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?


1 Answer 1


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.

  • $\begingroup$ Thanks! So $\alpha_i$ was the missing link in my solution. $\endgroup$ Apr 12, 2018 at 13:10

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