<p>We are given that <span class="math-container">$V_1, V_2$</span> are i.i.d <span class="math-container">$\text{Unif}(0,1)$</span>. We can look for a Bayesian Nash equilibrium where players will play a strategy of the following type:
<span class="math-container">\begin{eqnarray*} b_i(v_i) = \begin{cases} \frac{1}{4} &amp; \text{if } v_i &gt; c_i \\ 0 &amp; \text{if } v_i \leq c_i \end{cases} \end{eqnarray*}</span></p>
<p>i.e. they will bid <span class="math-container">$\frac{1}{4}$</span> if and only if their valuations are high enough.</p>
<p>We just need to find these cut-offs <span class="math-container">$c_1 \in [0,1]$</span> and <span class="math-container">$c_2 \in [0,1]$</span> for the two players.</p>
<p>Given that player <span class="math-container">$2$</span>'s cutoff is <span class="math-container">$c_2$</span> and player <span class="math-container">$1$</span>'s valuation is <span class="math-container">$v_1$</span>,</p>
<ul>
<li>player <span class="math-container">$1$</span>'s expected payoff from bidding <span class="math-container">$0$</span> equals <span class="math-container">$\frac{1}{4}\Pr(V_2 &gt; c_2)+v_1\left(\frac{1}{2}\Pr(V_2 \leq c_2)\right) = \frac{1}{4}(1-c_2) + \frac{1}{2}v_1c_2 = \frac{1}{4}-\frac{1}{4}c_2+\frac{1}{2}v_1c_2$</span></li>
<li>player <span class="math-container">$1$</span>'s expected payoff from bidding <span class="math-container">$\frac{1}{4}$</span> equals <span class="math-container">$\frac{1}{4}\left(\frac{1}{2}\Pr(V_2 &gt; c_2)\right)+\left(v_1-\frac{1}{4}\right)\left(\frac{1}{2}\Pr(V_2 &gt; c_2)+\Pr(V_2 \leq c_2)\right) = \frac{1}{8}(1-c_2) + \left(v_1-\frac{1}{4}\right)\left(\frac{1}{2}(1+c_2)\right) = -\frac{1}{4}c_2+\frac{1}{2}v_1+\frac{1}{2}v_1c_2$</span></li>
</ul>
<p>So, we get that player <span class="math-container">$1$</span> is indifferent between the two bids at <span class="math-container">$v_1=\frac{1}{2}$</span>. Therefore, his best response choice of cutoff is <span class="math-container">$c_1=\frac{1}{2}$</span>. By symmetry, <span class="math-container">$c_2=\frac{1}{2}$</span>. Therefore, Bayesian Nash Equilibrium strategy of each player <span class="math-container">$i\in\{1,2\}$</span> is as follows:
<span class="math-container">\begin{eqnarray*} b_i(v_i) = \begin{cases} \frac{1}{4} &amp; \text{if } v_i &gt; \frac{1}{2} \\ 0 &amp; \text{if } v_i \leq \frac{1}{2} \end{cases} \end{eqnarray*}</span></p>
<p>Please note that at the cutoff valuation <span class="math-container">$\frac{1}{2}$</span>, since the players are indifferent between the available actions, so they can choose any of the two actions at that valuation and therefore, there are a total of <span class="math-container">$4$</span> pure-strategy Bayesian Nash equilibrium profiles.</p>

CC BY-SA 4.0
2023-06-04T12:12:44.830Z
2023-06-04T12:34:16.617Z
2023-06-04T12:34:16.617Z
Amit
<p>I really couldn't solve this problem. Can someone help me with this?</p>
<blockquote>
<p>Consider the sealed-bid auction for a single item, where there are only two
allowable bids. There are two risk-neutral bidders who have valuations that are private info and which are drawn from i.i.d random variables that are uniformly distributed over <span class="math-container">$[0,1]$</span>. After observing their own valuation, each of the two bidders simultaneously and individually submit a bid selected from the two-element set <span class="math-container">$\{0,1/4\}$</span>. The high bidder wins the item and pays the amount of her bid. However, instead of paying his price to the seller, he pays to the other bidder. So the losing bidder receives the other bidder's payment. When tied, each bidder takes the role of the winning bidder with probability <span class="math-container">$1/2$</span> and takes the role of the losing bidder with prob. <span class="math-container">$1/2$</span>. Solve for Bayesian Nash Equilibrium.</p>
</blockquote>

