<p>It is actually assumed that $b_i(v_i)$ is of the form $\alpha_i+\beta_i \cdot v_i$. So it is an affine function. Linearity only works if the bottom of the uniform distribution is 0.</p>

<p>A somewhat intuitive reasoning is that if the valuations are uniformly distributed over $[c,d]$ and $b_i(v_i)$ is a symmetric equilibrium, then $b_i(v_i)+k$ should define a symmetric equilibrium when valuations are uniform over $[c+k,d+k]$ as neither the probability of winning nor the gain changes.</p>

<p>Assuming a strategy of the form $\alpha_i+\beta_i \cdot v_i$ you get</p>

<p>$$
=Pr[\alpha_1 + \beta_1 \cdot v_1 &lt; \alpha_2 + \beta_2 \cdot v_2] \cdot ((1- \beta_2) \cdot v_2 - \alpha_2) 
$$</p>

<p>$$
=Pr\left[v_1 &lt; \frac{\alpha_2 - \alpha_1 + \beta_2 \cdot v_2}{\beta_1} \right] \cdot ((1- \beta_2) \cdot v_2 - \alpha_2) 
$$</p>

<p>$$
=\frac{\frac{\alpha_2 - \alpha_1 + \beta_2 \cdot v_2}{\beta_1}-c}{d-c}\cdot ((1- \beta_2) \cdot v_2 - \alpha_2)
$$</p>

<p>Taking derivative w.r.t. $\alpha_2$ and $\beta_2$ you get your first order conditions. Assuming symmetry you get a unique solution. If $c = 0$ then $\alpha_1 = \alpha_2 = 0$.</p>

<p>By the way, I prefer treating $b_2(v_2)$ as the decision variable, then you can
$$
\max_{b_2(v_2)} 
Pr\left[v_1 &lt; \frac{b_2(v_2) - \alpha_1}{\beta_1} \right] \cdot (v_2 - b_2(v_2)) 
$$
and the last fraction is somewhat simpler.</p>

CC BY-SA 3.0
2018-04-12T13:10:34.460Z
Thanks! So $\alpha_i$ was the missing link in my solution.
Erel Segal-Halevi
CC BY-SA 3.0
2018-04-12T11:05:11.170Z
2018-04-13T17:01:28.603Z
2018-04-13T17:01:28.603Z
Giskard
<p>In a famous textbook example of a Bayesian-Nash equilibrium, there is a first-price auction with two independent players. Each player $i$ values the item as $v_i$, which is  distributed uniformly in $[0,1]$. It is assumed that the strategy of each player $i$ is to bid a fraction of his value, i.e:</p>

<p>$$ b_i(v_i) := a_i \cdot v_i$$</p>

<p>for some constant $a_i$. Then, we can conclude that, for every $a_1$, the best response of player 2 is to pick $a_2=1/2$. Hence, the only Bayes-Nash equilibrium of this form is $a_1=1/2, a_2=1/2$.</p>

<p>Now, I tried to slightly generalize the example by assuming the values of the players are distributed uniformly in $[c,d]$, for some constants $d&gt;c&gt;0$. But I did not find an equilibrium. Suppose player 1 bids $ a_i \cdot v_i$ and player 2 bids $ $a_2 \cdot v_2$. Then the expected utility of player 2 is:</p>

<p>$$ProbOfWinning(a_2)\cdot GainWhenWinning(a_2)$$</p>

<p>$$=Pr[a_1 v_1 &lt; a_2 v_2]\cdot (1-a_2)v_2$$</p>

<p>$$=\frac{(a_2 v_2 /a_1)-c}{d-c}\cdot (1-a_2)v_2$$</p>

<p>Taking the derivative of this expression w.r.t $a_2$ gives that the best response of player 2 is:</p>

<p>$$
a_2 = {1\over 2} + {a_1 c\over 2 v_2}
$$</p>

<p>so the best $a_2$ depends on $v_2$. This means that the function from value to action is not linear as I initially assumed.</p>

<p>Do I have any mistake in the calculation? What is a Bayes-Nash equilibrium in this auction?</p>

