# Sealed bid first price auction with 2 players.

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].

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.