Sealed bid first price auction with 2 players.

577 Views Asked by At

I would like help with the following question; Consider a sealed bid first price auction with 2 players in which the valuation of each of the players is best described by a uniform distribution on [10, 30].
Identify a Nash equilibrium and show that this strategy profile is indeed a Nash equilibrium.

I know how to do it for [0,30], however I am unsure how to do it for [10,30].

1

There are 1 best solutions below

0
On

Let $b_A(x)$ and $b_B(x)$ denote the bids that players $A$ and $B$ make, respectively, when their valuation is $x$. We can assume that these are strictly monotonically increasing functions (and later check for consistency). Then the expected profit of player $A$ with valuation $x$ is

$$ \frac{b_B^{-1}(b_A(x))-10}{30-10}(x-b_A(x))\;. $$

Multiplying by $20$ and setting the derivative with respect to $b_A(x)$ to zero yields

$$ {b_B^{-1}}'(b_A(x))(x-b_A(x))-\left(b_B^{-1}(b_A(x))-10\right)=0\;. $$

In equilibrium, $b_B\equiv b_A\equiv b$, which yields

$$ \frac{x-b(x)}{b'(x)}-(x-10)=0\;, $$

or

$$ b'(x)+\frac{b(x)}{x-10}=\frac x{x-10}\;. $$

The general solution of the homogeneous equation is $b(x)=\frac c{x-10}$. The ansatz $b(x)=mx+n$ yields the particular solution $b(x)=\frac x2+5$ of the inhomogeneous equation, so the general solution is $b(x)=\frac x2+5+\frac c{x-10}$. Since $b(x)$ has a pole at $x=10$ for $c\ne0$, we choose $c=0$ to obtain $b(x)=\frac x2+5$. If player $B$ uses this strategy and player $A$ bids $b_A(x)$, the expected profit of player $A$ is

$$ \frac{2b_A(x)-20}{30-10}(x-b_A(x))\;, $$

which is quadratic in $b_A(x)$ with negative coefficient. Thus the stationary point found above is a global maximum, and the strategy profile $(b,b)$ with $b(x)=\frac x2+5$ is indeed a Nash equilibrium.