p>If $b$ is really an equilibrium then the optimal choice of $\tilde{v}$ should be the bidder's true value, $v$: $v-b(v)-vb'(v)=0$. This is a differential equation. To solve it we just need a boundary condition.</p>

<p>Anyone with $v&lt;r$ will not bid. So the set of bidders will be the truncated uniform distribution on $[r,1]$. Intuitively, a bidder with value $r$ cannot earn a positive surplus, so a reasonable boundary condition seems to be $b(r)=r$.</p>

<p>Solving the initial value problem
$$v-b(v)-vb'(v)=0,\ b(r)=r$$</p>

<p>yields the equilibrium $$b(v)=\frac{r^2+v^2}{2v}.$$</p>

CC BY-SA 4.0
2018-07-11T09:32:45.937Z
2018-07-11T09:32:45.937Z
Ubiquitous
<p>I'm trying to solve the following question: </p>

<p>"The seller wants to auction off a single item to two bidders, the valuation of each bidder is an iid draw from a uniform distribution on $[0,1]$ where the seller sets a reserve price, $r$, and has zero valuation for the item. </p>

<p>I was able to solve for the second price auction, such that the equilibrium bidding function is to bid your valuation for $v&gt;r$ and to not enter if $v&lt;r$. </p>

<p>For the first price auction, I usually go about solving for the bidding function by first calculating the probability that bidder i wins: </p>

<p>$\Pr(i\text{ wins})=\Pr(v_i &gt; r) * \Pr(v_i &gt; v_j)$.</p>

<p>Where I know that $\Pr(v_i &gt; v_j) = (1-b^{-1}(b))$ which I can then use to derive the bidding function as usual in an ordinary FPA $(0,0)$. </p>

<p>My question is, how do I work the reserve price into the bidding function. </p>

CC BY-SA 4.0
2018-07-10T18:53:19.497Z
The reserve price $r$ is already there in your equation for Pr(i wins). What exactly are you asking here?
Giskard
<microeconomics><game-theory><auctions>
First Price Auction with Reserve Price
A. D. 
# First Price Auction with Reserve Price

(Asked by: username_0 on 2018-07-10)

I'm trying to solve the following question:

"The seller wants to auction off a single item to two bidders, the valuation of each bidder is an iid draw from a uniform distribution on $[0,1]$ where the seller sets a reserve price, $r$, and has zero valuation for the item.

I was able to solve for the second price auction, such that the equilibrium bidding function is to bid your valuation for $v>r$ and to not enter if $v

For the first price auction, I usually go about solving for the bidding function by first calculating the probability that bidder i wins:

$\Pr(i\text{ wins})=\Pr(v\_i > r) \* \Pr(v\_i > v\_j)$.

Where I know that $\Pr(v\_i > v\_j) = (1-b^{-1}(b))$ which I can then use to derive the bidding function as usual in an ordinary FPA $(0,0)$.

My question is, how do I work the reserve price into the bidding function.

**Tags:** microeconomics, game-theory, auctions
> **Comments:**
> **username_1**: The reserve price $r$ is already there in your equation for Pr(i wins). What exactly are you asking here?  

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## 1 answers

### Answer by username_2 on 2018-07-11. Score: 2

Let's forget about the reserve price for a second.

Suppose there is a symmetric increasing equilibrium bidding strategy $b(v)$. We can think of a bidder with value $v$ as choosing which $\tilde{v}$ he will "pretend to be" by submitting bid $b(\tilde{v})$. The bidder wins iff $\tilde{v}>v\_j$ (where $v\_j$ is his rival's value). Thus, the payoff is

$$(v-b(\tilde{v}))\Pr(\tilde{v}>v\_j)=(v-b(\tilde{v}))\tilde{v}.$$

Now compute the FOC for choice of $\tilde{v}$:

$$v-b(\tilde{v})-\tilde{v}b'(\tilde{v})=0.$$

If $b$ is really an equilibrium then the optimal choice of $\tilde{v}$ should be the bidder's true value, $v$: $v-b(v)-vb'(v)=0$. This is a differential equation. To solve it we just need a boundary condition.

Anyone with $v

Solving the initial value problem
$$v-b(v)-vb'(v)=0,\ b(r)=r$$

yields the equilibrium $$b(v)=\frac{r^2+v^2}{2v}.$$


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