<nash-equilibrium><auctions><bayesian-game>
Bayesian-Nash equilibrium in a first-price auction
Erel Segal-Halevi
# Bayesian-Nash equilibrium in a first-price auction

(Asked by: username_0 on 2018-04-12)

In a famous textbook example of a Bayesian-Nash equilibrium, there is a first-price auction with two independent players. Each player $i$ values the item as $v\_i$, which is distributed uniformly in $[0,1]$. It is assumed that the strategy of each player $i$ is to bid a fraction of his value, i.e:

$$ b\_i(v\_i) := a\_i \cdot v\_i$$

for some constant $a\_i$. Then, we can conclude that, for every $a\_1$, the best response of player 2 is to pick $a\_2=1/2$. Hence, the only Bayes-Nash equilibrium of this form is $a\_1=1/2, a\_2=1/2$.

Now, I tried to slightly generalize the example by assuming the values of the players are distributed uniformly in $[c,d]$, for some constants $d>c>0$. But I did not find an equilibrium. Suppose player 1 bids $ a\_i \cdot v\_i$ and player 2 bids $ $a\_2 \cdot v\_2$. Then the expected utility of player 2 is:

$$ProbOfWinning(a\_2)\cdot GainWhenWinning(a\_2)$$

$$=Pr[a\_1 v\_1 < a\_2 v\_2]\cdot (1-a\_2)v\_2$$

$$=\frac{(a\_2 v\_2 /a\_1)-c}{d-c}\cdot (1-a\_2)v\_2$$

Taking the derivative of this expression w.r.t $a\_2$ gives that the best response of player 2 is:

$$
a\_2 = {1\over 2} + {a\_1 c\over 2 v\_2}
$$

so the best $a\_2$ depends on $v\_2$. This means that the function from value to action is not linear as I initially assumed.

Do I have any mistake in the calculation? What is a Bayes-Nash equilibrium in this auction?

**Tags:** nash-equilibrium, auctions, bayesian-game


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## 1 answers

### Answer by username_1 on 2018-04-12. Score: 6 | Accepted answer

It is actually assumed that $b\_i(v\_i)$ is of the form $\alpha\_i+\beta\_i \cdot v\_i$. So it is an affine function. Linearity only works if the bottom of the uniform distribution is 0.

A somewhat intuitive reasoning is that if the valuations are uniformly distributed over $[c,d]$ and $b\_i(v\_i)$ is a symmetric equilibrium, then $b\_i(v\_i)+k$ should define a symmetric equilibrium when valuations are uniform over $[c+k,d+k]$ as neither the probability of winning nor the gain changes.

Assuming a strategy of the form $\alpha\_i+\beta\_i \cdot v\_i$ you get

$$
=Pr[\alpha\_1 + \beta\_1 \cdot v\_1 < \alpha\_2 + \beta\_2 \cdot v\_2] \cdot ((1- \beta\_2) \cdot v\_2 - \alpha\_2)
$$

$$
=Pr\left[v\_1 < \frac{\alpha\_2 - \alpha\_1 + \beta\_2 \cdot v\_2}{\beta\_1} \right] \cdot ((1- \beta\_2) \cdot v\_2 - \alpha\_2)
$$

$$
=\frac{\frac{\alpha\_2 - \alpha\_1 + \beta\_2 \cdot v\_2}{\beta\_1}-c}{d-c}\cdot ((1- \beta\_2) \cdot v\_2 - \alpha\_2)
$$

Taking derivative w.r.t. $\alpha\_2$ and $\beta\_2$ you get your first order conditions. Assuming symmetry you get a unique solution. If $c = 0$ then $\alpha\_1 = \alpha\_2 = 0$.

By the way, I prefer treating $b\_2(v\_2)$ as the decision variable, then you can
$$
\max\_{b\_2(v\_2)}
Pr\left[v\_1 < \frac{b\_2(v\_2) - \alpha\_1}{\beta\_1} \right] \cdot (v\_2 - b\_2(v\_2))
$$
and the last fraction is somewhat simpler.

> **Comments:**
> **username_0**: Thanks! So $\alpha_i$ was the missing link in my solution.  

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