<p>There is another way to compute the symmetric BNE <strong>in increasing strategy</strong>.</p>
<p>Let <span class="math-container">$U(v)$</span> denote the expected utility of a player in equilibrium when his type is <span class="math-container">$v$</span>:  Given that the bidding strategy is increasing, a player with type <span class="math-container">$0$</span> will get the good with probability zero.</p>
<p>Thus he/she must bid zero and <span class="math-container">$U(0) = 0$</span>. For any other <span class="math-container">$v &gt; 0$</span>, the probability that the player
gets the good is <span class="math-container">$Q (v) = v^{n-1}$</span> (this is the probability that all the other players have a type
lower than <span class="math-container">$v$</span>) From the classes on mechanism design, we know that:</p>
<blockquote>
<p><span class="math-container">$U(v) = U(0) +  \int_{0}^{v} Q(x) \,dx = \int_{0}^{v} x^{n-1} \,dx  = 
 \frac{v^{n}}{n}$</span></p>
</blockquote>
<p>On the other hand, we can write <span class="math-container">$U(v)$</span> as</p>
<blockquote>
<p><span class="math-container">$U(v)= (v-\alpha b(v)))v^{n-1}$</span></p>
</blockquote>
<p>Therefore</p>
<blockquote>
<p><span class="math-container">$(v-\alpha b(v))v^{n-1} = \frac{v^{n}}{n}$</span></p>
</blockquote>
<p>and</p>
<p><strong><span class="math-container">$b(v) = \frac{n-1}{n \alpha} v$</span></strong></p>

CC BY-SA 4.0
2021-07-23T17:05:09.483Z
"*Thus he/she must bid zero*" Why though? If everyone with valuation $0$ would bid $1/10$ and follow an increasing strategy, bidding $1/10$ would still get you the item with probability zero. It seems like this line is not necessary to declare $U(0) = 0$. The player wins with probability $0$, thus they get an *expected* utility of $0$.
Giskard
CC BY-SA 4.0
2021-07-23T11:37:16.927Z
2021-07-23T17:08:10.907Z
2021-07-23T17:08:10.907Z
Lorenzo Castagno
<p>My problem.</p>
<p>Consider the following auction for a single object. There are <span class="math-container">$n \geq 2$</span>
bidders. They submit their bids simultaneously. The object is allocated to the player
who submits the largest bid. If the winner's bid is <span class="math-container">$b$</span> he pays the amount <span class="math-container">$\alpha b$</span> where <span class="math-container">$\alpha$</span> is a positive number. The losers do not pay anything. Ties are broken randomly, with
equal probabilities among all the players who submit the largest bid.
The bidders' valuations for the object are private information. In particular, each
player <span class="math-container">$i$</span> knows his own valuation <span class="math-container">$v_{i}$</span> which is distributed uniformly over the unit interval.
The valuations are distributed independently across the players.</p>
<p>Construct the symmetric BNE of the game. (Assume that the bidding strategy
<span class="math-container">$b :[0,1]$</span> <span class="math-container">$\rightarrow$</span> <span class="math-container">$\mathbb{R}$</span> is increasing).</p>
<p>Solution</p>
<p>Let <span class="math-container">$b :[0,1]$</span> <span class="math-container">$\rightarrow$</span> <span class="math-container">$\mathbb{R}$</span>  R denote the equilibrium bidding strategy. Then for every <span class="math-container">$v \in [0, 1]$</span>, we must have:</p>
<blockquote>
<p><span class="math-container">$v=arg$</span> <span class="math-container">$max_{w}((v-\alpha b(w))w^{n-1}$</span></p>
</blockquote>
<p>We compute the first order conditions at <span class="math-container">$v$</span> and obtain:</p>
<blockquote>
<p><span class="math-container">$-\alpha b^{'}(v)v^{n-1}+(n-1)(v-\alpha b(v))v^{n-2}=0$</span></p>
</blockquote>
<p>Which we can simplify as:</p>
<blockquote>
<p><span class="math-container">$-\alpha b^{'}(v)v+(n-1)(v-\alpha b(v))=0$</span></p>
</blockquote>
<p>The solution to this differential equation is linear: <span class="math-container">$b (v) = Av$</span> where <span class="math-container">$A$</span> satisfies</p>
<blockquote>
<p><span class="math-container">$-\alpha Av+(n-1)(v-\alpha Av)=0$</span></p>
</blockquote>
<p>Thus, the equilibrium bidding strategy is</p>
<blockquote>
<p><span class="math-container">$b(v)=\frac{n-1}{n \alpha}v$</span></p>
</blockquote>
<p>Is there ant alternative way to calculate the BNE for the problem presented above</p>

