 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 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 $n\ge 2$ 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 utilityTherefore 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 isFor $b>0$ its derivative isFor $v>0$ 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 $v-b\le v-\alpha$, 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, provingIts 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 $n\ge 2$.