= Solution
In a <second-price sealed-bid auction>, fix the other players' bids and let $a$ be their highest bid. A player of valuation $v$ who wins pays $a$, obtaining <quasilinear utility> $v-a$; losing gives zero. If $v>a$, winning is optimal and bidding $v$ wins. If $v<a$, losing is optimal and bidding $v$ loses. If $v=a$, winning and losing both give zero, regardless of the fixed tie-breaking rule. Thus truthful bidding is a <dominant strategy>. In particular \b[the profile in which every bidder bids its valuation is a Nash equilibrium], for every valuation profile, not merely on average.
A <symmetric independent private values model> gives each bidder a private valuation determined by its own type. The types are independent draws from a common distribution, and the bidders have identical roles and preferences. In the standard bidding calculation bidders are <risk-neutral> with <quasilinear utility>: they maximize expected valuation minus payment. Strategies map each bidder's privately observed value to a bid. Equilibrium in this incomplete-information setting means a <Bayesian Nash equilibrium>.
Take $n\geq2$ with independent uniform values on $[0,1]$. In a <first-price sealed-bid auction>, if all players bid their valuations, a type $v$ earns zero whenever it wins and zero whenever it loses. But for $v>0$, bidding $v/2$ instead yields positive <expected utility>
$$
\left(v-\frac v2\right)\Pr\{\text{every opposing value}<v/2\}
=\frac v2\left(\frac v2\right)^{n-1}>0.
$$
Therefore \b[truthful first-price bidding is not an equilibrium].
For the candidate <uniform private-value first-price bidding equilibrium>, put $\alpha=(n-1)/n$ and suppose every opponent bids $\alpha$ times its valuation. For a deviation $b\in[0,\alpha]$, winning means that every opposing value is less than $b/\alpha$. Independence gives winning <probability> $(b/\alpha)^{n-1}$, so the type-$v$ <expected utility> is
$$
u(v,b)=(v-b)\left(\frac b\alpha\right)^{n-1}.
$$
For $b>0$ its derivative is
$$
\frac{\partial u}{\partial b}
=\frac{b^{n-2}}{\alpha^{n-1}}\bigl((n-1)v-nb\bigr).
$$
For $v>0$ it is positive before $b=\alpha v$ and negative after; for $v=0$, every positive bid has nonpositive utility and zero is optimal. Thus the maximum on $[0,\alpha]$ is $b=\alpha v$. A bid above $\alpha$ wins with <probability> one but gives $v-b\leq v-\alpha$, which is no better than the already considered endpoint $b=\alpha$, hence no better than $b=\alpha v$. Ties have <probability> zero under the continuous opposing distributions. All admissible deviations are therefore covered.
Each bidder's best response is the proposed rule, proving
$$
\boxed{\beta(v)=\frac{n-1}{n}v\quad\text{is a symmetric Bayesian Nash equilibrium}.}
$$
Its conditional <expected utility> is $u(v,\alpha v)=v^n/n$. With only one bidder, bidding zero is optimal and agrees with the same formal rule; the derivative argument is needed only for $n\geq2$.
Back to article page