<microeconomics><mathematical-economics><game-theory><bayesian-game>
Alternative way to calculate the symmetric BNE of the game
DanIsabel
# Alternative way to calculate the symmetric BNE of the game

(Asked by: username_0 on 2021-07-23)

My problem.

Consider the following auction for a single object. There are $n \geq 2$
bidders. They submit their bids simultaneously. The object is allocated to the player
who submits the largest bid. If the winner's bid is $b$ he pays the amount $\alpha b$ where $\alpha$ is a positive number. The losers do not pay anything. Ties are broken randomly, with
equal probabilities among all the players who submit the largest bid.
The bidders' valuations for the object are private information. In particular, each
player $i$ knows his own valuation $v\_{i}$ which is distributed uniformly over the unit interval.
The valuations are distributed independently across the players.

Construct the symmetric BNE of the game. (Assume that the bidding strategy
$b :[0,1]$ $\rightarrow$ $\mathbb{R}$ is increasing).

Solution

Let $b :[0,1]$ $\rightarrow$ $\mathbb{R}$ R denote the equilibrium bidding strategy. Then for every $v \in [0, 1]$, we must have:

> $v=arg$ $max\_{w}((v-\alpha b(w))w^{n-1}$

We compute the first order conditions at $v$ and obtain:

> $-\alpha b^{'}(v)v^{n-1}+(n-1)(v-\alpha b(v))v^{n-2}=0$

Which we can simplify as:

> $-\alpha b^{'}(v)v+(n-1)(v-\alpha b(v))=0$

The solution to this differential equation is linear: $b (v) = Av$ where $A$ satisfies

> $-\alpha Av+(n-1)(v-\alpha Av)=0$

Thus, the equilibrium bidding strategy is

> $b(v)=\frac{n-1}{n \alpha}v$

Is there ant alternative way to calculate the BNE for the problem presented above

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

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## 1 answers

### Answer by username_1 on 2021-07-23. Score: 4 | Accepted answer

There is another way to compute the symmetric BNE **in increasing strategy**.

Let $U(v)$ denote the expected utility of a player in equilibrium when his type is $v$: Given that the bidding strategy is increasing, a player with type $0$ will get the good with probability zero.

Thus he/she must bid zero and $U(0) = 0$. For any other $v > 0$, the probability that the player
gets the good is $Q (v) = v^{n-1}$ (this is the probability that all the other players have a type
lower than $v$) From the classes on mechanism design, we know that:

> $U(v) = U(0) + \int\_{0}^{v} Q(x) \,dx = \int\_{0}^{v} x^{n-1} \,dx =
> \frac{v^{n}}{n}$

On the other hand, we can write $U(v)$ as

> $U(v)= (v-\alpha b(v)))v^{n-1}$

Therefore

> $(v-\alpha b(v))v^{n-1} = \frac{v^{n}}{n}$

and

**$b(v) = \frac{n-1}{n \alpha} v$**

> **Comments:**
> **username_2**: "*Thus he/she must bid zero*" Why though? If everyone with valuation $0$ would bid $1/10$ and follow an increasing strategy, bidding $1/10$ would still get you the item with probability zero. It seems like this line is not necessary to declare $U(0) = 0$. The player wins with probability $0$, thus they get an *expected* utility of $0$.  

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