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I'm trying to solve the following question:

"The seller wants to auction off a single item to two bidders, the valuation of each bidder is an iid draw from a uniform distribution on $[0,1]$ where the seller sets a reserve price, $r$, and has zero valuation for the item.

I was able to solve for the second price auction, such that the equilibrium bidding function is to bid your valuation for $v>r$ and to not enter if $v<r$.

For the first price auction, I usually go about solving for the bidding function by first calculating the probability that bidder i wins:

$\Pr(i\text{ wins})=\Pr(v_i > r) * \Pr(v_i > v_j)$.

Where I know that $\Pr(v_i > v_j) = (1-b^{-1}(b))$ which I can then use to derive the bidding function as usual in an ordinary FPA $(0,0)$.

My question is, how do I work the reserve price into the bidding function.

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  • $\begingroup$ The reserve price $r$ is already there in your equation for Pr(i wins). What exactly are you asking here? $\endgroup$ – Giskard Jul 10 '18 at 18:53
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Let's forget about the reserve price for a second.

Suppose there is a symmetric increasing equilibrium bidding strategy $b(v)$. We can think of a bidder with value $v$ as choosing which $\tilde{v}$ he will "pretend to be" by submitting bid $b(\tilde{v})$. The bidder wins iff $\tilde{v}>v_j$ (where $v_j$ is his rival's value). Thus, the payoff is

$$(v-b(\tilde{v}))\Pr(\tilde{v}>v_j)=(v-b(\tilde{v}))\tilde{v}.$$

Now compute the FOC for choice of $\tilde{v}$:

$$v-b(\tilde{v})-\tilde{v}b'(\tilde{v})=0.$$

If $b$ is really an equilibrium then the optimal choice of $\tilde{v}$ should be the bidder's true value, $v$: $v-b(v)-vb'(v)=0$. This is a differential equation. To solve it we just need a boundary condition.

Anyone with $v<r$ will not bid. So the set of bidders will be the truncated uniform distribution on $[r,1]$. Intuitively, a bidder with value $r$ cannot earn a positive surplus, so a reasonable boundary condition seems to be $b(r)=r$.

Solving the initial value problem $$v-b(v)-vb'(v)=0,\ b(r)=r$$

yields the equilibrium $$b(v)=\frac{r^2+v^2}{2v}.$$

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