<microeconomics><game-theory><bayesian-game>
First price sealed-bid auction with discrete bids
Hssktr
# First price sealed-bid auction with discrete bids

(Asked by: username_0 on 2023-05-22)

I really couldn't solve this problem. Can someone help me with this?

> Consider the sealed-bid auction for a single item, where there are only two
> allowable bids. There are two risk-neutral bidders who have valuations that are private info and which are drawn from i.i.d random variables that are uniformly distributed over $[0,1]$. After observing their own valuation, each of the two bidders simultaneously and individually submit a bid selected from the two-element set $\{0,1/4\}$. The high bidder wins the item and pays the amount of her bid. However, instead of paying his price to the seller, he pays to the other bidder. So the losing bidder receives the other bidder's payment. When tied, each bidder takes the role of the winning bidder with probability $1/2$ and takes the role of the losing bidder with prob. $1/2$. Solve for Bayesian Nash Equilibrium.

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

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## 1 answers

### Answer by username_1 on 2023-06-04. Score: 2

We are given that $V\_1, V\_2$ are i.i.d $\text{Unif}(0,1)$. We can look for a Bayesian Nash equilibrium where players will play a strategy of the following type:
\begin{eqnarray\*} b\_i(v\_i) = \begin{cases} \frac{1}{4} & \text{if } v\_i > c\_i \\ 0 & \text{if } v\_i \leq c\_i \end{cases} \end{eqnarray\*}

i.e. they will bid $\frac{1}{4}$ if and only if their valuations are high enough.

We just need to find these cut-offs $c\_1 \in [0,1]$ and $c\_2 \in [0,1]$ for the two players.

Given that player $2$'s cutoff is $c\_2$ and player $1$'s valuation is $v\_1$,

* player $1$'s expected payoff from bidding $0$ equals $\frac{1}{4}\Pr(V\_2 > c\_2)+v\_1\left(\frac{1}{2}\Pr(V\_2 \leq c\_2)\right) = \frac{1}{4}(1-c\_2) + \frac{1}{2}v\_1c\_2 = \frac{1}{4}-\frac{1}{4}c\_2+\frac{1}{2}v\_1c\_2$
* player $1$'s expected payoff from bidding $\frac{1}{4}$ equals $\frac{1}{4}\left(\frac{1}{2}\Pr(V\_2 > c\_2)\right)+\left(v\_1-\frac{1}{4}\right)\left(\frac{1}{2}\Pr(V\_2 > c\_2)+\Pr(V\_2 \leq c\_2)\right) = \frac{1}{8}(1-c\_2) + \left(v\_1-\frac{1}{4}\right)\left(\frac{1}{2}(1+c\_2)\right) = -\frac{1}{4}c\_2+\frac{1}{2}v\_1+\frac{1}{2}v\_1c\_2$

So, we get that player $1$ is indifferent between the two bids at $v\_1=\frac{1}{2}$. Therefore, his best response choice of cutoff is $c\_1=\frac{1}{2}$. By symmetry, $c\_2=\frac{1}{2}$. Therefore, Bayesian Nash Equilibrium strategy of each player $i\in\{1,2\}$ is as follows:
\begin{eqnarray\*} b\_i(v\_i) = \begin{cases} \frac{1}{4} & \text{if } v\_i > \frac{1}{2} \\ 0 & \text{if } v\_i \leq \frac{1}{2} \end{cases} \end{eqnarray\*}

Please note that at the cutoff valuation $\frac{1}{2}$, since the players are indifferent between the available actions, so they can choose any of the two actions at that valuation and therefore, there are a total of $4$ pure-strategy Bayesian Nash equilibrium profiles.


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